Reeds Marine Engineering and Technology Volume 13: Ship Stability, Powering and Resistance 9781472987563, 9781408176122

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For my darling wife Linny & For Leanne

Published by Adlard Coles Nautical an imprint of Bloomsbury Publishing Plc 50 Bedford Square, London WC1B 3DP www.adlardcoles.com Copyright © Chris J Patterson, Jonathan D Ridley and Adlard Coles Nautical, 2014 First published by Adlard Coles Nautical in 2014 Print ISBN 978-1-4081-7612-2 ePDF ISBN 978-1-4081-7614-6 ePub ISBN 978-1-4081-7613-9 All rights reserved. No part of this publication may be reproduced in any form or by any means – graphic, electronic or mechanical, including photocopying, recording, taping or information storage and retrieval systems – without the prior permission in writing of the publishers. The right of the author to be identified as the author of this work has been asserted by them in accordance with the Copyright, Designs and Patents Act, 1988. A CIP catalogue record for this book is available from the British Library. Note: while all reasonable care has been taken in the publication of this book, the publisher takes no responsibility for the use of the methods or products described in the book.

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ACKNOWLEDGEMENTS The authors would like to thank their colleagues, families and friends who have provided advice during the development of this book. They would also like to thank Andy Phillips at Formsys, for his permission to use sample design data. They would also like to sincerely thank Kirsty Schaper and Jenny Clark at Bloomsbury and Srikanth Srinivasan at Newgen for their patience and encouragement.

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NOMENCLATURE Symbol

Units

AC

–

AP

–

Name Admiralty coefficient Aft perpendicular

2

Waterplane area

AWP

m

b

m

Tank beam

b

m

Compartment beam

B

–

Centre of buoyancy

B

m

Waterline beam of a box shaped vessel

BAR

–

Propeller blade area ratio

BM

m

Vertical position of the metacentre above B

BM

tonne metres

BML

m

Vertical position of the longitudinal metacentre above B

BP

–

Propeller power coefficient

BWL

m

Waterline beam

CB

–

Block coefficient

CF

–

Frictional resistance coefficient

CF

m FOAP

Longitudinal centre of flotation

CM

–

Amidships area coefficient

CR

–

Residuary resistance coefficient

CT

–

Total resistance coefficient

CW

–

Waterplane area coefficient

d

m

Distance of a mass from the centreline

d

m

Transverse shift in the centre of buoyancy due to bilging in a side compartment

D

m

Draught

D

m

Propeller diameter

DA

m

Draught at the aft perpendicular

DB

m

Bilged draught

δ

–

Propeller speed coefficient

Δ

tonnes

DF

m

Draught at the forward perpendicular

DHeeled

m

Heeled draught

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Bending moment

Displacement or ship mass including contents

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xvi • Nomenclature

Symbol

Units

Name

DI

m

Initial draught

DLCF

m

True mean draught (draught at the LCF)

DM

m

Mean draught

DTMD

m

True mean draught (draught at the LCF)

DW

–

Subscript used to identify dock water

DWA

mm

ηDUCT

–

Waterjet duct efficiency

ηH

–

Hull efficiency

ηJET

–

Waterjet jet efficiency

ηJS

–

Waterjet system efficiency

ηM

–

Engine mechanical efficiency

ηO

–

Propeller open water efficiency

ηP

–

Propeller efficiency

ηPUMP

–

Waterjet pump efficiency

ηR

–

Propeller relative rotative efficiency

ηT

–

Transmission efficiency

FC

–

Fuel coefficient

Fn

Dock water allowance

Froude number

FOAP

–

Forward of aft perpendicular

FP

–

Forward perpendicular

FSC

m

Free surface correction (loss in GM due to FSM)

4

Free surface moment (not corrected for fluid density)

FSM

m

FSM

tonne metres

FWA

mm 2

Free surface moment (corrected for fluid density) Fresh water allowance Acceleration due to gravity (taken as 9.81 m/s2)

G

m/s

G

–

Centre of gravity

GM

m

Metacentric height

GMI

m

Initial metacentric height (at zero heel)

GML

m

Longitudinal metacentric height

GZ

m

Righting lever, righting arm, arm of statical stability, lever of statical stability

h

m

Height of a watertight flat or double bottom depth

I or Inertia

4

m

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Transverse second moment of area of the waterline measured through the centreline

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Nomenclature • xvii

Symbol ICentroid IEDGE IGG

Units m4

Name Second moment of area measured through the centre of a shape

4

Second moment of area measured through the edge of the shape

4

Second moment of area measured through the centre of a shape

4

m m

IL or InertiaL

m

Longitudinal second moment of area of the waterline measured through the LCF

IROLL AXIS

m4

Second moment of area of the waterplane measured through the centre of the waterplane

IXX or IX, IYY or IY

m4

Second moment of area measured through a point away from centre of a shape

J

–

K

Propeller advance coefficient The intersection of the centreline and the keel

KQ

–

Propeller torque coefficient

KT

–

Propeller thrust coefficient

KB

m

Vertical position of the centre of buoyancy

KM

m

Vertical position of the metacentre above the keel

KML

m

Vertical position of the longitudinal metacentre above the keel

KG

m

Vertical position of the centre of gravity

1+k

–

Frictional form factor

l

m

Tank length

l

m

Compartment length

L

m

Waterline length of a box shaped vessel

λ

m

Heeling arm

λ0

m

Grain heeling arm at 0 degrees

λ40

m

Grain heeling arm at 40 degrees

LBP

m

Length between perpendiculars

LCB

m FOAP

Longitudinal position of the centre of buoyancy

LCF

m FOAP

Longitudinal centre of flotation

LCG

m FOAP

Longitudinal position of the centre of gravity

LWL

m

Waterline length

m

kg

Mass

m ˙

kg/s

M

–

MCTC

tm/cm

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Mass flowrate Metacentre Moment to change the trim by 1 cm

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xviii • Nomenclature

Symbol

Units

Name

MCTC2

tm/cm

Used in draught surveys – the MCTC at the corrected midship draught (row 13) plus 0.50 m

MCTC1

tm/cm

Used in draught surveys – the MCTC at the corrected midship draught (row 13) minus 0.50 m

MHM

tonne metres

Mass heeling moment, used in the grain regulations

MSS

tonne metres

Moment of statical stability, or righting moment

μ

–

Compartment permeability

μ

kg/ms

Dynamic viscosity

n

rev/s

Propeller revolution speed

N

rpm

Propeller revolution speed

N

–

∇

m3

v

2

m /s

Pv

Pa

P

tonnes

P

m

Propeller pitch

PD

W

Delivered power

PE

W

Effective power

PEN

W

Effective naked power

PI

W

Installed power

PS

W

Shaft power

PT

W

Thrust power

Q

Nm

r

m

Intersection of the line of action of buoyancy and a horizontal line from the keel Underwater or submerged volume Kinematic viscosity Vapour pressure Up-thrust during dry-docking

Propeller torque Turn radius

3

Fluid density

ρ

t/m

RF

N

Frictional resistance

Rn

–

Reynold’s number

RR

N

Residuary resistance

RT

N

Total resistance

s

various

S SCF

2

m –

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Ordinate spacing when using Simpson’s Rule Wetted surface area Ship correlation factor

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Nomenclature • xix

Symbol

Units

Name

SF

m2/t

Stowage factor

SF

tonnes

sfc

kg/kW hr

σ

–

Cavitation number

SW

–

Subscript used to identify sea water (taken as 1.025 m3)

t

–

Thrust deduction factor

T

N

Propeller thrust

TCB

m

Transverse position of the centre of buoyancy

TCG

m

Transverse position of the centre of gravity

∴

–

Mathematical symbol meaning ‘therefore’

θ

degrees

~ TPC

Shear force Specific fuel consumption

Angle of inclination, list, heel or loll Difference between

t/cm

Tonnes per centimetre immersion

v

m/s, knots

Ship speed

vA

m/s

Advance speed

vM

m/s

Model speed

vS

m/s

Ship speed

vW

m/s

Wake speed

VHM

m4

Volumetric heeling moment, used in the grain regulations

w

tonnes

Mass of an item of cargo

x

various

Generic term for an unknown variable

Z

–

Number of propeller blades

Z

–

Intersection of the line of action of buoyancy and a horizontal line from the centre of gravity

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INTRODUCTION These notes are written with the intention of helping develop an understanding of stability and propulsion for students of nautical science based degree courses, Merchant Navy Officers undertaking their Officer of the Watch or Master/Chief Mate Stability and Structures examinations, and Marine Engineers’ Naval Architecture examinations with the UK Maritime and Coastguard Agency (MCA). Some topic areas are required for some examinations but not others. For each topic area, the titles are marked (OOW), (MCM) or (ENG) indicating the level of the topic. For each section, the learning aims and objectives are listed at the start of the section, and again at the end of the section as a checklist. You should ensure that you can tick off all of the relevant boxes before the end of the course. These lists are for guidance only – they are not an exhaustive list of all skills, abilities and knowledge required. Questions within the notes are designed to test knowledge up to that point in the notes, and outline solutions to all of these questions are given in Chapter 14. Try and attempt the questions before looking at the solutions. Questions suitable for the Officer of the Watch level course syllabus are marked (OOW) alongside the question number, while questions suitable for the Master/Chief Mate level course are marked (MCM). Questions suitable for the Marine Engineers’ courses are marked (ENG). Background mathematical proofs and assumptions are covered in the appendices of these notes, so that you can develop an understanding of the underlying physics. These are indicated by arrow symbols. The most important formulae, needed to solve problems, are marked in bold and are highlighted in boxes, for example: a+b=c ▲ Formula Example

It is recommended that you make a separate list of these, and note down what the terms mean, the units used and sign conventions.

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1

FLOTATION AND BUOYANCY AIMS AND OBJECTIVES

At the end of this section, you should be able to: Explain why a ship floats Understand and complete calculations based on the relationship between displacement and draught for a box shaped vessel Understand the relationship between underwater volume, fluid density and displacement or mass Understand the difference between displacement and tonnage Understand and use block coefficients, waterplane area coefficients and amidship area coefficients Calculate underwater volume, waterplane area and amidships areas Determine the displacement of a ship from the hydrostatics Determine the draught of a ship from the hydrostatics Determine the displacement of a ship from the hydrostatics at intermediate draughts Determine the draught of a ship from the hydrostatics at intermediate displacements Determine the displacement of a ship from the displacement and deadweight (DW) scales Draw, label and dimension a Load Line mark Understand the Load Line zones Use the TPC to calculate the mass to add to cause a sinkage or the mass to remove to cause a rise

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Flotation and Buoyancy • 3 Use the TPC to calculate the sinkage or rise caused by adding or removing a mass Understand and use the relationship between and calculate the TPC from the dimensions of the ship and the form coefficients Understand the relationship between density and TPC, and correct the sea water TPC value to a dock water value Correct the dock water TPC value to a sea water value Understand the relationship between draught and water density Calculate the fresh water allowance (FWA) of a ship, and apply it to the draught Calculate the dock water allowance (DWA) of a ship, and apply it to the draught Calculate the mass to add or remove to a ship so that the vessel floats in accordance with the Load Line regulations Understand the limitations of the TPC with respect to changes in draught Understand the difference between normal Load Lines and Lumber Load Lines Understand the process by which the position of the Load Line on a ship is determined (assignment of Load Line)

Archimedes’ Principle (OOW, MCM, ENG) Any object which floats will, when placed in water, have a certain proportion of its body immersed, or underwater. Archimedes’ principle states that: When an object is immersed or partially immersed in a fluid, it experiences an upwards thrust equal to the weight of water that it pushes aside, or displaces. Therefore, if an object, such as a ship, is placed in water it will experience an upwards force, known as buoyancy, as a result of displacing the water. Archimedes’ Principle tells us that the buoyancy force is equal to the weight of water displaced. The weight of water displaced can be found from the volume of water displaced (given the symbol ∇), the density of the water (given the symbol ρ) and the acceleration due to gravity (g): = ∇×ρ×g For a vessel to float, she must be in ‘equilibrium’. That means that the total forces acting on the object must be equal and opposite to each other, and effectively cancel each other out. As well as the buoyancy force acting on the vessel upwards, there is gravity

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4 • Ship Stability, Powering and Resistance acting downwards. The force of gravity on an object, in units of Newtons, is equal to the mass of the object in kilograms, multiplied by the acceleration due to gravity, or ‘standard gravity’: M Mass ass × g

Force due to grav r vitty

Therefore, the gravitational force acting downward can be given by: Force due to grav r vitttyy

Ship Ship ip mass × g

The mass of a ship is given the symbol ∆. Substituting this notation gives: Force due to grav r vitty

Δ×g

As previously stated, for the ship to be in equilibrium, the buoyancy forces acting up must equal the gravity forces acting down: Force due to grav r vitttyy

Buoyanc o cy forces

Substituting the expressions for the forces gives: Δ×g= ρ×g×∇ The gravitational acceleration, g, is constant on both sides, and so to simplify the equation it can be cancelled. This gives: Δ=r×∇ ▲ Formula 1.1 Archimedes' principle

X For a detailed proof of this for box shaped vessels, please see Appendix 2:

Derivation of Archimedes’ Principle for Box Shaped Vessels. This means that the mass of the ship is the same as the water displaced. For this reason, the term used to describe the mass of a ship is her ‘displacement’. If we were to use normal SI units, the mass should be in units of kilograms, and the density in units of kilograms per metres cubed (kg/m3). However, given the large mass

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Flotation and Buoyancy • 5 values used in the shipping industry, the common practice is to quote mass in units of tonnes, and density in units of tonnes per metres cubed (1.025 t/m3 for sea water, and 1.000 t/m3 for fresh water). Formula 1.1 allows us to directly calculate the mass and underwater volume of a ship. This is an important relationship which governs the flotation of ships. As the density of water in a fixed location can be considered constant, any increase in the mass of the vessel, as a result of loading cargo, will result in an increase in volume – that is, the draught increases. The pressure still acts at right angles to the hull, but can be resolved into horizontal and vertical components of force. The horizontal components act inwards on the hull, while the vertical components create the buoyancy force. X For a mathematical proof of this for vessels with a rounded section, please

see Appendix 3: Derivation of Archimedes’ Principle for Semi-Circular Section Vessels.

QUESTIONS Q1.1 (OOW, MCM, ENG) A box shaped barge has a length of 30 m, a beam of 5 m and floats at a draught of 2 m in salt water, as shown below. Calculate the displacement of the barge. (Hint – the volume of a box is given by the length multiplied by the beam multiplied by the depth.)

2m

30 m 5m

Q1.2 (OOW, MCM, ENG) A box shaped barge has a length of 50 m and a beam of 7 m. The displacement of the barge is 1,076.25 tonnes. Calculate the draught of the barge if it is floating in salt water.

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6 • Ship Stability, Powering and Resistance Q1.3 (OOW, MCM, ENG) A box shaped barge has a length of 60 m, a beam of 8 m and floats in an unloaded condition at a draught of 3 m in salt water. 524 tonnes of cargo is added to the barge. Determine the draught after loading. Q1.4 (OOW, MCM, ENG) A box shaped barge has a length of 50 m, a beam of 7 m and floats in a lightship condition at a draught of 2 m in salt water. Determine what the draught of the vessel would be in fresh water.

Displacement or Tonnage? (OOW, MCM, ENG) So far we have dealt with the displacement of the ship. There are a number of different definitions of displacement that need to be understood, and also definitions for ‘tonnage’, which is often confused with displacement. The ‘Light Displacement’ or ‘lightship’ is the mass of the vessel with no cargo, crew stores, fuel, and so on. It does, however, include water in boilers to working levels and hydraulic fluid. The ‘Load Displacement’ is the mass of the hull and everything aboard when floating at the summer waterline. The term displacement (Δ) is generally used to indicate any value between lightship and load. Deadweight (DWT) is the difference between lightship and the displacement – effectively the amount of cargo, people and stores carried. The total DWT is the difference between lightship and the load displacement, and is effectively the maximum amount of stores, cargo and people that can be carried. The ‘Gross Tonnage’ of a ship is a measure of the internal volume of a vessel. Ships were historically charged operating fees based on how much cargo they could carry, not necessarily the actual DWT. The cargo carrying capacity of ships is referred to as the ‘tonnage’. This possibly comes from an old measure of how much wine a vessel could carry – measured in units of Tuns. A ‘tun’ in this case was not a unit of mass but a unit of volume – 2.78 m3. Note that tonnage, as a measure of the total volume of the vessel, has no direct mathematical link with the mass of the vessel.

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Flotation and Buoyancy • 7

Form Coefficients (OOW, MCM, ENG) The simplified box shaped vessels allow some basic calculations to be undertaken, but obviously real ships have curved hull forms for hydrodynamic efficiency. For real ships, determining the underwater volume is more complicated than simply multiplying the length, beam and draught, as the curvature of the hull means that the underwater volume is less than simply the waterline length multiplied by the waterline beam multiplied by the draught. Form coefficients can be used to compare a box shaped vessel to a real ship with the same length, beam and draught.

Block coefficient For real ships, the underwater volume can be found using a value known as the block coefficient. The block coefficient measures the actual underwater volume compared to a box shaped vessel of the same length, beam and draught, as shown in Figure 1.1.

BWL

D LWL

▲ Figure 1.1 Block coefficient

The block coefficient, or CB, can be used to find the volume using the formula below:

CB =

∇ LWL × BWL × D

▲ Formula 1.2 Block coefficient

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8 • Ship Stability, Powering and Resistance Within this, CB is the block coefficient, ∇ is the underwater volume of the vessel in units of metres3, LWL and BWL are the waterline length and beam of the vessel in metres and D is the draught of the vessel in metres. The block coefficient is calculated by the Naval Architect, and supplied in a table which gives the block coefficient for a particular draught. Therefore, at any draught, the block coefficient can be found, and the underwater volume calculated.

Waterplane area coefficient Another form coefficient that allows us to describe the shape of the hull is the waterplane area coefficient. The waterplane area of a vessel is the two dimensional area enclosed by the waterline of the vessel, as shown in Figure 1.2.

LWL Waterplane area BWL

▲ Figure 1.2 Waterplane area coefficient

The waterplane area can be found from:

CW =

Waterplane area LWL × BWL

▲ Formula 1.3 Waterplane area coefficient

Within this, CW is the waterplane area coefficient, and LWL and BWL are the waterline length and beam of the vessel in metres. The waterplane area is measured in units of metres2. Again, the waterplane area coefficient is calculated by the Naval Architect, and supplied in a table which gives the waterplane area coefficient for a particular draught. Therefore, at any draught, the waterplane area coefficient can be found and the waterplane area calculated.

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Flotation and Buoyancy • 9

QUESTION Q1.5 (OOW, MCM, ENG) A vessel has a waterline length of 100.00 m and a waterline beam of 20.00 m. Her waterplane area coefficient is 0.65. Determine her waterplane area.

Amidships area coefficient Another form coefficient that allows us to describe the shape of the hull is the amidships area coefficient. The amidships area of a vessel is the shape enclosed by the hull from the waterline down at amidships, as shown in Figure 1.3. As a general rule, as the amidships area coefficient gets larger, the internal cargo carrying volume increases for a given ship length.

D

BWL

▲ Figure 1.3 Amidships area coefficient

The amidships area coefficient measures can be found from:

CM =

Amidships area BWL × D

▲ Formula 1.4 Amidships area coefficient

Within this, CM is the amidships area coefficient, BWL is the waterline beam of the vessel in metres and D is the draught of the vessel in metres. The amidships area is measured in units of metres2.

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10 • Ship Stability, Powering and Resistance

QUESTION Q1.6 (OOW, MCM, ENG) A vessel has a beam of 15 m and draught of 8 m. The amidships area coefficient is 0.95. Determine the amidships area.

Hydrostatics (OOW, MCM, ENG) Clearly the form coefficients, along with the underwater volume and displacement, will change if the draught of the vessel changes. Calculating the underwater volume and displacement values from the form coefficients can be time-consuming, so they are pre-calculated by the Naval Architect, and supplied in tables known as ‘hydrostatic tables’. These tables show how the displacement (and other hydrostatic parameters, which will be covered in more detail later) change with the draught of the vessel. Samples of these are in Appendix 1: MV Reed – Sample Stability Data. To use these tables, we need to look at the definition of draught. If a vessel is ‘trimmed’, that is, not evenly loaded fore and aft, then the draught of the vessel will vary along the length, as shown in Figure 1.4, where the vessel is shown with stern trim (i.e. the stern is deeper in the water than the keel), on an even keel (where the keel is horizontal) and bow trim (i.e. the bow is deeper in the water than the stern). The draught used in the tables is always the draught measured at a point along the ship known as the longitudinal centre of flotation, or LCF, and is known as the True Mean Draught. The LCF is the pivot point of the vessel when trimming, and so at any

AP

LCF Amidships

FP

DA

DLCF

DF

DM

▲ Figure 1.4 Variation in draught with trim

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Flotation and Buoyancy • 11 displacement the draught at this point is independent of the trim of the ship. The LCF, and variations between true mean and mean draught will be covered in more detail later. The hydrostatic values are always calculated assuming the vessel is in salt water. Calculations using these tables will be covered in more detail later in the book.

QUESTIONS Q1.7 (OOW, MCM, ENG) Using the MV Reed Sample Stability Data Book (see Appendix 1), determine the displacement of MV Reed when she has a draught of 5.30 m. Q1.8 (OOW, MCM, ENG) Using the MV Reed Sample Stability Data Book (see Appendix 1), determine the draught of MV Reed when she has a displacement of 6,008 tonnes.

It is a requirement of part 6 of the Statutory Instruments 1998 No. 2241 The Merchant Shipping (Load Line) Regulations 1998, Amended 2000 that this information be provided to the ship in book form. This is explained in more detail in MSN 1701(M).

Hydrostatics and Linear Interpolation (OOW, MCM, ENG) Often the rows in the tables do not match the exact values we need. In this case, linear interpolation can be used to determine the draught or displacement, or any other values required. For example, consider the data in Table 1.1. If we wanted to determine the displacement of the vessel at a draught between these two values, for example, at 4.12 m, then we would need to use linear interpolation. Table 1.1 Sample hydrostatic data showing draught and displacement Draught (m)

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Displacement (tonnes)

4.20

3,890

4.10

3,781

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12 • Ship Stability, Powering and Resistance The easiest way to understand linear interpolation is to sketch a graph showing the relationship between the numbers. On the x axis (horizontal axis) we plot the value we know, in this case the draught, and on the y axis we plot the value we are trying to find, in this case, the displacement. This is shown in Figure 1.5. 3,900

Displacement (tonnes)

3,880 3,860 3,840 3,820 3,800 3,780 3,760 4.08

4.10

4.12

4.14

4.16

4.18

4.20

4.22

Draught (m)

▲ Figure 1.5 Draught and displacement

We can read up from the known draught at 4.12 m, and across to the displacement to find the displacement at 4.12 m, as shown in Figure 1.6. This gives us a displacement of 3,803 tonnes. We can, however, speed up this process and interpolate mathematically, so that we don’t need to draw a graph. If we consider the graph, as shown in Figure 1.7, we can see that the proportion of distance a to distance b is the same as the proportion of distance A to distance B. Therefore:

a A = b B ▲ Formula 1.5 Linear interpolation

This numerical method is much faster and far more accurate than drawing a graph and reading off the value.

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Flotation and Buoyancy • 13 3,900 3,880

Displacement (tonnes)

3,860 3,840 3,820 3,800 3,780 3,760 4.08

4.10

4.12

4.14

4.16

4.18

4.20

4.22

4.18

4.20

4.22

Draught (m)

▲ Figure 1.6 Interpolating displacement

3,900 3,880

Displacement (tonnes)

B 3,860 3,840 3,820 3,800 A 3,780 3,760 4.08

a

4.10

4.12

4.14

4.16

b Draught (m)

▲ Figure 1.7 Linear interpolation

QUESTIONS Q1.9 (OOW, MCM, ENG) Determine the displacement of MV Reed when the vessel has a draught of 3.48 m.

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14 • Ship Stability, Powering and Resistance Q1.10 (OOW, MCM, ENG) Determine the draught of MV Reed when she has a displacement of 3,365 tonnes. (Hint – the known values go on the x axis, the unknown values go on the y axis.) X For an explanation of an alternative method of interpolation by using the

equation of a straight line, please see Appendix 4: Linear Interpolation Using the Equation of a Straight Line.

Displacement and Deadweight Scales (OOW, MCM, ENG) Displacement (tonnes) Draught (m) Deadweight (tonnes) 8.00

TF F

8.000

5.385

7.000

4.385

6.000

3.385

5.000

2.385

4.000

1.385

3.000

385

T S

7.00

W

6.00

5.00

4.00

3.00

▲ Figure 1.8 Displacement and deadweight scale

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Flotation and Buoyancy • 15 Occasionally the hydrostatic data may also be shown in terms of diagrams showing a scale of draught against displacement and deadweight, as shown in Figure 1.8. To use these scales, a horizontal line is drawn across the scale at the known displacement or draught, and the other values read from the horizontal line.

Hydrostatic Curves MCTC (tm/cm) 100 96 92 88 84 80 76 72 68 64 60 56

41 40

48 44 40

52

LCB & LCF (m FOAP) 48 47 45 43

9 7 6

8

10 9

3 2 1

5

2,000 1,000 2

3

4

5

6

KB 7

4

KM 8

KB (m)

4,000 3,000

F LC

KM (m) 10

5,000

11

44

12

6,000

B LC

42

c

TP

13

46

14

7,000

15

C t MCT men lace Disp

8,000 9,000 Displacement (tonnes) TPc (t/cm)

(ENG)

Draught (m)

▲ Figure 1.9 Hydrostatic curves

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16 • Ship Stability, Powering and Resistance Occasionally the hydrostatic data shown in the tables may instead be shown in graphical form. The draught of the vessel is plotted on the y axis, with the other parameters each plotted on their own scale along the x axis. The hydrostatic curves for MV Reed are shown in Figure 1.9.

100 96 92 88 84 80 76 72 68 64 60 56 48 44 40

52

41 40

MCTC (tm/cm)

LCB & LCF (m FOAP) 48 47 45 42

9 3 2 1

7

8

10

KM (m) 10

2,000 1,000

2

3

4

5

6

KB 7

6

3,000

KM 8

KB (m) 5

F

LC

4

4,000

9

5,000

11

44

12

6,000

B

LC

43

c

TP

13

46

14

7,000

15

C t men MCT lace p s i D

8,000 9,000 Displacement (tonnes) TPc (t/cm)

These graphs give the same data as the hydrostatic tables; however, as there is an element of subjectivity in reading graphs, they are less accurate than the tables. They

Draught (m)

▲ Figure 1.10 Hydrostatic curves and known draught

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Flotation and Buoyancy • 17

100 96 92 88 84 80 76 72 68 64 60 56

41 40

3 2

44

1

40

52

46 43

9

2,000 1,000

2

3

4

5

6

KB

7

48

KM

8

8 7 6

3,000

KB (m) 5

F

LC

4

4,000

9 10 KM (m) 10

5,000

11

44

45

12

6,000

B

LC

42

c

TP

13

47

14

7,000

C t MCT men lace Disp

8,000 9,000 Displacement (tonnes) TPc (t/cm) 15

LCB & LCF (m FOAP) 48 MCTC (tm/cm)

do have the advantage of not requiring interpolation to use. As with the tables, to use the curves, either the draught or displacement must be known. If the draught is known, then a horizontal line is drawn across all of the curves at the known draught. The intersection of this line and the curves can be read off the corresponding x axes to determine the hydrostatic parameters. An example of this for a draught of 5.50 m is shown in Figure 1.10.

Draught (m)

▲ Figure 1.11 Hydrostatic curves and known displacement

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18 • Ship Stability, Powering and Resistance Alternatively, if the displacement is known, then a vertical line is draw up from the displacement axis to the displacement curve. At the intersection of the line and the curve, a horizontal line is draw across the graph. The intersection of this horizontal line and the other curves can be read to determine the values of the hydrostatic parameters. An example of this is given in Figure 1.11, for a displacement of 4,600 tonnes.

Load Line (OOW, MCM, ENG) Note: Information in this section is based on the Statutory Instruments 1998 No. 2241 The Merchant Shipping (Load Line ) Regulations 1998, Amended 2000 (The Merchant Shipping (Load Line) Regulations 1998). The reality of commercial shipping is that ships carry cargo in order to make money. As a general rule, carrying more cargo results in making more money, so there is an economic argument to load as much cargo on a ship as is possible. However, as cargo is added to the ship, she sinks lower into the water, increasing her draught and reducing her freeboard (the height of the hull above the water). This reduction in freeboard reduces the ability of the vessel to survive waves, and also reduces the stability of the vessel (this will be covered in more detail later). All ships (with some exclusions detailed later) are therefore marked on the side of the hull with ‘Load Lines’, which show the minimum allowable freeboard for the vessel. The minimum freeboard is measured from the waterline of the vessel to the ‘freeboard deck’, which is the ‘the uppermost complete deck exposed to weather and sea, which has permanent means of closing all openings open to the weather, and below which all openings in the sides of the ship are fitted with permanent means of watertight closing’. This minimum freeboard is determined from a set of rules, and takes into account, among other factors, the size, block coefficient, hull sheer and construction of the vessel. A photo of these marks in shown in Figure 1.12, and a drawing of these marks is shown in Figure 1.13. These are marks on the side of the hull which show the highest allowable waterline for the vessel. When the ship is loaded so that the waterline is on these marks, she will be floating at her maximum draught, with her minimum freeboard. As these are international symbols, there are strict rules about the size of the marks. They must conform to the dimensions shown in Figure 1.14. To ensure that the marks

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Flotation and Buoyancy • 19

▲ Figure 1.12 Photograph of a Load Line mark

TF

F

T

S

W

▲ Figure 1.13 Diagram of a Load Line mark

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20 • Ship Stability, Powering and Resistance are permanent, they are painted onto strips of metal welded to the hull, so that they can be found even if painted over.

Statutory freeboard

300

230 TF 230

300 F

T

540 S

W 450 All lines are 25 mm thick. All dimensions shown are in millimetres. Freeboards are measured from the top of each line.

▲ Figure 1.14 Dimensions of a Load Line mark

The ‘TF’, ‘F’, ‘T’, ‘S’, ‘W’ and ‘WNA’ markings refer to ‘zones’.

Designation

Meaning

TF

Tropical Freshwater

F

Freshwater

T

Tropical – always

S

Summer

W

Winter– always

WNA

Winter North Atlantic (only used on vessels less than 100 m, and located 50 mm below the winter mark)

9781408176122_Ch01_Rev_txt_prf.indd 20

1 th of the summer draught above the summer mark 48

1 th of the summer draught below the summer mark 48

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Flotation and Buoyancy • 21 All of the world’s seas are broken down into a series of zones, based on the prevailing weather conditions and water density. These zones are named after the seasons, to indicate the overall weather conditions – the main zones do not change during the year. Not all zones are always marked on the Load Line. The lowest zone that the ship is sailing in, or through, controls which line the vessel can load down to. For example, in a winter zone, she can only load so that the top of the W line is on the waterline, but in a tropical zone, she could load additional cargo so that the top of the T line is on the 1

waterline. The tropical and winter lines are always equal to th of the summer 48 draught from the summer mark. The letters by the circular mark (e.g. AB in Figure 1.12) are the signature letters of what is known as the ‘Assigning Authority’. This is the organisation which analyses the design of the ship, and determines what the minimum safe freeboard is. The black bar above the circular mark represents the point at which the freeboard is measured from. Vessels which are exempt from the Load Line are ‘Ships of war’, ‘ships solely engaged in fishing’, ‘pleasure vessels’ and ‘ships which do not go to sea’. Also exempt are the following ships under 80 tonnes register, which do not carry cargo (unless specified authorised on the Passenger Certificate). Engaged in coastal trade, such as: Tugs and salvage ships Hopper barges or dredgers Ships used by lighthouse authorities Fisheries protection Scientific research Military firing range control vessels Passenger vessels with passenger certificates specifying regions where the vessel may sail Ships carrying not less than 12 passengers within certain limits from shore Under certain circumstances, the Secretary of State may allow vessels to sail with exemptions, detailed in Paragraph 5 of the act. Some passenger vessels may also be allocated ‘combination’ Load Lines. These are used when there may be spaces which are alternatively used as passenger or cargo spaces. These are marked as ‘C’ on the Load Line marks and are always below the summer Load Line.

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22 • Ship Stability, Powering and Resistance

The TPC (OOW, MCM, ENG) To safely load a vessel, we need to know how much cargo can be added to the vessel. While the hydrostatics give us an idea of the draught for a certain displacement, it is often useful to be able to ‘fine tune’ the draught to get the maximum possible cargo aboard, and therefore make the maximum possible profit. When loading the vessel, it is vital that the relevant Load Line is not submerged when the vessel proceeds to sea. To ensure this, calculations are required during loading to ensure that the maximum cargo is carried, but that the loading is legal. To do this, before completion of loading, the vertical distance from the waterline to the relevant Load Line mark is measured. This distance can be used with a hydrostatic value known as the tonnes per centimetre immersion, or TPC, to calculate the mass to load. This is shown in the hydrostatic tables. The TPC tells us how many tonnes of mass need to be added to the vessel to make her sink down by 1 cm, or how many tonnes of cargo need to be removed from the vessel to make her rise up by 1 cm. The TPC, sinkage or rise and mass added or removed are linked by:

Sinkag k e (o or rrise ) =

Mass added ( or removed ) TPC

▲ Formula 1.6 Sinkage or rise and the TPC

As the vessel is loaded, the distance from the waterline to the Load Line can be used to determine the allowable sinkage until the top of the relevant Load Line touches the water, when loading should stop. If the allowable sinkage is known, then the amount of cargo that can still be loaded can be found from the TPC.

QUESTIONS Q1.11 (OOW, MCM, ENG) If the mass added or removed has units of tonnes, and the TPC has units of tonnes per centimetre, what are the units of sinkage or rise?

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Flotation and Buoyancy • 23 Q1.12 (OOW, MCM, ENG) A ship has a TPC value of 35. An additional 70 tonnes of cargo is added to the vessel. How much further in the water will the ship sink? Q1.13 (OOW, MCM, ENG) A ship has a TPC value of 25. How much cargo must be removed from the vessel to make her rise by 0.2 m? Q1.14 (OOW, MCM, ENG) A ship is floating with the waterline 50 mm below the bottom of the summer Load Line, in a summer zone. The TPC is 26. How much additional cargo can be loaded? Q1.15 (OOW, MCM, ENG) A ship is floating with the waterline 90 mm above the bottom of the summer Load Line, in a summer zone. The TPC is 22. How much cargo must be removed? Q1.16 (OOW, MCM, ENG) A ship is floating with the waterline on the bottom of the summer Load Line, in a tropical zone. The TPC is 20. How much cargo can be added if the summer draught of the vessel is 9.80 m?

The TPC can be directly calculated from the waterplane area, water density, and the length and beam of the vessel:

TPC = CW × L × B × 0 01× ρ ▲ Formula 1.7 The TPC

This can also be written as:

TPC =

(CW × L × B ×ρ) 100

▲ Formula 1.8 The TPC (alternative)

Within these, the TPC is measured in units of tonnes per centimetre, the CW is the waterplane area coefficient, L and B are the vessel length and beam (on the waterline) in metres and ρ is the water density in units of tonnes per metre3.

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24 • Ship Stability, Powering and Resistance X To see a proof of how the TPC can be directly calculated, please see

Appendix 5: The Derivation of the TPC Formula.

QUESTION Q1.17 (OOW) A ship has a waterplane area coefficient of 0.80. The waterline length is 110 m, the waterline beam is 15 m and she is floating in sea water. Determine the TPC.

As seen in Formulae 1.7 and 1.8, the exact value of the TPC depends on the density of the water that the vessel is floating in. Often when loading in a port, the water around the ship will not be pure sea water. Depending on the location, the water may be fresh water (with a density of 1.000 t/m3), or it may be a mix of sea water and fresh water known as dock water or brackish water. This will have a density somewhere between sea water and fresh water. Unless you are told otherwise, the values for TPC quoted in hydrostatic tables are always for salt water. If the water is not salt water, then corrections are required to obtain the ‘dock water value’ or ‘actual’ value of the TPC:

TPCDW = TPC T SW ×

ρDDW ρSW

▲ Formula 1.9 Dock water TPC

Within this formula, the subscript SW refers to the sea water values, and the subscript DW refers to the dock water values.

QUESTIONS Q1.18 (OOW, MCM, ENG) Your ship has a TPC of 25. The vessel moves into fresh water. What is the dock water TPC value?

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Flotation and Buoyancy • 25 Q1.19 (OOW, MCM, ENG) Your ship has a TPC in sea water of 30. The vessel is in dock water with a density of 1.010 t/m3. What is the dock water TPC value?

Fresh Water Allowance (OOW, MCM, ENG) In fresh water, the vessel will float lower in the water than in salt water. This is because fresh water is less dense, and therefore less buoyant. This means that (assuming displacement remains constant) as ships move from fresh to salt water, they rise up slightly. The Load Line rules which govern ship loading require that the vessel must be on her correct marks when she proceeds to sea, but do not mention what happens when she is alongside her berth or manoeuvring in a harbour. This allows us to ‘overload’ the vessel in a fresh water port, as she will rise up to the correct Load Line when she proceeds into salt water at sea. The distance between the fresh water mark and the summer mark is known as the ‘fresh water allowance’, or FWA, as it is the amount that the vessel is allowed to be ‘overloaded’ by in fresh water. The FWA (in units of millimetres) can be found from:

FWA W =

Δ in sea w water ate at tthe e relevant mark r 4 × TPC in i sea wate water at a the relevant mark r

▲ Formula 1.10 Fresh water allowance

Therefore in a tropical zone, the FWA can be found from:

FWA W =

Δ Tropical 4 × TPC SW at the tropica t l mark r

▲ Formula 1.11 Tropical fresh water allowance

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26 • Ship Stability, Powering and Resistance In a summer zone, the FWA can be found from:

FWA W =

Δ Summer 4 × TPC SW at the su summer e mark r

▲ Formula 1.12 Summer fresh water allowance

In a winter zone, the FWA can be found from:

FWA W =

ΔWinter 4 × TPC SW at the winter mark

▲ Formula 1.13 Winter fresh water allowance

Within these, Δ is the displacement of the vessel when floating at the appropriate mark, in units of tonnes, the TPC is the sea water value at the appropriate mark, and the FWA is the fresh water allowance in units of millimetres. X For a mathematical proof of the FWA, please see Appendix 6: The Derivation

of the Fresh Water Allowance Formula.

QUESTION Q1.20 (OOW, MCM, ENG) A ship with a summer displacement of 10,000 tonnes has a TPC in salt water of 30. What is the summer draught FWA in units of millimetres and centimetres?

Dock Water Allowance (OOW, MCM, ENG) If the vessel is in dock water, then there will still be a change in draught and freeboard when moving into sea water. When the vessel is in dock water, a dock water allowance

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Flotation and Buoyancy • 27 (DWA) is used instead of the FWA. The principle is the same – the vessel can be overloaded by a certain amount so that she floats on her marks when proceeding to sea.

DWA W = FWA FWA ×

(ρSW ρDDW ) (ρSW ρFFW ) S

▲ Formula 1.14 Dock water allowance

The DWA is measured in units of millimetres, with the sea water and dock water densities measured in units of tonnes per metre3. As the densities of salt water and fresh water are constant, the formula is often simply written as one of:

DWA W FWA FWA ×

DWA W FWA FWA ×

(

(

SW

−

DW D

( .025)

SW

(

−

DW D

)

) with t ρ in units t of tonnes / metre 3

) with ρ in unitst of kkilograms / metre 3

QUESTION Q1.21 (OOW, MCM, ENG) A ship with a summer displacement of 15,000 tonnes has a TPC in salt water of 35. What is the DWA at the summer draught in water with a density of 1.015 t/m3? (Hint – you will need to calculate the FWA first.)

Together, the FWA, the DWA and the dock water or actual TPC can be used to determine how much additional cargo can be loaded onto a vessel to get her to her maximum possible legal displacement and draught when loading in dock water. To determine how much additional cargo can be loaded aboard, the FWA (and the DWA, if in dock water) must be calculated, and therefore the distance from the Load Line mark to the allowable waterline can be found. The distance from the actual waterline to the Load Line mark can be measured, and therefore the distance from the actual waterline to the allowable waterline can be found. This is the ‘allowable sinkage’.

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28 • Ship Stability, Powering and Resistance The allowable sinkage can be used with the TPC (suitably corrected for fresh or dock water) to determine how much mass can be loaded aboard to cause the allowable sinkage.

QUESTIONS Q1.22 (OOW, MCM, ENG) A ship with a summer displacement of 11,000 tonnes with a salt water TPC of 29 t/cm floats in dock water with a density of 1.005 t/m3. The waterline is 5 cm below the lower edge of the summer Load Line. Determine the amount of cargo the vessel can load. Q1.23 (OOW, MCM, ENG) A ship with a summer displacement of 14,000 tonnes and a TPC of 35 t/cm floats in dock water with a density of 1.020 t/m3. The waterline is 15 cm above the upper edge of the summer Load Line. Determine the amount of cargo the vessel must discharge. Q1.24 (OOW, MCM, ENG) A vessel has a summer draught of 10.6 m. The FWA at the Winter draught is 290 mm. The vessel has a TPC, adjusted for dock water, of 33 t/cm. The density of the dock water is 1,018 kg/m3. The waterline is 490 mm below the summer Load Line. What is the Winter displacement of the vessel, and how much more cargo can the vessel load so that she is on her winter marks at sea? Q1.25 (OOW, MCM, ENG) MV Reed is floating in a summer zone, with the waterline 40 cm below the top of the summer mark. The density of the water is 1.010 t/m3. Determine the amount of cargo to load to bring the vessel to her summer displacement using the mean TPC.

Accuracy of the TPC (OOW, MCM, ENG) As seen in Formulae 1.7 and 1.8, the TPC depends on the waterplane area coefficient. This changes as the draught changes, so as we load cargo onto the vessel, the TPC changes. This is also shown in the hydrostatics, where TPC varies with draught. This change in TPC with draught introduces a small error into the mass values found in Formula 1.6.

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Flotation and Buoyancy • 29 Table 1.2 Errors due to using the TPC Draught (m)

Displacement (tonnes)

Actual mass to load (tonnes)

TPC based mass to load (tonnes)

7

7,329

0

0.0

6.9

7,192

137

137.2

6.8

7,056

273

273.3

6.7

6,921

408

408.3

6.6

6,787

542

542.1

6.5

6,654

675

674.6

6.4

6,522

807

806.0

6.3

6,392

937

936.2

6.2

6,263

1,066

1,065.2

6.1

6,135

1,194

1,193.3

6

6,008

1,321

1,320.2

To minimise this error, if the sinkage is large (more than a few centimetres) and the full hydrostatic data is available, then the TPC value can be interpolated at the initial draught and the permitted draught, and the average of the two found. This average TPC can then be used to determine the mass to load. Even this process introduces small errors into the mass to load, which are shown in Table 1.2, which shows the mass to add found using the mean TPC against the actual mass to add found from the hydrostatics, for MV Reed in sea water, in a summer zone.

Lumber or Timber Load Lines (OOW, MCM, ENG) Vessels which are designed to carry timber deck cargoes may have special Load Lines, known as Lumber Load Lines, marked onto the hull. If the vessel has a securely packaged timber deck cargo, then that cargo is assumed to be part of the vessel, and therefore contributes to the freeboard. This is because timber is buoyant. Please see the section on Large Angle Stability for more details. The vessel must have the bow protected by a forecastle (which must be at least the ‘standard height’, which is 1.80 m high for a vessel of 75 m or less in length, or 2.30 m high for a vessel over 125 m in length, with linear interpolation used in between) of at least 7% of the Load Line length, and a raised

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30 • Ship Stability, Powering and Resistance quarter deck of at least standard height if the vessel is less than 100 m in length. She must also have efficient railings or bulwarks to help secure the cargo, and have double bottom tanks along the half-mid length of the vessel, with adequate longitudinal subdivision. Lumber Load Lines are very similar to normal Load Lines, except that the freeboard of the vessel is less because of the reasons outlined in the previous paragraph, and the winter mark is closer to the summer mark. Each of the zones is prefixed with an ‘L’. The zones are as follows: Designation Meaning LTF

Timber Tropical Freshwater

LF

Timber Freshwater

LT

Timber Tropical – always

LS

Timber Summer

LW

Timber Winter– always

LWNA

Timber Winter North Atlantic (only used on vessels less than 100 m, and located 50 mm below the winter mark)

1 th of the summer draught above the summer mark 48

1 th of the summer draught below the summer mark 36

Assigning a Load Line (MCM, ENG) The intentions of the Load Line Act are to ensure that a vessel is strong enough for a proposed loading condition, to ensure that she has sufficient stability for probable loading conditions, to ensure that the crew are sufficiently protected when working on deck in bad weather, and to ensure that the vessel has sufficient reserve buoyancy. The current Load Line Regulations, SI 2241 (1998) as amended by SI 1335 (2000), stem from the 1966 International convention. In order to understand the Load Line requirements, some basic definitions are required. Within the regulations, watertight means ‘capable of preventing the passage of water in any direction’, while weathertight for fittings means ‘water will not penetrate and enter the hull in the worst sea and weather conditions’. In the case of a bulkhead door weathertight means ‘permanently attached, made of steel or equivalent material

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Flotation and Buoyancy • 31 which, together with its frame, is of equivalent strength to the unpierced bulkhead’. It must be closed by permanently attached gaskets and clamping devices. Weathertight bulkhead doors must be capable of being operated from either side of the bulkhead and should normally open outwards. The freeboard deck is the uppermost complete deck exposed to weather and sea which has permanent means of closing all openings in its weather portions and below which all side openings have permanent watertight closing appliances. In effect, this is the deck which forms the top of the watertight compartments. The superstructure itself is defined as ‘decked structure on the freeboard deck extending from side to side or such that its side plating is not inboard of the shell by more than 4% of the ship’s breadth’. An enclosed superstructure is one where ‘the end bulkheads are of efficient construction and access openings have sills and weathertight doors. All other openings in the sides and ends (such as ports) must be fitted with weathertight closing arrangements.’ A superstructure deck is a deck forming the top of part of the superstructure. Narrower deck structures which open onto lower decks are known as trunks, which are defined as ‘a structure having at least 60% of the ship’s breadth at the position in which it is situated and which has the same strength as a superstructure which opens directly into the space below the freeboard deck’. In order for a vessel to operate, she must be ‘assigned’ a freeboard. What this means is that the design (and the built ship herself ) are examined, and based on the shape and fittings, given, or ‘assigned’, a minimum freeboard that is considered safe. It is this freeboard which is marked on the hull as the summer Load Line or ‘Plimsoll mark’. To do this, cargo ships are classed into two main types. ‘Type A’ vessels are those which are designed to carry liquid cargo in bulk, while ‘Type B’ vessels are designed to carry any other type of cargo vessel. Given the type of cargo carried, normally Type A vessels will have strong, watertight decks, while Type B vessels will only have weathertight decks. In addition, due to the free surface effects of liquid cargoes, Type A vessels normally have greater subdivision than Type B vessels.

Tabular freeboard Within the Load Line Act, there is a table of data which shows, either for a Type A or Type B vessel, a minimum freeboard for the length of the vessel. Type A vessels have smaller minimum freeboards than Type B vessels, as the deck of Type A vessels has an inherent greater strength and integrity due to watertight decks, stronger deck structure; greater

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32 • Ship Stability, Powering and Resistance Table 1.3 Sample tabular freeboards Type A vessel

Type B vessel

Length (m)

Freeboard (mm)

Length (m)

Freeboard (mm)

140

1,803

140

2,109

143

1,853

143

2,171

internal subdivision than dry cargo ships and low permeability of loaded cargo spaces. The freeboard listed in Table 1.3 is known as the ‘tabular freeboard’. An example extract from the tables is shown in Table 1.3. If a Type B vessel meets the following requirements, then it is considered to have better structural integrity of the deck than a normal Type B vessel: Over 100 m in length. Adequate protection for crew working on deck and adequate water freeing arrangements (such as railings rather than bulwarks). Steel, gasketed and clamped hatch covers in exposed locations with adequate strength and sealing arrangements. The ability to remain afloat (when loaded to the summer Load Line ) with one compartment (other than the machinery space) bilged with a permeability of 95%. For vessels over 225 m in length, then this also includes the machinery space, but with a permeability of 85%. Bilging means that the compartment is open to the sea, this is covered in more detail later. If the vessel meets the above requirements, it is possible to reduce the tabular freeboard by 60% of the difference between a Type B and Type A vessel of the same size, in which case the vessel is known as a Type B-60 vessel. If it meets these Type B-60 requirements and also meets the following requirements, then a further reduction in freeboard is allowed: Suitable freeing arrangements for trapped water. Suitable machinery case protection. Safe and suitable means of access forward for crew working on deck. The ability to remain afloat (when loaded to the summer Load Line ) with any two adjacent compartments (other than the machinery space) bilged with a permeability of 95%. For vessels over 225 m in length, then this also includes the machinery space bilged in isolation, but with a permeability of 85%.

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Flotation and Buoyancy • 33 In this case, the tabular freeboard can be reduced by 100% of the difference between a Type B and Type A vessel of the same length. This is known as a Type B-100 vessel. As the reduction is 100% of the difference in tabular freeboard between a Type A vessel and a Type B vessel of the same size, a Type B-100 vessel would have the same tabular freeboard as a Type A vessel. If the vessel is under 100 m in length, and the effective length of the superstructure is less that 35% of the length, then an additional increase is made to the tabular freeboard to increase it for safety. In both the B-60 and B-100 requirements, the vessel must be designed so that in the event of the vessel being bilged as per the requirements, the water must not enter the vessel through any opening in the hull, the heel angle must not exceed 15 degrees, the metacentric height must be at least 50 mm (see the chapters on small angle stability and bilging for an explanation of metacentric height) and there must be adequate residual stability.

Block coefficient correction The tabular freeboard values are based on a standard shape of hull, with a block coefficient of 0.68, and a length to depth ratio of 15. If the actual block coefficient is different to the standard vessel, then a correction to the freeboard is made. If the actual block coefficient is greater than the standard, then the freeboard is increased by formula. This is now known as the Basic freeboard. The reasoning for this is that vessels with a large block coefficient are ‘fatter’ or ‘fuller’ than vessels with a small block coefficient. They therefore have more buoyancy below the waterline in proportion to the reserve buoyancy (the watertight volume above the waterline). This means that if the vessel is damaged and floods there is a greater resulting sinkage, and therefore a greater freeboard is required to ensure that the ship has sufficient freeboard after damage. The concept of sinkage after damage is explained in more details in the chapter covering bilging.

Length and depth correction Once the basic freeboard is found, corrections must be made for the length to depth ratio. As the depth of the vessel increases for a fixed length, the sinkage due to the same amount of damage along the ship will increase, as more volume will be lost as a result

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34 • Ship Stability, Powering and Resistance of damage. Therefore, vessels with large depth values for their length require additional freeboard, and conversely vessels with small depth values for their length can have their freeboard reduced. As the standard vessel used in the assignment process has a length to depth ratio of 15, any vessel with a depth in excess of the length divided by 15 must have the freeboard increased according to formula. If the depth is less than the length divided by 15, and the vessel meets additional requirements related to the size of the superstructure, then the freeboard can be reduced according to formula. If the length to depth ratio of the actual vessel is less than 15, and the vessel has an enclosed superstructure running at least 60% of the length of the vessel, then the basic freeboard can be decreased by formula.

Deck line correction A further correction to the freeboard is made if the deck line (the line which marks the point at the top of the hull at which the freeboard is measured) is not placed on the hull at a depth corresponding to the depth for freeboard. If the depth (measured up from the inside of the keel plates) to the freeboard deck is more than the depth to the deck line, the difference is subtracted from the freeboard. This effectively allows the freeboard to be measured to the deck plates. Conversely, if the depth (measured up from the inside of the keel plates) to the freeboard deck is less than the depth to the deck line, the difference is added to the freeboard.

Superstructure correction The standard vessel has no superstructure. Enclosed superstructure can have a beneficial effect in increasing reserve buoyancy, and so a reduction in freeboard is applied based on the extent of the superstructure. This reduction is complicated, and is found from tables based on vessel length and superstructure length.

Sheer and bow height correction The standard vessel has a certain sheer profile along the length of the hull. Increasing sheer will increase bow height, and hence influence the sea keeping of the vessel. Increasing the sheer profile above the standard vessel results in an allowable reduction in freeboard, while reducing the sheer profile below the standard vessel results in an

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Flotation and Buoyancy • 35 increase in freeboard. If the final bow height is less than a specified minimum value, then an additional correction is required to bring the bow height to a safe value. This freeboard, resulting from the corrections to the basic freeboard, is known as the ‘assigned freeboard’. It is this assigned freeboard that is then used to mark the summer Load Line on the vessel. The tropical and winter Load Line marks can then be found from calculation using the summer draught of the vessel. In order for the vessel to be ‘assigned’ or given a freeboard, she must also meet structural requirements.

FLOTATION AND BUOYANCY – LEARNING CHECKLIST Objective

Level

Explain why a ship floats

OOW, MCM, ENG

Understand and complete calculations based on the relationship between displacement and draught for a box shaped vessel

OOW, MCM, ENG

Understand the relationship between underwater volume, fluid density and displacement or mass

OOW, MCM, ENG

Understand the difference between displacement and tonnage

OOW, MCM, ENG

Understand and use block coefficients, waterplane area coefficients and amidship area coefficients

OOW, MCM, ENG

Calculate underwater volume, waterplane area and amidships areas

OOW, MCM, ENG

Determine the displacement of a ship from the hydrostatics

OOW, MCM, ENG

Completed

Determine the draught of a ship from the hydrostatics OOW, MCM, ENG Determine the displacement of a ship from the hydrostatics at intermediate draughts

OOW, MCM, ENG

Determine the draught of a ship from the hydrostatics OOW, MCM, ENG at intermediate displacements Determine the displacement of a ship from the displacement and DW scales

OOW, MCM ENG

Draw, label and dimension a Load Line mark

OOW, MCM, ENG

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36 • Ship Stability, Powering and Resistance

Objective

Level

Understand and use the Load Line zones

OOW, MCM, ENG

Use the TPC to calculate the mass to add to cause a sinkage or the mass to remove to cause a rise

OOW, MCM, ENG

Use the TPC to calculate the sinkage or rise caused by adding or removing a mass

OOW, MCM, ENG

Completed

OOW, MCM, ENG Understand and use the relationship between and calculate the TPC from the dimensions of the ship and the form coefficients Understand the relationship between density and TPC, and correct the sea water TPC value to a dock water value

OOW, MCM, ENG

Correct the dock water TPC value to a sea water value

OOW, MCM, ENG

Understand the relationship between draught and water density

OOW, MCM, ENG

Calculate the FWA of a ship, and apply it to the draught

OOW, MCM, ENG

Calculate the DWA of a ship, and apply it to the draught

OOW, MCM, ENG

Calculate the mass to add or remove to a ship so that the vessel floats in accordance with the Load Line regulations

OOW, MCM, ENG

Understand the limitations of the TPC with respect to changes in draught

OOW, MCM

Understand the difference between normal Load Lines and Lumber Load Lines

OOW, MCM

Understand the process by which the position of the Load Line on a ship is determined (assignment of Load Line )

MCM

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2

SMALL ANGLE OR INITIAL METACENTRIC

STABILITY  WILL MY SHIP FLOAT UPRIGHT? AIMS AND OBJECTIVES At the end of this section, you should be able to: Understand what is meant by the centre of buoyancy Understand what is meant by the vertical centre of buoyancy (KB) Understand what is meant by the longitudinal centre of buoyancy (LCB) Understand what is meant by the transverse centre of buoyancy (TCB) Calculate the position of the centre of buoyancy for a box shaped vessel Determine the position of the centre of buoyancy using hydrostatic data Understand what is meant by the centre of gravity Understand what is meant by the vertical centre of gravity (KG) Understand what is meant by the longitudinal centre of gravity (LCG) Understand what is meant by the transverse centre of gravity (TCG or GGH) Calculate the position of the centre of gravity when loads are added

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38 • Ship Stability, Powering and Resistance Calculate the position of the centre of gravity when loads are removed Calculate the position of the centre of gravity when loads are moved up or down Calculate the position of the central gravity when loads are suspended Calculate the position to load a known mass of cargo to obtain a required centre of gravity Calculate the mass of cargo to load at a known position to obtain a required centre of gravity Understand what is meant by the metacentre or transverse metacentre Understand why the centre of buoyancy moves as a vessel inclines Understand how a righting effect may be generated as a vessel inclines Understand how a capsizing effect may be generated as a vessel inclines Understand the definition of the metacentre Understand what is meant by metacentric height Understand what is meant by positive, neutral and negative metacentric heights Understand the significance of positive, neutral and negative metacentric heights on the stability of the vessel Calculate the metacentric height of a ship using the hydrostatic data and the centre of gravity Understand what is meant by transverse second moment of area or transverse inertia Calculate the metacentric height of a ship using the hydrostatic data and the centre of gravity after moving masses aboard Calculate the position to load cargo to achieve a required metacentric height Understand and know the minimum allowable metacentric heights for vessels carrying a range of cargo types Understand the dangers of excessive metacentric height Calculate the waterplane transverse second moment of area or waterplane transverse inertia for a box shaped vessel Calculate BM and GM for a box shaped vessel Understand and complete calculations based on the metacentric height and the geometry of a box shaped vessel Understand what is meant by an angle of list Calculate angles of list using the metacentric height Understand the limitations of calculating the angle of list using metacentric height Calculate the angle of list after loading operations if the vessel has an initial angle of list Calculating the required distribution of mass to achieve a final list angle or the upright condition Calculate angles of list for vessels with zero metacentric height Understand what is meant by free surface effect (FSE) and free surface moment (FSM)

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Small Angle or Initial Metacentric Stability • 39 Understand what is mean by, and calculate, the fluid, virtual or effective centre of gravity Understand what is meant by, and calculate, the free surface correction (FSC) Calculate the loss in metacentric height due to FSE Understand the effect of compartment subdivision on FSE and metacentric height Calculate the metacentric height after operations involving solid and fluid masses Calculate the loss in stability due to FSE in rectangular compartments Understand what is meant by an angle of loll Calculate the angle of loll using hydrostatics Calculate the angle of loll for a box shaped vessel Understand the procedure to recover a vessel from an angle of loll Understand why a vessel heels when turning and calculate the angle of heel when turning Calculate the increase in draught when inclined Calculate the effect of small load changes on GM

The Centre of Buoyancy (OOW, MCM, ENG) As previously shown, any object which floats is supported by buoyant forces pushing upwards. Although the buoyancy force acts on the underwater surface of the hull, the overall resultant buoyancy force acts upwards through a point known as the centre of buoyancy, B, which is located at the overall centre of the underwater volume. As this is a position within the vessel, it is described in terms of Cartesian coordinates, along a vertical, longitudinal and transverse measurement axis. The position of centre of buoyancy in a vertical sense is measured as a distance above the keel, and is abbreviated to KB. The position of centre of buoyancy in a longitudinal sense is measured as a distance along the ship as a distance from the aft perpendicular (the aft end of the summer waterline, which is usually the same as the rudder post, and is abbreviated to LCB, and quoted as a distance FOAP (Forward of the Aft Perpendicular). The position of centre of buoyancy in a transverse sense is measured as a distance from the ship centreline, and is abbreviated to TCB. These are shown in Figure 2.1. For a symmetrical ship in an upright condition, the TCB should always be on the centreline, and hence zero. For ships, determining the centre of the underwater volume is timeconsuming, and so the values for KB and LCB are given in the hydrostatics tables for the vessel.

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40 • Ship Stability, Powering and Resistance

LCB Aft perpendicular TCB

KB

▲ Figure 2.1 The centre of buoyancy

QUESTIONS Q2.1 (OOW, MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10 m, and floats at a draught of 2.00 m. Determine the LCB, KB and TCB values. Q2.2 (OOW, MCM, ENG) Determine the LCB and KB for MV Reed if her displacement is 3,563 tonnes.

The Centre of Gravity (OOW, MCM, ENG)

Adding loads In the same way that all the buoyant forces act through the centre of buoyancy, the mass forces can be assumed to act through a point known as the centre of gravity. The centre of gravity is measured longitudinally (LCG), vertically (KG) and transversely (TCG) in the same way as the centre of buoyancy. The TCG is sometimes also referred to as GGH. Distances to port are taken as positive, while distances to starboard are taken as negative (however, this may vary between ships – provided it is constant through any calculation this does not make a difference to results). The position of the centre of buoyancy is dependent on the shape of the vessel and her displacement, but the position of the centre of gravity depends on the distribution of the mass on board (which is directly under the control of the ship’s officers, and is the responsibility of the officers), and can be found by ‘loading tables’. Loading tables are a simple, tabular calculation.

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Small Angle or Initial Metacentric Stability • 41 A loading table lists the mass and the individual longitudinal, vertical or transverse position of all the items on the vessel, as shown in Tables 2.1, 2.2 and 2.3. Also listed is the ‘lightship’ displacement and centres of gravity. The lightship values for a ship are available from the stability data book. The mass and each position are multiplied together to find the longitudinal, vertical or transverse moments of mass, often just referred to as ‘moments’. The moment of mass is totalled up and divided by the total mass to find the centre of gravity in each direction (Formulae 2.1, 2.2 and 2.3). Overall l KG after loading =

Total v vertica ertical moment Total mass

▲ Formula 2.1 KG from a loading table

The KG value has units of metres.

Table 2.1 Sample loading table for KG Item

Mass (tonnes)

KG (m)

Vertical moment (tonne metres)

Lightship

Lightship mass

Lightship KG

Lightship mass × lightship KG

Cargo

Cargo mass

Cargo KG

Cargo mass × cargo KG

Fuel

Fuel mass

Fuel KG

Fuel mass × fuel KG

Crew/stores

Crew & stores mass

Crew & stores KG

Crew & stores mass × crew & stores KG

Total

Total mass

Total vertical moment

Table 2.2 Sample loading table for LCG Item

Mass (tonnes)

LCG (m FOAP)

Longitudinal moment (tonne metres)

Lightship

Lightship mass

Lightship LCG

Lightship mass × lightship LCG

Cargo

Cargo mass

Cargo LCG

Cargo mass × cargo LCG

Fuel

Fuel mass

Fuel LCG

Fuel mass × fuel LCG

Crew/stores

Crew & stores mass

Crew & stores LCG

Crew & stores mass × crew & stores LCG

Total

Total mass

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Total longitudinal moment

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42 • Ship Stability, Powering and Resistance Table 2.3 Sample loading table for TCG Item

Mass (tonnes)

TCG (m)

Transverse moment (tonne metres)

Lightship

Lightship mass

Lightship TCG

Lightship mass × lightship TCG

Cargo

Cargo mass

Cargo TCG

Cargo mass × cargo TCG

Fuel

Fuel mass

Fuel TCG

Fuel mass × fuel TCG

Crew/stores

Crew & stores mass

Crew & stores TCG

Crew & stores mass × crew & stores TCG

Total

Total mass

Overall l LCG after loading =

Total transverse moment

Total longitudinal ongitudina moment Total mas a s

▲ Formula 2.2 LCG from a loading table

The LCG value has units of metres. Overall l TCG after loading =

Total tra t nsverse moment Total mass

▲ Formula 2.3 TCG from a loading table

The TCG value has units of metres. Values for KG and LCG are normally calculated to the nearest centimetre, while values for TCG are usually calculated to the nearest millimetre or centimetre to reduce rounding accuracy. This will be discussed more later.

QUESTIONS Q2.3 (OOW, MCM, ENG) MV Reed starts loading from the lightship condition. 250 tonnes of cargo is loaded onto the vessel, 5.00 m above the keel, 55.00 m FOAP and 0.50 m to port of the centreline. Determine the overall KG, LCG and TCG of the vessel after loading the cargo.

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Small Angle or Initial Metacentric Stability • 43 Q2.4 (OOW, MCM, ENG) MV Reed starts loading from the lightship condition. 500 tonnes of cargo is loaded onto the vessel, at a KG of 8.00 m, an LCG of 48.00 m FOAP, on the ship centreline. Determine the overall KG, LCG and TCG of the vessel after loading the cargo.

Removing loads So far we have dealt with loading the vessel from the lightship condition. Loading tables also work if we are unloading cargo from a loaded condition. Instead of starting with the lightship, we start with the loaded ship. When masses are removed, the process is the same, but the removed masses are treated as negative masses.

QUESTION Q2.5 (OOW, MCM, ENG) MV Reed has a displacement of 2,915 tonnes, with a KG of 6.79 m, an LCG of 45.13 m FOAP and a TCG of 0.05 m. 150 tonnes of cargo is unloaded from the vessel, from a KG of 8.00 m, an LCG of 45.00 m FOAP and a TCG of –0.10 m. Determine the displacement and the overall KG, LCG and TCG of the vessel after unloading the cargo.

Moving loads Occasionally a mass which is aboard may need to be moved. A loading table is again used to calculate the KG after a mass has been moved. The mass is removed from the old location (treated as a negative mass) and replaced at the new location (treated as a positive mass).

QUESTION Q2.6 (OOW, MCM, ENG) MV Reed has a loaded displacement of 5,000 tonnes and a loaded KG of 6 m. Cargo, weighing 500 tonnes, is removed from a KG of 2 m and re-stowed at a KG of 4 m. Calculate the overall KG of the vessel after re-stowing.

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44 • Ship Stability, Powering and Resistance When a mass is moved, we can use a simple shortcut. The initial starting position and final position are not actually important – what controls the change in KG of the ship is the distance the mass moves. Try Question 2.6 again, but remove the mass from a KG of 10 m and replace it at a KG of 12 m – the final answer should be the same, as the actual distance the mass is moved through is the same. You can use which ever KG values you like, but make sure the distance between them is correct.

QUESTION Q2.7 (OOW, MCM, ENG) MV Reed has a loaded displacement of 6,615 tonnes and a loaded KG of 5.79 m. Cargo, weighing 500 tonnes, is moved downwards through a distance of 3 m. Calculate the overall KG of the vessel after re-stowing.

The movement of the centre of gravity of the ship, when a mass aboard is moved, can also be found using:

Change in G =

w × Distance t moved Δ

▲ Formula 2.4 Change in G due to moving masses

The change in G is measured in units of metres, w is the mass moved, in units of tonnes, the distance that the mass is moved is measured in metres and Δ is the displacement of the vessel, including the moved load, in tonnes. Note that the centre of gravity of the ship moves in the same direction as the mass which is moved. X For a mathematical proof of Formula 2.4, please see the explanation in

Appendix 7: The Derivation of the Change in KG Formula. Therefore, it can be see that the change in KG is dependent on the distance the mass is moved, rather than the actual positions of the mass.

Suspended loads Masses which are suspended (e.g. from a derrick during unloading) act at their point of suspension from the ship, not the actual position of the mass. Dealing with suspended

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Small Angle or Initial Metacentric Stability • 45 loads is similar to moved loads, with the mass being removed from the table at the original position and replaced in the table at the point of suspension.

QUESTION Q2.8 (OOW, MCM, ENG) MV Reed has a loaded displacement of 6,000 tonnes, and a loaded KG of 7 m. Cargo, weighing 50 tonnes, is lifted from a stowed position at KG of 3 m and suspended from a derrick head 20 m above the keel. Calculate the overall KG of the vessel with the load suspended at a position 4.00 m below the derrick head. (Hint – there is one piece of information in this question that you do not need.)

Occasionally it might be necessary to load an item of cargo so that the vessel has a particular final KG value. In cases such as this, algebra is needed to determine the position of the cargo. A loading table is used as before, except x is used as the KG of the cargo. As the final KG is known, the loading table can be completed in terms of x, and solved for x.

QUESTION Q2.9 (OOW, MCM, ENG) A vessel has a displacement of 8,000 tonnes and a KG of 6.00 m. 4,000 tonnes of cargo is to be loaded so that the final KG of the vessel and contents is 7.00 m. Determine the KG of the cargo.

Alternatively it might be necessary to load an item of cargo so that the vessel has a particular final KG value, where the KG of the cargo is known, but the mass of the cargo is unknown. In cases such as this, algebra is needed to determine the mass of the cargo. A loading table is used as before, except x is used as the mass of the cargo. As the final required KG is known, the value for x is the only unknown, and can be solved.

QUESTIONS Q2.10 (OOW, MCM, ENG) A vessel has a displacement of 10,000 tonnes and a KG of 12.00 m. Cargo is to be loaded at a KG of 7 m so that the final KG of the vessel and contents is 11.00 m. Determine the mass of the cargo.

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46 • Ship Stability, Powering and Resistance Q2.11 (OOW, MCM, ENG) A vessel has a displacement of 20,000 tonnes and a KG of 9.00 m. Cargo is to be unloaded from a KG of 5 m so that the final KG of the vessel and contents is 10.00 m. Determine the mass of the cargo to unload.

The Transverse Metacentre (OOW, MCM, ENG) The centres of buoyancy and gravity act together to determine the stability of the vessel. By analysing the relative positions of the centres of gravity and buoyancy an assessment can be made of the stability of the vessel. Remember – the mass forces, trying to sink the ship, act downwards through the centre of gravity (G), and the buoyancy forces, trying to float the ship, act upwards through the centre of buoyancy (B). When the buoyant forces equal the mass forces, the vessel floats. Viewed from the bow, the transverse centres of gravity (G) and buoyancy (B) lie on the centreline when the ship is upright, as shown in Figure 2.2.

G

B

▲ Figure 2.2 Upright equilibrium

The centre of buoyancy remains at the centre of the overall underwater volume. As the vessel rolls, the amount of underwater volume remains the same, but the shape of the underwater volume changes, therefore the overall centre of the underwater volume, which is the centre of buoyancy, moves as the vessel rolls, as shown in Figure 2.3. Please note that in Figure 2.3 the movements have been exaggerated for clarity. As long as the

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Small Angle or Initial Metacentric Stability • 47 cargo does not shift, G can be considered to be a fixed point relative to the vessel. This is a valid assumption for solid cargoes at small angles of inclination. Fluid cargoes, and large angles, will be dealt with later.

G

Old B New B

▲ Figure 2.3 The movement of B as the vessel rolls

As can be seen in Figure 2.4, the overall effect of the two misaligned forces is to create a torque which acts to try and roll the vessel upright – known as a righting moment.

Overall effect

G

Overall effect New B

▲ Figure 2.4 The righting moment

It is vital to be able to quantify the torque and forces acting to roll the vessel upright. If a vertical line is drawn up from the centre of buoyancy, then the point where it crosses the centreline is known as the metacentre, M, as shown in Figure 2.5. At small angles of heel (typically less than 10 degrees) the position of the metacentre relative to the centre of gravity can be assumed to remain constant. For this reason,

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48 • Ship Stability, Powering and Resistance

M

G

B

▲ Figure 2.5 The metacentre

Distance along the centreline between G and M (m)

1.4 1.3

Distance along the centreline between G and M

1.2

Initial distance along the centreline between G and M % variation

1.1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.0

5.0

10.0

15.0

60 55 50 45 40 35 30 25 20 15 10 5 0 –5 –10 –15 –20 –25 –30 –35 –40 20.0

Variation in distance along the centreline between G and M (%)

the (assumed to be) fixed metacentre at small angles of heel is often referred to as the ‘initial metacentre’. In practice, the metacentre moves, but only by a very small amount. Above around 10 degrees (depending exactly on hull form) the metacentre moves significantly. The change with heel in the position of G relative to M, for MV Reed, is shown in Figure 2.6.

Heel (degrees)

▲ Figure 2.6 The change in GM with heel for MV Reed

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Small Angle or Initial Metacentric Stability • 49 The position of the metacentre relative to the centre of gravity, measured along the centreline of the ship (known as the initial metacentric height or simply the metacentric height, abbreviated to GM) is the measure of what is known as the ‘small angle stability’ or ‘initial stability’ of the vessel. As previously stated, the OOW has direct control over the final centre of gravity of the vessel. Therefore, the OOW has control over the stability of the ship. If the vessel is loaded so that G is below M relative to the centreline, then the vessel is said to be ‘stable’ (as shown in Figure 2.7). This means that if she is inclined over, the torque will act to roll her back upright. The vessel is said to be stable, and the GM distance is taken to be positive.

M GM

G

B

▲ Figure 2.7 Metacentric height

If the vessel is loaded so that M coincides with G, as shown in Figure 2.8, then the vessel is said to be ‘neutral’. This means that if the vessel is inclined over, there is no overall torque trying to bring the vessel back upright. GM is zero. This is very important: If M is above G, the vessel is stable, and will return upright if a force causing it to incline is removed. GM is said to be positive. If M coincides with G, the vessel is neutral, and will remain at any small heel angle it is inclined to. GM is said to be zero. If M is below G, the vessel will be unstable and, if inclined even slightly, will incline further over. GM is said to be negative.

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50 • Ship Stability, Powering and Resistance

M

G

B

▲ Figure 2.8 Neutral stability

If the vessel is loaded so that M is below G, as shown in Figure 2.9, then the vessel is unstable, and will try and roll over further when inclined. The GM value is taken to be negative.

G

GM

M

B

▲ Figure 2.9 Negative metacentric height

A summary of the relative positions of G and M is shown in Figure 2.10 – note that as the position of the metacentre is assumed to be constant at small angles, we also can consider the metacentre to be in the same position when upright. If the positions of B, G and M above the keel are known, then GM can be found, as shown in Figure 2.11 and Formula 2.5.

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Small Angle or Initial Metacentric Stability • 51

M GM G

G M

B

B

Stable

G M B

Neutral

GM

Unstable

▲ Figure 2.10 Conditions of stability

M GM G KM

KG

▲ Figure 2.11 Measuring the metacentric height

GM = KM − KG ▲ Formula 2.5 GM from KM

GM, KM and KG are all measured in units of metres, normally rounded to the nearest centimetre. KG can be found using loading tables, as shown in the previously. KM can be calculated by dividing a value known as the ‘transverse inertia of the waterplane through the geometric centre’, which for obvious reasons is often just abbreviated to ‘inertia’, by the underwater volume of the vessel. The inertia is a mathematical measure of the distribution of the waterplane. For curved hull forms, this is difficult to calculate, so KM is usually pre-calculated by the Naval Architect and supplied to the ship in the hydrostatic data. The values for KM and KG can be used to find GM, using Formula 2.5. If the resulting GM is positive, the vessel is stable. If the resulting GM is zero, the vessel is neutral. If the resulting GM is negative, the vessel is unstable.

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52 • Ship Stability, Powering and Resistance QUESTIONS Q2.12 (OOW, MCM, ENG) MV Reed has a draught of 4.8 m and a KG of 5 m. Using the MV Reed Sample Stability Data Book (see Appendix 1), determine the metacentric height of the vessel, and comment on the answer. Q2.13 (OOW, MCM, ENG) MV Reed has a draught of 3.66 m and a KG of 7.32 m. Using the MV Reed Sample Stability Data Book (see Appendix 1), determine the metacentric height of the vessel. You will need to use linear interpolation. Comment on the answer.

These questions are easier with practice. The following questions tie together the work you have completed on loading tables and metacentric height.

QUESTIONS Q2.14 (OOW, MCM, ENG) MV Reed starts loading from the lightship condition. 1,386 tonnes of cargo is loaded at a KG of 8.00 m. Determine the final KG, KM and GM, and hence determine if the vessel will be stable. Q2.15 (OOW, MCM, ENG) MV Reed has a displacement of 6,008 tonnes with a KG of 6.50 m. 374 tonnes of cargo is unloaded from a KG of 7.00 m. Determine the final KG, KM and GM, and hence determine if the vessel will be stable. Q2.16 (OOW, MCM, ENG) MV Reed has a displacement of 7,192 tonnes, with a KG of 6.90 m. 500 tonnes of cargo is to be moved from a position 9.00 m above the keel to a position 6.00 m above the keel. Determine the GM after the cargo is re-stowed. Q2.17 (OOW, MCM, ENG) MV Reed is floating with a draught of 5.00 m and a KG of 6.00 m. An item of cargo, with a mass of 70 tonnes, is to be lifted from the hold (KG = 4.00 m) to the deck (KG = 12.00 m) using the ship’s own heavy lift crane, with a derrick head 30 m above the keel. Calculate GM before, during and after the cargo movement.

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Small Angle or Initial Metacentric Stability • 53 Q2.18 (OOW, MCM, ENG) MV Reed is being loaded. She starts in her lightship condition. She is to be loaded to a displacement of 5,150 tonnes, and a GM of 1.50 m. Determine the amount of cargo to load, and the KG to load the cargo.

Minimum Initial Metacentric Heights (OOW, MCM, ENG) As has been seen, the value of the metacentric height is very important. Minimum values for GM are described in the International Maritime Organisation’s (IMO) publication Code in Intact Stability 2008, known as the ‘2008 IS Code’ (International Maritime Organisation, 2008). These are partially incorporated into the Merchant Shipping (Load Line) Regulations 1998, as amended by the Merchant Shipping (Load Line) (Amendment) Regulations 2000, via Merchant Shipping Notice 1752(M) (Maritime and Coastguard Agency, 2000). It is important to note, however, that the IS Code is certainly not a guarantee of safety. Simply complying with the code is no absolute confirmation of safety. The 2002 edition of the code, in paragraph 2.5.1, states: Compliance with the stability criteria does not ensure immunity against capsizing, regardless of the circumstances, or absolve the master from his responsibilities. (International Maritime Organisation, 2002)

Vessels carrying grain cargoes For a vessel carrying a grain cargo, the minimum legal value of GM is 0.30 m, as set out in the Grain Regulations (International Maritime Organisation, 1991). This will be discussed in more detail later.

Vessels carrying timber deck cargoes For a vessel carrying a timber deck cargo, alternative criteria may apply. This is because of the reasons outlined in the section on Lumber and Timber Load Lines.

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54 • Ship Stability, Powering and Resistance Provided that the timber is securely packaged on deck, extends between the superstructures, and water absorption in the cargo is taken into account, the minimum GM at all times is 0.10 m, or 10 cm (see the 2008 IS Code, Part A, paragraph 3.3.2) (International Maritime Organisation, 2008). For UK flagged vessels, the MCA allow a reduction in the metacentric height to 0.05 m, as detailed in Schedule 2, paragraph 2.2.d of MSN 1752(M) (Maritime and Coastguard Agency, 2000). These small margins for error are why accurate stability calculations are so important. There are additional large angle stability requirements which must be complied with as well to enable the vessel to sail with a reduced GM, please see the information on this in the large angle stability section of the book.

Other vessels For vessels not covered above, such as general cargo vessels and passenger vessels, the minimum required value of GM is 0.15 m or 15 cm (see the 2008 IS Code, Part A, paragraph 2.2.4) (International Maritime Organisation, 2008).

Large Metacentric Heights (OOW, MCM, ENG) Obviously the vessel will be more stable with larger GM values. However, as GM increases, the torque trying to roll the vessel upright also increases, resulting in a faster roll. If GM is too large, then the vessel will roll too quickly (the roll period is said to be too short), and it may be difficult to stand upright or move around, and motion sickness may occur. In extreme conditions, it is possible to cause structural damage to the ship, or cause cargo to break loose and move around the ship, or injury to the crew. For this reason, GM values are normally kept within a certain range to ensure that the vessel is stable, but not too stable. Vessels with small GM values are said to be ‘tender’, as they roll slowly and incline under a small force. Vessels with large GM values are said to be ‘stiff ’, as they roll quickly and incline under a large force. The 2008 IS Code suggests that an appropriate maximum GM values for vessels with a timber deck cargo is 3% of the beam (see the 2008 IS Code, Part B, paragraph 3.7.5) (International Maritime Organisation, 2008); however, there is no overall set maximum GM value for all ships.

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Small Angle or Initial Metacentric Stability • 55

Calculating KB, BM and GM for a Box Shaped Vessel (MCM, ENG) For a simplified box shaped vessel, the KM value can be easily calculated. Figure 2.12 shows the arrangement of B, G and M for a box shaped vessel. It can be seen that: GM = KM − KG = KB + BM − KG ▲ Formula 2.6 GM from KB, BM and KG

All the values are measured in units of metres. CL

M GM G

BM

KM

KG

B KB

▲ Figure 2.12 KB, BM, KG and GM

For a box shaped vessel, the KB can be found directly from the geometry. As the centre of buoyancy is the centre of the underwater volume, for a box shaped vessel it must be at half the length, half the draught and on the centreline (see Question 2.1). Therefore, for a box shaped vessel: KB =

D 2

▲ Formula 2.7 KB for a box shaped vessel

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56 • Ship Stability, Powering and Resistance Finding the distance from B to M, or BM, is more complicated. BM depends on the underwater volume of the vessel, and also on a value known as the ‘transverse inertia of the waterplane through the geometric centre’, often simply referred to as ‘inertia’. This ‘transverse inertia of the waterplane through the geometric centre’, or ‘inertia’, is a property of the geometry of the waterplane. To complicate matters further, inertia is also sometimes referred to as second moment of area. The inertia of a shape is essentially a measure of the distribution of area relative to an axis running through a point within a shape. Consider the two shapes shown in Figure 2.13. Both shapes have exactly the same length and surface area, but taking a horizontal axis through the geometric centre of the shape as the measuring point, the right hand shape has slightly more height, and therefore more of the area of the shape distributed further away from the centreline of the shape. The right hand shape would therefore have greater inertia, even though it has exactly the same total area as the left hand shape.

▲ Figure 2.13 Inertia

The value (valid only for box shaped vessels) for the inertia for a rectangular waterplane is found using:

Inertia =

LB 3 12

▲ Formula 2.8 Waterplane inertia for a box shaped vessel

X For a discussion and mathematical proof of the inertia of a rectangle, and

a ship shaped waterplane, please see Appendix 8: The Derivation of the Formulae Giving the Transverse Inertia of a Rectangular Waterplane and a Ship Shape Waterplane Measured through the Centreline. Once the inertia has been found, the vertical distance from B to M can be found. For a box shaped vessel, BM can be found from:

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Small Angle or Initial Metacentric Stability • 57

BM =

Inertia ∇

▲ Formula 2.9 BM for a box shaped vessel

X For a mathematical proof of this for a box shaped vessel, see Appendix

9: The Derivation of the Relationship between Transverse Inertia, BM and Underwater Volume for a Box Shaped Vessel. For a mathematical proof of this for a ship, see Appendix 10: The Derivation of the Relationship between Transverse Inertia, BM and Underwater Volume for a Ship. The values from Formulae 2.8 and 2.9 can be combined to give KM for a box shaped vessel: KM = KB + BM =

D Inertia + 2 ∇

Given that the inertia and the volume for a box shaped vessel can be expressed in terms of the length, beam and draught of a vessel, Formula 2.9 can be simplified: LB 3 Inertia 12 LB 3 1 B2 BM = = = × = ∇ LBD 12 LBD 12D

QUESTIONS Q2.19 (MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10 m, and floats with a draught of 2.00 m. Determine GM if KG is 4 m. Q2.20 (MCM, ENG) A box shaped vessel has a length of 60 m, a beam of 8 m, and floats with a draught of 1.00 m. Determine the required KG if GM is to be 0.15 m. Q2.21 (MCM, ENG) A box shaped vessel has a beam of 9.00 m. The vessel has a draught of 3.00 m. The KG is 1.25 m. Determine the GM. Note that you do not require the length.

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58 • Ship Stability, Powering and Resistance Q2.22 (MCM, ENG) A box shaped vessel has a length of 60 m and a beam of 8.00 m. The vessel has a draught of 2.00 m in sea water. The KG is 3.00 m. 492 tonnes of cargo is then loaded on to the vessel at a KG of 5.00 m. Determine, and comment on, the GM of the vessel after loading the cargo. Q2.23 (MCM, ENG) For the vessel in Question 2.22, determine the maximum KG of the cargo so that the vessel finishes loading with a positive metacentric height.

Metacentric Diagrams (ENG) The values for KB and KM for a range of draughts can be plotted on a graph, with the draught on the y axis and KB and KM on the x axis. This is known as the metacentric diagram. The metacentric diagram allows BM to be determined for any draught (as it is the difference between KB and BM). The metacentric diagram for MV Reed is shown in Figure 2.14.

KB

KM

8

7

Draught (m)

6

5

4

3

2 1

2

3

4

5

6

7

8

9

10

KB (m) and KM (m)

▲ Figure 2.14 Metacentric diagram

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Small Angle or Initial Metacentric Stability • 59

Angles of List (OOW, MCM, ENG) If masses are moved transversely or loaded asymmetrically, then the overall centre of gravity of the vessel will move off of the centreline of the vessel, as shown in Figure 2.15. In this condition the vessel will have a non-zero TCG value, meaning that the centre of gravity will no longer line up with the centre of buoyancy. This will result in a torque trying to incline the vessel over, known as a listing moment. The vessel will incline over until the centre of buoyancy is again directly under the centre of gravity, as shown in Figure 2.16. In this condition, the vessel is said to have an ‘angle of list’.

M

G TCG

B

▲ Figure 2.15 Shift in the TCG

The processes and calculations used so far can be combined to give a method of calculating the ‘angle of list’ if a mass already aboard is moved transversely. This has a major limiting factor – the theory assumes GM is constant (i.e. the metacentre does not move). As seen earlier, this is only valid up to a list angle of around 10 degrees. List angles above 10 degrees will be covered later. The geometry of the metacentre and the TCG can be used to determine the angle of list. Figure 2.17 shows these in detail. From the TCG dimension, the line between B and M, and the centreline of the vessel, we can form a right angle triangle. The tip of the triangle, between the centreline of the ship and the vertical line of action of buoyancy, is the angle of list.

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60 • Ship Stability, Powering and Resistance

M

G

B

▲ Figure 2.16 Angle of list

M GM List angle TCG G

▲ Figure 2.17 Finding the angle of list

From basic trigonometry, it is known that for a right angle triangle: tanθ =

Opposite Adja d cent

From the right angle triangle in Figure 2.17, it can be seen that the opposite side of the triangle is given by the TCG, and the adjacent side is given by the GM. Substituting these values gives:

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Small Angle or Initial Metacentric Stability • 61

tanθ =

TCG GM

▲ Formula 2.10 Small angle of list (TCG based)

QUESTION Q2.24 (OOW, MCM, ENG) MV Reed has been loaded so that her metacentric height is 0.81 m. Her TCG value after loading is 0.043 m. Determine her angle of list.

When using the TCG (also sometimes known as GGH), and the convention that distances to port are positive, and starboard negative, angles from Formula 2.10 to port are positive, and angles to starboard are negative. It can be seen that the angle of list is inversely proportional to GM. Therefore, for a given TCG, a reduction in GM will result in an increase in list.

QUESTION Q2.25 (MCM, ENG) A boxed shaped vessel has a length of 70 m, and a beam of 9 m. She floats with a draught of 2 m. In this condition, she is upright, and her KG is 4.00 m. 100 tonnes, which is already aboard, is moved transversely 0.5 m to starboard. Determine the resulting list angle.

Alternatively, the list may be found by the use of listing moments. Assuming that the vessel is upright initially, if a mass, ‘w’ tonnes, already aboard, is moved transversely a distance ‘d’ metres, then the resulting total of the moment column in the loading table would be ‘w × d’ (see Table 2.3). The overall TCG can be found by dividing the total moment by the displacement, as given by Formula 2.3:

TCG =

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Total transverse moment w × d = Δ Δ

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62 • Ship Stability, Powering and Resistance Substituting this into Formula 2.10 gives: w ×d w ×d tanθ = Δ = GM Δ × GM Therefore, for movements of mass, when the vessel is initially upright:

tanθ =

w×d Δ × GM

▲ Formula 2.11 Small angle of list (mass and distance based)

The value w × d is often referred to as the heeling moment, even though it creates an angle of list. All the theory learnt so far can be used to analyse the loading or change in loading of a ship. This seems complex initially, but is easier with practice. The objective is to find GM, and then a list angle.

QUESTIONS Q2.26 (OOW, MCM, ENG) MV Reed starts loading from her lightship condition. 2,654 tonnes of cargo is added at a KG of 5.00 m, 0.050 m to port of the centreline. Determine the GM and the angle of list after loading the cargo. Q2.27 (OOW, MCM, ENG) MV Reed floats in her lightship condition. 1,000 tonnes of hold cargo is loaded at a KG of 5.00 m, 0.30 m to port of the centreline. 497 tonnes of deck cargo is loaded at a KG of 8.00 m, 0.50 m to starboard of the centreline. After loading the vessel, determine the metacentric height and the list.

If the vessel is loaded from a condition with an initial list, then the TCG of the vessel must be found before loading. This can be found using Formula 2.10. This initial TCG value must then be used in the first line of the TCG loading table.

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Small Angle or Initial Metacentric Stability • 63

QUESTIONS Q2.28 (OOW, MCM, ENG) MV Reed starts loading from a displacement of 4,001 tonnes, and a KG of 6.00 m. In this condition she has a list of 4 degrees to port. Determine the final list angle if 1,000 tonnes is added at a KG of 5.50 m, 0.50 m to port of the centreline, and 500 tonnes is added at a KG of 5.00 m, 1.00 m to starboard of the centreline. Q2.29 (OOW, MCM, ENG) MV Reed starts loading from a displacement of 3,781 tonnes, and a KG of 6.10 m. In this condition she has a list of 3 degrees to starboard. Determine the list if 109 tonnes is added at a KG of 9.00 m, 1 m to port of the centreline. Q2.30 (OOW, MCM, ENG) MV Reed has a displacement of 5,000 tonnes and is listing by 4 degrees to starboard. The GM is 1.00 m. Space is available to load cargo 4.5 m to port of the centreline. Determine the amount of cargo to load so that the vessel finishes loading upright, assuming that GM remains positive.

Remember – all of these calculations for determining the list of a vessel are based on the metacentric height, and are therefore only valid if the angle of list is small (less than approximately 10 degrees). If the angle of list is larger, a different method, known as large angle stability, must be used. The tangent function in the list formula can result in what appears to be large rounding errors in the list angle, depending on the rounding accuracy of the TCG. This is shown in Table 2.4, which shows the variation in the calculated angle of list for a vessel with Table 2.4 Effect of rounding on list angles Rounding (decimal places)

TCG (m)

List angle (degrees)

6

0.123456

14.441697

5

0.12346

14.44219

4

0.1235

14.4471

3

0.123

14.386

2

0.12

14.02

1

0.1

11.6

0

0

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0

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64 • Ship Stability, Powering and Resistance a TCG of 0.123456 m, and a GM of 0.50 m, as the TCG is rounded to lower levels of precision. As can be seen, the rounding error increases if the TCG is rounded to below three decimal places. Keeping the TCG value to three decimal places helps to ensure that any rounding errors are not too excessive, however the accuracy of the solution will be dependent on the accuracy of data given to calculate the TCG. X For passenger vessels, there is also a requirement to determine the angle

of list as a result of passengers crowding to one side of the vessel. This is explained in Appendix 11: Maximum Angles of List on Passenger Vessels due to Passenger Crowding.

QUESTION Q2.31 (OOW, MCM, ENG) MV Reed is alongside a berth with a displacement of 3,781 tonnes. In this condition, she has a GM of 1.50 m and a list of 3.0 degrees to starboard. Cargo is to be moved transversely across a deck, from a position 4 m to starboard of the centreline to a position 6 m to port of the centreline. Determine the amount of cargo to move so that the vessel returns upright.

Angles of List for Vessels with Zero GM (MCM, ENG) When a vessel has zero GM, she has neutral stability. If a weight is then moved across the vessel so that the centre of gravity moves off of the centreline of the vessel, then the vessel will incline until the centre of buoyancy lines up with the moved centre of gravity and the vessel goes back into a condition of equilibrium. The list angle of a vessel with neutral stability, caused by a mass w a distance d from the centreline, can be found from:

3

w×d Δ = tanθ BM

2×

▲ Formula 2.12 The angle of list for a vessel with zero GM

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Small Angle or Initial Metacentric Stability • 65 X For a mathematical proof of this, see Appendix 12: The Derivation of the

Formula Giving the Angle of List for a Neutrally Stable Vessel. Note that BM is the same as KM – KB, as shown in Figure 2.12.

QUESTION Q2.32 (MCM, ENG) MV Reed starts loading from her lightship condition. 1,386 tonnes of cargo is loaded at a KG of 7.10 m, 0.02 m to port of the centreline. Determine the resulting angle of list.

Fluids and Free Surface Effects (OOW, MCM, ENG) So far we have considered loading solid items of cargo onto a ship. When fluid is added to the vessel, either as cargo such as crude oil, or as water ballast or fuel, then the hydrostatic tank data, given in the stability data book, must be used to determine the mass and centre of gravity of the fluid. An example of hydrostatic tank data is given in Figure 2.18. No. 3 Double Bottom Port Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

metres

metres

metres

m4

metres

metres

%

m3

2.000

0.000

100.000

245.966

49.375

4.159

1.057

0.000

1.961

0.038

97.900

240.800

49.369

4.153

1.037

539.733

1.900

0.098

94.600

232.574

49.360

4.144

1.005

536.830

1.800

0.198

89.000

219.031

49.345

4.128

0.953

531.593

1.700

0.298

83.600

205.546

49.330

4.111

0.900

525.930

1.600

0.398

78.100

192.126

49.313

4.092

0.848

519.630

▲ Figure 2.18 Tank hydrostatic data

For any sounding (depth of fluid in the tank), the row of data at that sounding shows the ullage of the tank (depth of the air gap at the top of the tank); the capacity of the tank at that sounding in metres3; and the longitudinal, transverse and vertical centres of gravity of the fluid in the tank. An additional value, known as the free surface moment, or FSM is also shown. This will be discussed later.

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66 • Ship Stability, Powering and Resistance To determine the mass of fluid in a tank at any sounding, the capacity must be multiplied by the density of the fluid in the tank.

QUESTIONS Q2.33 (OOW, MCM, ENG) Determine, to two decimal places, the mass, KG, LCG and TCG of sea water ballast in the No. 3 Double Bottom Port tank at a sounding of 1.80 m. Q2.34 (OOW, MCM, ENG) MV Reed starts in the lightship condition. The forepeak and after peak tanks are pressed full with sea water ballast, and both bunkers are pressed full of oil fuel, relative density 0.95. Determine the displacement, LCG, TCG and KG of the vessel after loading.

Sounding

Sounding pipe

Soun

ding

▲ Figure 2.19 Variation in sounding with ship inclination

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Small Angle or Initial Metacentric Stability • 67 If the vessel is heeled or listed, then the sounding measured at the sounding pipe in the tank will not be the actual sounding, as shown in Figure 2.19. For each tank, tables of corrections for soundings and list or heel angles are provided so that the actual sounding can be determined. For the purposes of this book, the corrections will be neglected. So far we have only considered the mass and centre of mass of the fluids. Carrying fluids can, under certain circumstances, have a very significant effect on the stability of ships. This effect is known as ‘free surface effect’, and is a reduction in the metacentric height, GM, as a result of the fluid sloshing around the tanks or compartments. FSE has been the primary cause of numerous losses of ships, especially Ro-Ro vessels and fishing vessels. Consider an amidships cross-section of a simple vessel with a tank containing fluid, as shown in Figure 2.20.

G of the fluid

▲ Figure 2.20 Centre of gravity of a pressed tank

As the vessel heels under the action of wind and waves, the fluid retains the shape of the tank, and the overall centre of gravity of the fluid remains at the tank centre – there is no movement of fluid and no FSE. However, if the tank is partially full, then the fluid can move within the tank as the ship moves. As the fluid moves, the centre of gravity of the fluid will also move, in a transverse and vertical direction. If the vessel lists, the fluid will flow ‘downhill’, causing the vessel to list further, as shown in Figure 2.21. As the fluid moves ‘downhill’, a listing moment will be generated, and the vessel will list further over. In effect, the vessel is less stable than she would be with a solid cargo, as a result of the change of shape of the fluid. This reduction in stability is accounted for by a rise in the centre of gravity of the vessel, and hence a reduction in the metacentric

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68 • Ship Stability, Powering and Resistance

Fluid ballast

Solid ballast

▲ Figure 2.21 Free surface effect

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Small Angle or Initial Metacentric Stability • 69

Fluid ballast

Solid ballast

▲ Figure 2.21 Continued

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70 • Ship Stability, Powering and Resistance

Fluid ballast

Solid ballast

▲ Figure 2.21 Continued

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Small Angle or Initial Metacentric Stability • 71 height of the vessel. The left hand column in Figure 2.21 shows a vessel inclining to an angle of list with fluid ballast in the double bottom. The right hand column shows the same vessel, with the same initial TCG and GM, with the same mass and location of ballast, but assuming the ballast is solid. As can be seen from the diagram, the vessel with the fluid ballast will list significantly further than the vessel with the solid ballast, and can therefore be considered to be less stable. This is because the movement of the fluid ‘downhill’ within the vessel exaggerates the list angle by moving the TCG of the ship further outboard. The centre of gravity of the vessel is therefore raised from a ‘solid’ position, to a position known either as the ‘fluid’, ‘effective’ or ‘virtual’ position. The rise in the centre of gravity from the solid to the effective or virtual position can be explained by looking at Figure 2.22.

M Original GM

Effective, fluid, or virtual GM

GV

G TCG

B

▲ Figure 2.22 The effective, fluid or virtual centre of gravity

As the vessel heels, the centre of gravity of the fluid moves outboard, and as a result, the overall TCG of the vessel moves further towards the low side of the vessel. Note that this is different from the previous list case, as shown in Figure 2.16, as the TCG of the tank contents and vessel depends on the list angle. This effect on the stability can be modelled by looking at the intersection of the line of action of gravity with the centreline. This point is the ‘fluid’, ‘effective’ or ‘virtual’ position

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72 • Ship Stability, Powering and Resistance of the centre of gravity. It can be seen that the effect of this is to reduce GM from the original value to the effective, fluid or virtual value. The upwards movement of the centre of gravity, and effective reduction in the metacentric height of the vessel, which is effectively the same as the reduction in GM, is known as the free surface effect (FSE) or free surface correction (FSC). The FSE depends on a value known as the ‘transverse inertia of the free surface, measured through the centre of the surface’, but often simply known as the ‘free surface moment’, or FSM. This is a measure of the shape of the free surface, and is precalculated by the Naval Architect, or for a rectangular tank, can be directly calculated based on the length and beam of the tank. For curved tanks, it can be found from the ship’s hydrostatic data book. The FSM is normally quoted in units of either metres4, or tonne metres. If units of metres4 are quoted, then the FSM must be multiplied by the density of the fluid in the tank to get the actual value. If units of tonne metres are quoted, then no correction is required. Once the FSM value corrected for density is known, then an extra line for each free surface is added to the KG loading table, as shown in Table 2.5. In this line for each free surface, the mass values are set to zero, the KG value is set to zero and the corrected FSM value is put in the ‘vertical moment’ column. The resulting KG value calculated from the loading table will then be the ‘fluid’, ‘effective’ or ‘virtual’ KG, and the GM value calculated using this will be the ‘fluid’, ‘effective’ or ‘virtual’ GM. FSMs are not used in the TCG or LCG loading tables, as the rise of the KG from the solid to the fluid value models all of the aspects of the FSE.

Table 2.5 Sample KG loading table including FSM Item

Mass (tonnes)

KG (m)

Vertical moment (tonne metres)

Lightship

Lightship mass

Lightship KG

Lightship mass × lightship KG

Cargo

Cargo mass

Cargo KG

Cargo mass × cargo KG

Swimming pool

Pool water mass

Pool water KG

Pool water mass × pool water KG

Swimming pool FSM

ZERO

ZERO

Pool water FSM from the hydrostatics, corrected for density

Crew/stores

Crew & stores mass Crew & stores KG

Crew & stores mass × crew & stores KG

Total

Total mass

Total vertical moment, including FSM

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Small Angle or Initial Metacentric Stability • 73

Overall l KG after loading =

Total vertica ve t cal moment ( iincluding c ud g FS F M) Total mass

▲ Formula 2.13 Effective KG from a loading table

It is the ‘fluid’, ‘effective’ or ‘virtual’ KG, and the ‘fluid’, ‘effective’ or ‘virtual’ GM values that must be used in all subsequent stability calculations to determine the stability of the vessel. In summary, the only difference between loading solid items and fluid items in stability calculations is to include an FSM row after the item in the KG loading table. This increases KG, and reduces GM.

QUESTION Q2.35 (OOW, MCM, ENG) MV Reed starts loading from the lightship condition. 282 tonnes of deck cargo is loaded at a KG of 11 m. All of the No. 3 and No. 4 Double Bottoms are filled to a sounding of 0.10 m with sea water ballast. Determine, and comment on, the effective GM of the vessel after ballasting.

The actual loss in GM (known as the FSC) due to FSEs can be directly calculated. The FSE, or FSC, is then found by dividing the total FSM, corrected for density, by the displacement of the vessel: FSC =

FSM Δ

▲ Formula 2.14 Free surface correction

X For a mathematical proof of the effects of free surfaces, see Appendix 13:

The Derivation of the Formulae Describing the Effect of Free Surfaces on Metacentric Height. The FSE (or FSC) can then be directly added to the solid KG to find the effective or fluid KG, or subtracted from the GM to find the effective or fluid GM. However, it is better to use the FSM in the loading tables, as there are fewer opportunities to forget to apply it,

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74 • Ship Stability, Powering and Resistance incorrectly apply it, or make a mistake in the calculation. Both methods of dealing with FSE should give the same final answer. FSE reduces stability, and is clearly dangerous. FSE can be reduced by efficient tank design. If the tank is subdivided by longitudinal baffles (fore and aft vertical plates in the tank), then the FSM, and hence the FSC is reduced as the fluid cannot flow as far transversely. Further reduction in FSC can be achieved by placing longitudinal bulkheads within the tank, which again reduces the mass of fluid which can move, and the distance it can move through transversely. When dividing a tank like this, the FSC is reduced by the number of equal size compartments squared. For example, if the tank is longitudinally split into two equal size compartments, then the FSC will be reduced by a factor of 22, or 4. If the tank is split into three equal size compartments, then the FSC will be reduced by a factor of 32, or 9. If the tank is split into four equal size compartments, then the FSC will be reduced by a factor of 42, or 16.

QUESTION Q2.36 (OOW, MCM, ENG) MV Reed floats in her lightship condition. The port bunker is filled to a sounding of 6.00 m with fuel oil, relative density 0.96. The starboard bunker is filled to an ullage of 2.04 m with fuel oil, relative density 0.96. Cargo, with a mass of 535 tonnes, is loaded at a KG of 0.50 m, 0.20 m to port of the centreline. No. 3 Double Bottom starboard is filled with sea water ballast to a sounding of 0.50 m. After loading the cargo, bunkering and ballasting, determine the displacement, metacentric height and the list.

Clearly FSEs are very dangerous – it is the fluid, or effective GM and KG values that should be used with all stability calculations.

QUESTION Q2.37 (OOW, MCM, ENG) MV Reed starts in the lightship condition. Both Bunkers are then filled to a sounding of 6.00 m with fuel, relative density 0.96. 3,209.1 tonnes of cargo is then loaded, at a KG = 5.50 m, with a TCG = 0.145 m to Port of the centreline. No. 3 Double Bottom Port is then filled to a sounding of 1.50 m with SW ballast, and No. 3 Double Bottom Starboard is filled to a sounding of 1.70 m with SW ballast. Determine the resulting GM and list.

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Small Angle or Initial Metacentric Stability • 75 If a compartment is partially full, or slack, and the fluid is completely removed from the tank, then the free surface of the fluid, which creates the FSE, must also have been removed. The FSM row must still be included in the KG loading table, but the FSM value should be treated as negative, as the free surface has been removed.

QUESTION Q2.38 MV Reed has a displacement of 5,150 tonnes, with an effective KG of 6.80 m. In this condition, the No. 4 Port Double Bottom is partially filled to a sounding of 1.00 m. Determine the effective KG of the vessel if the No. 4 Port Double Bottom is then completely emptied.

If a compartment is full, or pressed, and some (but not all) of the fluid is removed from the tank, then a free surface, which creates the FSE, must have been created. An FSM row must therefore be included in the KG loading table, with a positive value, as the free surface has been created. The easiest way to do this is to completely ‘empty’ the tank in the KG loading table, and then ‘refill’ the tank to the new sounding.

QUESTION Q2.39 MV Reed has a displacement of 5,150 tonnes, with an effective KG of 6.80 m. In this condition, the No. 4 Port Double Bottom is partially filled to a sounding of 1.00 m. Determine the effective KG of the vessel if the No. 4 Port Double Bottom is then de-ballasted to a sounding of 0.50 m.

Free Surface Effects in Rectangular Tanks (OOW, MCM, ENG) For rectangular tanks, the FSM can be directly calculated using the dimensions of the tank. The FSM depends on the ‘transverse inertia of the free surface measured about the centre of the surface’, again more commonly referred to as ‘inertia’. This is found from the length and beam of the free surface, as is shown in Formula 2.15. This gives the moment in units of m4, and so the density correction is still required.

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76 • Ship Stability, Powering and Resistance

FSM =

lb3 12

▲ Formula 2.15 Free surface moment for a rectangular tank

QUESTIONS Q2.40 (MCM, ENG) A boxed shaped vessel has a length of 60 m, and a beam of 7 m. She floats in a lightship condition with a draught of 3 m. In this condition, her KG is 2.60 m. A double bottom tank, with a length of 10 m, and a beam of 6 m, is then partially filled to a depth of 1.00 m with oil fuel, relative density 0.97. Determine the GM after bunkering.

A Summary of Formulae for Real Ships and Box Shaped Vessels (MCM, ENG) Parameter

Real ships

Draught

Interpolate from hydrostatics

Displacement

Interpolate from hydrostatics

Underwater volume

∇=

Δ ρ

Centres of gravity Loading tables

Box shaped vessels

=

Δ from o Δ = ∇ ×ρ LB ρ

Δ=∇×ρ Δ = LBD

Loading tables

KB

Interpolate from hydrostatics

BM

BM = KM − KB

LB 3 Inertia 12 BM = = ∇ ∇

KM

Interpolate from hydrostatics

KM = KB + BM

GM = KM − KG

GM = KB + BM − KG

Interpolate from hydrostatics

FSM =

GM 4

FSM (m )

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KB =

D 2

lb3 (Rectangular ecta gu gula a tanks k only ly ) 12

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Small Angle or Initial Metacentric Stability • 77

Instability and Angles of Loll (OOW, MCM, ENG) As previously discussed, if M is below G (negative GM), the vessel will be unstable and will incline over further. As the vessel inclines over to larger angles, she will enter a new equilibrium condition known as an ‘angle of loll’. In the unstable (negative GM) condition, the centres of gravity and buoyancy act to roll the vessel from the upright over to a larger angle, as shown in Figure 2.23. Overall effect

G

M

B Overall effect

▲ Figure 2.23 Unstable vessel

As the vessel rolls, the shape of the underwater volume changes, and for a typical ship form, the centre of the underwater volume (the centre of buoyancy) moves further outboard, away from the original position. For typical ship forms, the rolling as a result of instability continues until a point where the centre of buoyancy has moved sufficiently far that it is under the centre of gravity, as shown in Figure 2.24.

G

G

Wind and waves

M

G M

M

B Upright but unstable G is above M – any small disturbance will cause the vessel to incline

B Capsizing moment The action of G and B causes a capsizing moment

B Angle of loll G and B realign in equilibrium

▲ Figure 2.24 Movement of B

Once in this condition, the forces line up and the vessel is in equilibrium – at an angle of inclination known as the angle of loll, as shown in Figure 2.25.

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78 • Ship Stability, Powering and Resistance

G

B

▲ Figure 2.25 Angle of loll

When in this condition, the vessel is in an extremely dangerous state, and while it will appear to be steady at an angle of loll, it may only be seconds away from a complete capsize. Recovering the vessel from this condition requires careful calculations and redistribution of ballast and cargo in a particular sequence. The angle of loll may be so large that the vessel starts to ‘down-flood’. This is the term used to describe uncontrollable flooding through fittings on the deck which are not watertight. If the vessel down-floods, she may capsize or sink under the additional mass of water. This risk means that GM must always be kept positive under all circumstances. If GM becomes negative, then it must be returned to a positive value very carefully. A specific feature of an angle of loll is that they can occur both to port or starboard. A very small change in mass distribution is needed to make it roll from port loll to starboard loll or vice versa. As the vessel rolls from port to starboard, it gains momentum, and can roll past the angle of loll on the opposite side, as shown in Figure 2.26. It is possible that it will roll far enough for down-flooding or capsize. If the vessel is assumed to be wall sided, and has not inclined so that the deck edge is under, then the angle of loll can be predicted:

tan =

−2 × GM KM − KB

▲ Formula 2.16 Angle of loll

Note that KM-KB is the equivalent of BM. The proof of Formula 2.16 (which is examinable at MCM level) is covered in the discussion of the wall-sided formula later.

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Small Angle or Initial Metacentric Stability • 79 Additional force

1. Lolling to port

Momentum

4. Rolling to starboard

Momentum

2. Rolling to starboard

Momentum

5. Lolling to starboard

Momentum

3. Through upright

Momentum

6. Roll past loll due to momentum

▲ Figure 2.26 Capsize due to loll

QUESTIONS Q2.41 (MCM, ENG) MV Reed starts loading from her lightship condition. 1,386 tonnes of cargo is loaded at a KG of 7.20 m, on the centreline. Determine the resulting angle of loll. Q2.42 (MCM, ENG) A box shaped vessel has length of 50.00 m, and a beam of 8.00 m. In her lightship condition, she has a displacement of 1,110.0 tonnes, and a KG of 3.15 m. The vessel contains a rectangular swimming pool, with a length of 10.00 m, and a width of 8.00 m. The base of the pool is 5.00 m above the keel. The vessel starts in her lightship condition. The pool is then filled with fresh water until the true mean draught of the vessel is 3.00 m in sea water. Determine the resulting angle of loll.

To recover the vessel from an unstable condition, the GM needs to be increased. This can be achieved by moving G downwards. However, for safety, we cannot risk the vessel

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80 • Ship Stability, Powering and Resistance

1. Lolling to port

2. Moving G

3. Positive GM and list

6. Find out why it happened

4. Correct list

5. Stable and upright

▲ Figure 2.27 Recovering loll

rolling to the opposite side, so we must also move G towards the low side of the vessel, as shown in Figure 2.27. This will make a combined loll and list angle. As GM increases, the loll will reduce and eventually become pure list as GM becomes positive again. This will be covered in more detail later.

Heel Angle in a Turn (MCM, ENG) As a vessel turns, centripetal force acts inwards towards the centre of the turn, through the centre of buoyancy of the vessel. The equal and opposite force is the centrifugal force, which acts through the centre of gravity. This creates a couple which generates a heeling moment about the centre of buoyancy of the vessel. The heel due to the turn can be found from:

tanθ =

v 2 × (K KG KB) g × r × GM

▲ Formula 2.17 Heel in a turn

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Small Angle or Initial Metacentric Stability • 81 In this, g is the acceleration due to gravity (9.81 m/s2), r is the turn radius in metres, and v is the speed of the ship in metres per second. To get the speed in metres per second, multiply the speed in knots by 0.514. Occasionally, the (KG − KB) term in the formula may be written as (B ~ G). This has the same meaning.

QUESTION Q2.43 (MCM, ENG) MV Reed has a draught of 5.05 m, and is travelling at a speed of 16 knots. Her KG is 6.20 m. At this speed, she enters a turn to starboard with a diameter of 400 m. Determine the angle of heel due to the turn.

X The proof of the formula giving the heel angle in a turn requires some more

knowledge about stability, and is therefore shown in the section on large angle stability. X For passenger vessels, there is also a requirement to determine the angle

of list as a result of passengers crowding to one side of the vessel. This is explained in Appendix 14: Passenger Vessel Heel in a Turn.

Increase in Draught due to Inclining (MCM, ENG) As a vessel inclines over, the draught of the vessel increases. Figure 2.28 shows the increased draught compared to the original draught for a heeling vessel.

▲ Figure 2.28 Increase in draught with inclination

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82 • Ship Stability, Powering and Resistance If a box shaped vessel is considered (which is similar to most modern vessels in the midship section) then the increase in draught can be found from: DHEELED = DUPRIGHT × cos θ +

B × sin i θ 2

▲ Formula 2.18 Increase in draught due to heel X For a derivation of this formula, see Appendix 15: Derivation of the Increase

in Draught When Heeling Formula. The increase in draught as a result of heeling can be quite large, especially when small under-keel clearances are considered. Table 2.6 shows the calculated draught increase for MV Reed as she heels. Table 2.6 Draught increase due to heeling Angle (degrees)

Draught increase (%)

0

0

2

4

4

8

6

11

8

15

10

18

12

22

14

25

16

28

18

31

20

33

Note that this formula is only valid up to a heel angle at which the deck is submerged, or the keel emerges. At this point, the underwater shape of the vessel changes considerably. Neither is the effect of trim with heel included, which is discussed in detail later in the book.

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Small Angle or Initial Metacentric Stability • 83 QUESTION Q2.44 (MCM, ENG) A box shaped barge has a length of 90.00 m, and a beam of 9.00 m, as shown below. Initially the draught of the vessel is 2.00 m in sea water, and the metacentric height is 0.48 m. The vessel has no trim or list, and all tanks are empty. An amidships rectangular ballast tank, extending across the full beam of the vessel, with a length of 10.00 m, is then partially filled to a sounding of 1.50 m with sea water ballast. The base of the tank is on the keel. 10 m

9m 2m

90 m

The barge is then towed at a speed of 10 knots in water with a depth of 4.00 m, in a turn with a diameter of 400 m. Ignoring the effect of squat, determine the under-keel clearance of the vessel during the turn.

Small Changes in GM due to Small Changes in Loading (ENG) If a small load is applied to the vessel, so that the sinkage is small and the TPC and waterplane area can be assumed to be constant, then an approximation can be used to determine the new GM, rather than recalculating or interpolating the hydrostatics. If the waterplane area is assumed to be constant, therefore any additional volume gained (or lost) due to adding (or removing) a load can be assumed to be a thin ‘slice’ of volume on the waterline of the vessel. KB can therefore be determined from a table of moments of volume. In the case of adding mass:

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84 • Ship Stability, Powering and Resistance Table 2.7 KB after adding a small load Item

Volume (m3)

KB (m)

Moment (m4)

Ship at the original draught

Original underwater volume

Original KB

Volume column × KB column

Added slice of volume

Added mass divided by water density

Original draught + half the sinkage

Volume column × KB column

Total

Total volume column

Total moment column

In the case of removing mass: Table 2.8 KB after removing a small load Item

Volume (m3)

KB (m)

Moment (m4)

Ship at the original draught

Original underwater volume

Original KB

Volume column × KB column

Added slice of volume

Removed mass divided by water density

Original draught – half the sinkage

Volume column × KB column

Total

Total volume column

Total moment column

If the waterplane area is assumed to be constant, then the waterplane inertia can also be assumed to remain constant. Therefore, BM will vary with the overall volume before adding or removing the mass. As seen previously (Formula 2.9): BM =

I ∇

Therefore in the initial condition: ∴ BMOrigin r al =

IOrigin r al ∇Origin r al

Therefore in the final condition: BMNew =

INew ∇New

Transposing Formula 2.9 gives: I

BM × ∇

Therefore: IOrigin r al

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BMOrigin r al × ∇Origin r al

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Small Angle or Initial Metacentric Stability • 85 However, assuming the waterplane area is constant: IOrigin r al

INew

Therefore combining these equations gives: BMNew × ∇New = BMOrigin r al × ∇Origin r al Transposing for the new BM gives: BMNew =

BMOrigin r al × ∇Origin r al ∇New

KG can be found using a loading table, as seen previously.

QUESTIONS Q2.45 (ENG) MV Reed is loaded to a draught of 6.00 m and a displacement of 6,008 tonnes in sea water, with a KG of 6.50 m. 100 tonnes of cargo is then added to the vessel at a KG of 13.00 m. Assuming that the waterplane area remains constant, determine the new GM. Q2.46 (ENG) MV Reed is loaded to a draught of 7.00 m and a displacement of 7,329 tonnes in sea water, with a KG of 6.90 m. 50 tonnes of cargo is then removed from the vessel at a KG of 8.00 m. Assuming that the waterplane area remains constant, determine the new GM.

SMALL ANGLE OR INITIAL METACENTRIC STABILITY  LEARNING CHECKLIST Objective

Level

Understand what is meant by the centre of buoyancy

OOW, MCM, ENG

Understand what is meant by the vertical centre of buoyancy (KB)

OOW, MCM, ENG

Understand what is meant by the longitudinal centre of buoyancy (LCB)

OOW, MCM, ENG

Understand what is meant by the transverse centre of buoyancy (TCB)

OOW, MCM, ENG

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86 • Ship Stability, Powering and Resistance

Objective

Level

Calculate the position of the centre of buoyancy for a box shaped vessel

OOW, MCM, ENG

Determine the position of the centre of buoyancy using hydrostatic data

OOW, MCM, ENG

Understand what is meant by the centre of gravity

OOW, MCM, ENG

Understand what is meant by the vertical centre of gravity (KG)

OOW, MCM, ENG

Understand what is meant by the longitudinal centre of gravity (LCG)

OOW, MCM, ENG

Understand what is meant by the transverse centre of gravity (TCG or GGH)

OOW, MCM, ENG

Calculate the position of the centre of gravity when loads are added

OOW, MCM, ENG

Calculate the position of the centre of gravity when loads are removed

OOW, MCM, ENG

Calculate the position of the centre of gravity when loads are moved up or down

OOW, MCM, ENG

Calculate the position of the central gravity when loads are suspended

OOW, MCM, ENG

Calculate the position to load a known mass of cargo to obtain a required centre of gravity

OOW, MCM, ENG

Calculate the mass of cargo to load at a known position to obtain a required centre of gravity

OOW, MCM, ENG

Understand what is meant by the metacentre or transverse metacentre

OOW, MCM, ENG

Understand why the centre of buoyancy moves as a vessel inclines

OOW, MCM, ENG

Understand how a righting effect may be generated as a vessel inclines

OOW, MCM, ENG

Understand how a capsizing effect may be generated as a vessel inclines

OOW, MCM, ENG

Understand the definition of the metacentre

OOW, MCM, ENG

Understand what is meant by metacentric height

OOW, MCM, ENG

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Completed

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Small Angle or Initial Metacentric Stability • 87

Objective

Level

Understand what is meant by positive, neutral and negative metacentric height

OOW, MCM, ENG

Understand the significance of positive, neutral and negative metacentric height on the stability of the vessel

OOW, MCM, ENG

Calculate the metacentric height of a ship using the hydrostatic data and the centre of gravity

OOW, MCM, ENG

Understand what is meant by transverse second moment of area or transverse inertia

MCM, ENG

Calculate the metacentric height of a ship using the hydrostatic data and the centre of gravity after moving masses aboard

OOW, MCM, ENG

Calculate the position to load cargo to achieve a required metacentric height

OOW, MCM, ENG

Understand and know the minimum allowable metacentric heights for vessels carrying a range of cargo types

OOW, MCM, ENG

Understand the dangers of excessive metacentric height

OOW, MCM, ENG

Calculate the waterplane transverse second moment of area or waterplane transverse inertia for a box shaped vessel

MCM, ENG

Calculate BM and GM for a box shaped vessel

MCM, ENG

Completed

MCM, ENG Understand and complete calculations based on the metacentric height and the geometry of a box shaped vessel Understand what is meant by an angle of list

OOW, MCM, ENG

Calculate angles of list using the metacentric height

OOW, MCM, ENG

Understand the limitations of calculating the angle of list using metacentric height

OOW, MCM, ENG

Calculate the angle of list after loading operations if the vessel has an initial angle of list

OOW, MCM, ENG

Calculating the required distribution of mass to achieve a final list angle or the upright condition

OOW, MCM, ENG

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88 • Ship Stability, Powering and Resistance

Objective

Level

Calculate angles of list for vessels with zero metacentric height

MCM, ENG

Understand what is meant by FSE and FSM

OOW, MCM, ENG

Understand what is mean by, and calculate, the fluid, virtual or effective centre of gravity

OOW, MCM, ENG

Understand what is meant by, and calculate, the FSC

OOW, MCM, ENG

Calculate the loss in metacentric height due to FSE

OOW, MCM, ENG

Completed

Understand the effect of compartment subdivision on OOW, MCM, ENG FSE and metacentric height Calculate the metacentric height after operations involving solid and fluid masses

OOW, MCM, ENG

Calculate the loss in stability due to FSE in rectangular MCM, ENG compartments Understand what is meant by an angle of loll

OOW, MCM, ENG

Calculate the angle of loll using hydrostatics

MCM, ENG

Calculate the angle of loll for a box shaped vessel

MCM, ENG

Understand the procedure to recover a vessel from an OOW, MCM, ENG angle of loll Understand why a vessel heels when turning, and calculate the angle of heel when turning

MCM, ENG

Calculate the increase in draught when inclined

MCM, ENG

Calculate the effect of small load changes on GM

ENG

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3

LARGE ANGLE STABILITY  JUST HOW FAR CAN SHE ROLL? AIMS AND OBJECTIVES At the end of this section, you should be able to: Understand how the centre of buoyancy moves at large angles Understand how a righting moment or capsizing moment may be generated Understand what is meant by the terms GZ, righting lever, righting arm, statical stability or lever of statical stability Understand the significance of positive and negative GZ values Understand the meaning of KN, and determine KN at any angle or displacement using the hydrostatics Calculate GZ for any angle or displacement using the hydrostatics Plot a GZ curve Determine the range of stability

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90 • Ship Stability, Powering and Resistance Determine, and know the importance of, the angle of vanishing stability Compare the peak value of the GZ against the current regulations Calculate GZ at small angles using the metacentric height Determine GM from the GZ curve Understand the accuracy of GZ at angles where cargo shifts and down-flooding may occur Determine the down-flooding angle using the hydrostatics, and comment on the accuracy of calculated down-flooding angles Understand how changes in KG influence the GZ curve Correct GZ values for a change in KG at a constant displacement Understand how the TCG influences the GZ curve Identify a vessel which is listing from the shape of the GZ curve Correct GZ values for the TCG Determine the angle of list at large angles using the corrected GZ curve Correct GZ for a cargo shift or change in cargo position Identify a vessel which is lolling from the shape of the GZ curve Determine the angle of loll from a GZ curve Identify a vessel which is initially neutrally stable from the shape of the GZ curve Identify a vessel which is both listing and lolling from the shape of the GZ curve Determine the combined angle of list and loll from the shape of the GZ curve Understand the process by which a vessel must be recovered from an angle of loll Calculate and plot the GZ at stages of loll recovery to ensure that a vessel can be safely recovered Determine the deck edge immersion (DEI) angle using the GZ curve Calculate the moment of statical stability or righting moment Plot the moment of statical stability or righting moment curve Use the righting moment curve to determine the list angle by correcting the righting moment values Use the righting moment curve to determine the list angle by superimposing the heeling moment values Use the GZ curve to determine the list angle by superimposing the heeling arm values Understand what is meant by dynamic stability and understand the importance of the area under a righting moment or GZ curve Calculate the dynamic stability of ship in tonne metre degrees and tonne metre radians Calculate the area under GZ curves and compare them against the current regulations Correct the area under GZ curves for down-flooding, and compare them against the current regulations

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Large Angle Stability • 91 Understand the accuracy of Simpson’s Rule and ensure that they are suitably used for the shape of the curve Calculate approximate GZ values at larger angles, and comment on the accuracy of the approximation Determine the approximate angle of loll using the wall-sided formula Determine the effective GM at the angle of loll Understand how the freeboard affects GZ values and curves Understand how the beam affects GZ values and curves Understand how the symmetrical and asymmetrical ice accretion affects GZ values and curves Understand how GZ changes with trim Understand how free surface effect (FSE) influences GZ Understand the differences between free to trim and fixed to trim GZ curves Understand how wind heels ships, and determine the heel angle under a beam wind Understand and explain the current recommendations for ships and wind heeling Understand synchronous and parametric roll, and actions to be taken to avoid them Understand and explain the link between stability and roll period Understand and explain the problems associated with the carriage of grain or similar cargoes Understand the assumptions and cargo shift model upon which the grain regulations are based Determine the ability of a vessel to meet requirements of the grain regulations by plotting the GZ curve and determining the residual stability Determine the ability of a vessel to meet requirements of the grain regulations by using the maximum grain heeling moment tables, and determine the approximate list angle in the event of a grain cargo shift

As previously seen, the basic, or initial, metacentric stability theory breaks down at around 10 to 15 degrees of heel. At larger angles of heel, we can no longer assume that the metacentre is stationary, and our use of GM as a measure of stability is not valid. Instead, we have to switch to another method of assessing the stability, known as large angle stability. The centres of gravity and buoyancy are still used, with gravity acting down through G and buoyancy acting up through B. This can be seen in Figures 3.1, 3.2, 3.3 and 3.4, which show the position of the centres and forces of buoyancy and gravity as a vessel rolls to large angles. This assumes that there is no cargo shift or water ingress at large angles.

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92 • Ship Stability, Powering and Resistance

0°

5°

10°

15°

20°

25°

▲ Figure 3.1 Large angle movement of B (0 to 25 degrees)

30°

35°

40°

45°

50°

55°

▲ Figure 3.2 Large angle movement of B (30 to 55 degrees)

At larger angles of heel, what is important is the magnitude and direction in which the torque created by the misalignment of buoyancy and gravity is acting. In Figure 3.5, it can be seen that at smaller angles, the upwards buoyancy force and the downwards gravity force are acting to right the vessel. In Figure 3.6, it can be seen that at larger

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Large Angle Stability • 93

60°

65°

70°

75°

80°

85°

▲ Figure 3.3 Large angle movement of B (60 to 85 degrees)

90°

95°

100°

▲ Figure 3.4 Large angle movement of B (90 to 100 degrees)

25°

▲ Figure 3.5 Righting action

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94 • Ship Stability, Powering and Resistance

85°

▲ Figure 3.6 Capsizing action

angles, the upwards buoyancy force and the downwards gravity force are acting to incline the vessel further. The overall moment is controlled by the transverse distance between G and B. A horizontal line is drawn from G out to the line of action of B. The point at which these intersect is called Z. The distance between G and Z is known as the righting lever, righting arm, statical stability arm, or just GZ. If the vessel is trying to right, GZ is said to be positive. If the vessel is trying to capsize, GZ is said to be negative. Clearly as the vessel inclines over, the GZ value will change.

G

Z B

▲ Figure 3.7 Righting lever, righting arm or GZ

As GZ varies with heel angle, it is normally shown as a graph, known as the GZ curve, with GZ on the y axis, and heel angle on the x axis (Figure 3.8). Positive GZ means the vessel wants to return upright, negative GZ means she will capsize. The range of the graph where GZ is positive is known as the range of stability. The point where GZ changes from positive to negative is known as the angle of vanishing stability, and is the point of no return for the vessel. There is a requirement for UK flagged vessels, detailed in MSN1752 (Maritime and Coastguard Agency, 2000) to load a vessel so that the peak of the GZ curve is greater than 0.20 m and occurs at an angle of 30 degrees or greater (note that the 2008 IS Code (International Maritime Organisation, 2008) requires this to occur at 25 degrees or more).

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Large Angle Stability • 95 0.4 0.3 0.2

GZ (m)

0.1 0

10

20

30

40

50

60

70

–0.1 –0.2 –0.3 –0.4 Heel (degrees)

▲ Figure 3.8 Sample GZ curve

Calculating GZ and Drawing GZ Curves (OOW, MCM, ENG) Calculating GZ is actually very simple – the process requires some additional hydrostatic curves, known as the cross curves of stability, an example of which is shown in Figure 3.9. These are unique to each ship, and calculated by the Naval Architect and supplied as part of the ship’s hydrostatic data. This data may also be given in tabular form. These show a value known as KN, as shown in Figure 3.10. A horizontal line is drawn out from the keel. Where this intersects the line of action of buoyancy, a new imaginary point is created, called N. This point depends on the position of the centre of buoyancy, which itself depends only on the displacement and heel. If KN is known, and KG for the ship is found using loading tables, then GZ can be found. Figure 3.11 shows the vessel inclined, with all of the dimensions annotated. Using Figure 3.11, it can be seen that: Distance K KN

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Dis ista ance (KG sinθ ) + Dista ce G GZ

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96 • Ship Stability, Powering and Resistance 5.00 40 4.50

35

4.00

30

3.50 KN (m)

25 3.00 2.50

20

2.00

15 12 10

1.50 1.00

5

0.50

0.00 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000 7500 8000 Displacement (tonnes)

▲ Figure 3.9 Cross curves of stability for MV Reed (0 to 40 degrees)

B N

K

▲ Figure 3.10 KN

θ

KG

G

Z

θ B K

N

KG sin θ

▲ Figure 3.11 Finding GZ from KN

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Large Angle Stability • 97 This equation can be transposed to give GZ: GZ = KN − KG sin i θ ▲ Formula 3.1 GZ from KN

QUESTIONS Q3.1 (OOW, MCM, ENG) Using the KN data in the MV Reed data books, determine the KN values at a displacement of 3,200 tonnes, and hence draw the GZ curve for angles between 0 and 90 degrees, in steps of 10 degrees, if the KG of the vessel is 7.18 m. Q3.2 (OOW, MCM, ENG) MV Reed starts loading from her lightship condition. 352 tonnes of cargo are loaded at a KG of 7.00 m, 1,000 tonnes of cargo is loaded at a KG of 6.00 m, and both the bunkers are pressed full of fuel oil, relative density 0.95. Draw the GZ curve at intervals of 10 degrees for the vessel in this condition, up to an angle of 80 degrees.

Calculating GZ at Small Angles (OOW, MCM, ENG) At small angles of heel, up to around 10 degrees, the GZ values can be directly calculated from GM. This assumes that the metacentre remains fixed, and GM is therefore constant. Figure 3.12 shows the geometry of G, Z and M at a small angle

M G

Heel angle Z

B

▲ Figure 3.12 GZ and GM

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98 • Ship Stability, Powering and Resistance of heel. From Figure 3.12 it can be seen that at small angles of heel GZ is given by Formula 3.2. GZ ≈ GM × sin i θ ▲ Formula 3.2 GZ approximation from GM

This formula allows GZ to be calculated at small angles where the metacentre can be assumed to remain fixed.

Finding GM from the GZ Curve (OOW, MCM, ENG) The GZ and the initial GM of the vessel are related, and the GZ curve can be used to determine the initial GM. Consider the GZ curve shown in Figure 3.13. Formula 3.2 showed that GM and GZ are approximately linked at small angles: GZ ≈ GM × sinθ

1.000

0.800

GZ (m)

0.600

0.400

0.200

0.000

0

10

20

30

40

50

60

70

80

90

–0.200 Heel (degrees)

▲ Figure 3.13 Sample GZ curve

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Large Angle Stability • 99 Therefore, if we read off GZ at a small angle, we can use it to find GM. For example, using Figure 3.13, at 6 degrees, the GZ value is 0.04 m. Substituting these values into Formula 3.2 gives: 0 04 GM GM ≈

i 6

0 04 = 0.38 38 m sin6

This gives us an approximate GM value, which is useful as a check against our other calculations. However, it can be difficult to accurately read off the GZ value at small angles. To reduce this error, we can use an alternative method. The first stage is to draw a straight line which is tangential to the initial slope of the GZ curve. What this effectively means is to draw a straight line through the GZ curve at zero and a very small angle, such as 2 degrees, as shown in Figure 3.14. 1.000

0.800

GZ (m)

0.600

0.400

0.200

0.000

0

10

20

30

40

50

60

70

80

90

–0.200 Heel (degrees)

▲ Figure 3.14 Gradient of a GZ curve

The next stage is to draw a vertical line up at 57.3 degrees. This equates to one radian, converted to degrees. This is constant for any GZ curve. This line is shown in Figure 3.15. At the intersection of the vertical line and the diagonal line, we can draw a horizontal line, as shown in Figure 3.16. The intersection of this horizontal line and the y axis gives us the initial GM value. In this case, this gives a GM of 0.40 m.

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100 • Ship Stability, Powering and Resistance 1.000

0.800

GZ (m)

0.600

0.400

0.200

0.000

0

10

20

30

40

50

60

70

80

90

60

70

80

90

–0.200 Heel (degrees)

▲ Figure 3.15 Vertical mark at 1 radian

1.000

0.800

GZ (m)

0.600

0.400

0.200

0.000

0

10

20

30

40

50

–0.200 Heel (degrees)

▲ Figure 3.16 Finding GM from a GZ curve

The steeper the initial slope, the greater the GM. As will be seen later, a downward slope indicates a negative GM, with the angle of loll at the point of intersection of the curve and the x axis. The greater the initial slope, the greater GM. If the initial slope of the curve is negative (i.e. it goes down), then the GM will be negative – this will be covered in more detail later.

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Large Angle Stability • 101 X For a mathematical derivation showing how this process works, see

Appendix 16: The Mathematical Proof of the Determination of GM from the GZ Curve.

Cargo Shifts and Down-Flooding (OOW, MCM, ENG) The GZ curves are constructed assuming that the position of the centre of gravity remains constant. Clearly as the vessel inclines to larger and larger angles, the contents of the vessel are likely to move (known as a cargo shift), with the result that the centre of gravity will move, and KG will change. Therefore, GZ values should be treated with caution, particularly if the cargo is prone to shifting, such as grain cargo. Special rules exist to deal with these cargoes, and will be covered later. The GZ curve also assumes that the vessel remains watertight. It is impossible to build large ships which are completely watertight through 180 degrees of roll. At some angle of heel, fittings such as engine room vents and exhausts, which cannot be made watertight, will submerge. At this point, known as the down-flooding angle, water will progressively flood down into the vessel, moving the centre of gravity, lowering the vessel into the water and introducing FSE, significantly changing the stability of the vessel. The heel angle at which the vessel down-floods is given in the hydrostatic data for the vessel. This gives an indication of the angle at which the vessel will down-flood; however, it assumes that the water is calm, and so should be treated with caution.

Corrections to GZ due to Changes in the Vertical Centre of Gravity (Assuming Displacement Remains Fixed) (OOW, MCM, ENG) As KG changes, GZ will also change, as shown in Figure 3.17. As KG increases, GZ decreases, and as KG reduces, GZ increases. This assumes that the displacement is constant, and therefore KN will be constant at any heel angle. Plotting GZ curves for the original, reduced KG and increased KG conditions shows a specific trend in the GZ curves, as shown in Figure 3.18. As KG increases, the GZ decreases, and the angle of vanishing stability decreases. As KG reduces, GZ increases,

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Z

al K G

G

Z

G

ced KG

Z

G

du

Ori

gin

Inc rea sed KG

102 • Ship Stability, Powering and Resistance

Re

B

▲ Figure 3.17 Change in GZ with a change in KG 0.300 0.200

GZ (m)

0.100 0.000

0

10

20

30

40

50

60

–0.100 –0.200 –0.300 –0.400 –0.500

Original GZ Increased KG Reduced KG Heel (degrees)

▲ Figure 3.18 GZ curve variation due to KG variation

and the angle of vanishing stability increases. As can be seen from the previous section, the initial slope of the GZ curve increases as KG decreases, and hence GM increases. If the displacement is kept constant, and cargo adjusted so that KG changes, then a shortcut can be used which avoids the complete procedure of calculating GZ from KN. The change in GZ can be expressed in terms of the change in KG. This all assumes the displacement is constant, as when the displacement is constant, KN will be also constant at any heel angle.

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KG

Large Angle Stability • 103

Z

e in

G

Ch

ang

θ

Z

G

B

Change in GZ

▲ Figure 3.19 Change in GZ with KG changes

As can be seen in Figure 3.19, the change in GZ can be found from the change in KG: Change in GZ = Change in i KG × sin i θ ▲ Formula 3.3 Change in GZ due to change in KG

This is only valid if the displacement remains constant. The change in GZ varies at each angle of heel, and must be recalculated. If KG increases, the vessel becomes less stable, and so GZ must reduce. Therefore, the change in GZ must be subtracted from the original GZ. If KG decreases, the vessel becomes more stable, and so GZ must increase. Therefore, the change in GZ must be added to the original GZ.

QUESTION Q3.3 (MCM, ENG) In the lightship condition, MV Reed has the following GZ values: Angle

degrees

0

10

20

30

40

50

60

70

80

GZ

metres

0.00

0.15

0.31

0.39

0.39

0.46

0.46

0.21

–0.17

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104 • Ship Stability, Powering and Resistance Determine (and plot) the new GZ values and curves at intervals of 10 degrees, from 0 to 80 degrees, if the displacement is kept constant, but: 1

The lightship KG is increased by 0.10 m;

2

The lightship KG is decreased by 0.10 m.

Corrections to GZ due to a Transverse Centre of Gravity (OOW, MCM, ENG) If the centre of gravity moves off of the centreline of the vessel, so that the vessel has a TCG (or GGH) value, as shown in Figure 3.31, then there will also be a change in the GZ values.

G

B

▲ Figure 3.20 The TCG

In this condition, the centres of buoyancy and gravity are both acting to incline the vessel over further (see the section on angles of list in the small angle stability). If we draw the GZ on to Figure 3.31, we can see that it still runs from the centre of gravity to the line of action of buoyancy, as shown in Figure 3.21. In Figure 3.21, the centres of buoyancy and gravity are acting to try and roll the vessel further over, so GZ would be said to be negative. The vessel will incline over, until the centres of buoyancy and gravity are aligned, at which point the vessel will be in equilibrium at the angle of list, as shown in Figure 3.22. As GZ measures the separation

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Large Angle Stability • 105

Z

G

B

▲ Figure 3.21 Negative GZ when upright

Z

G

B

▲ Figure 3.22 Angle of list and GZ

between the centre of gravity and the line of action of buoyancy, at the angle of list, the GZ must be zero, as the line of action of buoyancy and the centre of gravity are aligned in equilibrium. If the vessel is then heeled past the angle of list, then the centre of buoyancy will move further outboard, and create a positive GZ, as shown in Figure 3.23. As the vessel inclines further, the centre of buoyancy will continue to move, as shown in Figure 3.24. As the angle of heel increases, and the keel emerges and deck submerges, the centre of buoyancy will move back towards the line of action of gravity. When the centre

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106 • Ship Stability, Powering and Resistance

G

Z

B

▲ Figure 3.23 Inclined past list

G

Z

B

▲ Figure 3.24 Inclined past list to a larger angle

of buoyancy crosses back under the line of action of gravity (the angle of vanishing stability), the action of buoyancy and gravity will be such that the vessel will try and roll over further, as shown in Figure 3.25. GZ is said to be negative again. Therefore, a vessel with a non-zero TCG will have a GZ curve that starts negative, is zero at the angle of list, then positive to the angle of vanishing stability, and then negative, as shown in Figure 3.29. In comparison to the vessel with the centre of gravity on the centreline, all GZ values will be reduced proportional to the TCG and the cosine of the heel angle. The complete sequence of heeling for a vessel with a centre of gravity off of the centreline is shown in Figure 3.26.

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Large Angle Stability • 107

Z

G B

▲ Figure 3.25 Inclined past list with negative GZ

0°

5°

10°

15°

20°

25°

▲ Figure 3.26 A vessel with an off-centre TCG rolling to large angles (0 to 25 degrees)

The GZ curve for a listing vessel, as shown in Figure 3.29, can be used to determine the angle of list for large angles, by reading off the first angle at which GZ is zero, which in this case is 13.5 degrees. The previous formulae using GM to determine list are only valid at small angles (where the metacentre is assumed to remain constant). GZ curves are the only valid method to accurately determine angles of list above 10 degrees.

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108 • Ship Stability, Powering and Resistance

30°

35°

40°

45°

50°

55°

▲ Figure 3.27 A vessel with an off-centre TCG rolling to large angles (30 to 55 degrees)

60°

65°

70°

75°

80°

85°

▲ Figure 3.28 A vessel with an off-centre TCG rolling to large angles (60 to 85 degrees)

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Large Angle Stability • 109

90°

95°

100°

▲ Figure 3.29 A vessel with an off-centre TCG rolling to large angles (90 to 100 degrees)

0.800

0.600

GZ (m)

0.400

0.200

0.000 0

10

20

30

40

50

60

70

80

90

–0.200 Heel (degrees)

▲ Figure 3.30 GZ curve of a listing vessel

Calculating the GZ for a vessel with a TCG value is similar to the normal calculations for GZ. GZ is found as before, and then corrected for a loss in GZ due to the TCG. This loss in GZ due to a TCG value can be found, as shown in Figures 3.31 and 3.32. From this, it can be seen that the loss in GZ can be found from the TCG of the vessel: Loss in GZ = TCG × cosθ ▲ Formula 3.4 Loss in GZ due to a TCG

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110 • Ship Stability, Powering and Resistance

TCG

Z Z

G G

B Loss of GZ

▲ Figure 3.31 Loss of GZ due to a TCG

θ

TC

G

G

Z

θ

Z G

B Loss of GZ = TCG cos θ

▲ Figure 3.32 Geometry of GZ loss

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Large Angle Stability • 111 QUESTION Q3.4 (MCM, ENG) MV Reed is loaded to a displacement of 3,000 tonnes, with a KG of 6.77 m and a TCG of 0.289 m. Calculate the corrected GZ at 0, 5, 10, 15 and 20 degrees, and hence determine the angle of list. Compare your answer against that given by small angle list theory.

Changes in GZ due to Vertical and Transverse Changes in the Centre of Gravity (OOW, MCM, ENG) In the event of a cargo shift, it is possible that the centre of gravity of the cargo will move both sideways and upwards. This results in a big reduction in the GZ values of the vessel, as there is a reduction in GZ due to the transverse shift (giving G2Z2), and a reduction due to the vertical shift (giving G3Z3), as shown in Figure 3.33. This results in a reduction in GZ due to an increase in KG and a list angle. As a cargo shift is simply a movement of internal cargo, then the displacement remains constant, and Formulae 3.3 and 3.4 can be used to determine the overall change in GZ of the vessel.

G3 G1 G2

Z3 Z1 Z2 B

Change in GZ due to transverse shift

Change in GZ due to vertical shift

▲ Figure 3.33 Loss of GZ due to KG and TCG shifts

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112 • Ship Stability, Powering and Resistance

QUESTIONS Q3.5 (MCM, ENG) MV Reed is loaded so that she has the GZ values as shown below: Angle degrees

0

10

20

30

40

50

60

70

80

GZ

0.00

0.10

0.25

0.50

0.72

0.69

0.49

0.20

–0.12 –0.46

metres

90

Plot the new GZ curve for the vessel if the cargo shifts, resulting in the centre of gravity of the ship moving upwards by 0.24 m, and transversely to port by 0.40 m. Q3.6 (MCM, ENG) MV Reed is to unload a 100 tonne item of cargo, from a stowed position on the centreline. Initially the vessel has a true mean draught of 4.10 m in sea water. In this condition, she has a list of 1.10 degrees to starboard, and an effective KG of 5.49 m. When the mass is unloaded, it is lifted using the ship’s own heavy lift crane, with the crane derrick 11.00 m above the initial stowed position of the load, and swung out 17.00 m to port from the centreline. By plotting the GZ curves, determine the maximum angle of list during the unloading operation.

GZ Curves and Unstable Vessels (OOW, MCM, ENG) The GZ curve can also be used to determine the angle of loll of an unstable vessel. Figure 3.34 shows a vessel with a sufficiently large KG to give a negative GM inclining to an angle of loll. As can be seen, the GZ distance would be described as negative, as the forces of buoyancy and gravity are acting to roll the vessel over further. Therefore, vessels which are unstable, and have negative GM, will also have negative GZ values between the upright condition, where GZ will be zero, and the angle of loll, where GZ will also be zero. The curve will therefore have an initial negative slope. This downwards initial slope indicates a negative GM, with the angle of loll at the point of intersection of the curve and the x axis. The complete sequence of heeling for a vessel with a centre of gravity off of the centreline is shown in Figure 3.35.

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Large Angle Stability • 113

Z

G

M

B

▲ Figure 3.34 GZ at an angle below an angle of loll

0°

5°

10°

15°

20°

25°

▲ Figure 3.35 A vessel with a negative GM rolling to large angles (0 to 25 degrees)

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114 • Ship Stability, Powering and Resistance

30°

35°

40°

45°

50°

55°

▲ Figure 3.36 A vessel with a negative GM rolling to large angles (30 to 55 degrees)

60°

65°

70°

75°

80°

85°

▲ Figure 3.37 A vessel with a negative GM rolling to large angles (60 to 85 degrees)

90°

95°

100°

▲ Figure 3.38 A vessel with a negative GM rolling to large angles (90 to 100 degrees)

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Large Angle Stability • 115

QUESTIONS Q3.7 (MCM, ENG) MV Reed is loaded from the lightship condition. 1,985 tonnes of cargo is loaded at a KG of 7.50 m. Draw the GZ curve after loading, in steps of 10 degrees, up to an angle of 70 degrees, and hence determine the angle of loll of the vessel. Q3.8 (OOW, MCM, ENG) MV Reed is loaded to a displacement of 4,000 tonnes with a KG of 7.23 m. 1. Plot the GZ curve at 0, 5, 10, 15, 20, 30, 40, 50, 60 and 70 degrees, and hence determine the angle of loll (19 degrees); 2. Use the GZ curve to determine GM; 3. Verify the GM value from the curve by calculating GM using the hydrostatics.

GZ Curves and Neutrally Stable Vessels (OOW, MCM, ENG) When a vessel is loaded so that GM is zero (neutral stability), any small force heeling the vessel will not generate a righting moment, as shown in Figure 3.39. Clearly at this point GZ will also be zero. As the vessel heels to large angles, the centre of buoyancy will move further, and the assumptions of small angle stability breakdown. At this point, GZ will become positive. This results in a GZ curve as shown in Figure 3.40.

M

G

B

▲ Figure 3.39 Heeling with neutral GM

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116 • Ship Stability, Powering and Resistance 0.12 0.1 0.08 0.06

GZ (m)

0.04 0.02 2 0

5

10

15

20

25

30

35

40

–0.02 –0.04 –0.06 –0.08 Angle (degrees)

▲ Figure 3.40 GZ curve for a neutrally stable vessel

QUESTIONS Q3.9 (OOW, MCM, ENG) MV Reed is loaded to a displacement of 6,000 tonnes with a KG of 7.01 m. Plot the GZ curve at 0, 5, 10, 20, 30, 40 and 50 degrees, and hence determine the GM of the vessel from the GZ curve. Q3.10 (OOW, MCM, ENG) MV Reed is loaded so that she has the following KN values:

Angle (degrees)

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KN (m)

0

0.000

5

0.611

10

1.220

15

1.840

20

2.471

25

3.102

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Large Angle Stability • 117 Working to three decimal places: 1. Plot the GZ curve at 0, 5, 15, 20 and 25 degrees if KG is 6.810 m; 2. Plot the GZ curve at 0, 5, 15, 20 and 25 degrees if KG is 7.010 m; 3. Plot the GZ curve at 0, 5, 15, 20 and 25 degrees if KG is 7.210 m; 4. For each of the above graphs, use the GZ curve to determine the GM of the vessel.

GZ Curves and Combined Angles of List and Loll (OOW) In certain circumstances, it is possible that a vessel may be both lolling and listing at the same time. For this to happen the vessel would have to be unevenly loaded and be unstable. The only way to directly find the combined angle of list and loll is to use a GZ curve. The process is similar to that for finding list, except the curve will have a negative initial slope.

QUESTION Q3.11 (MCM, ENG) MV Reed starts in her lightship condition. 2,985 tonnes of cargo is added at a KG of 7.50 m, at a position 0.10 m to port of the centreline. Determine the combined angle of list and loll by plotting the GZ curve, in steps of 5 degrees, up to an angle of 35 degrees.

Instability and Recovering Loll Using GZ Curves (MCM, ENG) The large angle stability theory covered so far is useful in analysing the steps required to recover a vessel from an angle of loll. As previously seen, to recover a lolling vessel, mass must be added to the low side of the vessel. The effect of adding mass to the low

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118 • Ship Stability, Powering and Resistance side is to reduce KG, and hence increase GM to a safe, positive value. At this point, the vessel can be counter-ballasted to an upright condition. The easiest way to add mass to the low side of the ship is to ballast a double bottom. This will create asymmetric loading, and hence cause the centre of gravity to move off of the centreline of the vessel, causing a combined angle of loll and list. The GZ curves can be created for this combined loll and list condition, to determine the actual combined angle. Clearly the greatest reduction in KG would occur when the double bottom tank is pressed, as at that point there is the greatest mass added low in the vessel, and there will be no FSE; however, it is important to remember that as a tank is filled, there will a point when it is 10% full, 20% full, 30% full, and so on. The stability of the vessel must be considered at all stages of filling the tank. As the tank is filled, the initial gradient of the GZ curve will slowly become positive as the GM increases, and the first intersection of the curve and the x axis will give the combined angle of list and loll. Figure 3.41 shows the variation in the GZ curve over time as a double bottom tank in a lolling vessel is progressively filled.

0.4 0.3

GZ (m)

0.2 0.1

0 Sounding = 0.00 m 0 Sounding = 0.40 m Sounding = 0.80 m –0.1

5

10

15

20

25

30

35

40

Sounding = 1.20 m Sounding = 1.40 m –0.2 Sounding = 2.00 m –0.3 –0.4

Angle (degrees)

▲ Figure 3.41 GZ and recovering loll

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Large Angle Stability • 119

QUESTION Q3.12 (MCM, ENG) MV Reed starts loading from her lightship condition. The following loading operations are undertaken: Both bunkers are pressed full with fuel oil, relative density 0.96 Both No. 3 Double Bottoms are partially filled to a sounding of 0.30 m with sea water ballast Both No. 4 Double Bottoms are partially filled to a sounding of 0.30 m with sea water ballast 3,245 tonnes of cargo is loaded at a KG of 7.50 m, on the centreline As a result of the above loading, the vessel suddenly inclines to port. The master proposes to recover the vessel by pressing both of the No. 3 Port and No. 4 Port Double Bottoms. Determine the reason for the vessel inclining, and in sounding steps of 0.20 m, determine if pressing both of these Port Double Bottoms at the same time will recover the vessel to a stable condition, and determine the maximum angle of inclination during the recovery. You should plot the GZ values in steps of 5 degrees to 30 degrees.

Deck Edge Immersion Angles (OOW, MCM, ENG) As has been seen, the shape of the GZ curve varies depending on the displacement, heel angle and centre of gravity. Another feature of the vessel which has an influence on the shape of the GZ curve is the ‘deck edge immersion’ (DEI) angle. This is the angle of heel at which the edge of the deck starts to be immersed. With most vessels, GZ gets progressively larger as the vessel starts to heel. This is seen as a gently increasing slope, or gradient, on the GZ curve. When the deck edge is immersed, the rate at which GZ grows reduces as the underwater geometry starts to change quickly. This is seen as a ‘point of inflection’ on the curve, or the point at which the curve stops increasing in steepness, as shown in Figure 3.42. The DEI angle is important, as this point of inflection has an influence on the angle of vanishing stability, and the range of stability.

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120 • Ship Stability, Powering and Resistance 0.60

Approximate point where gradient stops increasing – ‘point of inflection’, or DEI

0.50

GZ (m)

0.40

Gradually increasing slope or gradient

0.30

Gradually reducing gradient

0.20

0.10

0.00 0

5

10

15

20 25 30 Heel (degrees)

35

40

45

50

▲ Figure 3.42 DEI and the point of inflection

Moment of Statical Stability (MSS) or Righting Moments (OOW, MCM, ENG) As has been seen, the GZ indicates the transverse distance between the centres of buoyancy and gravity, and therefore gives us an idea of the direction the torque, or righting moment, is acting in. The value of the righting moment at any angle of heel (sometimes known as the Moment of Statical Stability, or MSS) can be found by multiplying the righting level by the displacement, as shown in Formula 3.5: Righting moment = GZ × Δ ▲ Formula 3.5 Righting moment

For any vessel, a curve of righting moments can be produced, which shows how the righting moment varies with heel angle. As the righting moment is the GZ multiplied by a constant value, the righting moment curve looks identical to the GZ curve, the only difference being the values on the y axis. This is shown in Figure 3.43.

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Large Angle Stability • 121 GZ curve 3 2

GZ (m)

1 0

0

10

20

30

40

50

60

70

80

90 100 110 120 130 140 150 160 170 180

–1 –2 –3 –4

Angle (degrees)

Righting moment (tonne metres)

Righting moment curve 3,000 2,000 1,000 0

0

10

20

30

40

50

60

70

80

90 100 110 120 130 140 150 160 170 180

–1,000 –2,000 –3,000 – 4,000

Angle (degrees)

▲ Figure 3.43 GZ curve and righting moment curve

QUESTION Q3.13 (OOW, MCM) MV Reed is loaded to a displacement of 4,800 tonnes, with a KG of 6.80 m. In steps of 10 degrees, from 0 to 70 degrees, plot the GZ and the righting moment curves.

Using Righting Moment Curves to Find Angles of List (MCM, ENG) Righting moment curves can also be used to determine the angle of list in the same way that GZ curves can. The curves are plotted as before and the intersection with the x axis gives the list angle.

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122 • Ship Stability, Powering and Resistance

QUESTION Q3.14 (MCM, ENG) MV Reed starts loading from her lightship condition. A total of 4,185 tonnes of cargo are loaded at a KG of 7.00 m, at a position 0.10 m to port of the centreline. In steps of 10 degrees, from 0 to 70 degrees, plot the GZ and the righting moment curves, and hence determine the angle of list using both curves.

As an alternative to this method, righting moment curves can be used to predict the angle of list of a vessel, using a method where the TCG correction is not required. As previously explained in the section on small angles of list, when a vessel lists, a listing or heeling moment is generated. Heeling or o listing moment = w × d ▲ Formula 3.6 Heeling or listing moment

When this equals the righting moment created by the vessel, a steady state of equilibrium exists and the vessel heels at a steady angle. To determine the angle where this occurs, the righting moment curve (with no correction for the TCG effect of the load causing the list) at a range of angles can be plotted, and the heeling moment (found from the mass causing the list, multiplied by the distance of the mass from the centreline) superimposed onto the righting moment curve, and the first point of intersection can be read off. At this point, the listing or heeling moment is equal to the righting moment, and the vessel is in a condition of equilibrium.

QUESTION Q3.15 (MCM, ENG) Using the scenario in Question 3.14, determine the angle of list by plotting the righting moment curve (without a TCG correction), and superimposing the heeling moment.

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Large Angle Stability • 123 Under certain circumstances, this method is much easier than using the TCG correction, particularly if the heeling or listing moment varies with the heel angle, as is the case for heel due to wind, or cargo shifts due to the heeling moment lever changing with the geometry of the vessel as she heels. Again, the righting moment and heeling moment curves can be superimposed, and the heel angle read off the graph, as shown in Figure 3.44. 4,000

Heeling and righting moments (tonne metres)

3,500

Righting moment 3,000

2,500

Heeling or listing moment

2,000

1,500

1,000

500

Heel or list angle

0 0

5

10

15

20

25

30

35

40

45

50

Heel (degrees)

▲ Figure 3.44 Heeling moment and righting moment

As can be seen from the graph (Figure 3.44), the heeling moment reduces with the angle of heel. The definition of a moment is a ‘force multiplied by the perpendicular distance to the pivot point’. As the vessel inclines over, the distance to the pivot point reduces, and the heeling or listing moment reduces. Consider a vessel heeling under the effect of the wind (which will be covered in more detail later). As the vessel heels, the point at which the heeling force acts rotates with the vessel, reducing the perpendicular distance from the overall wind force to the rotation point on the centreline at the waterline, as shown in Figure 3.45.

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124 • Ship Stability, Powering and Resistance

L C

5.91

L C

5.98

6.00

L C

L C

L C 5.44

5.64

5.80

L C

▲ Figure 3.45 Reduction in lever during heeling

X The equating of heeling and righting moments can be used to determine

the angle of heel in a turn, as previously discussed. For a full proof, see Appendix 17: The Derivation of the Angle of Heel in a Turn Formula. It is also possible to use the GZ curve in a similar method, so that the list angle can be determined without the need to apply a TCG correction to the basic GZ values. The basic GZ values area is calculated as previously, with no TCG correction. These are plotted to give the GZ curve. The heeling or listing moment is then divided by the displacement of the vessel to give the heeling or listing arm:

Heeling or o listing arm r =

w×d Δ

▲ Formula 3.7 Heeling or listing arm

This heeling or listing arm can be plotted as a line onto the basic GZ curve, and the intersection between the arm and the GZ curve can be read off to give the list angle.

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Large Angle Stability • 125 QUESTION Q3.16 (MCM, ENG) Using the scenario in Question 3.14, determine the angle of list by plotting the GZ curve (without a TCG correction), and superimposing the listing arm.

The other important function of the righting moment curve is related to energy. The area under the righting moment curve up to a specific angle is equal to the energy required to roll the vessel to that angle. Therefore, the greater the area under the righting moment curve, the more energy it will take to roll the vessel to a given angle. As the righting moment curve is proportional to the GZ curve, the area under the GZ curve up to a specific angle is proportional to the energy required to roll the vessel to that angle. Therefore, the greater the area under the GZ curve, the more energy it will take to roll the vessel.

QUESTION Q3.17 (OOW, MCM, ENG) Which of the following vessels will require the most energy to roll it to an angle of 40 degrees?

GZ

5 4 GZ (m)

3 2 1

–0

10

20

30

40

50

60

70

–1 –2

Angle (degrees)

This is even more evident when we look at unstable vessels. Figure 3.46 shows GZ curves for a vessel in a stable and unstable condition of loading. Clearly, the unstable vessel would require significantly less energy to roll it to the angle of vanishing stability.

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126 • Ship Stability, Powering and Resistance 0.800

New GZ Old GZ

0.600

GZ (m)

0.400 0.200 0.000

0

10

20

30

40

50

60

–0.200 –0.400 Angle (degrees)

▲ Figure 3.46 Areas under GZ curves for stable and unstable vessels

Dynamic Stability (OOW, MCM, ENG) If a mass is moved a certain linear distance, then the work done in moving the mass is given by: Work done = Force Distance The rotational equivalent of this is: Work done = Torque × Angle When a vessel rolls, the torque resisting the roll is the righting moment; therefore, the total righting moment multiplied by the total angle rolled through gives the work done in rolling a vessel. This value is equivalent to the area under the righting moment curve. Therefore, the area under the righting moment curve, up to any angle, gives the energy required to roll the vessel to that angle. The area under the righting moment curve is known as the dynamic, or dynamical stability of the vessel. As the units of the axes are tonnes and metres, the area of the curve has units of tonne metres. The mathematical measurement for angles is radians; therefore, the area under the curve must be converted to metre radians. To determine the area under the righting moment curve, Simpson’s Rule can be used – this is a mathematical method of determining areas under curved graphs.

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Large Angle Stability • 127 Table 3.1 Generic Simpson’s Rule tabular calculation Ordinate

Offset

Simpson’s Multiplier

Area product

X1

Y1

1

1 × Y1

X2

Y2

4

4 × Y2

X3

Y3

2

2 × Y3

X4

Y4

4

4 × Y4

X5

Y5

2

2 × Y5

X6

Y6

4

4 × Y6

X7

Y7

1

1 × Y7

Total

Sum of above

Simpson’s Rule requires us to know the x values of a graph (known as the ordinates), and the y values for each of the x values (known as the offsets). To use Simpson’s Rule, we must have an odd number of ordinates, and they must be evenly spaced apart. The first stage is to draw up a table (as shown in Table 3.1), with columns headed Ordinate, Offset, Simpson’s Multiplier and Area Product. The ordinate column lists each ordinate (the total of which must always be an odd number). The offset column lists each offset measured at each ordinate. The Simpson’s Multiplier column always follows the same pattern, starting and finishing with a 1, and alternating 4, 2, 4, 2 in between. The Area Product column is the product of the offset and Simpson’s Multiplier columns. The total area under the curve is found from:

Area =

Ordinate spacing × Σ( Area product ) 3

▲ Formula 3.8 Area by Simpson’s Rule

In this formula, the Σ symbol (capital sigma in the Greek alphabet) means the ‘total of’. X For a mathematical proof of Simpson’s Rule, see Appendix 18: The

Mathematical Proof of Simpson’s Rule.

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128 • Ship Stability, Powering and Resistance

QUESTIONS Q3.18 The table below shows a series of x and y coordinates forming a curve. The units of the axes are metres. Determine the area between the curve and the x axis, as shown shaded on the graph below the table. X

Y

0

5

2

6

4

5.25

6

4

8

2

7

6

5

4 Y 3

2

1

0 0

1

2

3

4

5

6

7

8

X

Q3.19 (MCM, ENG) MV Reed starts loading from her lightship condition. 3,985 tonnes of cargo is loaded at a KG of 6.00 m, on the centreline. Determine the dynamic stability of the vessel up to an angle of 40 degrees.

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Large Angle Stability • 129 In a similar way to the metacentric height criteria, minimum values for areas under the GZ curve are described in the International Maritime Organisation’s (IMO) publication Code in Intact Stability 2008, known as the ‘2008 IS Code’. (International Maritime Organisation, 2008). These are partially incorporated into the Merchant Shipping (Load Line) Regulations 1998, as amended by the Merchant Shipping (Load Line) (Amendment) Regulations 2000, via Merchant Shipping Notice 1752(M) (Maritime and Coastguard Agency, 2000). These require that between certain angles the area under the GZ curve have minimum values, and that the peak of the curve is in a certain region of the graph. The GZ curve is used for these regulations, rather than the righting moment curve, as it is proportional to the righting moment curve; however, the values for GZ are broadly independent of ship size, and so more generic criteria can be used. As well as the metacentric height requirements described previously, the GZ curve must be analysed to check compliance with the regulations before sailing.

Requirement

Details

Peak GZ value

Must be greater than 0.20 m, and occur at 30 degrees heel or greater

Area under the graph between 0 degrees and 401 degrees

Must be greater than 0.09 m radians

Area under the graph between 30 degrees and 401 degrees

Must be greater than 0.03 m radians

Area under the graph between 0 degrees and 30 degrees

Must be greater than 0.055 m radians

Initial GM

Minimum value for a general cargo ship is 0.15 m, for a vessel carrying grain is 0.30 m, and for timber carriers meeting certain requirements is 0.05 m. (Note that this increases to 0.10 m under the 2008 IS Code.)

1

If the angle of down-flooding (the angle of heel at which the vessel will progressively flood through weather-tight fittings) is less than 40 degrees, then that angle should be used instead.

X For details of alternative criteria for vessels carrying timber deck cargo,

given in the 2008 IS Code, but not permitted by the MCA, see Appendix 19: Alternative Criteria for Large Angle Stability (Timber Deck Cargo).

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130 • Ship Stability, Powering and Resistance Simpson’s Rule can be used with the GZ curve to determine the areas as required. The easiest way to do this is to first use Simpson’s Rule to determine the area under the curve between 0 degrees and 40 degrees, using steps of 10 degrees. Then use Simpson’s Rule to determine the area under the curve between 30 degrees and 40 degrees, using steps of 5 degrees. Finally, to find the area between 0 degrees and 30 degrees, subtract the area between 30 degrees and 40 degrees from the area between 0 degrees and 40 degrees. If the down-flooding angle is less than 40 degrees, this process should be followed, and the final values adjusted for the loss of area between the down-flooding angle and 40 degrees. The area between 40 degrees and the down-flooding angle can be found using trapezoidal integration from the GZ values at 40 degrees and at the downflooding angle: Area of a trapezoid

Mean Mea height Length t

QUESTIONS Q3.20 (MCM, ENG) MV Reed is in her lightship condition. 785 tonnes of cargo is loaded onto the vessel at a KG of 7.20 m. In this condition, her angle of down-flooding is 54 degrees. Determine if the vessel meets all of the large angle stability requirements. Q3.21 (MCM, ENG) MV Reed is in her lightship condition. 4,585 tonnes of cargo is loaded onto the vessel at a KG of 7.10 m. In this condition, her angle of down-flooding is 36 degrees. Determine if the vessel meets all of the large angle stability requirements.

This process of determining the areas under the GZ curves can be time-consuming, and it is easy to introduce errors into the calculations. For this reason, the stability data book for ships often includes ‘simplified stability data’ which shows the maximum allowable KG at a range of displacements. This is calculated on the basis of working backwards from all of the stability criteria at each displacement to determine the KG value which would just cause any one of the criteria to be failed.

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Large Angle Stability • 131

How Accurate Is Simpson’s Rule in Determining the Area under a GZ Curve? (MCM, ENG) As will be discussed later, a generic GZ curve shape can be generated using Formula 3.9: GZ = GM ssinθ cosθ Assuming that a vessel has a GM of 1.00 m, the GZ values would be as given in Table 3.2. Table 3.2 Angle and GZ Angle (degrees)

GZ (m)

0

0

10

0.17

20

0.32

30

0.43

40

0.49

Simpson’s Rule can be applied to these values to determine the area under the curve up to 40 degrees ( Table 3.3). Table 3.3 Simpson’s Rule and GZ Angle (degrees) 0

GZ (m)

Simpson’s Multiplier

Area (product)

0

1

0

10

0.17

4

0.68

20

0.32

2

0.64

30

0.43

4

1.72

40

0.49

1

0.49

Total

3.53

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132 • Ship Stability, Powering and Resistance

Area =

Spacing p g 10 × Σ Area product = × 3.53 = 11 11.7667 m degrees 3 3 11.7667 = 0.2054 m radians 57.3

Increasing the number of measuring points should theoretically increase the accuracy of Simpson’s Rule. Table 3.4 shows the number of measuring points and the resulting calculated area. As can be seen, increasing the number of points has little overall effect on the calculated area values. Table 3.4 Variation in area with number of measuring points Number of measuring points

Area (m radians)

3

0.2059

5

0.2054

9

0.2062

11

0.2059

21

0.2054

41

0.2060

X For a mathematical proof of how the area under the curve can be directly

calculated, please see Appendix 20: The Direct Calculation of the Area under a Generic GZ Curve. Using the directly calculated value, the values in Table 3.4 can be used to determine the percentage error in area using Simpson’s Rule at varying degrees of resolution ( Table 3.5). Table 3.5 Accuracy of Simpson’s Rule Number of measuring points

Area by Simpson’s Rule (m radians)

Accuracy (%)

3

0.2059

100

5

0.2054

99

9

0.2062

100

11

0.2059

100

21

0.2054

99

41

0.206

100

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Large Angle Stability • 133 The accuracy of Simpson’s Rule will reduce as the curvature between the measuring points increases. Consider the curve shown in Figure 3.47. While this is a very extreme example of a curve, it does illustrate the concept. The accurate use of Simpson’s Rule assumes that there is a smooth curve between the measuring points. If Simpson’s Rule were used with the minimum of only 3 measuring points, as shown 14 12

GZ (m)

10 8 6 4 2 0

0

5

10

15

20 25 Heel (degrees)

30

15

20 25 Heel (degrees)

30

35

40

▲ Figure 3.47 Accuracy and curvature 14 12

GZ (m)

10 8 6 4 2 0

0

5

10

35

40

▲ Figure 3.48 3 measuring points

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134 • Ship Stability, Powering and Resistance in Figure 3.48, then the assumed curve (dashed line) and the actual curve (solid line) can be seen to be different. As the number of measuring points are increased, then the assumed line can be seen to be closer to the actual line, as shown in Figures 3.49 to 3.54. 14 12

GZ (m)

10 8 6 4 2 0

0

5

10

15

20 25 Heel (degrees)

30

35

40

15

20 25 Heel (degrees)

30

35

40

▲ Figure 3.49 5 measuring points 14 12

GZ (m)

10 8 6 4 2 0

0

5

10

▲ Figure 3.50 7 measuring points

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Large Angle Stability • 135

14 12

GZ (m)

10 8 6 4 2 0

0

5

10

15

20 25 Heel (degrees)

30

35

40

15

20 25 Heel (degrees)

30

35

40

▲ Figure 3.51 9 measuring points

14 12

GZ (m)

10 8 6 4 2 0

0

5

10

▲ Figure 3.52 11 measuring points

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136 • Ship Stability, Powering and Resistance

14 12

GZ (m)

10 8 6 4 2 0

0

5

10

15

20 25 Heel (degrees)

30

35

40

15

20 25 Heel (degrees)

30

35

40

▲ Figure 3.53 21 measuring points

14 12

GZ (m)

10 8 6 4 2 0

0

5

10

▲ Figure 3.54 41 measuring points

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Large Angle Stability • 137 In a similar way to the generic GZ curve shown in Table 3.5, the area under these can be calculated using Simpson’s Rule and compared to the actual value determined from integration. This gives the following values (note that the error at 41 points is due to rounding in the tabular Simpson’s Rule calculation) for accuracy ( Table 3.6). Table 3.6 Accuracy of Simpson’s Rule (less even curve shape) Number of measuring points

Area by Simpson’s Rule (m radians)

Accuracy (%)

3

3.939

94.0

5

4.297

102.5

7

3.946

94.2

9

4.139

98.8

11

4.119

98.3

21

4.209

100.5

41

4.208

100.4

Approximating GZ at Larger Angles (MCM, ENG) So far we have seen that GZ can be found using the cross curves of stability, and at small angles can be approximated using: GZ = GM sinθ The cross curves of stability should be available – but to double check values, or for small angles where approximations are acceptable, GZ can be estimated. For older designs of rounded hull forms, this often overestimates the GZ value as angles increase, with the error growing as angles increase. Therefore, it is often modified by a cosine term to try and reduce the error, giving: GZ = GM si sin i θ cosθ ▲ Formula 3.9 GZ approximation

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138 • Ship Stability, Powering and Resistance However, the accuracy of Formula 3.9 at anything more than small angles is questionable and this should be used with extreme caution. For vessels which have vertical sides, GZ can be estimated up to the angle of DEI using a formula known as the ‘wall-sided formula’: BM ⎛ ⎞ GZ = sin i θ GM + tan2 θ ⎝ ⎠ 2 ▲ Formula 3.10 The wall-sided formula

The wall-sided formula can be considered as accurate for most vessels at small angles of heel, and is theoretically accurate up to angles of DEI or keel emergence for wallsided vessels. Figure 3.55 shows the actual GZ against that predicted by the various methods for MV Reed. X For a derivation of the wall-sided formula, see Appendix 21: The Derivation

of the Wall-Sided Formula for Approximating the Righting Lever.

1.5

Actual GZ GM sin θ GM sin θ cos θ

1

GZ (m)

Wall-sided formula

0.5

0

0

5

10

15

20 25 30 Heel (degrees)

35

40

45

50

▲ Figure 3.55 Approximate GZ curves

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Large Angle Stability • 139

QUESTIONS Q3.22 (MCM, ENG) A box shaped vessel has a length of 40 m, a beam of 7 m and a draught of 3 m. Determine GZ at 5, 10 and 15 degrees of heel if KG is 2 m. Q3.23 (MCM, ENG) The GZ curve below is for MV Reed at her summer displacement with a metacentric height of 1.55 m. In steps of 5 degrees, up to 30 degrees, draw onto the curve the GZ values as predicted by the wall-sided formula. 1.100 1.000 0.900 0.800

GZ (m)

0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.000 0

5

10

15 Heel (degrees)

20

25

30

Q3.24 (MCM, ENG) A box shaped barge has a length of 40 m, a beam of 6 m, and floats upright on an even keel at a draught of 1.00 m in sea water. In this condition, the vessel has a GM of 1.20 m and a freeboard of 3.00 m. The vessel has a double bottom tank, 1 m deep, which extends the full beam of the vessel. The length of the tank is 8 m. A mass of 40 tonnes is lowered onto the freeboard deck of the vessel, 0.50 m to port of the centreline. At the same time, the double bottom tank is filled with sea water to an ullage of 0.8 m. Using a suitable method, determine the resulting angle of list of the vessel.

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140 • Ship Stability, Powering and Resistance

Finding the angle of loll using the wall-sided formula (MCM, ENG) It is the wall-sided formula that allows us to determine the angle of loll of a vessel, as shown in Formula 2.16. At the angle of loll, the GZ must be zero, as the lines of action of buoyancy and gravity are aligned. Therefore, Formula 2.16 can be written as: 0 = sinθ((GM +

BM tan2 θ ) 2

The right hand side of this equation is made of two sections multiplied together, sinθ BM 2 and ( ) . For these to be multiplied to get zero, one of them must be equal 2 to zero. We know that the angle of loll, θ, is a real angle, therefore sinθ cannot be zero, BM 2 therefore ( ) must be equal to zero. This gives: 2 0 = GM +

BM tan2 θ 2

This can be transposed to give: −GM =

BM tan2 θ 2

−GM × 2 = BM tan2 θ −GM × 2 = tan2 θ BM GM × −2 = tan2 θ BM ⎛ GM × −2 ⎞ = tanθ ⎝ BM ⎠ It can be seen that this equation gives the same result as Formula 2.16.

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Large Angle Stability • 141

Effective GM at the Angle of Loll (MCM, ENG) The wall-sided formula can also be used to determine the effective GM of the vessel at the angle of loll. In order for the vessel to loll, the initial GM must be negative, as shown in Figure 3.56; however, as the angle of loll is a large angle, the metacentre moves, until it coincides with the centre of gravity at the angle of loll, as shown in Figure 3.56.

G M

B

▲ Figure 3.56 Negative metacentric height

G

M

Initial M

B

▲ Figure 3.57 The movement of M at the angle of loll

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142 • Ship Stability, Powering and Resistance If a heeling moment is then applied to the vessel, she will incline past the angle of loll, and the centre of buoyancy will move out further, creating a righting moment which will try and roll the vessel back to the angle of loll. The metacentre will move further up above the centre of gravity, and the vessel will effectively have a positive metacentric height, as shown in Figure 3.58. Note that the initial metacentric height is still negative, but the metacentre has moved as the angle of loll is almost always a large angle.

G Initial M

M

B

▲ Figure 3.58 Effective GM past the angle of loll

The effective GM at small angles past the angle of loll can be found from the following: −2GMI = GM at the angle of loll cosθ ▲ Formula 3.11 Effective GM at the angle of loll

X A proof of the formula for the effective metacentric height when lolling can

be seen Appendix 22: The Derivation of the Formula Giving the Effective Metacentric Height at an Angle of Loll.

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Large Angle Stability • 143

QUESTION Q3.25 MV Reed has a negative metacentric height of 10 cm, and is lolling at 10 degrees. Determine the effective metacentric height at the angle of loll.

Variations in GZ with Changes in Freeboard (Including Timber Deck Cargo) (MCM, ENG) As has been seen, the shape of the GZ curve varies depends on a number of factors including the shape of the vessel. As the freeboard of the vessel increases, and assuming that KG remains constant, the DEI angle will increase, as shown in Figure 3.59. Therefore, the point of inflection of the GZ curve occurs at a larger angle, and the resulting change of gradient means that there is more area under the GZ curve, and hence more dynamic stability. This is one of the reasons why preserving suitable freeboard is an important aspect of the Load Line rules.

▲ Figure 3.59 Increasing DEI with increasing freeboard

The resulting GZ curves will be as shown in Figure 3.60. The vessel with the double line has the greatest freeboard. As the vessel inclines, the centre of buoyancy of all three vessels will move outboard at the same rate, and

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144 • Ship Stability, Powering and Resistance 2 Lower freeboard Normal freeboard Higher freeboard

1.5 1

GZ (m)

0.5 0 0

10

20

30

40

50

60

70

80

90

100

110

120

–0.5 –1 –1.5 –2 Heel (degrees)

▲ Figure 3.60 Variation in GZ with increasing freeboard

(assuming GM is constant between the three vessels) the GZ values will increase at the same rate. For the vessel with the larger freeboard, it can also be seen that the point of inflection, and therefore the DEI angle, is larger, as DEI will occur at a larger angle. It can be seen from Figure 3.60 that the following occurs as the freeboard INCREASES: Same initial gradient (due to a constant GM) Increase in DEI angles (due to extra freeboard) Increase in GZ values (due to higher DEI angle) Increase in the area under the curve (due to higher GZ values) Increase in the angle of vanishing stability (due to higher GZ values) It is this increase in the large angle stability due to an increase in freeboard which enables vessels with a timber deck cargo to sail with smaller GM values, as the vessel will have greater stability at larger angles. The timber deck cargo, which is buoyant, is considered to add additional freeboard to the vessel. This is accounted for in special KN curves for these vessels, which assume 75% of the timber volume contributes to buoyancy. The effect of this for MV Reed is shown in Figure 3.61, where the two GZ curves show the vessel at an identical GM and displacement, but with and without the buoyancy effects of a timber deck cargo. The vessel may only sail with a reduced GM (see the section on small angle stability) if she meets the area requirements for the GZ curve created using these adjusted KN values.

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Large Angle Stability • 145 1 0.75 0.5

Main DEI angle

GZ (m)

0.25 0 0

10

20

30

40

50

60

70

80

Without With

–0.25 –0.5 –0.75 –1 Heel (degrees)

▲ Figure 3.61 The effect of timber deck cargo on GZ

Figure 3.62 shows the GZ curves for MV Reed at a draught of 7.00 m (the summer draught) and 8.00 m (overloaded by 1 m). Both conditions have been loaded so that the initial stability is the same (using the minimum allowable GM at the summer draught). The solid line shows the summer condition, and the dashed line shows the overloaded condition. For an increase in displacement of 19%, the area under the GZ curve up to 40 degrees has been reduced by 61%. 0.35 0.3

GZ (m)

0.25 0.2 0.15 0.1 0.05 0 0

10

20

30  Heel (degrees)

40

50

60

▲ Figure 3.62 Summer and overloaded GZ curves

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146 • Ship Stability, Powering and Resistance

Changes in GZ with Variations in Beam (MCM, ENG) The beam of a vessel also has an influence on the stability of the vessel, and therefore the shape of the GZ curve. As the beam of a vessel increases, the BM, and hence GM, also increases. Therefore, for a similar KG value, the initial slope will be greater if the beam is larger. However, an increase in beam actually reduces the DEI angle, as shown in the cross-section sketches in Figure 3.63. Therefore, the point of inflection on the graph will occur at a smaller angle.

▲ Figure 3.63 Reducing DEI with increasing beam

The resulting GZ curves will be as shown in Figure 3.64. The vessel with the double line has the greatest beam. As the vessel inclines, the centre of buoyancy will move outboard faster, and hence the GZ value will increase quicker if the beam is larger. For the vessel with the larger beam, it can also be seen that the point of inflection, and therefore the DEI angle, is lower, as DEI will occur at a smaller angle. It can be seen from Figure 3.64 that the following occurs as the beam increases: Larger initial gradient (due to a greater GM) Reduction in DEI angles (due to extra beam) Increase in GZ values (due to higher GM and wider beam) Increase in the area under the curve (due to higher GZ values) No variation in the angle of vanishing stability (due to constant vessel depth)

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Large Angle Stability • 147 2 Narrow beam Wider beam Normalbeam

1.5

GZ (m)

1

0.5

0 0

10

20

30

40

50

60

70

80

90

100

–0.5

–1 Heel (degrees)

▲ Figure 3.64 Variations in GZ with increasing beam

Changes in GZ with Symmetrical Ice Accretion (MCM, ENG) In extremes of cold weather, it is possible for water to freeze to the hull, decks, superstructure and rigging of the vessel in large quantities. This can have a problematic effect on the stability of the vessel. The addition of the ice increases the displacement of the vessel, and hence reduces the freeboard of the vessel. As seen previously, a reduction in the freeboard reduces the DEI angle. The ice is also invariably formed above the centre of gravity of the vessel, and therefore raises the centre of gravity, reducing the metacentric height (and hence initial slope) and GZ values, also as seen previously. The precise calculation of icing and centres of mass of ice is complex. Part 8 of the MCA’s Guidelines for Surveyors (Maritime and Coastguard Agency) stipulate an assumed icing of 30 kg/m2 on all exposed horizontal surfaces, and 15 kg/m2 on the exposed vertical surfaces on one side of the ship. For wires, rails, booms and rigging, the weight of the ice on lateral areas should be increased by 5% to account for icing on the rigging, and the moment in the loading table increased by 10%. The KG values should be calculated based on the vertical position of each surface.

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148 • Ship Stability, Powering and Resistance Figure 3.65 shows the GZ curves for MV Reed before and after ice accretion. The initial displacement used was the summer displacement, with a KG at the maximum permitted to meet the requirements of the SOLAS criteria applicable to all ships. 200 tonnes of ice was then added at a KG of 12 m to simulate ice accretion over the deck and superstructure. 0.4 0.3

Before After

0.2 0.1 0

0

10

20

DEI 30

40

50

60

70

– 0.1 – 0.2 – 0.3 – 0.4

▲ Figure 3.65 GZ curve and ice accretion

It can be seen from Figure 3.65 that the following occurs as a result of symmetrical ice accretion: Smaller initial gradient (due to a smaller GM) Reduction in DEI angles (due to a reduction in freeboard) Reduction in GZ values (due to lower GM and reduced freeboard) Reduction in the area under the curve (due to lower GZ values) Reduction in the angle of vanishing stability (due to lower GZ values)

Changes in GZ with Asymmetric Ice Accretion (MCM, ENG) When ice accretion occurs, it is rarely evenly across the vessel. The direction of the apparent wind will result in a greater build-up of ice on one side of the vessel. This

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Large Angle Stability • 149 results in a similar set of changes to the symmetrical case, but in addition, the centre of gravity of the vessel moves towards the ice accretion, resulting in a further reduction in GZ values and an angle of list. Figure 3.66 shows the same scenario as in Figure 3.65, but with the ice centred 2.00 m off of the centreline. 0.4 0.3 Before After

0.2

GZ (m)

0.1 0

0

10

20

DEI 30

40

50

60

70

– 0.1 – 0.2 – 0.3 – 0.4 Heel (degrees)

▲ Figure 3.66 GZ curve and asymmetric ice accretion

It can be seen from Figure 3.66 that the following occurs as a result of asymmetrical ice accretion: Smaller initial gradient (due to a smaller GM) Reduction in DEI angles (due to a reduction in freeboard) Reduction in GZ values (due to lower GM and reduced freeboard) Initial negative GZ values and an angle of list (due to the centre of gravity moving off of the ship’s centreline) Reduction in the area under the curve (due to lower GZ values) Reduction in the angle of vanishing stability (due to lower GZ values)

Changes in GZ with Changes in Trim (MCM, ENG) As a vessel trims, the waterplane area will vary. For some ships, this difference is very small, but for vessels with large overhanging sterns and large bow flare, such as offshore

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150 • Ship Stability, Powering and Resistance supply vessels, or container ships, the change in trim can have a large effect on the waterplane area. For these vessels, the typical hull shape results in a widening of the waterplane area with stern trim, as shown in Figures 3.67 and 3.68. The result of this is an increase (for stern trim) or a decrease (for bow trim) of the transverse waterplane inertia, and hence an increase (for stern trim) or a decrease (for bow trim) in the BM. This has a corresponding increase (for stern trim) or a decrease (for bow trim) in the GM, and the slope of the GZ curve.

Even keel Stern trim

Bow trim

▲ Figure 3.67 Underwater volume shape with trim

▲ Figure 3.68 Waterplane area for stern trim (top), zero trim (middle) and bow trim (bottom)

Figure 3.69 shows the GZ curves for MV Reed at her load displacement and maximum KG, for 1 m of stern trim, an even keel, and 1 m of head trim. It can be seen that in the

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Large Angle Stability • 151 0.4 0.3

1 m stern trim Even keel 1 m bow trim

0.2 0.1 0

0

10

20

30

40

50

60

70

– 0.1 – 0.2 – 0.3 – 0.4

▲ Figure 3.69 Variation in GZ with trim

case of stern trim, the initial gradient is steeper, and GZ values are increased, and in the case of bow trim, the initial gradient is shallower and GZ values are reduced.

Changes in GZ with FSE (MCM, ENG) The effects of free surfaces have already been discussed in relation to small angle stability. However, at large angles, the effects of free surfaces must be considered. As the vessel inclines to large angles, the shape of the free surface in the tank changes considerably, which changes the influence of the FSE on the stability of the vessel. This calculation is complex, and therefore not routinely undertaken. Instead, GZ values are calculated based on the effective of fluid KG. The effect of this on GZ is shown in Figure 3.70, for MV Reed in a condition where all of the Nos 2, 3 and 4 Double Bottom tanks are filled to a sounding of 1.00 m to create a large FSE. The dashed-dot line shows the GZ values ignoring any FSEs, and the dashed line shows the GZ values using the FSM data from the tank hydrostatics to determine the effective KG, and then GZ. The solid line shows the GZ values based on the simulation of fluid moving in the tanks at each angle of heel, and is therefore the most accurate calculation of the FSE. As can be seen, the use of the tank data to determine the effective KG and therefore GZ gives the

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152 • Ship Stability, Powering and Resistance 1.4 Tank FSM Fluid simulation

1.2

No FSM

1 0.8

GZ (m)

0.6 0.4 0.2 0

0

10

20

30

40

50

60

70

80

90

100

–0.2 –0.4 –0.6 –0.8 Heel (degrees)

▲ Figure 3.70 The effect of free surfaces on GZ

most pessimistic, and therefore safest, GZ values. The actual movement of the fluid as the vessel inclines to large angles results in slightly larger GZ values; therefore, the use of the effective KG value is a safe assumption.

Fixed and Free to Trim Curves (MCM, ENG) As the vessel rolls, she may also change her trim. Figure 3.71 shows the trim of a container ship, initially on an even keel, as the vessel rolls. This change in trim is because the underwater shape changes significantly as she rolls, mainly as a result of large stern overhangs and large bow flare. This makes the centre of buoyancy move forward or aft, and hence a trimming moment is generated. Historically, this movement in B was excessively time-consuming to calculate, so the trim was assumed to be fixed. Older stability books will give cross curves and large angle stability information assuming that the trim was fixed. The introduction of more computer processing power over the past few decades has allowed the effects of trim to be included in the curves. These are known as ‘free to trim’ curves. Figure 3.72 shows a GZ curve for a container ship, with the trim fixed, and also with the vessel free to trim.

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Large Angle Stability • 153 10

0

20

30

40

50

60

70

0

Trim (m)

–1

–2

–3

–4

–5 Heel (degrees)

▲ Figure 3.71 Change in trim of a container ship with list

0.4 0.3

Free to trim Fixed trim

0.2

GZ (m)

0.1 0 0

10

20

30

40

50

60

70

–0.1 –0.2 –0.3 –0.4 Heel (degrees)

▲ Figure 3.72 Fixed and free to trim curves for a container ship

As can be seen, there is little difference in the values for this ship, especially in the initial values and the angle of vanishing stability. However, for vessels with high freeboard forward, such as offshore supply vessels, the change in trim during heeling is large (see Figure 3.73), and therefore the difference in the GZ curves is significant.

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154 • Ship Stability, Powering and Resistance 10 9 8 7

Trim (m)

6 5 4 3 2 1 0

0

10

20

30

40

50

60

70

–1 Heel (degrees)

▲ Figure 3.73 Change in trim of an OSV with list

Figure 3.74 shows the fixed and free GZ curves for an offshore supply vessel. As can be seen, there is a large difference in the GZ values at larger angles of heel. A certain amount of cynicism must be used if fixed trim information is used to create GZ curves for vessels with large freeboard forward. 0.4 0.3

Free to trim Fixed  trim

0.2

GZ (m)

0.1 0

0

10

20

30

40

50

60

70

–0.1 –0.2 –0.3 –0.4

Heel (degrees)

▲ Figure 3.74 Fixed and free to trim curves for an OSV

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Large Angle Stability • 155

GZ Curves for Vessels with Large Freeboards and Small Draughts (MCM, ENG) For normally proportioned vessels loaded towards the maximum displacement, the draught will be larger than the freeboard. However, for some vessels, such as small ferries, the freeboard may be significantly larger than the draught, with a wide beam. In these circumstances, as the vessel heels, the keel may emerge from the water before the deck immerses in the water. In this situation, the point of inflection of the curve does not give the DEI angle. Figure 3.75 shows the GZ curve for a vessel with a freeboard of 12 m, and a draught of 8 m, with a beam of 25 m. Although the curve looks similar to those seen previously, the keel of the vessel emerges from the water at 30 degrees of heel, with the deck edge immersing at 45 degrees. 2 1.8 1.6 1.4 1.2

GZ (m)

1 0.8 0.6 0.4 0.2 0 –0.2

0

10

20

30

40

50

60

70

80

90

100

–0.4 –0.6 –0.8

Angle (degrees)

▲ Figure 3.75 GZ curve for a shallow draught vessel

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156 • Ship Stability, Powering and Resistance

GZ Curves for Vessels Showing Port and Starboard Angles (OOW, MCM, ENG)

Overview (OOW, MCM, ENG) In general GZ curves or righting moment curves only show the inclination of the vessel in one direction, normally using the positive x axis to show either port or starboard. However, sometimes it may be necessary to show both port and starboard on one GZ curve. In this case, the GZ curve is shown, but with a significant change to the direction of the values on one side of the curve. Consider the vessel in Figure 3.76, heeled to port at 20 degrees. The positive GZ value of approximately 0.22 m would indicate that the vessel is trying to right itself. In this context, a righting moment is acting to roll the vessel back in a starboard direction, and hence back towards upright – therefore looking at the vessel from ahead, an observer would see this righting moment acting in a counter-clockwise direction. If the righting moment were to be calculated, there would be a positive GZ multiplied by a positive displacement to give a positive righting moment. Now consider the vessel in Figure 3.76, heeled to starboard at 20 degrees. If the GZ curve were simply a mirror image around the y axis, as shown, then the positive GZ value of approximately 0.22 m to starboard would indicate that the vessel is trying to right itself. In this context, the righting moment is acting to roll the vessel back in a port direction, and hence back towards upright – if an observer were looking at the vessel from ahead, they would see this righting moment acting in a clockwise direction. As before, if the righting moment were to be calculated, there would be a positive GZ multiplied by a positive displacement to give a positive righting moment. This causes problems from the point of view of a sign convention for the moments, as there would be two positive GZ values at opposite angles, each creating a positive righting moment, but which act in opposite directions – for example, when heeled to port, the righting moment acts counter-clockwise to starboard, and when heeled to starboard, the righting moment acts clockwise to port. While the GZ values make sense, this can cause problems with the righting moments, as there is now no clear reference to indicate which way the moment is acting. To overcome this problem of the sign convention for the righting moments, the GZ curve is plotted with a GZ causing a righting moment as positive on one side, and a GZ causing a righting moment as negative on the other side, as shown in Figure 3.77.

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Large Angle Stability • 157 0.6

0.4

0.2

GZ (m)

–90

–80

–70 –60

–50

–40

–30

–20

–10

0

0

10

20

30

40

50

60

70

80

90

–0.2 Port

Stb –0.4

–0.6

–0.8

–1 Heel (degrees)

▲ Figure 3.76 Simplified GZ to port and starboard

This means that when heeling to both sides is considered, the previous GZ convention which said that positive GZ was a righting lever, and negative GZ was a capsizing lever needs adjusting when heeling to the opposite direction. Consider the vessel in Figure 3.77, heeled to port at 20 degrees. The positive GZ value of approximately 0.22 m would indicate that the vessel is trying to right itself. In this context, a righting moment is acting to roll the vessel back in a starboard direction, and hence back towards upright – if an observer were looking at the vessel from ahead, they would see this righting moment acting in a counter-clockwise direction. As before, if the righting moment were calculated, there would be a positive GZ multiplied by a positive displacement to give a positive righting moment. Now consider the vessel in Figure 3.77, heeled to starboard at 20 degrees. Now that the GZ curve have been adjusted as described, the negative GZ value of approximately –0.22 m to starboard would now still indicate that the vessel is trying to right itself. In this context, the righting moment is acting to roll the vessel back in a port direction, and hence back towards upright – if an observer were looking at the vessel from ahead, they would see this righting moment acting in a clockwise direction. If the righting moment were calculated, there would be a negative GZ multiplied by a positive displacement to give a negative righting moment. While the righting moment when heeled to port was positive, the righting moment when heeled to starboard is negative. This indicates that the righting moments are acting in opposite directions, which must be correct as they are both acting to roll the vessel upright.

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158 • Ship Stability, Powering and Resistance 1 0.8 0.6 0.4

GZ (m)

0.2 Stb –90

–80

–70 –60

Port 90

0 –50

–40

–30

–20

–10

0

10

20

30

40

50

60

70

80

–0.2 –0.4 –0.6 –0.8 –1 Heel (degrees)

▲ Figure 3.77 GZ to port and starboard

This shows that the GZ sign convention must be changed from the initial one where positive GZ was a righting lever and negative GZ was a capsizing lever. The new sign convention is that GZ values in the upper right and lower left quadrants are righting levers, and GZ values in the upper left and lower right quadrants are capsizing moments, as shown in Figure 3.78. 1 0.8

Capsizing levers

0.6

Righting levers

0.4

GZ (m)

0.2

Stb 0 – 90 – 80 – 70 – 60 – 50 – 40 – 30 – 20 – 10 0

Port 10

20

30

40

50

60

70

80

90

– 0.2 – 0.4 Righting levers

Capsizing levers

– 0.6 – 0.8 –1 Heel (degrees)

▲ Figure 3.78 Sign convention for righting GZ values to port and starboard

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Large Angle Stability • 159 1 0.8 0.6 0.4

GZ (m)

0.2 Stb 0 –90 –80 –70 –60 –50 –40 –30 –20 –10 0

Port 10 20 30 40 50 60 70 80 90

–0.2 –0.4 –0.6 –0.8 –1 Heel (degrees)

▲ Figure 3.79 GZ to port and starboard for a vessel with port list

This process can be applied to all of the types of GZ curve that have been seen previously, as shown in Figure 3.79. This would occur if the TCG was to port of the centreline, causing a list to port of 22 degrees. At any angle of inclination between 68 degrees to starboard, and 0 degrees, the negative GZ values indicate that the vessel will try and roll back towards the upright position of 0 degrees. At any angle of inclination between 0 degrees and 22 degrees to port, the negative GZ values indicate that the vessel will try and roll further towards the list angle of 22 degrees. For any angle between 22 degrees to port and 59 degrees to port, the positive GZ values indicate that the vessel will try and roll back to the list angle of 22 degrees. Again, the process works for an angle of loll, as shown in Figure 3.80, and for a vessel with combined list and loll, as shown in Figure 3.81.

Effects on GZ approximation and dynamic stability (MCM, ENG) Initially this seems confusing, however, as the angles to the left of the graph, or starboard, are negative, and the angles to the right, or port, are positive, this new convention starts to make sense when calculating other aspects of large angle stability. For example, consider the wall-sided formula. If we were calculating the GZ values at an angle θ to port, the angle θ would be positive, therefore giving a positive sinθ term and a positive tan2θ term, and so we would get a positive GZ for a stable vessel. However,

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160 • Ship Stability, Powering and Resistance 2 1.5 1

GZ (m)

0.5 Stb 0 –90 –80 –70 –60 –50 –40 –30 –20 –10 0

10

20

30

40

50

60

70

Port 80 90

–0.5 –1 –1.5 –2 Heel (degrees)

▲ Figure 3.80 GZ to port and starboard for a vessel with negative GM and an angle of loll 2 1.5 1

GZ (m)

0.5 Stb 0 –90 –80 –70 –60 –50 –40 –30 –20 –10 0

10

20

30

40

50

60

70

Port 80 90

–0.5 –1 –1.5 –2 Heel (degrees)

▲ Figure 3.81 GZ to port and starboard for a vessel with a combined angle of list and loll

if we were calculating the GZ values at an angle θ to starboard, the angle θ would be negative, therefore giving a negative sinθ term and a positive tan2θ term, and so we would get a negative GZ for a stable vessel. This sign convention also works for the area under the GZ curve. Consider the area under the GZ curve in the upper right quadrant. The positive GZ, multiplied by a positive angle, gives a positive area, indicating that the vessel can ‘absorb’ energy. In the lower left quadrant, the negative GZ, multiplied by a negative angle, gives a positive area, again indicating that the vessel can ‘absorb’ energy. In the top left and bottom right quadrants, the multiplication of GZ and angle will always result in negative areas, indicating that the vessel cannot absorb further energy.

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Large Angle Stability • 161

Heeling under the Effect of Wind (MCM, ENG) A vessel which heels due to the wind will be in equilibrium when the heeling moment is equal to the righting moment. The heeling moment due to wind is made up of the force of the wind acting to rotate the vessel. The force of the wind, abbreviated to ‘F’, acts through a lever ‘L’. L is the vertical distance from the centre of the exposed lateral area, ‘A’, to the half draught, as shown in Figure 3.82.

L

▲ Figure 3.82 Wind heeling moment definitions

The force, F, is found from the wind pressure multiplied by the exposed lateral area. Values of wind pressure against wind speed are shown in Figure 3.83. 140.00 130.00 120.00

Wind pressure (kg/m2)

110.00 100.00 90.00 80.00 70.00 60.00 50.00 40.00 30.00 20.00 10.00 0.00 0

10

20

30

40 50 Wind speed (knots)

60

70

80

90

▲ Figure 3.83 Wind speed and wind pressure

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162 • Ship Stability, Powering and Resistance The force F is then multiplied by the lever L, which gives the heeling moment caused by the wind. Therefore: Wind heeling ee g moment Win W dp pressure essu e × Lateral ate a area Lever × cosθ Note that the cosθ term comes from the reduction in L as the vessel heels – see Figure 3.45 for details. If this is plotted on top of the righting moment curve, then the point where the curves first intersect gives the point where the heeling moment and righting moment are in equilibrium, and therefore gives the heel angle due to the wind (see Figure 3.44 for details). We have previously seen that the wind heeling moment reduces as the vessel heels; however, by assuming that it remains constant we can simplify the process. This is valid as the assumption is a ‘worst-case’ scenario and will overestimate the resulting heel angle, and allows us to ignore the cosθ term to simplify the process.

QUESTION Q3.26 (MCM, ENG) MV Reed is loaded to a displacement of 4,600 tonnes, with a KG of 6.74 m. She has an exposed lateral area of 617 m2, with a wind heeling lever of 5.17 m. By plotting the righting moment in steps of 5 degrees to 20 degrees, determine the angle of heel that would result from a beam wind of 75 knots. 700

Heeling and righting moments (tonne metres)

600

500

400

300

200

100

0 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 Heel (degrees)

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Large Angle Stability • 163 Chapter 3 of the IS Code 2008 includes recommended criteria for ensuring stability in severe wind and weather. For UK flagged ships, the MCA has similar, but slightly simplified guidelines specific for container ships, where the height from the load waterline to the top of the container stacks is more than 30% of the beam. These are detailed in the MCA’s Guidance to Surveyors (Maritime and Coastguard Agency). These MCA guidelines assume the vessel is already loaded to her worst case for stability. The lateral windage area is assumed to be subjected to a steady wind load of 48.5 kg/m2 (which is approximately equal to Beaufort force 10). The heeling moment created by the wind is given by: λ 0 = Steady wind load × Lateral w winda indage ge area × Lever

Righting moment (tonne metres)

This is superimposed over the righting moment curve for the vessel in the worst possible loading condition. The resulting angle of heel, given by the intersection of the curve and the λ0 line is known as θ1. This should be less that 65% of the DEI angle. The heeling moment due to wind is then increased by 50%, to account for gusts, and the vessel is rolled 15 degrees into the wind. The area between the GZ curve and the increased heeling moment line to this 15 degree back roll is known as S1, and is shown in Figure 3.84. This area represents the energy which would be ‘stored’ in the vessel as she rolls under the effect of wind and waves. A vertical line is then drawn on the curve so that the area S2 is equal to S1. The point of intersection of this line with the x axis is known as θdy , and can be considered to be the worst-case heel angle. This should be less than the down-flooding angle of the vessel.

15° S2

1.5λ0 λ0

S1

θ1

θ dy

Heel (degrees)

▲ Figure 3.84 Wind heeling moments

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164 • Ship Stability, Powering and Resistance

Righting moment (tonne metres)

Typically the angle θ1 will be less than 15 degrees; therefore the area S1 will also include area to the left of the y axis. In this case, the GZ curve should be mirrored, as explained previously. This is shown in Figure 3.85.

15°

S2

1.5λ0 λ0

S1

θ1 θ dy Heel (degrees)

▲ Figure 3.85 Wind heeling moments to port and starboard

Grain Regulations (MCM) The loading and carriage of grain (along with wheat, maize, rye, barley, rice, pulses and seeds which behave in bulk in a similar manner to grain) differs from the usual rules and regulations for stability. Grain cargoes are particularly prone to shifting with ship motion. This can have problematic effects on the metacentric height and the large angle stability of the vessel, as shown previously in the section covering cargo shifts. The exact movement of the grain is subject to a large amount of research, and is an extremely complicated problem which we will not attempt to model. To ensure safe carriage, we have a simplistic set of rules to follow which allow us to ensure the loading pattern is safe. These rules use a simple, ‘worst-case’ approach to analyse the effect of the cargo distribution on the stability. They are detailed in the International Code for the Safe Carriage of Grain in Bulk (International Maritime Organisation, 1991), known as the International Grain Code.

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Large Angle Stability • 165 During the voyage, it is assumed that: The grain settles, reducing the initial volume and opening a void in the hold above the cargo, allowing a transverse shift in mass. At angles of heel of less than 30 degrees, a void appears at the top of the grain. This allows the grain, and the centre of mass of the grain, to shift transversely. In all circumstances an average 0.15 m deep void is assumed to exist at the top of the grain. Grain compartments which are initially fully loaded are assumed to have a maximum surface slope of 15 degrees. Grain compartments which are initially partially loaded are assumed to have a maximum surface slope of 25 degrees. The capacity and details of each space used for grain will be tabulated in the ships data in a similar way to tank data, and these can be used in a similar manner to determine the mass and KG of the grain cargo. A normal loading table can then be used to determine the loaded KG, and hence GM, of the vessel. The grain regulations require that a vessel carrying a grain cargo has a minimum GM of 0.30 m or 30 cm – double the normal requirement. Before sailing the Master must also ensure that the ship is upright. In addition to the normal hold data, there will be a ‘volumetric heeling moment’ or ‘horizontal heeling moment’ column. This lists, for any ullage of grain, the capacity of the grain hold and the worst-case volume shift of grain cargo, based on the above assumptions. This volumetric heeling moment gives the equivalent volume of grain that would move sideways in the event of a cargo shift. For example, a volumetric heeling moment of 100 m4 would be the equivalent of 100 m3 of grain moving 1 m sideways, or 50 m3 of grain moving 2 m sideways, or 25 m3 of grain moving 4 m sideways, or even 10 m3 moving 10 m sideways. Each of these shifts would generate the same heeling moment. The grain mass can be found from the capacity and the stowage factor.

Grain mass =

Grain volum v e Stowag w e ffactor

▲ Formula 3.12 Grain mass

From the volumetric heeling moment, the actual mass heeling moment can be found by dividing the volumetric moment by the stowage factor of the grain.

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166 • Ship Stability, Powering and Resistance

Grain mass ass heeling moment =

Volumetric heeling ee g moment Stowag w e factor

▲ Formula 3.13 Grain mass heeling moment

QUESTION Q3.27 (MCM) MV Reed is loaded with grain to a sounding of 4.00 m. The stowage factor of the grain is 1.40 m3/t. Determine the mass and the mass heeling moment of the grain.

Summing up all of the mass heeling moments together gives the overall worst-case heeling moment in the event of a grain cargo shift. This can be used to determine the list in the event of a cargo shift. The grain regulations state that the maximum allowable list in the event of a grain cargo shift is lesser of 12 degrees or the DEI angle. Dividing the total mass heeling moments by the overall mass displacement gives a value called the grain heeling arm, known as λ0, which has units of metres.

λ0 =

Total mass ass heeling moment Δ

▲ Formula 3.14 Grain heeling arm at zero degrees

On the GZ curve (shown in Figure 3.86) of the vessel, a line, known as the grain heeling line is drawn between the λ0 point on the x axis, and a new point, λ40, which is at 40 degrees on the GZ curve, at a value of 80% of λ0. λ 40 = λ 0 × 0 8 ▲ Formula 3.15 Grain heeling arm at 40 degrees

The residual area (see Figure 3.85) between this line and the GZ curve up to 40 degrees (or the peak GZ, or the down-flooding angle, whichever is least) must not be less than 0.075 m radians. Cross curves of stability are provided to the master for 12 degrees and 40 degrees to enable these GZ calculations.

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Large Angle Stability • 167 0.900 0.800 0.700

GZ (m)

0.600 0.500 0.400 0.300 0.200 0.100 0.000

0

5

10

15

20 25 Heel (degrees)

30

35

40

▲ Figure 3.86 Grain regulation residual area

Determining the shaded area requires a mix of Simpson’s Rule, and calculating the area of a triangle. To determine the residual area, Simpsons’ Rule must first be used to determine the overall area under the curve between 0 and 40 degrees, or the down-flooding angle. The area bounded by the GZ curve and the grain heeling arm, as shown in Figure 3.86, can be found by approximating the area to a triangle and trapezoid, with the split at 0.900 0.800 0.700

GZ (m)

0.600 0.500 0.400 0.300 0.200 0.100 0.000

0

5

10

15 20 Heel (degrees)

25

30

35

40

▲ Figure 3.87 Area under the grain heeling line and the GZ curve

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168 • Ship Stability, Powering and Resistance the intersection. This area can be subtracted from the area found by Simpson’s Rule to leave the residual area. Although this approximates the GZ curve to a straight line up to the list angle, the resulting error is small, and will generally result in the residual area being larger than that which is calculated.

QUESTION Q3.28 (MCM) MV Reed starts loading from the lightship condition. The following loading takes place: Both bunkers are pressed with fuel, relative density 0.960 Both No. 2 Double Bottom Tanks are filled to a sounding of 1.00 m with sea water ballast The grain hold is filled to a sounding of 5.00 m with grain, stowage factor 1.45 m3/t After loading the above, determine GM and plot the GZ curve (in 5 degree steps up to 40 degrees) for the vessel, and hence determine the ability of the vessel to meet all of the requirements of the grain regulations (including pre-1994 and post-1994 criteria).

In order to simplify the whole procedure, tables of maximum allowable mass heeling moments may be supplied to the ship, as shown in the sample stability data for MV Reed. These are the maximum mass heeling moments at given ship KG values and displacements that would just meet the GZ curve and list requirements (which will vary with the age of the vessel). To ensure that the vessel complies with the GZ curve and list requirements of the grain regulations, all that is necessary is to check the actual mass heeling moment against the maximum allowable mass heeling moment. GM must still be calculated and checked separately. In the event of a grain shift, the heel angle can be estimated from:

Approximate list = 12 ×

Actual mass ass heeling moment Allowable l mass s heeling moment

▲ Formula 3.16 Approximate list due to grain shift

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Large Angle Stability • 169

QUESTIONS Q3.29 (MCM) MV Reed has a displacement of 4,050 tonnes, with a KG of 5.35 m. The grain mass heeling moment after loading is 1,550 tonne metres. Using the tables of maximum allowable grain mass heeling moment, determine the ability of the vessel to pass the list and residual area requirements of the grain regulations. Q3.30 (MCM) MV Reed has a displacement of 6,300 tonnes, with a KG of 5.05 m. The grain mass heeling moment after loading is 1,400 tonne metres. Using the tables of maximum allowable grain mass heeling moment, determine the ability of the vessel to pass the list and residual area requirements of the grain regulations. Q3.31 (MCM) MV Reed has a displacement of 4,700 tonnes, with a KG of 5.85 m. The grain mass heeling moment after loading is 1,300 tonne metres. Using the tables of maximum allowable grain mass heeling moment, determine the ability of the vessel to pass the list and residual area requirements of the grain regulations. Q3.32 (MCM) MV Reed has a displacement of 7,100 tonnes, with a KG of 6.74 m. The grain mass heeling moment after loading is 400 tonne metres. Using the tables of maximum allowable grain mass heeling moment, determine the ability of the vessel to pass the list and residual area requirements of the grain regulations. Q3.33 (MCM) MV Reed loads from the lightship condition. The port and starboard bunkers are each pressed with 120.22 tonnes of fuel, with a KG of 7.81 m. The grain hold is filled to a sounding of 6.00 m with grain, stowage factor 1.38 m3/t. Determine if the vessel meets the GZ curve and list requirements of the grain regulations using the maximum allowable grain heeling moment data, and find the approximate list angle in the event of a grain cargo shift.

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170 • Ship Stability, Powering and Resistance

LARGE ANGLE STABILITY  LEARNING CHECKLIST Objective

Level

Understand how the centre of buoyancy moves at large angles

OOW, MCM, ENG

Understand how a righting moment or capsizing moment may be generated

OOW, MCM, ENG

Understand what is meant by the terms GZ, righting lever, righting arm, statical stability or lever of statical stability

OOW, MCM, ENG

Understand the significance of positive and negative GZ values

OOW, MCM, ENG

Understand the meaning of KN, and determine KN at any angle or displacement using the hydrostatics

OOW, MCM, ENG

Calculate GZ for any angle or displacement using the hydrostatics

OOW, MCM, ENG

Plot a GZ curve

OOW, MCM, ENG

Determine the range of stability

OOW, MCM, ENG

Determine, and know the importance of, the angle of vanishing stability

OOW, MCM, ENG

Compare the peak value of the GZ against the current regulations

OOW, MCM, ENG

Calculate GZ at small angles using the metacentric height

OOW, MCM, ENG

Determine GM from the GZ curve

OOW, MCM, ENG

Understand the accuracy of GZ at angles where cargo shifts and down-flooding may occur

OOW, MCM, ENG

Determine the down-flooding angle using the hydrostatics, and comment on the accuracy of calculated down-flooding angles

OOW, MCM, ENG

Understand how changes in KG influence the GZ curve

OOW, MCM, ENG

Correct GZ values for a change in KG at a constant displacement

MCM, ENG

Understand how the TCG influences the GZ curve

OOW, MCM, ENG

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Large Angle Stability • 171

Objective

Level

Identify a vessel which is listing from the shape of the GZ curve

OOW, MCM, ENG

Correct GZ values for the TCG

MCM, ENG

Determine the angle of list at large angles using the corrected GZ curve

OOW, MCM, ENG

Correct GZ for a cargo shift or change in cargo position

MCM, ENG

Identify a vessel which is lolling from the shape of the GZ curve

OOW, MCM, ENG

Determine the angle of loll from a GZ curve

OOW, MCM, ENG

Identify a vessel which is initially neutrally stable from the shape of the GZ curve

OOW, MCM, ENG

Identify a vessel which is both listing and lolling from the shape of the GZ curve

OOW, MCM, ENG

Determine the combined angle of list and loll from the shape of the GZ curve

OOW, MCM, ENG

Understand the process by which a vessel must be recovered from an angle of loll

OOW, MCM, ENG

Calculate and plot the GZ at stages of loll recovery to ensure that a vessel can be safely recovered

MCM, ENG

Determine the DEI angle using the GZ curve

OOW, MCM, ENG

Calculate the moment of statical stability or righting moment

OOW, MCM, ENG

Plot the moment of statical stability or righting moment curve

OOW, MCM, ENG

Use the righting moment curve to determine the list angle by correcting the righting moment values

MCM, ENG

Use the righting moment curve to determine the list angle by superimposing the heeling moment values

MCM, ENG

Use the GZ curve to determine the list angle by superimposing the heeling arm values

MCM, ENG

Understand what is meant by dynamic stability and understand the importance of the area under a righting moment or GZ curve

OOW, MCM, ENG

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172 • Ship Stability, Powering and Resistance

Objective

Level

Calculate the dynamic stability of ship in tonne metre degrees and tonne metre radians

MCM, ENG

Calculate the areas under GZ curves and compare them against the current regulations

MCM, ENG

Correct the areas under GZ curves for down-flooding, and compare them against the current regulations

MCM, ENG

Understand the accuracy of Simpson’s Rule and ensure that they are suitably used for the shape of the curve

MCM, ENG

Calculate approximate GZ values at larger angles, and comment on the accuracy of the approximation

MCM, ENG

Determine the approximate angle of loll using the wall-sided formula

MCM, ENG

Determine the effective GM at the angle of loll

MCM, ENG

Understand how the freeboard affects GZ values and curves

MCM, ENG

Understand how the beam affects GZ values and curves

MCM, ENG

Understand how the symmetrical and asymmetrical ice accretion affects GZ values and curves

MCM, ENG

Understand how GZ changes with trim

MCM, ENG

Understand how FSE influences GZ

MCM, ENG

Understand the differences between free to trim and fixed to trim GZ curves

MCM, ENG

Understand how wind heels ship, and determine the heel angle under a beam wind

MCM, ENG

Understand and explain the current recommendations for ships and wind heeling

MCM, ENG

Understand synchronous and parametric roll, and actions to be taken to avoid them

MCM, ENG

Understand and explain the link between stability and roll period

OOW, MCM, ENG

Understand and explain the problems associated with the carriage of grain or similar cargoes

MCM

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Large Angle Stability • 173

Objective

Level

Understand the assumptions and cargo shift model upon which the grain regulations are based

MCM

Determine the ability of a vessel to meet requirements of the grain regulations by plotting the GZ curve and determining the residual stability

MCM

Determine the ability of a vessel to meet requirements of the grain regulations by using the maximum grain heeling moment tables, and determine the approximate list angle in the event of a grain cargo shift

MCM

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4

LONGITUDINAL STABILITY – FORE AND AFT BALANCE AIMS AND OBJECTIVES

At the end of this section, you should be able to: Understand the concept of longitudinal metacentric height Understand why the longitudinal metacentric height can normally be assumed to be constant Calculate the longitudinal metacentric height for a box shaped vessel Understand what is meant by the terms trim, stern trim and bow trim Understand what is meant by the MCTC Calculate the MCTC for a box shaped vessel Determine the MCTC from hydrostatic data Understand the units of MCTC Calculate the trim of a ship Calculate the trim of a box shaped vessel Calculate the approximate change in trim for small changes in loading Understand the four positions at which the draught is measured, and the associated terminology Understand what is meant by the LCF, and the importance of the LCF in trimming the ship Determine the mean draught from the end draughts Determine the true mean draught from the hydrostatics Calculate the end draughts of the vessel

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Longitudinal Stability • 175 The stability of a vessel in the fore and aft sense is measured in a similar way to the transverse method. As previously seen, in a longitudinal sense, the buoyancy forces act at the longitudinal centre of buoyancy, abbreviated to the LCB, and normally measured as a distance Forward of the Aft Perpendicular (FOAP). The gravitational forces act at the longitudinal centre of gravity, abbreviated to the LCG. Consider the vessel on an even keel, as shown in Figure 4.1.

G B

▲ Figure 4.1 G and B on an even keel

If the pitch of the vessel is changed, then the centre of buoyancy will move to the new centre of underwater volume. The intersection of the line of action of buoyancy to a line which was originally drawn vertically through the original centre of buoyancy is known as the longitudinal metacentre, as shown in Figure 4.2. In a similar way to finding the transverse metacentric height, the longitudinal metacentric height can be found, as shown in Figure 4.3. GML = KML KG = KB + BML − KG ▲ Formula 4.1 GML from KB, BML and KG

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176 • Ship Stability, Powering and Resistance

Longitudinal metacentre, ML

G New B

Old B

▲ Figure 4.2 The longitudinal metacentre

As can be seen in Figure 4.2, GML is typically very large (often in the region of the length of the vessel), which indicates that ships are normally very stable fore and aft. This is the reason that intact ships tend to capsize transversely rather than bow over stern. As GML is so large, free surface effects are sometimes not included, as their influence is small in comparison to the transverse GM. Instead, the lightship or a worst case loaded value is used, and GML is assumed to be constant.

ML

GML

BML

G KG

B

KB

▲ Figure 4.3 Longitudinal metacentric height

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Longitudinal Stability • 177

Calculating GML for a Box Shaped Vessel (MCM, ENG) For a box shaped vessel, the calculations for KB and KG are the same as the transverse case, while BML requires a new calculation. BML is based on the ‘longitudinal inertia of the waterplane measured through the longitudinal centre’, or ‘longitudinal inertia’ which is a mathematical measure of the fore and aft distribution of the waterplane area. Once the inertia has been found, the vertical distance from B to ML can be found. For a box shaped vessel, the longitudinal waterplane inertia is given by the formula:

IL =

BLL3 12

▲ Formula 4.2 Calculating longitudinal waterplane inertia for a box shaped vessel

X For a mathematical proof of the longitudinal waterplane inertia of a box

shaped vessel, see Appendix 23: The Derivation of the Formula Giving the Longitudinal Inertia of a Rectangle. This is very similar to the transverse case, only length and beam have been reversed. Once the longitudinal inertia is known, the distance BML, and therefore GML, can be found:

BML =

IL ∇

▲ Formula 4.3 BML

X For a mathematical proof of BML for a box shaped vessel, see Appendix 24:

The Derivation of the Formula Giving the Longitudinal BM for a Box Shaped Vessel.

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178 • Ship Stability, Powering and Resistance

QUESTION Q4.1 (MCM, ENG) A box shaped vessel has a length of 70 m, a beam of 11 m, and floats at a draught of 6 m with a KG of 4 m. Calculate the longitudinal metacentric height, GML, and hence comment on the longitudinal stability of the barge.

The GML influences the trim of the vessel, which is defined as the difference between the draught measured at the after perpendicular and the forward perpendicular. If the vessel is deeper aft than forward, then she is said to be ‘trimmed by the stern’, or positive trim, and if the vessel is deeper forward than aft (which is unusual in most cases) then she is said to be ‘trimmed by the head’ or ‘trimmed by the bow’, or negative trim. While in the transverse case, we are mainly interested in calculating angles of list, in the longitudinal sense, trim is used instead of an angle of pitch, as the angles involved are very small.

MCTC – The Moment to Change Trim by 1 cm (OOW, MCM, ENG) The trim of the vessel is based on the relative positions of the LCB and the LCG. The upward, buoyant forces act through the LCB, and the downward, mass forces act through the LCG. The distance between the LCG and the LCB controls the trim of the vessel. The trim can be calculated using one of two methods; however, each of these requires a new hydrostatic value, known as the MCTC, or moment to change trim by 1 cm, to be found. The MCTC is a measure of the ‘trimming moment’, the longitudinal equivalent of a heeling or listing moment, needed to change the trim of the vessel by 1 cm. The MCTC can be interpolated at any draught or displacement from the hydrostatics, or it can be calculated from the following formula:

MCTC =

Δ × GML 100 × LBP

▲ Formula 4.4 MCTC X A mathematical proof of the MCTC can be seen in Appendix 25: The

Derivation of the Moment to Change Trim by 1 cm Formula.

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Longitudinal Stability • 179 The units of MCTC are often confusing. The ‘100’ value is a conversion from metres to centimetres, and so the correct units are: ⎡ ⎤ ⎢ tonnes metre et e ⎥ ⎡ to tonnes es metre r ⎤ MCTC = ⎢ ⎥=⎢ ⎥ centimetre ⎢ metre r ⎥ ⎣ centimetre ⎦ r ⎣ metre ⎦ However, as the ‘per cm’ is implicit in the title, the units are often quoted in units of tonne metres.

QUESTION Q4.2 (MCM, ENG) A box shaped vessel has a length of 74 m, a beam of 13 m, and floats at a draught of 5 m with a KG of 3.50 m. Calculate the longitudinal metacentric height, GML, and the MCTC.

Calculating the Trim (OOW, MCM, ENG) Once the MCTC has been found the trim (in units of cm) can be found using the following formula:

Trim =

C − LCG LCG ) Δ ( LCB MCTC

▲ Formula 4.5 Trim

This method is known as ‘taking moments about the aft perpendicular’. This formula is sometimes written as:

Trim =

LCB ) Δ MCTC

( LCG

▲ Formula 4.6 Trim (alternative)

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180 • Ship Stability, Powering and Resistance In this, the ‘~’ means ‘distance between’, and therefore LCG ~ LCB = LCB – LCG. However, using LCB – LCG results in positive answers being by the stern, and negative answers by the bow. The LCB can be found from the hydrostatic tables of the vessel, or by direct calculation for box shaped vessels, and the LCG can be found via a longitudinal loading table. X For a mathematical proof of the trim formula, see Appendix 26: The

Derivation of the Trim Equation.

QUESTION Q4.3 (MCM, ENG) A box shaped vessel has a length of 40 m, a beam of 8 m, and floats at a mean draught of 2 m with a KG of 2.50 m. The LCG is 19 m FOAP. Calculate the trim of the vessel.

For real ships, the MCTC and LCB values are pre-calculated, and can be found in the hydrostatic data for the ship.

QUESTIONS Q4.4 (OOW, MCM, ENG) Determine the trim of MV Reed in the lightship condition.

Changes in Trim (OOW, MCM, ENG) For a small (a few percentage of ship displacement) change in loading, or for a transfer of mass aboard so that the displacement remains constant, the change in trim can be APPROXIMATED using another method:

Change in trim =

Trimmin i g moment MCTC

▲ Formula 4.7 Change in trim

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Longitudinal Stability • 181 Please note that as this is an approximation, and with the exception of transferring mass when displacement is kept constant, and dry-docking, which will be covered later, it is not an acceptable method when better methods are available. The MCTC is found in exactly the same way as before. The trimming moment is found from one of three formulae, which need a value known as the LCF. The LCF is the longitudinal centre of floatation. This is the centre of area of the waterplane area, and is the pivot point of the vessel when she trims. As the LCF depends on the waterplane area, which depends on the draught and displacement, it will vary as masses are added or removed from the vessel. This method assumes the LCF does not move with loading – in reality it will move a small amount. If a load is added or removed from the vessel, then the trimming moment is given by: Trimmin i g moment o e

Mass added Mass dd d o or removed e o ed d × Dis Dii ta ce to LCF

▲ Formula 4.8 Trimming moment due to adding or removing mass

If a load is moved within the vessel, then the trimming moment is given by: Trimmin i g moment o e t

Mass Mas moved o ed d × Dis Dii ta ce moved

▲ Formula 4.9 Trimming moment due to moving mass

Trimming moments which push the stern down are positive, and trimming moments which push the bow down are negative. Therefore, a positive change in trim indicates a change in trim by the stern, and a negative change in trim indicates a change in trim by the bow.

QUESTIONS Q4.5 (OOW, MCM, ENG) A vessel has an MCTC of 4.613 tonne metres. She is initially floating on an even keel. A 5 tonne mass, already aboard, is moved aft a distance of 10 m. Determine the final trim. Q4.6 (OOW, MCM, ENG) A vessel has an MCTC of 7.324 tonne metres, which can be assumed to remain constant. She is initially floating on an even keel. A 7 tonne mass is added to the vessel 5 m aft of the LCF, and a 6 tonne mass is added 4 m forward of the LCF. Determine the APPROXIMATE final trim of the vessel.

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182 • Ship Stability, Powering and Resistance

Draughts and Drafts (OOW, MCM, ENG) The trim is only part of the problem for longitudinal stability. While it is important to know the trim, it is equally important to know the draughts at the ends of the vessel, which will vary with trim. The draught of a vessel can be measured at a number of positions, as shown in Figure 4.4. The draught at the aft perpendicular is known as the draught aft, or DA, and the draught at the forward perpendicular is known as DF, or the draught forward. The draught at amidships is known as the mean draught, or DM, as it is the average of the draughts forward and aft. The draught at the LCF is known as the true mean draught, or DTMD, or DLCF. It is important to know this value as the LCF is the rotation point in trim, as shown in Figure 4.5. Therefore, for a fixed displacement, the true mean draught will be constant, irrespective of trim. It is for this reason that the hydrostatic tables for ships use the true mean draught as their

AP

LCF Amidships

FP

DA

DLCF DM

DF

AP

LCF Amidships

FP

DA

DLCF DM

DF

▲ Figure 4.4 Draught definitions

▲ Figure 4.5 The LCF as the pivot point

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Longitudinal Stability • 183 measure of draught. If the displacement is known, then the true mean draught can be interpolated or read off. If the true mean draught is known, then the draught aft can be found using the following formula:

DA

DLCF LCF + Trim

LCF C LBP

▲ Formula 4.10 Draught aft and the true mean draught

Note that the numerical order of precedence (BODMAS) is critical when using this formula. The division must be calculated first, then the multiplication with trim, and then the addition to the true mean draught. Forgetting this will introduce serious errors into the solution. In this formula, the trim is in metres, and the LCF is measured as a distance forward of the aft perpendicular. Again, this is available from the hydrostatics. Since the trim is defined as the difference between the draught aft and draught forward, the draught forward can be found from:

F

DA − Trim

▲ Formula 4.11 End draughts and trim

For both of these formulae, care must be taken to ensure that positive trim values are used for stern trim, and negative trim values are used for bow trim. The trim must always be in units of metres.

QUESTIONS Q4.7 (MCM, ENG) A box shaped vessel has a length of 100 m and floats with a draught at the LCF of 3 m. The vessel has a trim of 110 cm by the stern. Calculate the end draughts of the vessel. Q4.8 (OOW, MCM, ENG) Determine the end draughts of MV Reed in the lightship condition, if the lightship trim is 190 cm by the stern.

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184 • Ship Stability, Powering and Resistance Q4.9 (OOW, MCM, ENG) MV Reed starts loading from her lightship condition. Both bunkers are filled to a sounding of 6.50 m with oil fuel, relative density 0.95. 212.71 tonnes of cargo is loaded aboard at a LCG of 90 m FOAP. Determine the end draughts of the vessel after loading the cargo.

LONGITUDINAL STABILITY – LEARNING CHECKLIST Objective

Level

Understand the concept of longitudinal metacentric height

OOW, MCM, ENG

Understand why the longitudinal metacentric height can normally be assumed to be constant

OOW, MCM, ENG

Calculate the longitudinal metacentric height for a box shaped vessel

MCM, ENG

Understand what is meant by the terms trim, stern trim and bow trim

OOW, MCM, ENG

Understand what is meant by the MCTC

OOW, MCM, ENG

Calculate the MCTC for a box shaped vessel

MCM, ENG

Determine the MCTC from hydrostatic data

OOW, MCM, ENG

Understand the units of MCTC

OOW, MCM, ENG

Calculate the trim of a ship

OOW, MCM, ENG

Calculate the trim of a box shaped vessel

MCM, ENG

Calculate the approximate change in trim for small changes in loading

MCM, ENG

Understand the four positions at which the draught is measured, and the associated terminology

OOW, MCM, ENG

Understand what is meant by the LCF, and the importance of the LCF in trimming the ship

OOW, MCM, ENG

Determine the mean draught from the end draughts

OOW, MCM, ENG

Determine the true mean draught from the hydrostatics

OOW, MCM, ENG

Calculate the end draughts of the vessel

OOW, MCM, ENG

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5

ADDITIONAL CALCULATIONS AND PROCESSES AIMS AND OBJECTIVES At the end of this section, you should be able to: Correct hydrostatic data for water density Calculate the hydrostatic values for a box shaped vessel in fresh and dock water Determine the hydrostatics of a vessel in fresh water if the draught is known Determine the hydrostatics of a vessel in fresh water if the displacement is known Determine the hydrostatics of a vessel in fresh water or dock water if the draught is known Determine the hydrostatics of a vessel in fresh water or dock water if the displacement is known Use the layer correction process to determine the true mean draught in sea water, dock water and freshwater Determine the LCG of a vessel from the trim Determine the mass and location of cargo so that the vessel finishes loading in a specified condition Determine the mass and distribution of cargo between two holds so that the vessel finishes loading in a specified condition Understand the accuracy of layer correction

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186 • Ship Stability, Powering and Resistance Understand the purpose of an inclining test Conduct an inclining test and determine the lightship KG and lightship displacement Understand how hydrostatic values change with trim and heel Understand how dry-docking or grounding will change the transverse stability of the ship Calculate the metacentric height at the critical instant Calculate the limiting KG before dry-docking to ensure that the vessel is stable during the dry-docking Calculate the limiting trim before dry-docking to ensure that the vessel is stable during the dry-docking Understand the accuracy of assuming that hydrostatics remain constant during drydocking Determine the true mean draught and the end draughts at the critical instant Calculate if it is safe to re-float following a dry-docking Understand the purpose of a draught survey Complete a draught survey

‘Real Ship’ Hydrostatic Data and Loading in Different Densities (MCM, ENG) We have previously seen that ship’s hydrostatics are presented in a series of tables or charts, as shown in the MV Reed Sample Stability Data Book (see Appendix 1), so that they can be read or interpolated for any true mean draught. Remember – the draught in the tables is always the true mean draught. This is because the vessel rotates about the LCF when trimming, and so the draught at the LCF is independent of the trim of the vessel. Unless indicated otherwise, the values in the hydrostatics are always undertaken assuming the vessel is in salt water. Often they will need to be used for fresh water or dock water. Corrections have to be applied to some of the values. Consider two vessels floating at the same draught, one in sea water, and one in fresh water. Both vessels, if loaded to float at the same draught, would have identical underwater volumes, regardless of fluid density. The displacements would, however, be different. Other hydrostatic values, which are related to mass, will change. This is easiest shown by calculation.

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Additional Calculations and Processes • 187

QUESTIONS Q5.1 (MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10 m and a draught of 2 m in sea water. KG is 3.00 m. Determine the displacement, KB, BM, KM, LCB, LCF, TPC and MCTC. Q5.2 (MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10 m and a draught of 2 m in fresh water. KG is 3.00 m. Determine the displacement, KB, BM, KM, LCB, LCF, TPC and MCTC.

It is this concept which underpins hydrostatic tables – they give information for a fixed draught, which allows us to modify certain values to get the actual hydrostatics. The LCB, LCF, KB and KM are based on the underwater volume, so at a fixed draught, regardless of density, they are the same. As previously seen, the displacement, TPC and MCTC are linked to the fluid density, so at any draught they will vary according to fluid density, and will need correcting. To remember which values need correcting, look at the units. Any hydrostatic value with tonnes in the units will depend on the displacement, and hence fluid density, and must be corrected. Any hydrostatic value without tonnes in the units will not depend on the density, and therefore does not need correcting. Table 5.1 Density effects on hydrostatics Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

Sea water

2,050

50.00

50.00

1.00

5.17

10.25

85.01

Fresh water

2,000

50.00

50.00

1.00

5.17

10.00

82.94

Table 5.1 shows the displacement, LCB, LCF, KB, KM, TPC and MCTC for a box shaped vessel with a length of 100 m, a beam of 10 m, a KG of 3 m, at a draught of 2 m in sea water and a draught of 2 m in fresh water. To correct the displacement, TPC and MCTC, the salt water values need to be multiplied by the actual water density divided by the density of sea water: ρAACTUAL Actual value = Salt S lt w water t value l × ρ SW S ▲ Formula 5.1 Dock water hydrostatic values from the sea water hydrostatics

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188 • Ship Stability, Powering and Resistance Quite often, ship hydrostatic data will give both the fresh water and sea water values for the appropriate hydrostatics, if the fresh water values are given, then the corrections are as follows: Actual value a ue = F Fresh esh water t v value l × ρAACTUAL ▲ Formula 5.2 Dock water hydrostatic values from the freshwater hydrostatics

Note that for both these formulae, the density should be in units of tonnes per metres cubed (t/m3). Table 5.2 shows the hydrostatic data for MV Reed at a draught of 6.00 m, corrected for fresh water and dock water at a relative density of 1.010. Table 5.2 Corrected hydrostatics Draught Displacement (m)

(tonnes)

LCB

LCF

(m FOAP) (m FOAP)

KB

KM

TPC

MCTC

(m)

(m)

(t/cm)

(tonne metres)

Sea water

6.00

6,008

46.15

43.043

3.277

7.006

12.634

68.639

Fresh water

6.00

5,861.5

46.15

43.043

3.277

7.006

12.326

66.965

Dock water

6.00

5,920.1

46.15

43.043

3.277

7.006

12.450

67.635

When dealing with fresh water or dock water, the easiest way to approach the problem is to take the appropriate rows of the hydrostatic tables, correct them for density, and then use the corrected tables as before for all calculations.

QUESTIONS Q5.3 (MCM, ENG) Using the hydrostatics for MV Reed, determine the hydrostatic values for the vessel at a draught of 4.00 m in dock water, relative density 1.010. Q5.4 (MCM, ENG) Using the hydrostatics for MV Reed, determine the hydrostatic values for the vessel at a draught of 4.15 m in dock water, relative density 1.007. Q5.5 (MCM, ENG) Using the hydrostatics for MV Reed, determine the hydrostatic values for the vessel at a displacement of 3,600 tonnes in dock water, relative density 1.010.

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Additional Calculations and Processes • 189 Q5.6 (MCM, ENG) MV Reed floats at a draught of 4.00 m in salt water. How much additional cargo could be added so that her draught would be 5.00 m in fresh water? Q5.7 (MCM, ENG) MV Reed floats at a draught of 5.00 m in dock water, density 1.005 t/m3. How much additional cargo could be added so that she would float at her summer draught in sea water? Q5.8 (MCM, ENG) MV Reed is floating in a summer zone, with the waterline 20 cm below the top of the summer mark. The density of the water is 1.015 t/m3. Determine the amount of cargo to load to bring the vessel to her summer displacement using both the mean TPC and the density corrected hydrostatics. Which one is more accurate? Q5.9 (MCM, ENG) MV Reed is alongside in fresh water in her lightship condition. 3,000 tonnes of cargo, stores and fuel is then loaded onto the ship. The vessel then sails. At the harbour mouth, the water depth is 10.00 m, and the water density is 1.020 t/m3. Assuming that the vessel is on an even keel, determine the under-keel clearance at the harbour mouth. Q5.10 (MCM, ENG) MV Reed arrives at a harbour mouth in sea water with a true mean draught of 6.00 m. She then sails up-river to a fresh water berth, using 100 tonnes of fuel. Determine the true mean draught at the fresh water berth.

Layer Correction (MCM, ENG) In all of the problems considered so far, we have known either the draft of the vessel or her displacement. This is the starting point for all of our calculations. Quite often, in practical loading calculations, the true mean draught and the displacement of the vessel are not known. However, the draughts at the fore and aft perpendiculars can be measured, and then the hydrostatic tables can be used to determine the true mean draught.

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190 • Ship Stability, Powering and Resistance The trim of the vessel can be found by taking the difference in the draughts forward and aft. Trim = DA DF ▲ Formula 5.3 Trim

We have previously seen in Formula 4.10 that the draught aft and the true mean draught are linked by the following formula: DA

DLCF + Trim

LCF LBP

There is, however, a problem in applying this formula to find the true mean draught (DLCF) using the end draughts. We know the draught aft (DA), and the trim, and the LBP, but both the true mean draught and the LCF position are unknown. To get the LCF position, we need to use the hydrostatic tables. Herein lies the problem – we cannot use the hydrostatic tables without knowing the true mean draught, and we don’t know the true mean draught. The method of solving this, and determining the true mean draught from the end draughts, is known as layer correction. We know that the true mean draught and the mean draught are very similar to each other. We can therefore find the mean draught, and use that in the hydrostatic tables to find an approximate position for the LCF:

DM =

DA DF 2

▲ Formula 5.4 Mean draught

Once we have an approximate value for the LCF position, we can use the following formula to get an approximate value for the true mean draught: DA

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DLCF + Trim

LCF LBP

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Additional Calculations and Processes • 191 This gives us a very good idea of the true mean draught, but this is still an approximation, and is not potentially accurate enough for stability calculations. To improve the accuracy, we can use our approximate true mean draught and read off the LCF from the tables. This will be more accurate than the LCF obtained from the mean draught. We can then use this formula again to get a better estimate of the true mean draught: DA

DLCF + Trim

LCF LBP

This should result in an accurate enough value for the true mean draught. Once this is known, then the displacement and other hydrostatic values can be found. In theory, we could continue this process of determining the LCF and the true mean draught, and then finding a more accurate value for the LCF indefinitely.

QUESTIONS Q5.11 (MCM, ENG) MV Reed has a draught aft of 7.00 m and a draught forward of 3.00 m. Determine her true mean draught to the nearest millimetre. Q5.12 (MCM, ENG) MV Reed has a draught aft of 7.00 m and a draught forward of 4.00 m. Determine her true mean draught (to two decimal places) and displacement. Q5.13 (MCM, ENG) MV Reed has a draught aft of 6.00 m, and a draught forward of 5.31 m. Determine her true mean draught, displacement and GM if KG is 6.00 m. Q5.14 (MCM, ENG) MV Reed has a draught aft of 5.00 m and a draught forward of 5.23 m. Determine the displacement of the vessel.

In scenarios where a layer correction is required, but the vessel is not in sea water, then the process is very similar. The LCF values can be taken directly from the hydrostatic tables without correction, as LCF is independent of water density Therefore, the only stage in the process which changes when not in sea water is the final stage where the hydrostatics are interpolated. At this stage, corrections as shown in Table 5.2 are required.

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192 • Ship Stability, Powering and Resistance

QUESTION Q5.15 (MCM, ENG) MV Reed has a draught aft of 6.00 m, and a draught forward of 5.31 m. Determine her true mean draught and displacement if she is floating in dock water, density 1.020 t/m3.

Once a layer correction has been undertaken, normal loading calculations can be used to determine the stability and trim of the vessel after loading additional cargo. To determine the trim after loading, the LCG must be known. This can be found from transposing the trim equation and using the hydrostatics. Trim =

−

LCG ) Δ MCTC

(LCB

⎛ Trim × MCTC ⎞ − LCB = LCG ⎝ ⎠ Δ

Note that if the trim is zero to start with, then the LCB must be equal to the LCG.

QUESTIONS Q5.16 (MCM, ENG) MV Reed has a draught aft of 4.00 m and a draught forward of 3.15 m. Determine the displacement of the vessel and the position of the LCG. Q5.17 (MCM, ENG) MV Reed has a draught aft of 4.50 m and is floating on an even keel. Determine the end draughts of the vessel if 1,000 tonnes of cargo is loaded at a position 35 m FOAP. Q5.18 (MCM, ENG) MV Reed has a draught aft of 4.00 m and is on an even keel. She is to be loaded to a draught aft of 6.23 m, with a trim of 1.00 m by the stern. Determine the amount of cargo to be loaded and the required LCG of the cargo. Q5.19 (MCM, ENG) MV Reed has a draught aft of 4.00 m and a draught forward of 2.71 m. Space is available in an aft hold, which is 30 m FOAP, and in a forward hold 80 m FOAP. Determine the

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Additional Calculations and Processes • 193 distribution of cargo so that the final true mean draught is 4.46 m and the trim is 0.34 m by the stern. Q5.20 (MCM, ENG) MV Reed is alongside a berth in dock water, with a density of 1.005 t/m3. The draught aft is 5.00 m with a draught forward of 3.90 m. The vessel is to be loaded so that when she reaches sea water, she will float at the summer draught on an even keel. Space is available in an aft hold, 25 m FOAP, and a forward hold, 75 m FOAP. Determine the required distribution of cargo between the two holds. Q5.21 (MCM, ENG) MV Reed is floating alongside in fresh water. Her draught aft is 4.00 m and her draught forward is 3.57 m. After loading, she will sail in sea water with a depth of 8.00 m. The required UKC is 2.00 m, and the vessel requires a stern trim of 0.47 m. Space is available in holds 30 m FOAP and 70 m FOAP. Determine the required cargo distribution. Q5.22 (MCM, ENG) MV Reed floats at a true mean draught of 3.00 m with a trim of 0.00 m by the stern. She is loaded with fuel and stores, and the crew is aboard. The only remaining mass to load is the cargo. The density of the water is 1.008 t/m3. The vessel is to be loaded so that when she leaves the harbour and enters sea water, she will be at draught of 7.00 m, on an even keel. Space is available in two holds. The after hold is 25 m FOAP. The forward hold is 75 m FOAP. Determine the distribution of the cargo so that the vessel sails in the required final condition. Q5.23 (MCM, ENG) MV Reed floats at a true mean draught of 3.60 m with a trim of 0.85 m by the stern. She is loaded with fuel and stores, and the crew is aboard. The only remaining mass to load is the cargo. The density of the water is 1.015 t/m3. The vessel is to be loaded so that when she leaves the harbour and enters sea water, she will be at draught of 6.00 m, on an even keel. Space is available in two holds. The after hold is 30 m FOAP. The forward hold is 50 m FOAP. Determine the distribution of the cargo so that the vessel sails in the required final condition.

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194 • Ship Stability, Powering and Resistance

The Effect of Hogging and Sagging on Layer Correction (MCM, ENG) The layer correction process does not take into account any distortion in the hull. For most ships, the distribution of cargo, fuel and ballast will be such that the hull will either sag (deform so that the middle of the vessel bends downwards), as shown in Figure 5.1, or hog (deform so that the middle of the vessel bends upwards), as shown in Figure 5.2. In both of these, the deformation has been exaggerated for clarity.

▲ Figure 5.1 Sagging distortion

▲ Figure 5.2 Hogging distortion

Clearing hogging and sagging will have an effect on the relationship between the true mean draught and the end draughts. Tables 5.3 and 5.4 show the error in the actual displacement from a layer correction process as a result of hogging and sagging. As can be seen, the percentage error is small. Table 5.3 Error in calculated displacement due to hogging (positive) and sagging (negative) at the summer displacement Hog/sag (cm)

Displacement (tonnes)

Error (tonnes)

Error (%)

–10

7,291

38

0.52

–9

7,294

35

0.48

–8

7,298

31

0.42

–7

7,302

27

0.37

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Additional Calculations and Processes • 195 Table 5.3 Continued Hog/sag (cm)

Displacement (tonnes)

Error (tonnes)

Error (%)

–6

7,306

23

0.31

–5

7,310

19

0.26

–4

7,314

15

0.21

–3

7,314

15

0.21

–2

7,322

7

0.10

–1

7,326

3

0.04

0

7,329

0

0.00

1

7,334

5

0.07

2

7,337

8

0.11

3

7,341

12

0.16

4

7,345

16

0.22

5

7,349

20

0.27

6

7,353

24

0.33

7

7,357

28

0.38

8

7,361

32

0.43

9

7,365

36

0.49

10

7,369

40

0.54

Table 5.4 Error in calculated displacement due to hogging (positive) and sagging (negative) at the lightship displacement Hog/sag (cm)

Displacement (tonnes)

Error (tonnes)

Error (%)

–10

2,598

17

0.65

–9

2,600

15

0.58

–8

2,602

13

0.50

–7

2,604

11

0.42

–6

2,606

9

0.35

–5

2,607

8

0.31

–4

2,609

6

0.23

–3

2,611

4

0.15

–2

2,613

2

0.08

–1

2,615

0

0.00

0

2,615

0

0.00

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196 • Ship Stability, Powering and Resistance Table 5.4 Continued Hog/sag (cm)

Displacement (tonnes)

Error (tonnes)

Error (%)

1

2,618

3

0.11

2

2,620

5

0.19

3

2,622

7

0.27

4

2,624

9

0.34

5

2,626

11

0.42

6

2,627

12

0.46

7

2,629

14

0.53

8

2,631

16

0.61

9

2,633

18

0.68

10

2,635

20

0.76

QUESTION Q5.24 (MCM, ENG) MV Reed is floating along a berth in fresh water. Her bunkers are pressed full with fuel oil, relative density 0.96, and there are some stores and some cargo aboard. In this condition, the draught aft is 4.00 m and the draught forward is 3.00 m. The vessel departs on a voyage, with a predicted voyage fuel consumption of 59.7 tonnes from the port bunker and 59.7 tonnes from the starboard bunker. Water ballast will not be used to compensate for the effect of the fuel consumption on the stability of the vessel. The destination is depth restricted, with a predicted minimum depth in the harbour at low water of 7.00 m, with water of relative density 1.010. The passage plan is arranged so that the vessel arrives at low water. The Company’s standing orders require a minimum under-keel clearance of 1.00 m, and a trim of a m by the stern at the arrival port to ensure that the manoeuvring characteristics are acceptable. Space is available in the No. 2 hold, 75 m FOAP, and the No. 4 hold, 35 m FOAP. Determine the cargo distribution between the two holds so that the vessel carries the maximum possible cargo, and arrives at the destination port with suitable draughts and trim. You may assume that there are no depth restrictions in the starting port.

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Additional Calculations and Processes • 197

Inclining Tests (OOW, MCM, ENG) All the stability calculations undertaken so far need an accurate lightship displacement, lightship KG and LCG to ensure an accurate calculation. These values can be theoretically calculated from the design of the vessel, but the actual values depend on the accuracy of the build, the accuracy of the design calculations, and any modifications made to the vessel when she is in service. To determine the lightship centre of gravity, the exact mass of every single item of structure on the vessel and the position of the centre of mass of the item need to be known. These are then tabulated into very large loading tables, which give the lightship KG, TCG and LCG. The potential for errors is large, considering the required accuracy for GM. A more accurate alternative is to conduct an experiment to determine the position of the centre of gravity and the lightship displacement. This is known as the inclining test or inclining experiment. In an inclining test, the end draughts are read, and a layer correction process is used to determine the true mean draught and therefore the displacement. A known mass is then moved a known distance to create an angle of list. The angle of list is measured, usually with a series of long pendulums. By measuring the deflection of a pendulum, the angle of list of the vessel can be determined with a high level of accuracy. Tanθ =

Pendulum deflection Pendulum length

The list, mass, distance moved and the displacement of the vessel can be used to find GM. We have previously seen in Formula 2.11 that list can be calculated using: tanθ =

Σ( × d ) Δ × GM

This can be transposed for GM: GM =

Σ(w d ) Δ × tanθ

By substituting the pendulum deflection and length for the list angle, this equation becomes:

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198 • Ship Stability, Powering and Resistance

GM =

Σ(w d ) ⎛ Pendulum deflection ⎞ Δ×⎜ ⎝ Pendulum length ⎟⎠

▲ Formula 5.5 GM from inclining tests

Once GM has been found, direct calculations can be used to find KB and BM, or the hydrostatics can be used to find KM, and therefore KG can be found using: GM = KM − KG = KB + BM − KG This can be transposed to give: KG = − (GM − KB − BM ) = −(GM − KM ) This can be transposed to give: KG = −GM + KB + BM = −GM + KM The KG found from this process is not the lightship KG. It will include the masses used to incline the ship, and any persons and equipment aboard as part of the test. A loading table must be used to ‘unload’ these items to get the actual lightship KG, and the lightship displacement. Any items aboard which are not considered to be part of lightship must be included in the table, and FSE accounted for. Any items not aboard, such as hatch covers, which are part of lightship must be added into the loading table. From the end draughts, the LCG can be determined. Again, this is not the lightship value. A loading table must be used to ‘unload’ any items aboard that are not part of lightship to get the actual lightship LCG of the vessel. If the vessel has an initial list, then this can be used to determine the TCG of the vessel in the experiment condition. Again, this is not the lightship value. A loading table must be used to ‘unload’ any items aboard that are not part of lightship to get the actual lightship TCG of the vessel. To reduce experimental error, a number of weights (normally 6) are used in the experiment. These are sequentially moved from port to starboard, as shown in Figure 5.3, with the list and heeling moment recorded after each move. The overall mean GM is used (excluding upright conditions). The Load Line regulations (The Merchant Shipping (Load Line) Regulations 1998) require ships to undergo an inclining test when construction has finished. Details of the process of inclining tests are given in detail in the MCA’s guidelines for surveyors (Maritime and Coastguard Agency, 2007).

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9781408176122_Ch05_Rev_txt_prf.indd 199

3

2

2

1

2

1

3

3

3

4

5

4

5

5

4

4

5

4

4

4

5

5

5

6

6

6

6

6

6

6

PORT STB

PORT STB

PORT STB

PORT STB

PORT STB

PORT STB

PORT STB

Step 7 – Check step

Step 6

Step 5

Step 4

Step 3

Step 2

Step 1 – Check step

▲ Figure 5.3 Mass movement in an inclining test

2

1

1

3

2

1

3

2

1

3

2

1

1

1

1

1

1

1

2

2

2

2

2

2

3

3

3

3

3

3

4

4

4

4

4

4

5

5

5

5

5

5

6

6

6

6

6

6

PORT STB

PORT STB

PORT STB

PORT STB

PORT STB

PORT STB

Step 13 – Check step

Step 12

Step 11

Step 10

Step 9

Step 8

Additional Calculations and Processes • 199

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200 • Ship Stability, Powering and Resistance To ensure a high level of accuracy the vessel must be floating in absolutely calm water in windless conditions. All mooring lines must be sufficiently slack to allow the vessel to take up heel angles unimpeded. All tanks and bilge spaces must be either pressed full or dry. The condition of all tanks should be recorded (25% lightship recommended limit). All loose equipment must be secured in its normal stowage and any items that are not part of the vessel’s normal outfit should be removed. Ideally there should be no consumable stores aboard. If this is not possible then the position and weight of all such stores must be recorded. The number of people aboard must be minimised. All those aboard must be in a defined position while measurements are taken and these positions and the weights of each person noted. The weights and positions, in both the vertical and horizontal planes of all inclining weights must be recorded. If an inclinometer is used, then the inclinometer used must be capable of measuring to 0.01 degrees. If a plumb bob is used, the pendulum length must be at least 1,800 mm. At least 2 pendulums must be used. The pendulum length should be sufficient to allow a deflection of at least 35 mm on each mass shift. The angle of list should not exceed 4 degrees, and usually be less than 2 degrees. The density of the water, and the drafts forward and aft, and any initial list angle, must be recorded. This will enable accurate calculation of the displacement, the LCB and the LCG. If an identical sister ship (other than passenger vessels) has been built, then a weight survey can be undertaken after construction has finished to determine the actual lightship displacement of the vessel. If there is a deviation from the sister ship of more than 1% of the lightship displacement for ships over 160 m in length, or more than 2% of the lightship displacement for ships under 50 m in length, then a full inclining test is required. Linear interpolation can be used to determine the maximum allowable deviation from the sister ship displacement for vessels between 50 m and 160 m in length. In addition, if there is a deviation of more than 0.5% of the subdivision length in the lightship LCG, then a full inclining test is required. For all passenger ships, where GM may be low in order to keep motions acceptable, SOLAS requires that lightship surveys (a check to determine the actual lightship displacement from the draughts) must be undertaken at intervals not exceeding five years. This check indicates if the lightship displacement has changed, and the lightship trim can be used to see of the lightship centre of gravity has moved. If there is a deviation from the stability book of more than 2% of the lightship displacement, or more than 1% of the subdivision length in the lightship LCG, then a full inclining test is required. After the inclining test, and following approval from the classification society and the Flag State, the new lightship values must be entered into the stability computer.

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Additional Calculations and Processes • 201 QUESTIONS Q5.25 (MCM, ENG) MV Reed is undergoing an inclining test to determine her lightship displacement and KG. The vessel is to be inclined with six inclining masses, each with a mass of 10.00 tonnes. These masses are placed 8.000 m above the keel. The masses, when moved, are moved a distance of 2.500 m from the centreline of the vessel. The list of the vessel is measured with a 6.000 m long pendulum. With the masses aboard, and no other masses, cargo or stores present, the true mean draught of the vessel is 3.085 m. The recorded results from the inclining test are as follows: Step

Mass 1

Mass 2

Mass 3

Mass 4

Mass 5

Mass 6

Pendulum position (m)

1

Port

Port

Port

Starboard

Starboard

Starboard

0.000

2

Port

Port

Port

Port

Starboard

Starboard

0.150

3

Port

Port

Port

Port

Port

Starboard

0.299

4

Port

Port

Port

Port

Port

Port

0.454

5

Starboard

Port

Port

Port

Port

Port

0.307

6

Starboard

Starboard

Port

Port

Port

Port

0.146

7

Starboard

Starboard

Starboard

Port

Port

Port

0.000

8

Starboard

Starboard

Starboard

Starboard

Port

Port

–0.148

9

Starboard

Starboard

Starboard

Starboard

Starboard

Port

–0.303

10

Starboard

Starboard

Starboard

Starboard

Starboard

Starboard

–0.448

11

Port

Starboard

Starboard

Starboard

Starboard

Starboard

–0.307

12

Port

Port

Starboard

Starboard

Starboard

Starboard

–0.148

13

Port

Port

Port

Starboard

Starboard

Starboard

0.000

Determine the lightship displacement and KG from the above data and the hydrostatics. You may assume that the mass of the staff undertaking the test is negligible, and that the masses are moved by a shipyard crane. Q5.26 (MCM, ENG) A box shaped vessel has a length of 50 m, a beam of 6 m and a draft of 2.033 m in her lightship condition, but with a 10 tonne inclining mass aboard, 4.00 m above the keel. In this condition she floats upright. The mass is moved 2 m to port, resulting in a 219 mm deflection of a 2.50 m pendulum. Determine the lightship KG.

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202 • Ship Stability, Powering and Resistance It is usual practice to display the results of the experiment graphically, with heeling moment on the x axis plotted against the list, or tangent of the list on the y axis, as shown in Figure 5.4, errors will show a deviation from a straight line. During the inclining test, the GM and displacement should remain constant. Therefore the ratio of the listing moment causing the incline to the tangent of the angle inclined to should remain constant. This constant ratio is represented by the gradient of the straight line through the points.

Tan (list)

Heeling moment

▲ Figure 5.4 Inclining test graph

QUESTION Q5.27 (MCM, ENG) MV Reed undergoes an inclining test. She is in the lightship condition, plus the inclining masses. The mass of the staff undertaking the test can be assumed to be negligible. She has a draught of 3.029 m. Six inclining masses are used, each with a mass of 5.000 tonnes. They are located 11.000 m above the keel. To incline the vessel, the masses are sequentially moved 3.000 m from the centreline of the vessel. The inclination is measured using a 7.500 m long pendulum. The measured deflection at each step is shown: Step Mass 1

Mass 2

Mass 3

Mass 4

Mass 5

Mass 6

Pendulum position (m)

1

Port

Port

Port

Starboard

Starboard

Starboard

0.000

2

Port

Port

Port

Port

Starboard

Starboard

0.120

3

Port

Port

Port

Port

Port

Starboard

0.234

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Additional Calculations and Processes • 203

Step Mass 1

Mass 2

Mass 3

Mass 4

Mass 5

Mass 6

Pendulum position (m)

4

Port

Port

Port

Port

Port

Port

0.335

5

Starboard

Port

Port

Port

Port

Port

0.240

6

Starboard

Starboard

Port

Port

Port

Port

0.115

7

Starboard

Starboard

Starboard

Port

Port

Port

0.000

8

Starboard

Starboard

Starboard

Starboard

Port

Port

–0.112

9

Starboard

Starboard

Starboard

Starboard

Starboard

Port

–0.215

10

Starboard

Starboard

Starboard

Starboard

Starboard

Starboard

–0.400

11

Port

Starboard

Starboard

Starboard

Starboard

Starboard

–0.230

12

Port

Port

Starboard

Starboard

Starboard

Starboard

–0.123

13

Port

Port

Port

Starboard

Starboard

Starboard

0.000

Determine the lightship displacement and lightship KG of the vessel and draw a graph showing the tangent of list against the heeling moment.

Provided that the masses are identical, and the distance moved is constant, a slightly alternative method of determining GM from an inclining test is to consider the deflection of the pendulum due to a shift of a mass. For each shift in the mass, the deflection of the pendulum is recorded. At the end of the test the total deflection of the pendulum is calculated, and divided by the number of shifts to determine the mean pendulum deflection. The mass and shift can then be used with the mean deflection to determine the GM by using Formula 5.5.

QUESTIONS Q5.28 (MCM, ENG) MV Reed is undergoing an inclining test to determine her lightship displacement and KG. The inclining masses are 10.50 tonnes each. When the masses are each shifted through a distance of 5.100 m, the average deflection of a 6.000 m long pendulum is 0.150 m. During the inclining test, the displacement is 2,710.00 tonnes. During the test, the vessel is in her lightship condition, but has a total of six inclining masses aboard. The masses are 8.100 m above the keel. Also aboard during the inclining test are the staff and some crew, with a total mass of 0.50 tonnes, and a location 7.900 m above the keel. Determine the lightship displacement and KG.

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204 • Ship Stability, Powering and Resistance Q5.29 (MCM, ENG) MV Reed is undergoing an inclining test to determine her lightship displacement and KG. The inclining masses are 9.50 tonnes each. When the masses are each shifted through a distance of 5.000 m, the average deflection of an 8.000 m long pendulum is 0.195 m. During the inclining test, the displacement is 2,700.00 tonnes. During the test, the vessel is in her lightship condition, but has a total of six inclining masses aboard. The masses are 8.500 m above the keel. Also aboard during the inclining test are the staff and some crew, with a total mass of 0.75 tonnes, at a location 8.000 m above the keel. Determine the lightship displacement and KG.

Changes in Hydrostatics with Trim and Heel (MCM, ENG)

KM (m)

As the trim of a vessel changes, the hydrostatic values also change. For vessels which are not wall sided, such as those with large overhangs, or big changes in waterplane area, these changes in the hydrostatics may be significant. The result of these is that as the vessel trims or heels, the waterplane area changes. This change in the waterplane area changes both BM and BML. These changes in turn change the TPC, KM and KML, and the MCTC. 8 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7 6.9 6.8 6.7 6.6 6.5

2 m Stern 1 m Stern Even keel 1 m Bow 2 m Bow

3

3.5

4

4.5 5 5.5 True mean draught (m)

6

6.5

7

▲ Figure 5.5 Variation in KM versus true mean draught with trim

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Additional Calculations and Processes • 205 100 2 m Stern 1 m Stern Even Keel 1 m Bow 2 m Bow

MCTC (tonne metres)

90 80 70 60 50 40

3

3.5

4

4.5

5

5.5

6

6.5

7

True mean draught (m)

▲ Figure 5.6 Variation in MCTC versus true mean draught with trim

14.5 2 m Stern 1 m Stern Even Keel 1 m Bow 2 m Bow

TPC (tonne metres)

14 13.5 13 12.5 12 11.5 11 10.5 10

3

3.5

4

4.5 5 5.5 True mean draught (m)

6

6.5

7

▲ Figure 5.7 Variation in TPC versus true mean draught with trim

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206 • Ship Stability, Powering and Resistance 60

12

55

11.5 MCTC

TPC

11.25 50

11 10.75

TPC (t/cm)

LCB, LCF (m FOAP) and MCTC (tonne metres)

11.75

LCB

45 LCF

10.5 10.25

40 0

5

10 15 Heel angle (degrees)

10 20

▲ Figure 5.8 Variation in hydrostatics with heel

Dry-Docking and Grounding (MCM, ENG) The theory of dry-docking can be broken down into two main subject areas: process and transverse stability. The process of dry-docking concerns the operational aspects of bringing a vessel into a dry-dock and effectively grounding her in a controlled manner. The transverse stability aspect concerns the change in stability of the vessel during dry-docking. This can be significant and directly endanger the safety of the ship. The exact process varies between different dry-docks, companies and ships, but the basic structure is similar. Data about the vessel is first provided to the dry-dock technical staff, who plan the positioning of the blocks to take into account the hull fittings and any areas requiring additional support. The dry-dock technical staff specify a trim (normally for undamaged vessels this is a small stern trim) and draft to enter the dock – you should check their stability calculations. The vessel enters the dry-dock, and is generally warped in and held with lines over the position of the blocks, as shown in Figure 5.9.

▲ Figure 5.9 Before de-ballasting the dry-dock

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Additional Calculations and Processes • 207 The gates are secured (or caissons placed) and the water drained until the stern touches the aft blocks, as shown in Figure 5.10.

▲ Figure 5.10 Stern contact

At this point, the vessel is checked to ensure she is properly aligned, and divers may check the blocks to ensure they are correct and in position. As the vessel is grounded at the stern only, due to the stern trim, the heading of the vessel can be changed by gently pivoting the vessel about the stern so that the vessel is exactly aligned with the blocks. The remaining water is then pumped out, and shore connections and gangways can be rigged. All these events should be logged, along with the soundings around the vessel in the dock, to assist in stability checks. Dry-docking, or grounding a vessel, can have a significant influence on GM, and hence the transverse stability of the vessel. As the vessel makes contact with the blocks, an upwards force is generated, known as the ground reaction force or up-thrust, which is given the symbol P. This force acts upwards at the stern, as shown in Figure 5.11.

PP

▲ Figure 5.11 Up-thrust or ground reaction force

As more water is pumped from the dry-dock, the up-thrust increases, as the vessel pushes down onto the blocks, as shown in Figure 5.12.

P

▲ Figure 5.12 Increasing ground reaction force

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208 • Ship Stability, Powering and Resistance Clearly as the water is pumped from the dock, the trim of the vessel changes, so that when the bow touches the blocks, the trim is effectively zero. After the bow has touched, the up-thrust will be spread over the entire length of the ship at each contact point, as shown in Figure 5.13, and the ship can be supported by side blocks and shores.

▲ Figure 5.13 After the bow grounds

The up-thrust force is significant, and can be calculated as a function of the change of trim, using Formula 4.7. Note that although this method is an approximation, in this circumstance any errors generated are likely to have a positive impact on the final answer. Change in trim =

Trimming moment MCTC

The change in trim of the vessel during dry-docking can be considered to be caused by the up-thrust. A bow-down trimming moment is caused by the up-thrust, which is assumed to act at the aft perpendicular. Using the trimming moment formula given in Formula 4.8: Trimming moment = P LCFFFOAP Combining these equations gives: Change in trim =

P LCFFFOAP MCTC

This can be transposed to give the up-thrust in terms of the change in trim, which can be measured during dry-docking using the draught marks: Change in trim × MCTC =P LCF CFFOAP ▲ Formula 5.6 Up-thrust or ground reaction force

Therefore, if the hydrostatics and the change in trim during dry-docking (which we can take to be the initial trim, as we can assume the blocks are level) are known, we can find

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Additional Calculations and Processes • 209 the up-thrust. The up-thrust varies as the trim changes. The maximum up-thrust occurs at the maximum trim change. As soon as the stern touches down, the trim will begin to change. This time while the trim is changing as water is being pumped out is known as the critical period. The worst part of the critical period is at the maximum change in trim, when the bow touches down on the blocks. This is known as the critical moment, as this is the point when the vessel is most likely to capsize.

QUESTION Q5.30 (MCM, ENG) MV Reed has a draught aft of 4.00 m and a draught forward of 2.94 m. Determine the up-thrust at the critical moment during dry-docking. This up-thrust has an influence on the stability of the vessel. The up-thrust can be thought of as a negative load acting at the keel. This can be dealt with in a number of ways, but the simplest is to use a loading table ( Table 5.5). Table 5.5 Loading table for KG during dry-docking Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Ship

Δ

KGSTART

Δ × KGSTART

Up-thrust

−P

0

0

Δ−P

Total

KG at the ccritical tica moment =

Δ × KGSTART

Moment Δ × KGSTART = Mass Δ−P

Clearly the up-thrust changes KG of the vessel and hence changes GM. The effect can be quite substantial. During dry-docking, the GM must always be kept positive.

QUESTION Q5.31 (MCM, ENG) MV Reed has a draught aft of 4.00 m and a draught forward of 3.15 m. She has a KG of 7.19 m. Determine the GM before dry-docking and at the critical moment during drydocking, assuming that KM remains constant. If the vessel is going to be unstable during the dry-docking process, then action must be taken to ensure that the stability is changed to a safe value. One method of

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210 • Ship Stability, Powering and Resistance achieving this is to reduce the initial KG of the vessel. This has the effect of increasing the metacentric height to start with, so any subsequent reduction in stability still results in a larger metacentric height at the critical instant. By determining the required KG to ensure that GM is positive at the critical instant, the process can be worked backwards to determine the required initial KG of the ship.

QUESTION Q5.32 (MCM, ENG) MV Reed has a displacement of 4,112 tonnes with a trim of 1.00 m by the stern. Determine the maximum initial KG so that the GM remains positive during dry-docking.

Alternatively, the initial trim can be reduced. By reducing the initial trim, the resulting change in trim will be less, and therefore the up-thrust will be reduced. This will have less effect on the KG of the vessel, and therefore result in a smaller loss of GM during dry-docking. To determine the limiting initial trim, the calculations must be undertaken in reverse, starting from the required KG, and working backwards to find the allowable up-thrust, and then the allowable initial trim.

QUESTIONS Q5.33 (MCM, ENG) MV Reed has a displacement of 4,112 tonnes, with an initial KG of 6.85 m. Determine the maximum initial trim so that the GM remains positive during dry-docking. Q5.34 (MCM, ENG) MV Reed is to be dry-docked. The density of the water in the dock is 1.007 t/m3. She has a true mean draught in the dock water of 3.60 m. In this condition, she has a KG of 7.18 m and a trim of 1.00 m by the stern. KM may be assumed to remain constant throughout the dry-docking process. Determine, and comment on, the GM at the critical instant during the dry-docking process, and then calculate the maximum initial trim of the vessel so that GM remains above 5 cm throughout the dry-docking process.

Throughout this process the LCF, MCTC and KM have assumed to be constant. In reality, as the up-thrust increases, the effective displacement and draught of the vessel decreases, and therefore the hydrostatics vary. In practice, this variation is small. Figure 5.14 shows the actual calculated up-thrust during dry-docking for MV Reed,

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Additional Calculations and Processes • 211 including the effects of variations in hydrostatics, against those predicted using the method above, assuming that the hydrostatics are constant. This assumes that the initial KG is 7.18 m, with a draught aft of 4 m and 1 m of stern trim. Figure 5.15 shows the same information, but for GM. Note that for the actual condition, the initial GM is larger as a result of a larger KM caused by stern trim. 120.00 P (Assuming constant hydrostatics) P Actual

Up-thrust (tonnes)

100.00

80.00

60.00

40.00

20.00

0.00

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Change in trim (m)

▲ Figure 5.14 Result of assuming constant hydrostatics on P during dry-docking

0.50 GM (Assuming constant hydrostatics) GM Actual

0.40 0.30

GM (m)

0.20 0.10 0.00 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

–0.10 –0.20 –0.30 –0.40 –0.50 Change in trim (m)

▲ Figure 5.15 Result of assuming constant hydrostatics on GM during dry-docking

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212 • Ship Stability, Powering and Resistance To improve the accuracy of the calculations, the KM can be interpolated at the ‘effective displacement’ of the vessel. This is the displacement of the vessel of the vessel found from the loading table, and is effectively the displacement as the up-thrust is considered to be a negative load.

QUESTION Q5.35 (MCM, ENG) MV Reed is to be dry-docked. The density of the water in the dock is sea water. She has a displacement of 3,136 tonnes. In this condition, she has a KG of 7.25 m and a trim of 1.10 m by the stern. Determine, and comment on, the difference in the GM at the critical instant during the dry-docking process, if KM is assumed to be constant, and if KM is assumed to vary.

As previously seen, during the dry-docking process, the up-thrust can be considered to be a negative load. At the critical moment, when the bow makes contact with the blocks, the up-thrust can be found using the methods already described. If the loading table is used to determine the effective KG at the critical instant, then the effective displacement is also found. If this value is used in the hydrostatics, then the interpolated true mean draught will be the true mean draught of the vessel at the point where the bow grounds.

QUESTION Q5.36 (MCM, ENG) MV Reed is undergoing a dry-docking in sea water. The initial displacement of the vessel is 3,241 tonnes and the calculated up-thrust is 83.60 tonnes. Determine the true mean draught of the vessel at the point when the bow grounds.

Clearly after a dry-docking period has finished, the ship must be re-floated. The process will then work in reverse, with water flooding the dock until the bow lifts, at which point the up-thrust will be greatest. As the dock continues to flood, the vessel will trim by the stern and the up-thrust will reduce. As with the initial docking calculations, it is vital to ensure that the vessel is safe to re-float. The easiest way of doing this is to assume that the vessel is being dry-docked, but in her new condition that she would be in after dry-docking has been completed.

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Additional Calculations and Processes • 213

QUESTIONS Q5.37 (MCM, ENG) MV Reed has undergone a dry-docking. She is ready to be re-floated, with a displacement of 3,563 tonnes and a KG of 6.90 m. The LCG of the vessel is 44.60 m FOAP. Determine if it will be safe to re-float the vessel in the above condition. (Note: you may assume that KM remains constant.) Q5.38 (MCM, ENG) MV Reed is undergoing a dry-docking. During the dry-docking, structural alterations are made so that, just before the dock is flooded, her LCG is 46.00 m FOAP, and her KG is 7.05 m and her displacement is 2,897.46 tonnes. The water in the dock is dock water with a density of 1.015 t/m3. Determine the metacentric height, the draught aft and forward at the critical moment during re-floating. (Note: You may assume that the KM remains constant during the dry-docking process.)

Draught Surveys (MCM) Draught surveys are a standardised method to determine the displacement of a vessel. While a normal layer correction is accurate enough to determine the displacement and true mean draught for hydrostatic and stability calculations, it does not take account of hogging and sagging. This is accounted for in draught surveys, which need to be undertaken to determine the exact amount of cargo carried. The draught survey process takes into account hogging and sagging in a standardised method. While the actual displacement values obtained via a draught survey may not be correct (see Table 5.7) in comparison, the standardised process means that any changes in the displacement as a result of loading are accurately determined. This is obviously important as the economics of the vessel depend on carrying the right amount of cargo. The process of a draught survey is very simple, and simply requires a form to be completed showing the relevant calculations, as shown in Figure 5.16. Filling in the form is relatively simple, with the formulae given, but care must be taken to use the correct values, accuracy and units. When using the draught survey process, the sign convention is very important. As before, stern trim is treated as positive,

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214 • Ship Stability, Powering and Resistance

Metres 1

Draught Forward

2

FP Correction

3

Draught at FP

4

Draught Aft

5

AP Correction

6

Draught at AP

7

True Trim

8

Draught (M) Port

9

Draught (M) Stb

Dist . ma marks ks displaced × Observed trim Dist . bet between ee marks k

Dist . marks a ks displaced × Observed trim Dist . between marks k

10

Draught Midships Mean

11

Amidship Line Correction

12

Draught at Amidships

13

Corrected Midship Draught

14

TPC

15

Displacement

16

1st Trim Correction (layer)

Dist . CF from o midships × Trim × TPC LBP

17

2nd Trim Correction (form)

50 × True trim2 × ( MCTC 2 − MCTC1 ) LBP

18

Corrected Displacement

19

Dock Water Displacement

Dist . marks k displaced × True trim LBP

dFP

(

dM ) + d AP 8 LCF FOAP

Δ×

R.D. Doc Dock w water 1.025

▲ Figure 5.16 Draught survey form

and bow trim is treated as negative. Distances AFT of a point are treated as positive, distances FORWARD of a point are treated as negative. For example, if the forward draughts marks are aft of the forward perpendicular, and we knew the distance of the perpendicular FROM the marks, then it would be treated as a negative distance, as shown in Figure 5.17, or if the aft draughts marks are forward of the aft perpendicular,

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Additional Calculations and Processes • 215 and we knew the distance of the perpendicular FROM the marks, then it would be treated as a positive distance, as shown in Figure 5.18.

Positive distance

Negative distance

▲ Figure 5.17 Sign convention for the FP AP

Positive Negative distance distance

▲ Figure 5.18 Sign convention for the AP

The draught forward is the draught measured at the forward draught mark and entered into row 1. The forward draught marks are not normally exactly on the forward perpendicular (the distance from the perpendicular is known as the displacement of the mark), so a correction must be made to the measured draught to get the actual draught forward. Using the sign convention, if the perpendicular is forward of the marks, the distance is negative, if the perpendicular is aft of the marks, the distance is positive. The trim must also be taken as positive for stern trim, and negative for bow trim. The formula for the correction is shown in row 2 with the result entered into row 2. The measured draught forward is corrected to the actual draught forward and entered into row 3. The actual draught forward is found by taking the measured draught and adding the correction, respecting the positive or negative sign of the correction.

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216 • Ship Stability, Powering and Resistance The draught aft is the draught measured at the aft draught mark and entered into row 4. The aft draught marks are not normally exactly on the aft perpendicular, so a correction must be made to the measured draught to get the actual draught aft. As with the forward perpendicular, using the sign convention, if the perpendicular is forward of the marks, the distance is negative, if the perpendicular is aft of the marks, the distance is positive. The formula for the correction is shown in row 5 with the result entered into row 5. The measured draught aft is corrected to the actual draught aft and entered into row 6. The actual draught forward is found by taking the measured draught and adding the correction, respecting the positive or negative sign of the correction. The true trim of the vessel is the actual trim of the vessel based on the draughts at the perpendiculars (found by the difference in values in rows 6 and 3). This is entered into row 7. Again, positive values indicate stern trim, negative values indicate bow trim. Rows 8 and 9 are for the draughts measured at amidships, port and starboard. By taking the draught each side, any effects of list can be included. Row 10 is the average of the port and starboard draught values. The amidships draught marks are not normally exactly at amidships (the distance from the perpendicular is known as the displacement of the mark), so a correction must be made to the measured draught to get the actual draught at amidships (the mean draught). The formula for the correction is shown in row 11, with the result entered into row 11. The sign convention that applies to the forward and aft perpendicular also applies – if the amidships point is aft of the amidships draught marks, then the distance is positive, if the amidships point is forward of the amidships draught marks, then the distance is negative. The measured draught at amidships is corrected to the actual draught at amidships (the mean draught) and entered into row 12. The actual draught amidships is found by taking the measured draught and adding the correction, respecting the positive or negative sign of the correction. The corrected midship draught, as found by formula in row 13, is a method of averaging the draughts. The result of the formula is entered in row 13. Using the hydrostatics, the displacement, TPC and the LCF are found for the vessel for the corrected midship draught. The salt water values are used, regardless of the actual density of water that the vessel is in. The result for the TPC and the LCF, which for intermediate values can be found by linear interpolation from the hydrostatics, are put into row 14, and the result for displacement is put into row 15.

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Additional Calculations and Processes • 217 In row 16, a correction is calculated for the difference in the mean draught and the true mean draught (although this is very small). Within this correction, the trim is the true trim of the vessel, in units of centimetres. It is important to use the correct sign convention. If the LCF is aft of amidships, the distance of the LCF from amidships is positive, if the LCF is forward of amidships, the distance of the LCF from amidships is negative. As before, stern trim is positive, and bow trim is negative. The answer is entered in row 16. In row 17, another correction is made for the form of the vessel. Again, within this correction, the trim is the true trim of the vessel, in units of metres. As the trim is squared, the value will always be positive, regardless of the direction of trim of the vessel. This is because a positive number squared always gives a positive value, and a negative number squared always gives a positive value. The MCTC1 is the MCTC of the vessel interpolated for a draught of 0.50 m less than the actual mean draught, and the MCTC2 is the MCTC of the vessel interpolated for a draught of 0.50 m more than the actual corrected midship draught in row 13. This variation in the MCTC values approximately accounts for the variation in the MCTC as a result of trim. Row 18 is the corrected displacement of the vessel. This is equal to the displacement found in row 15 added to the corrections found in rows 16 and 17. Finally, the displacement, corrected for dock water, is found, using the formula in row 19. This is taken as the actual displacement of the vessel. In HND, FdSc and SQA exams, you will be given the form as shown in Figure 5.16. Within draught surveys, accuracy is important. You should work draughts in metres with an accuracy of three decimal places (i.e. the nearest millimetre), and displacements in tonnes to two decimal places (i.e. the nearest 10 kilograms).

QUESTIONS Q5.39 (MCM) MV Reed has a draught of 7.000 m measured at the aft draught marks. These are 2.500 m forward of the aft perpendicular. She has a draught of 6.000 m measured at the forward draught marks. These are 3.000 m aft of the forward perpendicular. The amidships marks are 2.000 m aft of amidships. The port amidships mark shows a draught of 6.530 m, and the starboard amidships mark shows a draught of 6.514 m. Complete a draught survey, and hence determine the displacement of the vessel if she is in dock water, density 1.010 t/m3. Q5.40 (MCM) MV Reed is floating in dock water with a density of 1.006 t/m3. Table 5.6 shows a partially completed draught survey for the vessel in the above condition. The forward draught

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218 • Ship Stability, Powering and Resistance Table 5.6 Draught survey 1

Draught forward

4.000

2

FP correction

3

Draught at FP

4

Draught aft

5

AP correction

6

Draught at AP

7

True trim

8

Draught (M) port

4.486

9

Draught (M) Stb

4.500

Dist . ma marks ks displaced × Observed trim Dist . bet between wee marks k

5.000 Dist . marks a ks displaced × Observed trim Dist . between w marks k

10

Draught midships mean

11

Amidship line correction

12

Draught at amidships

13

Corrected midship draught

14

TPC

15

Displacement

16

1st trim correction (layer)

Dist . CF from midships × Trim × TPC LBP

17

2nd trim correction (form)

50 × True trim2 × ( MCTC 2 − MCTC1 ) LBP

18

Corrected displacement

19

Dock water displacement

Dist . marks k displaced × True trim LBP

dFP

(

d M ) + d AP 8 LCF FOAP

Δ×

R.D.Dock w water 1.025

marks are 2.000 m aft of the forward perpendicular. The aft draught marks are 3.000 m forward of the aft perpendicular. The amidships draught marks are 0.500 m forward of amidships. Complete the remainder of the draught survey form, and hence determine the displacement of the vessel via a draught survey.

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Additional Calculations and Processes • 219 Q5.41 (MCM) MV Reed is to undergo a draught survey. The draughts and positions of the draught marks are as shown below, where the trim is exaggerated for clarity. The water relative density is 1.010. Using a draught survey form, complete the draught survey and hence find the displacement. Note that the vessel is trimmed by the bow, and that the aft draught marks are aft of the aft perpendicular. AP FP Actual waterline

1.500 m

1.500 m Draught at port amidship marks = 5.517 m

Draught at aft marks = 5.000 m

2.000 m Draught at fwd marks = 6.000 m

Draught at stb amidship marks = 5.500 m

The draught survey process is in itself not an accurate way to exactly determine the displacement. However, it is a consistent method which allows the displacement to be determined before and after loading, with the change in displacement between the two conditions giving a reliable indication of the mass loaded. The accuracy of the actual displacement found from the draught survey process for MV Reed is shown in Table 5.7. Table 5.7 Accuracy of the draught survey process Hog/sag (cm)

Actual displacement (tonnes)

Displacement via draught survey

Error (tonnes)

Error (%)

–10

7,291

7,432.5

141.5

1.93

–9

7,294

7,422.84

128.84

1.76

–8

7,298

7,411.8

113.8

1.55

–7

7,302

7,402.14

100.14

1.37

–6

7,306

7,391.1

85.1

1.16

–5

7,310

7,381.44

71.44

0.97

–4

7,314

7,370.4

56.4

0.77

–3

7,314

7,360.74

46.74

0.64

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220 • Ship Stability, Powering and Resistance Table 5.7 Continued Hog/sag (cm)

Actual displacement (tonnes)

Displacement via draught survey

Error (tonnes)

Error (%)

–2

7,322

7,349.7

27.7

0.38

–1

7,326

7,340.04

14.04

0.19

0

7,329

7,329

1

7,334

7,319.41

14.59

0.20

2

7,337

7,308.45

28.55

0.39

3

7,341

7,298.86

42.14

0.57

4

7,345

7,287.9

57.1

0.78

5

7,349

7,278.31

70.69

0.96

6

7,353

7,267.35

85.65

1.17

7

7,357

7,257.76

99.24

1.35

8

7,361

7,246.8

114.2

1.56

9

7,365

7,237.21

127.79

1.74

10

7,369

7,226.25

142.75

1.95

0

0.00

ADDITIONAL CALCULATIONS AND PROCESSES  LEARNING CHECKLIST Objective

Level

Correct hydrostatic data for water density

MCM, ENG

Calculate the hydrostatic values for a box shaped vessel in fresh and dock water

MCM, ENG

Determine the hydrostatics of a vessel in fresh water if the draught is known

OOW, MCM, ENG

Determine the hydrostatics of a vessel in fresh water if the displacement is known

OOW, MCM, ENG

Determine the hydrostatics of a vessel in fresh water or dock water if the draught is known

MCM, ENG

Determine the hydrostatics of a vessel in fresh water or dock water if the displacement is known

MCM, ENG

Use the layer correction process to determine the true mean draught in sea water, dock water and freshwater

MCM, ENG

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Additional Calculations and Processes • 221

Objective

Level

Determine the LCG of a vessel from the trim

MCM, ENG

Determine the mass and location of cargo so that the vessel finishes loading in a specified condition

MCM, ENG

Determine the mass and distribution of cargo between two holds so that the vessel finishes loading in a specified condition

MCM, ENG

Understand the accuracy of layer correction

MCM, ENG

Understand the purpose of an inclining test

MCM, ENG

Conduct an inclining test and determine the lightship KG and lightship displacement

MCM, ENG

Understand how hydrostatic values change with trim and heel

MCM, ENG

Understand how dry-docking or grounding will change the transverse stability of the ship

MCM, ENG

Calculate the metacentric height at the critical instant

MCM, ENG

Calculate the limiting KG before dry-docking to ensure that the vessel is stable during the dry-docking

MCM, ENG

Calculate the limiting trim before dry-docking to ensure that the vessel is stable during the dry-docking

MCM, ENG

Understand the accuracy of assuming that hydrostatics remain constant during dry-docking

MCM, ENG

Determine the true mean draught and the end draughts at the critical instant

MCM, ENG

Calculate if it is safe to re-float following a dry-docking

MCM, ENG

Understand the purpose of a draught survey

MCM

Complete a draught survey

MCM

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6

BILGING AND DAMAGED STABILITY AIMS AND OBJECTIVES

At the end of this section, you should be able to: Understand the definition of bilging Calculate the lost volume in a bilged compartment Determine the sinkage as a result of lost volume in a bilged compartment at any point along the length of the vessel Determine the sinkage as a result of lost volume in a bilged double bottom at any point along the length of the vessel Determine the sinkage as a result of lost volume in a bilged compartment above a double bottom at any point along the length of the vessel Determine the sinkage as a result of lost volume in a bilged compartment with a watertight flat above the initial waterline Determine the sinkage as a result of lost volume in a bilged side compartment Apply the effects of permeability to sinkage calculations Calculate the permeability of a compartment Calculate KB after bilging a whole compartment, a double bottom or a compartment above a double bottom, including the effects of permeability Calculate the LCB after bilging a whole compartment, a double bottom or a compartment above a double bottom, including the effects of permeability Calculate the TCB after bilging a side compartment Calculate the transverse waterplane inertia after bilging a full beam compartment, a double bottom or a compartment above a double bottom, located at amidships, including the effects of permeability Calculate the transverse waterplane inertia after bilging a full beam compartment, a double bottom (including the effects of permeability) or a compartment above a double bottom, located at the end of a ship

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Bilging and Damaged Stability • 223 Calculate the longitudinal waterplane inertia after bilging a full beam compartment, a double bottom (including the effects of permeability) or a compartment above a double bottom, located at the end of a ship Calculate the longitudinal waterplane inertia after bilging a full beam compartment, a double bottom (including the effects of permeability) or a compartment above a double bottom, located away from the end of a ship Calculate the position of the roll axis after bilging a side compartment Calculate the transverse inertia of a waterplane with a bilged side compartment Calculate BM and BML using the waterplane inertia after bilging Calculate GM and GML after bilging Calculate the trim of a vessel after bilging Calculate the end draughts of a vessel after bilging Calculate the list after bilging Understand how bilging influences the GZ curve Calculate the change in stability as a result of bilging Understand and explain the current and historical requirements for minimum damaged stability of passenger vessels, and Type A and Type B vessels

Bilging is the term used to describe the uncontrolled flow of water in and out of a compartment in a vessel. A compartment which is bilged is considered to be damaged so that water is free to flow in and out of the compartment. There are several methods of determining the stability and trim of a vessel after bilging. The ‘standard’ IMO method is known as the ‘lost volume’ method. For reasons that will be clearer later, this name is a bit confusing. During bilging, the underwater shape of the vessel changes, and as a result the stability of the vessel changes. To determine the effect of bilging, the bilging process needs to be considered in a series of steps: Parallel sinkage New position of the centre of buoyancy New BM and BML New GM and GML Resulting list and trim

Parallel Sinkage (MCM, ENG) When a vessel is bilged, it will sink further into the water, and possibly then, depending on the position of the bilged compartment, trim or list. The initial sinkage is referred to as the parallel sinkage.

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224 • Ship Stability, Powering and Resistance Before a compartment is bilged, the compartment contributes to the overall underwater volume of the vessel, and hence provides some of the buoyancy supporting the vessel. When a compartment is bilged, it no longer contributes to the ‘dry’ underwater volume of the vessel, and therefore does not provide buoyancy. This means that when a compartment is bilged, the overall buoyancy force reduces. As the buoyancy force reduces, the vessel is no longer in vertical equilibrium, as the force of gravity will exceed the force of buoyancy. Therefore, the vessel will start to move vertically downwards. As she does so, the underwater volume produced by the remaining intact compartments starts to increase as the draught increases. As the underwater volume increases again, the buoyancy force starts to increase, until such point that the buoyancy force increases to be back in equilibrium with the force of gravity. At this point, the vessel stops sinking. Overall the underwater volume gained through sinkage is equal to the underwater volume lost through bilging. This principle can be used to determine the parallel sinkage as a result of bilging. Therefore, overall during bilging, as a result of losing volume in the bilged compartment and gaining volume in the remaining intact compartments during sinkage: Underwater volume lost = Underwater volume gained ▲ Formula 6.1 Underwater volume change due to bilging

QUESTIONS Q6.1 (MCM, ENG) 20 m

2m

100 m 10 m

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Bilging and Damaged Stability • 225 For the vessel shown above, determine the final draught of the vessel if the 20 m long amidships compartment is bilged, and the overall underwater ‘dry’ volume and displacement before and after bilging. Q6.2 (MCM, ENG)

15 m

1m

1.1 m 100 m 10 m

For the vessel shown above, determine the final draught of the vessel if the amidships compartment is bilged below the watertight flat.

As seen in the calculations, you should find that the overall underwater volume and displacement of the vessel is the same before and after bilging – that is, the amount of ‘dry’ vessel is constant. There may be some small variation due to rounding. For most scenarios of box shaped vessels, the parallel sinkage can also be found in terms of the lost volume and the final waterplane area. If the barge has a length L, beam B and initial draught DI, with a compartment of length l, then the underwater volume lost as a result of bilging can be found from: Lost ∇ = l B × DI The underwater volume gained as a result of the sinkage can be found by: Gained ∇ = ((L B ) − (l B )) × Sinkage

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226 • Ship Stability, Powering and Resistance l

DI

L B

l

DI

DB

L B

▲ Figure 6.1 Bilging definitions

Equating the lost underwater volume and the gained underwater volume gives: l B × DI

(( L × B ) ( l × B ))

Sinkage

Transposing this for the sinkage gives: Sinkage =

l B × DI

(( L B ) − ( l

B ))

In this, the numerator is the underwater volume lost as a result of bilging, while the denominator is the waterplane area of the vessel after bilging.

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Bilging and Damaged Stability • 227 Therefore, this could also be written as:

Parallel si kag k e=

Lost volume o u e in tthe e biilged compartment Final waterplan aterrplane area

▲ Formula 6.2 Parallel sinkage due to bilging

This formula is valid providing that the waterplane area remains constant during the parallel sinkage. If this is not the case, then Formula 6.1 must be used instead.

QUESTION Q6.3 (MCM, ENG) A box shaped vessel has a length of 50 m, a beam of 10 m and a draught of 2 m. A full beam amidships compartment is bilged. The final draught is 2.857 m. Determine the length of the compartment.

The theory so far assumes that the compartment completely floods. In practice, this may not happen, as solid objects such as machinery, cargo and structure cannot flood. This is modelled using a factor known as compartment permeability. This is a decimal or percentage value which tells us how much of the compartment floods. For example, a permeability of 0.6 means 60% of the compartment can flood. Permeability has the symbol μ. Permeability affects both the volume flooded and the waterplane area lost. This means that we have to adjust the parallel sinkage formula to take into account the effects of permeability on the lost volume and the waterplane area.

QUESTIONS Q6.4 (MCM, ENG) A box shaped vessel has a length of 50 m, a beam of 6 m and a draught of 1 m. A full beam amidships compartment, with a length of 5 m, is bilged. Determine the final draught of the vessel if the permeability is 95%. Q6.5 (MCM, ENG) A box shaped vessel has a length of 70 m, a beam of 10 m and a draught of 3 m. A full beam amidships compartment, 8 m long, is bilged. The final draught is 3.34 m. Determine the permeability of the compartment.

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228 • Ship Stability, Powering and Resistance The permeability of a compartment, as a decimal, can be directly calculated if the density of the cargo and the stowage factor of the cargo is known. ⎛ 1 ⎞ SF − ⎜ ⎟ ⎝ ρc go ⎠ μ= SF ▲ Formula 6.3 Compartment permeability and stowage factor

X For a mathematical proof of this, see Appendix 12: The Derivation of the

Formula Giving the Angle of List for a Neutrally Stable Vessel. So far we have considered an amidships compartment, so there will be no list or trim after bilging. Even in cases where the vessel will list or trim, parallel sinkage is still the first stage. For end compartments or side compartments, the process of calculation for parallel sinkage is exactly the same as the previous cases, but the parallel sinkage is applied to the true mean draught of the vessel to find the new draught.

QUESTION Q6.6 (MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10 m and a draught of 2 m. A full beam forepeak compartment, 20 m long, is bilged. Determine the parallel sinkage due to bilging.

The parallel sinkage can be complicated by the addition of watertight flats within compartments. These restrict flooding vertically, for example, in a double bottom. These problems have to be analysed logically – look at the actual lost volume, and look at the final waterplane area. A good method is to draw the shape of the vessel after bilging, and try and determine the lost volume and waterplane area. A series of bilging scenarios is shown in Figures 6.2 to 6.8, along with the appropriate parallel sinkage formula. L and B are the length of the vessel, DI is the initial draught, l and b are the length and beam of the bilged compartment, and h is the depth of the double bottom.

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Bilging and Damaged Stability • 229 l

DI

L B

l

DI

DB L B

▲ Figure 6.2 Full beam amidship compartment

Full beam amidship compartment Lost volume (m3)

l × B × DI × μ

Final waterplane area (m2)

(L × B) − (l × B μ)

Parallel sinkage (m)

(L B × DI × μ ) (L B ) − (l B × μ )

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230 • Ship Stability, Powering and Resistance l

DI h

L B

l

DI h DB L B

▲ Figure 6.3 Full beam amidship compartment bilged in a double bottom

Full beam amidship compartment bilged in a double bottom Lost volume (m3)

l×B×h×μ 2

Final waterplane area (m )

(L × B)

Parallel sinkage (m)

(L

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B × h ×μ) (L B )

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Bilging and Damaged Stability • 231

l

DI h

L B

l

DI h DB

L B

▲ Figure 6.4 Full beam amidship compartment bilged above a double bottom

Full beam amidship compartment bilged in a double bottom Lost volume (m3)

l × B × (DI − h) × μ

Final waterplane area (m2)

(L × B) − (l × B × μ)

Parallel sinkage (m)

l B × (DI h) × μ (L B ) − (l B ×μμ )

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232 • Ship Stability, Powering and Resistance

l

DI

L B

l

DI

DB

L B

▲ Figure 6.5 Full beam end compartment

Full beam end compartment Lost volume (m3)

l × B × DI

Final waterplane area (m2)

(L × B) − (l × B)

Parallel sinkage (m)

(L

l B × DI B ) − (l B )

Note that for this scenario, the effects of permeability complicate the problem significantly, so no permeability is used to assume a worst-case scenario.

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Bilging and Damaged Stability • 233

l

DI

h L B

l

DI

DB h L B

▲ Figure 6.6 Full beam end compartment bilged in a double bottom

Full beam end compartment bilged in a double bottom l×B×h×μ

Lost volume (m3) 2

Final waterplane area (m )

(L × B)

Parallel sinkage (m)

l B ×h× μ (L B )

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234 • Ship Stability, Powering and Resistance

l

DI

h L B

l

DB

DI

h L B

▲ Figure 6.7 Full beam end compartment bilged above a double bottom

Full beam end compartment bilged above a double bottom Lost volume (m3)

l × B × (DI − h) 2

Final waterplane area (m )

(L × B) − (l × B)

Parallel sinkage (m)

l B × (DI h) (L B ) − ( l B )

Note that for this scenario, the effects of permeability complicate the problem significantly, so no permeability is used to assume a worst-case scenario.

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Bilging and Damaged Stability • 235 l

b DI

L B

l

b DI

DB L B

▲ Figure 6.8 Side compartment

Side compartment Lost volume (m3)

l × b × DI 2

Final waterplane area (m )

(L × B) − (l × b)

Parallel sinkage (m)

l b × DI (L B ) − ( l b )

Note that for this scenario, the effects of permeability complicate the problem significantly, so no permeability is used to assume a worst-case scenario.

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236 • Ship Stability, Powering and Resistance l

DI

DB L B

l

DI

DB L B

▲ Figure 6.9 Amidships compartment with a watertight flat above the initial waterline

Amidships compartment with a watertight flat above the initial waterline Lost volume (m3)

l × b × DI × μ

Final waterplane area (m2)

Variable with draught

Parallel sinkage (m)

Lost ∇ = Gained ∇

Note that for this scenario, the waterplane varies with the sinkage; therefore, the sinkage must be directly calculated using the lost volume and gained volume.

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Bilging and Damaged Stability • 237

QUESTIONS Q6.7 (MCM, ENG) A vessel is as shown below: 20 m

2m

1.5 m 100 m 10 m

Determine the parallel sinkage if the amidships double bottom is then bilged. Q6.8 (MCM, ENG) The vessel shown below is bilged, and is shown in the bilged condition. 15 m

3m

1m

80 m 12 m

Determine the parallel sinkage if the amidships double bottom compartment is then bilged, with a permeability of 95%.

When bilging, particularly as a result of damage from collisions, it is possible to be in a situation where the vessel is bilged in a compartment above a water-tight flat.

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238 • Ship Stability, Powering and Resistance In situations like these, the approach is the same, but care must be taken with the waterplane area. Again, a good method is to draw the shape of the vessel after bilging, and try and determine the lost volume and waterplane area.

QUESTIONS Q6.9 (MCM, ENG) The vessel shown below has an initial draught of 2.00 m. The vessel is then bilged in the amidships compartment, above the double bottom. Determine the parallel sinkage. 20 m

2m

1.5 m

100 m 10 m

Q6.10 (MCM, ENG) The vessel shown below is bilged in the amidships compartment, above the double bottom. 25 m

5m

3m 70 m 8m

If the initial draught is 5.00 m, determine the parallel sinkage if the permeability of the compartment is 77%.

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Bilging and Damaged Stability • 239

The Centre of Buoyancy after Bilging (MCM, ENG) Once a compartment has been bilged, it is important to be able to determine both GM and GML in order to assess the stability of the vessel. GM does not always reduce after bilging. Finding GM involves finding KB, BM and KG. Finding KB after bilging requires a table of moments of volume. The vessel, in the bilged condition, needs to be analysed to determine the overall centre of underwater volume after bilging. To do this, a table of moments of volume is used. This is similar to a loading table, except volume and centre of volume are used instead of mass and KG. There are a number of ways of analysing the volume, but the simplest is to determine the total volume of the vessel at the bilged draught, ignoring the lost volume, and then determine the volume of the bilged compartment. Treating the volume of the bilged compartment as negative means that the total of the underwater volume column in the table will be correct. Note that this value should be the same as the initial underwater volume, as the overall underwater volume should remain constant during bilging. Table 6.1 Tabular calculation to find KB after bilging Item

Underwater volume (m3)

Ship at the bilged draught

L × B × bilged draught Bilged draught × 0.5

L × B × bilged draught × bilged draught × 0.5

Bilged compartment

d of bilged –(l × b × d of bilged compartment after compartment after bilging × permeability) bilging × 0.5

–(l × b × d of bilged compartment after bilging × permeability) × d of bilged compartment after bilging × 0.5

Total

Total of volume column

Total of moment column

Overall l KB after biilging =

Centre from keel (m)

Vertical moment of volume (m4)

Total v vertica ertical moment o of v volume Tota t l volum v e

▲ Formula 6.4 Bilged KB from a table of moments of volume

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240 • Ship Stability, Powering and Resistance

QUESTIONS Q6.11 (MCM, ENG) Determine KB for the vessel shown below.

3.379 m

22 m

2m 80 m 9m

If permeability is involved, then the calculation becomes slightly more complex. Previously, we have seen that permeability (μ) is the measure of the amount of the compartment which floods. Therefore, part of a permeable compartment must still be providing buoyancy. This must be accounted for in the calculations. Q6.12 (MCM, ENG) Determine the KB for the vessel shown below if the bilged amidships compartment has a permeability of 70%.

3.379 m

22 m

2m 80 m 9m

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Bilging and Damaged Stability • 241 If the compartment is a side or end compartment, the process is exactly the same, but be careful with the dimensions. The fact that the vessel may trim or list as a result of bilging is not important at this stage. If the bilged compartment runs all of the way from the keel to the final waterline (i.e. there are no watertight flats), then there is a shortcut to finding KB for box shaped vessels. In these scenarios, KB is equal to half of the final draught of the vessel. Table 6.2 Tabular calculation to find LCB after bilging Item

Underwater volume (m3)

Centre from AP (m)

Longitudinal moment of volume (m4)

Ship at the bilged draught

L × B × bilged draught

L × 0.5

L × B × bilged draught × L × 0.5

Bilged compartment

–(l × b × d of bilged compartment after bilging × permeability)

Centre of bilged compartment from AP

–(l × b × d of bilged compartment after bilging × permeability) × centre of bilged compartment from AP

Total

Total of volume column

Total of moment column

Finding the LCB after bilging is very similar to finding KB; however, the centres of each compartment are measured from the aft perpendicular.

Overall LCB CB after biilging i =

Total longitudinal o g tud a moment o of v volum u e Total v volume

▲ Formula 6.5 Bilged LCB from a table of moments of volume

QUESTION Q6.13 (MCM, ENG) Determine the LCB of the vessel shown if the permeability of the compartment is 70%.

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242 • Ship Stability, Powering and Resistance

3.379 m

22 m

80 m 9m

If the bilged compartment runs all of the way from the keel to the final waterline (i.e. there are no watertight flats), and is at the extreme end of the vessel, and there is no permeability, then there is a shortcut to finding LCB for boxed shaped vessels. In these scenarios, the LCB is equal to half of the final waterplane length of the vessel. Finding the TCB after bilging is very similar to finding KB and LCB, however the centres of each compartment are measured from the centreline of the vessel. Distances to port are treated as positive, and distances to starboard are treated as negative. Table 6.3 Tabular calculation to find TCB after bilging Item

Underwater volume (m3)

Ship at the bilged draught

L × B × bilged draught 0

0

Bilged compartment

Distance from –(l × b × d of bilged original centreline compartment after bilging × permeability) to centre of bilged compartment (port positive, stb negative)

–(l × b × d of bilged compartment after bilging × permeability × distance from original centreline to centre of bilged compartment)

Total

Total of volume column

Total of moment column

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Centre from centreline (m)

Transverse moment of volume (m4)

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Bilging and Damaged Stability • 243

Overall l TCB after biilging i =

Total ttra ansverse s e se moment o of v volume Total volum v e

▲ Formula 6.6 Bilged TCB from a table of moments of volume

QUESTION Q6.14 (MCM, ENG) For the vessel below, determine the TCB if the mean draught after bilging is 4 m and the damage is on the port side of the vessel. You may assume that the compartment permeability is 100%.

8m

4m

2m 100 m 10 m

BM after Bilging a Full Beam Compartment (MCM, ENG) To determine GM, BM is required. Again, this needs consideration as to what the final waterplane area looks like. If the bilged compartment runs right across the vessel, then determining BM is straight forward. The vessel has effectively ‘lost’ the waterplane

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244 • Ship Stability, Powering and Resistance inertia provided by the bilged compartment. Therefore, if the bilged compartment has a length l, the BM calculation becomes: LB 3 lb3 −μ I 12 BM = = 12 ∇ ∇ ▲ Formula 6.7 BM after bilging a full beam compartment where waterplane area is lost

QUESTION Q6.15 (MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10 m and floats at a draught of 2 m. The vessel has an amidships compartment formed by two watertight transverse bulkheads spaced 20 m apart. Determine BM after bilging this amidships compartment if the permeability is 100%.

The process is identical if the full beam compartment is amidships, at the ends or anywhere along the length of the vessel. However, for side compartments, the process is a bit different.

BM after Bilging a Side Compartment (MCM, ENG) When a side compartment is bilged, the final waterplane is no longer symmetrical about the centreline of the ship. The result of this is that the centre of area of the waterplane moves away from the damage, off of the centreline of the ship. The ship rolls about an axis which goes through the centre of area of the waterplane in a transverse direction (just in the same way that she trims about the centre in the longitudinal direction – the LCF), therefore the ship will roll around an axis which will be away from the centreline of the vessel, as shown in Figure 6.10. This shows the waterplane of a box shaped vessel before and after bilging a side compartment with the resulting movement of the centre of area of the waterplane and the roll axis.

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Bilging and Damaged Stability • 245 Before bilging

After bilging Waterplane centre Waterplane centre

Roll axis through ship centreline

Roll axis through waterplane centre

▲ Figure 6.10 Waterplane geometry and side compartment bilging

To determine BM, we need to determine the transverse inertia of the waterplane when measured about an axis running through the centre of the waterplane. Previously, we have been able to do this by directly calculating the transverse inertia of the waterplane, and hence find BM, as the waterplane has been rectangular or symmetrical about the centreline of the ship. However, we cannot directly calculate the transverse inertia through the new roll axis after bilging a side compartment, as the shape of the waterplane after bilging is not rectangular or symmetrical about the centreline. To calculate the transverse inertia for this scenario, we need to look at the concept of inertia, or second moment of area, in a bit more detail. Inertia is more complicated than geometric measures such as area, in that it varies depending on the axis it is measured about. This is because inertia is a measure of the distribution of area from an axis running through a point (see Figure 2.13). The closer the area of the shape is to that point, the smaller the inertia of the shape. Conversely, the greater the area of the shape away from the point, the greater the inertia.

The parallel axis theory Before we look at inertia in more detail, we need to define some terminology. This can be confusing, as there are a number of terms which mean the same thing. The value of inertia depends on where it is measured from. When inertia is measured through an axis running through the geometric centre of a shape, it is referred to as InertiaCENTROID, InertiaCENTRE, InertiaNEUTRAL AXIS or InertiaGG, often abbreviated to ICENTROID, ICENTRE, INEUTRAL AXIS or IGG. These terms all mean the same thing.

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246 • Ship Stability, Powering and Resistance When inertia is measured through an axis running through a point away from the geometric centre of a shape, it is referred to as InertiaREMOTE (or InertiaX or InertiaY when measured through the x or y axis of a graph), often abbreviated to IR, IXX or IYY. An example of these measurement axes is shown in Figure 6.11, with the centre axis running through the centre of the shape and the remote axis running through the bottom of the shape.

Centre of area Centroid axis

Remote axis

▲ Figure 6.11 Centroid and remote axes

There is a mathematical link between the inertia measured through an axis at the centre of the shape (which for the sake of clarity, we’ll call ICENTROID), the inertia measured through a parallel axis at a remote point away from the centre of the shape (which again for the sake of clarity, we’ll call IREMOTE), the area of the shape and the distance between the two axes. This link is known as the parallel axes formula: IREMOTE = ICENT i E E TROID + ( Area Dis

ce 2 )

▲ Formula 6.8 Parallel axes formula

Provided that the area of the shape and the distance between the axes is known, along with one of the inertia values, then the other inertia value can be determined using the parallel axes formula. X For a mathematical proof of this formula, see Appendix 29: Derivation of

the Parallel Axes Theorem (Huygens-Steiner Theorem).

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Bilging and Damaged Stability • 247

QUESTIONS Q6.16 (MCM, ENG) The shape in Figure 6.11 has an area of 6,853 m2. The remote axis is 46 m from the centroid axis. If the inertia of the shape measured at the centroid axis is 2,814,730.8 m4, determine the inertia measured at the remote axis. Q6.17 (MCM, ENG) A shape has an area of 1,000 m2. The remote axis is 5 m from the centroid axis. If the inertia of the shape measured at the remote axis is 33,333.33 m4, determine the inertia measured at the centroid axis.

For rectangular shape of length L metres and width B metres, we have previously seen that the transverse inertia measured through the centre of the shape can be determined using Formula 2.8: Inertia =

LB 3 12

Note that this is an InertiaCENTROID value, as it is measured through the centre of the shape. For rectangular shape of length L metres and width B metres, the remote inertia, measured along one of the long edges of the rectangle, can be determined using:

Inertia =

LB 3 3

▲ Formula 6.9 Remote inertia of a rectangle measured along the long edge

Note that this is an InertiaREMOTE value, as it is measured through an axis away from centre of the shape. This formula allows us to determine the transverse waterplane inertia after bilging a side compartment. X For a mathematical proof of this formula, see Appendix 28: Derivation of

the Formula Giving the Transverse Inertia of a Rectangular Waterplane Measured from the Edge.

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248 • Ship Stability, Powering and Resistance

QUESTION Q6.18 (MCM, ENG) A rectangular waterplane has a length of 60 m and a beam of 20 m. Determine the inertia measured along the side of the vessel (the long edge of the rectangle).

Calculating the transverse inertia of the waterplane after bilging a side compartment (MCM, ENG) The theory shown in the previous section is needed to determine the transverse inertia of the waterplane after bilging. To determine this, the centre of the waterplane after bilging must be found. This will tell us where the roll axis of the vessel is. We can find the centre of area of the waterplane area by taking moments of area about the centreline of the vessel. This is very similar to finding the centre of buoyancy, except that the tables are based on waterplane area. Table 6.4 Tabular calculation to find the roll axis after bilging Item

Waterplane area (m2)

Centre from centreline Transverse moment (m) of area (m3)

Waterplane of the ship at the bilged draught

L×B

0

0

Waterplane of the bilged compartment

–(l × b)

Distance from original centreline to centre of bilged compartment (port positive, stb negative)

–(l × b) × distance from original centreline to centre of bilged compartment

Total

Total of area column

Waterplane centre after biilgiing =

Total of moment column

Transverse moment of area Wa W aterrplane area a

▲ Formula 6.10 Bilged centre of waterplane area from a table of moments of area

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Bilging and Damaged Stability • 249 QUESTION Q6.19 (MCM, ENG) A box shaped vessel has a length of 100 m, a beam of 10.00 m, and has a port side bilged side compartment 8.00 m long and 2.00 m wide. The permeability of the compartment is 100%. Determine the distance from the centreline of the vessel to the roll axis after bilging.

The next step is to determine the transverse inertia of the vessel, measured at the damaged edge. In Formula 6.9, we saw the formula for determining the inertia of a rectangle along the long edge. This can be adapted for a vessel with a rectangular waterplane and a bilged rectangular side compartment. For a vessel with a length L, and beam B, and a damaged side compartment with a length l and a beam b, the inertia measured at the damaged edge is given by:

IEDGE =

LB 3 lb3 − 3 3

▲ Formula 6.11 Inertia at the long edge of a rectangle

As seen previously, the parallel axes theory can be used to determine the inertia measured through the centre of the shape (which is the roll axis) using the inertia measured about the edge: IREMOTE = ICENTRO Distance 2 ) E ID + ( Area ICENTROID is the inertia at the centre of the shape (the new roll axis), the area is the area of the shape (the final waterplane area) and the distance is the distance from the damaged edge to the new roll axis. Therefore, this can be transposed and written as: IROLL AXIS X

IEDGE − (Waterplane area a ea × New e roll lll axiis to edge dg 2 )

▲ Formula 6.12 Parallel axis formula

Note that with different notation, this may also be written as: IGG

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I XX − ( Area h2 )

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250 • Ship Stability, Powering and Resistance Or: ICentroid

IRRemote − ( Area h2 )

QUESTION Q6.20 (MCM, ENG) A vessel with an initial draught of 2 m is bilged in a starboard side compartment as shown. Find BM. 20 m

4m 100 m 10 m

BML after Bilging a Full Beam End Compartment (MCM, ENG) For full beam end compartments with a length l, the BML must also be found so that the GML and MCTC can be found, and the trim determined. If the bilged compartment is at the end of the vessel, waterplane area is lost as a result of bilging, and the compartment permeability is one, then the longitudinal inertia of the waterplane can be found from:

IL =

( L l )3 B 12

▲ Formula 6.13 Longitudinal waterplane inertia after bilging an end compartment where waterplane area is lost

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Bilging and Damaged Stability • 251 Therefore the bilged BML can be found with: (( L l )3 B) I 12 BML = L = ∇ ∇ ▲ Formula 6.14 BML after bilging an end compartment where waterplane area is lost

If there is a watertight flat, so that the waterplane area is not reduced as a result of bilging, then BML can be found as before with: (L3 B) BML = 12 ∇ ▲ Formula 6.15 BML after bilging an end compartment where waterplane area is lost

QUESTIONS Q6.21 (MCM, ENG) A box shaped vessel has a length of 80.00 m and a beam of 15.00 m, with a draught of 3.00 m. The vessel has a full beam end compartment, running the full depth of the vessel, which is 10.00 m long. Determine the longitudinal BM of the vessel if the end compartment is bilged. Q6.22 (MCM, ENG) A box shaped vessel has a length of 90.00 m and a beam of 20.00 m, with a draught of 5.00 m. The vessel has a 15.00 m long full beam end compartment, with a watertight flat 2.00 m above the keel. Determine the longitudinal BM of the vessel if the end compartment is bilged below the watertight flat.

BML after Bilging a Full Beam Compartment (ENG) For full depth, full beam end compartments with a length l, where the bilged compartment is not at the end of the ship, a different approach must be taken. This

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252 • Ship Stability, Powering and Resistance is because the waterplane contributing to the longitudinal BM is effectively in two sections. This complicates the BML calculation, as the parallel axes theory must be used to determine the overall longitudinal inertia about the LCF, which is required to calculate the longitudinal BM. The first stage is to determine the position of the LCF after bilging. This can be done by using a table of moments of area, which is the same approach as seen in Table 6.4, however this time the distances measured must be taken from the aft perpendicular. Table 6.5 Tabular calculation to find the LCF after bilging Item

Waterplane area (m2)

Centre from AP (m FOAP)

Longitudinal moment of area (m3)

Waterplane area of the ship at the bilged draught

L×B

L/2

0

Bilged compartment waterplane area

–(μ × l × b)

Distance from AP to centre of bilged compartment

–(μ × l × b) × distance from AP to centre of bilged compartment

Total

Total of area column

CF after te biilging =

Total of moment column

Longitudinal moment oment of area Total water w rpllan a e area

▲ Formula 6.16 Bilged centre of waterplane area from a table of moments of area

QUESTION Q6.23 (ENG) A box shaped vessel has a length of 80 m and a beam of 15 m. A full depth, full beam watertight compartment, with an aft bulkhead located 40 m FOAP and a forward bulkhead located 50 m FOAP, with a permeability of 0.75, is bilged. Determine the LCF after bilging.

The next stage in determining the BML is to examine the longitudinal inertia of the waterplane area measured about the LCF of the vessel, as this is the pivot point when trimming. To do this, first the longitudinal waterplane inertia of each section of

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Bilging and Damaged Stability • 253 waterplane must be determined, measured about the centre of area of the section of waterplane area. The parallel axes theory can be used to then determine the longitudinal inertia of each section of waterplane, measured about the LCF.

QUESTIONS Q6.24 (ENG) A box shaped vessel has a length of 80 m and a beam of 15 m, with an initial draught of 2.00 m. A full beam, full depth watertight compartment, with an aft bulkhead located 40 m FOAP and a forward bulkhead located 50 m FOAP, with a permeability of 0.75, is then bilged. Determine the BML after bilging, if the LCF is 39.48 m FOAP. Q6.25 (ENG) A box shaped vessel has a length of 100 m and a beam of 20 m, with an initial draught of 3.00 m. A full beam, full depth watertight compartment, with an aft bulkhead located 30 m FOAP and a forward bulkhead located 40 m FOAP, with a permeability of 0.80, is then bilged. Determine the LCF and the BML after bilging.

KG after Bilging (MCM, ENG) Finding KG after bilging is extremely simple. The lost buoyancy method works by changing the distribution of buoyancy of the vessel. This in turn changes the stability of the vessel. We have seen that overall the total displacement is constant. We can also make the assumption that the distribution of mass has not changed – the water in the bilged compartment is not technically adding any mass to the vessel. Although there may be a small change due to damage, we can assume that KG REMAINS CONSTANT.

GM and GML after Bilging (MCM, ENG) As has been seen, for a variety of situations, KB, BM and BML can be found, and as KG is constant, GM and GML can be found. These can be used to determine the transverse stability and list, as well as longitudinal stability and trim, after bilging.

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254 • Ship Stability, Powering and Resistance GM = KB + BM − KG GML = KB + BML − KG

Trim after Bilging (MCM, ENG) When a compartment is bilged, the centre of gravity remains constant. The centre of buoyancy, which is the centre of the underwater volume, moves longitudinally away from the original position, which creates a longitudinal imbalance between buoyancy and gravity. The vessel trims as a result of this. The MCTC of a bilged vessel can be found using Formula 4.4: MCTC =

ΔGML 100LBP

Note that the GML is in the bilged condition. The trim can then be found using: Trim =

LCG ) Δ MCTC

(LCB

The LCB and MCTC must be in the bilged condition.

QUESTIONS Q6.26 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, with an initial draught of 2 m, and is initially floating on an even keel. The vessel has a 20 m long forward end compartment, with a watertight flat creating a double bottom 1.00 m deep. Determine the GML, MCTC and hence the trim if the forward end compartment is bilged above the watertight flat. KG is 1.40 m. Q6.27 (ENG) A box shaped vessel has a length of 60.00 m and a beam of 10.00 m, with a draught before bilging of 4.00 m, and floats on an even keel before bilging. The vessel has a 5.00 m long compartment, running between bulkheads 40.00 m FOAP and 45.00 m FOAP. The compartment runs the full depth of the vessel.

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Bilging and Damaged Stability • 255 Determine the GML, MCTC and hence the trim and end draughts after bilging. You may assume that the compartment permeability is 60% and that the KG is 2.00 m. Q6.28 (ENG) A box shaped vessel has a length of 120.00 m and a beam of 20.00 m, with a draught before bilging of 6.00 m, and floats on an even keel before bilging. The vessel has a 30.00 m long compartment, running between bulkheads 80.00 m FOAP and 110.00 m FOAP. The compartment runs the full depth of the vessel. Determine the GML, MCTC and hence the trim and end draughts after bilging. You may assume that the compartment permeability is 70% and that the KG is 5.00 m. Q6.29 (ENG) A box shaped vessel has a length of 100.00 m and a beam of 10.00 m, with a draught before bilging of 2.00 m, and floats on an even keel before bilging. The vessel has a 10.00 m long compartment, running between bulkheads 60.00 m FOAP and 70.00 m FOAP. There is a watertight flat 1.50 m above the keel. The compartment is then bilged below the watertight flat. Determine the GML, MCTC and hence the trim and end draughts after bilging. You may assume that the compartment permeability is 90% and that the KG is 3.50 m. Q6.30 (ENG) A box shaped vessel has a length of 80.00 m and a beam of 15.00 m, with a draught before bilging of 3.00 m, and floats on an even keel before bilging. The vessel has a 15.00 m long compartment, running between bulkheads 55.00 m FOAP and 70.00 m FOAP. There is a watertight flat 2.00 m above the keel. The compartment is then bilged below the watertight flat. Determine the GML, MCTC and hence the trim and end draughts after bilging. You may assume that the compartment permeability is 75% and that the KG is 4.00 m. Q6.31 (ENG) A box shaped vessel has a length of 80.00 m and a beam of 10.00 m, with a draught before bilging of 5.00 m, and floats on an even keel before bilging. The vessel has a 15.00 m long compartment, running between bulkheads 40.00 m FOAP and 55.00 m FOAP. There is a watertight flat 1.50 m above the keel. The compartment is then bilged above the watertight flat. Determine the GML, MCTC and hence the trim and end draughts after bilging. You may assume that the compartment permeability is 60% and that the KG is 3.00 m.

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256 • Ship Stability, Powering and Resistance Q6.32 (ENG) A box shaped vessel has a length of 140.00 m and a beam of 20.00 m, with a draught before bilging of 6.00 m, and floats on an even keel before bilging. The vessel has a 30.00 m long compartment, running between bulkheads 70.00 m FOAP and 100.00 m FOAP. There is a watertight flat 2.00 m above the keel. The compartment is then bilged above the watertight flat. Determine the GML, MCTC and hence the trim and end draughts after bilging. You may assume that the compartment permeability is 75% and that the KG is 6.00 m.

List after Bilging (MCM, ENG) When a compartment is bilged, the centre of gravity remains constant. The centre of buoyancy, which is the centre of the underwater volume, moves transversely away from the side compartment if a side compartment is bilged. The mass of the vessel acts down from the centre of gravity, and the buoyancy acts up through the centre of buoyancy, as shown in Figure 6.12.

G

B

B

▲ Figure 6.12 Movement of B due to bilging a side compartment

The list can be determined from Formula 2.11: tanθ =

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d) Δ × GM

(w

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Bilging and Damaged Stability • 257 The force causing the vessel to list is the buoyancy force, which has moved transversely across the vessel. Therefore, the w term, the mass causing the list, is equal to the buoyancy force, which equals the mass of the vessel (see the notes on Archimedes’ Principle for the proof of this). Therefore, Formula 2.11 can be written in terms of the mass of the vessel: tanθ =

Δ×d Δ × GM

The displacement values cancel, giving Formula 6.17, in which d is the shift in the centre of buoyancy. For a box shaped vessel, this is equal to the shift from the original centreline to the new roll axis.

tanθ =

d GM

▲ Formula 6.17 List after bilging a side compartment

QUESTIONS Q6.33 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, with an initial draught of 2 m, and initially floats upright. The vessel has a 20 m long starboard side compartment, 3 m wide. The vessel then bilges the starboard side compartment. Determine the transverse stability and the list if the permeability is 100% and KG is 4 m. Q6.34 (MCM, ENG) A box shaped vessel has a length of 50 m, a beam of 8 m and floats at an initial draught of 2 m in sea water. The vessel is on an even keel and is upright. The vessel has a full beam, amidships compartment which is 10 m long. The compartment has a watertight flat in it, 1.20 m above the keel, which forms a double bottom and a partially filled cargo hold. As a result of a collision, the cargo hold is then bilged, but the double bottom remains intact. The vessel sinks until the draught is 2.15 m. Determine the change in metacentric height as a result of bilging.

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258 • Ship Stability, Powering and Resistance Q6.35 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight transverse bulkheads form an amidships compartment with a length of 20 m. Determine the metacentric height of the vessel if the amidships compartment is bilged with a permeability of 100%. Q6.36 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight transverse bulkheads form an amidships compartment with a length of 20 m. Determine the metacentric height of the vessel if the amidships compartment is bilged with a permeability of 80%. Q6.37 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight transverse bulkheads form an amidships compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the metacentric height of the vessel if the amidships double bottom is bilged with a permeability of 100%. Q6.38 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight transverse bulkheads form an amidships compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the metacentric height of the vessel if the amidships double bottom is bilged with a permeability of 80%. Q6.39 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight transverse bulkheads form an amidships compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the metacentric height of the vessel if the amidships compartment is bilged above the double bottom, with a permeability of 100%. Q6.40 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight

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Bilging and Damaged Stability • 259 transverse bulkheads form an amidships compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the metacentric height of the vessel if the amidships compartment is bilged above the double bottom, with a permeability of 80%. Q6.41 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. A watertight transverse bulkhead forms a forward end compartment with a length of 20 m. Determine the end draughts of the vessel if the forward end compartment is bilged with a permeability of 100%. Q6.42 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. A watertight transverse bulkhead forms a forward end compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the end draughts of the vessel if the forward end compartment is bilged in the double bottom with a permeability of 100%. Q6.43 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. A watertight transverse bulkhead forms a forward end compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the end draughts of the vessel if the forward end compartment is bilged in the double bottom with a permeability of 80%. Q6.44 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. A watertight transverse bulkhead forms a forward end compartment with a length of 20 m. Within the compartment, a watertight flat forms a double bottom 0.90 m deep. Determine the end draughts of the vessel if the forward end compartment is bilged in the compartment above the double bottom, with a permeability of 100%. Q6.45 (MCM, ENG) A box shaped vessel has a length of 100 m and a beam of 10 m, and an initial draught of 2.00 m. She floats upright, on an even keel, with a KG of 4.00 m. Two watertight transverse bulkheads and a longitudinal bulkhead form an amidships compartment

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260 • Ship Stability, Powering and Resistance with a length of 20 m and a width of 1.00 m. Determine the list of the vessel if the amidships compartment is bilged with a permeability of 100%.

GZ Curves for Vessels after Bilging a Side Compartment (MCM, ENG) If a side compartment is bilged, then the centre of buoyancy will move transversely away from the damaged area. This creates a misalignment between the centre of gravity and buoyancy, which is analogous to a transverse shift in the centre of gravity for an intact vessel. Therefore, when the vessel is upright, she will have a negative GZ value, and all the GZ values at larger angles of heel will be reduced by the cosine of the initial offset of the centre of buoyancy from the centreline of the ship. The loss in waterplane area will also result in a reduction in the transverse waterplane inertia, and hence a reduction in BM and GM, reducing the initial slope of the curve. The parallel sinkage as a result of bilging will result in a reduction of freeboard, and hence a reduction in the DEI angle, and subsequent reduction in GZ values past the DEI angle. These features can be seen in Figure 6.13. 0.4 0.3

Intact   wing tank Bilged

0.2

0



GZ (m)

0.1

0

10

20

30

40

50

60

70

–0.1

–0.2 –0.3 –0.4 Heel (degrees)

▲ Figure 6.13 GZ curve after bilging a side compartment

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Bilging and Damaged Stability • 261

Damaged Stability Requirements Ships are designed so that in the event of damage, the vessel will retain sufficient freeboard (i.e. the parallel sinkage is acceptable), and sufficient stability, both in terms of the initial and large angle stability. These methods are designed to improve chances of survival. However, it should be remembered that there is no such thing as an unsinkable ship, only unsinkable materials. We can only minimise the probability of losing a vessel – not exclude it completely. The requirements for different regulations are complicated, and are only summarised here. The appropriate publications should be read for the full requirements.

Passenger vessels pre-2009 (MCM, ENG) These are described in detail in MSN 1698M (Maritime and Coastguard Agency, 1998). The freeboard requirements are written using an imaginary datum line known as the margin line. This runs all around the vessel, 76 mm below the freeboard deck. A vessel may be assumed to be lost when the margin line touches the water, so the requirements are written to ensure that, for a reasonable amount of damage, this does not happen. Depending on the type of vessel, ships must be designed so that the margin line does not touch the water if one, two or three compartments are bilged. This is achieved by keeping the compartments small enough that in the event of bilging the parallel sinkage is small, or by the use of double bottom and wing tanks, or a combination of the above. In order to do this, the Naval Architect analyses the design, and determines the ‘floodable length’ of the potential compartments in the vessel. The floodable length is the length of a compartment, centred at a point on the vessel, which would cause the margin line to just touch the water when flooded. To ensure maximum safety, a number of factors are taken into consideration when determining how large compartments can actually be, in relation to the maximum size that would cause the margin line to touch the water surface. The first factor is the ‘criterion of service numeral’, known as CS. This is effectively a measure of the passenger carrying capability and distribution in relation to the size of the vessel. This measures, by formulae, the amount of the vessel occupied by

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262 • Ship Stability, Powering and Resistance passengers above the margin line, the accommodation space below the margin line and the machinery spaces. This can be thought of as a measure of the ‘vulnerability’ of the design of the ship. The next variable is the ‘factor of subdivision’, known as F. This value, which is always less than one, is determined by formulae from CS and the length of the vessel. CS and F are inversely proportional to each other, so an increase in CS would result in a decrease in F. These values are used to determine the ‘permissible length’ of a compartment: issible lengt length t Floodable Floodable l d bl length × Factor of subdivision What this effectively means is that the compartments in a vessel are smaller than that which is required to cause the margin line to touch the water in the event of bilging, and therefore a factor of safety is included in the design of the compartment. ‘Curves of Permissible Length’ are then created by the Naval Architect, which show the allowable length of compartment at any point. The reciprocal (i.e. 1/F) of the ‘factor of subdivision’ is known as the ‘compartment standard’. This is effectively the number adjacent compartments which could be flooded while still keeping the margin line dry. The compartment standard is rarely a whole number – it should be rounded down to find the number of compartments. The permissible length must be subdivided by watertight bulkheads so that it contains the same number of compartments as the compartment standard. The vessel must also retain suitable stability characteristics after damage. The stability rules are based on what is known as a ‘deterministic’ approach. This approach applies a predetermined amount of damage to the ship, and the ship must have suitable stability characteristics after receiving this damage. Under this approach, the vessel is assumed to have been damaged so that a section of the ship is destroyed as follows: Over a length of 3 m plus 3% of the ship length, or 11 m, whichever is less (if the vessel is a three compartment standard, this damage must cover three compartments). 20% of the beam. The entire depth of the vessel. The vessel is assumed to be in the worst-case loading scenario. Permeability is set (e.g. 0.95 for accommodation, 0.85 for machinery spaces). If lesser damage causes a more serious condition, then the worst case must be used.

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Bilging and Damaged Stability • 263 After this damage is inflicted on the vessel (and using the lost volume calculation method for bilging), and she has reached her equilibrium condition, she must meet the requirements as follows: A metacentric height of at least 0.05 m. An angle of list of no more than 7 degrees for one compartment flooding, or 12 degrees for two or more adjacent compartments. The area under the GZ curve must be at least 0.015 metre radians, measured from the equilibrium angle to the lesser of progressive flooding, 22 degrees for one compartment flooding, or 27 degrees for simultaneous flooding of two or more adjacent compartments, or a major progressive flooding point where there is a rapid reduction in GZ of 0.04 m or more. The range of stability must be at least 15 degrees. If the area under the curve as described above is increased by the ratio of 15/range (in degrees), then the range of stability may be reduced from 15 to 10 degrees. The peak GZ value must be no less than 0.10 m, and at least the value given by: GZ =

Heeling g moment + 0 04 Δ

Within this, the heeling moment is the maximum of the heeling moment generated by the following scenarios: All passengers crowding to one side of the ship (this assumes that each passenger has a mass of 75 kg and that passengers can crowd with 4 to a square metre, on deck areas on one side of the vessel where the muster stations are located). The launching of all fully loaded, davit launched survival craft on one side of the vessel. Wind pressure of 120 N/m2 on one side of the ship. For passenger vessels not carrying vehicles on the bulkhead deck, the margin line may be submerged during the flooding and equalisation process (however it must not be submerged in the final equilibrium condition), provided that there is partial subdivision above the bulkhead deck stops the spread of water along the bulkhead deck, and results in a heel angle below 15 degrees. This method (known as a deterministic method, as the ship must survive a predetermined amount of damage) is sufficient to model a range of scenarios, however, in reality any accidents are unlikely to cause damage which neatly fits the scenario planned for, so

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264 • Ship Stability, Powering and Resistance they are slowly being replaced by ‘probabilistic’ methods. These probabilistic methods look at the probability of damage occurring at a particular location. For example, damage may be more likely under the vessel forward than on the side shell forward. The possibility of the damage running through more than one compartment, and the probability of different types of incident, is then determined. These are put together to determine the probability of flooding a particular compartment in the vessel. Each compartment in turn is then assumed to be flooded. The possible GM, trim, weather, permeability and other statistical factors are taken into account to determine the possibility of the vessel sinking or capsizing. The overall probability of survival in each case is then assessed, and the overall chance of losing the vessel is determined. The layout of the vessel can then be designed in such a way that a large range of incidents can be dealt with. The overall probability of survival of any incident is then compared against a minimum allowable, which itself is based on the ship length and passenger numbers (or the length for cargo vessels). Probabilistic rules are more expensive and harder to use (from the point of view of the Naval Architect and shipyard) as they give less initial guidance to placement of bulkheads, and are computationally expensive (specialist software is required), however they give a much better assessment of the ‘overall’ problem than simply assuming damage scenarios.

The Stockholm Agreement (MCM) Large Ro-Ro passenger vessels have a number of potential risks associated with them: Lack of watertight division on the car decks, with the potential for significant free surface effects High compartment permeability allowing greater potential for flooding Large holes for bow and stern doors for fast turn-around Wide sterns and stern doors cause slamming in heavy seas Significant windage and wind heeling angles Significant danger of shifting cargo (vehicles and vehicle’s own cargo) Large heel angles when turning (especially evasive manoeuvres in congested waters)

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Bilging and Damaged Stability • 265 Probabilistic rules are already being used on these vessels. These address not only the danger of water on vehicle decks, but the often rapid sinking of this type of vessel after flooding. MSC/Circ. 574 describes the Stockholm Agreement, which is now in force. This assesses the probability that the vessel will survive after damage with water ingress, with the probability of damage varying according to location on the ship. The overall probability of survival takes into account the chance of a compartment flooding, and the chance of surviving that compartment flooding. The agreement requires the calculation of an A/Amax value. A is a subdivision index, and Amax is maximum value of survivability of the vessel. The agreement also takes into account the hypothetical effect of 0.5 m (or less, depending on freeboard) of water on the vehicle deck closest to the waterline.

Safe return to port regulations (MCM, ENG) In 2006, IMO approved a new set of regulations for passenger vessels. These were written on the basis that the safest lifeboat is the ship itself. As the name suggests, these regulations are intended to ensure that a vessel can survive damage (up to a realistic threshold) and be able to safely return to port. These regulations are applicable to all passenger ships built on or after 1 July 2010, which have a length of 120 m or more, or have three or more main vertical fire zones. These regulations require relevant ships to be able to return to port in the event of the flooding of any one single watertight compartment. This includes the engine room, so twin engine vessels will have to have subdivided engine rooms to ensure that one engine can operate even if the other engine is in a flooded space. In the event of more than one compartment flooding, the vessel must be designed to be evacuated safely, with 3 hours available for abandonment.

Damaged stability requirements – ‘Type A’ and ‘Type B’ vessels (pre-2009) (MCM) Type A vessels, as defined in the Load Line Regulations (i.e. carrying liquid cargo in bulk) and Type B (all other vessels) have to be able to withstand a certain amount of damage and remain stable and afloat. The depth of damage is assumed to be over the entire depth of the vessel. The transverse extent of the damage is assumed to be B/5 or 11.5 m, whichever is lesser. The flooding shall longitudinally be confined to a single compartment between transverse bulkheads, provided any longitudinal bulkheads are outside of the transverse damage limits – this changes for B-60 and B-100 vessels.

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266 • Ship Stability, Powering and Resistance A B-60 vessel must be able to withstand the flooding of any compartment or compartments, with a permeability of 95%, consequent on the damage assumptions given, so that she remains afloat in an equilibrium position (detailed later). A B-100 vessel must be able to withstand the flooding of any two adjacent compartments (not including the machinery space if the vessel is less than 150 m). All vessels must meet the minimum requirements shown in Figure 6.14. GZ

Max GZ must be greater than 0.1 m

Total list must be less than 15º (17 if the deck edge is not immersed)

Total area must be greater than 0.0175 m Rad

Range must be greater than 20º

Heel

Min GM must be greater than 0.05 m

▲ Figure 6.14 Type A and Type B damaged GZ curves

SOLAS damaged stability rules post-2009 IMO has introduced new regulations for damaged stability requirements for passenger and dry cargo vessels in SOLAS 2009. These new regulations differ from the previous regulations, which were based on a deterministic approach, as the new regulations are based on a mix of deterministic and probabilistic approaches. The new rules apply to vessel with a keel laying date on or after the first of January 2009 respectively for vessels which undergo a major conversion on or after that date. The regulations apply to cargo vessels over 80 m in length, and all passenger vessels. The rules require double bottoms to be fitted, unless it can be proved that alternative arrangements are as safe. The double bottom must be the minimum of 0.76 m or 5% of the maximum beam, but need not be deeper than 2 m.

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Bilging and Damaged Stability • 267 Cross flooding should be, as far as possible, automatic. Manual systems must be operated from above the bulkhead deck. Cross flooding arrangements must bring the vessel to an acceptable condition within 10 minutes, and must be detailed in the damage control plans in the stability book. For all ships, a value known as the Required Subdivision Index, R is found. For cargo vessels, this depends on the length of the vessel, and for passenger ships, this depends on the length of the vessel, the capacity of the lifeboats and the number of people the vessel is permitted to carry in relation to the capacity of the lifeboats. The vessel is considered at three draught and trim combinations: DS – Deepest subdivision draught, on an even keel DP – Partial subdivision draught, on an even keel DL – Light service draught, which is the ballast condition for dry cargo vessels or the arrival condition for passenger vessels, with a trim corresponding to that condition, but no more than 1% of the length. For each draught, the safety of the vessel is measured by a partial safety index AS, AP or AL. These partial safety index values are mathematically derived from the probability of compartments, or combination of compartments flooding, and the condition of stability of the vessel after flooding those compartments or combination of compartments. This takes into account the angle of list after flooding, the down-flooding angle after flooding, the peak GZ values and the range of stability after flooding. For passenger vessels, this also includes a factor to allow for passenger crowding (at 75 kg per person, with 4 passengers per square metre), wind heeling pressure of 120 Newtons per metre squared and the launching of fully laden survival craft. For different types of cargo, permeability values for compartments are specified. For passenger vessels carrying over 36 passengers, there are additional deterministic requirements: vertical damage from the baseline to 12.5 m above the deepest draught if more than 400 passengers are carried, the damage is assumed to be greater than 3 m or 3% of the length of the ship, at any point along the side shell, with a hull penetration of the greater of 0.75 m or 10% of the maximum beam of the vessel if less than 400 passengers are carried, the damage is assumed to occur at any point along the side shell between watertight bulkheads where 36 passengers are carried, the damage is assumed to be greater than 3 m or 1.5% of the length of the ship, with a hull penetration greater than 0.75 m or 5% of the maximum beam

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268 • Ship Stability, Powering and Resistance between 36 and 400 passengers, linear interpolation may be used to determine the damage length and penetration. These partial safety indices are combined to give an overall safety index A, which for dry cargo vessels must be more than 0.5 R, or for passenger vessels, more than 0.9 R.

BILGING AND DAMAGED STABILITY – LEARNING CHECKLIST Objective

Level

Understand the definition of bilging

MCM, ENG

Calculate the lost volume in a bilged compartment

MCM, ENG

Determine the sinkage as a result of lost volume in a bilged compartment at any point along the length of the vessel

MCM, ENG

Determine the sinkage as a result of lost volume in a bilged double bottom at any point along the length of the vessel

MCM, ENG

Determine the sinkage as a result of lost volume in a bilged compartment above a double bottom at any point along the length of the vessel

MCM, ENG

Determine the sinkage as a result of lost volume in a bilged compartment with a watertight flat above the initial waterline

MCM, ENG

Determine the sinkage as a result of lost volume in a bilged side compartment

MCM, ENG

Apply the effects of permeability to sinkage calculations

MCM, ENG

Calculate the permeability of a compartment

MCM, ENG

Calculate KB after bilging a whole compartment, a double bottom or a compartment above a double bottom, including the effects of permeability

MCM, ENG

Calculate the LCB after bilging a whole compartment, a double bottom or a compartment above a double bottom, including the effects of permeability

MCM, ENG

Calculate the TCB after bilging a side compartment

MCM, ENG

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Completed

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Bilging and Damaged Stability • 269

Objective

Level

Calculate the transverse waterplane inertia after bilging a full beam compartment, a double bottom or a compartment above a double bottom, located at amidships, including the effects of permeability

MCM, ENG

Calculate the transverse waterplane inertia after bilging a full beam compartment, a double bottom (including the effects of permeability) or a compartment above a double bottom, located at the end of a ship

MCM, ENG

Calculate the longitudinal waterplane inertia after bilging a full beam compartment, a double bottom (including the effects of permeability) or a compartment above a double bottom, located at the end of a ship

MCM, ENG

Calculate the longitudinal waterplane inertia after bilging a full beam compartment, a double bottom (including the effects of permeability) or a compartment above a double bottom, located away from the end of a ship

ENG

Calculate the position of the roll axis after bilging a side compartment

MCM, ENG

Calculate the transverse inertia of a waterplane with a bilged side compartment

MCM, ENG

Calculate BM and BML using the waterplane inertia after bilging

MCM, ENG

Calculate GM and GML after bilging

MCM, ENG

Calculate the trim of a vessel after bilging

MCM, ENG

Calculate the end draughts of a vessel after bilging

MCM, ENG

Calculate the list after bilging

MCM, ENG

Understand how bilging influences the GZ curve

MCM, ENG

Calculate the change in stability as a result of bilging

MCM, ENG

Understand and explain the current and historical requirements for minimum damaged stability of passenger vessels, and Type A and Type B vessels

MCM, ENG

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7

CALCULATING HULL SHEAR FORCE AND BENDING MOMENT AIMS AND OBJECTIVES

At the end of this section, you should be able to: Plot the shear force diagram for a simplified vessel Plot the bending moment diagram for a simplified vessel Analyse the shear force and bending moment diagrams for a simplified vessel

Basic Structural Calculations So far we have considered the buoyancy forces to act through a single point – the centre of buoyancy. We have also considered the mass forces to act through a single point – the centre of gravity. When the longitudinal centres of buoyancy and gravity are aligned and the forces are equal in magnitude, the vessel floats at a static waterline (and a static trim). While these assumptions are valid for the ship as a whole, for structural calculations we need to look at the distribution of buoyancy and mass in the vessel in more detail.

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Calculating Hull Shear Force and Bending Moment • 271 While the overall amount of buoyancy supporting the ship will be equal to the displacement of the vessel, the distribution of buoyancy is not equal along the length of the vessel. At any point along the vessel, the buoyancy is proportional to the underwater cross-section of the ship at that point. Therefore, at the ends of the vessel, where the beam narrows at the bow and stern, there will be less buoyancy force acting on the vessel than at amidships, where the beam and draught are maximum. The distribution of mass also varies along the length of the vessel. The amount of mass at any point along the vessel will depend on the structure of the ship, and the distribution of cargo, fuel and stores. If we imagine the vessel sliced transversely into a series of sections, the buoyancy force acting upwards on that section of the ship is unlikely to be equal to the mass of that section of the ship. The difference between the buoyancy acting upwards on the section and the mass acting downwards on the section is known as the load. On some sections, there will be less buoyancy acting upwards than mass acting downwards, so there will be a net downwards (or negative) load acting on the section, while in other sections there will be more buoyancy acting upwards than mass acting downwards, so there will be a net upwards (or positive) load acting on the section. The overall buoyancy acting on all of the sections will equal the overall mass of the ship and contents, however it is the difference between mass and buoyancy on each individual section that creates stress in the structure of the vessel. This stress must be kept within acceptable limits to avoid damage to the structure and possible structural failure. The mechanisms by which the buoyancy and mass create stress in the structure are known as ‘shear force’ and ‘bending moments’. The shear force at any point along the vessel is the sum of the loads acting aft of the point under consideration. This is effectively the total imbalance in vertical forces up to the point under consideration. This must be kept within acceptable limits to ensure that the shear stress acting on the structure is within acceptable limits. The bending moment at any point along the vessel is the sum of the loads acting aft of the point under consideration, multiplied by the distance to the point under consideration. This must be kept within acceptable limits to ensure that the bending stresses acting on the structure, which causes hogging or sagging distortion of the structure, is kept within acceptable limits. The normal method of presenting load, shear force and bending moment distribution is to plot them on a graph, with the shear force and bending moment plotted on the y axis, against the ship length on the x axis. These are known as the load, shear force and bending moment diagrams. To determine the shear force and bending moment, first the ‘load’ in each section must be determined. As mentioned previously, this is the difference between the total mass of the section, being acted on by gravity, and the total buoyancy of the section. The mass of each section can be determined from the lightship displacement and the mass of cargo carried.

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272 • Ship Stability, Powering and Resistance Assuming that the structural mass is evenly distributed along the length of the vessel, the structural mass per metre of ship length can be found from: Δ LIGHTSHIP L

Structural r mass per metre =

▲ Formula 7.1 Structural mass per metre of ship length

Assuming that the cargo mass in each hold is evenly distributed along the length of each of the holds, the cargo mass per metre of hold length can be found from: C go mass ass per metre =

Mass of hold c go Hold length

▲ Formula 7.2 Cargo mass per metre of hold length

As the buoyancy force (in units of tonnes force) must be equal overall to the total mass of the vessel (see Archimedes), and assuming that the buoyancy force is evenly distributed along the length of the vessel (which is valid for box shaped vessels), the buoyancy force per metre of ship length can be found from:

Buoyancy force per metre =

Δ LOAD O ED L

▲ Formula 7.3 Buoyancy force per metre of ship length

The overall load per metre of ship length can then be found at any point along the vessel from the sum of the structural, cargo and buoyancy forces: Load pe per metre etre = Buoyancy force pe − C go mass ass per metre

Structural mass per metre

▲ Formula 7.4 Load acting per metre of ship length

This can also be written as: Load =

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Δ LOADED Δ Mass of hold c go O − LIGHTSHIP − L L Hold length

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Calculating Hull Shear Force and Bending Moment • 273 A graph can then be drawn, showing the position along the ship on the x axis, and the load on the y axis. This is known as the load diagram. The load diagram is a ‘square’ shaped diagram, as shown in Figure 7.1. 10 5

Load (tonnes)

0 0

10

20

30

40

50

60

70

80

90

100

–5 –10 –15 –20 –25 Position (m FOAP)

▲ Figure 7.1 Sample load diagram

The definition of shear force at any point along the vessel is ‘the sum of the vertical forces acting to one side of that point’. This is equivalent to the area under the load diagram up to that point. Therefore, to determine the shear force at any point, the area under the load diagram is found up to that point. This allows another diagram to be plotted, showing the variation in shear force along the length of the vessel. This is the shear force diagram. The shear force diagram is a ‘saw tooth’ shape, as shown in Figure 7.2. The definition of bending moment at any point along the vessel is ‘the sum of the vertical forces acting to one side of that point, multiplied by the distance of the force to the point of consideration’. This is equivalent to the area under the shear force diagram up to that point. Therefore, to determine the bending moment at any point, the area under the shear force diagram is found up to that point. This allows another diagram to be plotted, showing the variation in bending moment along the length of the vessel. This is the bending moment diagram. This is a smooth curve, as shown in Figure 7.3. Positive bending moment values indicate sagging, negative values indicate hogging.

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274 • Ship Stability, Powering and Resistance 300 250 200

Shear force (tonnes)

150 100 50 0

0

20

10

30

40

50

60

70

80

90

100

–50

–100 –150 –200 –250 Position (m)

▲ Figure 7.2 Sample shear force diagram

5,000

Bending moment (tonne metres)

4,000

3,000

2,000

1,000

0

–1,000

0

10

20

30

40

50

60

70

80

90

100

Position (m FOAP)

▲ Figure 7.3 Sample bending moment diagram

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Calculating Hull Shear Force and Bending Moment • 275 QUESTION Q7.1 (MCM, ENG) A box barge is 60 m long with a beam of 10 m. The barge floats at a lightship draft of 0.524 m in salt water. The barge is separated into three equal size compartments, each 20 m long. The two end compartments are each loaded with 300 tonnes of cargo, evenly distributed along the length and beam of the compartment. Draw the load, shear force and bending moment diagrams. Load diagram 12.00 10.00 8.000

Load (t/m)

6.000 4.000 2.000 0.000 0

5

10

15

20

25

30

35

40

45

50

55

60

–2.000 –4.000 –6.000 Position (m)

Shear force (tonnes)

Shear force diagram 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 –10 0 –20 –30 –40 –50 –60 –70 –80 –90 –100 –110 –120 –130 –140 –150

5

10

15

20

25

30

35

40

45

50

55

60

Position (m)

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276 • Ship Stability, Powering and Resistance Bending moment diagram 0 0

5

10

15

20

25

30

35

40

45

50

55

60

Bending moment (t/m)

–200 –400 –600 –800 –1,000 –1,200 –1,400 –1,600 Position (m)

Note that the overall area under the load diagram must be zero (as the vessel is floating in equilibrium the overall vertical loads must be zero), the shear force value must also therefore start and finish at zero, and the peak bending moment value occurs at the point of minimum shear force with the steepest gradient. X The shear force and bending moment can also be determined directly using

calculus. For a demonstration of this, see Appendix 30: Direct Calculation of Shear Force and Bending Moment.

QUESTION Q7.2 (MCM, ENG) A box shaped vessel has a length of 100 m and a loaded displacement of 5,840 tonnes. In the lightship condition, the vessel floats on an even keel. The vessel is split by transverse bulkheads into five equal size compartments. Compartment 5 is loaded with 300 tonnes of cargo; compartment 4 is loaded with 700 tonnes of cargo; compartment 3 is loaded with 400 tonnes of cargo; compartment 2 is loaded with 200 tonnes of cargo; and compartment 1 is loaded with cargo so that the vessel finishes loading on an even keel. Construct the curves of load, shear force and bending moment, and hence determine the position, direction and magnitude of the maximum bending moment.

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Calculating Hull Shear Force and Bending Moment • 277 CALCULATING HULL SHEAR FORCE AND BENDING MOMENT  LEARNING CHECKLIST Objective

Level

Plot the shear force diagram for a simplified vessel

MCM, ENG

Plot the bending moment diagram for a simplified vessel

MCM, ENG

Analyse the shear force and bending moment diagrams for a simplified vessel

MCM, ENG

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8

CALCULATING HYDROSTATICS USING SIMPSON’S RULE AIMS AND OBJECTIVES

At the end of this section, you should be able to: Calculate the area under a graph Calculate the moment of area under a graph, relative to the x axis Calculate the centre of area under a graph, relative to the x axis Calculate the moment of area under a graph, relative to the y axis Calculate the centre of area under a graph, relative to the y axis Calculate the inertia of area under a graph, relative to the x axis Calculate the inertia of area under a graph, relative to the y axis Use the parallel axis theory to determine the inertia through the centre of the graph, parallel to the x axis Use the parallel axis theory to determine the inertia through the centre of the graph, parallel to the y axis Use Simpson’s Rule as appropriate with half stations Understand the importance of the hull lines plan

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Calculating Hydrostatics Using Simpson’s Rule • 279 Calculate the submerged volume and displacement of a vessel from the section area curve Calculate the submerged volume and displacement of a vessel from the draught waterplane area curve Calculate the KB of a vessel from the draught waterplane area curve Estimate KB using Morrish’s method Calculate the transverse inertia of the waterplane area from the waterline half beam curve Calculate GM using data from the lines plan data Calculate the waterplane area and the TPC from the lines plan data Calculate the free surface effect caused by a non-rectangular tank Calculate the GM for a multihull Determine the effect of sponsons on metacentric height Calculate the position of the LCB from the lines plan data Calculate the position of the LCF from the lines plan data Calculate the longitudinal inertia of the waterplane area from the waterline half beam curve

So far all the calculations have been based on box shaped vessels, or on given hydrostatic data. This significantly simplifies the calculation process. For real ships with curved hull forms, the hydrostatic values can be calculated from graphs derived from the hull shape. All of these calculations are based on Simpson’s Rule, which we have previously used to find the area under a graph. In addition to the calculation of area, Simpson’s Rule also allow the analysis of the following geometric properties of graphs: Area under a graph (integration, as seen previously) Moments of area of the area under a graph Centroids of area of the area under a graph Second moment of area (inertia) of the area under a graph

Finding the Moment of Area about the X Axis, and the Centre of Area from the X Axis (ENG) The next parameter to be found is the moment of area, measured relative to (or ‘about’) the x axis. This is a measure of the distribution of the area under the graph relative to (or

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280 • Ship Stability, Powering and Resistance ‘about’) the x axis, as shown in Figure 8.1. This can be used to find the centre, or centroid of the graph, as a distance from the x axis, also shown in Figure 8.1. 7 6 5

y

4 3 2

Centre of area from X axis

1 0 0

1

2

3

4 x

5

6

7

8

▲ Figure 8.1 Moment of area about the X axis and the centre of area from the X axis

Again, this is found using a tabular calculation. The first stage is to draw up a table, with columns headed Ordinate, Offset, Offset2, Simpson’s Multiplier and MomentX product.

Offset

Offset2

Simpson’s Multiplier

MomentX product

X1

Y1

(Y1)2

1

1(Y1)2

X2

Y2

(Y2)2

4

4(Y2)2

X3

Y3

(Y3)2

2

2(Y3)2

X4

Y4

(Y4)2

4

4(Y4)2

X5

Y5

(Y5)2

2

2(Y5)2

X6

Y6

(Y6)2

4

4(Y6)2

X7

Y7

(Y7)2

1

1(Y7)2

Total

Sum of above

Ordinate

As before, the ordinate column lists each ordinate. The offset column lists each offset measured at each ordinate, with the offset2 column listing the squared value. The Simpson’s Multiplier column is as before. The MomentX product column is the product of the offset2 and Simpson’s Multiplier columns. Once the MomentX product column

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Calculating Hydrostatics Using Simpson’s Rule • 281 has been completed, the summation of the column can be found. This can be used to determine the moment of area about the x axis using: Moment o of area about tthe X a axiis =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ ( Momen o t X product ) ⎝ ⎠ ⎝ 2⎠ 3

▲ Formula 8.1 Moment of area about the X axis

Once the moment of area about the x axis has been found, the centre of the graph from the x axis can be found using:

Centre of area from the X axis =

Moment of area about tthe X axis a Area u under de tthe graph

▲ Formula 8.2 Centre of area from the X axis

Note that the area under the graph can be found using the methods seen previously in Chapter 3, ‘Dynamic Stability (OOW, MCM, ENG)’.

QUESTION Q8.1 (ENG) A graph has the following coordinates: X

Y

0

5

2

6

4

5.25

6

4

8

2

Determine the first moment of area about the x axis, and hence determine the centre of area from the x axis.

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282 • Ship Stability, Powering and Resistance

Finding the Moment of Area about the Y Axis, and the Centre of Area from the Y Axis (ENG) The next parameter to be found is the moment of area, measured relative to (or ‘about’) the y axis, as shown in Figure 8.2. This is a measure of the distribution of the graph from the y axis, and can be used to find the centre, or centroid of the graph, as a distance from the y axis, also shown in Figure 8.2. 7 6 5

y

4 3

Centre of area from Y axis

2 1 0

0

1

2

3

4

5

6

7

8

x

▲ Figure 8.2 Moment of area about the Y axis and the centre of area from the Y axis

Again, this is found using a tabular calculation. The first stage is to draw up a table, with columns headed Ordinate, Offset, Lever, Simpson’s Multiplier and MomentY product: Ordinate

Offset

Lever

Simpson’s Multiplier

MomentY product

X1

Y1

0

1

0 × 1(Y1)

X2

Y2

1

4

1 × 4(Y2)

X3

Y3

2

2

2 × 2(Y3)

X4

Y4

3

4

3 × 4(Y4)

X5

Y5

4

2

4 × 2(Y5)

X6

Y6

5

4

5 × 4(Y6)

X7

Y7

6

1

6 × 1(Y7)

Total

Sum of above

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Calculating Hydrostatics Using Simpson’s Rule • 283 As before, the ordinate column lists each ordinate and the offset column lists each offset measured at each ordinate. The lever column lists the number of ‘spacings’ of the ordinate from the y axis. This is the same as the ordinate. The Simpson’s Multiplier column is as before. The MomentY product column is the product of the offset, lever and Simpson’s Multiplier columns. Once the MomentY product column has been completed, the summation of the column can be found. This can be used to determine the moment of area about the y axis using: Moment o of area ⎛ Spacing ⎞ = × Spacing about the t Ya axiis ⎝ ⎠ 3

e tY ( Momen

product )

▲ Formula 8.3 Moment of area about the Y axis

Once the moment of area about the y axis has been found, the centre of the graph from the y axis can be found using:

Centre of area from the Y axis =

Moment of area about tthe Y axis a Area u under de tthe graph

▲ Formula 8.4 Centre of area from the Y Axis

QUESTION Q8.2 (ENG) A graph has the following coordinates: X

Y

0

5

2

6

4

5.25

6

4

8

2

Determine the first moment of area about the y axis, and hence determine the centre of area from the y axis.

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284 • Ship Stability, Powering and Resistance

Finding the Second Moment of Area, or Inertia, Relative to the X Axis (ENG) The concept of second moment of area, or inertia, has already been discussed in the chapter explaining the transverse metacentre. Simpson’s Rule can be used to determine the inertia of the area under the graph relative to the x axis of the graph.

Offset

Offset3

Simpson’s Multiplier

InertiaX product

X1

Y1

(Y1)3

1

1(Y1)3

X2

Y2

(Y2)3

4

4(Y2)3

X3

Y3

(Y3)3

2

2(Y3)3

X4

Y4

(Y4)3

4

4(Y4)3

X5

Y5

(Y5)3

2

2(Y5)3

X6

Y6

(Y6)3

4

4(Y6)3

X7

Y7

(Y7)3

1

1(Y7)3

Total

Sum of above

Ordinate

The first stage is to draw up a table, with columns headed Ordinate, Offset, Offset3, Simpson’s Multiplier and InertiaX product: As before, the ordinate column lists each ordinate. The offset column lists each offset measured at each ordinate, with the offset3 column listing the cubed value. The Simpson’s Multiplier column is as before. The InertiaX product column is the product of the offset3 and Simpson’s Multiplier columns. Once the InertiaX product column has been completed, the summation of the column can be found. This can be used to determine the inertia of area about the x axis using:

Inertia about the X axis =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ ( Inertia X product ) ⎝ ⎠ ⎝ 3⎠ 3

▲ Formula 8.5 Inertia about the X axis

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Calculating Hydrostatics Using Simpson’s Rule • 285

QUESTION Q8.3 (ENG) A graph has the following coordinates: X

Y

0

5

2

6

4

5.25

6

4

8

2

Determine the inertia about the x axis.

Finding the Second Moment of Area, or Inertia, Relative to the Y Axis (ENG) Simpson’s Rule can be used to determine the inertia of the area under the graph relative to the y axis of the graph. The first stage is to draw up a table, with columns headed Ordinate, Offset, Lever2, Simpson’s Multiplier and InertiaY product. Offset

Lever2

Simpson’s Multiplier

InertiaY product

X1

Y1

02

1

02 × 1(Y1)

X2

Y2

12

4

12 × 4(Y2)

X3

Y3

22

2

22 × 2(Y3)

X4

Y4

32

4

32 × 4(Y4)

X5

Y5

42

2

42 × 2(Y5)

X6

Y6

52

4

52 × 4(Y6)

X7

Y7

62

1

62 × 1(Y7)

Total

Sum of above

Ordinate

As before, the ordinate column lists each ordinate and the offset column lists each offset measured at each ordinate. The lever2 column lists the number of ‘spacings’ squared of

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286 • Ship Stability, Powering and Resistance the ordinate from the y axis. This is the same as the ordinate squared, and the Simpson’s Multiplier column is as before. The InertiaY product column is the product of the offset, lever2 and Simpson’s Multiplier columns. Once the InertiaY product column has been completed, the summation of the column can be found. This can be used to determine the inertia of area about the y axis using:

Inertia about the Y axis =

⎛ Spacing ⎞ × Spacing 2 ⎝ ⎠ 3

(InertiaY

product )

▲ Formula 8.6 Inertia about the Y axis

QUESTION Q8.4 (ENG) A graph has the following coordinates: X

Y

0

5

2

6

4

5.25

6

4

8

2

Determine the inertia about the y axis.

Graphs and Parallel Axes Theory (ENG) As discussed in the section on bilging a side compartment, the inertia of a shape varies depending on where it is measured through. So far we have measured the inertia of the area under the graph about the x and y axes. The values of inertia that we have calculated about the x and y axes have been the InertiaREMOTE values, as they are not measured through an axis running through the centroid of the shape. The parallel axes theory allows us to calculate the inertia through axes running through the centre of the

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Calculating Hydrostatics Using Simpson’s Rule • 287

y InertiaY remote

shape (as shown in Figure 8.3) if we know the remote inertia. This will be needed for further calculations later. 7 InertiaY centroid

6 5 4

Inertiax centroid

3 2

Centre of area from Y axis

1 0

0

1

2

Centre of area from X axis

3

4

5 x

6 7 8 Inertiax remote

9

▲ Figure 8.3 Remote and centroid inertia through the X and Y axes and axes through the centre parallel to the X and Y axes

QUESTION Q8.5 (ENG) For the graph formed by the coordinates in Question 8.4, the area is 38.333 units2, the centre of the graph is 2.540 m from the x axis and 3.513 m from the y axis. The inertia measured about the x axis is 342.758 units4, and the inertia measured about the y axis is 645.34 units4. Determine the Inertia measured through an axis parallel to the X axis, but running through the centre of area of the graph, and determine the inertia measured through an axis parallel to the Y axis, but running through the centre of area of the graph.

These techniques are all required to determine the hydrostatics from graphs which are derived from the shape of the hull. This will be covered in more detail later.

QUESTION Q8.6 (ENG) A graph has the following coordinates:

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288 • Ship Stability, Powering and Resistance

X (m)

Y (m)

0

0

4

8

8

15

12

20

16

17

20

10

24

0

Using Simpson’s Rule, determine: The area under the graph The first moment of area relative to the x axis The centre of area relative to the x axis The first moment of area relative to the y axis The centre of area relative to the y axis The inertia relative to the x axis, measured through the x axis The inertia relative to the y axis, measured through the y axis The inertia relative to the x axis, measured through the centre of the graph The inertia relative to the y axis, measured through the centre of the graph

Half Stations (ENG) Simpson’s Rule are accurate enough for normal use in Naval Architecture. However, the accuracy can be further improved by the use of ‘half stations’. As seen previously in the section on dynamic stability, the accuracy of Simpson’s Rule depends on the curvature of the graph and the number of ordinates. Increasing the curvature of the graph decreases accuracy, and increasing the number of ordinates increases accuracy. However, increasing the number of ordinates improves the accuracy, but also significantly increases the time it takes to complete the calculation. For virtually all the graphs used in the calculation of hydrostatics, there is more curvature at the ends of the graph than in the middle. To improve the accuracy of the calculation, additional ordinates are used at the start and end of the curve. These are known as ‘half stations’. These half stations are placed exactly halfway between the first set of

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Calculating Hydrostatics Using Simpson’s Rule • 289 ordinates, and halfway between the last set of ordinates. It is not necessary to have half stations at each end, they can be used either at the start or finish of the curve. When using half stations, the tabular calculations for half stations are very similar to the normal calculations – the only difference is the Simpson’s Multiplier pattern of numbers. If a half station is used, the Simpson’s Multiplier column for the ordinates at that end of the curve start with 0.5, 2, 1.5, and then alternate 4, 2, 4, 2 in between, and finishing with a 1.5, 2, 0.5 if a half station is used at the end of the curve.

QUESTIONS Q8.7 (ENG) A graph has the following coordinates, including half stations: X (m)

Y (m)

0

5

1

5.6

2

6

4

5.25

6

4

7

3

8

2

Determine (including using the half stations) the area of the graph. Q8.8 (ENG) A graph has the following coordinates, including half stations: X (m)

Y (m)

0

0

2

4

4

8

8

15

12

20

16

17

20

10

22

5

24

0

Determine (including using the half stations) the area of the graph.

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290 • Ship Stability, Powering and Resistance Q8.9 (ENG) A graph has the following coordinates: X (m)

Y (m)

0

0

2

1

4

2

8

3

12

4

16

5

20

5

24

4

28

2

30

1

32

0

Using Simpson’s Rule, determine: The area under the graph The first moment of area relative to the x axis The centre of area relative to the x axis The first moment of area relative to the y axis The centre of area relative to the y axis The inertia relative to the x axis, measured through the x axis The inertia relative to the y axis, measured through the y axis The inertia relative to the x axis, measured through the centre of the graph The inertia relative to the y axis, measured through the centre of the graph Q8.10 (ENG) A graph has the following coordinates: X (m)

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Y (m)

0

4

3

5

6

6

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Calculating Hydrostatics Using Simpson’s Rule • 291

X (m)

Y (m)

12

7

18

7

24

6

30

5

36

4

42

3

45

2

48

0

Using Simpson’s Rule, determine: The area under the graph The first moment of area relative to the x axis The centre of area relative to the x axis The first moment of area relative to the y axis The centre of area relative to the y axis The inertia relative to the x axis, measured through the x axis The inertia relative to the y axis, measured through the y axis The inertia relative to the x axis, measured through the centre of the graph The inertia relative to the y axis, measured through the centre of the graph

The Lines Plan (ENG) Most of the information required to directly calculate the hydrostatics can be determined from a plan of the hull known as the lines plan. This shows the shape of the hull from three viewpoints (ahead, above and from the side), using a series of contours to show the shape. The view from the bow or stern is called the body plan or section plan, as shown in Figure 8.4. The contours on this view, which effectively show the shape of transverse slices across the hull, are called the stations or sections. These are normally numbered, with zero at the aft perpendicular, and increase in number going forward – note that

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292 • Ship Stability, Powering and Resistance

▲ Figure 8.4 Body plan view

some sign conventions number the station at the forward perpendicular as zero, and then increase the station number going aft. The view from above is called the plan view, as shown in Figure 8.5. The contours on this view, which show slices through the hull parallel to the waterline, are called waterlines. These show the shape of the waterplane area at a range of draughts.

▲ Figure 8.5 Plan view

The view from the side is called the profile view, as shown in Figure 8.6. The contours on this view, which show slices through the hull parallel to the centreline, are called buttock lines.

▲ Figure 8.6 Profile view

By using each of the three views, it is possible to determine the coordinates of the hull surface at any point on the hull. More usefully from a calculation point of view, it is possible to determine the shape of the hull for calculations.

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Calculating Hydrostatics Using Simpson’s Rule • 293

Calculating the Underwater Volume and Displacement from the Lines Plan (ENG) To determine the underwater volume, a graph known as the Section Area Curve (or SAC) is required. This is a graph which shows waterline length along the x axis, and the underwater cross-section area up the y axis. The underwater cross-section area can be determined from analysing the underwater cross-section of the vessel at a number of points along the vessel (which can be found on the body plan view on the lines plan) using Simpson’s Rule. An example of the SAC for the hull shown in the previous diagrams is shown in Figure 8.7. 4 Immersed section area (m2)

3.5 3 2.5 2 1.5 1 0.5 0

0

2

4

6

8

10

12

14

16

18

20

Length along the waterline (m FOAP)

▲ Figure 8.7 Example section area curve

The area of the SAC (which can be determined using Simpson’s Rule) at any given draught is equal to the underwater volume of the vessel at that draught. To determine the displacement, the underwater volume needs to be multiplied by the density of the water which the vessel is floating in.

QUESTIONS Q8.11 (ENG) A ship’s lifeboat, which has a waterline length of 9.414 m, has the following underwater cross-section area at each station when floating on her designed waterline:

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294 • Ship Stability, Powering and Resistance

Station

Immersed section area (m2)

0 (AP)

0

1

0.115

2

0.392

3

0.702

4

0.933

5

1.008

6

0.904

7

0.660

8

0.353

9

0.097

10 (FP)

0

Using the data, determine the design displacement of the lifeboat in sea water. Q8.12 (ENG) A ship with a waterline length of 120 m has the following values for immersed underwater section areas when floating at her summer waterline: Position (m FOAP)

Immersed area (m2)

0

0.0

12

25.6

24

65.9

36

79.4

48

80.0

60

80.0

72

80.0

84

79.4

96

74.2

108

42.9

120

0.0

Using the data, determine the underwater volume at the summer draught, and the summer displacement in sea water.

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Calculating Hydrostatics Using Simpson’s Rule • 295

Finding KB (ENG) To determine the KB value, a graph known as the draught waterplane curve is required. This is a graph which shows the draught of the vessel along the x axis, and the waterplane area up the y axis, as shown in Figure 8.8 (occasionally the axes may be switched). The waterplane area can be determined from Simpson’s Rule and the waterplanes of the vessel, which can be found on the plan view on the lines plan. 70

Waterplane area (m2)

60 50 40 30 20

1

1.05

0.9

0.95

0.8

0.85

0.7

0.75

0.6

0.65

0.5

0.55

0.4

0.45

0.3

0.35

0.2

0.25

0.1

0.15

0

0

0.05

10

Draught (m)

▲ Figure 8.8 Draught waterplane area curve

As with the SAC, the area under the draught waterplane curve (when plotted with the draught along the x axis) gives the underwater volume of the vessel. When plotted with the draught along the x axis, the distance from the y axis to the centre of the area under the graph is equal to the KB value. Note that if the axes are swapped, so that the draught is plotted up the y axis, then the area between the curve and the y axis gives the underwater volume, and the distance from the x axis to the centre of area of the graph between the curve and the y axis gives the KB value.

QUESTIONS Q8.13 (ENG) A ship’s lifeboat has the following waterplane areas at each draught, when loaded to a draught of 0.625 m:

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296 • Ship Stability, Powering and Resistance

Draught (m)

Waterplane area (m2)

0

0

0.063

1.797

0.125

3.544

0.188

5.199

0.250

6.754

0.313

8.208

0.375

9.559

0.438

10.807

0.500

11.951

0.563

12.989

0.625

13.92

Using the data, and assuming that the draught is plotted along the x axis, determine the underwater volume of the lifeboat and the KB value when at a draught of 0.625 m. Q8.14 (ENG) A vessel with an underwater volume of 24,693.12 m3 has the following waterplane areas at a range of draughts up to the summer draught: Draught (m)

Waterplane area (m2)

0.00

0.0

0.90

1,807.2

1.80

2,620.8

2.70

2,808.0

3.60

2,880.0

4.50

2,952.0

5.40

3,024.0

6.30

3,096.0

7.20

3,168.0

8.10

3,240.0

9.00

3,312.0

Using the data, determine the KB at the summer draught.

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Calculating Hydrostatics Using Simpson’s Rule • 297

Estimating KB Using Morrish’s Method (ENG) The KB can be estimated by using Morrish’s method, which gives an estimation of KB:

KB =

⎞⎞ 1 ⎛ 5D ⎛ ∇ × − 3 ⎜⎝ 2 ⎜⎝ Waterplane area ⎟⎠ ⎟⎠

▲ Formula 8.7 Morrish’s formula to approximate KB

QUESTION Q8.15 (ENG) A ship has a draught of 7.00 m and a displacement in sea water of 7,000.00 tonnes. The waterplane area of the vessel is 1,500.00 m2. Using Morrish’s method, calculate KB.

Transverse Inertia of the Waterplane and BM (ENG) The waterplane shapes on the plan view of the lines plans can also be used to determine the transverse inertia of the waterplane, and hence BM. To do this, we need to create a graph of the waterline half beams at each station. This is the transverse distance from the centreline to the waterline at each station, plotted on a graph on the y axis, with the length of the vessel plotted on the x axis, as shown in Figure 8.9. This is known as the curve of waterline half beams. The inertia of the area under the waterline half beam curve about the x axis is half of the total transverse inertia of the waterplane.

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298 • Ship Stability, Powering and Resistance

Waterline half beam (m)

3 2.5 2 1.5 1 0.5 0

0

2

4

6

8

10

12

14

16

18

20

Length along the waterline (m FOAP)

▲ Figure 8.9 Waterline half beam curve

QUESTIONS Q8.16 (ENG) A ship’s lifeboat has a waterline length of 9.414 m and an underwater volume of 4.904 m3 at her design draught. At that draught, the vessel has the following waterline half beams at each station: Ordinate

Half beam (m)

0 (AP)

0

1

0.366

2

0.712

3

0.976

4

1.129

5

1.179

6

1.108

7

0.935

8

0.661

9

0.305

10 (FP)

0

Determine the BM of the vessel in the above condition.

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Calculating Hydrostatics Using Simpson’s Rule • 299 Q8.17 (ENG) A vessel with an underwater volume of 48,228.75 m3 and a waterline length of 225 m has the following waterline half beams: Position (m FOAP)

Waterplane half beam (m)

0

0.00

23

9.45

45

12.90

68

13.80

90

13.80

113

13.80

135

13.80

158

13.80

180

12.75

203

8.85

225

0.00

Using the above data, determine the BM.

Calculating GM Using the Lines Plan Data The processes seen in the previous sections can be combined to determine the GM of a vessel if KG is known, and if the data from the lines plan is available.

QUESTIONS Q8.18 (ENG) A vessel with a waterline length of 150 m and a KG of 7.07 m has the following values for immersed underwater section areas when floating at her summer waterline:

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300 • Ship Stability, Powering and Resistance

Position (m FOAP)

Immersed area (m2)

0

0.0

15

40.0

30

103.0

45

124.0

60

125.0

75

125.0

90

125.0

105

124.0

120

116.0

135

67.0

150

0.0

Her waterplane areas at a range of draughts up to the summer draught are as follows:

Draught (m)

Waterplane area (m2)

0.00

0.0

0.75

1,255.0

1.50

1,820.0

2.25

1,950.0

3.00

2,000.0

3.75

2,050.0

4.50

2,100.0

5.25

2,150.0

6.00

2,200.0

6.75

2,250.0

7.50

2,300.0

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Calculating Hydrostatics Using Simpson’s Rule • 301 At the summer draught, she has the following waterline half beams. Position (m FOAP)

Waterplane half beam (m)

0

0.00

15

6.30

30

8.60

45

9.20

60

9.20

75

9.20

90

9.20

105

9.20

120

8.50

135

5.90

150

0.00

Using the above data, determine the following: Underwater volume at the summer draught Summer displacement in sea water KB at the summer draught BM at the summer draught GM at the summer draught with the KG given Q8.19 (ENG) A vessel with a waterline length of 180 m and a KG of 9.55 m has the following values for immersed underwater section areas when floating at her summer waterline: Position (m FOAP)

Immersed area (m2)

0

0.0

18

57.6

36

148.3

54

178.6

72

180.0

90

180.0

108

180.0

126

178.6

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302 • Ship Stability, Powering and Resistance

Position (m FOAP)

Immersed area (m2)

144

167.0

162

96.5

180

0.0

Her waterplane areas at a range of draughts up to the summer draught are as follows: Draught (m)

Waterplane area (m2)

0.00

0.0

0.90

1,807.2

1.80

2,620.8

2.70

2,808.0

3.60

2,880.0

4.50

2,952.0

5.40

3,024.0

6.30

3,096.0

7.20

3,168.0

8.10

3,240.0

9.00

3,312.0

At the summer draught, she has the following waterline half beams: Position (m FOAP)

Waterplane half beam (m)

0

0.00

18

7.56

36

10.32

54

11.04

72

11.04

90

11.04

108

11.04

126

11.04

144

10.20

162

7.08

180

0.00

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Calculating Hydrostatics Using Simpson’s Rule • 303 Using the above data, determine the following: Underwater volume at the summer draught Summer displacement in sea water KB at the summer draught BM at the summer draught GM at the summer draught with the KG given

Finding the Tonne per Centimetre Immersion Value (ENG) Previously in the section on the TPC, we saw that: TPC

CW × L B × 0 01× ρ

In the section on form coefficients, Formula 1.3 can be transposed so that: Waterplane area = CW

L×B

Combining these two equations gives: TPC W Waterrplane area × 0 01× ρ ▲ Formula 8.8 TPC and waterplane area

If the lines plan data is available, then the waterplane area can be calculated using Simpson’s Rule. To do this, the waterline half beam curve is used (as shown in Figure 8.9). The area under the waterline half beam curve gives half of the waterplane area, so the overall waterplane area can be found by determining the area under the curve with Simpson’s Rule, and doubling the value.

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304 • Ship Stability, Powering and Resistance

QUESTIONS Q8.20 (ENG) A ship’s lifeboat, with a waterline length of 9.414 m, has the following values for the waterline half beam at each station along the vessel: Ordinate

Half beam (m)

0 (AP)

0

1

0.366

2

0.712

3

0.976

4

1.129

5

1.179

6

1.108

7

0.935

8

0.661

9

0.305

10 (FP)

0

Determine the TPC of the vessel. Q8.21 (ENG) A vessel with a LBP of 150 m has the following waterline half beam values: Position (m FOAP)

Waterplane half beam (m)

0

0.00

15

6.30

30

8.60

45

9.20

60

9.20

75

9.20

90

9.20

105

9.20

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Calculating Hydrostatics Using Simpson’s Rule • 305

Position (m FOAP)

Waterplane half beam (m)

120

8.50

135

5.90

150

0.00

Determine the TPC of the vessel in sea water.

Calculating the Free Surface Moment in a Non-Rectangular Tank (ENG) Previously, we have covered the basic concept of free surface effects, and calculated the effects of free surface moment using the ship’s hydrostatic data or by directly calculating the free surface moment for a rectangular free surface. As most spaces for fluids are built into the shape of the hull, we need to be able to determine the inertia of irregular shapes. Normally, the free surfaces in these tanks will have three straight sides formed by transverse and longitudinal bulkheads, and one curved side formed by the hull surface. The method for calculating the fluid stability is the same as seen previously, but when we calculate inertia of the free surface needs to be calculated differently. This requires a set of Simpson’s Rule calculations and the parallel axes theory formulae. The first stage is to plot the shape of the free surface on an x–y graph. The longitudinal edge of the tank should run along the x axis, the aft transverse bulkhead of the tank should run up the y axis. The curved edge of the free surface forms the curve of the graph, and the forward transverse bulkhead forms the left hand edge of the graph. To determine the free surface moment, first the area has to be determined using Simpson’s Rule. Then the first moment of area about the x axis must be found using Simpson’s Rule, which when divided by the area allows the centre of the free surface from the x axis to be found. The inertia about the x axis can then be found, again using Simpson’s Rule. Finally, the parallel axes theory can be used to convert the inertia value from the x axis to the centre axis. This value is the free surface moment in units of metres4, and can then be used in subsequent stability calculations.

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306 • Ship Stability, Powering and Resistance

QUESTIONS Q8.22 (ENG) A tank on a ship is formed by a longitudinal bulkhead, two transverse bulkheads and the hull plating. The tank has a planform (in units of metres) as shown below:

Edge formed by aft transverse bulkhead

Edge formed by forward transverse bulkhead 1.90

2.60

1.10

Edge formed by longitudinal bulkhead

1.25

1.25

1.25

0.10

3.00

Edge formed by the ship’s hull

1.25

5.00

Determine the transverse free surface moment of the tank in units of metres4. Q8.23 (ENG)

1.82

2.09

2.32

A ship has a displacement of 5,000 tonnes. In this condition, she has a KM of 8.00 m (which can be assumed to remain constant) and a KG of 7.00 m. She is upright in sea water. The vessel has a 10.00 m long starboard wing tank (which can be assumed to be a constant planform), formed by the side of the hull, two transverse bulkheads, and a longitudinal bulkhead 7.00 m to starboard of the centreline, as shown (all dimensions are in metres):

1.50

2.50

Hull side

Aft transverse bulkhead 7.00

10.00

Forward transverse bulkhead

Longitudinal bulkhead

Ship centreline

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Calculating Hydrostatics Using Simpson’s Rule • 307 The base of the tank is 2.00 m above the keel. Water ballast is pumped into the tank until the depth of water is 3.00 m. Determine the resulting angle of list.

Multihulls (ENG) In the past few years, large multihull vessels have started to become more common. Due to their beam, multihulls have greater initial stability than similar length mono-hulls and a greater working deck space. They tend to have lower hydrodynamic resistance than mono-hulls, and so require less fuel and power, or can operate at higher speeds. Multihulls have greater initial stability because the inertia of the waterplane is much large than a mono-hull. This increases BM and hence GM. Before looking at multihull vessels in detail, it might be beneficial to revise inertia and the parallel axes theory. When the inertia is measured through an axis running through the centre of a shape, it is referred to as InertiaCENTROID. When the inertia is measured through an axis running through a point away from the centre of a shape, it is referred to as InertiaREMOTE. As seen previously, the InertiaCENTROID and InertiaREMOTE are linked by the parallel axes theory given in Formula 6.8:

InertiaRemote

InertiaCentroid + ( Area Distance 2 )

In this formula, the area is the area of the shape under consideration, and the distance is the distance between the axis through the centre of the shape and the axis through the remote point. The formula can be used to determine the inertia about a remote point if the inertia through the centre is known.

QUESTIONS Q8.24 (ENG) A waterplane has an area of 800 m2. The transverse inertia measured through the centreline of the shape is 8,000 m4. Determine the inertia measured through an axis parallel to the centreline but 10 m to port of the centreline.

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308 • Ship Stability, Powering and Resistance Q8.25 (ENG) A rectangular waterplane has dimensions, as shown below. Calculate the area of the waterplane, the transverse inertia through the centre of the waterplane (InertiaCENTROID) and the remote inertia measured through the axis XX, parallel to the centreline of the vessel, but 30 m from the centreline.

60 m 20 m

30 m

X

X

The complication with multihulls is that the vessel has two waterplanes making up the total waterplane area. Each waterplane rolls not around the centreline of the hull, as a mono-hull would, but around the centreline of the vessel, which will be halfway between the centrelines of the two hulls. To determine the stability of multihulls, we find GM in the same way to a mono-hull, however we have two smaller hulls to deal with. Each hull can be analysed in isolation, using the techniques already show, to determine the underwater volume, KB and transverse inertia of the waterplane about the hull centreline. The overall volume of the vessel will be (assuming that the hulls are identical) double the volume of one hull. Assuming that the hulls are identical, the KB and LCB of the combined hulls will be the same as one single hull. The BM, however, is more complicated. For multihulls, the inertia in the BM formula is the overall transverse inertia of the two waterplanes (one for each hull) measured at the overall centreline of the ship, not the centreline of the single hull. The first stage in determining the BM of a multihull is to analyse a single hull in isolation. The area of the waterplane area of one hull, and the transverse inertia of the waterplane area of the hull, measured about the centreline of the hull (an InertiaCENTROID value), can be found as it has for mono-hulls previously. This InertiaCENTROID value can then be used, along with the waterplane area and the distance from the hull centreline to the ship centreline, in the parallel axes formula to determine the inertia of the waterplane of a single hull, about the centreline of the overall vessel (an inertiaREMOTE value). Doubling

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Calculating Hydrostatics Using Simpson’s Rule • 309 this value gives the overall transverse inertia of the complete waterplane measured about the centreline of the ship. This can be divided by the total underwater volume of the vessel to get the BM value. This can then be used with KB and KG as before to determine GM.

QUESTIONS Q8.26 (ENG) A catamaran is constructed from two identical box shaped hulls. Each hull is 40 m long and 4 m wide. The draught of each hull is 3 m and the overall KG of the vessel is 5 m. Determine the GM of the vessel if the distance from the centreline of each hull to the overall centreline of the vessel is 6 m. Q8.27 (ENG) A catamaran has a waterline length of 60 m and a hull centreline to hull centreline distance of 10 m. The KB is 1 m, the KG is 6 m and the total underwater volume is 530 m3. The vessel has the following waterline half beams at equally spaced stations along the hull: Station

Waterline half beam (m)

0

0.1

1

1.0

2

1.7

3

2.0

4

1.5

5

0.7

6

0.0

Determine GM of the vessel.

Sponsons (ENG) The stability of some vessels, particularly Ro-Ro vessels, can be improved by the addition of sponsons. These are additional sections of watertight volume, placed on

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310 • Ship Stability, Powering and Resistance the waterline of the vessel, to increase the beam of the vessel on the waterline without a significantly large increase in the underwater volume. As BM is related to beam3, even a small increase in the beam with a small increase in underwater volume can cause a large increase in BM, and therefore also in GM. Sponsons are often fitted during the lifetime of the ship, rather than during the initial build, to bring the vessel up to more stringent requirements as legislation gets tougher. To determine the effect of adding sponsons to the vessel, the shape of the waterplane area of the sponsons must be known. Using methods seen previously, the waterplane area of each sponson can be determined, along with the centre of the waterplane area as a distance from the centreline of the ship. Finally, the transverse inertia of the sponson’s waterplane area can be found. As the vessel rolls about the ship’s centreline, the parallel axes formula used previously must be used to determine the transverse inertia of the sponson’s waterplane area about the centreline of the ship. This can be added to the transverse inertia of the main waterplane to get the overall transverse inertia of the ship and sponsons.

QUESTION Q8.28 (ENG)

15 m 50 m

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1.5 m

10 m

A box shaped vessel has a main hull with a length of 50.00 m, a beam of 10.00 m, and floats upright on an even keel with a draught of 2.00 m, with a KG of 5.00 m. Rectangular sponsons with a length of 15.00 m and a beam of 1.50 m are to be added to the vessel to improve the stability, as shown below.

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Calculating Hydrostatics Using Simpson’s Rule • 311 Assuming that the draught and KG remain constant after adding the sponsons, determine the GM after adding the sponsons.

One of the drawbacks of sponsons is that they increase the hydrodynamic drag of the vessel, and hence increase the required power to maintain the service speed of the vessel (and also therefore increase the fuel consumption). To minimise this, the sponsons must be streamlined in shape. This requires Simpson’s Rule to analyse the shape of the sponson, using the x axis as the fore and aft straight side of the sponson attached to the hull surface, and the y axis as the local beam of the sponson measured from the hull surface, as shown in Figure 8.10.

Port sponson waterplane y

x Main hull waterplane

Stb sponson waterplane

▲ Figure 8.10 Sponson notation

This graph can be used to determine the area of the sponson waterplane, the moment of area about the x axis (and hence the centre of the sponson’s waterplane from the x axis) and the inertia of the sponson’s waterplane about the x axis. The parallel axis theory can then be used to determine the inertia of the sponson’s waterplane area about an axis parallel to the x axis but running through the centreline of the sponson’s waterplane. Finally, the parallel axes theory can be used again to determine the inertia of the sponson’s waterplane area about the centreline of the ship, which itself can be used to determine the overall waterplane inertia of the vessel, and hence BM.

QUESTIONS Q8.29 (ENG) A ship is to be retro-fitted at amidships with 4.00 m long sponsons to improve her stability. In the original condition, she has a displacement of 645.75 tonnes, with a transverse inertia of the waterplane area, measured about the centreline of the ship, of 857.43 m4. The beam of the vessel at amidships is 7.00 m. The sponsons, which are 2.10 m deep and can be considered to be a constant planform in shape, have the following beams, measured as a distance forward of the aft end of the sponsons:

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312 • Ship Stability, Powering and Resistance

Position forward of the aft end of the sponson (m)

Beam of the sponson (m)

0.00

0.00

1.00

1.00

2.00

1.50

3.00

1.00

4.00

0.00

After fitting the sponsons, the overall draught of the vessel remains constant, and the waterline lies at exactly half of the depth of the sponson. Determine the BM of the vessel after fitting the sponsons. Q8.30 (ENG) A ship is to be retro-fitted at amidships with 16.00 m long sponsons to improve her stability. In the original condition, she has a displacement of 3,000 tonnes, with a BM of 11.39 m. The beam of the vessel at amidships is 20.00 m. The sponsons, which are 4.00 m deep and can be considered to be a constant planform in shape, have the following beams, measured as a distance forward of the aft end of the sponsons: Position forward of the aft end of the sponson (m)

Beam of the sponson (m)

0.00

0.00

1.00

2.00

2.00

3.00

3.00

2.00

4.00

0.00

After fitting the sponsons, the overall draught of the vessel remains constant, and the waterline lies at exactly half of the depth of the sponson. Determine the BM of the vessel after fitting the sponsons.

The Longitudinal Centre of Buoyancy (ENG) To determine the position of the LCB, the SAC (see Figure 8.7) is again used. The centre of the area under the curve from the y axis represents the distance from the

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Calculating Hydrostatics Using Simpson’s Rule • 313 aft perpendicular to the centre of the underwater volume, which is the LCB value. Therefore, to determine the LCB, the moment of area about the Y axis must be found, and divided by the area under the graph (which is also equal to the underwater volume of the ship).

QUESTION Q8.31 (ENG) A ship has an LBP of 60 m. At the load draught in sea water, she has the following immersed section areas: Station 0

Section area (m2) 1.57

0.5

25.13

1

56.55

2

76.97

3

76.97

4

56.55

5

25.13

5.5

6.28

6

0.00

Using the above information, determine the underwater volume and the LCB.

The Longitudinal Centre of Flotation (ENG) The waterplane shapes on the plan view of the lines plans can be used to determine the longitudinal centre of flotation of the waterplane. To do this, the curve of waterline half beams (as shown in Figure 8.9) is used. The LCF is the longitudinal centroid of the area under the waterline half beam curve. This value can be found from the moment of area about the y axis of the area under the waterline half beam curve.

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314 • Ship Stability, Powering and Resistance QUESTIONS Q8.32 (ENG) A vessel has a waterline length of 120 m. She has the following waterline half beams at equally spaced stations: Station

Waterline half beam (m)

0

0

1

6

2

12

3

12

4

12

5

3

6

0

By analysing the curve of waterline half beams, calculate the waterplane area, the first moment of area of the curve about the y axis, and hence the LCF. Q8.33 (ENG) A vessel has a waterline length of 90 m. She has the following waterline half beams: Station

Waterline half beam (m)

0

0

0.5

4

1

8

2

10

3

10

4

9

5

6

5.5

3

6

0

By analysing the curve of waterline half beams, calculate the waterplane area, the first moment of area of the curve about the y axis, and hence the LCF.

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Calculating Hydrostatics Using Simpson’s Rule • 315

Longitudinal Inertia of the Waterplane and BML (ENG) The waterplane shapes on the plan view of the lines plans can also be used to determine the longitudinal inertia of the waterplane, and hence BML. To do this, the curve of waterline half beams (as shown in Figure 8.9) is again used. The second moment of area of the waterplane about the y axis can be found using Simpson’s Rule. The parallel axes formula can then be used to convert the value from about the y axis (a remote value) to be about the LCF (the centre of the waterplane area, and hence the centroid value).

QUESTIONS Q8.34 (ENG) A ship has the following waterline half beams at equally spaced stations: Station

Waterline half beam (m)

0

0

1

10

2

20

3

20

4

16

5

8

6

0

If the vessel has an LBP of 180 m, an underwater of 30,000 m3, a waterplane area of 4,480 m2 and an LCF of 86.79 m FOAP, determine the longitudinal inertia of the waterplane about a transverse axis running through the AP and the LCF, and hence BML. Q8.35 (ENG) A ship has an LBP of 90 m. At the load draught in sea water, she has the following waterplane half beams and immersed section areas:

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316 • Ship Stability, Powering and Resistance

Station

Waterline half beam (m) Section area (m2)

0

1.00

1.57

0.5

3.00

14.14

1

5.00

39.27

2

7.00

76.97

3

8.00

100.53

4

7.00

76.97

5

5.00

39.27

5.5

3.00

14.14

6

0.00

0.00

Using the above information, determine the underwater volume and displacement, the LCB, LCF, BM and BML. Q8.36 (ENG) A ship has an LBP of 120 m. At the load draught in sea water, she has the following waterplane half beams and immersed section areas: Station

Waterline half beam (m) Section area (m2)

0

0.00

0.00

0.5

4.00

25.13

1

7.00

76.97

2

10.00

157.08

3

11.00

190.07

4

10.00

157.08

5

6.00

56.55

5.5

3.00

14.14

6

0.00

0.00

Using the above information, determine the underwater volume and displacement, the LCB, LCF, BM and BML. Q8.37 (ENG) A ship has an LBP of 105 m. At the load draught of 9.00 m in sea water, she has the following waterplane half beams and immersed section areas:

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Calculating Hydrostatics Using Simpson’s Rule • 317

Waterline half beam (m)

Section area (m2)

0

3.00

14.14

0.5

6.00

56.55

1

8.00

100.53

2

9.00

127.23

3

9.00

127.23

4

8.00

100.53

5

5.00

39.27

5.5

3.00

14.14

6

0.00

0.00

Station

Using the above information, and Morrish’s method for KB, determine the metacentric height of the vessel if KG is 7.00 m, and the trim if the LCG is 45.00 m FOAP.

CALCULATING HYDROSTATICS USING SIMPSON’S RULE – LEARNING CHECKLIST Objective

Level

Calculate the area under a graph

ENG

Calculate the moment of area under a graph, relative to the x axis

ENG

Calculate the centre of area under a graph, relative to the x axis

ENG

Calculate the moment of area under a graph, relative to the y axis

ENG

Calculate the centre of area under a graph, relative to the y axis

ENG

Calculate the inertia of area under a graph, relative to the x axis

ENG

Calculate the inertia of area under a graph, relative to the y axis

ENG

Use the parallel axis theory to determine the inertia through the centre of the graph, parallel to the x axis

ENG

Use the parallel axis theory to determine the inertia through the centre of the graph, parallel to the y axis

ENG

Use Simpson’s Rule as appropriate with half stations

ENG

Understand the importance of the hull lines plan

ENG

Calculate the submerged volume and displacement of a vessel from the SAC

ENG

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318 • Ship Stability, Powering and Resistance

Objective

Level

Calculate the submerged volume and displacement of a vessel from the draught waterplane area curve

ENG

Calculate the KB of a vessel from the draught waterplane area curve

ENG

Estimate KB using Morrish’s method

ENG

Calculate the transverse inertia of the waterplane area from the waterline half beam curve

ENG

Calculate GM using data from the lines plan data

ENG

Calculate the waterplane area and the TPC from the lines plan data

ENG

Calculate the free surface effect caused by a non-rectangular tank

ENG

Calculate the GM for a multihull

ENG

Determine the effect of sponsons on metacentric height

ENG

Calculate the position of the LCB from the lines plan data

ENG

Calculate the position of the LCF from the lines plan data

ENG

Calculate the longitudinal inertia of the waterplane area from the waterline half beam curve

ENG

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9

SHIP RESISTANCE AIMS AND OBJECTIVES At the end of this section, you should be able to: Understand the causes of ship resistance Explain the components of ship resistance Calculate the frictional resistance of a ship using Froude’s method Calculate the total resistance of a ship from a model test using Froude’s method Calculate the frictional resistance of a ship using the 1957 ITTC Ship-Model Correlation Line Calculate the total resistance of a ship from a model test using the 1957 ITTC method Calculate the total resistance of a ship from a model test using the 1978 ITTC method

The driving force acting on a vessel comes from the thrust produced by the propellers, waterjets, sails, paddles or oars. This is opposed by the resistance or drag of the ship. When these forces are in balance, the system is in equilibrium and the vessel maintains a constant speed. The forces of thrust and drag are measured using Newtons (N) and the speed of the vessel is measured using either knots or metres per second (m/s). The conversion factors of knots and metres per second are: Commonly used: 1 knot = 0.514 m/s (used by Marine Engineers) More accurate: 1 knot = 0.5144 m/s (used by Undergraduate Students) Exact: 1 knot = 1,852 ÷ 3,600 m/s (1 nautical mile = 1,852 m) Determining the resistance of a vessel by direct calculation is so complex that it is not routinely used in ship design. Even the most powerful modern computers are limited in their ability to calculate ship resistance accurately.

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320 • Ship Stability, Powering and Resistance In practice, several approaches to calculating ship resistance are used which can be categorised as: Experimental Methods (Measurements) Towing Tank Testing Models Wind Tunnel Testing Models Ship Trials Empirical Methods (Estimates from Data) Parametric Studies Regression Analysis Standard Series Numerical Methods – Computational Fluid Dynamics (Computer Simulations) Potential Flow Solvers Navier-Stokes Solvers The first step is to understand the various causes of ship resistance.

Components of Ship Resistance (ENG) The resistance of a ship is commonly broken up into a number of components and subcomponents as shown in Figure 9.1.

Total ship resistance

Frictional resistance

Wave making drag

Form drag

Residuary resistance

Eddy making drag

Air resistance

▲ Figure 9.1 Components of ship resistance

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Ship Resistance • 321

Frictional resistance (ENG) The frictional resistance of a vessel is caused by the viscosity (stickiness) of the water flowing past the hull. Friction is generated by the contact between the hull and the water. As the hull moves through the water, a microscopic layer attaches itself to the hull and moves along at the same speed as the ship. Water flowing over this layer will also end up being dragged along with the vessel because water is viscous. Figure 9.2 shows the effect of viscosity on the flow over a surface such as the hull of a ship.

Distance from surface

Flow speed

Object surface

▲ Figure 9.2 Flow speeds in a boundary layer resulting from flow over a surface

The result of this effect is there is a layer of water near the hull that is no longer stationary but is being dragged along with the ship at some proportion of the ship speed. This layer of water is called the boundary layer. The overall thickness of the boundary layer gradually increases between the bow and the stern and will vary with the speed and length of the vessel. A large container vessel travelling at 20 knots will typically have a boundary layer thickness of around 10 cm near the stern. The viscous forces that cause the water in the boundary layer to speed up also have the effect of trying to slow down the vessel. This force is called the frictional resistance. There are numerous methods for calculating frictional resistance. More information about the flow in boundary layers and various methods for calculating frictional resistance can be found in later sections of the book.

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322 • Ship Stability, Powering and Resistance

Residuary resistance (ENG) As shown in Figure 9.1, the residuary resistance of a ship is made up of a number of subcomponents. These are more complex than the frictional resistance, and cannot be simply calculated. As only the frictional resistance can be calculated with relative ease the other components of resistance are often lumped together and termed the residuary resistance (literally – the remaining, or rest of the, resistance). As will be explained later the residuary resistance is often determined by subtracting the frictional resistance from a measurement of the total resistance.

Form drag or viscous pressure resistance (ENG) Form drag occurs because of pressure changes in the water flowing around the hull. All objects moving through a fluid experience form drag, for example, cars and aeroplanes moving through the air. As a body moves through a fluid, it creates pressure changes around the body. At the front of body the fluid is pushed aside and high pressure is created. At the sides of the body the fluid speeds up and low pressure is created. At the back of the body a wake is formed of fluid that is dragged along with the body. The pressure at the back of the body is always lower than the pressure at the front of the body causing an overall force that opposes the motion of the body. This is called form drag. Form drag, as its name suggests, is highly influenced by the shape (or form) of the object. One method of reducing form drag is to streamline the shape of the object, for example, car bodies are carefully designed to reduce form drag which results faster and/or more fuel efficient cars.

Wave making resistance (ENG) The wave making drag is the largest and most complex subcomponent of the residuary resistance. Wave making resistance, like form drag, is caused by pressure changes around the hull. Wave making resistance only occurs when bodies are travelling close to the surface of the water. Bodies travelling just beneath the surface, such as submarines, or just over the surface, such as landing seaplanes, also create waves. High pressure areas underwater push the water surface upwards and conversely low pressure sucks the water surface downwards. This causes waves to be created around

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Ship Resistance • 323 the hull. These waves are heavy and move at the same speed as the ship – potentially absorbing a vast amount of energy. The wave pattern created by ships is known as the ‘Kelvin’ wave pattern, and contains two types of waves as shown in Figure 9.3. Divergent stern wave system

Divergent bow wave system

Transverse wave system

▲ Figure 9.3 The features of the wave system generated by a vessel

Further information about wave making resistance can be found later in the book.

Bulbous bows reduce wave making resistance (ENG) The wave making resistance of a ship can be reduced by the use of a bulbous bow, as shown in Figure 9.4. Although bulbous bows can reduce residual resistance they also increase frictional resistance because the hull has more area in contact with the water (called the wetted surface area). When suitable vessels are fitted with bulbous bows, the increase in frictional resistance will be smaller than the decrease in residual resistance resulting in a reduction of the total resistance. The method by which a bulbous bow reduces resistance has been the topic of some debate but the most widely accepted explanation is that it works by creating a second bow wave system which interferes with the hull’s bow wave system. The superposition of the two waves results in a reduction of the bow wave which reduces the wave making resistance. A schematic representation is shown in Figure 9.5.

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324 • Ship Stability, Powering and Resistance

▲ Figure 9.4 Bulbous bow of a container ship

Bow without bulb

Bulb in isolation

Complete system

▲ Figure 9.5 Effect of a bulbous bow on the size of a ship’s bow wave

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Ship Resistance • 325 Bulbous bows tend to work best with vessels at moderate Froude numbers, with U and V shaped sections forward and with high block coefficients. Bulbous bows are not suitable for all vessels as demonstrated by several unsuccessful attempts to use bulbous bows on sailing yachts.

Eddy drag (ENG) Eddy drag is similar to form drag, but acts in a smaller scale. As fluid flows around underwater appendages and discontinuities (inlets, openings, shafts, etc.), it oscillates and breaks away into small eddies. The creation of these eddies adds to the resistance of the vessel. These eddies also create vibration which can sometimes be heard, for example, in the case of rudder or propeller ‘singing’.

Air drag (ENG) Most ships and boats do not typically operate at speeds where air drag is significant and they need to be ‘streamlined’. Air drag is very small in comparison with other components of resistance in normal operating conditions, and only becomes significant when a ship is operating in extreme weather conditions. Air drag is not generally considered in detail in the design process of a ship. However, the air flow around the exhaust stack on passenger vessels is often analysed to ensure that the exhaust fumes are carried clear of accommodation or passenger areas.

Calculating Frictional Resistance Using Froude’s Method (ENG) Frictional resistance is the only component of resistance that can accurately be found by direct calculation. The following method was developed over many years by William Froude and his son Robert Edmund Froude at the end of the nineteenth and beginning of the twentieth century. William Froude’s work on ship resistance resulted in the first accurate method for calculating ship resistance and the building of the first towing tanks for testing ships’ models. Froude’s analysis of frictional resistance began by towing planks of wood of various sizes and finishes and measuring their resistance. One of the most interesting results of his experiments was that he found that if he doubled the wetted surface area of a plank

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326 • Ship Stability, Powering and Resistance the resistance doubled, as common sense might suggest, but this was only true if its length stayed the same. If the wetted surface area was doubled by doubling the length of the plank he found that the resistance was somewhat less than doubled. Another important finding related to how resistance increased with speed. According to simple theory, the resistance should increase with the square of the speed, for example, doubling the speed should theoretically increase the resistance fourfold. Froude discovered that this was often not the case and again the resistance he measured was less than theory suggested. Froude derived an equation that calculated the frictional resistance of an object moving at a speed, V, with wetted surface area, S, as: Frictional resistance = RF

ffSv n

where f is a friction coefficient and n is the power index of ship speed. Over the years, Froude built up a set of values of f and n but in 1935, at the Paris conference of Tank Superintendents, it was agreed internationally that all towing tanks should use a value of n = 1.825 and also the formulae for calculating f were agreed. This results in the calculation for frictional resistance being: Frictional resistance t = RF

ffSv 1.825

▲ Formula 9.1 Frictional resistance by Froude’s method

Using the equation with SI units: Rf is the frictional resistance in N. f is the coefficient of friction. This can be found from tables of values for various ship lengths or the graphs shown in Figures 9.6 and 9.7. S is the wetted surface area of the vessel in m2. v is the speed of the ship, in m/s.

QUESTIONS Q9.1 A ship has a wetted surface area of 4,000 m2. The frictional coefficient, f, of the vessel is 1.392. Determine the frictional resistance of the vessel at a speed of 10 knots. Q9.2 A ship has a frictional resistance of 134,044 N at 10 knots. Determine the frictional resistance at 11 knots.

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Ship Resistance • 327 2.2

2.1

Friction coefficient (f )

2

1.9

1.8

1.7

1.6

1.5 0

1

2

3

4

5

6

7

8

9

10

Vessel length (m)

▲ Figure 9.6 Froude friction coefficient, f, values for ship models and small boats

1.57 1.55 1.53

Friction coefficient (f )

1.51 1.49 1.47 1.45 1.43 1.41 1.39 1.37 0

20

40

60

80

100 120 140 160 180 200 220 240 260 280 300 Vessel length (m)

▲ Figure 9.7 Froude friction coefficient, f, values for ships and large boats

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328 • Ship Stability, Powering and Resistance

The Froude Number (ENG) Naval Architects commonly refer to ship speed not in knots or metres per second, but using the ‘Froude Number’. This is a non-dimensional measure of speed, and measures how fast a vessel is travelling in relation to its length. For a vessel of length, L, travelling at a speed, v, the Froude number is:

Froude number u be , Fn =

v gL

▲ Formula 9.2 Froude number

The Froude number is important because wave making drag is the biggest component of resistance at higher speeds. As a vessel travels faster, the shape of the waves changes significantly around the vessel, with the bow and stern waves growing in length as the vessel accelerates. To understand the significance of the Froude number, it is necessary to consider the relationship between the speed, C, of a wave in deep water and it’s wavelength, LW, which is found by the following formula: C2 =

gLW 2π

As the speed of the vessel increases, and hence the speed of the waves it is generating, the wave length of the waves will also increase. At a Froude number of around 0.54, the wave system around a displacement hull form vessel is such that the bow wave is approaching double the ship length, resulting in the stern of the ship sitting in the trough of the bow wave. This causes the ship to ‘squat’ with the result that adding further power simply causes the stern to squat more and the bow to rise and does not give an appreciable extra speed. The speed of a vessel at a Froude number of 0.54 is known as ‘hull speed’ or ‘displacement speed’.

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Ship Resistance • 329 For smaller craft, such as yachts, the hull speed is the limiting factor on the speed that the vessel can realistically reach. For larger vessels, such as merchant ships, limiting the Froude number to values significantly lower than 0.54 is essential for economic running. A survey of the Froude numbers that relate to the service speed of many types of merchant vessels is shown in Figure 9.8. It should be noted that all the vessels have a Froude number between 0.15 and 0.33. A general rule of thumb is that slow vessels have a Froude number around 0.15, and as a result are very economical to run for their size, suiting large bulk carriers and VLCCs. Vessels where speed is important, and the income from faster speeds offsets the increase in running costs, will tend to have a Froude number of no more than 0.35 m. Vessels in this category tend to include ferries, fast container vessels and ocean liners. Froude number is also important in ship design as the form of a vessel underwater affects its resistance curve. A typical resistance curve for a vessel, as shown in Figure 9.9, has humps and hollows. The resistance of the vessel in the region of a hollow equates to more economical running of the vessel and ideally the service speed, or Froude number, should correspond to this area of the curve.

Froude number

0.55 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

0

50

100

150 200 Length (m)

250

300

350

400

▲ Figure 9.8 Froude number comparison of different merchant vessels

The humps and hollows are caused by the characteristics of the wave system around a vessel and this is considered later in the book.

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330 • Ship Stability, Powering and Resistance

Friction

Main hump

Resistance

Residual

0.1

Total

Prismatic hollow Prismatic hump

0.15

0.2

0.25

0.3 0.35 0.4 Froude number

0.45

0.5

0.55

0.6

▲ Figure 9.9 Resistance components versus Froude number

Corresponding Speed (ENG) When comparing the speed of two different length vessels, it is often important to determine their corresponding speeds. The corresponding speeds occur when each vessel has the same Froude number: Froude number Vessel A = Froude number Vessel B vA gLA

=

vB gLB

Simplifying and rearranging this gives: v A = vB ×

LA LB

When two geometrically similar vessels have the same Froude number, or corresponding speed, their wave systems will also be to scale, that is, if one vessel is twice the length of the other the wavelengths of its wave pattern will also be twice as long.

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Ship Resistance • 331

Model Testing Using Towing Tanks (ENG) As discussed previously, the mathematical process of determining the overall resistance of a ship is too complex for direct calculation. Instead, we rely on the testing of a model of the ship in a towing tank, as shown in Figure 9.10. Towing tanks are long, narrow tanks in which model ships are towed at a range of speeds. The resistance of the model is measured via a dynamometer, and is scaled up to predict the resistance of the ship at the corresponding speed.

▲ Figure 9.10 The towing tank at Southampton Solent University

To ensure accuracy, the model should be as large as possible. This makes it easier to build an accurate model, and the resistance data should therefore be more accurate. However – if the model is too large compared to the tank, then ‘blockage’ effects will occur. This means that the model may suffer from shallow water effects (squat), canal effects and bank effects which will increase the resistance of the model.

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332 • Ship Stability, Powering and Resistance Scale models of the ship are built from wood, GRP or wax. The accuracy is very important – tolerances should be ±1 mm for beam and depth, and the larger of ±1 mm or ±0.05% LBP for length. Often CNC machines are used to get the required accuracy – although this is very expensive. The surface finish should be smooth – the equivalent of 300 to 400 grit wet and dry. Often appendages such as rudders and bilge keels are not modelled, as it is difficult to accurately model the appendages at small scales (ITTC require ±0.2 mm). It can also be difficult to make model scale appendages strong enough to resist deforming or breaking in the flow. To ensure that the flow over the whole model is turbulent (as it would be at ship scale), ‘turbulent flow stimulators’ are fitted at the bow, around 5% of the LBP aft of the FP. These are in the form of a thin wire, studs or strip of sand paper (with a grain size of 0.5 mm). These ‘trip’ the flow, increasing turbulence and creating a turbulent boundary layer. The accuracy and surface finish mean that models are expensive – in the order of £3,000 for a professionally built 2 m long model. For larger models for final testing (length approx. 6 m), costs are much higher. A small model being tested is shown in Figure 9.11.

▲ Figure 9.11 Testing a model in a towing tank

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Ship Resistance • 333

Calculating the Resistance of a Ship from a Model Test Using Froude’s Method (ENG) Froude’s method splits the total resistance of both the vessel and its model into two parts: the frictional resistance, RF , and the residuary resistance, RR . RT

RF + RR

▲ Formula 9.3 Total resistance components in Froude’s method

The frictional resistance of both the model and the vessel are calculated using Froude’s formula for frictional resistance. For geometrically similar vessels, such as a ship and its model, that are travelling at corresponding speeds, the residuary resistance is considered to scale on displacement:

RRS

ΔS ΔM

RRM

▲ Formula 9.4 Froude’s method for scaling residuary resistance

The steps involved in determining the full sized vessel’s resistance from a model test is as follows: The model is towed down the tank and the towing speed, vM, and the total resistance, RTM, of the model are measured and recorded The frictional resistance of the model is calculated using Froude’s formula, and is known as RFM. RFM fM SM v Mn The difference between the total model resistance and the frictional model resistance is the model residual resistance known as RRM. RRM RTM − RFM The RRM is scaled up by the ratio of the displacement of the model, ΔM, and the ship, ΔS ΔS, to get the ship residual resistance known as RRS. RRS RRM ΔM The speed of the ship that corresponds to the model test is calculated from the LS model speed. v S = × vM LM

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334 • Ship Stability, Powering and Resistance The frictional resistance of the ship is then calculated using Froude’s formula and is known as RFS. RFS fS SS v Sn The ship frictional resistance is added to the ship residual resistance to get the total ship resistance known as RTS. RTS = RFS + RRS The process is typically repeated for a range of speeds to build up the ship’s resistance curve.

QUESTION Q9.3 A 4 m long, 179.2 kg model of a ship is tested at a speed of 2.2 m/s in a towing tank. At this speed, the total measured resistance of the model is 70 N. The f coefficient for the model is 1.609. The wetted surface area of the model is 4.5 m2. Determine the resistance and speed of the ship, length 120 m, if the displacement is 4,959 tonnes, the wetted surface area is 4,050 m2 and the f coefficient is 1.483.

Scaling of Geometrically Similar Vessels (ENG) The scale of a model can be used to determine its wetted surface area and displacement from those of the ship if these values are not known. If the scale between the model and the ship has the symbol λ such that: LSHIP =λ LMODEL Then areas and volumes will scale according to the principles shown in Figure 9.12. These leads to the scaling of the wetted surface area and volume of displacement being:

9781408176122_Ch09_Rev_txt_prf.indd 334

⎛ L ⎞ SSHIP = λ 2 = ⎜ SHIP ⎟ SMODEL ⎝ LMODEL ⎠

2

⎛ L ⎞ ∇ SHIP = λ 3 = ⎜ SHIP ⎟ ∇ MODEL ⎝ LMODEL ⎠

3

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Ship Resistance • 335 As volume × density = mass, we can express the ratio of the displacements in terms of the lengths: 3

⎛ L ⎞ Δ SHIP ∇ SHIP × ρSSEAWATER ρ ρ = = λ 3 × SSEAWATER = ⎜ SHIP ⎟ × SEAWATER Δ MODEL ∇ MODEL × ρFRESHWATEER ρFRESHWATER ⎝ LMODEL ⎠ ρFRESHWATER

Area scaling considerations 1 2 1 Area = 1

λ 2 Area = 22 = 4 λ Area = λ2 Volume scaling considerations

2 2 1

1

1

2

Volume = 1 Volume = 23 = 8 Volume = λ3

▲ Figure 9.12 Scaling areas and volumes of geometrically similar shapes

QUESTIONS Q9.4 A ship with a length of 150 m, a wetted surface area of 6,500 m2 and a displacement of 9,800 tonnes is to be tested using a scale model, with a scale of 1:20. Determine the length, wetted surface area and displacement of the model.

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336 • Ship Stability, Powering and Resistance Q9.5 A 5 m long model of a ship, scale 1:20, is used to determine the resistance of a ship. The model is tested at a speed of 1.4 m/s. The f coefficient for the model is 1.655. The f coefficient of the ship is 1.401. The wetted surface area of the model is 5.60 m2. The total resistance of the model is 25 N. Determine the total resistance of the ship at the same Froude number.

Laminar and Turbulent Boundary Layers When fluid flows over a surface a boundary layer is created as discussed earlier. When considering the flow over a flat surface, such as a simple plate rudder, as shown in Figure 9.13, the boundary layer thickness increases. The edge of the boundary layer is usually defined where the flow speed is 99% of the actual flow speed (referred to as the free stream velocity). The boundary layer thickness at any point is given the symbol δ. The values of drag predicted by Froude’s formula for frictional resistance are good but in 1883, Professor Osborne Reynolds conducted a set of experiments which led to a

Flow direction

Solid surface δ

Boundary layer

Boundary layer edge

▲ Figure 9.13 Boundary layer on a flat plate

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Ship Resistance • 337 much greater understanding of boundary layers. This work showed that, depending on the speed of flow and length of the surface over which it was flowing, the boundary layers behaved in two separate manners – laminar and turbulent flow. In laminar flow, each layer within the boundary layer ‘slides’ smoothly over the previous layer as shown in Figure 9.14. When the roughness, speed or object size is such that turbulence is introduced into the flow, the ‘layers’ within the boundary layer mix together and the velocity profile changes dramatically, as shown in Figure 9.15. Free stream velocity

y Local velocity

Flow speed Object surface

▲ Figure 9.14 Flow in a laminar boundary layer

Free stream velocity

y Local velocity Equilibrium turbulent layer

Buffer layer Laminar sublayer

Flow speed

▲ Figure 9.15 Flow in a turbulent boundary layer

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338 • Ship Stability, Powering and Resistance

The Reynolds Number The actual flow type, and hence velocity profile depends on the Reynolds number, named after Osborne Reynolds. The Reynolds number, based on the overall length of a plate, is found from: Rn =

L ρLv = ν μ

where v is the kinematic viscosity (for FW @ 20°C taken as 1.004 × 10–6 m2 s–1. μ is the dynamic viscosity, for FW @ 20°C taken as 1.002 × 10–3 N s m–2. ρ is the fluid density in kg/m3, L is the length of the surface in metres and v is the free stream speed, in m/s. At low Reynolds numbers (below approx. 4.5 × 105), the flow can be considered to be laminar, and at high Reynolds numbers, the flow can be considered to be turbulent. As the Reynolds number depends on the length of flow, at the leading edge of the body, the flow is normally laminar up to a point where the Reynolds number is approximately 4.5 × 105, then turbulent as shown in Figure 9.16. Once the difference between laminar and turbulent flow was more understood, scientists looked at a method of predicting friction drag based on the Reynolds number, as turbulent flow clearly created more friction than laminar flow. At transition speeds between laminar and turbulent flow, the drag will jump from low values in laminar flow, to higher values in turbulent flow as shown in Figure 9.17. Once turbulent flow was understood, more efforts were made to accurately predict the friction drag based on the Reynolds number. Lots of different formulae were

Laminar

Transition

Turbulent

▲ Figure 9.16 Effect of laminar and turbulent flow on boundary layer thickness

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Ship Resistance • 339 Drag on a 10 m by 1 m plate 0.1 Turbulent flow drag Laminar flow drag

0.08

Drag (N)

0.06

0.04

0.02

0 0

100,000

200,000

300,000 400,000 500,000 Reynolds number

600,000

700,000

▲ Figure 9.17 Change of drag at the transition from laminar to turbulent flow

predicted, all based around different assumed boundary layers. Different shipyards and designers used different methods – and hence got lots of different results. From around 1932 onwards, the industry started to standardise the approach used, and in 1957 settled on the ‘1957 ITTC Friction Line’. This is an experimentally derived regression line.

Calculating Frictional Resistance Using the 1957 ITTC Friction Line This is a method based on non-dimensionalised force coefficients. The coefficient of friction, known as CF, is found from: CF =

0.075

(

Rn −

)2

This can then be used to determine the frictional resistance, in Newtons, using: RF

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1 Sv 2 C F 2

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340 • Ship Stability, Powering and Resistance where ρ is the water density in kg/m3, S is the surface area in contact with the fluid, in m2, V is the flow speed in m/s, and CF is the friction coefficient as found using: CF =

0.075

(

Rn −

)2

QUESTION Q9.6 A plate is 10 m long and 1 m wide. The plate has water flowing past one side of it. Calculate the drag of the plate at flow speeds of 0.1, 1 and 10 m/s using the 1957 ITTC Friction Line.

Form Factors and Viscous Resistance With the ITTC 1957 friction calculation, it became possible to use a simple method to calculate the drag of a flat plate based on the Reynolds number. Assuming that the underwater surface area of a hull could be realistically ‘unrolled’ into a flat plate, then the ITTC 1957 method could be used to determine the frictional drag of that plate, and hence give the frictional drag of the ship. However, ships are not flat plates – so how realistic is this? Any curved shaped will, when placed in a water flow, create changes in pressure around it. This is caused by the flow accelerating around the shape – increases in flow speed cause reductions in pressure, and vice versa. Figures 9.18 and 9.19 show the pressure and flow speed around a simple ellipse, calculated using computational fluid dynamics (CFD). It can be seen that at the leading edge, the pressure is high and the flow speed is low. Around the sides, the pressure is lower, and the flow speed is higher. This can change the local Reynolds number, but more importantly, compress or extend the boundary layer. In Figure 9.20 the effect of a curved body on the boundary layer is shown. Towards the trailing edge, it can be seen that the boundary layer thickens significantly as the pressure changes around the body distort the boundary layer – the effect of the pressure changes around the shape is to change the boundary layer.

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Ship Resistance • 341

Flow direction

▲ Figure 9.18 Pressure field around a curved body

Flow direction

▲ Figure 9.19 Velocity field around a curved body

As can be seen, these pressure effects thicken and reverse the boundary layer. Calculating the actual change in the boundary layer, and hence the drag, is beyond the scope of this book. It can be dealt with practically by the use of ‘form factors’. These are coefficients which are used to increase the ‘flat plate’ friction drag coefficient to a value which is appropriate for certain shapes, to model the effects of the viscous flow on the

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342 • Ship Stability, Powering and Resistance

Actual boundary layer

Note that the boundary layer thicknesses have been exaggerated for clarity

‘Flat plate’ boundary layer

▲ Figure 9.20 Boundary layer development around a curved body

boundary layer. Most of these form factors are based on experimental data. Form factors are given the notation (1+k). When the frictional resistance coefficient is adjusted using a form factor, it is referred to as the viscous resistance coefficient. When the frictional resistance is calculated using the viscous resistance coefficient, it is referred to as the ‘viscous’ resistance. Viscous rresistanc es sta cee coefficient , CV = (1+ k ) C F 1 Viscous rresistanc es sta ce, e RV = ρSv 2 CV 2

QUESTIONS Q9.7 A ship has a length of 100 m and a wetted surface area of 4,000 m2. The vessel is in sea water, with a kinematic viscosity of 1.05 ×10–6 m2 s–1. The k value in the form factor of the ship is 0.10. Determine the viscous drag of the ship at a speed of 15 knots. Q9.8 A rudder has a chord length, c, of 2 m and a span of 5 m. The thickness, t, of the rudder is 0.40 m. Using the 1957 ITTC friction line, determine the viscous resistance of the rudder in sea water, if the flow speed is 8 m/s.

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Ship Resistance • 343 Note: The form factor of a rudder of thickness, t, and chord length, c, can be found from: t (1 k ) = 1+ 2 ⎛⎝ ⎞⎠ c

⎛t⎞ 60 ⎝ c⎠

4

Manufacturer’s data shows that the rudder wetted surface area can be found from: Area Chord C × Span × 2.24 The vessel is in sea water, with a kinematic viscosity, v, of 1.05 × 10–6 m2s–1.

Wave Interference Effects on Ship Resistance The largest component of the residual resistance is the wave making resistance. As the name suggests, this is the resistance due to the creation of waves around the vessel as it moves through the water. As a singular object moves along the water surface, a specific wave pattern is created as shown in Figure 9.21. Ships normally create two separate wave systems – one from the bow and one from the stern, normally starting from the stern shoulders where the aft hull curvature starts at the aft end of the parallel mid-body. The effect of this is that there is a disturbance effect between the two wave systems, as shown in Figure 9.22.

Divergent system Transverse system

▲ Figure 9.21 Wave pattern due to a singular body

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344 • Ship Stability, Powering and Resistance Stern wave system Bow wave system

Transverse system – main influence at normal service speed

Wave interference

▲ Figure 9.22 Wave interference effect in the wave pattern about a ship

The transverse waves travel along at the same speed as the ship. As wave speed can be expressed in terms of wave length, the wave length can be expressed in terms of the ship speed, V. Therefore, it can be seen that the wave lengths are proportional to the ship speed squared. LW =

2πV 2 g

As the ship speed increases, this shows that the wave length of the bow and stern transverse wave systems increases. At varying speeds, these waves either superimpose or cancel each other, as shown in Table 9.1, with each scenario showing an increasing speed. The shape of the hull means that the bow wave always starts as a crest, normally just aft of the bow. The stern wave always starts as a trough, just ahead of the stern. As seen, the waves from the bow and stern transverse wave systems either superimpose or cancel each other, and increasing or reducing the wave making resistance. This results in the wave making resistance varying with speed in a nonlinear manner. The Froude number of the vessel as waves form along the hull can be theoretically determined. Due to the high pressure regions and hull form, the waves are assumed to develop along around 90% of the hull length – known as the characteristic length. The shape of the vessel’s hull changes the characteristic length but for most vessels this will be between 85% and 95% of waterline length.

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Ship Resistance • 345 Table 9.1 Bow and stern wave interference at various speeds Scenario

Wave pattern (Solid line shows the bow wave system, dashed line shows the stern wave system)

Bow and stern wave interaction

1

Cancellation

2

Superposition

3

Cancellation

4

Superposition

5

Cancellation

6

Superposition

At varying speeds where the wave systems either superimpose or cancel, the wavelength of the wave systems can be expressed in terms of the LBP of the vessel, as shown in Table 9.2. Table 9.2 Wave length of bow and stern waves against characteristic length of 90% LBP at various speeds Scenario

Wave pattern

Wave lengths

1

0.9LBP LW = ______ 3

2

0.9LBP LW = ______ 2.5

3

0.9LBP LW = ______ 2

4

0.9LBP LW = ______ 1.5

5

0.9LBP LW = ______ 1

6

0.9LBP LW = ______ 0.5

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346 • Ship Stability, Powering and Resistance The length and speed of these waves can be related to the Froude number. As seen previously: LW =

g LW 2π V 2 ∴ =V g 2π

Substituting in the constants gives an expression for wave length and wave speed: g LW 9 81× LW = = 1.561LW 2π 2π

V

At each of these speeds, the wave speed can be expressed in terms of the LBP, by combining the expressions for wave speed in Table 9.2 and this equation. This gives the wave speed expressions shown in Table 9.3. Table 9.3 Ship speed versus ship length Wave speed

Wave pattern 1

V=

1.561LW =

0.9LBP 1.561 X ______ = 3

0.468LBP

2

V=

1.561LW =

0.9LBP 1.561 X ______ = 2.5

0.562LBP

3

V=

1.561LW =

0.9LBP 1.561 X ______ = 2

0.702LBP

4

V=

1.561LW =

0.9LBP 1.561 X ______ = 1.5

0.937LBP

5

V=

1.561LW =

0.9LBP 1.561 X ______ = 1

1.405LBP

6

V=

1.561LW =

0.9LBP 1.561 X ______ = 0.5

2.801LBP

Finally the Froude number for each wave pattern can be determined, as shown in Table 9.4. Therefore, we can determine the Froude number, and hence ship speed, when the waves either superimpose or cancel. A graph of wave making resistance against Froude number shows a series of humps and hollows occurring at the Froude numbers found in Table 9.5. An exaggerated example graph is shown in Figure 9.23.

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Ship Resistance • 347 Table 9.4 Ship wave pattern at various Froude numbers V Froude number = ________ g x LBP

Wave pattern

1

0.468LBP V FN = ________ = ___________ = 0.22 gLBP g x LBP

2

0.562LBP V FN = ________ = ___________ = 0.24 gLBP g x LBP

3

0.702LBP V FN = ________ = ___________ = 0.27 gLBP g x LBP

4

0.937LBP V FN = ________ = ___________ = 0.31 gLBP g x LBP

5

1.405LBP V FN = ________ = ___________ = 0.38 gLBP g x LBP

6

2.801LBP V FN = ________ = ___________ = 0.54 gLBP g x LBP

Most higher-speed ships are designed to operate either at a Froude number around 0.35 known as the prismatic hollow as this gives the best speed for wave making resistance, or around 0.25, below a peak known as the prismatic hump.

Wave making drag  

Superposition at F N = 0.54

Cancellation at FN = 0.38 Superposition at F  N = 0.31 Cancellation at FN = 0.27 Superposition at F N = 0.24 Cancellation at FN = 0.22

0

0.1

0.2

0.3 Froude number

0.4

0.5

0.6

▲ Figure 9.23 Humps and hollows (exaggerated) in a wave making resistance curve

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348 • Ship Stability, Powering and Resistance

Modern Methods for Calculating the Resistance of a Ship from a Model Test Current towing tank test methods are agreed by the International Towing Tank Conference (ITTC) and are based on the 1957 ITTC method. The 1957 ITTC method makes use of ‘non-dimensional coefficients’. Subscripts M and S are used to refer to model and ship scale. Table 9.5 Force coefficients used in the 1957 ITTC method CTM =

Model total resistance 0 5ρSM v M2

CTS =

Ship tota total rresistance 0 5ρSS v S2

CTM =

RT M 0 5ρ SM v M2

CTS =

RT S 0 5ρ SS v S2

CWM =

Model w wav a e resistance 0 5ρ SM v M2

CWS =

Ship wave a e resistance 0 5ρ SS v S2

CWM =

RW M 0 5ρ SM v M2

CWS =

RW S 0 5ρ SS v S2

CVM =

Model vvisco iscous us resistance 0 5ρ SM v M2

CVS =

Ship viscous rresistance 0 5ρ SS v S2

CVM =

RVM

CVS =

0 5ρ S v

2 M M

RVS 0 5ρ SS v S2

For each speed tested, the ‘viscous drag coefficient’ is determined, using the 1957 ITTC Friction Line calculation: V

(

k )CF

As seen earlier the frictional resistance is based around the concept of equivalent drag of a flat plate. The calculation of frictional drag can be made more accurate by incorporating a ‘form factor’, which accounts for variations in the boundary layer caused by the form of the ship, which in turn affects the frictional resistance.

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Ship Resistance • 349 The method used to determine the form factor is known as the ‘Prohaska Method’. This is a method of determining the form factor – it assumes that at very low Froude numbers, the wave making resistance is zero, and therefore the only resistance is the actual frictional resistance. To determine the form factor, a plot of low speed values is made of Fr4/CF on the x axis, against C TM/CFM on the y axis. A line of best fit is used, and the intercept on the y axis gives (1 + k). An example of this is shown in Figure 9.24. 1.28 Prohaska 1.26 Linear (Prohaska) 1.24 1.22 CTM /CFM

1.2 1.18 1.16 1.14 1.12 1.1 1.08

–0.1

–0.05

0

0.05

0.1

0.15

0.2

0.25

Fr 4/CFM

▲ Figure 9.24 Prohaska plot

The total resistance of the model is assumed to consist of the viscous resistance and the wave making resistance. In terms of coefficients: CT M

(

k )CF M

CW M = C V M

CW M

The coefficient of total resistance, CTM, can be found. Therefore, for each speed that the model is tested, the wake making coefficient, CWM, can be determined. For each speed, the wave making coefficient is assumed to be constant for the ship and the model, therefore: WM

WS

For each speed tested, the ‘viscous drag coefficient’ for the ship is determined. The form factor is assumed to be the same for the ship and model, therefore: CV S

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(

k )CF S

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350 • Ship Stability, Powering and Resistance CFS is found using the ITTC 1957 Friction Line and the total resistance of the ship is assumed to consist of the viscous resistance and the wave making resistance. In terms of coefficients: CT S

(

k )CF S

CW S = C V S

CW S

The total hull resistance can then be found. The air drag of the vessel can be estimated for a ship in a head wind, using: Air drag

AT × VR2

In this equation, the AT term can be found from the projected area of the hull and superstructure, as shown in Figure 9.25, and found from: AT

0 3 A1 + A2

Area 1, A1

Area 2, A2

▲ Figure 9.25 Projected area of a hull

For more accurate determination of air drag, specific ship hull and superstructure models are tested in wind tunnels. This is also undertaken to measure and check that exhaust gases are safely ventilated away from accommodation areas. To calculate the drag caused by appendages, published data is used to determine ‘drag coefficients’, CD. These can then be used to determine the appendage drag due to eddy making and form of appendages: = 0 5 × ρ × S V 2 × CD The friction drag from appendages can be found using the friction line. In summary, for each tested speed, the flow chart shown in Figure 9.26 can be used.

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Ship Resistance • 351

Measure RTM using a towing tank

Measure VM when towing

Calculate CTM

Calculate the model Reynolds number

Calculate the model Froude number

Calculate CFM

This will also be the ship Froude number

Calculate CWM

Calculate CVM

This will also be the CWS

Calculate the speed of the ship

Calculate CTS

Calculate the ship Reynolds number

Calculate RTS and add appendage and air drag

Calculate CFS

Calculate CVS

▲ Figure 9.26 1957 ITTC method flow chart

QUESTION Q9.9 A 5 m long 1:50 scale model ship, with a wetted surface area of 6.40 m2, is tested at a speed of 1.10 m/s in fresh water. At that speed the total resistance is measured at 25.000 N. Data for the hull form indicates that the form factor, (1 + k), is 1.15. Calculate the total resistance for the full size ship at the corresponding speed in sea water,

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352 • Ship Stability, Powering and Resistance assuming that air drag and appendage drag are negligible. You may assume the following: ρFW

1 000 k /

ν FW 1 00 × 10 −6

2

3

ρSW 0 5 kg/m3 S = 1, 025

1

ν SW = 1.06 × 10 −6 m2 s-1

SHIP RESISTANCE  LEARNING CHECKLIST Objective

Level

Understand the causes of ship resistance

ENG

Explain the components of ship resistance

ENG

Calculate the frictional resistance of a ship using Froude’s method

ENG

Calculate the total resistance of a ship from a model test using Froude’s method

ENG

Completed

Calculate the frictional resistance of a ship using the 1957 ITTC Ship-Model Correlation Line Calculate the total resistance of a ship from a model test using the 1957 ITTC method Calculate the total resistance of a ship from a model test using the 1978 ITTC method

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10

SHIP PROPULSION AIMS AND OBJECTIVES At the end of this section, you should be able to: Explain the types of power that exist in a ship propulsion system Identify the efficiencies of the components of a ship propulsion system Estimate changes in the required shaft power of a vessel due to loading Estimate changes in the required shaft power of a vessel due to speed changes Estimate the powering requirements of geometrically similar ships Estimate changes to daily and voyage fuel consumption

Converting Resistance to Effective Power (ENG) The resistance of a ship is only part of the problem of determining the propulsive requirements to propel a ship. The resistance values need to be converted to power requirements, so the performance of the vessel can be predicted for the size of engine installed, and the speed known for any power setting. The power required to tow a vessel through the water at a particular speed is known as the effective power, PE. The first step in determining this is to calculate the effective naked power, PEN.

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354 • Ship Stability, Powering and Resistance This is calculated directly from the scaled up model resistance, which generally measures the resistance without appendages (as they do not give reliable results at small scales) or wave drag. RT × v

PEN

The naked effective power does not include the power requirements to overcome the appendage drag or additional allowances for wind or waves. A factor, known as the Ship Correlation Factor, or SCF, can be applied to the effective naked power, PEN, to get the effective power, PE. Typically the SCF is between 1.1 and 1.2. PE

PEN × SCF

The SCF is made up of two further subfactors, which model the additional drag due to appendages and a safety margin for heavy weather: Ship co correlation e at o factor ( SCF SC ) Weather Weat e allowance Appendage allowanc a e

QUESTION Q10.1 A ship has a total resistance of 850 kN at a speed of 10 m/s. The additional drag from appendages is 5%. The additional drag due to weather is 12%. Determine the effective power required.

Powering a Vessel (ENG) If we took the effective power required and installed an engine of the same size, we would find that the losses over the whole propulsion system would mean that the vessel would achieve significantly less than the intended service speed. Typically, the installed engine power needs to be around double the effective power. To determine

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Ship Propulsion • 355 the required power from the engines the various powers occurring in the system need to be considered as shown in Figure 10.1.

Effective power

Installed power Thrust power

Delivered power

Shaft power

▲ Figure 10.1 Powers in a ship propulsion system

Losses in a Ship Propulsion System (ENG) A schematic representation of the losses in a typical propulsion system is shown in Figure 10.2.

Mechanical efficiency (ENG) The overall output of the engine at the drive shaft is less than the indicated power of the engine. The losses occur through noise, heat, and so on, but also through the engine driving its ancillary services such as the fuel system. For a modern diesel engine the mechanical efficiency is around 80–85%. Mechanical efficiency =

ηM =

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Shaft f power Installed power

PS PI

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356 • Ship Stability, Powering and Resistance Installed (indicated) power PI

Mechanical efficiency ηM

Engine

Shaft power PS

Transmission

Transmission efficiency ηT

Delivered power PD

Propeller

Propeller efficiency ηP

Hull

Hull efficiency ηH

Thrust power PT

Effective power PE

▲ Figure 10.2 Powers and efficiencies in a ship propulsion system

Transmission efficiency (ENG) The losses in the transmission system such as gearboxes, shaft bearings and thrust blocks are termed the transmission efficiency. Without a gearbox the transmission losses will typically be 2–3% giving a transmission efficiency of 97–98%. Transmission efficiency = ηT =

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Delivered r power Shaft f power

PD PS

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Ship Propulsion • 357

Propeller efficiency (ENG) The job of the propeller is to convert rotational force into linear force. This will be covered in more detail later. Propellers are generally pretty inefficient – this is one of the major problems still to be tackled in ship design. Typically propeller efficiencies vary from 50 to 70%. Propeller efficiency =

ηp =

Thrust power Delivered r power

PT PD

Hull efficiency (ENG) Ship models are towed without the propellers rotating as the scale is too small to give reliable results. As the propeller rotates, it creates areas of low pressure ahead of the propeller, and areas of high pressure astern of the propeller. The area of low pressure has the effect of creating a small aft force on the hull, which itself means the thrust power, PT, needed to push the vessel is slightly greater than the effective power, PE. Hull efficiency =

Effective v power Thrust power

ηH =

PE PT

It is important to note that the hull efficiency is not a true efficiency measurement as it compares the power produced by the propeller when driving the vessel with the power required to tow the vessel. For this reason, hull efficiency values will sometimes be greater than 100% which is impossible for a true efficiency value. Methods for calculating the hull efficiency are considered later in the propeller section of the book.

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358 • Ship Stability, Powering and Resistance

QUESTION Q10.2 A ship has a bare hull total resistance of 900 kN at a speed of 11 m/s. The ship correlation factor is 1.15. The hull efficiency is 98%. The propeller efficiency is 57%. The transmission efficiency is 97%. The mechanical efficiency is 80%. Determine the installed power required.

Propulsive coefficients (ENG) Several propulsive coefficients are used that combine various efficiencies together in order that different propulsion systems can be compared. The overall efficiency of the propulsion system is measured by the propulsive coefficient, PC: PC =

PE =η T × PS

P

×ηH

The overall efficiency of the propulsion system, excluding the transmission, is measured by the quasi-propulsive coefficient, QPC: QPC =

PE = ηP ×ηH PD

Estimating Ship Power Requirements (ENG) As has been seen, the overall process for determining the required power for a ship from model testing is time consuming and expensive – possibly costing over £250,000 for a simple analysis.

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Ship Propulsion • 359

Admiralty coefficient (ENG) For preliminary design purposes, and for assessing small changes in operating conditions, the ‘admiralty coefficient’ can be used. This process is an estimation, but is good for typical ship hull forms operating at moderate Froude numbers. 2

AC =

Δ 3v 3 PS

▲ Formula 10.1 Admiralty coefficient based on shaft power

The power used is the shaft power, PS in kW, the ship speed, v, has units of knots, and the displacement, Δ, is measured in tonnes. For small changes in either displacement, speed or power, the admiralty coefficient can be assumed to remain constant. Note: Care must be taken with units when using published admiralty coefficient values. The formula has been routinely used, particularly in the past, with the displacement measured in tons and shaft power measured in horsepower. The admiralty coefficient values calculated with these units will be of a similar magnitude but differ by around 36% from those calculated using metric units.

QUESTIONS Q10.3 A ship has a shaft power of 20,000 kW and travels at 15 knots. The displacement of the vessel is 90,000 tonnes. Estimate the power required if the speed is increased by 2 knots. Q10.4 A ship has a shaft power of 10,000 kW and travels at 8 knots. Estimate the power required if the speed is decreased by 1 knot. For scaled vessels travelling at the same Froude number, the admiralty coefficient can also be assumed to be constant.

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360 • Ship Stability, Powering and Resistance

QUESTION Q10.5 A 100 m long vessel has a displacement of 7,000 tonnes and travels at a speed of 14 knots with a shaft power of 8,000 kW. What would the shaft power be for a scaled up vessel which is 10% larger, travelling at the same Froude number?

Estimating Ship Fuel Requirements (ENG) In a very similar way to the admiralty coefficient, we have a method of estimating changes in fuel consumption. This is based on the assumption that over the normal power range, the specific fuel consumption of the engine (kg/kWh), remains constant.

Fuel coefficient (ENG) For small changes in either displacement, speed or power, the fuel coefficient can be assumed to remain constant. 2

Δ 3v 3 Fuel coefficient = Daily fuel consumption ▲ Formula 10.2 Fuel coefficient

2

Δ 3v 3 Daily fuel consumption = Fuel coefficient

QUESTION Q10.6 A ship has a displacement of 90,000 tonnes, a shaft power of 20,000 kW and travels at 15 knots. At this speed, the daily fuel requirement is 35 tonnes. Estimate the daily fuel required if the speed is increased by 2 knots.

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Ship Propulsion • 361

Estimating voyage fuel requirements (ENG) Changes in ship speed also have the effect of increasing the voyage time. Changes to the voyage fuel consumption are usually required and can be found using the following formula: Voyage fuel consumption = Daily fuel consumption ×

Voyage distance a 24 × Ship speed

QUESTIONS Q10.7 A ship has a displacement of 80,000 tonnes. The fuel coefficient is 130,000. Estimate the amount of fuel required to complete a 2,000 mile voyage at 16 knots. Q10.8 Using the information in Question 10.7, estimate the amount and percentage of fuel saved by reducing the ship speed to 14 knots?

SHIP PROPULSION  LEARNING CHECKLIST Objective

Level

Explain the types of power that exist in a ship propulsion system

ENG

Identify the efficiencies of the components of a ship propulsion system

ENG

Estimate changes in the required shaft power of a vessel due to loading

ENG

Estimate changes in the required shaft power of a vessel due to speed changes

ENG

Estimate the powering requirements of geometrically similar ships

ENG

Estimate changes to daily and voyage fuel consumption

ENG

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Completed

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11

SCREW PROPELLERS AIMS AND OBJECTIVES

At the end of this section, you should be able to: Understand the basic geometry and terminology associated with a propeller Understand how the performance of a propeller can be measured Calculate propeller slip

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Screw Propellers • 363 The job of the propeller is to convert torque from the propeller shaft into linear thrust to propel the vessel. The mathematical process is complex, so this section covers simple analysis methods and techniques to understand the basic aspects of propeller performance.

Screw Propeller Terminology (ENG) The terminology used to describe the features of a screw propeller is identified in Figures 11.1 and 11.2. Leading edge

Trailing edge

Hub

Blade tip

Blade root

Propeller blade back (unseen side)

Propeller blade face

▲ Figure 11.1 Features of a propeller as viewed from astern of a vessel

Rake Skew Trailing edge

Rotation Leading edge

Looking forward

▲ Figure 11.2 Rake and skew of a propeller blade

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364 • Ship Stability, Powering and Resistance

Blade back and face The back face of the propeller blade is the face seen from astern. This is the high pressure side of the propeller. The forward face of the propeller blade is the face seen from ahead. This is the low pressure side of the propeller.

Propeller rotation or walk There is no set direction of rotation for all propellers. Each propeller is designed for a particular direction of rotation. A propeller which is designed to rotate clockwise when viewed from astern is known as a right handed propeller. This type of propeller will tend to push the stern to starboard as it creates thrust ahead. Conversely, a propeller which is designed to rotate anticlockwise when viewed from astern is known as a left handed propeller. This type of propeller will tend to push the stern to port as it creates thrust ahead. As the propeller rotates it creates lift over the blades, moving it forward. The ‘pitch’ of the propeller is the distance travelled forward in one revolution in an unyielding fluid – effectively the distance that the propeller would ‘screw’ though if it were rotated through a solid – just like turning a wood screw into a block of wood. The pitch and the diameter of a propeller are linked by the ‘pitch ratio’, which is a nondimensional ratio which measures the pitch of the propeller to the diameter of the propeller Pitch rrat atio o,

P Propeller pitch = D Propeller diameter

The blade area ratio (BAR) is the actual area of all of the blades compared to the area of the ‘propeller disc’, as shown in Figure 11.3, which is a solid disc of the same diameter as the propeller. BAR =

Blade area Disc area

Disc area =

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π D2 4

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Screw Propellers • 365

Propeller diameter

Propeller disc

▲ Figure 11.3 Propeller disc area

Propeller Performance (ENG)

Theoretical speed (ENG) As a propeller rotates, it tries to move forwards. In an unyielding material, the propeller would move forward by the pitch of the propeller. The propeller could, in theory, not travel faster than this speed under its own rotation. The maximum theoretical speed, vT, of the propeller can be found in terms of the pitch and the rotation speed, n, in revolutions per second: ν T = Pn ▲ Formula 11.1 Propeller theoretical speed

QUESTION Q11.1 A propeller has a pitch of 5 m and rotates at 2.5 rev/s. Determine the theoretical speed of the propeller.

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366 • Ship Stability, Powering and Resistance

Apparent slip (ENG) In practice, water is a yielding fluid, which means that the propeller moves forward through the water with a certain amount of ‘slip’. This is the maritime equivalent of ‘wheel spin’. The propeller blades do not move the vessel by ‘gripping’ the water as a traditional screw would, instead they create forward motion by creating a pressure differential and hence lift across the propeller blades. This is not an entirely efficient process, hence the actual distance moved in one rotation is less than the theoretical distance that the propeller would move in one rotation. The difference between the theoretical propeller speed and the actual ship speed is known as the apparent slip, abbreviated to SA. SA =

vT v vT

▲ Formula 11.2 Apparent slip ratio

The apparent slip is almost always referred to as the apparent slip ratio, which measures the apparent slip as a ratio to the theoretical speed. Typical values are around 4 to 6%.

QUESTION Q11.2 A ship has a speed of 12 m/s. The theoretical speed of the propeller is 12.6 m/s. Determine the apparent slip ratio.

Advance speed and the Taylor wake fraction (ENG) As the vessel travels through the water, a certain amount of water (‘entrained flow’) is pulled along behind the vessel, as shown in Figure 11.4. This is the region in which the propeller typically operates. Effectively, the speed through the water of the propeller is less than the ship speed. The actual speed of the propeller through the water is known as the advance speed, vA.

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Screw Propellers • 367 The difference between the ship speed and the advance speed is the wake speed, vW. v = v A + vW

Entrained flow moving forward close to the ship speed

▲ Figure 11.4 Propeller operating in the ship's wake

The ship speed and the advance speed are linked by a factor known as the ‘Taylor wake fraction’, often abbreviated to ω. This is very hard to calculate exactly for ships and standard values are normally used. For normal merchant vessel hull forms, it can be approximated by: = 0.5C 5 B − 0.05 The ship speed and the advance speed are linked by: vW

vA

ω ×v v ( 1−ω )

▲ Formula 11.3 Advance speed using the Taylor wake fraction

QUESTION Q11.3 A ship has a block coefficient of 0.6. The speed of the ship is 13 m/s. Determine the advance speed and wake speed.

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368 • Ship Stability, Powering and Resistance

Real slip (ENG) The difference between the theoretical speed and the advance speed is the real slip speed. The real slip is almost always referred to as the real slip ratio, which measures the real slip as a ratio to the theoretical speed. Typical values are around 30 to 40%.

SR =

vT

vA vT

▲ Formula 11.4 Propeller real slip ratio

QUESTION Q11.4 A ship has a theoretical speed of 10 m/s and an advance speed of 7 m/s. Determine the real slip.

Speed relationships It is helpful to note that the three speeds of interest are related by three factors as shown in Figure 11.5. V

ω

SA

VT

SR

VA

▲ Figure 11.5 Relationship between ship, advance and theoretical speed

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Screw Propellers • 369

QUESTION Q11.5 A ship has a speed of 11 m/s, with a propeller with a pitch of 4.5 m, rotating at a speed of 2.55 rev/s. The block coefficient of the ship is 0.66. Determine the theoretical speed, the apparent slip ratio, the advance speed, the wake speed and the real slip ratio.

Propeller thrust and torque (ENG) Previously, we looked at the powering of ships and found the thrust power, PT, produced by the propeller and the delivered power, PD, required to turn the propeller. The thrust, T, and torque, Q, of the propeller are linked to the thrust power and the delivered power: PT

T Tv vA

▲ Formula 11.5 Propeller thrust

PD

Qn

▲ Formula 11.6 Propeller torque

Manufacturers data can be used to compare propeller designs against thrust and torque values obtained from tests. This data can be used to specify a propeller which, for a given vA, will deliver enough thrust. This can be a bit of a trial and error process – often propellers are adjusted or changed after sea trials, as predicting values such as the Taylor Wake Fraction can be difficult. To try and reduce uncertainty, previous trial data on similar vessels, modern computing methods and computational fluid dynamics (CFD) are used to get good estimates

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370 • Ship Stability, Powering and Resistance for VA – however these are not 100% fool-proof or reliable, and a certain amount of judgement and experience is needed to predict these values. We have seen that the propeller itself increases the resistance of the vessel. Therefore, the thrust generated by the propeller must be more than the total resistance of the ship. This is modelled by the thrust deduction factor, t:

RTS

T ( 1− t )

▲ Formula 11.7 Thrust deduction factor

Typical values for t are around 0.1 to 0.3.

Hull efficiency (ENG) We have previously seen that the hull efficiency is given by: ηH =

PE PT

We know that the effective power can be expressed in terms of the total ship resistance and the ship speed, and the thrust power can be expressed in terms of the thrust. ηH =

PE RT v = PT Tv A

The total ship resistance and thrust are linked by the thrust deduction factor: RTS

T (1 − t )

This allows the hull efficiency to be expressed in terms of the thrust: ηH =

PE RT v T ( t ) v v (1 t ) = = = PT Tv A Tv A vA

The advance speed can be linked to the ship speed using the Taylor wake fraction: vA

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v (1 − ω )

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Screw Propellers • 371 This allows the hull efficiency to be expressed in terms of the thrust deduction factor and the Taylor wake fraction: ηH =

PE RT v T ( t ) v v (1 t ) v (1 t ) = = = = PT Tv A Tv A vA v (1 − ω )

Giving:

hH =

1− t 1−w

▲ Formula 11.8 Hull efficiency

Propeller Design Propeller design principles As with most aspects of naval architecture, the design of the propeller is a compromise between a range of different requirements and conflicting parameters. The initial design stages when selecting a propeller are to determine the maximum available diameter, based on the space available and a suitable tip clearance from the hull, the required number of blades and the BAR. In general, the diameter should be as large as possible, however the tip clearance should be no less than around 10 to 15% of the propeller diameter, to ensure that the flow around the blades does not cause vibration or cavitation issues with the hull surface. It may also be advantageous to keep the diameter below the maximum possible if it also keeps the propeller tips clear of disturbances in the wake field around the vessel, and hence increase vibration. The choice of number of blades depends on the individual vessel. Increasing the number of blades typically allows ‘tuning’ of vibration issues if the propeller is operating in a varied and turbulent in-flow, however increasing the number of blades generally results in reduced efficiency. The desired BAR must then be selected. If the BAR is too small, the loading on the propeller blades will be excessive and performance will be reduced through the process of cavitation. Excessive BAR can result in reduced efficiency, so it is important to select a propeller with the correct characteristics. Propeller selection often requires a certain amount of trial and error with iterative improvement.

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372 • Ship Stability, Powering and Resistance

Propeller efficiency The efficiency of a propeller operating behind a vessel differs from that obtained from open water testing. The propeller efficiency ηP is found by multiplying the open water efficiency η0 by the relative rotative efficiency ηR: P

ηR η0

The relative rotative efficiency is close to unity and typical values are single-screw vessels – 1.0 to 1.1 twin-screw vessels – 0.95 to 1.0 The values for single-screw propellers may at first glance appear surprising as the influence of the ship’s wake is generally responsible for a deterioration of performance. However, this can be offset by the added turbulence in the flow about the propeller due to the hull boundary layer and can have a beneficial effect.

BP-δ propeller design charts These are charts showing the performance of a range of propellers tested systematically for a range of blade numbers and BARs. Once the number of blades and BAR have been selected, the charts allow the selection of a suitable pitch diameter ratio based on the advance speed, rotation speed and delivered power. These values are used to determine the power coefficient, BP: BP =

NP 0 5 VA2 5

The design charts (see Figure 11.4) show the power coefficient along the x axis. To use these charts, the predicted BP value is determined, and marked on the x axis. A vertical line is then drawn up to the line of optimum efficiency (shown as the thick solid line in Figure 11.4), The efficiency lines (long dashes on Figure 11.4) on the chart can be used to determine the actual propeller efficiency. A horizontal line can then be drawn across from this point to the y axis, which gives the optimum pitch diameter ratio for the propeller.

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Screw Propellers • 373 The intersection between the vertical BP line and the optimum efficiency line can also be used to determine a value known as the speed coefficient, or delta, shown in Figure 11.4 by the short dashed line. The speed coefficient is given by: δ=

ND VA

This can then be used to determine the optimum diameter of the propeller.

ηP

‘delta’ lines

nD VA

P D

Optimum ηP

BP

▲ Figure 11.4 Sample BP δ graph

KT-KQ-J diagrams An alternative method of presenting propeller data is in the form of thrust and torque curves. These are graphs which again make use of non-dimensionalised data to allow the performance of a propeller to be predicted. In a similar way to the previous method, a range of charts exist, with specific charts for propellers of a given BAR and number of blades. These charts show three values on the y axis, known as the Thrust Coefficient, KT, and the Torque Coefficient, KQ, and open water efficiency η0 . The thrust and torque coefficients can be found from: Torque coefficient: KQ =

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Q ρ n2 D 5

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374 • Ship Stability, Powering and Resistance Thrust coefficient: KT =

T ρ n2 D 4

The x axis of the chart shows a non-dimensional value known as the advance coefficient, or J. This is found from: J=

VA nD

The open water efficiency of a propeller has previously been seen as the ratio of the thrust power to the delivered power. By expressing the thrust and torque in terms of their coefficients, the open water efficiency can be expressed in terms of the thrust and torque coefficients and the advance ratio: η0 =

PT T × VA K T ρ n2 D 4 × VA K T ρn ρ n2 D 4 × VA = = = PD 2π πQ 2π × K Q ρ n2 D 5 × n 2π × K Q ρ n2 D 5 n η0 =

K T × VA KT V JK T = × A = 2π K Q D 2π K Q nD 2π K Q

A typical KT-KQ chart is shown below. The thick solid lines show the open water efficiency for a range of P/D ratios, while the thin solid lines show the thrust coefficient for a range of P/D ratios. The dashed lines show the torque coefficient (multiplied by 10 to allow the same y axis to be used for all of the variables), again for a range of P/D ratios. At a simple level, the KT-KQ charts can be used to determine the propeller characteristics for a propeller with a given BAR and number of blades. If the propeller rotation speed is known (typically from the machinery and gearbox specification), and the advance speed is known (which can be estimated from the Taylor Wake Fraction and the ship speed), and diameter is known, then the advance coefficient J can be determined. For the calculated value of J, the KT and KQ curves can be used to determine the torque delivered and the thrust produced, along with the open water efficiency. Iterative calculations can be used to determine the most suitable propeller specification. More advanced methods exist to remove some of the iterative nature of this process, however they are beyond the scope of this text.

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Screw Propellers • 375 1.2

P/D lines

1.1 1

Torque coefficient lines

0.9 Efficiency lines

KT, 10KQ, η

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

1.1 1.2 1.3 1.4 1.5 1.6

Advance coefficient (J) Thrust coefficient lines

▲ Figure 11.5 Features of KT-KQ-J diagram

Cavitation Fresh water boils at 100°C in normal circumstances at sea level but at lower pressures will boil at lower temperatures. Hence, the climbers lament that you cannot get a hot cup of tea up a mountain due to the drop in atmospheric pressure at altitude. The pressure drop over the blades of a propeller can be large enough to cause the water to boil and this phenomena is called cavitation. The temperature and salt content of the water will vary the pressure at which it boils and is called the vapour pressure, pv. Typical values for salt and fresh water are shown in Table 11.1. Table 11.1 Vapour pressure of fresh and salt water at various temperatures Temperature (°C)

0.01

5

10

15

20

25

30

Fresh water pv (Pa)

611

872

1,228

1,704

2,377

3,166

4,241

Sea water pv (Pa)

590

842

1,186

1,646

2,296

3,058

4,097

Cavitation can cause extensive damage to a propeller as the collapse of bubbles of water vapour is surprisingly violent. To reduce the potential for cavitation the blade area of the propeller needs to be increased which reduces the loading and therefore

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376 • Ship Stability, Powering and Resistance pressure acting over the blade. An empirical method for working out the minimum blade area is: AE (1.3 + 0.3 Z )T = +K A0 ( p0 pv ) D 2

Vessel type

K value

Fast twin-screw

0

Other twin-screw

0.1

Single-screw

0.2

SCREW PROPELLERS  LEARNING CHECKLIST Objective

Completed

Understand the basic geometry and terminology associated with a propeller Understand how the performance of a propeller can be measured Calculate propeller slip

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12

PROPELLER ALTERNATIVES FOR HIGHSPEED CRAFT AIMS AND OBJECTIVES At the end of this section, you should be able to: Discuss the use of surface piercing propellers for ship propulsion Identify the role of waterjet propulsion systems for ship propulsion

At high speeds it becomes impractical to use a traditional screw propeller, this is largely due to the loss of efficiency and problems overcoming cavitation. The most widely used solutions to the problem are to fit surface piercing propellers or waterjets.

Surface Piercing Propellers A surface piercing propeller is positioned so that when the vessel is underway the waterline passes right through the propeller’s hub. The normal arrangement is to extend the propeller shaft out through the transom of the vessel as shown in Figure 12.1. In the case of articulated surface drive systems, the propeller shaft is driven through a double universal joint inside an oil-tight ball joint, allowing the shaft to

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378 • Ship Stability, Powering and Resistance

▲ Figure 12.1 Typical articulated surface piercing propeller drive system

rotate athwartships for steering and to trim up and down for control of propeller submergence. Fixed-shaft surface drives can use conventional shafts and stern tube bearings, but require rudders. In many racing applications, outboards and outdrives can be positioned sufficiently high on the vessel for the propellers to operate in a surface piercing mode. The important operating feature is that each propeller blade is out of the water for half of each revolution. A summary of the principal reasons for the high performance of surface propeller systems relative to conventional installations follows.

Propeller efficiency Traditional propeller design and selection is almost always an exercise in trading off diameter against several other performance-limiting parameters. Basic momentum theory tells us that for a given speed and thrust, the larger the propeller, the higher the efficiency. While there are exceptions, most notably the effects of frictional resistance on large, slow-turning propellers, it is generally borne out in practice that a larger propeller with a sufficiently deep gear ratio will be more efficient than a small one. A number of design considerations conspire to limit the maximum feasible propeller diameter to something considerably smaller than the optimal size. These include blade tip clearance from the hull, maximum vessel draft, shaft angle and engine location. While this may at times make life easy for the designer – the propeller diameter specified is simply the maximum that fits – it can also result in a considerable sacrifice of propulsive

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Propeller Alternatives for High-Speed Craft • 379 efficiency. And if these geometric limits on propeller diameter are exceeded, the result can be excessive vibration and damage due to low tip clearances, or a steep shaft angle with severe loss of efficiency and additional parasitic drag, or deep navigational draft that restricts operation or requires a protective keel and its associated drag. The surface piercing propeller frees the designer from these limitations. There is virtually no limit to the size of propeller that will work. The designer is able to use a much deeper reduction ratio, and a larger, lightly loaded, and more efficient propeller.

Cavitation When a submerged propeller blade cavitates, the pressure on part of the blade becomes so low that a near vacuum is formed. If the suction on the low-pressure side of the propeller blade dips below ambient pressure – atmospheric plus hydrostatic head – then a cavity forms. When these cavities collapse, water impacts on the blade surface with a local pressure singularity – that is, a point with theoretically infinite velocity and pressure. The effect can approximate that of hitting the blade with a hammer on each revolution. Cavitation is a major source of propeller damage, vibration, noise and loss of performance. And although high-speed propellers are often designed to operate in a fully cavitating (supercavitating) mode, problems associated with cavitation are frequently a limiting factor in propeller design and selection. The surface propeller effectively eliminates cavitation by replacing it with ventilation. With each stroke, the propeller blade brings a bubble of air into what would otherwise be the vacuum cavity region. The water ram effect that occurs when a vacuum cavity collapses is suppressed because the air entrained in the cavity compresses as the cavity shrinks in size. Although the flow over a superventilating propeller blade bears a superficial resemblance to that over a supercavitating blade, most of the vibration, surface erosion and underwater noise are absent. In theory, there is a slight performance penalty for allowing surface air into the lowpressure cavities but in practice, this effect is not significant considering the total thrust pressures involved in high-speed propellers. Note that cavitation can also be associated with sudden loss of thrust and high propeller slip, often caused by a sharp manoeuvre or resistance increase. This can still occur with surface propellers, although the propeller is ventilating rather than cavitating and the result is not as damaging.

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380 • Ship Stability, Powering and Resistance

Appendage drag Exposed shafts, struts and propeller hubs all contribute to parasitic drag. Inclined the exposed shafts not only produces form and frictional drag, but there is also induced drag associated with the Magnus-effect lift caused by their rotation. There is a surprising amount of power loss resulting from the friction of the shaft rotating in the water flow. In fact, for conventional installations a net performance increase can often be realised by enclosing submerged shafts in non-rotating shrouds, despite the increase in diameter. Surface propellers virtually eliminate drag from all of these sources, as the only surfaces to contact the water are the propeller blades and a skeg or rudder.

Variable geometry When a surface propeller is used in conjunction with an articulated drive system, the vessel operator then has the ability to adjust propeller submergence underway. This has roughly the same effect as varying the diameter of a fully submerged propeller, and allows for considerable tolerance in selecting propellers – or it allows one propeller to match a range of vessel operating conditions. This capability is somewhat analogous to adjusting pitch on a controllable pitch propeller. When the articulated drive is used for steering, the result can be exceptionally good high-speed manoeuvring characteristics. On single-shaft applications, drive steering can also be used to compensate for propeller-induced side force, without resorting to an excessively large rudder or skeg.

Shallow draft The vessel’s navigational draft can be as low as half a propeller diameter. Compared with other options for shallow water propulsion, most notably waterjets, surface propellers enjoy a very significant efficiency advantage. In the case of articulated drives, the propellers can be trimmed up until just the tips are submerged for intermittent operation in very shallow water, including beaching. Sometimes the design allows the propellers to trim sufficiently above the baseline so that the vessel can ‘dry out’ with the props well clear of the bottom. These are the intrinsic performance advantages of surface propellers. Other desirable characteristics include:

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Propeller Alternatives for High-Speed Craft • 381 Flexibility in machinery arrangement Ease of maintenance and repair Simplified installation In some applications involving hybrid propulsion systems, such as the combination of diesel cruise engines with a gas turbine sprint engine, the ability to retract one set of propellers completely clear of the water when not in use is an overriding consideration.

System selection Having elected to investigate the surface propulsion option, the builder or designer is faced with a series of major decisions and a very limited amount of reliable data. First is the issue of fixed versus articulated. As outlined above, articulated drives have the advantage of variable propeller submergence superior manoeuvrability extreme shallow draft capabilities. Fixed systems, on the other hand, do not require the hydraulic cylinders and associated pumps, control devices and high pressure plumbing. Furthermore, fixed systems are often designed to work with conventional solid shafts and stern tubes, rather than the more complex universal-joint drivelines found in articulated systems. It should also be noted that articulated surface drives should not be relied upon to control vessel trim angle. Trimming the drive up and down will have only a small effect on vessel running trim, and separate trim tabs or other devices may still be desirable. For fixed surface drives, the Levi Drive Unit is the most popular worldwide. This system is distinctive for its inverted U-shaped rudder that encloses the propeller. A handful of other fixed drive manufactures compete in certain areas. For articulated surface drives, the Arneson Surface Drive is the dominant product, thanks to the ‘universal joint inside a ball joint’ configuration patented by Howard Arneson.

Surface piercing propeller types The distinguishing features of a surface piercing propeller are that the pressure face of the blade is always concave, the leading edge is relatively sharp with a narrow entry angle, and the hub and blade root are built to withstand heavy eccentric and alternating loads. There is major incentive to keep the blade section thin. Nearly all successful designs have moderate to heavy trailing edge cupping.

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382 • Ship Stability, Powering and Resistance Surface propellers are usually associated with the stainless steel ‘cleaver’ style, shown in Figure 12.2, common to race boat applications. These propellers have straight trailing edges, razor-sharp leading edges, and sometimes as many as eight blades. The cleaver propeller is typically manufactured out of stainless steel, with some larger configurations made out of NiBrAl Bronze. Heavy cambered wedge sections, similar to super cavitating propellers, are standard with a blunt, squared-off trailing edge. Up to 8-blade cleaver propellers have been used to allow smoother operation and an increase in efficiency with reduced propeller submergence. These propellers typically are used for vessel speeds in excess of 50 knots. Depending on the size of the craft, the performance gains fall off using cleavers for the slower speed applications. As a general rule, there are better styles for the slower speed applications.

▲ Figure 12.2 Cleaver style surface piercing propeller

Low rake propellers have a very similar appearance to conventional propellers. Typically blade areas range from 0.75 for a 4-blade to 1.25 for a 6-blade. These propellers have a cambered blade section with a trailing edge cup. Sometimes, for slow speed applications, the cup is extended forward of the maximum radius on the propeller. The design and application for this style of propeller has come a long way recently. This is the propeller of choice for most applications under 50 knots. Propeller geometry can be altered on these propellers to enhance the slow and mid-speed performance. The high rake propeller was developed primarily as a ‘poor man’s cleaver’. This is a propeller that is easily manufactured and has evolved from the standard wedge-shaped cleaver propeller. These propellers generally are for use on the light, high-speed, small vessel applications.

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Propeller Alternatives for High-Speed Craft • 383

Vibration One of the amazing features of surface propulsion is its smoothness at high speed, due mainly to the suppression of cavitation. This is contrary to intuition, and must be experienced to be fully believed. However, some installations have experienced serious vibration problems. In most cases, this is due to improper design or alignment of the shafting between the gearbox and drive input shaft. When double universaljoint drivelines are required, as is the case with articulated systems, it is especially important to plan the driveline geometry so that operating angles of the two joints are approximately equal and within accepted tolerances. This is because a universal joint does not transmit rotational velocity evenly, causing angular acceleration and deceleration twice with each shaft revolution. The less common vibration problems that are not driveline-related can almost always be solved by using propellers with a larger number of blades, although there is some cost penalty involved.

Backing performance Surface propulsion has a reputation for very poor performance in reverse. A certain amount of this reputation is based on the fact that until very recently, nearly all surface propeller installations were on very high-speed vessels using ‘cleaver’ style propellers. These propellers, due to the thick trailing edges, concave pressure face and often heavy trailing edge cupping, are notoriously poor performers in reverse. And this is true whether they are used as surface propellers or as cavitating fully submerged propellers. However, there is an occasional problem with backing performance of surface propulsion systems, regardless of propeller style. Part of the slipstream of the propellers is directed right into the vessel’s transom, with an obvious loss of net astern thrust. Side curtains (hull side extensions aft of the transom) can seriously aggravate this condition. In fact, there has been at least one installation in which the vessel was actually propelled forward when the propellers were turning backwards at certain speeds. The aft overhang and side curtains combined to work like the reversing bucket on a waterjet, except that in this case reverse thrust was being ‘reversed’ to forward thrust. Fortunately there is an easy fix. The addition of baffle plates between the transom and the propeller that direct the slipstream down and forward (the plates are dry when the vessel is operating ahead at speed) has proved extremely effective. But for the majority of applications, no such hardware is required to provide adequate, although not outstanding, performance in reverse.

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384 • Ship Stability, Powering and Resistance

Transitional speeds Most planing hull designs, especially moderately low-powered or heavy designs, are subject to problems getting through ‘hump’ speed. High vessel resistance at pre-planing speeds, high propeller slip and reduced engine torque output at less than full RPM can sometimes combine to make it impossible to reach design speed, even though the vessel may be perfectly capable of operating at design speed once it gets there. The boat that ‘can’t get out of the hole’ is a phenomenon that should be quite familiar to many designers and builders. With surface propulsion systems there is an additional factor which may make the situation worse – the propeller is designed to operate with only half of the blade area immersed. But at low speeds, before the transom aerates or ‘drys out’, the propeller must operate fully submerged. Not only is the submerged area doubled, but the top half is operating in very strong wake turbulence right behind the transom. The result is that it takes much more torque to spin the propeller at a given RPM, and sometimes the engine is not capable of providing the torque necessary to turn the propeller fast enough to get the boat up to the speed which allows the transom to aerate and unload the top half of the propeller. To reduce this potential problem, various methods of aerating the top half of the propeller have been employed. The Levi Drive, for example, directs engine exhaust into the water in front of the propeller. On some installations, passive ‘aeration pipes’ leading from above the static waterline to the forward side of the propeller have been effective. When the lower surface of the aft overhang is below the static waterline, it is sometimes advisable to leave cut-outs through the overhang to let air get to the propellers. With articulated drives, maximum up-trim can sometimes reduce propeller submergence sufficiently to achieve required RPM for take-off power.

Waterjet Propulsion Waterjet propulsion may seem to be a recent innovation but production on a large scale began in the late 1950s and they were used on a Great Lakes steamboat as far back as the 1800s. In recent years, large jet units have been installed in medium size ships with powers up to 40 MW per unit.

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Propeller Alternatives for High-Speed Craft • 385 High speeds can be reached and the ideal speed range for jets is around 30 to 55 knots. Typical applications include: Fast Passenger Ferries Tourist Excursion Vessels Crew Boats Coast Guard & Police Patrol Vessels Military & Combat Support Boats Pilot Boats Pleasure Cruisers & Recreational Craft Search & Rescue Vessels Fishing & Other Work Boats

Waterjet Operating Principles Main features of waterjet system A single or multi-stage pump is connected directly to a gas turbine, diesel, petrol or other type of engine. Water is drawn through an opening in the bottom of the hull and discharged at high velocity out of the transom. The discharge nozzle acts as the rudder and is turned to port or starboard through conventional steering controls.

Advantages of waterjet propulsion Engine reversing gear is not required since a deflector plate provides reverse thrust. Boats can be operated in very shallow depths. No external shafts, struts, propellers or rudders protruding below the hull. Can have good acceleration characteristics.

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386 • Ship Stability, Powering and Resistance

Disadvantages of waterjet propulsion Limited application for moderate and slow speed craft. In certain cases waterjets can generate excessive noise and be less efficient than conventional propulsion systems. Installation tends to be heavier than the equivalent propeller based system. At higher speeds this is the major limitation that promotes the use of surface piercing propellers.

Waterjet System Components A waterjet system can be broken into three subsystems: Jet system Duct system Pump system The components of a waterjet system can be considered as in Figure 12.3.

Jet system Generally the jet diameter is smaller than the equivalent propeller’s slipstream. For the mechanical energy only, the theoretical efficiency is given by the ratio of useful work done to the work input. Efficiency =

Useful work done Work input

Propulsion force (thrust) = Rate of change of momentum T

ρ Av j v abs

where ρ – water density (kg/m3) A – nozzle CSA (m2) vabs – the absolute velocity of the waterjet (m/s) = vj – vs vs – the boat speed (m/s) vj – the velocity of the waterjet relative to the boat (m/s)  ρ Av j Note: mass flow rate, m

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Propeller Alternatives for High-Speed Craft • 387

Engine Service margin usually 10% Brake power, PB

Transmission Transmission efficiency, ηtransmission Delivered power, PD Pump Pump efficiency, ηpump Jet system Pump power, PP

Jet system efficiency, η js

Duct Inlet, head, friction, η duct Jet system power, PJS Nozzle Jet efficiency, η jet

Thrust power, PT = T.vs

Hull Thrust deduction factor, 1 – t

Effective power, PE = PT ·Vs

▲ Figure 12.3 Schematic diagram of a waterjet propulsion system

Now the work done on the water by the pump is to increase the kinetic energy of the water (no losses – ideal efficiency) Work do donee by pu pump p per

9781408176122_Ch12_Rev_txt_prf.indd 387

o ond d ( pump p p power p )

(

1 ρ Av j v j − v s 2

)

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388 • Ship Stability, Powering and Resistance Then the ideal efficiency of waterjet is defined by Tv s 1 ρ Av j v j 2 2v s η jet = v j + vs η jet =

(

vs

)

In order to increase the jet efficiency, the jet speed must be as small as possible, that is, big nozzle diameter.

QUESTION Q12.1

The resistance of a boat is 50,000 N, the boat speed is 60 knots and the jet efficiency is 60%. Calculate the waterjet speed, flow rate and waterjet diameter.

With additional enlarging of the jet pipe and the pumping of more water with less velocity change, the efficiency can be further increased. The type of pump best suited for large flows at small head is the axial flow propeller pump. Increasing the size of the pump and jet pipe would increase the weight greatly and take up useful space in the boat.

Duct system Due to the losses in the duct, the energy losses occurring in the ducting system can be accounted for by a ducting efficiency, ηduct. The duct losses will include the losses due to the inlet, friction within the duct, change of height of the fluid and flow through any flow conditioning devices in the jet system.

Pump system The pump system converts the delivered mechanical power PD into hydraulic power. η pump =

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PP PD

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Propeller Alternatives for High-Speed Craft • 389 where PP – Pumping power PD – Delivered power The pumping power required = increase in kinetic energy of water + energy losses in system PP

(

)

1 Av A j v j − v s + ρ gAv j H 2

where H is the head representing energy losses (m).

Interaction between the waterjet system and hull The waterjet system affects the hull performance. The bare hull resistance is changed due to a distortion in the flow about the aft body. At higher speeds this distortion can furthermore cause a distinct equilibrium position of the hull, thus causing another change in resistance. In an equilibrium situation, the actual resistance of the hull balances the net thrust delivered by the jet system. The change in hull resistance from the known bare hull resistance, RT, can be expressed by a thrust deduction factor, 1 – t RTS PT

T (1 − t ) TTvv s

Similarly, the jet system performance is affected by the flow distortion caused by the hull. Due to this distortion, a boundary layer is ingested through the intake area and the local flow velocity is likely to differ from the free stream or the hull’s velocity. As a result, the ingested momentum and energy fluxes and thus the delivered thrust and power will be affected. Analogous to interaction effects in the intake region, interaction effects may also occur in the nozzle region. This occurs, for instance, when the nozzle is submerged in the transom flow instead of in air. Correction on these fluxes can be applied by introducing a momentum interaction and an energy interaction efficiency respectively.

Overall efficiency of the waterjet system

ηoa =

9781408176122_Ch12_Rev_txt_prf.indd 389

PE RT v s = = η pump η duct η jet (1 − t ) d PD PD

js

jet

(1 t )

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390 • Ship Stability, Powering and Resistance

QUESTION Q12.2

A high-speed motor boat is designed to reach a speed of 60 knots. It is to be fitted with two waterjet units. The shaft power of each engine is 1,300 kW and the effective power of the boat is 1,600 kW. The efficiency of the axial pump of the waterjet unit is 87% and the internal and kinetic energy losses in the inlet manifold of the waterjet is 7%. Calculate the jet outlet diameter assuming an equivalent thrust deduction factor of 0.08.

PROPELLER ALTERNATIVES FOR HIGHSPEED CRAFT  LEARNING CHECKLIST Objective

Completed

Discuss the use of surface piercing propellers for ship propulsion Identify the role of waterjet propulsion systems for ship propulsion

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13

RUDDERS AIMS AND OBJECTIVES At the end of this section, you should be able to: Understand how a rudder generates forces Understand how the forces can be resolved into a transverse (turning) force and the drag force Calculate the forces generated by a rudder Calculate the stresses in the rudder stock caused by the rudder forces

Rudder Forces The role of the rudder is to see that all the vessels have directional control. To do this, it must generate a force acting transversely to the direction of travel. The simplest way to do this efficiently is to use an aerofoil section to generate lift forces, in a similar way to an aircraft wing. These forces are transmitted into the ship’s structure via the rudder stock, creating stress in the rudder. In order to understand and calculate the stress in the rudder stock, it is important to understand the forces generated. In the central position (and assuming that the vessel has no leeway) the rudder generates hydrodynamic drag, which acts directly astern:

Drag Water flow

Rudder blade

9781408176122_Ch13_Rev_txt_prf.indd 391

Rudder stock

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392 • Ship Stability, Powering and Resistance When the rudder is moved so that it is at an angle of attack to the water flow, it generates a lift force, which acts at right angles to the rudder blade. This is referred to as the Normal Force, or FN. Normal force at right angles to the rudder blade

Water flow

This ‘normal force’ effect acts both transversely and astern relative to the ship itself. In order to determine the effects of these, the normal force must be resolved into the transverse and drag components, relative to the centreline of the ship: FN

Transverse force

Drag Water flow

If the angle of attack of the rudder (given the symbol alpha, α) is known, then the transverse and drag forces can be found from: Transverse force Drag force

FN cos α

FN sinα

QUESTION Q13.1 (ENG) A rudder generates 20kN of normal force at an angle of 15 degrees. Determine the drag and the transverse force.

The normal force itself depends on a number of factors, such as the shape of the rudder, the area of the rudder, the location of the rudder relative to the propellers and the flow into the rudder around the hull. This means that there is no single formula that can be used to determine the normal force. Instead, empirically determined formula can be used, normally in the form of: FN

Coefficient Rudder plan area SSpeed 2 × Rudder angle

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Rudders • 393

QUESTION Q13.2 (ENG) A rudder has an plan area of 20m2. Calculate the normal force at a speed of 10 knots and a rudder angle of 15 degrees if the normal force is given by: FN

560 A × V 2

sinα

(Note that V is in units of metres per second.)

Rudder Torque and Bending Moment The rudder forces act at the centre of pressure of the rudder. This is also known as the centre of effort. The actual centre of effort varies with dynamic factors such as speed and angle, but can be approximated by taking the centre of area of the rudder. The hydrodynamic drag and transverse forces act together through the centre of effort to create two moments of force on the rudder stock, rudder torque and rudder bending moment:

FN

Torque FN

Bending moment

Viewed from above Viewed from astern

The distance from the centre of effort, and hence the normal force, and the centre of the rudder stock is known as the Torque Lever, while the vertical distance from the centre of effort to the lower rudder bearing is known as the bending lever: FN Bending lever

Torque lever

9781408176122_Ch13_Rev_txt_prf.indd 393

FN

11/16/2013 2:47:33 AM

394 • Ship Stability, Powering and Resistance The Torque acting on the rudder can therefore be found from: T

FN × Torque lever

The bending moment acting on the rudder can be found from: M

FN × Bending lever

QUESTION Q13.3 (ENG) A rudder has an plan area of 10 m2. The torque lever is 1.00 m, and the bending moment lever is 4.00 m. Calculate the torque and bending moment acting on the rudder stock at a speed of 8 knots and a rudder angle of 10 degrees. You may assume that the normal force is given by: FN

500 A × V 2

sinα

Equivalent Torque As seen, the rudder forces generate both a torque in the rudder shaft and a bending moment. The combined effect of these two stress generating mechanics is known as the equivalent torque, and is given by: TE = M + M 2

T2

Where TE is the equivalent torque, T is the actual torque, and M is the bending moment.

QUESTION Q13.4 (ENG) A rudder has an plan area of 15 m2. The torque lever is 0.50 m, and the bending moment lever is 2.60 m. Calculate the equivalent torque acting in the rudder stock at a speed of 10 knots and a rudder angle of 25 degrees. You may assume that the normal force is given by: FN

9781408176122_Ch13_Rev_txt_prf.indd 394

80 A × V 2

sinα

11/16/2013 2:47:34 AM

Rudders • 395

Rudder Stock Stress Once the equivalent torque has been determined, the stress in the rudder stock (q) can be determined. This depends on both the equivalent Torque, TE, and the stock diameter, D: q=

16TE π D3

QUESTIONS Q13.5 (ENG) A rudder has an plan area of 12 m2. The torque lever is 0.60 m, and the bending moment lever is 2.30 m. Calculate the stress acting in the rudder stock at a speed of 13 knots and a rudder angle of 35 degrees, if the rudder stock is 0.50 m in diameter. You may assume that the normal force is given by: FN

505 A × V 2

sinα

Q13.6 (ENG) A rudder has an area of 6 m2. At the maximum rudder angle of 35 degrees, the centre of effort of the rudder is 0.15 m aft of the axis of rotation, and 1.65 m below the lower edge of the rudder stock bearing. The stress in the stock must be limited to 77 MN/m2. Determine the required diameter of the rudder stock for a ship speed of 18 knots. The normal force is given by: FN

8.0 AV 2α

Q13.7 (ENG) Due to corrosion, the rudder stock in Question 13.6 is reduced in diameter to 0.400 m. Determine the limiting ship speed to ensure the rudder stock is not overstressed.

RUDDERS  LEARNING CHECKLIST Objective

Completed

Understand how a rudder generates forces Understand how the forces can be resolved into a transverse (turning) force and the drag force Calculate the forces generated by a rudder Calculate the stresses in the rudder stock caused by the rudder forces

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14

SOLUTIONS TO QUESTIONS These solutions provide a brief outline of the numerical solution to the questions in the text. Q1.1 Archimedes’ Principle tells us that: ∇×ρ = Δ For a box shaped vessel, the underwater volume can be found from the length, beam and volume: ∇= L×B×D Therefore: ∇ = 30 × 5 × 2 = 300 m3 Substituting this into the formula for Archimedes’ Principle gives: ∇ × ρ = Δ ∴ 300 × 1.025 = 307.5 tonnes Q1.2 ∇×ρ = Δ ∴L × B × D × ρ = Δ ∴ 50 × 7 × D × 1.025 = 1, 076.25 D = 3.00 00 m Q1.3 Before loading: ∇×ρ = Δ

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Solutions to Questions • 397 ∴ 60 × 8 × 3 × 1.025 = 1, 476 tonnes After loading: Δ = 1, 476 + 524 = 2, 000 tonnes ∇×ρ = Δ ∴ 60 × 8 × D × 1.025 = 2, 000 D = 4.07 07 m Q1.4 In sea water: ∇×ρ = Δ ∴ 50 × 7 × 2 × 1.025 = 717.5 tonnes In fresh water: ∇×ρ = Δ ∴ 50 × 7 × D × 1.000 = 717.5 D = 2.05 m Q1.5 CW =

Waterplane area LWL BWL

0 65 =

Waterplane area 100 × 20

∴Waterplane area = 0.65 × 100 × 20 = 1, 300 m2 Q1.6 CM =

Amidships area BWL D

0 95 =

Amidships area 15 × 8

∴ Amidships area = 0 95 × 15 × 8 = 114 m2 Q1.7 From the 5.30 m draught row, the displacement can be read as 5,150 tonnes.

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398 • Ship Stability, Powering and Resistance Q1.8 From the 6,008 tonne row, the draught can be read as 6.00 m. Q1.9 At a draught of 3.48 m, the displacement will be between 3,136 and 3,030 tonnes, as shown in the hydrostatics, and reproduced here: Draught (m) Displacement (tonnes) 3.50

3,136

3.40

3,030

From this we can determine the ratio of a/b and A/B: 3.48 3.40 A = 3.50 3.40 3,136 − 3, 030 A = 84.8 The actual displacement will be: 3, 030 + 84 8 .8 3,114.8 tonnes Q1.10 At a displacement of 3,365 tonnes, the draught will be between 3.70 and 3.80 m, as shown in the hydrostatics, and reproduced here: Draught (m) Displacement (tonnes) 3.80

3,455

3.70

3,348

From this we can determine the ratio of a/b and A/B: 3, 365 − 3, 348 A = 3, 455 − 3, 348 3.80 3.70 A = 0.016 The actual draught will be: 3.70 0.016 = 3.716 m = 3.72 m to 2 decimal places

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Solutions to Questions • 399 Q1.11 Sinkage =

Mass TPC

Sinkage =

t t cm = = cm t t cm

Sinkage =

Mass 70 = = 2 cm TPC 35

Q1.12

Q1.13 Sinkage = 20 =

Mass TPC

Mass ∴ Mass = 500 tonnes 25

Q1.14 Sinkage = 5 2.5 =

Mass TPC

Mass ∴ Mass = 195 tonnes 26

Q1.15 Sinkage = 9 2.5 =

Mass TPC

Mass ∴ Mass = 143 tonnes 22

Q1.16 Tropical to summer mark ar = 20.4 2.5 =

DSummer 9 80 = = 0.204 m 20 0.4 cm 48 48

Mass ∴ Mass = 458 tonnes 20

Q1.17 TPC

CW × L B × 0 01× ρ

TPC = 0.8 × 110 × 15 × 0.01× 1.025 = 13.52 t/cm

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400 • Ship Stability, Powering and Resistance Q1.18 TPC DW

T SW × TPC

ρDDW 1.000 = 25 × = 24.39 t/cm ρSSW 1.025

TPC DW

T SW × TPC

ρDDW 1.010 = 30 × = 29.56 t/cm ρSSW 1.025

Q1.19

Q1.20 FWA W =

Δ SUMMER E 4 × TPC SW AT THE SU SUMMER E MARK

=

10 , 000 = 83.3 4 × 30

= 8.3 8. cm

Q1.21 The fresh water allowance can be found: FWA W =

Δ SUMMER E 4 × TPC SW AT THE SU SUMMER E MARK

=

15, 000 = 107 mm 4 × 35

The dock water allowance can then be found: DWA W FWA FWA ×

( (

S SW

−

SW

−

) = 107 × (1.025 − 1.015) = 43 4 mm (1.025 − 1.000 ) FFW )

DW D

Q1.22 The fresh water allowance can be found: FWA W =

Δ SUMMER E 4 × TPC SW AT THE SU SUMMER E MARK

=

11, 000 = 95 mm 4 × 29

The dock water allowance can then be found: DWA W FWA FWA ×

( (

S SW

−

S SW

−

) = 95 × (1.025 − 1.005) = 76 (1.025 − 1.000 ) FFW )

DW D

7.6 cm

The TPC can be adjusted for dock water: TPC DW

T SW × TPC

ρDDW 1.005 = 29 × = 28.43 43 t/cm ρSSW 1.025

The sinkage can be found and therefore the mass: Sinkage =

9781408176122_Ch14_1_Rev_txt_prf.indd 400

Mass TPC

11/16/2013 6:57:25 PM

Solutions to Questions • 401 Mass ∴ Mass = 429.3 tonnes 28.43

5 2.5 7.6 = Q1.23

Δ SUMMER 14 , 000 E = = 100 mm 4 × TPC SW AT THE SUMME 4 × 35 SUMMER E MARK

FWA W = DWA W FWA FWA ×

( (

SW

−

SW

−

) = 100 × (1.025 − 1.020) = 20 2 mm (1.025 − 1.000 ) FFW )

DW D

T SW × TPC

TPC DW

2 cm

ρDDW 1.020 = 35 × = 34.83 83 t/cm ρSSW 1.025

Sinkage =

Mass TPC

Mass ∴ Mass = 452.8 8 tonnes 34.83

15 − 2 = Q1.24

The winter displacement can be found from the TPC in sea water and the FWA: TPC DW 33 = TPC SW × FWA W = 290 =

T SW × TPC

ρDDW ρSSW

1.018 ∴ TPC SW = 33.23 t/cm 1.025 ΔWinter

4 × TPC SW AT THE WWINTEER MARK

ΔWinter ∴ ΔWinter = 38 , 546.8 tonnes 4 33 23

The Load Line data can be used to determine the required change in draught, and hence the mass to load. Tropical to summer mark ar = DWA W FWA FWA ×

( (

SW

−

SW −

DSummer 10.60 = = 0.221 m 22.1 cm 48 48

) = 290 × (1.025 − 1.018) = 81 8 .2 (1.025 − 1.000 ) FFW )

DW D

Sinkage =

9781408176122_Ch14_1_Rev_txt_prf.indd 401

8.1 cm

Mass TPC

11/16/2013 6:57:27 PM

402 • Ship Stability, Powering and Resistance

49 − 22.1 8.1 =

Mass ∴ Mass = 1155 tonnes 33

Q1.25 The fresh water allowance can be determined: FWA W =

Δ SUMMER 7329 E = = 133 mm 4 × TPC SW 4 × 13 77

The dock water allowance can be determined: ⎛ ρ −ρ ⎞ ⎛ 1.025 − 1.010 ⎞ DWA W FWA ⎜ SW DDW ⎟ = 133 = 80 mm ⎝ 1.025 − 1.000 ⎠ ⎝ ρSW − ρFFW ⎠ Therefore, the vessel can have her Load Lines submerged by 80 mm, or 8 cm. The vessel has a summer draught of 7.00 m, therefore the final draught will be 7.08 m. As the starting waterline is 40 cm, or 0.4 m below the summer Load Line, the initial draught must be 6.60 m. The TPC in seawater at 7.08 m can be interpolated as 13.85 t/cm, and the TPC in seawater at 6.60 m is 13.33 t/cm. The mean of these values can be found: Mean TPC C=

13.85 + 13.33 = 13.59 t/cm 2

This is the seawater value, and must therefore be converted to the dock water value: TPC DW

T SW × TPC

ρDDW 1.010 = 13.59 × = 13.39 39 t/cm ρSSW 1.025

The allowable sinkage is 40 cm to the summer Load Line, and then 8 cm to the allowable waterline, giving a total allowable sinkage of 48 cm. Therefore, the mass to add can be determined: Sinkage = 48 =

Mass TPC

Mass 13.39

Mass = 642.72 72 tonnes Therefore, using the mean TPC value, the mass to load is 642.72 tonnes. Q2.1 The centre of buoyancy is at the overall centre of underwater volume. For a simple box shape, this will be at half of the length, on the centreline, at half of the draught.

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Solutions to Questions • 403 Therefore, the LCB will be 50 m FOAP, the TCB will be zero (as it is measured from the centreline) and the KB will be 1.00 m. Q2.2 In the Stability Data Book (see Appendix 1), look at the pages detailing the Upright Hydrostatics. Find the 3,563 tonnes row, and read across to the LCB and KB columns. The LCB value is 47.06 m FOAP, and the KB value is 2.11 m. Q2.3 Item

Mass

KG

Moment

Lightship

2,615

7

18,305

Cargo

250

5

1,250

Total

2,865

Item

Mass

LCG

Moment

Lightship

2,615

44

115,060

Cargo

250

55

13,750

Total

2,865

Item

Mass

TCG

Lightship

2,615

0

Cargo

250

Total

2,865

KG =

Moment 19 , 555 = = 6.83 83 m Mass 2, 865

19,555

LCG =

Moment 128 , 810 = = 44.96 m 2, 865 Mass

128,810

0.5

Moment 0 125

TCG =

Moment 125 = = 0.044 Mass 2, 865

(port of centreline )

125

Q2.4 Item

Mass

KG

Moment

Lightship

2,615

7

18,305

Cargo

500

8

4,000

Total

3,115

Item

Mass

LCG

Moment

Lightship

2,615

44

115,060

Cargo

500

48

24,000

Total

3,115

9781408176122_Ch14_1_Rev_txt_prf.indd 403

KG =

Moment 22, 035 = = 7.16 16 m Mass 3,115

22,305

LCG =

Moment 139 , 060 = = 44.64 m Mass 3,115

139,060

11/16/2013 6:57:31 PM

404 • Ship Stability, Powering and Resistance

Item

Mass

TCG

Moment

Lightship

2,615

0

0

Cargo

500

Total

3,115

0

Moment 0 = = 0.000 Mass 3,115

TCG =

0

(on centreline)

0

Q2.5 Item

Mass

KG

Moment

Ship

2,915

6.79

19,792.9

Cargo

–150

8

–1,200

Total

2,765

Item

Mass

LCG

Ship

2,915

45.13

Cargo

–150

45

Total

2,765

Item

Mass

TCG

Moment

Ship

2,915

0.05

145.75

Cargo

–150

Total

2,765

KG =

Moment 18 , 592.9 = = 6.72 72 m Mass 2, 765

LCG =

Moment 124 , 814 = = 45.14 m Mass 2, 765

18,592.9 Moment 131,554 –6,750 124,804

–0.1

15

TCG =

Moment 160.75 = = 0.058 Mass 2, 765

(port of centreline )

160.75

Q2.6 Item

Mass

KG

Moment

Ship

5,000

6

30,000

Cargo off

–500

2

–1,000

Cargo on

500

4

2,000

Total

KG =

9781408176122_Ch14_1_Rev_txt_prf.indd 404

5,000

31,000

Moment 31, 000 = = 6.20 20 m Mass 5, 000

11/16/2013 6:57:33 PM

Solutions to Questions • 405 Alternative solution for removing the mass from 10 m and replacing it at 12 m: Item

Mass

KG

Moment

Ship

5,000

6

30,000

Cargo off

–500

10

–5,000

Cargo on

500

12

6,000

Total

5,000

KG =

31,000

Moment 31, 000 = = 6.20 m Mass 5, 000

Q2.7 Item

Mass

Ship

6,615

Cargo off

–500

100

–50,000

Cargo on

500

97

48,500

Total

KG =

KG 5.79

6,615

Moment 38,300.9

36,800.9

Moment 36 , 800.9 = = 5.56 56 m Mass 6 , 615

Q2.8 Item

Mass

KG

Moment

Ship

6,000

7

42,000

Cargo off

–50

3

–150

Cargo on

50

20

1,000

Total

KG =

9781408176122_Ch14_1_Rev_txt_prf.indd 405

6,000

42,850

Moment 42, 850 = = 7.14 14 m Mass 6 , 000

11/16/2013 6:57:34 PM

406 • Ship Stability, Powering and Resistance Q2.9 Item

Mass

KG

Moment

Ship

8,000

6

48,000

Cargo

4,000

x

4,000x

Total

12,000

48,000 + 4,000x

KG =

Moment Mass

As the final KG is known, the equation for KG only contains one unknown, x, therefore x can be found: 7 00 =

48 , 000 + 4 , 000 x 12, 000

7.00 12, 000 = 48 , 000 + 4 , 000 x 7.00 12, 000 − 48 , 000 = 4 , 000 x 7.00 12, 000 − 48 , 000 =x 4 , 000 x = 9.00 m Q2.10 Item

Mass

KG

Moment

Ship

10,000

12

120,000

x

7

7x

Cargo on Total

10,000 + x

KG = 11.00 = Bilged drau r ght

120,000 + 7x

Moment Mass

120 , 000 + 7 x 10 , 000 + x

Original O iginal ginal draught + Parallel sinkage = 2 + 0.14 = 2.14 1 m 110 , 000 + 11 = 120 , 000 + 7 x

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Solutions to Questions • 407

KB =

Moment 2, 060.81 = = 1.04 m Volume 1, 975.2 4

x=

10 , 000

10 , 000 = 2,500 500 tonnes 4

Q2.11 Item

Mass

KG

Moment

Ship

20,000

9

180,000

–x

5

––5x

Cargo off Total

20,000 – x

KG = 10.00 =

180,000 – 5x

Moment Mass

180 , 000 − 5 x 20 , 000 − x

10.00 (20 ,000 , 000

) = 180 , 000 − 5 x

200 , 000 −10x 10 = 180 , 000 − 5 x 200 , 000 = 180 , 000 − 5

10 x

200 , 000 − 180 , 000 = −5 5

10 x

20 , 000 = 5 x 20 , 000 = x = 4 , 000 tonnes 5 Q2.12 At a draught of 4.80 m, the KM in the hydrostatics is 6.94 m. GM = KM − KG = 6.94 − 5.00 = 1.94 m As GM is positive, the vessel is stable at small angles.

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408 • Ship Stability, Powering and Resistance Q2.13 At a draught of 3.66 m, the KM in the hydrostatics can be interpolated to be 7.30 m. GM = KM − KG = 7.30 − 7.32 = −0.02 02 m As GM is negative, the vessel is unstable at small angles. Q2.14 A loading table can be used to determine KG: Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Lightship

2,615

7.00

18,305

Cargo

1,386

8.00

11,088

Total

4,001

KG =

29,393

Moment 29 , 393 = = 7.35 35 m Mass 4 , 001

From the hydrostatics, at 4,001 tonnes displacement the KM is 7.03 m. This allows GM to be found: GM = KM − KG = 7.03 − 7.35 = −0.32 m GM is negative, therefore the vessel is unstable. Q2.15 A loading table can be used to determine KG: Item

Mass (tonnes)

KG (m)

Ship

6,008

6.50

39,052

Cargo

–374

7.00

–2,618

Total

5,634

KG =

Moment (tonne metres)

36,434

Moment 36 , 434 = = 6.47 47 m Mass 5, 634

From the hydrostatics, at 4,001 tonnes displacement the KM is 6.96 m. This allows GM to be found: GM = KM − KG = 6.96 − 6.47 = 0.49 m GM is positive, therefore the vessel is stable.

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Solutions to Questions • 409 Q2.16 A loading table can be used to determine KG: Item

Mass (tonne)

KG (m)

Moment (tonne metres)

Ship

7,192

6.90

49,624.8

Remove cargo

–500

9.00

–4,500

Replace cargo

+500

6.00

3,000

Total

7,192

KG =

48,124.8

Moment 48 ,124.80 = = 6.69 m Mass 7,192

Alternatively, the distance moved can be used in the loading table: Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Ship

7,192

6.90

49,624.8

Remove cargo

–500

3.00

–1,500

Replace cargo

+500

0.00

0

Total

7,192

KG =

48,124.8

Moment 48 ,124.80 = = 6.69 m Mass 7,192

Alternatively, the change in KG can be found: Change in KG =

Mass moved o ed Distance D sta ce moved 500 × 3 = 0 21 m = Δ 7,192

As the cargo has moved down, the KG must reduce, therefore: Final K KG G = 6.90 − 0.21 = 6 69 m At 7,192 tonnes, KM is 7.23 m. GM = KM − KG = 7.23 − 6.69 = 0.54 m Q2.17 At a draught of 5.00 m, KM is 6.93 m. Therefore, in the starting condition GM can be found: GM = KM − KG = 6.93 − 6.00 = 0.93 93 m ← Starting GM

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410 • Ship Stability, Powering and Resistance With the mass suspended, the KG can be found with a loading table: Item

Mass (tonnes)

Starting ship

KG (m)

Moment (tonne metres)

4,796

6.00

28,776

Remove cargo

–70

4.00

–280

Suspend cargo

70

30.00

2,100

Total

4,796

KG =

30,596

Moment 30 , 596 = = 6.38 m Mass 4 , 796

Alternatively, the distance moved can be used in the loading table: Item

Mass (tonnes)

KG (m)

4,796

6.00

Starting ship

Moment (tonne metres) 28,776

Remove cargo

–70

0

0

Suspend cargo

70

26

1,820

Total

4,796

KG =

30,596

Moment 30 , 596 = = 6.38 m Mass 4 , 796

Alternatively, the change in KG can be found: Change in KG =

Mass moved o ed Distance D sta ce moved 70 × 26 = = 0 38 m Δ 4 , 796

As the cargo has moved up, the KG must increase, therefore: Final K KG G = 6.00 + 0.38 = 6 38 m The displacement is still 4,796 tonnes, so KM is still 6.96 m. Therefore, GM can be found with the mass suspended: GM = KM − KG = 6.93 − 6.38 = 0.55 m ← GM with tthee cargo r suspended With the mass suspended, the displacement is 4,796 tonnes with a KG of 6.38 m. When the mass is stowed, the KG can be found with a loading table that removes the mass from the suspended position and stows the mass on deck:

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Solutions to Questions • 411

Item Ship with mass suspended Remove cargo from suspension Replace cargo on deck Total

Mass (tonnes)

KG (m)

Moment (tonne metres)

4,796

6.38

30,598.48

–70

30

–2,100

70

12

840

4,796

KG =

29,338.48

Moment 29 , 338.48 = = 6.12 m Mass 4 , 796

Alternatively, the same solution can be found by directly moving the mass from the stowed position to the suspended position: Item

Mass (tonnes)

Starting ship Remove cargo from hold Replace cargo on deck Total

KG (m)

Moment (tonne metres)

4,796

6.00

28,776

–70

4.00

–280

70

12.00

840

4,796

KG =

29,336

Moment 29 , 336 = = 6.12 m Mass 4 , 796

The displacement is still 4,796 tonnes, so KM is still 6.96 m. Therefore, GM can be found with the mass stowed on deck: GM = KM − KG = 6.93 − 6.12 = 0 81

← GM with the cargo r stowed on deck

Q2.18 The mass of cargo to load can be determined from the initial displacement and the final displacement: Cargo = Δ FINAL − Δ START Cargo = 5,150 − 2, 615 = 2, 535 tonnes At a displacement of 5,150 tonnes, the hydrostatic data shows that the KM is 6.93 m. Therefore, for a GM of 1.50 m, the required final KG can be found: GM = KM − KG −10.500 × 6 Final K KG G = 5 43 m

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412 • Ship Stability, Powering and Resistance Using the final KG as a target value, a loading table can be created, with the KG of the cargo as x: Item

Mass

KG

Moment

Lightship

2,615

7

18,305

Cargo on

2,535

x

2,535x

Total

5,150

18,305 + 2,535x

KG = 5 43 =

Moment Mass

18 , 305 + 2, 535 x 5,150 x = 3.81 m

Q2.19 GM = KB + BM − KG LB 3 D Inertia LB KB = B BM = Inertia = ∴ BM = 12 2 ∇ 12 ∇ 3

The individual formulae for KB and BM can be substituted: LB 3 100 × 103 D 2 12 GM = + 12 − KG = + − 4 = 1.16 m 2 ∇ 2 100 × 10 × 2 Q2.20 LB 3 D GM = KB + BM − KG = + 12 − KG 2 ∇

h 0 15

1 2

60 × 83 12 K KG 60 × 8 × 1

KG = 5.68 m

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Solutions to Questions • 413 Q2.21 LB 3 D GM = KB + BM − KG = + 12 − KG 2 ∇ L × B3 3 ∴ GM = + 12 − 1.25 2 L×9×3 ∴ GM = 1.5 +

B2 92 − 1.25 = 1.5 + − 1.25 = 2.50 m 12 × 3 12 × 3

Q2.22 The initial displacement can be found: ∇ × ρ = Δ ∴ Δ = 60 × 8 × 2 × 1.025 025 = 984 tonnes A loading table can be used to determine the KG after loading: Item

Mass

KG

Moment

Start

984

3

2,952

Cargo on

492

5

2,460

Total

1,476

KG =

5,412

Moment 5, 412 = = 3.67 m Mass 1, 476

The new draught of the vessel must be found: ∇ × ρ = Δ ∴ Δ = 60 × 8 × D × 1.025 0 5 = 1, 476 6∴D =

1, 476 = 3.00 m 60 × 8 × 1.025

The new metacentric height can be found: LB 3 D GM = KB + BM − KG = + 12 − KG 2 ∇ L × B3 3 ∴ GM = + 12 − 3 67 2 L×9×3 60 × 83 12 ∴ GM = 1.50 + − 3.67 = 1.50 + 1.78 − 3.67 = −0.39 m 60 × 8 × 3

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414 • Ship Stability, Powering and Resistance The GM is negative, therefore the vessel will be unstable. Q2.23 As has been seen previously, for this vessel after loading KB is 1.50 m and BM is 1.78. If GM is to remain positive, the final maximum KG of the vessel can be found: GM = KB + BM − KG ∴ 0 = 1.50 + 1.78 − KG ∴ KG = 3.28 28 m maximum A loading table can be used to determine the KG of the cargo to give an overall KG of 3.28 m: Item

Mass

KG

Moment

Start

984

3

2,952

Cargo on

492

x

492x

Total

KG =

1,476

2,952 + 492x

Moment 2, 952 + 492 x ∴ 3 28 = ∴ x = 3.84 84 m Mass 1, 476

Q2.24 tanθ

TCG GM

tanθ =

0.043 0 81

θ = 3.039 = 3.0 degrees (port) Q2.25 LB 3 D GM = KB + BM − KG = + 12 − KG 2 ∇ 70 × 93 2 12 ∴ GM = + − 4 = 0.38 38 m 2 70 × 9 × 2 tanθ =

w ×d 100 × −0 5 = Δ × GM (70 × 9 × 2 × 1.025) × 0.38

θ = −5.8 = 5.8 degrees to starboard Alternatively, a loading table for TCG could be used: ∇×ρ = Δ 70 × 9 2 × 1.025 = 1, 291.5 5 tonnes

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Solutions to Questions • 415

Item

Mass

Ship

1,291.5

0

0

Cargo off

–100

0

0

Cargo on

100

–0.5

Total

Moment

–50

1,291.5

TCG =

–50

Moment −50 = = −0.039 039 m Mass 1, 291.5 tanθ =

θ = tan−1

TCG

TCG GM

−0.039 = −5.8 degrees (starboard) 0 38

Q2.26 A loading table can be used to determine the KG after loading: Item

Mass

KG

Moment

Lightship

2,615

7

18,305

Cargo on

2,654

5

13,270

Total

5,269

KG =

31,575

Moment 31, 575 = = 5.99 99 m Mass 5, 269

At 5,269 tonnes, the KM is 6.93 m. GM = KM − KG GM = 6.93 − 5.99 = 0.94 m A loading table can be used to determine the TCG after loading:

9781408176122_Ch14_1_Rev_txt_prf.indd 415

Item

Mass

TCG

Lightship

2,615

0

Cargo on

2,654

0.050

Total

5,269

Moment 0 132.700 132.700

11/16/2013 6:57:51 PM

416 • Ship Stability, Powering and Resistance

TCG =

Moment 132.7 = = 0.025 m Mass 5, 269 tanθ =

θ

1

0.025 0 94

TCG GM

θ = 1 5 degrees to port

Q2.27 Item

Mass

KG

Moment

Lightship

2,615

7

18,305

Cargo on

1,000

5

5,000

Cargo on

497

8

3,976

Total

4,112

27,281

Moment 27, 281 = = 6.63 63 m Mass 4 ,112

KG =

At 4,112 tonnes, the KM is 7.01 m. GM = KM − KG GM = 7.01 − 6.63 = 0.38 38 m Item

Mass

TCG

Moment

Lightship

2,615

0

Cargo on

1,000

0.300

300.000

Cargo on

497

–0.500

–248.500

Total

4,112

TCG =

51.500

Moment 51.5 = = 0.013 m Mass 4 ,112 tanθ =

θ

9781408176122_Ch14_1_Rev_txt_prf.indd 416

1

0

0.013 0 38

TCG GM

θ = 2 0 degrees to port

11/16/2013 6:57:53 PM

Solutions to Questions • 417 Q2.28 At 4,001 tonnes, the KM is 7.03 m. GM = KM − KG GM = 7.03 − 6.00 = 1.03 03 m The initial list can be used to determine the TCG in the initial condition: tanθ = tan4 =

TCG GM

TCG ∴ TCG CG = 0 0 07 m 1 03

A loading table can be used to determine the KG in the final condition: Item

Mass

KG

Moment

Ship

4,001

6.00

24,006.00

Cargo on

1,000

5.50

5,500.00

Cargo on

500

5.00

2,500.00

Total

5,501

KG =

32,006

Moment 32, 006 = = 5.82 m Mass 5, 501

At 5,501 tonnes, the KM is 6.95 m. GM = KM − KG GM = 6.95 − 5.82 = 1.13 13 m A loading table can be used to determine the TCG in the final condition: Item

Mass

TCG

Moment

Ship

4,001

0.07

280.07

Cargo on

1,000

0.50

500

Cargo on

500

–1.00

–500

Total

TCG =

5,501

Moment 280.07 = = 0.05 m Mass 5, 501 tanθ =

9781408176122_Ch14_1_Rev_txt_prf.indd 417

280.07

TCG GM

11/16/2013 6:57:55 PM

418 • Ship Stability, Powering and Resistance

θ

1

0 05 1.13

θ = 2.5 = 2.5 degrees to port

Q2.29 At 3,781 tonnes, the KM is 7.10 m. GM = KM − KG GM = 7.10 − 6.10 = 1.00 00 m The initial list can be used to determine the TCG in the initial condition: tanθ = tan − 3 =

TCG GM

TCG ∴ TCG = −0..052 052 m 1 00

A loading table can be used to determine the KG in the final condition: Item

Mass

KG

Moment

Ship

3,781

6.1

23,064.1

Cargo on

109

Total

9

3,890

KG =

981 24,045

Moment 24 , 045 = = 6.18 m Mass 3, 890

At 3,890 tonnes, the KM is 7.06 m. GM = KM − KG GM = 7.06 − 6.18 = 0.88 88 m A loading table can be used to determine the TCG in the final condition: Item

Mass

TCG

Moment

Ship

3,781

–0.052

–196.61

Cargo on Total

TCG =

9781408176122_Ch14_1_Rev_txt_prf.indd 418

109 3,890

1

109 –87.61

Moment −87.61 = = −0.023 m Mass 3, 890

11/16/2013 6:57:57 PM

Solutions to Questions • 419

tanθ = θ

1

−0.023 0 88

TCG GM

θ = −1.5 = 1.5 degrees to starboard

Q2.30 The initial TCG can be found from the initial GM and list: TCG GM

tanθ

ta − 4 = tan

TCG 1 00

TCG = −0.070 070 m For the vessel to finish loading upright, the overall TCG must be zero. A TCG loading table can be used, with x as the cargo TCG. The table can then be used to determine the value of x so that the final TCG is zero: Item

Mass

Ship Cargo on Total

TCG

Moment

5,000

–0.070

–350

x

4.50

4.5x

5,000 + x

TCG = 0=

–350 + 4.5x

Moment Mass

−350 + 4 5 x 5, 000 + x

0(5, 000 + )

350 4.5 x

0 = −350 + 4 5 x 350 = 4 5 x x=

350 = 77.78 tonnes 45

Q2.31 The initial list and GM can be used to determine the starting TCG: tanθ

TCG GM

9781408176122_Ch14_1_Rev_txt_prf.indd 419

TCG

tanθ × GM = tan ( −3) × 1.50 = −0.079 m

11/16/2013 6:57:59 PM

420 • Ship Stability, Powering and Resistance To finish loading upright, the overall TCG must be zero. Therefore, a loading table can be used to determine the amount of cargo to move to obtain a zero TCG: Item

Mass (tonnes)

Ship

3,781

−0.079

−298.699

Cargo off

−x

−4.000

4x

Cargo on

x

6.000

6x

Total

TCG (m) Moment (tonne metres)

3,781

−298.699 + 10x

To get a TCG of zero, the overall total of the moment column must be zero. Therefore: −298.699 + 10 x = 0 Both sides can have 298.699 added to them, which gives: 10 x = 298.699 Both sides can be divided by 10, which gives: ∴ = 29.87 87 tonnes Q2.32 A loading table can be used to determine the KG after loading: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Cargo

1,386.00

7.10

9,840.60

Total

4,001

KG =

28,145.60

Moment 28 ,145.60 = = 7.03 m Mass 4 , 001

At a displacement of 4,001 tonnes, the KM, from the hydrostatics, is 7.03 m. Therefore, GM can be found: GM = KM − KG = 7.03 − 7.03 = 0.00 m The angle of list can be found from the formula, noting that BM is the difference in KM and KB. From the hydrostatics, at 4,001 tonnes displacement the KB is 2.33 m:

3

2×

1, 386 × 0.02 w ×d 2× 3 4 001 Δ = tanθ ∴ = tanθ BM 7.03 2.33 θ = 8.2 degrees to port

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11/16/2013 6:58:01 PM

Solutions to Questions • 421 Q2.33 Ballast mass = Ballast volume × Ballast density t Ballast mass ass = 219.03 × 1.025 = 224.51 tonnes KG = 0.95 m TCG = 4.13 m LCG = 49.35 m Q2.34 The mass and centre of mass of the fluid in the tanks can be determined from the tank data in the stability book. These values can then be used in loading tables to find the KG, LCG and TCG of the vessel: Item

Mass

KG

Moment

2,615.00

7.00

18,305.00

Port bunker

116.54

7.81

910.15

Stb bunker

116.54

7.81

910.15

Fore peak

71.52

2.96

211.71

After peak

109.66

5.51

604.25

Lightship

Total

3,029.26

KG =

Item

20,941.27

Moment 20 , 941.27 = = 6.91 m Mass 3, 029.26 Mass

TCG

Moment

2,615.00

0.00

0.00

Port bunker

116.54

6.93

807.60

Stb bunker

116.54

–6.93

–807.60

Fore peak

71.52

0.00

0.00

After peak

109.66

0.00

0.00

Lightship

Total

3,029.26

TCG =

9781408176122_Ch14_1_Rev_txt_prf.indd 421

0.00

Moment 0 = = 0.000 m Mass 3, 029.26

11/16/2013 6:58:03 PM

422 • Ship Stability, Powering and Resistance

Item

Mass

LCG

Moment

2,615.00

44.00

115,060.00

Port bunker

116.54

13.67

1,593.05

Stb bunker

116.54

13.67

1,593.05

Fore peak

71.52

98.04

7,012.26

After peak

109.66

2.40

263.20

Lightship

Total

3,029.26

LCG =

125,521.57

Moment 125, 521.57 = = 41.44 44 m FOAP Mass 3, 029.26

Therefore, the displacement is 3,029.26 tonnes Q2.35 The mass and centre of mass of the fluid in the tanks can be determined from the tank data in the stability book. For each partially filled tank, the FSM must be read from the tables and corrected for fluid density. These values can then be used in a loading table to find the KG of the vessel: Item

Mass

KG

Moment

2,615.00

7.00

18,305.00

No. 3 DB port

8.39

0.06

0.50

No. 3 DB port FSM

0.00

0

No. 4 DB port

5.96

0.06

No. 4 DB port FSM

0.00

0

No. 3 DB Stb

8.39

0.06

No. 3 DB Stb FSM

0.00

0

No. 4 DB Stb

5.96

0.06

No. 4 DB Stb FSM

0.00

0

172.12

Cargo

282.00

11

3,102.00

Total

2,925.70

Lightship

KG =

238.14 0.36 172.12 0.50 238.14 0.36

22,229.23

Moment 22, 229.23 = = 7.60 m Mass 2, 925.70

At 2,925 tonnes, the KM from the tables is 7.55 m. This allows GM to be found: GM = KM − KG = 7.55 − 7.60 = −0.05 m The GM is negative, therefore the vessel is unstable.

9781408176122_Ch14_1_Rev_txt_prf.indd 422

11/16/2013 6:58:04 PM

Solutions to Questions • 423 Q2.36 A loading table can be used to find the effective KG: Item

Mass

KG

Moment

2,615.00

7.00

18,305.00

Port bunker

86.99

7.22

628.04

No. 3 DB Stb

53.29

0.27

14.39

Stb bunker

58.07

6.60

383.26

Lightship

Port bunker FSM

0.00

0

11.01

No. 3 DB Stb FSM

0.00

0

378.99

Stb bunker FSM

0.00

0

8.96

Cargo

535.00

Total

3,348.35

KG =

0.5

267.50 19,997.15

Moment 19 , 997.15 = = 5.97 97 m Mass 3, 348.35

At 3,348 tonnes, the KM from the tables is 7.28 m. This allows GM to be found: GM = KM − KG ∴ GM = 7.28 − 5.97 = 1.31 31 m A loading table can be used to find the TCG: Item

Mass

TCG

Moment

2,615.00

0.00

0.00

Port bunker

86.99

6.88

598.46

No. 3 DB Stb

53.29

–3.73

–198.77

Stb bunker

58.07

–6.81

–395.46

Cargo

535.00

0.2

107.00

Total

3,348.35

Lightship

TCG =

111.23

Moment 111.23 = = 0.033 m Mass 3, 348.35

The TCG and the GM can be used to determine the list: tanθ

TCG GM

tanθ =

0.033 1 31

θ = 1.4 degrees (port )

9781408176122_Ch14_1_Rev_txt_prf.indd 423

11/16/2013 6:58:05 PM

424 • Ship Stability, Powering and Resistance Q2.37 The tank data and fluid density can be used to determine the mass and centre of mass of each of the fluid items, and a loading table can be used to determine KG: Item

Mass (tonnes)

Lightship

2,615

Port bunker

86.99

Port bunker FSM

0

Stb bunker

86.99

Stb bunker FSM

0

No. 3 DB P

183.25

No. 3 DB P FSM

0

No. 3 DB S

210.68

No. 3 DB S FSM

0

Cargo

3,209.1

Total

6,392.01

KG (m)

Moment (tonne metres)

7.00

18,305.00

7.22

628.07

0

11.01

7.22

628.07

0

11.01

0.80

146.60

0

525.43

0.90

189.61

0

539.08

5.50

17,650.05 38,633.93

Moment 38 , 633.93 = = 6.04 m Mass 6 , 392.01

KG =

At 6,392 tonnes displacement, KM from the hydrostatics is 7.07 m. Therefore, GM can be found: GM = KM − KG = 7.07 − 6.04 = 1.03 m Item Lightship

Mass (tonnes) 2,615

TCG (m)

Moment (tonne metres)

0.000

0.000

Port bunker

86.99

6.878

598.317

Stb bunker

86.99

–6.878

–598.317

No. 3 DB P

183.25

4.072

746.194

No. 3 DB S

210.68

–4.111

–866.105

0.145

465.320

Cargo

3,209.1

Total

6,392.01

TCG =

9781408176122_Ch14_1_Rev_txt_prf.indd 424

345.409

Moment 345.409 = = 0.054 m Mass 6 , 392.01

11/16/2013 6:58:07 PM

Solutions to Questions • 425 This allows the list to be found: tanθ

TCG GM

θ = tan−1

TCG 0.054 = tan tan−1 = 3 0 degrees (port) GM 1 03

Q2.38 A loading table can be used to determine the KG after removing all of the ballast: Item

Mass

Ship

5,150

No. 4 DB P

–94.21

No. 4 DB P FSM Total

0

KG

Moment

6.80

35,020.00

0.54

–50.87

0

–380.85

5,055.79

KG =

34,588.28

Moment 34 , 588.28 = = 6.84 84 m Mass 5, 055.79

Q2.39 A loading table can be used to determine the KG after de-ballasting and re-ballasting: Item

Mass

Ship

5,150

Completely empty the No. 4 DB Port

No. 4 DB P @ 1.00 m

Refill the No. 4 DB Port to the new sounding

No. 4 DB P @ 0.50 m

No. 4 DB P FSM @ 1.00 m

No. 4 DB P FSM @ 0.50 m Total

KG =

–94.21 0 41.45 0 5,097.24

KG

Moment

6.80

35,020.00

0.54

–50.87

0 0.27 0

–380.85 11.19 291.63 34,891.10

Moment 34 , 891.10 = = 6.85 m Mass 5, 097.24

Q2.40 The lightship displacement can be found: ∇×ρ = Δ 60 7 3 × 1.025 = 1, 291.5 5 tonnes The mass of fluid in the tank can be found: Fluid mass Fluid F volume × Fluid density

9781408176122_Ch14_1_Rev_txt_prf.indd 425

11/16/2013 6:58:07 PM

426 • Ship Stability, Powering and Resistance Oil mass ass = (10 × 6 × 1) × 0.97 = 58 58.2 tonnes The centre of mass of the oil will be at half of the depth of the oil, or 0.50 m. As the tank is partially filled, it will create free surface effect, so the free surface moment must be found: FSM =

lb3 10 × 63 = = 180 m4 12 12

This must be corrected by density to get the moment into units of tonne metres: 180 × 0.97 = 174.6 tonne metres A loading table can be used to determine the effective KG: Item Lightship Oil Oil FSM

Mass

KG

Moment

1,291.5

2.60

3,357.9

58.2

0.50

29.1

0

Total

0

1,349.7

KG =

174.6 3,561.6

Moment 3, 561.6 = = 2.64 m Mass 1, 349.7

The draught can be found in the final condition: ∇×ρ = Δ L B×D×ρ = Δ 60 × 7 D 1.025 = 1, 349.7 D = 3.14 14 m This can be used with the equations for KB and BM to determine the GM: LB 3 D GM = KB + BM − KG = + 12 − KG 2 ∇ 60 × 73 3 14 12 ∴ GM = + − 2.64 = 0.23 m 2 60 × 7 × 3.14

9781408176122_Ch14_1_Rev_txt_prf.indd 426

11/16/2013 6:58:09 PM

Solutions to Questions • 427 Q2.41 A loading table can be used to determine the KG after loading: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Cargo

1,386.00

7.20

9,979.20

Total

4,001

KG =

28,284.20

Moment 28 , 282.40 = = 7.07 07 m Mass 4 , 001

At a displacement of 4,001 tonnes, the KM, from the hydrostatics, is 7.03 m. Therefore, GM can be found: GM = KM − KG = 7.03 − 7.07 = −0.04 m The angle of list can be found from the formula, noting that BM is the difference in KM and KB. From the hydrostatics, at 4,001 tonnes displacement the KB is 2.33 m: tanθ =

−2 × GM −2 × −0 04 = = 0.131 KM − KB 7.02 − 2.33 θ = 7.4 4 degrees

Q2.42 The displacement after filling the pool can be found from the length, beam and draught: Δ FINAL = Δ FINAL × ρ = 50 × 8 × 3 × 1.025 = 1, 230 tonnes Therefore, the mass of pool water can be determined: Mass loaded = Δ FINAL − Δ LIGHTSHIP = 1, 230 − 1 1110 , = 120 tonnes The depth of pool water can be determined from the mass of the water: Mass = Volume × Density t

(

d ) × 1.000 ∴ d = 1.50 m

The KG of the pool water can be determined from the depth of the water: Water K KG G

Height Height of the base above K +

9781408176122_Ch14_1_Rev_txt_prf.indd 427

Water depth 1 50 = 5+ = 5..75 m 2 2

11/16/2013 6:58:11 PM

428 • Ship Stability, Powering and Resistance The FSM of the pool water can be determined from the geometry of the pool: FSM =

lb3 10 × 83 ×ρ = × 1.000 = 426.67 tonne metres 12 12

The overall effective KG after loading can be found from a loading table: Item

Mass (tonnes)

Ship

1,110

3.15

3,496.5

120

5.75

690

0

426.67

Water FSM

KG (m)

0

Total

Moment (tonne metres)

1,230

KG =

4,613.17

Moment 4 , 613.17 = = 3.75 m Mass 1, 230

The KB can be found from the final draught: KB =

D 3 00 = = 1.50 m 2 2

The BM can be found from the final draught: 50 × 83 I 12 BM = = = 1.78 m ∇ 50 × 8 × 3 This allows the GM to be found: GM = KB + BM − KG = 1.50 + 1.78 − 3.75 = −0.47 m This allows the loll angle to be found: tanθ

−2 × GM KM − KB

tanθ =

−2 × −0 47 ∴θ = 36 degrees 1.78

Q2.43 At a draught of 5.05 m, the KM can be interpolated as 6.93 m. This allows GM to be found: GM = KM − KG ∴ 0.740 = 7.746 − KG ∴ KG = 7.006 006 m GM = 6.93 − 6.20 = 0.73 m

9781408176122_Ch14_1_Rev_txt_prf.indd 428

11/16/2013 6:58:13 PM

Solutions to Questions • 429 At a draught of 5.05 m, the KB can be interpolated as 2.74 m. This allows the heel in the turn to be found, remembering that the speed must be in metres per second, and the turn radius is half of the turn diameter. tanθ =

V 2 × B ~ G (16 × 0.514 )2 × (6.20 − 2.74 ) = 400 g × r × GM 9.81× × 0.73 2 ∴θ = 9.3 degrees to port

Q2.44 The initial data can be used to determine the initial KG. For a box shaped vessel, KB is half of the draught: KB =

D 2 00 = = 1.00 m 2 2

For a box shaped vessel, BM is found from the length and beam of the waterplane and the underwater volume: LB 3 90 × 93 Inertia 12 12 BM = = = = 3.38 38 m ∇ ∇ 90 × 9 × 2.00 This allows the initial KG to be found: GM = KB + BM − KG 0.48 1+ 3.38 − KG KG = 3.90 90 m The initial displacement of the vessel can be found: Δ = ∇×ρ Δ = LBD × ρ Δ = 90 × 9 × 2 × 1.025 = 1660.5 tonnes The mass of the ballast can be found: Mass = Volume × ρ = 10 × 9 × 1.5 × 1.025 = 138.38 tonnes The KG of the ballast will be at half of the depth of the ballast measured from the keel: Ballast KG =

9781408176122_Ch14_1_Rev_txt_prf.indd 429

Ballast depth t 15 = = 0 75 m 2 2

11/16/2013 6:58:15 PM

430 • Ship Stability, Powering and Resistance The tank is partially filled, therefore there will be a free surface moment: FSM

IFS × ρFS =

lb3 10 × 93 × ρFFS = × 1.025 = 622.69 tonne metres 12 12

These values can be used in a loading table to determine the fluid KG after loading. Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Ship

1,660.50

3.90

6,475.95

138.38

0.75

103.78

0.00

0.00

622.69

Ballast Ballast FSM Total

1,798.88

KG =

7,202.42

Total moment 7, 202.42 = = 4.00 00 m Total mass 1, 798.88

The new displacement can be used to determine the draught after ballasting: ∇ ρ

∴

×ρ= Δ

90 × 9 D 1.025 0 5 = 1, 798 98.88 ∴ D =

1, 798.88 = 2.17 17 m 90 × 9 1.025

For a box shaped vessel, KB is half of the draught: KB =

D 2.17 = = 1.09 09 m 2 2

For a box shaped vessel, BM is found from the length and beam of the waterplane and the underwater volume: LB 3 90 × 93 Inertia 12 12 BM = = = = 3.11 m ∇ ∇ 90 × 9 × 2.17 This allows GM to be found: GM = KB + BM − KG = 1.09 + 3.11 − 4.00 = 0.20 m The heel in the turn can be found. Note that the speed needs to be in metres per second, and the turn radius is needed: tanθ =

(

×(

−

g × r × GM

9781408176122_Ch14_1_Rev_txt_prf.indd 430

))

=

(

×( − 400 9.81× × 0.20 2 2

×

))

θ = 11.1 degrees

11/16/2013 6:58:17 PM

Solutions to Questions • 431 The heeled draught can now be found: D cosθ +

Heeled drau r ght

B 9 sinθ = 2.17cos11.1+ sin11.1 = 3.00 m 2 2

This allows the under-keel clearance to be determined: UKC K

Water dept Water depth t Greatest eatest draught UKC K = 4 − 3 = 1.00 m

Q2.45 At 6,008 tonnes, the hydrostatics can be used to determine the KB, which gives 3.28 m, and the KM which is 7.01 m. This allows BM to be found: BM = KM − KB = 7.01 − 3.28 = 3.73 m The new KB can be found using a table of moments of volume: Item

Volume (m3)

Ship

5,861.46

3.28

19,225.59

Slice

97.56

6.04

589.26

Total

KB (m) Moment (m4)

5,959.02

KB =

19,814.85

Moment 19 , 814.85 = = 3.33 m Volume 5, 959.02

The new BM can be found: BMNew =

BMOrigin r al × ∇Origin r al ∇New

=

3.73 × 5, 861.46 = 3.67 m 5, 959.02

The new KG can be found using a table of moments of mass: Item Ship

Mass (tonnes) 6,008

Added mass Total

100 6,108

KG =

9781408176122_Ch14_1_Rev_txt_prf.indd 431

KG (m) 6.50

Moment (tonne metres) 39,052

13

1,300 40,352

Moment 40 , 352 = = 6.61 61 m Mass 6 ,108

11/16/2013 6:58:20 PM

432 • Ship Stability, Powering and Resistance This allows the new GM to be found: GM = KB + BM − KG = 3.33 + 3.67 − 6.61 = 0.39 m Q2.46 At 7,329 tonnes, the hydrostatics can be used to determine the KB, which gives 3.86 m, and the KM which is 7.26 m. This allows BM to be found: BM = KM − KB = 7.26 − 3.86 = 3.40 m The new KB can be found using a table of moments of volume: Item

Volume

KB

Moment

Ship

7,150.24

3.86

27,599.93

Slice

–48.78

6.98

–340.48

Total

KB =

7,101.46

27,259.45

Moment 27, 259.45 = = 3.84 m Volume 7,101.46

The new BM can be found: BMNew =

BMOrigin r al × ∇Origin r al ∇New

=

3.40 × 7,150.24 = 3.42 m 7,101.46

The new KG can be found using a table of moments of mass: Item

Mass

Ship

7,329

Added mass Total

–50 7,279

KG =

KG

Moment

6.90

50,570.1

8

–400 50,170.1

Moment 50 ,170.1 = = 6.89 m Mass 7, 279

This allows the new GM to be found: GM = KB + BM − KG = 3.84 + 3.42 − 6.89 = 0.37 m

9781408176122_Ch14_1_Rev_txt_prf.indd 432

11/16/2013 6:58:21 PM

Solutions to Questions • 433 Q3.1 The KN values can be read from the data book at 3,200 tonnes: Angle (degrees)

KN (m)

0

0.00

10

1.29

20

2.57

30

3.79

40

4.86

50

5.86

60

6.51

70

6.74

80

6.66

90

6.32

At each angle, GZ can be found using: GZ = KN − KG × sinθ For example, at 10 degrees: GZ = KN − KG × sinθ 1 29 7 18 × sin10 = 0.04 m This gives: Angle (degrees)

9781408176122_Ch14_1_Rev_txt_prf.indd 433

GZ (m)

0

0.00

10

0.04

20

0.11

30

0.20

40

0.24

50

0.36

60

0.29

70

–0.01

80

–0.41

90

–0.86

11/16/2013 6:58:23 PM

434 • Ship Stability, Powering and Resistance

GZ (m)

These values can be plotted on a GZ curve: 0.500 0.400 0.300 0.200 0.100 0.000 0 –0.100 –0.200 –0.300 –0.400 –0.500 –0.600 –0.700 –0.800 –0.900 –1.000

10

20

30

40

50

60

70

80

90

Heel (degrees)

Q3.2 A loading table can be used to determine the KG after loading and bunkering: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Port bunker

116.54

7.81

910.15

Stb bunker

116.54

7.81

910.15

Cargo 1

352.00

7

2,464.00

Cargo 2

1,000.00

6

6,000.00

Total

4,200.07

KG =

28,589.30

Moment 28 , 589.3 = = 6.81 m Mass 4 , 200

The KN values can be read from the data book at 4,200 tonnes, and the GZ values calculated: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

10

1.22

0.04

9781408176122_Ch14_1_Rev_txt_prf.indd 434

11/16/2013 6:58:23 PM

Solutions to Questions • 435

Angle (degrees)

KN (m)

GZ (m)

20

2.47

0.14

30

3.71

0.31

40

4.89

0.51

50

5.85

0.63

60

6.41

0.52

70

6.62

0.22

80

6.54

–0.16

These values can be plotted on a GZ curve: 0.700 0.600 0.500

GZ (m)

0.400 0.300 0.200 0.100 0.000

0

10

20

30

40

50

60

70

80

–0.100 –0.200 Heel (degrees)

Q3.3 Part 1: The increase in KG will reduce the GZ values. The loss in GZ can be found from: Change in GZ = Change in KG × sinθ For example, at 10 degrees, the correction is: Change in GZ = 0.1× sin10 = 0.02 m The increase in KG reduces GZ, therefore: New G GZ = 0.15 − 0.02 = 0 13 m

9781408176122_Ch14_1_Rev_txt_prf.indd 435

11/16/2013 6:58:24 PM

436 • Ship Stability, Powering and Resistance This process can be applied to all of the angles: Angle (degrees)

GZ (m)

Correction (m)

New GZ (m)

0

0

0

0

10

0.15

0.02

0.13

20

0.31

0.03

0.28

30

0.39

0.05

0.34

40

0.39

0.06

0.33

50

0.46

0.08

0.38

60

0.46

0.09

0.37

70

0.21

0.09

0.12

80

–0.17

0.1

–0.27

Part 2: The reduction in KG will increase the GZ values. The increase in GZ can be found from: Change in GZ = Change in KG × sinθ For example, at 10 degrees, the correction is: Change in GZ = 0.1× sin10 = 0.02 m The decrease in KG increases GZ, therefore: New G GZ = 0.15 5 + 0.02 = 0 17 m This process can be applied to all of the angles: Angle (degrees)

GZ

Correction

New GZ

0

0

0

0

10

0.15

0.02

0.17

20

0.31

0.03

0.34

30

0.39

0.05

0.44

40

0.39

0.06

0.45

50

0.46

0.08

0.54

60

0.46

0.09

0.55

70

0.21

0.09

0.3

80

–0.17

0.1

–0.07

9781408176122_Ch14_1_Rev_txt_prf.indd 436

11/16/2013 6:58:25 PM

Solutions to Questions • 437 The curves from parts 1 and 2 can be plotted and compared: 0.6 Original GZ Increased KG Reduced KG

0.5 0.4 0.3

GZ (m)

0.2 0.1 0.0

0

10

20

30

40

50

60

70

80

–0.1 –0.2 –0.3 Heel (degrees)

Q3.4 At 3,000 tonnes, the KN values can be read from the hydrostatics, and the GZ values calculated: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

5

0.65

0.06

10

1.31

0.13

15

1.96

0.21

20

2.61

0.30

The GZ at each angle can be corrected for the TCG value. This correction always reduces GZ. The correction is found from: Loss in GZ = TCG cosθ For example, at 5 degrees: Loss in GZ = TCG cosθ Loss in GZ = 0.289cos 289 5 = 0.29 GZ at 5 degrees 0.06 − 0.29

9781408176122_Ch14_1_Rev_txt_prf.indd 437

0.23 m

11/16/2013 6:58:26 PM

438 • Ship Stability, Powering and Resistance

Angle (degrees)

KN (m)

GZ (m)

Loss in GZ (m)

Corrected GZ (m)

0

0.00

0.00

0.29

–0.29

5

0.65

0.06

0.29

–0.23

10

1.31

0.13

0.29

–0.15

15

1.96

0.21

0.28

–0.07

20

2.61

0.30

0.27

0.03

The corrected GZ values can be plotted: 0.100

0.000

GZ (m)

0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

–0.100

–0.200

–0.300 Heel (degrees)

The intersection with the heel axis shows the list will be 18.7 degrees to port. From interpolation of the hydrostatics, the KM is 7.49 m at 3,000 tonnes. Therefore, GM can be found: GM = KM − KG GM = 7.49 − 6.77 = 0.72 m This allows the list to be found: tanθ

TCG GM

tanθ =

0.289 0 72

θ = 21.9 degrees (port)

9781408176122_Ch14_1_Rev_txt_prf.indd 438

11/16/2013 6:58:27 PM

Solutions to Questions • 439 Q3.5 The GZ values can be corrected for the change in KG and the change in TCG, using: Change in GZ = Change in KG × sinθ Loss in GZ = TCG × cosθ Angle (degrees)

GZ (m)

Loss in GZ due to KG increase (m)

Loss in GZ due to TCG increase (m)

0

New GZ (m)

0

0.00

0.40

–0.40

10

0.1

0.04

0.39

–0.33

20

0.25

0.08

0.38

–0.21

30

0.5

0.12

0.35

0.03

40

0.72

0.15

0.31

0.26

50

0.69

0.18

0.26

0.25

60

0.49

0.21

0.20

0.08

70

0.2

0.23

0.14

–0.17

80

–0.12

0.24

0.07

–0.43

90

–0.46

0.24

0.00

–0.70

GZ (m)

The GZ curve can be plotted after the cargo shift: 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 –0.1 0 –0.2 –0.3 –0.4 –0.5 –0.6 –0.7 –0.8

Original GZ New GZ

10

20

30

40

50

60

70

80

90

Heel (degrees)

9781408176122_Ch14_1_Rev_txt_prf.indd 439

11/16/2013 6:58:28 PM

440 • Ship Stability, Powering and Resistance Q3.6 The initial draught can be used to determine the displacement and KM. At 4.10 m draught the displacement is 3,781 tonnes. At this displacement, the KM is 7.10 m. This can be used to determine the GM: GM = KM − KG GM = 7.10 − 5.49 = 1.61 m The initial list can be used to determine the initial TCG: tanθ

TCG GM

tan − 1.1 =

TCG 1 61

TCG = −0.031 m Loading tables can be used to determine the KG and TCG with the load suspended and swung out: Item

Mass

KG

Moment 20,757.69

Ship

3,781

5.49

Un-stowed cargo

–100

0

0

Suspended cargo

100

11

1,100

Total

3,781

KG =

21,857.69

Moment 21, 857.69 = = 5.78 m Mass 3, 781

Item

Mass

TCG

Moment

–0.031

–117.211

Ship

3,781

Un-stowed cargo

–100

0

0

Suspended cargo

100

17

1,700

Total

3,781

TCG =

1,582.789

Moment 1, 582.789 = = 0.419 m Mass 3, 781

At 3,781 tonnes, and a KG of 5.78 m, the KN values can be interpolated, and the GZ values calculated, including the correction for the TCG:

9781408176122_Ch14_1_Rev_txt_prf.indd 440

11/16/2013 6:58:28 PM

Solutions to Questions • 441

Angle (degrees)

KN (m)

GZ (m)

TCG correction (m) Corrected GZ (m)

0

0.00

0.00

0.42

–0.42

5

0.62

0.12

0.42

–0.30

10

1.24

0.24

0.41

–0.18

15

1.87

0.37

0.41

–0.03

20

2.50

0.52

0.39

0.13

These corrected GZ values can be plotted and the list angle read from the graph: 0.200 0.100 0.000

GZ (m)

0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

–0.100 –0.200 –0.300 –0.400 –0.500 Heel (degrees)

It can be seen that the list angle is 16 degrees. Q3.7 A loading table can be used to determine the KG after loading: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Cargo

1,985.00

7.5

14,887.50

Total

4,600.00

9781408176122_Ch14_1_Rev_txt_prf.indd 441

33,192.50

11/16/2013 6:58:30 PM

442 • Ship Stability, Powering and Resistance

KG =

Moment 33,192.5 = = 7.22 22 m Mass 4 , 600

At 4,600 tonnes, and a KG of 7.22 m, the KN values can be interpolated, and the GZ values calculated: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

10

1.22

–0.04

20

2.45

–0.02

30

3.71

0.10

40

4.91

0.27

50

5.83

0.30

60

6.36

0.11

70

6.57

–0.21

These corrected GZ values can be plotted and the list angle read from the graph: 0.400 0.300

GZ (m)

0.200 0.100 0.000 0

10

20

30

40

50

60

70

–0.100 –0.200 –0.300 Heel (degrees)

It can be seen that the loll angle is 22.5 degrees.

9781408176122_Ch14_1_Rev_txt_prf.indd 442

11/16/2013 6:58:30 PM

Solutions to Questions • 443 Q3.8 The KN values can be read off at 4,000 tonnes and GZ calculated: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

5

0.61

–0.02

10

1.23

–0.03

15

1.85

–0.02

20

2.48

0.01

30

3.72

0.11

40

4.88

0.23

50

5.86

0.32

60

6.44

0.18

70

6.65

–0.15

These values can be plotted: 0.350 0.300 0.250 0.200

GZ (m)

0.150 0.100 0.050 0.000 –0.050

0

5

10

15

20

25

30

35

40

45

50

55

60

65

70

–0.100 –0.150 –0.200 –0.250 Heel (degrees)

9781408176122_Ch14_1_Rev_txt_prf.indd 443

11/16/2013 6:58:31 PM

444 • Ship Stability, Powering and Resistance A vertical line can be drawn at 57.3 degrees, a diagonal line can be drawn between 0 and 5 degrees, and extended to the vertical line at 57.3 degrees, and a horizontal line can be drawn across the graph at the intersection. The intersection of this line and the y axis gives the GM of the vessel: 0.350 0.300 0.250 0.200

GZ (m)

0.150 0.100 0.050 0.000 –0.050

0

5

10

15

20

25

30

35

40

45

50

55

60

65

70

–0.100 –0.150 –0.200 –0.250 Heel (degrees)

As can be seen, this occurs at –0.20 m. From the hydrostatics, KM at 4,000 tonnes is 7.03 m. As KG is 7.23 m, this also gives a GM of –0.20 m. Q3.9 The KN values can be read from the hydrostatics and GZ calculated: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

5

0.61

0.00

10

1.23

0.01

20

2.47

0.07

30

3.75

0.25

40

4.93

0.42

50

5.72

0.35

9781408176122_Ch14_1_Rev_txt_prf.indd 444

11/16/2013 6:58:31 PM

Solutions to Questions • 445 These can be plotted: 0.45 0.40 0.35 0.30

GZ (m)

0.25 0.20 0.15 0.10 0.05 0.00 0

10

20

30

40

50

60

–0.05 Angle (degrees)

If a vertical line is drawn at 57.3 degrees, and a diagonal line through 0 and 5 degrees, then the intersection between these lines will be at a y value of zero (as the diagonal line will actually be horizontal). Therefore, the GM of the vessel is zero. Q3.10 Task 1: Angle

9781408176122_Ch14_1_Rev_txt_prf.indd 445

KN (m)

GZ (m)

0

0.000

0.000

5

0.611

0.017

10

1.220

0.037

15

1.840

0.077

20

2.471

0.142

25

3.102

0.224

11/16/2013 6:58:31 PM

446 • Ship Stability, Powering and Resistance Task 2: Angle

KN (m)

GZ (m)

0

0.000

0.000

5

0.611

0.000

10

1.220

0.003

15

1.840

0.026

20

2.471

0.073

25

3.102

0.139

KN (m)

GZ (m)

0

0.000

0.000

5

0.611

–0.017

10

1.220

–0.032

15

1.840

–0.026

20

2.471

0.005

25

3.102

0.055

Task 3: Angle

0.300 7.21 7.01 6.81

0.200

GZ (m)

0.100

0.000 0

5

10

15

20

25

30

35

40

45

50

55

60

–0.100

–0.200

–0.300 Angle (degrees)

9781408176122_Ch14_1_Rev_txt_prf.indd 446

11/16/2013 6:58:31 PM

Solutions to Questions • 447 A vertical line can be drawn at 57.3 degrees, a diagonal line through the initial slope and a horizontal line through the intersection of the previous two lines. This gives: 0.300 7.21 7.01 6.81

0.200

GZ (m)

0.100

0.000 0

5

10

15

20

25

30

35

40

45

50

55

60

–0.100

–0.200

–0.300 Angle (degrees)

The GM values can be found at the intersection of the horizontal line and the y axis. For task 1, this gives 0.16 m for task 1, 0.00 m for task 2 and –0.18 m for task 3. Q3.11 A loading table can be used to determine the KG after loading: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Cargo

2,985.00

7.5

22,387.50

Total

5,600.00

KG =

40,692.50

Moment 40 , 692.5 = = 7.27 27 m Mass 5, 600

A loading table can be used to determine the TCG after loading: Item

Mass

TCG

Lightship

2,615.00

0.00

Cargo

2,985.00

0.1

Total

5,600.00

9781408176122_Ch14_1_Rev_txt_prf.indd 447

Moment 0.00 298.50 298.50

11/16/2013 6:58:31 PM

448 • Ship Stability, Powering and Resistance

TCG =

Moment 298.5 = = 0.053 053 m Mass 5, 600

At 5,600 tonnes, and a KG of 7.27 m, the KN values can be interpolated, and the GZ values calculated, including the correction for the TCG: Angle (degrees)

KN (m)

GZ (m)

TCG correction (m)

Corrected GZ (m)

0

0.00

0.00

0.05

–0.05

5

0.61

–0.03

0.05

–0.08

10

1.22

–0.05

0.05

–0.10

15

1.83

–0.05

0.05

–0.10

20

2.45

–0.03

0.05

–0.08

25

3.09

0.01

0.05

–0.04

30

3.73

0.09

0.05

0.05

35

4.37

0.20

0.04

0.16

These corrected GZ values can be plotted and the combined list and loll angle read from the graph: 0.200 0.150 0.100 0.050 GZ (m)

0

5

10

15

20

25

30

35

0.000 –0.050 –0.100 –0.150 –0.200 Heel (degrees)

It can be seen that the combined list and loll angle is 27.5 degrees.

9781408176122_Ch14_1_Rev_txt_prf.indd 448

11/16/2013 6:58:32 PM

Solutions to Questions • 449 Q3.12 For each sounding, the KG and TCG must be calculated, along with the corrected GZ values. These can be plotted to determine the combined list and loll angle. As an example, for a sounding of 0.50 m in the No. 3 and No. 4 Port Double Bottoms a loading table can be used to determine the KG: Item

Mass

KG

Moment

2,615.00

7.00

18,305.00

No. 3 DB Port

53.29

0.27

14.39

No. 4 DB Port

41.45

0.27

11.19

Port bunker

117.76

7.81

919.73

No. 3 DB Stb

29.92

0.16

4.79

No. 4 DB Stb

22.59

0.17

3.84

Stb bunker

117.76

7.81

919.73

Lightship

No. 3 DB Port FSM

0.00

0

378.99

No. 4 DB Port FSM

0.00

0

291.63

No. 3 DB Stb FSM

0.00

0

323.23

No. 4 DB Stb FSM

0.00

0

243.14

Cargo

3,245.00

Total

6,242.78

KG =

7.5

24,337.50 45,753.17

Moment 45, 753.17 = = 7.33 33 m Mass 6 , 242.78

A loading table can be used to determine the TCG after loading: Item

Mass

TCG

2,615.00

0.000

0.000

No. 3 DB Port

53.29

3.730

198.771

No. 4 DB Port

41.45

3.430

142.177

Port bunker

117.76

6.930

816.099

No. 3 DB Stb

29.92

–3.590

–107.412

No. 4 DB Stb

22.59

–3.300

–74.550

Stb bunker

117.76

–6.930

–816.099

Cargo

3,245.00

0.000

0.000

Total

6,242.78

Lightship

9781408176122_Ch14_1_Rev_txt_prf.indd 449

Moment

158.985

11/16/2013 6:58:32 PM

450 • Ship Stability, Powering and Resistance

TCG =

Moment 158.985 = = 0.025 025 m Mass 6 , 242.78

At 6,242.78 tonnes, and a KG of 7.33 m, the KN values can be interpolated, and the GZ values calculated (up to an angle where GZ becomes positive), including the correction for the TCG: Angle (degrees)

KN (m)

GZ (m)

TCG correction (m) Corrected GZ (m)

0

0.00

0.00

0.03

–0.03

5

0.61

–0.02

0.03

–0.05

10

1.23

–0.04

0.03

–0.07

15

1.85

–0.05

0.02

–0.07

20

2.48

–0.03

0.02

–0.05

25

3.12

0.02

0.02

–0.01

30

3.76

0.10

0.02

0.08

These corrected GZ values can be plotted and the combined list and loll angle read from the graph: 0.15 0.10 0.05 0.00 GZ (m)

0

5

10

15

20

25

30

–0.05 –0.10 –0.15 –0.20 –0.25 –0.30 Heel (degrees)

It can be seen that the combined list and loll angle is 25.5 degrees.

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Solutions to Questions • 451 This process can be repeated for each step in the range of soundings: Sounding (m)

KG (m)

TCG (m)

GZ @ 0 GZ @ 5 deg GZ @ 10 deg (m) (m) deg (m)

GZ @ 15 deg (m)

0.30

7.36

0.000

0.00

–0.03

–0.05

–0.05

0.50

7.33

0.025

–0.03

–0.05

–0.07

0.70

7.29

0.053

–0.05

–0.07

0.90

7.26

0.082

–0.08

–0.10

1.10

7.22

0.113

–0.11

1.30

7.18

0.144

1.50

7.14

1.70 1.90 2.00

GZ @ 20 deg (m)

GZ @ 25 deg (m)

GZ @ 30 deg (m)

–0.04

0.00

0.08

–0.07

–0.05

–0.01

0.08

–0.09

–0.08

–0.06

–0.01

0.08

–0.11

–0.10

–0.08

–0.02

0.07

–0.13

–0.13

–0.12

–0.09

–0.03

0.07

–0.14

–0.15

–0.15

–0.14

–0.10

–0.04

0.06

0.176

–0.18

–0.18

–0.17

–0.16

–0.12

–0.05

0.06

7.10

0.208

–0.21

–0.21

–0.20

–0.17

–0.13

–0.06

0.06

7.06

0.241

–0.24

–0.24

–0.22

–0.19

–0.14

–0.06

0.05

6.89

0.257

–0.26

–0.24

–0.21

–0.16

–0.10

–0.01

0.12

These GZ values can be plotted and the largest combined angle of list and loll read off: 0.15 0.10 0.05 0.00 GZ (m)

0

5

10

15

20

25

30

–0.05 –0.10 0.30 m Sounding 0.50 m Sounding 0.70 m Sounding 0.90 m Sounding 1.10 m Sounding 1.30 m Sounding 1.50 m Sounding 1.70 m Sounding 1.90 m Sounding 2.0 m Sounding

–0.15 –0.20 –0.25 –0.30 Heel (degrees)

As can be seen, the largest combined list and loll angle is 28 degrees, at a sounding of 1.90 m. Q3.13 The KN values can be found at 4,800 tonnes displacement. This allows the GZ values to be found using:

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452 • Ship Stability, Powering and Resistance GZ = KN − KG sinθ The righting moment can be found from: Righting moment = Δ × GZ This gives the following values: Angle (degrees)

KN (m)

GZ (m)

Righting moment (tonne metres)

0

0

0

0

10

1.21

0.03

20

2.45

0.12

576

30

3.71

0.31

1,488

40

4.93

0.56

2,688

50

5.82

0.61

2,928

60

6.34

0.45

2,160

70

6.54

0.15

720

80

6.48

–0.22

–1,056

144

Plotting the GZ values gives: GZ 0.7 0.6 0.5 0.4

GZ (m)

0.3 0.2 0.1 0 0

10

20

30

40

50

60

70

80

–0.1 –0.2 –0.3 Heel (degrees)

9781408176122_Ch14_1_Rev_txt_prf.indd 452

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Solutions to Questions • 453 Plotting the righting moment gives: Righting moment

3,500

Righting moment (tonne metres)

3,000 2,500 2,000 1,500 1,000 500 0 0

10

20

30

40

50

60

70

80

–500

–1,000 –1,500 Heel (degrees)

Q3.14 A loading table can be used to determine the displacement, TCG and KG after loading: Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Ship

2,615

7.00

18,305

Cargo

4,185

7.00

29,295

Total

6,800

KG =

47,600

Moment 47, 600 = = 7.00 m Mass 6 , 800

Item

Mass (tonnes)

Ship

2,615

0.00

0

Cargo

4,185

0.10

418.5

Total

6,800

TCG =

TCG (m) Moment (tonne metres)

418.5

Moment 418.5 = = 0.062 062 m Mass 6 , 800

For the displacement of 6,800 tonnes, the KN values can be found. The basic GZ values can be determined using: GZ = KN − KG sinθ

9781408176122_Ch14_1_Rev_txt_prf.indd 453

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454 • Ship Stability, Powering and Resistance The TCG correction can be determined using: TCG correction TCG T × cosθ The corrected GZ is then found by subtracting the TCG correction from the basic GZ. Finally, the corrected righting moment can be found by multiplying the corrected GZ by the displacement of the vessel. This gives the following values: Angle (degrees)

KN (m)

Basic GZ (m)

TCG correction (m)

Corrected GZ (m)

Corrected righting moment (tonne metres)

0

0.00

0.00

0.06

–0.06

–408.00

10

1.25

0.03

0.06

–0.03

–204.00

20

2.51

0.12

0.06

0.06

408.00

30

3.80

0.30

0.05

0.25

1,700.00

40

4.89

0.39

0.05

0.34

2,312.00

50

5.64

0.28

0.04

0.24

1,632.00

60

6.09

0.03

0.03

0.00

0.00

70

6.28

–0.30

0.02

–0.32

–2,176.00

The corrected GZ values can be plotted to give the GZ curve, and the intersection used to determine the angle of list, which is at 14. 5 degrees: 0.40 0.30 0.20

GZ (m)

0.10 0.00 0

10

20

30

40

50

60

70

–0.10 –0.20 –0.30 –0.40 Heel (degrees)

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Solutions to Questions • 455 The corrected righting moment values can be plotted to give the righting moment curve, and the intersection used to determine the angle of list, which is again at 14.5 degrees: 2,500.00 2,000.00

Righting moment (tonne metres)

1,500.00 1,000.00 500.00 0.00 0

10

20

30

40

50

60

70

–500.00 –1,000.00 –1,500.00 –2,000.00 –2,500.00 Heel (degrees)

Q3.15 The basic GZ values from Q3.14 can be used to determine the uncorrected righting moment, found by multiplying the basic GZ by the displacement. This gives: Angle (degrees)

Basic GZ (m)

Basic righting moment (tonne metres)

0

0.00

0

10

0.03

204

20

0.12

816

30

0.30

2,040

40

0.39

2,652

50

0.28

1,904

60

0.03

204

70

–0.30

–2,040

9781408176122_Ch14_1_Rev_txt_prf.indd 455

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456 • Ship Stability, Powering and Resistance The listing moment caused by the cargo can be found from the cargo mass and distance from the centreline: Listing moment

w × d = 4 ,185 × 0.10 = 418.5 tonne metres

This can be superimposed on the uncorrected righting moment curve, and the intersection read off to give the list angle, which is at 14.5 degrees: 3,000.00

Listing moment and righting moment (tonne metres)

2,500.00 2,000.00 1,500.00 1,000.00 500.00 0 10

0

20

30

40

50

60

70

–500.00 –1,000.00 –1,500.00 –2,000.00 –2,500.00 Heel (degrees)

Q3.16 The basic GZ values from Q3.14 can be plotted to give the uncorrected GZ curve. The listing arm can be determined from formula: Listing a arm =

w d 4 ,185 × 0.10 = = 0 06 m Δ 6 , 800

This can be superimposed on the uncorrected GZ curve, and the intersection read off to give the list angle, which again is at 14.5 degrees:

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Solutions to Questions • 457

0.40

Listing arm and GZ (m)

0.30 0.20 0.10 0.00 0

10

20

30

40

50

60

70

–0.10 –0.20 –0.30 –0.40 Heel (degrees)

Q3.17 The top of the three curves has the greatest area under the curve, and therefore will require the most energy to roll to 40 degrees. Q3.18 X

Y

Simpson’s Multiplier

Area product

0

5

1

5

2

6

4

24

4

5.25

2

10.5

6

4

4

16

8

2

1

2

Total

57.5

Area =

Spacing p g 2 × Σ Area product = × 57.5 = 38 38.3 m2 3 3

Q3.19 A loading table can be used to determine the displacement and KG after loading: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Cargo

3,985.00

6

23,910.00

Total

6,600.00

9781408176122_Ch14_1_Rev_txt_prf.indd 457

42,215.00

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458 • Ship Stability, Powering and Resistance

KG =

Moment 42, 215 = = 6.40 40 m Mass 6 , 600

At 6,600 tonnes, and a KG of 6.40 m, the KN values can be found, and the GZ values calculated: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0

10

1.24

0.13

20

2.50

0.31

30

3.79

0.59

40

4.91

0.80

Each of the GZ values can be multiplied by the displacement of the vessel to obtain the righting moment: Angle (degrees)

GZ (m)

Righting moment (tonne metres)

0

0.00

0

10

0.13

858

20

0.31

2,046

30

0.59

3,894

40

0.80

5,280

Simpson’s Rule can then be applied to the righting moment values to determine the area under the righting moment curve, and hence the dynamic stability: Angle (degrees) 0

Righting moment (tonne metres)

Simpson’s Multiplier

Area product

0

1

0

10

858

4

3,432

20

2,046

2

4,092

30

3,894

4

15,576

40

5,280

1

5,280

Total

28,380

9781408176122_Ch14_1_Rev_txt_prf.indd 458

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Solutions to Questions • 459

Area =

Spacing p g 10 × Σ Area product = × 28 , 380 = 94 , 600 tonne metre degrees 3 3

This can be divided by 57.3 to convert the units to tonne metre radians: 94 , 600 = 1, 650.96 tonne metre radians 57.3

Area =

Q3.20 A loading table can be used to determine the displacement and KG after loading: Item Lightship Cargo 1 Total

Mass

KG

Moment

2,615.00

7.00

18,305.00

7.2

5,652.00

785.00 3,400.00

KG =

23,957.00

Moment 23, 957 = = 7.05 05 m Mass 3, 400

At 3,400 tonnes, and a KG of 7.05 m, the KN values can be found, and the GZ values calculated: Angle (degrees)

9781408176122_Ch14_2_Rev_txt_prf.indd 459

KN (m)

GZ (m)

0

0.00

0.00

10

1.27

0.04

20

2.54

0.13

30

3.77

0.24

35

4.33

0.29

40

4.86

0.33

11/16/2013 2:43:41 AM

460 • Ship Stability, Powering and Resistance These can be plotted: 0.40 0.35 0.30

GZ (m)

0.25 0.20 0.15 0.10 0.05 0.00 0

5

10

15

20 25 Heel (degrees)

30

35

40

It can be seen that the peak GZ value is in excess of 0.20 m, and will occur above 30 degrees. The area can be found between 0 and 40 degrees: Angle (degrees)

GZ (m)

Simpson’s Multiplier

0

0.00

1

0

10

0.04

4

0.176

20

0.13

2

0.264

30

0.24

4

0.976

40

0.33

1

0.33

Total

1.746

Area =

Area product

Spacing p g 10 × Σ Area product = × 1.746 = 5.820 m degrees 3 3

This can be divided by 57.3 degrees to convert the units to metre radians: 5.820 = 0.102 m radians 57.3 The minimum area from 0 to 40 degrees is 0.09 m radians, therefore the vessel passes the 0 to 40 degrees criteria.

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Solutions to Questions • 461 The area can be found between 30 and 40 degrees: Angle (degrees)

GZ (m) Simpson’s Multiplier

Area product

30

0.24

1

0.244

35

0.29

4

1.14

40

0.33

1

0.33

Total

1.714

Area =

Spacing p g 5 × Σ Area product = × 1.714 = 2.857 m degrees 3 3

This can be divided by 57.3 degrees to convert the units to metre radians: 2.857 = 0.050 m radians 57.3 The minimum area from 30 to 40 degrees is 0.03 m radians, therefore the vessel passes the 30 to 40 degrees criteria. The area between 0 and 30 degrees can be found by subtracting the area between 30 and 40 degrees from the area between 0 and 40 degrees: AREA0 to 30 = AREA0 to 40 − AREA30 to 40 AREA0 to 30 = 0.102 − 0.05 0.052 052 m radians The minimum area from 0 to 30 degrees is 0.055 m radians, therefore the vessel fails the 0 to 30 degrees criteria. This means that the vessel is not in a legal condition. Q3.21 A loading table can be used to determine the displacement and KG after loading: Item

Mass

KG

Moment

Lightship

2,615.00

7.00

18,305.00

Cargo

4,585.00

7.10

32,553.50

Total

7,200.00

KG =

50,858.50

Moment 50 , 828.50 = = 7.06 06 m Mass 7, 200

At 7,200 tonnes, and a KG of 7.06 m, the KN values can be found, and the GZ values calculated:

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462 • Ship Stability, Powering and Resistance

Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

10

1.26

0.03

20

2.54

0.13

30

3.82

0.29

35

4.38

0.33

40

4.86

0.32

These can be plotted: 0.350 0.300

GZ (m)

0.250 0.200 0.150 0.100 0.050 0.00 0

5

10

15

20 25 Heel (degrees)

30

35

40

It can be seen that the peak GZ value is in excess of 0.20 m, and will occur above 30 degrees. The area can be found between 0 and 40 degrees: Angle (degrees)

GZ (m)

0

0

1

0

10

0.03

4

0.12

20

0.13

2

0.26

30

0.29

4

1.16

40

0.32

1

0.32

Total

1.86

9781408176122_Ch14_2_Rev_txt_prf.indd 462

Simpson’s Multiplier

Area product

11/16/2013 2:43:44 AM

Solutions to Questions • 463

Area =

Spacing p g 10 × Σ Area product = × 1.86 = 6.20 m degrees 3 3

This can be divided by 57.3 degrees to convert the units to metre radians: 6 20 = 0.108 m radians 57.3 However, the down-flooding angle is 36 degrees, therefore the area between 36 and 40 degrees must be subtracted from this value. From the GZ curve, at 36 degrees, the GZ is 0.33 m. Trapezoidal integration can be used to find the area between 36 and 40 degrees: AREA36 to 40 =

0.33 0.32 × ( 40 − 36 ) = 1.30 m degrees 2 1 30 = 0.023 m radians 57.3

Therefore, the area between 0 and 36 degrees can be found: AREA0 to 36 = AREA0 to 40 − AREA36 to 40 AREA0 to 36 = 0.108 − 0.023 = 0.085 085 m radians The minimum area from 0 to down-flooding is 0.09 m radians, therefore the vessel fails the 0 to down-flooding degrees criteria. The area can be found between 30 and 40 degrees: Angle (degrees)

GZ (m)

Simpson’s Multiplier

Area product

30

0.29

1

0.29

35

0.33

4

1.32

40

0.32

1

0.32

Total

1.93

Area =

Spacing p g 5 × Total area product = × 1.93 = 3.217 m degrees 3 3

This can be divided by 57.3 degrees to convert the units to metre radians: 3.217 = 0.056 m radians 57.3

9781408176122_Ch14_2_Rev_txt_prf.indd 463

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464 • Ship Stability, Powering and Resistance However, the down-flooding angle is 36 degrees, therefore the area between 36 and 40 degrees must be subtracted from this value. Therefore, the area between 30 and 36 degrees can be found: AREA30 to 36 = AREA30 to 40 − AREA36 to 40 AREA0 to 36 = 0.056 − 0.025 = 0.031 031 m radians The minimum area from 30 to down-flooding is 0.03 m radians, therefore the vessel passes the 30 to down-flooding degrees criteria. The area between 0 and 30 degrees can be found by subtracting the area between 30 and 40 degrees from the area between 0 and 40 degrees: AREA0 to 30 = AREA0 to 40 − AREA30 to 40 AREA0 to 30 = 0.108 − 0.056 = 0.052 052 m radians The minimum area from 0 to 30 degrees is 0.055 m radians, therefore the vessel fails the 0 to 30 degrees criteria. Q3.22 The KB can be found: KB =

D 3 = = 1.50 50 m 2 2

The BM can be found: LB 3 40 × 73 Inertia 12 12 BM = = = = 1.36 36 m ∇ LBD 40 × 7 × 3 The GM can be found: GM = KB + BM − KG = 1.50 + 1.36 − 2 = 0.86 86 m At 5 degrees: BM ⎛ ⎞ GZ = sinθ GM + tan2 θ ⎝ ⎠ 2

1 36 ⎛ ⎞ si 5 0.86 + sin tan2 5 = 0.08 08 m ⎝ ⎠ 2

At 10 degrees: BM ⎛ ⎞ GZ = sinθ GM + tan2 θ ⎝ ⎠ 2

9781408176122_Ch14_2_Rev_txt_prf.indd 464

1 36 ⎛ ⎞ si 10 0 86 + sin tan2 10 = 0.15 m ⎝ ⎠ 2

11/16/2013 2:43:46 AM

Solutions to Questions • 465 At 15 degrees: BM 1 36 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = siin15 0 86 + tan2 15 = 0.24 m ⎝ ⎠ ⎝ ⎠ 2 2 Q3.23 The KB, from the hydrostatics, is 3.86 m in the summer condition, and the KM is 7.26 m. This allows BM to be found: KM = KB + BM 7.26 3.86 + BM BM = 3.40 40 m At 0 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = si 0 1 55 + tan2 0 = 0.16 m ⎝ ⎠ ⎝ ⎠ 2 2 At 5 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = siin 5 1 55 + tan2 5 = 0.14 m ⎝ ⎠ ⎝ ⎠ 2 2 At 10 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = siin10 1 55 + tan2 10 = 0.28 m ⎝ ⎠ ⎝ ⎠ 2 2 At 15 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = sin in15 1 55 + tan2 15 = 0.43 m ⎝ ⎠ ⎝ ⎠ 2 2 At 20 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = si 20 1 55 + tan2 20 = 0.61 61 m ⎝ ⎠ ⎝ ⎠ 2 2 At 25 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = sin n 25 1 55 + tan2 25 = 0.81 m ⎝ ⎠ ⎝ ⎠ 2 2 At 30 degrees: BM 3 40 ⎛ ⎞ ⎛ ⎞ GZ = sinθ GM + tan2 θ = siin 30 1 55 + tan2 30 = 1.06 m ⎝ ⎠ ⎝ ⎠ 2 2

9781408176122_Ch14_2_Rev_txt_prf.indd 465

11/16/2013 2:43:48 AM

466 • Ship Stability, Powering and Resistance These values can be plotted onto the GZ curve: 1.100 Actual GZ Wall-sided formula GZ

1.000 0.900 0.800

GZ (m)

0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.000 0

5

10

15 Heel (degrees)

20

25

30

Q3.24 The displacement can be found from the dimensions: ∇×ρ = Δ 025 = 246 tonnes 40 × 6 1× 1.025 The original KB can be found from the draught: KB =

Draught 1 = = 0.50 m 2 2

The BM can be determined before ballasting: ⎛ LB 3 ⎞ ⎛ 40 × 63 ⎞ Inertia ⎜⎝ 12 ⎟⎠ ⎜⎝ 12 ⎟⎠ BM = = = = 3.00 m ∇ ∇ 40 × 6 × 1 This allows the KG to be determined, as GM is known: GM = KB + BM − KG

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11/16/2013 2:43:54 AM

Solutions to Questions • 467 1.2 = 0.5 + 3 − KG KG = 2.30 30 m The mass of ballast in the double bottom can be determined: Mass = Volume × ρ = 8 × 0.2 × 6 × 1.025 = 9.84 tonnes The KG of the ballast will be at half the depth of the ballast, which is 0.10 m. The free surface effect of the ballast can be determined: FSM

InertiaFS × ρFFS =

8 × 63 × 1.025 = 147.6 tonne metres 12

A loading table can be used to determine the effective KG: Item

Mass (tonnes)

Barge

KG (m)

246

Moment (tonne metres)

2.3

565.8

Ballast

9.84

0.1

Ballast FSM

0

0

147.6

40

4

160

Cargo Totals

295.84

KG =

0.984

874.384

Moment 874.384 = = 2.96 96 m Mass 295.84

The draught after loading can be determined: D=

Δ 295.84 = = 1.20 m LB ρ 40 × 6 × 1.025

The new KB can be found from the draught: KB =

Draught 1.20 = = 0.60 m 2 2

The new BM can be determined after ballasting: ⎛ LB 3 ⎞ ⎛ 40 × 63 ⎞ Inertia ⎜⎝ 12 ⎟⎠ ⎜⎝ 12 ⎟⎠ BM = = = = 2.50 m ∇ ∇ 40 × 6 × 1.2 This allows the KG to be determined, as GM is known: GM = KB + BM − KG

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468 • Ship Stability, Powering and Resistance GM = 0.6 + 2.5 − 2 96 GM = 0.14 m A loading table can be used to determine the final TCG: Item

Mass (tonnes)

Barge

TCG (m)

246

Ballast

9.84

Cargo

40

Total

Moment (tonne metres)

0

0

0

0

0.5

20

295.84

TCG =

20

Moment 20 = = 0.068 068 m Mass 295.84

This allows the list to be determined from the GM and heeling moment: tanθ =

TCG GM

tanθ =

0.068 0 14

θ = 25.9 9 degrees, above limits of small angle theory As the angle is above the limits of small angle theory, large angle theory must be used. As the vessel is wall-sided, the wall-sided formula can be used to determine GZ. Angle (degrees)

GZ (m)

TCG correction (m)

Final GZ (m)

0

0.00

0.07

–0.07

5

0.01

0.07

–0.05

10

0.03

0.07

–0.04

15

0.06

0.07

–0.01

20

0.10

0.06

0.04

25

0.17

0.06

0.11

30

0.28

0.06

0.22

By linear interpolation, the list angle is 16 degrees.

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Solutions to Questions • 469 Q3.25 The values for the loll and GM can be entered into the formula: −2GMI = GM at the angle of loll cosθ −2 × −0.1 = 0 20 m cos10 Q3.26 At 4,600 tonnes, and a KG of 6.74 m, the KN values can be used to determine GZ: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

5

0.61

0.02

10

1.22

0.05

15

1.83

0.09

These can be multiplied by the displacement to find the righting moment values: Angle (degrees)

GZ (m)

Righting moment (tonne metres)

0

0.00

0

5

0.02

87.4

10

0.05

207

15

0.09

391

From Figure 3.83, at a wind speed of 75 knots, the wind pressure would be 95 kg/m2. This equates to 0.095 t/m2. The wind heeling moment can be found by: Wind heeling moment Win W d pressure × Lateral area Lever Wind feeling moment = 0.095 × 617 × 5.17 = 303.14 tonne metres The curve of righting moments can be drawn, and the heeling moment value superimposed on that:

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470 • Ship Stability, Powering and Resistance

700

Righting moment Heeling moment

Heeling and righting moments (tonne metres)

600

500

400

300

200

100

0 0

1

2

3

4

5

6

7

8 9 10 11 12 13 14 15 16 17 18 19 20 Heel (degrees)

It can be seen that the heeling moment and righting moment are equal at 12.8 degrees, therefore the vessel will heel at 12.8 degrees. Q3.27 The mass can be determined from the volume of the grain and the stowage factor. At a sounding of 4.00 m, the grain hold shows a volume of 600 m3. Therefore: Grain mass =

Grain volume 600 = = 428.57 tonnes Stowage ffactor 1.4

At a sounding of 4.00 m, the grain hold shows a volumetric heeling moment of 1,006.8 m4. Therefore: Grain mass ass heeling moment =

Grain volumetric o u et c heeling moment Stowage t factor

Grain mass ass heeling moment o e t=

9781408176122_Ch14_2_Rev_txt_prf.indd 470

1, 006.8 = 719.14 tonne metres 1.4

11/16/2013 2:43:59 AM

Solutions to Questions • 471 Q3.28 The mass and volumetric heeling moment of the grain can be found from the stowage factor and the grain hold data. The mass can be determined from the volume of the grain and the stowage factor. At a sounding of 5.00 m, the grain hold shows a volume of 750 m3. Therefore: Grain mass =

Grain volume 750 = = 517.24 tonnes Stowage ffactor 1.45

The KG of the grain will be 4.50 m. At a sounding of 5.00 m, the grain hold shows a volumetric heeling moment of 1,006.8 m4. Therefore: Grain mass ass heeling moment =

Grain volumetric o u et c heeling moment Stowage t factor

Grain mass ass heeling moment =

1, 006.8 = 694.35 tonne metres 1.45

A loading table can be used to determine the KG of the vessel after loading: Item

Mass

KG

Moment

2,615.00

7.00

18,305.00

No. 2 DB Port

39.22

0.60

23.33

Port bunker

117.76

7.81

919.73

No. 2 DB Stb

39.22

0.60

23.53

Stb bunker

117.76

7.81

919.73

Lightship

No. 2 DB Port FSM

0.00

0

62.15

No. 2 DB Stb FSM

0.00

0

62.15

Grain

517.241

Total

4.5

3,446.20

KG =

2,327.59 22,643.20

Moment 22, 643.2 = = 6.57 57 m Mass 3, 446.2

At 3,446.2 tonnes, the KM can be interpolated. This gives a value of 7.23 m for KM. This allows GM to be found: GM = KM − KG = 7.23 − 6.57 = 0.66 m

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472 • Ship Stability, Powering and Resistance The minimum GM for a vessel carrying grain is 0.30 m, therefore the vessel is legal with respect to the initial GM. The KN values can be interpolated, and the GZ values found: Angle (degrees)

KN (m)

GZ (m)

0

0.00

0.00

5

0.63

0.06

10

1.26

0.12

15

1.90

0.20

20

2.54

0.29

25

3.16

0.39

30

3.76

0.48

35

4.33

0.56

40

4.86

0.64

The λ0 value can be found: λ0 =

Total gra grain mass heeling moment 694.35 = = 0.201 m Δ 3, 446.2

The λ40 value can be found: λ 40 λ 0 ×0.8 = 0.201× 0.8 = 0.161 m The GZ curve and the grain heeling line can be plotted: 0.700 0.600

GZ (m)

0.500 0.400 0.300 0.200 0.100 0.000 0

5

9781408176122_Ch14_2_Rev_txt_prf.indd 472

10

15

20 25 Heel (degrees)

30

35

40

11/16/2013 2:44:01 AM

Solutions to Questions • 473 It can be seen that the grain heeling line and the GZ curve intersect at 14.4 degrees, therefore the vessel fails the list criteria. The area under the curve from 0 to 40 degrees can be found using Simpson’s Rule: Angle (degrees)

GZ (m)

Simpson’s Multiplier

Area product

0

0.00

1

0.00

10

0.12

4

0.48

20

0.29

2

0.58

30

0.48

4

1.92

40

0.64

1

0.64

Total

3.62

Area =

Spacing p g 10 × Total area product = × 3.62 = 12 12.07 m degrees 3 3 Area =

12.07 = 0.211 m radians 57.3

Using the GZ, it can be seen that the GZ value at the intersection of the lines is 0.188 m. Therefore, the area under the grain heeling line can be found between 0 and 14.4 degrees, assuming that the area can be approximated by a triangle: Area

1 1 Length t Height Height = × 14.4 × 0.188 = 1.354 m degrees 2 2 1.354 = 0.024 m radians 57.3

The area under the grain heeling line can be found between 14.4 and 40 degrees, assuming that the area can be approximated by a trapezoid: Area

Mean height eight Lengt Length t =

0.188 + 0.161 × ( 40 − 14.4 ) = 4.467 m degrees r 2

4.467 = 0.078 m radians 57.3 Therefore, the residual area between the GZ curve and the grain heeling line can be found: Area = 0.211 − 0.024 − 0.078 = 0.109 m radians The vessel therefore passes the residual area criteria.

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474 • Ship Stability, Powering and Resistance Q3.29 There is no direct row in the tables for 4,050 tonnes displacement, or a column for a KG of 5.35 m. The row and column must be created using linear interpolation. The displacement of 4,050 tonnes lies 44.1% of the range between 4,001 and 4,112 tonnes, and the KG of 5.35 m lies 75% of the range between 5.20 and 5.40 m. Displacement\KG

5.20 m

4,112

1,635.9

4,050

1,623.7

4,001

1,614.1

5.35 m

5.40 m 1,464.9

1,497.6

1,455.55 1,447.7

The maximum allowable grain mass heeling moment is therefore 1,497.6 tonne metres. The actual grain mass heeling moment is 1,550 tonne metres, therefore the vessel fails the criteria as the actual value exceeds the allowable value. Q3.30 There is no direct row in the tables for 6,300 tonnes displacement, or a column for a KG of 5.05 m. The row and column must be created using linear interpolation. The displacement of 6,300 tonnes lies 28.7% of the range between 6,263 and 6,392 tonnes, and the KG of 5.05 m lies 25% of the range between 5.00 and 5.20 m. Displacement\KG

5.00 m

6,392

1,528.0

6,300

1,485.9

6,263

1,469.5

5.05 m

5.20 m 1,262.2

1,420.4

1,224.0 1,209.1

The maximum allowable grain mass heeling moment is therefore 1,420.4 tonne metres. The actual grain mass heeling moment is 1,400 tonne metres, therefore the vessel passes the criteria as the actual value does not exceed the allowable value. Q3.31 The appropriate values can be found in the table of maximum allowable grain mass heeling moments: Displacement\KG

5.80 m

4,796

1,219.7

4,700

?

4,680

1,197.4

9781408176122_Ch14_2_Rev_txt_prf.indd 474

5.85 m

6.00 m 1,020.3

??

? 1,002.8

11/16/2013 2:44:03 AM

Solutions to Questions • 475 Interpolation is not actually needed to answer this question. The interpolated value of the maximum allowable mass grain heeling moment cannot be greater than any of the four boundary values in the table above. Therefore, for any displacement between 4,680 and 4,796 tonnes, and any KG between 5.85 and 6.00 m, the vessel will not pass with an actual grain mass heeling moment of 1,300 tonne metres. Q3.32 The appropriate values can be found in the table of maximum allowable grain mass heeling moments: KG = 6.80 m 7,192 tonnes

758.8

7,100 tonnes

?

7,056 tonnes

701.4

KG = 6.74 m

KG = 7.00 m 459.8

??

? 408.0

As with the previous question, interpolation is not actually needed to answer this question. The interpolated value of the maximum allowable mass grain heeling moment cannot be less than any of the four boundary values in the table above. Therefore, for any displacement between 7,056 and 7,192 tonnes, and any KG between 6.80 and 7.00 m, the vessel will pass with an actual grain mass heeling moment of 400 tonne metres. Q3.33 The mass and volumetric heeling moment of the grain can be found from the stowage factor and the grain hold data. The mass can be determined from the volume of the grain and the stowage factor. At a sounding of 6.00 m, the grain hold shows a volume of 900 m3. Therefore: Grain mass =

Grain volume 900 = = 652.17 tonne t s Stowage ffactor 1 38

The KG of the grain will be 5.00 m. At a sounding of 6.00 m, the grain hold shows a volumetric heeling moment of 925.2 m4. Therefore: Grain mass ass heeling moment =

Grain volumetric olumet c heeling moment Stowage t factor

Grain mass ass heeling moment =

9781408176122_Ch14_2_Rev_txt_prf.indd 475

925.2 = 670.44 tonne metres 1 38

11/16/2013 2:44:03 AM

476 • Ship Stability, Powering and Resistance A loading table can be used to determine the KG of the vessel after loading: Item

Mass

Lightship

2,615

KG

Moment

7

18,305

Port bunker

120.22

7.81

938.92

Starboard bunker

120.22

7.81

938.92

Grain

652.17

5

Total

3,507.61

KG =

3,260.85 23,443.69

Moment 23, 443.69 = = 6.68 m Mass 3, 507.61

From the maximum permissible grain heeling moment tables, the following mass heeling moments are permitted: KG = 6.60 m

KG = 6.80 m

3,563 tonnes

512.2

364.1

3,455 tonnes

532.1

388.5

The values can be interpolated for the displacement of 3,507.6 tonnes. KG = 6.60 m

KG = 6.80 m

3,563 tonnes

512.2

364.1

3,507.6 tonnes

522.4

376.6

3,455 tonnes

532.1

388.5

The values can be then be interpolated for the KG of 6.68 m. KG = 6.60 m 3,563 tonnes

512.2

3,507.6 tonnes

522.4

3,455 tonnes

532.1

KG = 6.68 m

KG = 6.80 m 364.1

464.1

376.6 388.5

As can be seen, the limiting mass heeling moment is 464.1 tonne metres, while the actual mass heeling moment is 670.44 tonne metres, therefore the vessel fails to meet the criteria.

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Solutions to Questions • 477 The approximate list angle can be found from: Approximate list angle = 12 ×

Actual mass ass heeling moment Maximum mass heeling momemt

Approximate list angle = 12 ×

670.44 = 17.3 degrees 464.4

Q4.1 The KB can be found: KB =

D 6 = = 3.00 m 2 2

The BML can be found, remembering that it requires the longitudinal inertia: BLL3 11× 703 InertiaL 12 BML = = 12 = = 68.06 06 m ∇ LBD 70 × 11× 6 The GML can be found in a similar way to GM: GML = KB B + BML − KG = 3 + 68.06 − 4 = 67.06 m Clearly the very large value indicates great stability in the longitudinal sense. Q4.2 The KB can be found: KB =

D 5 = = 2.50 m 2 2

The BML can be found, remembering that it requires the longitudinal inertia: BLL3 13 × 74 3 InertiaL 12 BML = = 12 = = 91.28 m ∇ LBD 74 × 13 × 5 The GML can be found in a similar way to GM: GML = KB + BML − KG = 2.50 + 91.28 − 3.50 = 90.28 m The MCTC can be found (note the displacement for a box shaped vessel can be found from the water density and the dimensions): MCTC =

Δ × GML 74 × 13 × 5 × 1.025 × 90.28 = = 60.15 tonne mettres 100 × LBP 100 × 74

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478 • Ship Stability, Powering and Resistance Q4.3 The KB can be found: KB =

D 2 = = 1.00 00 m 2 2

The BML can be found, remembering that it requires the longitudinal inertia: BLL3 8 × 40 3 InertiaL 12 BML = = 12 = = 66.67 67 m ∇ LBD 40 × 8 × 2 The GML can be found in a similar way to GM: GML = KB B + BML − KG = 1.00 + 66.67 − 2.50 = 65.17 m The MCTC can be found (note the displacement for a box shaped vessel can be found from the water density and the dimensions): MCTC =

Δ × GML 40 × 8 × 2 × 1.025 × 65.17 = = 10.69 69 tonne metre r s 100 × LBP 100 × 40

The LCB will be at the centre of underwater volume, which for a box shaped vessel will be at half of the length: LCB =

L 40 = = 20 m FOAP 2 2

The trim can be found: Trim = Trim =

LCG ) Δ MCTC

(LCB

(20 − 19 )( 40 × 8 × 2 × 1.025) 10.69

= 61.4 4

= 0.61 m stern trim

Q4.4 At 2,615 tonnes, the hydrostatics can be read from the data, and substituted into the trim formula: Trim =

LCG ) Δ ( 47.18 − 44 ) 2, 615 = = 189.6 MCTC 43.87

(LCB

= 1.90 m sttern trim

Q4.5 Change in trim =

9781408176122_Ch14_2_Rev_txt_prf.indd 478

Trimming moment 5 × 10 = = 10.8 cm MCTC 4.613

11/16/2013 2:44:07 AM

Solutions to Questions • 479 This change of trim will be by the stern as the mass has moved aft. Q4.6 Trimming moment = Mass × Distance to LCF Trimming moments that push the stern down are treated as positive, while trimming moments that push the bow down are treated as negative. Therefore: Trimming moment = (7 × 5) + ( − ( × Change in trim =

))

Trimming moment (7 × 5) + ( − ( × = MCTC 7.324

))

= 1.50 0 cm

This will be by the stern as the change in trim is positive. Q4.7 The LCF is at the centre of the waterplane area, therefore for a box shaped vessel, this will be at half of the length of the vessel: DLCF + Trim

DA

DF

⎛ LCF ⎞ ⎛ 50 ⎞ = 3 + 1.10 = 3.55 m ⎝ LBP ⎠ ⎝ 100 ⎠

DA − Trim = 3.55 − 1.10 = 2.45 m

Q4.8 In the lightship condition, the hydrostatics show that the true mean draught is 3.00 m, and the LCF is 47.14 m FOAP. The LBP of the vessel is 100 m. DA

DLCF + Trim DF

⎛ LCF ⎞ ⎛ 47.14 ⎞ = 3.00 + 1.90 = 3.90 90 m ⎝ LBP ⎠ ⎝ 100 ⎠

DA − Trim = 3.90 − 1.90 = 2.00 m

Q4.9 A loading table can be used to determine the LCG and displacement after loading: Item

Mass

LCG

Moment

2,615.00

44.00

115,060.00

Port bunker

101.15

13.83

1,398.86

Stb bunker

101.15

13.83

1,398.86

Cargo 1

212.71

90

Lightship

Total

9781408176122_Ch14_2_Rev_txt_prf.indd 479

3,030

19,143.90 137,001.61

11/16/2013 2:44:09 AM

480 • Ship Stability, Powering and Resistance

LCG =

Moment 137, 001.61 = = 45.22 m FOAP Mass 3, 030

At 3,030 tonnes, the hydrostatics can be found from the data book, giving the LCB as 47.15 m FOAP, the LCF as 46.81 m FOAP, the MCTC as 45.16 tonne metres/cm and the true mean draught as 3.40 m. These can be used to find the trim and the end draughts: Trim =

LCG ) Δ ( 47.15 − 45.22) 3, 030 = = 129.5 5 MCTC 45.16

(LCB

DA

DLCF + Trim DF

= 1.30 m stern trim

⎛ LCF ⎞ ⎛ 46.81⎞ = 3.40 0 + 1.30 = 4.01 01 m ⎝ LBP ⎠ ⎝ 100 ⎠

DA − Trim = 4.01 − 1.30 = 2.71 m

Q5.1 The underwater volume can be found from the dimensions: ∇ = LBD = 100 × 10 × 2 = 2, 000 m3 The displacement can be found from the underwater volume: ∇ × ρ = Δ ∴ 2, 000 × 1.025 = 2, 050 tonnes The KB can be found from the draught: KB =

D 2 = = 1.00 00 m 2 2

The BM can be found: LB 3 100 × 103 Inertia 12 12 BM = = = = 4.17 17 m ∇ LBD 100 × 10 × 2 The KM can be found: KM = KB + BM = 1.00 + 4.17 = 5.17 17 m The LCF is the longitudinal centre of the waterplane area. The waterplane area is 100 m long, 10 m wide and rectangular. The centre of the waterplane will therefore be at half the length of the length of the vessel, or 50.00 m FOAP. The LCB is the longitudinal centre of the underwater volume. The underwater volume is 100 m long, 10 m wide, 2 m deep and rectangular. The centre of the underwater volume will therefore be at half the length of the length of the vessel, or 50.00 m FOAP. To determine the MCTC, the longitudinal metacentric height must be found:

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11/16/2013 2:44:10 AM

Solutions to Questions • 481 BLL3 D I D GML = KB + BML − KG = + L − KG = + 12 − KG 2 ∇ 2 LBD 10 × 1003 2 12 ∴ GML = + − 3 = 414.67 67 m 2 100 × 10 × 2 MCTC =

Δ × GML 2, 050 × 414.67 = = 85.01 tonne metres 100 × LBP 100 × 100

The TPC can be found by formula: TPC =

Waterplane p area 100 × 10 ×ρ= × 1.025 = 10.25 t/cm 100 100

Q5.2 The underwater volume can be found from the dimensions: ∇ = LBD = 100 × 10 × 2 = 2, 000 m3 The displacement can be found from the underwater volume: ∇ × ρ = Δ ∴ 2, 000 × 1.000 = 2, 000 tonnes The KB can be found from the draught: KB =

D 2 = = 1.00 m 2 2

The BM can be found: LB 3 100 × 103 Inertia 12 12 BM = = = = 4.17 17 m ∇ LBD 100 × 10 × 2 The KM can be found: KM = KB + BM = 1.00 + 4.17 = 5.17 17 m The LCF is the longitudinal centre of the waterplane area. The waterplane area is 100 m long, 10 m wide and rectangular. The centre of the waterplane will therefore be at half the length of the length of the vessel, or 50.00 m FOAP. The LCB is the longitudinal centre of the underwater volume. The underwater volume is 100 m long, 10 m wide, 2 m deep and rectangular. The centre of the underwater volume will therefore be at half the length of the length of the vessel, or 50.00 m FOAP.

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11/16/2013 2:44:12 AM

482 • Ship Stability, Powering and Resistance To determine the MCTC, the longitudinal metacentric height must be found: BLL3 D I D GML = KB B + BML − KG = + L − KG = + 12 − KG 2 ∇ 2 LBD 10 × 1003 2 12 ∴ GML = + − 3 = 414.67 67 m 2 100 × 10 × 2 MCTC =

Δ × GML 2, 000 × 414.67 = = 82.93 93 tonne metres 100 × LBP 100 × 100

The TPC can be found by formula: TPC =

Waterplane p area 100 × 10 ×ρ = × 1.000 = 10.00 t/cm 100 100

Q5.3 The hydrostatics at 4.00 m draught can be read from the data book. These are the sea water values: Draught (m)

4.00

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

3,671

47.04

46.17

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

2.16

7.13

10.88

47.70

Any value with tonnes in the units must be corrected to be the dock water value by multiplying by the ratio of the dock water density to the sea water density. For example, the displacement is corrected by: Δ DW

Δ SW ×

ρDDW 1.010 = 3, 671× = 3, 617.3 tonnes ρSSW 1.025

Draught (m) Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

4.00

2.16

7.13

10.72

47.00

3,617.3

9781408176122_Ch14_2_Rev_txt_prf.indd 482

47.04

46.17

11/16/2013 2:44:14 AM

Solutions to Questions • 483 Q5.4 The hydrostatics for sea water at 4.10 m and 4.20 m show: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

4.20

3,890

46.99

45.95

2.27

7.06

11.02

48.91

4.10

3,781

47.01

46.06

2.22

7.10

10.95

48.29

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

These can be corrected for dock water: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

4.20

3,821.7

46.99

45.95

2.27

7.06

10.83

48.05

4.10

3,714.6

47.01

46.06

2.22

7.10

10.76

47.44

Finally, the values can be interpolated for 4.15 m: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

4.20

3,821.7

46.99

45.95

2.27

7.06

10.83

48.05

4.15

3,768.15

47.00

46.01

2.25

7.08

10.80

47.86

4.10

3,714.6

47.01

46.06

2.22

7.10

10.76

47.44

Q5.5 The hydrostatics for sea water at the two displacements closest to 3,600 tonnes show: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

4.00

3,671

47.04

46.17

2.16

7.13

10.88

47.70

3.90

3,563

47.06

46.29

2.11

7.18

10.82

47.23

9781408176122_Ch14_2_Rev_txt_prf.indd 483

11/16/2013 2:44:16 AM

484 • Ship Stability, Powering and Resistance These can be corrected for dock water: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

4.00

3,617.3

47.04

46.17

2.16

7.13

10.72

47.00

3.90

3,510.9

47.06

46.29

2.11

7.18

10.66

46.54

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

Finally, the values can be interpolated for 4.15 m: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

4.00

3,617.3

47.04

46.17

2.16

7.13

10.72

47.00

3.98

3,600

47.04

46.19

2.15

7.14

10.70

46.92

3.90

3,510.9

47.06

46.29

2.11

7.18

10.66

46.54

Q5.6 The hydrostatics at 4.00 m draught can be read from the data book. These are the sea water values: Draught (m)

4.00

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP) 3,671

47.04

46.17

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

2.16

7.13

10.88

47.70

Therefore, at a draught of 4.00 m in sea water, the displacement is 3,671 tonnes. The hydrostatics can be read off at a draught of 5.00 m in sea water: Draught (m)

5.00

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP) 4,796

46.69

44.89

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

2.71

6.93

11.64

55.31

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

2.71

6.93

11.36

53.96

The hydrostatics can be corrected to be for fresh water: Draught (m)

5.00

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP) 4,679.0

9781408176122_Ch14_2_Rev_txt_prf.indd 484

46.69

44.89

11/16/2013 2:44:16 AM

Solutions to Questions • 485 It can be seen that at a draught of 5.00 m in sea water, the displacement is 4,679 tonnes. As the vessel started with a displacement of 3,671 tonnes, a total of 1,008 tonnes of cargo can be loaded. Q5.7 At 5.00 m, the displacement in sea water is 4,796 tonnes. Corrected for dock water gives a displacement of 4,702.4 tonnes at 5.00 m. At her summer draught, the displacement in sea water is 7,329 tonnes. Therefore, the total mass that can be added to the vessel is 2,626.6 tonnes. Q5.8 The fresh water allowance can be determined: FWA W =

Δ SUMMER 7, 329 E = = 133 mm 4 × TPC SW 4 × 13 77

The dock water allowance can be determined: ⎛ ρ −ρ ⎞ ⎛ 1.025 − 1.015 ⎞ DWA W FWA ⎜ SW DDW ⎟ = 133 = 53 mm ⎝ 1.025 − 1.000 ⎠ ⎝ ρSW − ρFFW ⎠ Therefore the vessel can have her Load Lines submerged by 53 mm. The vessel has a summer draught of 7.00 m, therefore the final draught will be 7.05 m. As the starting waterline is 20 cm, or 0.2 m below the summer Load Line, the initial draught must be 6.80 m. The TPC in seawater at 7.05 m can be interpolated as 13.82 t/cm, and the TPC in seawater at 6.80 m is 13.56 t/cm. The mean of these values can be found: Mean TPC C=

13.82 + 13.56 = 13.69 t/cm 2

This is the seawater value, and must therefore be converted to the dock water value: TPC DW

T SW × TPC

ρDDW 1.015 = 13.69 × = 13.56 56 t/cm ρSSW 1.025

The allowable sinkage is 20 cm to the summer Load Line, and then 5.3 cm to the allowable waterline, giving a total allowable sinkage of 25.3 cm. Therefore, the mass to add can be determined: Sinkage =

9781408176122_Ch14_2_Rev_txt_prf.indd 485

Mass TPC

11/16/2013 2:44:16 AM

486 • Ship Stability, Powering and Resistance

25.3 =

Mass 13.56

Mass = 343.07 tonnes Therefore, using the mean TPC value, the mass to load is 343.07 tonnes. The relevant hydrostatic values can be corrected for the dock water density: Sea water values: Draught (m) Displacement (tonnes) 6.8

7,056

Dock water values: Draught (m) Displacement (tonnes) 6.8

6,987.2

Therefore, at a draught of 6.80 m in dock water, the displacement is 6,987.2 tonnes. The summer displacement of the vessel is 7,329 tonnes, therefore the mass to load can be found: Mass to load = Δ SSUMMER ER − Δ START = 7 , 329 − 6 , 987.2 = 341.8 tonnes This is the more accurate solution, as the solution using the TPC assumes that the vessel is wall-sided over the change in draught. Q5.9 In the lightship condition, the displacement is 2,615 tonnes. After 3,000 tonnes is loaded, the displacement will be 5,615 tonnes. In sea water, the hydrostatics show the following close to the final displacement: Draught (m) Displacement (tonnes) 5.80

5,757

5.70

5,634

These can be corrected to the dock water values: Draught (m) Displacement (tonnes)

9781408176122_Ch14_2_Rev_txt_prf.indd 486

5.80

5,728.9

5.70

5,606.5

11/16/2013 2:44:17 AM

Solutions to Questions • 487 Interpolating between these displacements for the draught gives a draught of 5.71 m in dock water. This allows the UKC to be found: UKC K

Depth Depth − Draught = 10.00 − 5.71 = 4.29 m

Q5.10 In the starting condition, the displacement at 6.00 m in sea water is 6,008 tonnes. After 100 tonnes is unloaded, the displacement will be 5,908 tonnes. In sea water, the hydrostatics show the following close to the final displacement: Draught (m) Displacement (tonnes) 6.10

6,135

6.00

6,008

These can be corrected to the fresh water values: Draught (m)

Displacement (tonnes)

6.10

5,985.4

6.00

5,861.5

Interpolating between these displacements for the draught gives a draught of 6.04 m in fresh water. Q5.11 The mean draught can be found: DM =

DA

DF 2

=

7+3 = 5.000 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 44.889 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 7.000 = DLCF ⎝ LBP ⎠

.000

⎛ 44.889 ⎞ ⎝ 100 ⎠

DLC 204 m L F = 5.204

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 44.571 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

9781408176122_Ch14_2_Rev_txt_prf.indd 487

⎛ LCF ⎞ ∴ 7.000 = DLCF ⎝ LBP ⎠

.000

⎛ 44.571⎞ ⎝ 100 ⎠

DLC 217 m L F = 5.217

11/16/2013 2:44:18 AM

488 • Ship Stability, Powering and Resistance This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 44.550 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 7.000 = DLCF ⎝ LBP ⎠

.000

⎛ 44.550 ⎞ ⎝ 100 ⎠

DLC L F = 5.218 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 44.548 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 7.000 = DLCF ⎝ LBP ⎠

.000

⎛ 44.548 ⎞ ⎝ 100 ⎠

DLC L F = 5.218 m

Therefore, the true mean draught can be taken as 5.218 m. Q5.12 The mean draught can be found: DM =

DA

DF 2

=

7+4 = 5.50 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 44.05 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 7 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

44.05 ⎞ 100 ⎠

DLCF = 5.68 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 43.70 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 7 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

43.70 ⎞ 100 ⎠

DLCF = 5.69 m

Therefore, the true mean draught can be taken as 5.69 m. The displacement can be interpolated at this draught, giving a displacement of 5,621.7 tonnes Q5.13 The mean draught can be found: DM =

9781408176122_Ch14_2_Rev_txt_prf.indd 488

DA

DF 2

=

6 + 5.31 = 5.66 m 2

11/16/2013 2:44:19 AM

Solutions to Questions • 489 This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 43.74 m FOAP. This can then be used to determine the approximate true mean draught. DA

⎛ LCF ⎞ ∴ 6 = DLCF ⎝ LBP ⎠

DLCF + Trim

(

) ⎛⎝

43.74 ⎞ 100 ⎠

DLCF = 5.70 70 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 43.66 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

⎛ LCF ⎞ ∴ 6 = DLCF ⎝ LBP ⎠

DLCF + Trim

(

) ⎛⎝

43.66 ⎞ 100 ⎠

DLCF = 5.70 m

Therefore, the true mean draught can be taken as 5.70 m. The displacement can be read directly at this draught giving a displacement of 5,634 tonnes. The KM can be read directly giving a value of 6.96 m. This allows GM to be found: GM = KM − KG = 6.96 − 6.00 = 0.96 96 m Q5.14 The mean draught can be found: DM =

DA

DF 2

=

5 + 5.23 = 5.12 12 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 44.71 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 5 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

44.71⎞ 100 ⎠

DLCF = 5.10 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 44.74 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 5 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

44.76 ⎞ 100 ⎠

DLCF = 5.10 10 m

Therefore, the true mean draught can be taken as 5.10 m. The displacement can be read directly at this draught, giving a displacement of 4,913 tonnes.

9781408176122_Ch14_2_Rev_txt_prf.indd 489

11/16/2013 2:44:21 AM

490 • Ship Stability, Powering and Resistance Q5.15 The layer correction process can be used to determine the displacement of the vessel in the initial condition: DM =

DA

DF 2

=

6 + 5.31 = 5.66 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 43.74 m FOAP. Note that the LCF is independent of density, and therefore no density correction is needed. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

43.74 ⎞ 100 ⎠

DLCF = 5.70 70 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 43.66 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

43.66 ⎞ 100 ⎠

DLCF = 5.70 m

Therefore, the true mean draught can be taken as 5.70 m. At 5.70 m in sea water, the displacement is 5,634 tonnes. This can be corrected for dock water: Δ DW

Δ SW ×

ρDDW 1.020 = 5, 634 × = 5, 606.5 tonnes ρSSW 1.025

The mean draught can be found: DM =

DA

DF 2

=

4 + 3.15 = 3.58 58 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 46.63 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.63 ⎞ 100 ⎠

DLCF = 3.60 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 46.61 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

9781408176122_Ch14_2_Rev_txt_prf.indd 490

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.61⎞ 100 ⎠

DLCF = 3.60 60 m

11/16/2013 2:44:23 AM

Solutions to Questions • 491 Therefore, the true mean draught can be taken as 3.60 m. The hydrostatics can be read directly at this draught, giving a displacement of 3,241 tonnes, an LCB of 47.13 m FOAP and an MCTC of 45.84 tonne metres/cm. Trim =

LCG ) Δ (47.13 − LCG ) × 3, 241 ∴100 ( 4 − 3.15) = MCTC 45.84

(LCB

85 = ( 47.13 13 − LCG ) × 70 0.7 ∴1.20 0 = 47.13 3 − LCG ∴ LCG = 45.93 m FOAP Q5.17 On an even keel, the draught aft must be the same as all of the other draught values, therefore the true mean draught must also be 4.50 m. As the trim of the vessel is zero, the LCB must be the same as the LCG. Therefore, the LCG of the vessel is 46.89 m FOAP and the displacement is 4,224 tonnes. A loading table can be used to determine the LCG and displacement after loading: Item

Mass

LCG

Moment

Ship

4,224

46.89

198,063.36

Cargo

1,000

35.00

Total

5,224

LCG =

35,000 233,063.36

Moment 233, 063.36 = = 44.61 m FOAP Mass 5, 224

At 5,224 tonnes, the hydrostatics can be interpolated from the data book, giving the LCB as 46.52 m FOAP, the LCF as 44.30 m FOAP, the MCTC as 59.34 tonne metres/cm and the true mean draught as 5.36 m. These can be used to find the trim and the end draughts: Trim =

LCG ) Δ ( 46.52 − 44.64 ) 5, 224 = = 165.5 5 MCTC 59.34

(LCB

DA

DLCF + Trim DF

= 1.66 m stern trim

⎛ LCF ⎞ ⎛ 44.30 ⎞ = 5.36 + 1.66 = 6.10 10 m ⎝ LBP ⎠ ⎝ 100 ⎠

DA − Trim = 6.10 − 1.66 = 4.44 44 m

Q5.18 On an even keel, the draught aft must be the same as all of the other draught values, therefore the true mean draught must also be 4.00 m. As the trim of the vessel is zero, the LCB must be the same as the LCG. Therefore, the initial LCG of the vessel is 47.04 m FOAP, and the initial displacement is 3,671 tonnes.

9781408176122_Ch14_2_Rev_txt_prf.indd 491

11/16/2013 2:44:26 AM

492 • Ship Stability, Powering and Resistance For the final condition, the layer correction process can be used to find the required hydrostatics to achieve the required final condition. The mean draught can be found: DM =

DA

DF 2

=

6.23 + 5.23 = 5.73 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 43.58 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6.23 23 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

43.60 ⎞ 100 ⎠

DLCF C = 5.79 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 43.46 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6.23 23 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

43.48 ⎞ 100 ⎠

DLCF C = 5.80 m

Therefore, the final true mean draught can be taken as 5.80 m. The hydrostatics can be read directly at this draught, giving a displacement of 5,757 tonnes, an LCB of 46.28 m FOAP and an MCTC of 65.43 tonne metres/cm. These can be used to determine the required LCG in the final condition: Trim =

LCG ) Δ (46.27 − LCG ) × 5, 757 ∴100 = ∴ LCG = 45.13 13 m FOAP 65.43 MCTC

(LCB

The amount of cargo to load can be found from the difference between the starting displacement and the final displacement: Cargo = Δ FINAL − Δ INITIAL = 5, 757 − 3, 671 = 2, 086 tonnes A loading table can be used to determine the position of the cargo so that the vessel finishes loading with the correct final LCG: Item

Mass

LCG

Moment

Ship

3,671

47.04

172,683.84

Cargo

2,086

x

2,086x

Total

5,757

LCG =

172,683.84 + 2,086x

Moment 172, 683.84 + 2, 086 x = = 45.13 m FOAP Mass 5, 757

9781408176122_Ch14_2_Rev_txt_prf.indd 492

11/16/2013 2:44:27 AM

Solutions to Questions • 493 172, 683.84 + 2, 086 x = 45.13 .13 3 ∴172, 683.84 + 2, 086 = 45.13 × 5, 757 5, 757 ∴ 2, 086 = ( 45.13 × 5, 757) − 172, 683.84 ∴ =

(45.13 × 5, 757) − 172, 683.84 2, 086

= 41.77 m FOAP

Q5.19 For the initial condition, the layer correction process can be used to find the required hydrostatics to achieve the required final condition. The mean draught can be found: DM =

DA

DF 2

=

4.00 + 2.71 = 3.36 36 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 46.85 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.85 ⎞ 100 ⎠

DLCF = 3.40 40 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 46.81 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.81⎞ 100 ⎠

DLCF = 3.40 40 m

Therefore, the initial true mean draught can be taken as 3.40 m. The hydrostatics can be read directly at this draught, giving a displacement of 3,030 tonnes, an LCB of 47.15 m FOAP and an MCTC of 45.16 tonne metres/cm. These can be used to determine the required LCG in the initial condition: Trim =

LCG ) Δ (47.15 − LCG ) × 3, 030 ∴129 = ∴ LCG = 45.23 m FOAP MCTC 45.16

(LCB

For the final condition, the true mean draught is given. Therefore, layer correction is not required. The hydrostatics can be read directly at this draught, giving a displacement of 4,179.2 tonnes, an LCB of 46.90 m FOAP and an MCTC of 50.71 tonne metres/cm. These can be used to determine the required LCG in the final condition: Trim =

LCG ) Δ (46.90 − LCG ) × 4 ,179.2 ∴ 34 = ∴ LCG = 46.49 m FOAP MCTC 50.71

(LCB

9781408176122_Ch14_2_Rev_txt_prf.indd 493

11/16/2013 2:44:29 AM

494 • Ship Stability, Powering and Resistance The amount of cargo to load can be found from the difference between the starting displacement and the final displacement: Cargo = Δ FINAL − Δ INITIAL = 4 ,179.2 − 3, 030 = 1 1149 , .2 tonnes If x tonnes of cargo is loaded aft, then 1,149.2 − x tonnes of cargo must be loaded forward. A loading table can be used to determine the position of the cargo so that the vessel finishes loading with the correct final LCG: Item

Mass

LCG

Moment

Ship

3,030

45.23

137,046.9

Aft cargo

x

30

30x

Fwd cargo

1,149.2 – x

80

91,936 – 80x

Total

LCG =

4,179.2

228,982.9 – 50x

Moment 228 , 982.9 − 50 x = = 46.49 49 m FOAP ∴ x = 693.84 to t nnes aft f Mass 4 ,179.2

We know that we must load 1,149.2 − x tonnes forward, therefore: 1149 ,149.2 − 693.84 = 455.36 tonnes forward Q5.20 For the initial condition, the layer correction process can be used to find the required hydrostatics to achieve the required final condition. The mean draught can be found: DM =

DA

DF 2

=

5.00 + 3.90 = 4.45 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 45.65 m FOAP. This can then be used to determine the approximate true mean draught. Note that at a given draught the LCF value is not affected by water density, therefore no density corrections are required to the hydrostatics. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 5 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

45.65 ⎞ 100 ⎠

DLCF = 4.50 50 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 45.58 m FOAP. This can be used to calculate a more accurate value for the true mean draught. Again, at a given draught the LCF value is not affected by water density, therefore no density corrections are required to the hydrostatics.

9781408176122_Ch14_2_Rev_txt_prf.indd 494

11/16/2013 2:44:31 AM

Solutions to Questions • 495

DA

DLCF + Trim

⎛ LCF ⎞ ∴ 5 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

45.58 ⎞ 100 ⎠

DLCF = 4.50 50 m

Therefore, the initial true mean draught can be taken as 4.50 m. This can be used to determine the hydrostatics of the vessel. The hydrostatics at a true mean draught of 4.50 m give the following values for sea water: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

4.50

4,224

46.888

45.579

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

2.437

6.987

11.244

51.004

These must be corrected to dock water values, by multiplying the hydrostatic value by the ratio of dock water density to sea water density: rrected hydrostatic yd ostat c vvalue = Sea w water ate hydrostatic vvalue ×

ρDW DW ρSSW

Remember – only the hydrostatics which have tonnes in the units need correcting. Therefore, the corrected dock water values at 1.005 t/m3 are: Draught (m)

Displacement LCB LCF (tonnes) (m FOAP) (m FOAP)

4.50

4,141.6

46.888

45.579

KB (m)

KM (m)

TPC (t/cm)

MCTC (tonne metres)

2.437

6.987

11.025

50.009

These can be used to determine the LCG in the initial condition: Trim =

LCG ) Δ (46.89 − LCG ) × 4 ,141.6 ∴110 = ∴ LCG = 45.56 56 m FOAP MCTC 50.01

(LCB

For the final condition, the summer draught is 7.00 m (see the principle data in the stability book. The vessel is required to be on an even keel, therefore layer correction is not required. The hydrostatics can be read directly at this draught, giving a displacement of 7,329 tonnes, and an LCB of 45.40 m FOAP. The vessel will be on an even keel, therefore the LCG will be the same as the LCB. As the vessel will be in sea water in this condition, no corrections to the hydrostatics are required. The amount of cargo to load can be found from the difference between the starting displacement and the final displacement: Cargo = Δ FINAL − Δ INITIAL = 7, 329 − 4 ,141.6 = 3,187.4 tonnes

9781408176122_Ch14_2_Rev_txt_prf.indd 495

11/16/2013 2:44:33 AM

496 • Ship Stability, Powering and Resistance If x tonnes of cargo is loaded aft, then 3,187.4 − x tonnes of cargo must be loaded forward. A loading table can be used to determine the position of the cargo so that the vessel finishes loading with the correct final LCG: Item

Mass

LCG

Moment

Ship

4,141.6

45.56

188,691.3

Aft cargo

x

25

25x

Fwd cargo

3,187.4 – x

75

239,055 – 75x

Total

LCG =

7,329

427,746.3 – 50x

Moment 427, 746.3 − 50 x = = 45.40 0 m FOAP ∴ x = 1, 900.2 tonnes o afft Mass 7, 329

We know that we must load 3,187.4 − x tonnes forward, therefore: 3,187.4 1, 900.2 1, 287.2 tonnes forward Q5.21 The mean draught can be found: DM =

DA

DF 2

=

4.00 + 3.57 = 3.79 m 2

At 3.79 m, the LCF is 46.41 m FOAP. The approximate true mean draught can be found: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4.00 = DLCFF ⎝ LBP ⎠

⎛ 46.41⎞ ⎝ 100 ⎠

DLCF = 3.80 80 m

At 3.80 m, the LCF is 46.40 m FOAP. The true mean draught can be found: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4.00 = DLCFF ⎝ LBP ⎠

⎛ 46.40 ⎞ ⎝ 100 ⎠

DLCF = 3.80 80 m

Therefore, the true mean draught in the initial condition is 3.80 m. At this draught in sea water, the displacement is 3,455 tonnes, and the MCTC is 46.74 tonne metres. These must be corrected to the fresh water values, which is 3,370.73 tonnes displacement and an MCTC of 45.60 tonne metres. The LCB is 47.09 m FOAP. These values can be used to determine the initial LCG: Trim =

LCG ) Δ (47.09 − LCG )3, 370.73 ∴ 43 = ∴ LCG = 46.51 m FOAP. MCTC 45.60

(LCB

9781408176122_Ch14_3_Rev_txt_prf.indd 496

11/16/2013 2:36:07 AM

Solutions to Questions • 497 In the final condition, an under-keel clearance of 2.00 m in water of a depth of 8.00 m means that the navigation draught must be 6.00 m. As the vessel needs stern trim, the draught aft after loading must therefore be 6.00 m. The trim after loading must be 0.47 m, therefore the final draught forward must be 5.53 m. These values can be used with the layer correction process to determine the true mean draught after loading. The mean draught can be found: DM =

DA

DF 2

=

6.00 + 5.53 = 5.77 77 m 2

At 5.77 m, the LCF is 43.52 m FOAP. The approximate true mean draught can be found: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6.00 = DLCFF ⎝ LBP ⎠

⎛ 43.52 ⎞ ⎝ 100 ⎠

DLCF = 5.80 m

At 5.80 m, the LCF is 43.46 m FOAP. The true mean draught can be found: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6.00 = DLCFF ⎝ LBP ⎠

⎛ 43.46 ⎞ ⎝ 100 ⎠

DLCF = 5.80 80 m

Therefore, the true mean draught in the final condition is 5.80 m. At this draught in sea water, the displacement is 5,757 tonnes and the MCTC is 65.43 tonne metres. The LCB is 46.28 m FOAP. These values can be used to determine the final LCG: Trim =

LCG ) Δ (46.28 − LCG )5, 757 ∴ 47 = ∴ LCG = 45.75 m FO OAP. MCTC 65.75

(LCB

As the starting and final displacements are known, the total mass of cargo to load can be found: Cargo = Δ Finall − Δ Start = 5, 757 − 3, 370.73 = 2, 386.27 tonnes If x tonnes are loaded aft, then 2,386.27 − x tonnes must be loaded forward. A loading table can be used to determine the cargo distribution: Item

Mass (tonnes)

LCG (m FOAP)

Moment (tonne metres)

Ship

3,370.73

46.51

156,772.65

Aft cargo

x

30

30x

Fwd cargo

2,386.27 – x

70

167,038.9 – 70x

Total

9781408176122_Ch14_3_Rev_txt_prf.indd 497

5,757

323,811.55 – 40x

11/16/2013 2:36:10 AM

498 • Ship Stability, Powering and Resistance

LCG =

Moment 323, 811.55 − 40 x ∴ 45.75 = ∴ x = 1, 510.7 tonnes Mass 5, 757

Therefore 1,510.7 tonnes must be loaded in the aft hold, and 2,386.27 − 1,510.7 = 875.55 tonnes must be loaded forward. Q5.22 As the two conditions are based on the true mean draught of the vessel, no layer correction is required. The displacement of the vessel at a draught of 3.00 m in dock water can be determined from the hydrostatics. At 3.00 m in sea water, the displacement would be 2,615.00 tonnes. This can be converted to the dock water displacement: Δ DOCKWATER E

Δ SW ×

ρDDW 1.008 = 2, 615.00 × = 2, 571.63 tonnes ρSSW 1.025

The displacement of the vessel at a draught of 7.00 m in sea water can be determined from the hydrostatics. At 7.00 m in sea water, the displacement would be 7,329.00 tonnes. As the sea is salt water, no corrections are needed. The difference between the displacements must be the cargo to load: Cargo mass ass = 7, 329.00 − 2, 571.63 = 4 , 757.37 tonnes The conditions are both based on true mean draughts, so no draught correction is required. In the initial condition, the hydrostatics, corrected for density as appropriate, can be used to determine the initial LCG of the vessel. The SW LCB initially is 47.18 m FOAP. This does not require correction for density. The SW MCTC initially is 43.87 tonne metres. This must be corrected for density, which gives a value of 43.14 tonne metres. Trim =

0=

LCG ) Δ MCTC

(LCB

(47.18 − LCG )2, 571.63 43.14

∴ LCG = 47.18 m FOAP

In the final condition, the vessel is on an even keel, therefore the LCB must equal the LCG. No correction for density is required for the LCB. In the final condition, the LCB and hence LCG is 45.40 m FOAP.

9781408176122_Ch14_3_Rev_txt_prf.indd 498

11/16/2013 2:36:11 AM

Solutions to Questions • 499 Assuming that x tonnes of cargo is loaded in the aft hold, 4,757.37 – x tonnes of cargo must be loaded in the forward hold. A loading table can be used to determine the distribution of the cargo. Item

Mass (tonnes)

Ship

2,571.63

LCG (m FOAP) Moment (tonne metres) 47.18

121,329.5

Aft cargo

x

25

25x

Fwd cargo

4,757.37 – x

75

356,802.75 – 75x

Total

7,329.00

478,132.25 – 50x

LCG =

45.40 =

Moment Mass

478 ,132.25 − 50 x ∴ = 2, 907.91 tonnes 7, 329.00

Therefore, 2,907.91 tonnes must be loaded in the aft hold and in the forward hold: 4 , 757.37 − 2, 907.91 = 1, 849.46 tonnes Therefore, 1,849.46 tonnes must be loaded in the forward hold. Q5.23 The mean draught can be found: DM =

DA

DF 2

=

4.00 + 3.15 = 3.58 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 46.63 m FOAP. This can then be used to determine the approximate true mean draught. Note that at a given draught the LCF value is not affected by water density, therefore no density corrections are required to the hydrostatics. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.63 ⎞ 100 ⎠

DLCF = 3.60 60 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 46.61 m FOAP. This can be used to calculate a more accurate value for the true mean draught. Again, at a given draught the LCF value is not affected by water density, therefore no density corrections are required to the hydrostatics.

9781408176122_Ch14_3_Rev_txt_prf.indd 499

11/16/2013 2:36:12 AM

500 • Ship Stability, Powering and Resistance

DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.61⎞ 100 ⎠

DLCF = 3.60 60 m

Therefore, the initial true mean draught can be taken as 3.60 m. This can be used to determine the hydrostatics of the vessel. The hydrostatics at a true mean draught of 3.60 m gives the following values for sea water: Draught (m)

Displacement LCB (m LCF (m KB (m) KM (m) TPC (t/cm) (tonnes) FOAP) FOAP)

3.60

3,241

47.13

46.61

1.95

7.34

10.62

MCTC (tonne metres) 45.84

These must be corrected to dock water values, by multiplying the hydrostatic value by the ratio of dock water density to sea water density: Corrected hydrostatic vvalue = Sea w water hydrostatic vvalue ×

ρDW DW ρSSW

Remember – only the hydrostatics which have tonnes in the units need correcting. Therefore, the corrected dock water values at 1.015 t/m3 are: Draught (m)

Displacement LCB (m LCF (m KB (m) KM (m) TPC (t/cm) (tonnes) FOAP) FOAP)

3.60

3,209.4

47.13

46.61

1.95

7.34

10.51

MCTC (tonne metres) 45.39

These can be used to determine the LCG in the initial condition: Trim =

LCG ) Δ (47.13 − LCG ) × 3, 209.4 ∴ 85 = ∴ LCG = 45.93 93 m FOAP MCTC 45.39

(LCB

The displacement of the vessel at a draught of 6.00 m in sea water can be determined from the hydrostatics. At 6.00 m in sea water, the displacement would be 6,008.00 tonnes. As the sea is salt water, no corrections are needed. The difference between the displacements must be the cargo to load: Cargo mass = 6 , 008.00 − 3, 209.38 = 2, 798.62 tonnes In the final condition, the vessel is on an even keel, therefore the LCB must equal the LCG. No correction for density is required for the LCB. In the final condition, the LCB and hence LCG is 46.15 m FOAP.

9781408176122_Ch14_3_Rev_txt_prf.indd 500

11/16/2013 2:36:14 AM

Solutions to Questions • 501 Assuming that x tonnes of cargo is loaded in the aft hold, 2,798.62 – x tonnes of cargo must be loaded in the forward hold. A loading table can be used to determine the distribution of the cargo. Item

Mass (tonnes)

Ship

3,209.38

45.93

147,406.8234

Aft cargo

X

30

30x

Fwd cargo

2,798.62 – x

50

139,931 – 50x

Total

LCG (m FOAP) Moment (tonne metres)

6,008.00

287,337.8234 – 20x

LCG = 46.15 =

Moment Mass

287, 337.8 , 234 − 20 x ∴ = 503.43 tonnes 6 , 008.00

Therefore, 503.43 tonnes must be loaded in the aft hold and in the forward hold: 2, 798.62 − 503.43 = 2, 295.19 tonnes Therefore, 2,295.19 tonnes must be loaded in the forward hold. Q5.24 The layer correction process can be used to determine the displacement of the vessel in the initial condition: DM =

DA

DF 2

=

4+3 = 3.50 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 46.71 m FOAP. Note that the LCF is independent of density, and therefore no density correction is needed. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.71⎞ 100 ⎠

DLCF = 3.53 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 46.68 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

9781408176122_Ch14_3_Rev_txt_prf.indd 501

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.68 ⎞ 100 ⎠

DLCF = 3.53 m

11/16/2013 2:36:15 AM

502 • Ship Stability, Powering and Resistance Therefore, the true mean draught can be taken as 3.53 m. The hydrostatic values for draughts of 3.50 and 3.60 m can be corrected for fresh water, and interpolated for 3.53 m to give the displacement, LCB and MCTC in the initial condition, in fresh water: Sea water values: Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm)

MCTC (tonne metres)

3.60

3,241

47.125

46.605

1.947

7.335

10.619

45.838

3.50

3,136

47.141

46.713

1.893

7.399

10.557

45.498

Fresh water values: Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm)

MCTC (tonne metres)

3.60

3,161.95

47.125

46.605

1.947

7.335

10.360

44.720

3.50

3,059.51

47.141

46.713

1.893

7.399

10.300

44.388

At 3.53 m in fresh water: Δ = 3, 090.24 LCB = 47.14 .14 m FOAP O MCTC T = 44.49 tonne metres As an alternative method, it is possible to interpolate using the sea water values, and then convert the interpolated values to fresh water values. This will give the same result. These values can be used to determine the initial LCG of the vessel: Trim =

LCG ) Δ (47.14 − LCG ) × 3, 090.24 ∴100 = ∴ LCG = 45.70 m FOAP MCTC 44.49

(LCB

The arrival destination data can be used to determine the arrival condition. As the water depth is 7 m, with 1 m UKC required, the maximum possible draught is 6.00 m, with a trim of 1.00 m by the stern. Therefore, the arrival condition will be with a draught aft of 6.00 m, and a trim of 1.00 m by the stern, giving a draught forward of 5.00 m. The layer correction process can be used to determine the displacement of the vessel in the final condition: DM =

9781408176122_Ch14_3_Rev_txt_prf.indd 502

DA

DF 2

=

6+5 = 5.50 50 m 2

11/16/2013 2:36:17 AM

Solutions to Questions • 503 This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 44.05 m FOAP. This can be used to determine the approximate true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

44.05 ⎞ 100 ⎠

DLCF = 5.56 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives an interpolated value of 43.94 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 6 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

43.94 ⎞ 100 ⎠

DLCF = 5.56 m

Therefore, the true mean draught can be taken as 5.56 m. The hydrostatic values for draughts of 5.50 and 5.60 m can be corrected for dock water, and interpolated for 5.56 m to give the displacement, LCB and MCTC in the arrival condition, in dock water. Sea water values: Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm)

MCTC (tonne metres)

5.60

5,511

46.393

43.858

3.049

6.948

12.205

62.479

5.50

5,390

46.448

44.049

2.993

6.939

12.104

61.104

Dock water values: Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm)

5.60

5,430.35

46.393

43.858

3.049

6.948

5.50

5,311.12

46.448

44.049

2.993

6.939

12.03

MCTC (tonne metres) 61.57 60.21

At 5.56 m in dock water: Δ = 5, 382.66 LCB = 46 6.41 .41 m FOAP O P MCTC T = 61.03 tonne metres Again, as an alternative method, it is possible to interpolate using the sea water values, and then convert the interpolated values to dock water values. This will give the same result.

9781408176122_Ch14_3_Rev_txt_prf.indd 503

11/16/2013 2:36:18 AM

504 • Ship Stability, Powering and Resistance These values can be used to determine the target LCG of the vessel: Trim =

LCG ) Δ (46.41 − LCG ) × 5, 382.66 ∴100 = ∴ LCG = 45.28 8 m FOAP MCTC 61.03

(LCB

On arrival, the displacement will be 5,382.66 tonnes. The difference between the initial condition and the arrival condition gives the amount of cargo to load: Δ Final

Δ Start = 5, 382.66 − 3, 090.24 = 2, 292.42 tonnes

However, on passage, the ship will burn 59.7 tonnes from each bunker, and so the vessel can load additional cargo to compensate for the reduction in draught due to fuel consumption. Therefore, the final cargo to load can be found from: 2, 292.42 + 59.7 .7 59.7 = 2, 411.82 tonnes A loading table can be used to determine the cargo distribution, assuming that x tonnes are loaded in the aft hold, and 2,411.82 − x tonnes in the forward hold. Note that the fuel burn must also be accounted for. The simplest way to undertake this is to remove all of the fuel, and replace it with the fuel at the arrival condition. The tank data for the bunkers shows that when full, the bunkers have a capacity of 122.67 m3 (corresponding to 117.76 tonnes of fuel at a RD of 0.96), with an LCG of 13.67 m FOAP. Therefore, at the arrival port, each bunker will contain 58.06 tonnes of fuel, which corresponds to a fuel volume in each bunker of 60.48 m3. This corresponds to a sounding of 5.00 m, at which point the LCG of the tank will be 14.65 m FOAP. Therefore: Item

Mass (tonnes)

LCG (m FOAP)

Moment (tonne metres)

3,090.24

45.70

141,223.97

−117.76 × 2

13.67

−3,219.56

58.06 × 2

14.65

1,701.16

Initial condition Fuel out Fuel in Aft hold cargo

x

35

35x

Fwd hold cargo

2,411.82 − x

75

180,886.5 − 75x

Total

5,382.66

320,593.07 − 40x

The target LCG is 45.29 m FOAP, therefore: LCG =

Moment 320 , 592.07 − 40 x ∴ 45.28 = Mass 5, 382.66

∴ x = 1,921.63 tonnes in No. 4 and 2,411.82 − 1,921.63 = 490.19 tonnes in No. 2

9781408176122_Ch14_3_Rev_txt_prf.indd 504

11/16/2013 2:36:19 AM

Solutions to Questions • 505 Q5.25 At 3.085 m draught, the displacement can be interpolated. This gives a displacement of 2,702.55 tonnes. For each step in the inclining process, the net number of masses to port and starboard can be found, and hence the overall listing moment determined. This can be used with the pendulum length and deflection to find GM. For example, for step 2, four masses to port and two to starboard gives a net two masses to port. GM =

Σ(w d ) 2 × ((10 × 2.5) = = 0.740 m ⎛ Pendulum deflection ⎞ ⎛ 0 15 ⎞ 2 , 702 . 55 × Δ×⎜ ⎝ 6 00 ⎠ ⎝ Pendulum length ⎟⎠

Repeating this process for the other steps gives the following: Step

Port masses

Stb masses

Net masses

GM (m)

1

3

3

None

–

2

4

2

2 Port

0.740

3

5

1

4 Port

0.743

4

6

0

6 Port

0.734

5

5

1

4 Port

0.723

6

4

2

2 Port

0.760

7

3

3

None

–

8

2

4

2 Stb

0.750

9

1

5

4 Stb

0.733

10

0

6

6 Stb

0.743

11

1

5

4 Stb

0.723

12

2

4

2 Stb

0.750

13

3

3

None

–

MEAN

0.740

At 2,702.55 tonnes, the KM is 7.746 m. Therefore, KG can be found based on the mean GM: = KM − KG ∴ 0.740 = 7.746 − KG ∴ KG = 7.006 m

9781408176122_Ch14_3_Rev_txt_prf.indd 505

11/16/2013 2:36:20 AM

506 • Ship Stability, Powering and Resistance A loading table can be used to unload the inclining masses to obtain the lightship values: Item

Mass

KG

Moment

Ship

2,702.55

7.006

18,934.07

–60

8.000

–480.00

Inclining masses Total

2,642.55

KG =

18,454.07

Moment 18 , 454.07 = = 6.983 m 2, 642.55 Mass

Therefore, the lightship displacement is measured at 2,642.55 tonnes, and the lightship KG is measured at 6.983 m. Q5.26 The dimensions of the vessel can be used to determine the displacement: ∇ × ρ = Δ ∴ 50 × 6 × 2.033 × 1.025 = 625.15 15 tonnes KB and BM and be found to determine KM: 50 × 63 D 2.033 I 12 KB = = = 1.017 m BM = = = 1.476 m 2 2 ∇ 50 × 6 × 2.033 The GM can be found from the inclining data: GM =

Σ(w d ) ( × ) = = 0.365 m ⎛ Pendulum deflection ⎞ ⎛ 0.219 ⎞ 6 625 . 15 × Δ×⎜ ⎝ 2 50 ⎠ ⎝ Pendulum length ⎟⎠

The KG can be found: GM = KB + BM − KG 0.365 = 1.017 + 1.476 − KG ∴ KG = 2.128 m A loading table can be used to remove the inclining mass to determine the lightship condition: Item

Mass

KG

Moment

Ship

625.15

2.218

1,386.58

Inclining masses

–10

4.00

Total

615.15

9781408176122_Ch14_3_Rev_txt_prf.indd 506

–40.00 1,346.58

11/16/2013 2:36:21 AM

Solutions to Questions • 507

KG =

Moment 1, 346.58 = = 2.189 m Mass 615.15

Therefore, the lightship displacement is measured at 615.15 tonnes, and the lightship KG is measured at 2.189 m. Q5.27 At 3.029 m draught, the displacement can be interpolated. This gives a displacement of 2,644.87 tonnes. For each step in the inclining process, the net number of masses to port and starboard can be found, and hence the overall listing moment determined. This can be used with the pendulum length and deflection to find GM. For example, for step 3, five masses to port and one to starboard gives a net four masses to port. GM =

Σ(w d ) 4 × ((5 × 3)) = = 0.727 m ⎛ Pendulum deflection ⎞ ⎛ 0.234 ⎞ 2 , 644 . 87 × Δ×⎜ ⎝ 7.50 ⎠ ⎝ Pendulum length ⎟⎠

Repeating this process for the other steps gives the following: Step

Port masses

Stb masses

Net masses

GM (m)

1

3

3

None

–

2

4

2

2 Port

0.709

3

5

1

4 Port

0.727

4

6

0

6 Port

0.762

5

5

1

4 Port

0.709

6

4

2

2 Port

0.740

7

3

3

None

–

8

2

4

2 Stb

0.760

9

1

5

4 Stb

0.791

10

0

6

6 Stb

0.638

11

1

5

4 Stb

0.740

12

2

4

2 Stb

0.692

13

3

3

None

–

MEAN

0.727

9781408176122_Ch14_3_Rev_txt_prf.indd 507

11/16/2013 2:36:22 AM

508 • Ship Stability, Powering and Resistance At 2,644.87 tonnes, the KM is 7.804 m. Therefore, KG can be found based on the mean GM: GM = KM − KG ∴ 0.727 = 7.804 − KG ∴ KG = 7.077 077 m A loading table can be used to unload the inclining masses to obtain the lightship values: Item

Mass

Ship Inclining masses Total

KG

Moment

2,644.87

7.077

18,717.745

–30

11.000

2,614.87

KG =

–330 18,387.745

Moment 18 , 387.745 = = 7.032 032 m Mass 2, 614.87

Therefore, the lightship displacement is measured at 2,614.87 tonnes, and the lightship KG is measured at 7.032 m.

List angle

3.000

2.000

1.000 Heeling moment

0.000 –120

–90

–60

–30

0

30

60

90

120

–1.000

–2.000

–3.000

–4.000

Q5.28 The mean deflection, mass and distance moved can be used to determine GM: GM =

Σ(w d ) 10.50 × 5.100 = = 0.790 m ⎛ Pendulum deflection ⎞ ⎛ 0.150 ⎞ 2 , 710 . 00 × Δ×⎜ ⎝ 6.000 ⎠ ⎝ Pendulum length ⎟⎠

9781408176122_Ch14_3_Rev_txt_prf.indd 508

11/16/2013 2:36:23 AM

Solutions to Questions • 509 At 2,710.00 tonnes, the KM can be found from the hydrostatics. This gives a value of 7.738 m. This allows KG to be found during the inclining test: GM = KM − KG ∴ 0.790 = 7.738 − KG ∴ KG = 6.948 m A loading table can now be used to convert the vessel to the lightship condition: Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Ship

2,710.00

6.948

18,829.08

−10.500 × 6

8.100

−510.30

−0.50

7.900

−3.95

Inclining masses Staff and crew Total

2,646.50

KG =

18,314.83

Moment 18 , 314.83 = = 6.920 m Mass 2, 646.50

Therefore, the lightship displacement is 2,646.50 tonnes, and the lightship KG is 6.920 m. Q5.29 The mean deflection, mass and distance moved can be used to determine GM: GM =

Σ(w d ) 9.50 × 5.000 = = 0.722 m ⎛ Pendulum deflection ⎞ ⎛ 0.195 ⎞ 2 , 700 . 00 × Δ×⎜ ⎝ 8.000 ⎠ ⎝ Pendulum length ⎟⎠

At 2,700.00 tonnes, the KM can be found from the hydrostatics. This gives a value of 7.748 m. This allows KG to be found during the inclining test: GM = KM − KG ∴ 0.722 = 7.748 − KG ∴ KG = 7.026 m A loading table can now be used to convert the vessel to the lightship condition: Item

Mass (tonnes)

KG (m)

Moment (tonne metres)

Ship

2,700.00

7.026

18,970.20

−9.500 × 6

8.500

−484.50

−0.75

8.000

−6.00

Inclining masses Staff and crew Total

2,642.25

KG =

18,479.70

Moment 18 , 479.70 = = 6.994 994 m Mass 2, 642.25

Therefore, the lightship displacement is 2,642.25 tonnes, and the lightship KG is 6.994 m.

9781408176122_Ch14_3_Rev_txt_prf.indd 509

11/16/2013 2:36:24 AM

510 • Ship Stability, Powering and Resistance Q5.30 The mean draught can be found: DM =

DA

DF 2

=

4.00 + 2.94 = 3.47 47 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 46.74 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.74 ⎞ 100 ⎠

DLCF = 3.50 50 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 46.71 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.71⎞ 100 ⎠

DLCF = 3.50 50 m

Therefore, the true mean draught can be taken as 3.50 m. The MCTC can be read directly at this draught, giving a value of 45.50 tonne metres. This can be used to determine the up-thrust at the critical moment: Up − thrust =

Change in trim × MCTC 100 × ( 4 − 2.94 ) × 45.50 = = 103 0 .25 25 tonnes LCF 46.71

Q5.31 The mean draught can be found: DM =

DA

DF 2

=

4.00 + 3.15 = 3.58 m 2

This can be used to determine the approximate LCF position from the hydrostatics. This gives a value of 46.63 m FOAP. This can then be used to determine the approximate true mean draught. DA

DLCF + Trim

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.63 ⎞ 100 ⎠

DLCF = 3.60 m

This can be used to determine a more accurate LCF position from the hydrostatics. This gives a value of 46.61 m FOAP. This can be used to calculate a more accurate value for the true mean draught: DA

DLCF + Trim

9781408176122_Ch14_3_Rev_txt_prf.indd 510

⎛ LCF ⎞ ∴ 4 = DLCF ⎝ LBP ⎠

(

) ⎛⎝

46.61⎞ 100 ⎠

DLCF = 3.60 60 m

11/16/2013 2:36:25 AM

Solutions to Questions • 511 Therefore, the true mean draught can be taken as 3.60 m. The hydrostatics can be read directly at this draught, giving an MCTC value of 45.84 tonne metres, with a KM of 7.34 m and a displacement of 3,241 tonnes. The KM can be used to find the initial GM: GM = KM − KG = 7.34 − 7.19 = 0.15 15 m The hydrostatics can be used to determine the up-thrust at the critical moment: Up − thrust =

Change in trim × MCTC 100 × ( 4 − 3.15) × 45.84 = = 83.6 tonnes 46.61 LCF

A loading table can be used to determine the effect on KG: Item

Mass

KG

Moment

Ship

3,241

7.19

23,302.79

Up-thrust

–83.6

Total

0

0

3,157.4

KG =

23,302.79

Moment 23, 302.79 = = 7.38 38 m Mass 3,157.4

This can be used to determine GM at the critical moment: GM = KM − KG = 7.34 − 7.38 = −0.04 04 m Q5.32 At 4,112 tonnes, the MCTC is 50.27 tonne metres/cm. The LCF is 45.71 m FOAP. This allows the up-thrust to be found: Up − thrust =

Change in trim × MCTC 100 × 50.27 = = 110 tonnes LCF 45.71

The KM at 4,112 tonnes is 7.01 m. Therefore, to keep GM positive, KG at the critical moment cannot be above 7.01. Using this as the limiting value, a loading table can be used to determine the limiting initial ship KG:

9781408176122_Ch14_3_Rev_txt_prf.indd 511

Item

Mass

KG

Moment

Ship

4,112

x

4,112x

Up-thrust

–110

0

Total

4,002

0 4,112x

11/16/2013 2:36:27 AM

512 • Ship Stability, Powering and Resistance

KG =

Moment 4 ,112 x = = 7.01 01 m Mass 4 , 002

4 ,112 x = 7.01 4 , 002

6.82 m

Q5.33 The KM at 4,112 tonnes is 7.01 m. Therefore, to keep GM positive, KG at the critical moment cannot be above 7.01. Using this as the limiting value, a loading table can be used to determine the limiting up-thrust: Item

Mass

KG

Moment

Ship

4,112

6.85

28,167.2

Up-thrust Total

KG =

–x

0

4,112 – x

0 28,167.2

Moment 28 ,167.2 = = 7 01 m Mass 4 ,112 − x

28 ,167.2 = 7.01 4 ,112 − x

93.86 tonnes

At 4,112 tonnes, the MCTC is 50.27 tonne metres/cm. The LCF is 45.71 m FOAP. This allows the change in trim to cause the limiting up-thrust to be found: Up − thrust = 93.86 =

Change in trim × MCTC LCF

Change in trim × 50.27 ∴ Change in trim = 85 cm∴ Initial a trim = 0 85 m 45.71

Q5.34 The density corrected hydrostatics can be determined from the sea water hydrostatics in the Stability Data Book (see Appendix 1). Correcting for the dock water density gives: Sea water values: Draught (m)

3.60

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm) 3,241

9781408176122_Ch14_3_Rev_txt_prf.indd 512

47.13

46.61

1.95

7.34

10.62

MCTC (tonne metres) 45.84

11/16/2013 2:36:29 AM

Solutions to Questions • 513 Dock water values: Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm)

3.60

3,184.09

47.13

46.61

1.95

7.34

10.43

MCTC (tonne metres) 45.04

The up-thrust can be determined: Up − thrust =

Change in trim × MCTC 100 × 45.04 = = 96.63 tonn t es e LCF 46.61

A loading table can be used to determine the effect of the up-thrust on the KG of the vessel: Item

Mass (t)

Ship

3,184.1

Up-thrust Total

KG (m)

Moment (tonne metres)

7.18

–96.63

0

0

3,087.47

KG =

22,861.838

22,861.838

Moment 22, 861.84 = = 7.41 41 m Mass 3, 087.47

As KM is assumed to remain constant, GM can be found: GM = KM − KG = 7.34 − 7.41 = −0.07 m The GM is negative, and so is therefore not safe to dry-dock. In order for GM to remain above 5 cm, the KG must not exceed a certain value: GM = KM − KG 0.05 7.34 − KG KG = 7.29 m maximum Item

Mass (tonnes)

Ship

3,184.1

Up-thrust Total

7.18

−x

0

3,184.1 − x

KG =

9781408176122_Ch14_3_Rev_txt_prf.indd 513

KG (m) Moment (tonne metres) 22,861.838 0 22,861.838

Moment Mass

11/16/2013 2:36:30 AM

514 • Ship Stability, Powering and Resistance

7.29 =

22, 861.34 3,184.1 − x

x = 48.1 tonnes The change in trim that would result in an up-thrust of 48.1 tonnes can be found: Up − thrust = 48.1 =

Change in trim × MCTC LCF

Change in trim × 45.04 46.61

Change in trim = 49.8 cm Therefore, the initial trim should not exceed 50 cm by the stern. Q5.35 The hydrostatics can be determined from the hydrostatics in the Stability Data Book (see Appendix 1). Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (t/cm)

3.50

3,136

Up − thrust =

47.14

46.71

1.89

7.40

10.56

MCTC (tonne metres) 45.50

Change in trim × MCTC 110 × 45.50 = = 107.15 ton t n nes LCF 46.71

A loading table can be used to determine the effect of the up-thrust on the KG of the vessel: Item

Mass (tonnes)

KG (m) Moment (tonne metres)

Ship

3,136

7.25

UP-thrust

−107.15

0

Total

3,028.85

KG =

22,736 0 22,736

Moment 22, 736 = = 7.51 51 m Mass 3, 028.85

As KM is NOT assumed to remain constant, it must be interpolated at the effective displacement of 3,028.85 tonnes. This gives a KM of 7.47 m. This allows GM to be found: GM = KM − KG = 7.47 − 7.51 = −0.04 m The GM is negative, and so is therefore not safe to dry-dock.

9781408176122_Ch14_3_Rev_txt_prf.indd 514

11/16/2013 2:36:31 AM

Solutions to Questions • 515 If we had assumed KM to remain constant, we would have a KM of 7.40 m. This would give a GM of: GM = KM − KG = 7.40 − 7.51 = −0.11 m It can be seen that the assumption gives a ‘worst-case’ scenario. Q5.36 At the point when the bow grounds, the effective displacement can be found: EffectiveΔ = 3, 241 − 83.6 = 3,157.4 tonnes Interpolating in the hydrostatics for the above displacement gives a true mean draught of 3.52 m. This will be the true mean draught when the vessel grounds. Q5.37 At the displacement of 3,563 tonnes, the hydrostatics can be found from the data book. This gives the following values: KM = 7.18 m MCTC = 47.23 tonne metres LCB = 47.06 m FOAP These can be used to determine the trim of the vessel when re-floating: Trim =

LCG ) Δ ( 47.06 − 44.60 ) 3, 563 = = 186 cm MCTC 47.23

(LCB

This can be used to determine the up-thrust acting on the vessel during re-floating: P=

Change of trim × MCTC 186 × 47.23 = = 189.76 tonnes LCF 46.29

A loading table can be used to determine KG at the critical instant: Item

Mass (tonnes)

KG (m)

Ship

3,563

6.90

Up-thrust

–189.76

0

Total

3,373.24

KG =

Moment (tonne metres) 24,584.7 0 24,584.7

Moment 24 , 584.7 = = 7.29 m Mass 3, 373.24

This can be used to determine GM at the critical instant: GM = KM − KG = 7.18 − 7.29 = −0.11 11 m

9781408176122_Ch14_3_Rev_txt_prf.indd 515

11/16/2013 2:36:33 AM

516 • Ship Stability, Powering and Resistance Therefore, as GM at the critical instant is negative, it would be unsafe to re-float the vessel. Q5.38 The relevant hydrostatics can be rewritten at the correct density: Draught (m)

Displacement LCB (tonnes) (m FOAP)

3.30

2,897.46

LCF (m FOAP)

KM (m)

MCTC (tonne metres)

46.90

7.55

44.39

47.16

The final trim can be determined: Trim =

LCG ) Δ ( 47.16 − 46.00 ) 2, 897.46 = = 76 cm MCTC 44.39

(LCB

The up-thrust can be found: P=

CoT

MCTC T 76 × 44.39 = = 71.93 93 tonnes LCF 46.90

A loading table can be used to determine the KG at the critical moment: Item

Mass (tonnes)

Ship

2,897.46

P Total

–71.93

KG (m)

Moment (tonne metres)

7.05 0

0

2,825.53

KG =

20,427.09

20,427.09

Moment 20 , 427.09 = = 7.23 23 m Mass 2, 825.53

This allows GM to be found at the critical instant: GM = KM − KG = 7.55 − 7.23 = 0.32 32 m At the critical moment, when the bow lifts, the effective displacement will be 2,825.53 tonnes. The hydrostatics can be corrected for dock water: Draught (m) Displacement (tonnes)

9781408176122_Ch14_3_Rev_txt_prf.indd 516

3.30

2,897.45

3.20

2,794.47

11/16/2013 2:36:35 AM

Solutions to Questions • 517 The draught can be interpolated for the displacement of 2,825.53 tonnes. This gives a true mean draught of 3.23 m. As the trim is zero at this point, the draught aft and the draught forward would also be 3.23 m. Q5.39 Draught survey 1

Draught forward

2

FP correction

3

Draught at FP

4

Draught aft

5

AP correction

6

m

Dist .marks k displaced × Observed trim Dist .between marks k

6.000

6.000

−3 × (7.000 − 6.000 ) 100 − 2 5 − 3

−0.032

6.000 + −0.032

5.968

7.000

7.000

25 × ((7 7.000 − 6.000 ) 100 − 2 5 − 3

0.026

Draught at AP

7.000 + 0.026

7.026

7

True trim

7.026 − 5.968

1.058

8

Draught (M) port

6.530

6.530

9

Draught (M) Stb

6.514

6.514

6.530 + 6.514 2

6.522

−2 × 1.058 100

−0.021

6.522+ −0.021

6.501

5.968 968 + (6 6.501) + 7.026 8

6.500

13.214

LCF FOAP

41.953

16 1st trim corrn (layer)

Dist .CF CF from midships × Trim × TPC LBP

953) × 105.8 13.214 (50 − 41.953 100

112.50

17 2nd trim corrn (form)

50 × True trim2 × ( MCTC 2 − MCTC1 ) LBP

50 × 1.0582 × (86.961 − 68.639 ) 100

10.254

6,654.00+112.50+10.254

6,776.754

Dist .marks k displaced × Observed trim Dist .between marks k

10 Draught midships mean 11 Amidship line correction

Dist .marks marks k displaced × True trim LBP

12 Draught at amidships 13 Corrected midship draught

dFP

14 TPC

(

d M ) + d AP 8

15 Displacement

6,654

18 Corrected displacement 19 Dock water displacement

9781408176122_Ch14_3_Rev_txt_prf.indd 517

Δ×

R.D.Dock water 1.025

6776.754 ×

1.010 1.025

6,677.58

11/16/2013 2:36:36 AM

518 • Ship Stability, Powering and Resistance Q5.40 Draught survey 1

Draught forward

2

FP correction

3

Draught at FP

4

Draught aft

5

AP correction

6

m 4.000 Dist .marks k displaced × Observed trim Dist .between marks k

2.000 × (5.000 − 4.000 ) 100 − 2.000 − 3.000

0.021

4.000−0.021

3.979 5.000

3.000 × (5.000 − 4.000 ) 100 − 2.000 − 3.000

0.032

Draught at AP

5.000+0.032

5.032

7

True trim

5.032−3.979

1.053

8

Draught (M) port

4.500

9

Draught (M) Stb

4.486

Dist .marks k displaced × Observed trim Dist .between marks k

10 Draught midships mean 11 Amidship line correction

Dist .marks marks k displaced × True trim LBP

12 Draught at amidships 13 Corrected midship draught

dFP

14 TPC

(

d M ) + d AP 8

11.247(1)

4.500 + 4.486 2

4.493

2.000 × 1.053 100

–0.011

4.493 − 0.011

4.504

3.979 979 + (6 4.504 ) + 5.032 8

4.504

LCF FOAP

45.574

15 Displacement

4,228.52

16 1st trim corrn (layer)

Dist .CF CF from midships × Trim × TPC LBP

574 ) × 105.3 (50 − 45.574

11.247

52.420

17 2nd trim corrn (form)

50 × True trim2 × ( MCTC 2 − MCTC1 ) LBP

50 × 1.0532 × (55.345 − 47.727) 100

4.220

4,228.52 + 52.420 + 4.220

4,285.160

18 Corrected displacement 19 Dock water displacement

9781408176122_Ch14_3_Rev_txt_prf.indd 518

Δ×

R.D.Dock water 1.025

100

4 , 285.160 ×

1.006 1.025

4,205.730

11/16/2013 2:36:39 AM

Solutions to Questions • 519 Q5.41 Draught survey 1

Draught Forward

2

FP correction

3

Draught at FP

4

Draught aft

5

AP correction

6

m

Dist .marks k displaced × Observed trim Dist .between marks k

6.000

6.000

−2 × (5.000 − 6.000 ) 100 − 2 + 1.5

0.020

6.000 + 0.020

6.020

5.000

5.000

−1 5 × (5.000 − 6.000 ) 99.5

0.015

Draught at AP

5.000 + 0.015

5.015

Dist .marks k displaced × Observed trim Dist .between marks k

7

True trim

5.015 − 6.020

−1.005

8

Draught (M) port

5.517

5.517

9

Draught (M) Stb

5.500

5.500

5.517 + 5.500 2

5.509

15 × −1.005 100

−0.015

5.509 + −0.015

5.494

6.020 020 + (6 5.494 ) + 5.015 8

5.500

12.104

LCF FOAP

44.049

16 1st trim corrn (layer)

Dist .CF CF from midships × Trim × TPC LBP

+ (50 − 44.049 ) × −100.5 × 12.104 100

−72.39

17 2nd trim corrn (form)

50 × True trim2 × ( MCTC 2 − MCTC1 ) LBP

50 × ( −1.005)2 × (68.639 − 55.305) 100

6.73

5,390.00 − 72.39 + 6.73

5,324.34

10 Draught midships mean 11 Amidship line correction

Dist .marks k displaced × True trim LBP

12 Draught at amidships 13 Corrected midship draught

dFP

14 TPC

(

d M ) + d AP 8

15 Displacement

5,390

18 Corrected displacement 19 Dock water displacement

9781408176122_Ch14_3_Rev_txt_prf.indd 519

Δ×

R.D.Dock w water 1.025

5324.34 ×

1.010 1.025

5,246.43

11/16/2013 2:36:42 AM

520 • Ship Stability, Powering and Resistance Q6.1 20 m

2m

100 m 10 m

For the vessel shown above, determine the final draught of the vessel if the 20 m long amidships compartment is bilged, and the overall underwater ‘dry’ volume and displacement before and after bilging. Before bilging, the underwater volume can be found from the dimensions of the vessel: ∇ = L × B × D = 100 × 10 × 2 = 2, 000 m3 This can be used to find the initial displacement: Δ = ∇ × ρ = 2, 000 × 1.025 = 2, 050 tonnes The lost volume can be found from the original dimensions of the bilged compartment: Lost volume = Compartment length t Compartment beam × Compartment depth Lost volume o u e = 20 × 10 × 2 = 400 m3 The final waterplane area can be found from the original dimensions of the vessel and the bilged compartment: Final w water aterp pla anee area Initial waterplane ate p a e area − Bilged waterrpl p ane area Final w water aterpl rplane area = (100 × 10 ) − (20 × 10 ) = 800 m2

9781408176122_Ch14_3_Rev_txt_prf.indd 520

11/16/2013 2:36:46 AM

Solutions to Questions • 521 Therefore, the parallel sinkage can be found: Parallel sinkage =

Lost volume 400 = = 0..50 m Final w water aterpl rplane area 800

The initial draught was 2.00 m, and so the bilged draught will be 2.50 m. After bilging, the underwater volume can be found from the dimensions of the two remaining dry compartments: ∇ = (L − l ) × B × DBilged = (

−

) × 10 × 2.5 = 2, 000 m3

This can be used to find the final displacement: Δ = ∇ × ρ = 2, 000 × 1.025 = 2, 050 tonnes Q6.2

15 m

1m

1.1 m 100 m 10 m

For the vessel shown above, determine the final draught of the vessel if the amidships compartment is bilged below the watertight flat. The lost volume can be found from the dimensions of the bilged compartment to the original waterline: Lost volume

l b × D 15 × 10 0 × 1 150 m3

The volume gained as the vessel sinks lower in the water can be found. This needs to be determined in two steps, first between the original waterline and the watertight flat, and then from the watertight flat to the final waterline.

9781408176122_Ch14_3_Rev_txt_prf.indd 521

11/16/2013 2:36:47 AM

522 • Ship Stability, Powering and Resistance From the original waterline to the watertight flat, the gained volume will be in the two end compartments: Gained volume o u e to tthee WT ffla att = 2 × 42.5 × 10 × (1.1 − 1 0 ) = 85 m3 Therefore, when the vessel has sunk so that the waterline is on the watertight flat, a total of 85 m3 of volume has been regained. Therefore, from the original 150 m3 lost, another 65 m3 needs to be gained through sinkage until the vessel is back in equilibrium. Therefore, the further sinkage above the WT flat can be found: Sinkage S age = 65 m3

Gained volume o u e past tthee W WT fla att ∴ Sinkage =

65 = 0.065 m 100 × 10

Therefore, the sinkage to the watertight flat is 0.10 m, and the sinkage above that is 0.065 m, giving a total sinkage of 0.165 m. This gives a total draught after bilging of 1.165 m, or 1.17 m to the nearest centimetre. Q6.3 Parallel sinkage = 2.857 − 2 =

Lost volume Final w water aterp pla anee area

l × 10 × 2 20l ∴ 0.857 = 500 − 10l (50 × 10 ) − (l 10)

0.857 (500 − 10l ) = 20l ∴ 428.5 8.57l = 20l ∴428 428.5 20l 8.57l 428.5 28.57l ∴ l =

428.5 = 15.00 m 28.57

Q6.4 Parallel sinkage =

Lost volume 5 × 6 × 1× 0.95 9 = = 0.11 11 m Final w water aterpl rplane area (50 × 6 ) (5 × 6 0.95) ∴ Bilged d drau raug a ghtt = 1.11 m

Q6.5 Parallel sinkage =

Lost volume 8 × 10 × 3 × μ = Final w water aterp pla anee area (70 7 × 10 ) − (8 10 × μ )

3.34 3.00 =

9781408176122_Ch14_3_Rev_txt_prf.indd 522

8 10 3 × μ 240 × μ = (8 10 μ ) (700 ) − (80 (80 μ ) (70 × 10 ) − (8

11/16/2013 2:36:48 AM

Solutions to Questions • 523

0.34 =

240 μ ∴0 0.34 (700 − 80 μ ) = 240 μ∴ 238 − 27.2 μ = 240 μ 700 − 80 μ 238 − 27.2 2 μ = 240 μ ∴238 ∴ 238 = 240 μ + 27.2 μ

μ

0.89

Q6.6 Parallel sinkage =

Lost volume 20 × 10 × 2 = = 0.50 50 m Final w water aterpl rplane area (100 0 × 10 ) − (20 × 10 )

Q6.7 Parallel sinkage =

Lost volume 20 × 10 × 1.5 5 = = 0.30 30 m Final w water aterpl rplane area (100 × 10 ) − (0 )

Parallel sinkage =

Lost volume 15 × 12 × 1× 0 95 = = 0 18 m Final w water aterp pla anee area (80 × 12) − (0)

Q6.8

Q6.9 Parallel sinkage =

Lost volume 20 × 10 × (2 − 1.5) = = 0.13 m Final w water aterpl rplane area (100 × 10 ) − (20 × 10 )

Q6.10 Parallel sinkage =

25 × 8 × (5 − 3) × 0 77 Lost volume = = 0.76 m Final w water aterp pla anee area (70 × 8 ) − (25 × 8 × 0.77)

Q6.11 Item

Ship at the bilged draught

Underwater volume (m3)

Centre from keel (m)

Vertical moment of volume (m4)

80 × 3.379 × 9 = 2,432.88

3.379 = 1.690 2

4,111.57

−(22 × 9 × 2) = −396

2 = 1 00 1

−396

Bilged compartment

Total

9781408176122_Ch14_3_Rev_txt_prf.indd 523

2,036.88

3,715.57

11/16/2013 2:36:51 AM

524 • Ship Stability, Powering and Resistance Moment 3, 715.57 = = 1.82 m Mass 2, 036.88

KB =

Q6.12 Item

Underwater volume (m3)

Centre from keel (m)

Vertical moment of volume (m4)

Ship at the bilged draught

80 × 3.379 × 9 = 2,432.88

3.379 = 1.690 2

4,111.57

Bilged compartment

−(

× ×(

−

Total

)×

) = −191.13 ((3.379 − 2)

0.5) 2 = 2.690

−514.14

2,241.75

3,597.43

Moment 3, 597.43 = = 1.60 60 m Mass 2, 241.75

KB = Q6.13 Item

Underwater volume (m3)

Centre from AP (m)

Longitudinal moment of volume (m4)

Ship at the bilged draught

80 × 3.379 × 9 = 2,432.88

80 = 40 2

97,315.20

Bilged compartment

−(

× ×(

Total

−

)×

) = −191.13

80 − (

22 ) = 69.00 2

2,241.75

LCB =

–13,187.97

84,127.23

Moment 84 ,127.23 = = 37.53 m Mass 2, 241.75

Q6.14 Underwater volume (m3)

Centre from centreline (m)

Transverse moment of volume (m4)

Ship at the bilged draught

100 × 4 × 10 = 4,000

0

0

Bilged compartment

−(2 × 8 × 4) = −64

10 2 − =4 2 2

−256

Item

Total

9781408176122_Ch14_3_Rev_txt_prf.indd 524

3,936

−256

11/16/2013 2:36:53 AM

Solutions to Questions • 525

TCB =

Moment −256 = = −0.065 Mass 3, 936

(starboard)

Q6.15 LB 3 lb3 100 × 103 20 × 103 −μ − I 12 = 12 12 BM = = 12 = 3.33 33 m ∇ ∇ 100 × 10 × 2 Q6.16 The parallel axes formula can be used: IREMOTE = ICENTRO Distance 2 ) E ID + ( Area The values can be substituted in: IREMOTE

,8

, 30.8

(

)∴ I

REMOTE

= 17, 315, 678.8 m4

Q6.17 The parallel axes formula can be used: IREMOTE = ICENTRO Distance 2 ) E ID + ( Area The values can be substituted in:

(

33, 333.33 = ICENTRO 000 × 52 E ID + 1,000

)

CENTRO E ID

8 , 333.3 m4

Q6.18 Inertia =

LB 3 60 × 203 = = 160 ,000 000 m4 3 3

Q6.19 Item

Ship at the bilged draught Bilged compartment Total

9781408176122_Ch14_4_Rev_txt_prf.indd 525

Waterplane area (m2)

Centre from centreline (m)

Transverse moment of area (m3)

100 × 10 = 1,000

0

0

−(8 × 2) = −16

4

−64

984

−64

11/16/2013 6:55:21 PM

526 • Ship Stability, Powering and Resistance

Centre of waterplane area after bilging =

Total moment Total waterplane w area

Centre of waterplane area after bilging =

−64 = −0.065 m 984

Therefore, the centre of the waterplane is 0.065 m to starboard of the centreline of the ship. Q6.20 The new roll axis can be found by finding the transverse centre of the waterplane area after bilging: Waterplane area (m2)

Centre from centreline (m)

Transverse moment of area (m3)

Ship at the bilged draught

100 × 10 = 1000

0

0

Bilged compartment

−(20 × 4) = −80

−3

240

Item

Total

920

240

Centre of waterplane area after bilging =

Total moment Total waterplane w area

Centre of waterplane area after bilging =

240 = 0.261 m 920

The new roll axis is therefore 0.261 m to port of the centreline of the vessel. This means the new roll axis must be 5.261 m from the damaged starboard edge. The inertia can be found, measured at the starboard edge: IEDGE =

LB 3 lb3 100 × 103 20 × 4 3 − = − = 32, 906.67 m4 3 3 3 3

The parallel axes theorem can be used to determine the inertia measured at the new roll axis: IROLL AAXIS

IEDGE New rollll a Ne axis to edge dg 2 ) DG − (Waterplane area

IROLL AXIS = 32, 906.67 − A

(

× .

) = 7, 442.79 m

3

This can be used to determine BM: BM =

9781408176122_Ch14_4_Rev_txt_prf.indd 526

7, 442.79 Inertia = = 3.72 m ∇ 100 × 10 × 2

11/16/2013 6:55:23 PM

Solutions to Questions • 527 Q6.21 The longitudinal inertia can be found from: IL =

(L l )3 B (80 − 10 )3 × 15 12

=

= 428 , 750.00 m4

12

This allows the longitudinal BM to be found. Note that the overall underwater volume remains constant, therefore the original length, beam and draught can be used to find the underwater volume: BML =

IL 428 , 750.00 = = 119.10 m ∇ 80 × 15 × 3

Q6.22 As the bilging is contained below a watertight flat, which itself is below the waterline, the waterplane area will not change during bilging. Therefore: IL =

L3 B 903 × 20 = = 1, 215, 000.00 00 m4 12 12

This allows the longitudinal BM to be found. As in the previous examples, the overall underwater volume remains constant during bilging, so the original length, beam and draught can be used: BML =

IL 1, 215, 000 = = 135.00 00 m ∇ 90 × 20 × 5

Q6.23 The table of moments of area can be constructed: Waterplane area (m2)

Centre from AP (m FOAP)

Longitudinal moment of area (m3)

Waterplane area of the ship at the bilged draught

1,200.00

40

48,000.00

Bilged compartment waterplane area

–112.50

45

–5,062.50

Total

1,087.50

Item

LCF a after te bilging =

9781408176122_Ch14_4_Rev_txt_prf.indd 527

42,937.50

Longitudinal moment o e t of area 42, 937.50 = = 39.48 m FOAP Total w waterrplan a e area 1, 087.50

11/16/2013 6:55:25 PM

528 • Ship Stability, Powering and Resistance Q6.24 Using the question data, it can be seen that the waterplane can be split into three sections, the aft, bilged and forward sections. It is easiest to visualise the problem by drawing the sections of waterplane area and the overall LCF: Centre of bilged waterplane Centre of aft waterplane 40.00

30.00

15.00

Centre of forward waterplane 10.00

Bilged LCF

19.48

5.52 25.52

For each of these, the area can be determined: Aft section: Waterplane area = 40 × 15 = 600 m2 Bilged section: Waterplane area a ea = 10 × 15 × (1 − 0.75) = 37.50 m2 Forward section: Waterplane area = 30 × 15 = 450 m2 For each of these sections of waterplane, the longitudinal inertia about the centre of the section can be determined: Aft section: Longitudinal inertia about the centroid =

Bl 3 15 × 403 = = 80,000 0 m4 12 12

Bilged section: 1 Bl 3 (1 − 0.75) × 15 × 10 = = 312.5 m4 12 12 3

Longitudinal inertia about the centroid = Forward section:

Longitudinal inertia about the centroid =

9781408176122_Ch14_4_Rev_txt_prf.indd 528

Bl 3 15 × 303 = = 33,750 7 m4 12 12

11/16/2013 6:55:26 PM

Solutions to Questions • 529 For each of these sections of waterplane, the (remote) longitudinal inertia about the LCF can be determined using the parallel axes formula: Aft section: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

(

InertiaREMOTE = 80 , 000 +

) = 307, 682.24 m

×

4

Bilged section: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

InertiaREMOTE = 312.5 +

(

)

)

14 m ) = 1, 455.14

×

4

Forward section: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

InertiaREMOTE = 33, 750 +

(

×

)

) = 326, 821.68 m

4

The remote inertia values can be summed to give the overall longitudinal inertia measured about the LCF: Longitudinal inertia = 307, 682.24 + 1, 455.14 + 326 , 821.68 = 635, 958 8.06 06 m4 This can be divided by the underwater volume to determine the BML: BML =

IL 635, 958.06 = = 264.98 98 m ∇ 80 × 15 × 2

Q6.25 A table of moments of area of the waterplane can be used to determine the waterplane area and the LCF position after bilging: Item Aft waterplane section Bilged waterplane Forward waterplane section

Area (m2)

Lever (m FOAP)

Area (m3)

600

15

9,000

40

35

1,400

1,200

70

84,000

1,840

LCF =

9781408176122_Ch14_4_Rev_txt_prf.indd 529

94,400

Moment 94 , 400 = = 51.30 m FOAP Area 1, 840

11/16/2013 6:55:27 PM

530 • Ship Stability, Powering and Resistance For each of these sections of waterplane, the longitudinal inertia about the centre of the section can be determined: Aft section: Longitudinal inertia about the centroid =

Bl 3 20 × 303 = = 45,000 0 .00 m4 12 12

Bilged section: 1 Bl 3 (1 − 0.80 ) × 20 × 10 = = 333.33 m4 12 12 3

Longitudinal inertia e t a about tthee centroid = Forward section:

Longitudinal inertia about the centroid =

Bl 3 20 × 603 = = 360,, 000.00 m4 12 12

For each of these sections of waterplane, the (remote) longitudinal inertia about the LCF can be determined using the parallel axes formula: Aft section: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

)

2 ⎛ ⎛ ⎛ 30 ⎞ ⎞ ⎞ InertiaREMOTE = 45, 000 + ⎜ 600.00 × 51.30 − 3 , 614.00 m4 ⎟⎠ ⎟ = 835 ⎝ ⎠ ⎝ 2 ⎝ ⎠

Bilged section: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

InertiaREMOTE = 333.33 +

(

×(

−

)

)

) = 10, 960.93 m

4

Forward section: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

InertiaREMOTE = 360 , 000.00 +

(

×

−

)

) = 779, 628.00 m

4

The remote inertia values can be summed to give the overall longitudinal inertia measured about the LCF: Longitudinal inertia = 835, 6 614.00 + 10 , 960.93 + 779 , 6 628 8.00 = 1, 626, 626 , 202.90 m4 This can be divided by the underwater volume to determine the BML: BML =

9781408176122_Ch14_4_Rev_txt_prf.indd 530

IL 1, 626 , 202.90 = = 271.03 m ∇ 100 × 20 × 3

11/16/2013 6:55:30 PM

Solutions to Questions • 531 Q6.26 The parallel sinkage can be found: Parallel sinkage =

Lost volume 20 × 10 × (2 − 1) = = 0.25 m Final w water aterp pla anee area (100 × 10 ) − (20 × 10 )

Therefore, after bilging the true mean draught will be 2.25 m. The KB and LCB can be found after bilging: Item

Underwater volume (m3)

Centre from keel (m)

Vertical moment of volume (m4)

100 × 10 × 2.25 = 2250

2.25 = 1.125 2

2531.25

Ship at the bilged draught Bilged compartment

−(

×(

×

Total

((2.25 1) × 0.5) + 1 = 1.625

−406.25 2,125.00

Moment 2,125 = = 1.063 063 m Mass 2, 000

Underwater volume (m3)

Centre from AP (m)

Longitudinal moment of volume (m4)

100 × 10 × 2.25 = 2250

100 = 50 2

112,500.00

Ship at the bilged draught Bilged compartment

)) = −250

2,000.00

KB = Item

−

−(

Total

×

×(

−

)) = −250

100 − (20 × 0.5) = 90

2,000.00

LCB =

−22,500 90,000

Moment 90 , 000 = = 45.00 00 m Mass 2, 000

The BML can be found, using the longitudinal inertia: 10 × 803 I 12 BML = L = = 213.33 m ∇ 100 × 10 × 2

9781408176122_Ch14_4_Rev_txt_prf.indd 531

11/16/2013 6:55:32 PM

532 • Ship Stability, Powering and Resistance This can be used to determine the GML: GML = KB B + BML − KG = 1.063 + 213 3.33 − 1.40 = 212.99 m This can be used to determine the MCTC: MCTC =

ΔGML (100 × 10 × 2) × 1.025 × 212.99 = = 43.66 tonne metres 100LBP 100 × 100

As the vessel was initially floating on an even keel, the LCG before bilging must be equal to the LCB before bilging. Therefore before bilging, the LCG must have been 50 m FOAP. As the position of the centre of gravity remains constant during bilging, the LCG must also be 50 m FOAP after bilging. This can be used to determine the trim of the vessel: Trim =

LCG ) Δ ( 45 − 50 ) (100 × 10 × 2 × 1.025) = = −235 cm MCTC 43.66

(LCB

∴ Trim = 2.35 m bow trim Q6.27 The parallel sinkage can be found: Sinkage =

Lost ∇ 5.00 × 10.00 × 4.00 × 0.6 120.00 = = = 0.21 m Final w water aterp pla anee area 600.00 − 30 570.00

Therefore, the bilged true mean draught can be found: Bilged DLCF

Original O iginal i l DLCF + Sinkage = 4.00 + 0.21 = 4.21 m

The KB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

2,526.00

2.11

5,329.86

Bilged compartment

–126.30

2.11

–266.49

Total

2,399.70

Item

Bilged KB =

9781408176122_Ch14_4_Rev_txt_prf.indd 532

5,063.37

Moment 5, 063.37 = = 2.11 m Volume 2, 399.70

11/16/2013 6:55:34 PM

Solutions to Questions • 533 The LCB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

2,526.00

30.00

75,780.00

Bilged compartment

–126.30

42.50

–5,367.75

Total

2,399.70

Item

Bilged LCB CB =

70,412.25

Moment 70 , 412.25 = = 29.34 m FOAP Volume 2, 399.70

The LCF can be found after bilging: Area (m2)

LCF (m)

Moment (m3)

Whole vessel

600.00

30.00

18,000.00

Bilged compartment

–30.00

42.50

–1,275.00

Total

570.00

Item

Bilged LCF C =

16,725.00

Moment 16 , 725.00 = = 29.34 m FOAP Area 570.00

For each section of the waterplane, the area, the longitudinal inertia and the distance from the centre of the section to the overall LCF can be found. These can be used to determine the longitudinal inertia of each section of the waterplane measured about the overall LCF of the vessel, and hence the total longitudinal inertia of the waterplane, measured about the overall LCF of the vessel: Area (m2)

Item Aft waterplane section

Section IL (m4) Distance to LCF (m)

Section IL GG (m4)

400.00

53,333.33

9.34

88,227.57

20.00

41.67

13.16

3,505.38

Forward waterplane section

150.00

2,812.50

23.16

83,270.34

Total

570.00

Bilged compartment

175,003.29

This allows the bilged BML to be found: BML =

IL 175, 003.29 = = 72.92 92 m ∇ 2, 400.00

This allows the bilged GML to be found: GML = KB + BML − KG = 2.11+ 72.92 − 2.00 = 73.03 m

9781408176122_Ch14_4_Rev_txt_prf.indd 533

11/16/2013 6:55:36 PM

534 • Ship Stability, Powering and Resistance This allows the bilged MCTC to be found: MCTC =

Δ × GML 2, 460.00 × 73.03 = = 29.94 tonne metre es 100 × LBP 100 × 60.00

This allows the bilged trim to be found: Trim =

LCG ) Δ (29.34 − 30.00 ) × 2, 460.00 = = −54.23 cm = 0.54 54 m by the bow 29.94 MCTC

(LCB

Finally the bilged end draughts can be found: DA

DLCF + Trim DF

⎛ LCF ⎞ ⎛ 29.34 ⎞ = 4.21 1+ −0.54 = 3.95 95 m ⎝ LBP ⎠ ⎝ 60.00 ⎠

DA − Trim = 3.95 − −0.54 = 4.49 m

Q6.28 The parallel sinkage can be found: Sinkage =

Lost ∇ 30.00 × 20.00 × 6.00 × 0.7 2, 520.00 = = = 1.27 m Final w water aterp pla anee area 2, 400 0 .00 − 420 1, 980.00

Therefore, the bilged true mean draught can be found: Bilged DLCF

Original O iginal i l DLCF + Sinkage = 6.00 + 1.27 = 7.27 m

The KB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

17,448.00

3.64

63,510.72

Bilged compartment

–3,053.40

3.64

–11,114.38

Total

14,394.60

Item

Bilged KB =

52,396.34

Moment 52, 396.34 = = 3 64 m Volume 14 , 394.60

The LCB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

17,448.00

60.00

1,046,880.00

Bilged compartment

–3,053.40

95.00

–290,073.00

Total

14,394.60

Item

9781408176122_Ch14_4_Rev_txt_prf.indd 534

756,807.00

11/16/2013 6:55:37 PM

Solutions to Questions • 535

Bilged LCB CB =

Moment 756 , 807.00 = = 52.58 m FOAP Volume 14 , 394.60

The LCF can be found after bilging: Area (m2)

LCF (m)

Moment (m3)

Whole vessel

2,400.00

60.00

144,000.00

Bilged compartment

–420.00

95.00

–39,900.00

Total

1,980.00

Item

Bilged LC LCF =

104,100.00

Moment 104 ,100.00 = = 52.58 m FOAP Area 1, 980.00

For each section of the waterplane, the area, the longitudinal inertia and the distance from the centre of the section to the overall LCF can be found. These can be used to determine the longitudinal inertia of each section of the waterplane measured about the overall LCF of the vessel, and hence the total longitudinal inertia of the waterplane, measured about the overall LCF of the vessel: Item

Area (m2)

Section IL (m4)

Distance to LCF (m)

Section IL GG (m4)

1,600.00

853,333.33

12.58

1,106,543.57

Bilged compartment

180.00

13,500.00

42.42

337,402.15

Forward waterplane section

200.00

1,666.67

62.42

780,917.95

Aft waterplane section

Total

1,980.00

2,224,863.67

This allows the bilged BML to be found: BML =

IL 2, 224 , 863.67 = = 154.5 5m ∇ 14 , 400.00

This allows the bilged GML to be found: GML = KB B + BML − KG = 3.64 + 154 5 .5 − 5.00 = 153.14 m This allows the bilged MCTC to be found: MCTC =

Δ × GML 14 , 760.00 × 153.14 = = 188.36 36 tonne metres m 100 × LBP 100 × 120.00

9781408176122_Ch14_4_Rev_txt_prf.indd 535

11/16/2013 6:55:39 PM

536 • Ship Stability, Powering and Resistance This allows the bilged trim to be found: Trim =

LCG ) Δ (52.58 − 60.00 ) × 14 , 760.00 = = −581.44 44 cm c = 5 81 m by the bow MCTC 188.36

(LCB

Finally, the bilged end draughts can be found: DA

DLCF + Trim DF

⎛ LCF ⎞ ⎛ 52.58 ⎞ = 7.27 7 + −5.81 = 4.72 72 m ⎝ LBP ⎠ ⎝ 120.00 ⎠

DA − Trim = 4.72 − −5.81 = 10.53 53 m

Q6.29 The parallel sinkage can be found: Sinkage =

Lost ∇ 10.00 × 10.00 × 1.50 × 0.9 135.00 = = = 0 14 m Final w water aterpl rplane area 1, 000 0 .00 − 0 1, 000.00

Therefore, the bilged true mean draught can be found: Bilged DLCF

Original O igi al DLCF + Sinkage = 2.00 + 0.14 = 2.14 m

The KB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

2,140.00

1.07

2,289.80

Bilged compartment

–135.00

0.75

–101.25

Total

2,005.00

Item

Bilged KB =

2,188.55

Moment 2,188.55 = = 1 09 m Volume 2, 005.00

The LCB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

2,140.00

50.00

107,000.00

Bilged compartment

–135.00

65.00

–8,775.00

Total

2,005.00

Item

Bilged LCB =

9781408176122_Ch14_4_Rev_txt_prf.indd 536

98,225.00

Moment 98 , 225.00 = = 48.99 m FOAP Volume 2, 005.00

11/16/2013 6:55:40 PM

Solutions to Questions • 537 As the waterplane remains intact, the LCF will remain at the centre of the waterplane area, which is 50.00 m FOAP. For each section of the waterplane, the area, the longitudinal inertia and the distance from the centre of the section to the overall LCF can be found. These can be used to determine the longitudinal inertia of each section of the waterplane measured about the overall LCF of the vessel, and hence the total longitudinal inertia of the waterplane, measured about the overall LCF of the vessel: Item

Area (m2)

Section IL (m4)

Distance to LCF (m)

Section IL GG (m4)

Aft waterplane section

600.00

180,000.00

20.00

420,000.00

Bilged compartment

100.00

833.33

15.00

23,333.33

Forward waterplane section

300.00

22,500.00

35.00

390,000.00

Total

1,000.00

833,333.33

This allows the bilged BML to be found: BML =

IL 833, 333.33 = = 416.67 m ∇ 2, 000.00

This allows the bilged GML to be found: GML = KB B + BML − KG = 1.09 + 416 6.67 − 3.50 = 414.26 m This allows the bilged MCTC to be found: MCTC =

Δ × GML 2, 050.00 × 414.26 = = 84.92 92 tonne mettres 100 × LBP 100 × 100.00

This allows the bilged trim to be found: Trim =

LCG ) Δ ( 48.99 − 50.00 ) × 2, 050.00 = = −24.38 38 cm = 0.24 24 m by the bow MCTC 84.92

(LCB

Finally, the bilged end draughts can be found: DA

DLCF + Trim DF

9781408176122_Ch14_4_Rev_txt_prf.indd 537

⎛ LCF ⎞ ⎛ 50.00 ⎞ = 2.14 4 + −0.24 = 2.02 02 m ⎝ LBP ⎠ ⎝ 100.00 ⎠

DA − Trim = 2.02 − −0.24 = 2.26 26 m

11/16/2013 6:55:42 PM

538 • Ship Stability, Powering and Resistance Q6.30 The parallel sinkage can be found: Sinkage =

Lost ∇ 15.00 × 15.00 × 2.00 × 0.75 337.50 = = = 0.28 m Final w water aterpl rplane area 1, 200.00 − 0 1, 200.00

Therefore, the bilged true mean draught can be found: Bilged DLCF

Original O iginal i l DLCF + Sinkage = 3.00 + 0.28 = 3 28 m

The KB can be found after bilging: Volume (m3)

Item

KB (m) Moment (m4)

Whole vessel

3,936.00

1.64

6,455.04

Bilged compartment

–337.50

1.00

–337.50

Total

3,598.50

Bilged KB B=

6,117.54

Moment 6 ,117.54 = = 1.70 m Volume 3, 598.50

The LCB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

3,936.00

40.00

157,440.00

Bilged compartment

–337.50

62.50

–21,093.75

Total

3,598.50

Item

Bilged LCB =

136,346.25

Moment 136 , 346.25 = = 37.89 m FOAP Volume 3, 598.50

As the waterplane remains intact, the LCF will remain at the centre of the waterplane area, which is 40.00 m FOAP. For each section of the waterplane, the area, the longitudinal inertia and the distance from the centre of the section to the overall LCF can be found. These can be used to determine the longitudinal inertia of each section of the waterplane measured about the overall LCF of the vessel, and hence the total longitudinal inertia of the waterplane, measured about the overall LCF of the vessel:

9781408176122_Ch14_4_Rev_txt_prf.indd 538

11/16/2013 6:55:43 PM

Solutions to Questions • 539

Item

Area (m2)

Section IL (m4)

Distance to LCF (m)

Section IL GG (m4)

Aft waterplane section

825.00

207,968.75

12.50

336,875.00

Bilged compartment

225.00

4,218.75

22.50

118,125.00

Forward waterplane section

150.00

1,250.00

35.00

185,000.00

Total

1,200.00

640,000.00

This allows the bilged BML to be found: BML =

IL 640 , 000.00 = = 177.78 m ∇ 3, 600.00

This allows the bilged GML to be found: GML = KB + BML − KG = 1.70 + 177.78 − 4.00 = 175.48 m This allows the bilged MCTC to be found: MCTC =

Δ × GML 3, 690.00 × 175.48 = = 80.94 tonne metre r s 100 × LBP 100 × 80.00

This allows the bilged trim to be found: Trim =

LCG ) Δ (37.89 − 40.00 ) × 3, 690.00 = = −96.19 cm = 0.96 96 m by the bow MCTC 80.94

(LCB

Finally, the bilged end draughts can be found: DA

DLCF + Trim

DF

⎛ LCF ⎞ ⎛ 40.00 ⎞ = 3.28 8 + −0.96 = 2.80 m ⎝ LBP ⎠ ⎝ 80.00 ⎠

DA − Trim = 2.8 − −0.96 = 3.76 76 m

Q6.31 The parallel sinkage can be found: Sinkage =

Lost ∇ 15.00 × 10.00 × 3.50 × 0.6 315.00 = = = 0 44 m Final w water aterp pla anee area 800 00 90 710.00

Therefore, the bilged true mean draught can be found: Bilged DLCF

9781408176122_Ch14_4_Rev_txt_prf.indd 539

Original O iginal i l DLCF + Sinkage = 5.00 + 0.44 = 5 44 m

11/16/2013 6:55:45 PM

540 • Ship Stability, Powering and Resistance The KB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

4,352.00

2.72

11,837.44

Bilged compartment

–354.60

3.47

–1,230.46

Total

3,997.40

Item

Bilged KB B=

10,606.98

Moment 10 , 606.98 = = 2 65 m Volume 3, 997.40

The LCB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

4,352.00

40.00

174,080.00

Bilged compartment

–354.60

47.50

–16,843.50

Total

3,997.40

Item

Bilged LCB =

157,236.50

Moment 157, 236.50 = = 39.33 m FOAP Volume 3, 997.40

The LCF can be found after bilging: Area (m2)

LCF (m)

Moment (m3)

Whole vessel

800.00

40.00

32,000.00

Bilged compartment

–90.00

47.50

–4,275.00

Total

710.00

Item

Bilged LC LCF =

27,725.00

Moment 27, 725.00 = = 39.05 m FOAP Area 710.00

For each section of the waterplane, the area, the longitudinal inertia and the distance from the centre of the section to the overall LCF can be found. These can be used to determine the longitudinal inertia of each section of the waterplane measured about the overall LCF of the vessel, and hence the total longitudinal inertia of the waterplane, measured about the overall LCF of the vessel:

9781408176122_Ch14_4_Rev_txt_prf.indd 540

11/16/2013 6:55:47 PM

Solutions to Questions • 541

Item

Area (m2)

Section IL (m4)

Distance to LCF (m)

Section IL GG (m4)

400.00

53,333.33

19.05

198,494.33

60.00

1,125.00

8.45

5,409.15

Forward waterplane section

250.00

13,020.83

28.45

215,371.46

Total

710.00

Aft waterplane section Bilged compartment

419,274.94

This allows the bilged BML to be found: BML =

IL 419 , 274.94 = = 104.82 m ∇ 4 , 000.00

This allows the bilged GML to be found: GML = KB + BML − KG = 2.65 + 104.82 − 3.00 = 104.47 m This allows the bilged MCTC to be found: MCTC =

Δ × GML 4 ,100.00 × 104.47 = = 53.54 54 tonne metre r s 100 × LBP 100 × 80.00

This allows the bilged trim to be found: Trim =

LCG ) Δ (39.33 − 40.00 ) × 4 ,100.00 = = −51.31 31 cm = 0.51 51 m by the bow MCTC 53.54

(LCB

Finally, the bilged end draughts can be found: DA

DLCF + Trim

DF

⎛ LCF ⎞ ⎛ 39.05 ⎞ = 5.44 4 + −0.51 = 5.19 19 m ⎝ LBP ⎠ ⎝ 80.00 ⎠

DA − Trim = 5.19 − −0.51 = 5.70 m

Q6.32 The parallel sinkage can be found: Sinkage =

Lost ∇ 30.00 × 20.00 × 4.00 × 0.75 1, 800.00 = = = 0.77 m Final w water aterp pla anee area 2, 800.00 − 450 2, 350.00

Therefore, the bilged true mean draught can be found: Bilged DLCF

9781408176122_Ch14_4_Rev_txt_prf.indd 541

Original O iginal i l DLCF + Sinkage = 6.00 + 0.77 = 6 77 m

11/16/2013 6:55:47 PM

542 • Ship Stability, Powering and Resistance The KB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

18,956.00

3.39

64,260.84

Bilged compartment

–2,146.50

4.39

–9,423.14

Total

16,809.50

Item

Bilged KB B=

54,837.70

Moment 54 , 837.70 = = 3 26 m Volume 16 , 809.50

The LCB can be found after bilging: Volume (m3)

KB (m)

Moment (m4)

Whole vessel

18,956.00

70.00

1,326,920.00

Bilged compartment

–2,146.50

85.00

–182,452.50

Total

16,809.50

Item

Bilged LCB =

1,144,467.50

Moment 1144 ,144 , 467.50 = = 68.08 m FOAP Volume 16 , 809.50

The LCF can be found after bilging: Item

Area (m2)

LCF (m)

Moment (m3)

Whole vessel

2,800.00

70.00

196,000.00

Bilged compartment

–450.00

85.00

–38,250.00

Total

2,350.00

Bilged LCF C =

157,750.00

Moment 157, 750.00 = = 67.13 m FOAP Area 2, 350.00

For each section of the waterplane, the area, the longitudinal inertia and the distance from the centre of the section to the overall LCF can be found. These can be used to determine the longitudinal inertia of each section of the waterplane measured about the overall LCF of the vessel, and hence the total longitudinal inertia of the waterplane, measured about the overall LCF of the vessel:

9781408176122_Ch14_4_Rev_txt_prf.indd 542

11/16/2013 6:55:49 PM

Solutions to Questions • 543

Item

Area (m2)

Section IL (m4)

Distance to LCF (m)

Section IL GG (m4)

Aft waterplane section

1,400.00

571,666.67

32.13

2,016,938.33

Bilged compartment

150.00

11,250.00

17.87

59,150.54

Forward waterplane section

800.00

106,666.67

52.87

2,342,856.19

Total

2,350.00

4,418,945.06

This allows the bilged BML to be found: BML =

IL 4 , 418 , 945.06 = = 263.03 03 m ∇ 16 , 800.00

This allows the bilged GML to be found: GML = KB B + BML − KG = 3.26 + 263 63.03 − 6.00 = 260.29 m This allows the bilged MCTC to be found: MCTC =

Δ × GML 17, 220.00 × 260.29 = = 320.16 16 tonne metres m 100 × LBP 100 × 140.00

This allows the bilged trim to be found: Trim =

LCG ) Δ (68.08 − 70.00 ) × 17, 220.00 = = −103.27 27 cm c = 1 03 m by the bow MCTC 320.16

(LCB

Finally, the bilged end draughts can be found: DA

DLCF + Trim DF

⎛ LCF ⎞ ⎛ 67.13 ⎞ = 6.77 7 + −1.03 = 6.28 m ⎝ LBP ⎠ ⎝ 140.00 ⎠

DA − Trim = 6.28 − −1.03 = 7.31 31 m

Q6.33 The parallel sinkage can be found: Parallel sinkage =

Lost volume 20 × 3 × 2 = = 0.127 m Final w water aterpl rplane area (100 0 10 ) (20 3)

The bilged true mean draught will therefore be 2.127 m. As there are no watertight flats, the KB can be taken as half this value: KB =

9781408176122_Ch14_4_Rev_txt_prf.indd 543

DBilged 2

=

2.127 = 1.064 m 2

11/16/2013 6:55:50 PM

544 • Ship Stability, Powering and Resistance The new roll axis can be found by finding the transverse centre of the waterplane area after bilging: Item

Waterplane area (m2)

Centre from centreline (m)

Ship at the bilged draught

100 × 10 = 1000

0

Bilged compartment

−(20 × 3) = −60

−3.5

Total

Transverse moment of area (m3) 0 210

940

210

Centre of waterplane area after bilging =

Total moment Total waterplane w area

Centre of waterplane area after bilging =

210 = 0.223 m 940

The new roll axis is therefore 0.223 m to port of the centreline of the vessel. This means the new roll axis must be 5.223 m from the damaged starboard edge. The inertia can be found, measured at the starboard edge: IEDGE =

LB 3 lb3 100 × 103 20 × 33 − = − = 33,153.34 m4 3 3 3 3

The parallel axes theorem can be used to determine the inertia measured at the new roll axis: IROLL AXIS

IEDGE New rollll a axis to edge dg 2 ) DG − (Waterplane area × Ne

IROLL AXIS = 33,153.34 −

(

×

) = 7, 510.39 m

3

This can be used to determine BM: BM =

7, 510.39 Inertia = = 3.755 m ∇ 100 × 10 × 2

This can be used to determine the GM: GM = KB + BM − KG = 1.064 + 3.755 − 4.00 = 0.819 m The TCB after bilging can be found from the bilged dimensions:

9781408176122_Ch14_4_Rev_txt_prf.indd 544

11/16/2013 6:55:52 PM

Solutions to Questions • 545

Item

Ship at the bilged draught

Underwater volume (m3)

Centre from centreline (m)

Transverse moment of volume (m4)

100 × 2.127 × 10 = 2,127

0

0

Bilged compartment −(20 × 3 × 2.127) = −127.62

Total

−

⎛ 10 3 ⎞ − = −3 5 ⎝ 2 2 ⎟⎠

1,999.38

TCB =

446.67

446.67

Moment 446.67 = = 0.223 Mass 1, 999.38

(port )

Therefore the list can be found: tanθ

d ⎛ 0.223 ⎞ θ = tan−1 ∴θ = 15.2 2 degrees ⎝ 0.819 ⎠ GM

Q6.34 The original KB can be found from the draught: Overall K KB B=

Draught 2 = = 1 00 m 2 2

The original BM can be determined: ⎛ LB 3 ⎞ ⎛ 50 × 83 ⎞ Inertia ⎜⎝ 12 ⎟⎠ ⎜⎝ 12 ⎟⎠ BM = = = = 2.66 m ∇ ∇ 50 × 8 × 2 The KM can be found from KB and BM: KM = KB + BM = 1.00 + 2.66 = 3.66 66 m The permeability can be determined from the sinkage, lost volume and final waterplane area: Lost volume Sinkage = Final w water aterpl rplane area 2.15 2 =

10 × 8 (2 − 1.2) × μ (50 × 8 ) (10 8 μ )

0 15 =

64 μ 400 − 80 μ

0.15( 400 − 80μ 80 ) = 64 μ

9781408176122_Ch14_4_Rev_txt_prf.indd 545

11/16/2013 6:55:54 PM

546 • Ship Stability, Powering and Resistance 60 −12μ 12 = 64 μ 60 = 64μ 64 + 12 μ 60 = μ = 0 79 76

A table of moments of volume can be used to determine the KB after bilging: Underwater volume (m3)

Item

Ship at the bilged draught Bilged compartment

Centre from eel (m)

50 × 8 × 2.15 = 860 −(

× ×(

−

)×

= −60.04

Total

2.15 = 1.075 2

)

((2.15

1.2) 0.5 5) 1.2

Vertical moment of volume (m4) 924.50 −100.57

= 1.675 799.96

Overall K KB B=

823.93

Moment 823.93 = = 1 03 m ∇ 799.96

The BM can be determined after bilging: ⎛ LB 3 ⎞ ⎛ lB 3 ⎞ ⎛ 50 × 83 ⎞ ⎛ 10 × 83 ⎞ − μ − 0 79 ⎜⎝ 12 ⎟⎠ ⎜⎝ 12 ⎟⎠ ⎜⎝ 12 ⎟⎠ Inertia ⎜⎝ 12 ⎟⎠ BM = = = = 2.25 25 m ∇ ∇ 799.96 The KM can be found from KB and BM: KM = KB + BM = 1.03 + 2.25 = 3.28 m As KG is constant, the change in GM is the same as the change in KM. Therefore, the change in GM is: New K KM M Old Old KM = 3.28 − 3.66 = −0 38 m Therefore, GM reduces by 0.38 m. Q6.35 Parallel sinkage =

Lost volume 20 × 2 × 10 × 1 400 = = = 0.50 m Final w water aterp pla anee area (100 × 10 ) − (20 × 10 × 1) 800

9781408176122_Ch14_4_Rev_txt_prf.indd 546

11/16/2013 6:55:56 PM

Solutions to Questions • 547 Therefore, the bilged true mean draught can be found: Bilged d drau raug ghtt

Original O iginal g a draught d aught + Parallel sinkage = 2 + 0.50 = 2.50 5 m

This can be used to determine the KB. As there are no watertight flats creating a double bottom, the bilged KB can be taken as half of the bilged draught: KB =

Bilged drau r ght 2 50 = = 1.25 m 2 2

The waterplane area will have been reduced, therefore the inertia will have been reduced, and needs to be recalculated: Inertia =

LB 3 μ lb3 100 × 103 1× 20 × 103 − = − = 6 , 666.67 m4 12 12 12 12

This can be used to find BM in the bilged condition: BM =

Inertia 6 , 666.67 = = 3.33 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught. As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.25 + 3.33 − 4.00 = 0.58 m Q6.36 Parallel sinkage =

Lost volume 20 × 2 × 10 × 0 8 320 = = = 0 38 m Final w water aterpl rplane area (100 × 10 ) − (20 × 10 × 0 8 ) 840

Therefore, the bilged true mean draught can be found: Bilged d drau raug ghtt

Original O iginal g a draught d aught + Parallel sinkage = 2 + 038 = 2.38 38 m

This can be used to determine the KB. As there are no watertight flats creating a double bottom, the bilged KB can be taken as half of the bilged draught: KB =

Bilged drau r ght 2 38 = = 1.19 m 2 2

The waterplane area will have been reduced, therefore the inertia will have been reduced, and needs to be recalculated:

9781408176122_Ch14_4_Rev_txt_prf.indd 547

11/16/2013 6:55:59 PM

548 • Ship Stability, Powering and Resistance

Inertia =

LB 3 μllb3 100 × 103 0 8 × 20 × 103 − = − = 7, 000 m4 12 12 12 12

This can be used to find BM in the bilged condition: BM =

7, 000 Inertia = = 3.50 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught. As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.19 + 3.50 − 4.00 = 0.69 m Q6.37 Parallel sinkage =

Lost volume 20 × 0 9 × 10 0 1 180 = = = 0.18 18 m Final w water aterpl rplane area (100 × 10 ) − (0 ) 1, 000

Therefore, the bilged true mean draught can be found: Original O iginal ginal draught + Parallel sinkage = 2 + 0.18 = 2.18 1 m

Bilged drau r ght

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Volume (m3)

KB (m)

Moment (m4)

100 × 10 × 2.18 = 2,180

2.18 = 1 09 2

2,376.2

−(20 ×0.9 × 10 × 1) = −180

09 = 0 45 2

−81

Item Whole vessel at the bilged draught Bilged compartment Total

2,000

KB =

2,295.2

Moment 2, 295.2 = = 1.15 m Volume 2, 000

The waterplane area will not have been reduced, therefore the inertia remains constant, and can be calculated: Inertia =

9781408176122_Ch14_4_Rev_txt_prf.indd 548

LB 3 100 × 103 = = 8 , 333.33 m4 12 12

11/16/2013 6:56:00 PM

Solutions to Questions • 549 This can be used to find BM in the bilged condition: BM =

Inertia 8 , 333.33 = = 4.16 16 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught. As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.15 + 4.16 − 4.00 = 1.31 m Q6.38 Parallel sinkage =

Lost volume 20 × 0 9 × 10 0 0.8 144 = = = 0.14 m Final w water aterpl rplane area (100 × 10 ) − (0 ) 1, 000

Therefore, the bilged true mean draught can be found: Bilged drau r ght

Original O iginal ginal draught + Parallel sinkage = 2 + 0.14 = 2.14 1 m

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Volume (m3)

KB (m)

Moment (m4)

100 × 10 × 2.14 = 2,140

2.14 = 1 07 2

2,289.8

−(20 × 0.9 × 10 × 0.8) = −144

09 = 0 45 2

−64.8

Item Whole vessel at the bilged draught Bilged compartment Total

1,996

KB =

2,225.0

Moment 2, 225.0 = = 1.11 11 m Volume 2, 000

Note that the difference in the total volume column and the directly calculated volume is due to rounding, and gives a rounding error of 0.2%. The waterplane area will not have been reduced, therefore the inertia remains constant, and can be calculated: Inertia =

9781408176122_Ch14_4_Rev_txt_prf.indd 549

LB 3 100 × 103 = = 8 , 333.33 m4 12 12

11/16/2013 6:56:03 PM

550 • Ship Stability, Powering and Resistance This can be used to find BM in the bilged condition: BM =

8 , 333.33 Inertia = = 4.16 16 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught. As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.11+ 4.16 − 4.00 = 1.27 m Q6.39 Parallel sinkage =

Lost volume 20 × (2 − 0.9 ) 10 10 1 220 = = = 0 28 m Final w water aterpl rplane area (100 × 10 ) − (20 × 2 1) 800

Therefore, the bilged true mean draught can be found: Original O iginal ginal draught + Parallel sinkage = 2 + 0.28 = 2.28 2 m

Bilged drau r ght

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Volume (m3)

KB (m)

Moment (m4)

100 × 10 × 2.28 = 2,280

2.28 = 1.14 2

2,599.2

Item Whole vessel at the bilged draught Bilged compartment

− (20 × (2.28 − 0.9 ) × 10 × 1) = −276 ( .28 0. ) + 0.9 = 1.59 2

Total

2,004

KB =

−438.84 2,160.36

Moment 2,160.36 = = 1.08 08 m Volume 2, 004

The waterplane area will have been reduced, therefore the inertia will have been reduced, and needs to be recalculated: Inertia =

LB 3 μlb3 100 × 103 1× 20 × 103 − = − = 6 , 666.67 m4 12 12 12 12

This can be used to find BM in the bilged condition: BM =

9781408176122_Ch14_4_Rev_txt_prf.indd 550

Inertia 6 , 666.67 = = 3.33 m ∇ 100 × 10 × 2

11/16/2013 6:56:04 PM

Solutions to Questions • 551 Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught. As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.08 + 3.33 − 4.00 = 0.41 41 m Q6.40 Parallel sinkage =

Lost volume 20 × (2 − 0.9 ) 10 0.8 176 = = = 0 18 m Final w water aterpl rplane area (100 × 10 ) − (20 2 0.8 ) 968

Therefore, the bilged true mean draught can be found: Original O iginal ginal draught + Parallel sinkage = 2 + 0.18 = 2.18 1 m

Bilged drau r ght

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Volume (m3)

Item Whole vessel at the bilged draught

100 × 10 × 2.18 = 2,180

− (20 × (2.18 − 0.9 ) × 10 × 0.8 )

Bilged compartment

KB (m)

= −204.8

Total

2.18 = 1 09 2 ( .18 0. ) + 0.9 = 1.54 2

1,975.2

KB =

Moment (m4) 2,376.2

−315.39

2,060.81

Moment 2, 060.81 = = 1.04 m Volume 1, 975.2

The waterplane area will have been reduced, therefore the inertia will have been reduced, and needs to be recalculated: Inertia =

LB 3 μlb3 100 × 103 0 8 × 20 × 103 − = − = 7, 000 m4 12 12 12 12

This can be used to find BM in the bilged condition: BM =

Inertia 7, 000 = = 3.50 50 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught.

9781408176122_Ch14_4_Rev_txt_prf.indd 551

11/16/2013 6:56:07 PM

552 • Ship Stability, Powering and Resistance As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.04 + 3.50 − 4.00 = 0.54 m Q6.41 Parallel sinkage =

Lost volume 20 × 2 × 10 × 1 400 = = = 0.50 m Final w water aterp pla anee area (100 × 10 ) − (20 × 10 × 1) 800

Therefore, the bilged true mean draught can be found: Bilged drau r ght

Original O iginal ginal draught + Parallel sinkage = 2 + 0.50 = 2.50 5 m

This can be used to determine the KB. As there are no watertight flats creating a double bottom, the bilged KB can be taken as half of the bilged draught: KB =

Bilged drau r ght 2 50 = = 1.25 m 2 2

The waterplane length has been reduced to an effective length of 80 m due to the bilging. Therefore, the longitudinal inertia of the waterplane must be calculated based on this effective length: B (L − l ) 10 × (100 − 20 ) = = 426 , 666.67 m4 12 12 3

InertiaL =

3

This can be used to determine the longitudinal BM: BML =

IL 426 , 666.67 = = 213.33 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the longitudinal BM equation is to use the original length, beam and draught. This volume can also be used to determine the displacement for the MCTC equation later. The longitudinal BM can be used to determine the longitudinal metacentric height: GML = KB + BML − KG = 1.25 + 213.33 − 4.00 = 210.58 m This allows the MCTC to be found: MCTC =

Δ × GML (100 × 10 × 2 × 1.025) × (210.58 ) = = 43.17 17 to onne metres 100 × LBP 100 × 100

9781408176122_Ch14_4_Rev_txt_prf.indd 552

11/16/2013 6:56:09 PM

Solutions to Questions • 553 The LCB will be at the longitudinal centre of underwater volume. As the vessel now has an effective length of 80 m, the LCB will be at half this value: LCB =

Effective v length t 80 = = 40 m FOAP 2 2

As the vessel was initially on an even keel, the LCG must be at amidships. The position of the centre of gravity remains constant during bilging. Therefore, the LCG must be 50 m FOAP. This allows the trim to be found: Trim =

LCG ) Δ ( 40 − 50 ) × (100 × 10 × 2 × 1.025) = = −474.9 9 cm m MCTC 43.17

(LCB

4.75 m

The LCF will be at the longitudinal centre of the waterplane. As the vessel now has an effective length of 80 m, the LCF will be at half this value: LCF =

Effective length t 80 = = 40 m FOAP 2 2

This allows the draught aft to be found: DA

DLCF + Trim

⎛ LCF ⎞ ⎛ 40 ⎞ = 2.50 − 4.75 = 0.60 60 m ⎝ LBP ⎠ ⎝ 100 ⎠

This allows the draught forward to be found: DF

DA − Trim = 0.60 − −4.75 = 5.35 35 m

Q6.42 Parallel sinkage =

Lost volume 20 × 0 9 × 10 0 1 180 = = = 0.18 18 m Final w water aterpl rplane area (100 × 10 ) − (0 ) 1, 000

Therefore, the bilged true mean draught can be found: Bilged d drau raug ghtt

Original O iginal g a draught d aught + Parallel sinkage = 2 + 0.18 = 2.18 1 m

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Item Whole vessel at the bilged draught Bilged compartment Total

9781408176122_Ch14_4_Rev_txt_prf.indd 553

Volume (m3)

KB (m)

Moment (m4)

100 × 10 × 2.18 = 2,180

2.18 = 1 09 2

2,376.2

−(20 × 0.9 × 10 × 1) = −180

09 = 0 45 2

−81

2,000

2,295.2

11/16/2013 6:56:11 PM

554 • Ship Stability, Powering and Resistance

KB =

Moment 2, 295.2 = = 1.15 m Volume 2, 000

As the bilging is contained in the double bottom, the waterplane area is unchanged. Therefore, the longitudinal inertia and BM will be the same as the vessel before bilging: InertiaL =

BLL3 10 × 1003 = = 833, 333.33 m4 12 12

This can be used to determine the longitudinal BM: BML =

IL 833, 333.33 = = 416.67 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the longitudinal BM equation is to use the original length, beam and draught. This volume can also be used to determine the displacement for the MCTC equation later. The longitudinal BM can be used to determine the longitudinal metacentric height: GML = KB + BML − KG = 1.15 + 416.67 − 4.00 = 413.82 m This allows the MCTC to be found: MCTC =

Δ × GML (100 × 10 × 2 × 1.025) × (413.82) = = 84.83 83 to onne metres 100 × LBP 100 × 100

The LCB will be at the longitudinal centre of underwater volume. As there is a watertight flat creating a double bottom, the bilged LCB must be determined by a table of moments of volume: Item Whole vessel at the bilged draught Bilged compartment

Volume (m3)

LCB (m)

Moment (m4)

100 × 10 × 2.18 = 2,180

100 = 50 2

109,000

−(20 × 0.9 × 10 × 1) = −180

Total

100 −

20 = 90 2

2,000

LCB =

9781408176122_Ch14_4_Rev_txt_prf.indd 554

−16,200 92,800

Moment 92, 800 = = 46.40 m FOAP Volume 2, 000

11/16/2013 6:56:13 PM

Solutions to Questions • 555 As the vessel was initially on an even keel, the LCG must be at amidships. The position of the centre of gravity remains constant during bilging. Therefore, the LCG must be 50 m FOAP. This allows the trim to be found: Trim =

LCG ) Δ ( 46.40 − 50 ) × (100 × 10 × 2 × 1.025) = = −87 cm m MCTC 84.83

(LCB

0 87 m

The LCF will be at the longitudinal centre of the waterplane. As the waterplane has a length of 100 m, the LCF will be at half this value: LCF =

Length 100 = = 50 m FOAP 2 2

This allows the draught aft to be found: DA

DLCF + Trim

⎛ LCF ⎞ ⎛ 50 ⎞ = 2.18 − 0.87 = 1.75 m ⎝ LBP ⎠ ⎝ 100 ⎠

This allows the draught forward to be found: DF

DA − Trim = 1.75 − −0.87 = 2.62 m

Q6.43 Parallel sinkage =

Lost volume 20 × 0 9 × 10 0 0.8 144 = = = 0.14 m Final w water aterp pla anee area (100 × 10 ) − (0 ) 1, 000

Therefore, the bilged true mean draught can be found: Bilged drau r ght

Original O iginal ginal draught + Parallel sinkage = 2 + 0.14 = 2.14 1 m

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Volume (m3)

KB (m)

Moment (m4)

100 × 10 × 2.14 = 2,140

2.14 = 1 07 2

2,289.8

−(20 × 0.9 × 10 × 0.8) = −144

09 = 0 45 2

−64.8

Item Whole vessel at the bilged draught Bilged compartment Total

1,996

KB =

9781408176122_Ch14_4_Rev_txt_prf.indd 555

2,225

Moment 2, 225 = = 1.11 m Volume 1, 996

11/16/2013 6:56:15 PM

556 • Ship Stability, Powering and Resistance As the bilging is contained in the double bottom, the waterplane area is unchanged. Therefore, the longitudinal inertia and BM will be the same as the vessel before bilging: InertiaL =

BLL3 10 × 1003 = = 833, 333.33 m4 12 12

This can be used to determine the longitudinal BM: BML =

IL 833, 333.33 = = 416.67 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the longitudinal BM equation is to use the original length, beam and draught. This volume can also be used to determine the displacement for the MCTC equation later. The longitudinal BM can be used to determine the longitudinal metacentric height: GML = KB + BML − KG = 1.11+ 416.67 − 4.00 = 413.78 m This allows the MCTC to be found: MCTC =

Δ × GML (100 × 10 × 2 × 1.025) × (413.78 ) = = 84.83 to onne metres 100 × LBP 100 × 100

The LCB will be at the longitudinal centre of underwater volume. As there is a watertight flat creating a double bottom, the bilged LCB must be determined by a table of moments of volume: Volume (m3)

LCB (m)

Moment (m4)

100 × 10 × 2.14 = 2,140

100 = 50 2

107,000

Item Whole vessel at the bilged draught

Bilged compartment −(20 × 0.9 × 10 × 0.8) = −144 Total

100 −

20 = 90 2

1,996

LCB =

9781408176122_Ch14_5_Rev_txt_prf.indd 556

−12,960 94,040

Moment 94 , 040 = = 47.11 m FOAP Volume 1, 996

11/16/2013 2:41:46 AM

Solutions to Questions • 557 As the vessel was initially on an even keel, the LCG must be at amidships. The position of the centre of gravity remains constant during bilging. Therefore, the LCG must be 50 m FOAP. This allows the trim to be found: Trim =

LCG ) Δ ( 47.11 − 50 ) × (100 × 10 × 2 × 1.025) = = −69.7 7 cm MCTC 84.83

(LCB

0 70 m

The LCF will be at the longitudinal centre of the waterplane. As the waterplane has a length of 100 m, the LCF will be at half this value: LCF =

Length 100 = = 50 m FOAP 2 2

This allows the draught aft to be found: DLCF + Trim

DA

⎛ LCF ⎞ ⎛ 50 ⎞ = 2.14 − 0.70 = 1.79 79 m ⎝ LBP ⎠ ⎝ 100 ⎠

This allows the draught forward to be found: DF

DA − Trim = 1.79 − 0.70 = 2.49 m

Q6.44 Parallel sinkage =

20 × (2 − 0.9 ) 10 10 1 220 Lost volume = = = 0 28 m Final w water aterp pla anee area (100 × 10 ) − (20 × 10 × 1) 800

Therefore, the bilged true mean draught can be found: Bilged drau r ght

Original O iginal ginal draught + Parallel sinkage = 2 + 0.28 = 2.28 2 m

This can be used to determine the KB. As there is a watertight flat creating a double bottom, the bilged KB must be determined by a table of moments of volume: Item Whole vessel at the bilged draught Bilged compartment

Volume (m3)

KB (m)

Moment (m4)

100 × 10 × 2.28 = 2,280

2.28 = 1.14 2

2,599.2

− (20 × (2.28 − 0.9 ) × 10 × 1) = −276

Total

2,004

KB =

9781408176122_Ch14_5_Rev_txt_prf.indd 557

( .28 0. ) + 0.9 = 1.59 2

−438.84 2,160.36

Moment 2,160.36 = = 1.08 08 m Volume 2, 004

11/16/2013 2:41:47 AM

558 • Ship Stability, Powering and Resistance The waterplane length has been reduced to an effective length of 80 m due to the bilging. Therefore, the longitudinal inertia of the waterplane must be calculated based on this effective length: B (L − l ) 10 × (100 − 20 ) = = 426 , 666.67 m4 12 12 3

InertiaL =

3

This can be used to determine the longitudinal BM: BML =

IL 426 , 666.67 = = 213.33 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the longitudinal BM equation is to use the original length, beam and draught. This volume can also be used to determine the displacement for the MCTC equation later. The longitudinal BM can be used to determine the longitudinal metacentric height: GML = KB + BML − KG = 1.08 + 213.33 − 4.00 = 210.41 m This allows the MCTC to be found: MCTC =

Δ × GML (100 × 10 × 2 × 1.025) × (210.41) = = 43.13 to onne metres 100 × LBP 100 × 100

The LCB will be at the longitudinal centre of underwater volume. As there is a watertight flat creating a double bottom, the bilged LCB must be determined by a table of moments of volume: Item Whole vessel at the bilged draught Bilged compartment

Volume (m3)

LCB (m)

Moment (m4)

100 × 10 × 2.28 = 2,280

100 = 50 2

114,000

− (20 × (2.28 − 0.9 ) × 10 × 1) = −276

Total

100 −

20 = 90 2

2,004

LCB =

9781408176122_Ch14_5_Rev_txt_prf.indd 558

−24,840 89,160

Moment 89 ,160 = = 44.49 m FOAP Volume 2, 004

11/16/2013 2:41:50 AM

Solutions to Questions • 559 As the vessel was initially on an even keel, the LCG must be at amidships. The position of the centre of gravity remains constant during bilging. Therefore, the LCG must be 50 m FOAP. This allows the trim to be found: Trim =

LCG ) Δ ( 44.49 − 50 ) × (100 × 10 × 2 × 1.025) = = −262 cm c = −2 62 m MCTC 43.13

(LCB

The LCF will be at the longitudinal centre of the waterplane. As the waterplane has a length of 80 m, the LCF will be at half this value: LCF =

Length 80 = = 40 m FOAP 2 2

This allows the draught aft to be found: DA

DLCF + Trim

⎛ LCF ⎞ ⎛ 40 ⎞ = 2.28 − 2.62 = 1.23 m ⎝ LBP ⎠ ⎝ 100 ⎠

This allows the draught forward to be found: DF

DA − Trim = 1.23 − −2.62 = 3.85 85 m

Q6.45 Parallel sinkage =

Lost volume 20 × 1× 2 40 = = = 0.04 04 m Final w water aterp pla anee area (100 0 10 ) (20 1) 980

Therefore, the bilged true mean draught can be found: Bilged drau r ght

Original O iginal ginal draught + Parallel sinkage = 2 + 0.04 = 2.04 0 m

This can be used to determine the KB. As there are no watertight flats creating a double bottom, the bilged KB can be taken as half of the bilged draught: KB =

Bilged drau r ght 2 04 = = 1 02 m 2 2

The loss of waterplane will have moved the roll axis away from the centreline, therefore the new roll axis must be found via a table of moments of area: Item

Area (m2)

Whole waterplane 100 × 10 = 1,000 Bilged section Total

9781408176122_Ch14_5_Rev_txt_prf.indd 559

−(20 × 1) = −20 980

Lever (m)

Moment (m3)

10 =5 2

5,000

1 = 0 50 2

−10 4,990

11/16/2013 2:41:51 AM

560 • Ship Stability, Powering and Resistance

Roll axis x =

Moment 4 , 990 = = 5 09 m FOAP Area 980

The waterplane area will have been reduced, therefore the inertia will have been reduced, and needs to be recalculated. However, this must be done in two stages, using the parallel axes theorem. The inertia at the damaged edge can be found: InertiaEDGE =

LB 3 lb3 100 × 103 20 × 13 − = − = 33, 326.67 m4 3 3 3 3

The parallel axis theory can be used to convert this value to be about the new roll axis: IRoll Axis

(

IEDGE Lever DG − Area

) = 33, 326.67 − (

×

) = 7, 936.73 m

4

This can be used to find BM in the bilged condition: BM =

7, 936.73 Inertia = = 3.97 m ∇ 100 × 10 × 2

Note that as the overall underwater volume remains constant during bilging, the easiest way to determine the underwater volume in the BM equation is to use the original length, beam and draught. As KG remains constant during bilging, the metacentric height can now be found: GM = KB + BM − KG = 1.02 + 3.97 − 4.00 = 0.99 m Finally, the list can be determined: tanθ

d GM

θ = ta tan−1

5 09 − 5 = 5 2 degrees 0 99

Q7.1 The lightship displacement can be found: Δ = ∇ × ρ = 60 × 10 × 0.524 = 322.26 tonnes The structural loading rate can be found: Structual rrat atee =

Δ LS 322.26 = = 5 37 t/m L 60

The loaded displacement can be found: Δ Load = Δ Lightshipi + C g o = 314.4 + 300 + 300 = 922.26 tonnes

9781408176122_Ch14_5_Rev_txt_prf.indd 560

11/16/2013 2:41:53 AM

Solutions to Questions • 561 The buoyancy loading rate can be found: Buoyancy rate =

Δ LOAD 922.26 O = = 15.37 t/m L 60

The cargo loading rate can be found: Cargo load loading g rrate (hold 1) =

Cargo 300 = = 15 t/m Hold length 20

Cargo load loading g rrate (hold 2) = Cargo load loading g rrate (hold 3) =

Cargo 0 = = 0 t/m Hold length 20

Cargo 300 = = 15 t/m Hold length 20

These values can be tabulated to determine the overall load: Hold

Structure (t/m)

Cargo (t/m)

Buoyancy (t/m) Load (t/m)

1

–5.37

–15

15.37

–5

2

–5.37

0

15.37

10

3

–5.37

–15

15.37

–5

This allows the load diagram to be plotted: Load diagram 12.000 10.000

Load (tonnes/metre)

8.000 6.000 4.000 2.000 0.000 0

5

10

15

20

25

30

35

40

45

50

55

60

–2.000 –4.000 –6.000 Position (m)

Between 0 and 20 m, the total area under the load diagram is: −5 × 20 = −100 tonnes

9781408176122_Ch14_5_Rev_txt_prf.indd 561

11/16/2013 2:41:55 AM

562 • Ship Stability, Powering and Resistance Between 20 and 40 m, the total area under the load diagram is: 10 × 20 = 200 tonnes Between 40 and 60 m, the total area under the load diagram is: −5 × 20 = −100 tonnes Therefore at 0 m, the cumulative total area under the load diagram is 0 tonnes. At 20 m, the cumulative total area is: 0 − 100 = −100 tonnes At 40 m, the cumulative total area is: 0 − 100 + 200 = 100 tonnes At 60 m, the cumulative total area is: 0 − 100 + 200 − 100 = 0 tonnes These values can be plotted to give the shear force diagram:

Shear force (tonnes)

Shear force diagram 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 –10 0 –20 –30 –40 –50 –60 –70 –80 –90 –100 –110 –120 –130 –140 –150

5

10

15

20

25

30

35

40

45

50

55

60

Position (m)

9781408176122_Ch14_5_Rev_txt_prf.indd 562

11/16/2013 2:41:57 AM

Solutions to Questions • 563 Between 0 and 20 m, the total area under the load diagram is: −100 × 20 × 0.5 = −1, 000 tonne metres Between 20 and 30 m, the total area under the load diagram is: −100 × 10 × 0.5 = −500 m Between 30 and 40 m, the total area under the load diagram is: 100 × 10 × 0.5 5 = 500 tonne metres Between 40 and 60 m, the total area under the load diagram is: 100 × 20 × 0.5 = 1, 000 tonne metres Therefore at 0 m, the cumulative total area under the shear force diagram is 0 tonnes. At 20 m, the cumulative total area is: 0 1, 000 = −1, 000 tonne metres At 30 m, the cumulative total area is: 0 1, 000 − 500 = −1, 500 tonne metres At 40 m, the cumulative total area is: 0 1, 000 − 500 + 500 = −100 tonne metres At 60 m, the cumulative total area is: 0 1, 000 − 500 + 500 − 1, 000 = 0 tonne metres This can be used to plot the bending moment diagram: Bending moment diagram 0 Bending moment (tonnes metres)

0

5

10

15

20

25

30

35

40

45

50

55

60

–200 –400 –600 –800 –1,000 –1,200 –1,400 –1,600 Position (m)

9781408176122_Ch14_5_Rev_txt_prf.indd 563

11/16/2013 2:41:58 AM

564 • Ship Stability, Powering and Resistance Q7.2 The vessel floats on an even keel in the lightship condition. Therefore, for the vessel to finish loading on an even keel, the LCG of the combined cargo must also be at amidships, or 50 m FOAP. A loading table can be used to determine the cargo to load: Item

Mass (t)

LCG (m)

Moment (tm)

Hold 1 Cargo

300

10

3,000

Hold 2 Cargo

700

30

21,000

Hold 3 Cargo

400

50

20,000

Hold 4 Cargo

200

70

14,000

Hold 5 Cargo

x

90

90x

Totals

1,600 + x

LCG = 50 =

58,000 + 90x

Moment Mass

58 , 000 + 90 x 1, 600 + x

50 (1, 600 + x ) = 58 , 000 + 90 x 80 , 000 − 58 , 000 = 90 x − 50 x x=

22, 000 = 550 tonnes 40

Therefore, the lightship displacement can be found: Δ LS

Δ LOAD O D − Cargo

Δ LS = 5, 840 − 300 − 700 − 400 − 200 − 550 = 3, 690 tonnes The structural loading rate can be found: Structual rrat atee =

Δ LS 3, 690 = = 36.9 t/m L 100

The buoyancy loading rate can be found: Buoyancy rate ate =

Δ LOAD 5, 840 O = = 58.4 t/m L 100

The cargo loading rate can be found: Cargo loading oad g rrate (hold o d 1) =

9781408176122_Ch14_5_Rev_txt_prf.indd 564

Cargo 300 = = 15 t/m Hold length 20

11/16/2013 2:41:59 AM

Solutions to Questions • 565

Cargo load loading g rrate (hold 2) =

Cargo 700 = = 35 t/m Hold length 20

Cargo load loading g rrate (hold 3) =

Cargo 400 = = 20 t/m Hold length 20

Cargo load loading g rrate (hold 4 ) =

Cargo 200 = = 10 t/m Hold length 20

Cargo load loading g rrate (hold 5) =

Cargo 550 = = 27.5 t/m Hold length 20

These values can be tabulated to determine the overall load: Hold

Buoyancy (t/m)

Structure (t/m)

Cargo (t/m)

Load (t/m)

1

58.4

36.9

15

6.5

2

58.4

36.9

35

−13.5

3

58.4

36.9

20

1.5

4

58.4

36.9

10

11.5

5

58.4

36.9

27.5

−6

These can be plotted to form the load diagram: 15

10

Laod (dm)

5

0 0

10

20

30

40

50

60

70

80

90

100

–5

–10

–15 Position (m FOAP)

9781408176122_Ch14_5_Rev_txt_prf.indd 565

11/16/2013 2:42:02 AM

566 • Ship Stability, Powering and Resistance The areas for each hold can be determined from the graph: Hold

Load (t/m)

Area (t)

1

6.5

130

2

−13.5

−270

3

1.5

30

4

11.5

230

5

−6

−120

The cumulative total area can be determined at each bulkhead: Position (m FOAP)

Total Area (t)

0

0

20

130

40

−140

60

−110

80

120

100

0

These values can be plotted to form the shear force diagram: 150

100

Shear force (tonnes)

50

29.63 m FOAP

0 0

20

40

60

80

100

–50 69.57 m FOAP

–100

–150

–200 Position (m FOAP)

9781408176122_Ch14_5_Rev_txt_prf.indd 566

11/16/2013 2:42:03 AM

Solutions to Questions • 567 Linear interpolation can be used to determine the crossing points, as shown on the shear force diagram. The areas for each section can be determined from the graph: Hold

Load (t/m)

Area (t)

1

6.5

130

2

−13.5

−270

3

1.5

30

4

11.5

230

5

−6

−120

The cumulative total area can be determined at each bulkhead and crossing point: Position (m FOAP) 0

Total Area (tm) 0

20

1,300

29.63

1,925.95

40

1,200.05

60

−1,299.95

69.57

−1,826.3

80

−1,200.5

100

−0.5

These values can be plotted to form the bending moment diagram: 2,500

Bending moment (tonnes metres)

2,000 1,500 1,000 500 0 0

20

40

60

80

100

–500 –1,000 –1,500 –2,000 –2,500 Position (m FOAP)

9781408176122_Ch14_5_Rev_txt_prf.indd 567

11/16/2013 2:42:03 AM

568 • Ship Stability, Powering and Resistance The peak bending moment value is therefore 1,926 tonne metres hogging at 29.63 m FOAP. Q8.1 Ordinate

Offset

Offset2

Simpson’s Multiplier MomentX product

0

5

25

1

25

1

6

36

4

144

2

5.25

27.5625

2

55.125

3

4

16

4

64

4

2

4

1

4

Total

Moment o of area about the X a axis =

292.125

⎛ Spacing ⎞ ⎛ 1 ⎞ × × Σ ( Mom o ent X product ) ⎝ 3 ⎠ ⎝ 2⎠

Moment o of area about the X a axis =

⎛ 2 ⎞ ⎛ 1⎞ × × 292.125 = 97.375 375 units3 ⎝ 3⎠ ⎝ 2⎠

In Q3.18, we saw that the area under a graph with these coordinates, also found via Simpson’s Rule, was 38.333 units2. Centre of area from the X axis x =

Moment o of area about the X ax axis Area under the graph

Centre of area from the X axis x =

97.375 = 2.540 units 38.333

Q8.2 Ordinate

Offset

Lever

Simpson’s Multiplier MomentY product

0

5

0

1

0

1

6

1

4

24

2

5.25

2

2

21

3

4

3

4

48

4

2

4

1

8

Total

Moment o of area bout the Y axis x =

9781408176122_Ch14_5_Rev_txt_prf.indd 568

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

101

e tY (Momen

product )

11/16/2013 2:42:03 AM

Solutions to Questions • 569

Moment of area about the Y axi a s=

⎛ 2⎞ × 2 × 101 = 134.67 unit its3 ⎝ 3⎠

As before, in question Q3.18, we saw that the area under this graph, also found via Simpson’s Rule, was 38.333 units2. Centre of area from the Y axis x =

Moment o of area about the Y ax axis Area u under de the graph

Centre of area from the Y axis x =

134.67 = 3.513 units 38.333

Q8.3 Offset3

Ordinate

Offset

0

5

125

1

125

1

6

216

4

864

2

5.25

144.7

2

289.41

3

4

64

4

256

4

2

8

1

8

Total

1,542.41

Inertia about the X axis x =

Simpson’s Multiplier

InertiaX product

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 2 ⎞ ⎛ 1⎞ × × 1, 542.41 = 342.758 8 units 4 ⎝ 3⎠ ⎝ 3⎠

Q8.4 Ordinate

Offset

Lever2

Simpson’s Multiplier InertiaY product

0

5

0

1

0

1

6

1

4

24

2

5.25

4

2

42

3

4

9

4

144

4

2

16

1

32

Total

242

9781408176122_Ch14_5_Rev_txt_prf.indd 569

11/16/2013 2:42:04 AM

570 • Ship Stability, Powering and Resistance

Inertia about the Y axis x =

⎛ Spacing ⎞ × Spacing2 ⎝ 3 ⎠

Inertia about the Y axis x =

(InertiaY

product )

⎛ 2⎞ × 22 × 242 = 645.34 units 4 ⎝ 3⎠

Q8.5 In the x direction:

(

InertiaCENTRE + Area Distance E

InertiaREMOTE 342.758

CENTRE E

)

(38.333 × 2.5402 )

InertiaCENTRE = 95.449 449 units 4 E In the y direction:

(

InertiaCENTRE + Area Distance E

InertiaREMOTE 645.340

CENTRE E

)

(38.333 × 3.5132 )

InertiaCENTRE = 172.226 units 4 E Q8.6 The area can be found using: Ordinate

Offset

Simpson’s Multipliers Area product

0

0

1

0

4

8

4

32

8

15

2

30

12

20

4

80

16

17

2

34

20

10

4

40

24

0

1

0

Total

Area =

⎛ Spacing ⎞ × Σ ( Area product ) ⎝ 3 ⎠

Area =

9781408176122_Ch14_5_Rev_txt_prf.indd 570

216

⎛ 4⎞ × 216 = 288.00 m2 ⎝ 3⎠

11/16/2013 2:42:06 AM

Solutions to Questions • 571 The moment of area relative to the x axis can be found using: Offset

Offset2

0

0

0

1

0

4

8

64

4

256

8

15

225

2

450

12

20

400

4

1,600

16

17

289

2

578

20

10

100

4

400

24

0

0

1

0

Ordinate

Simpson’s Multipliers MomentX product

Total

Moment o of area about the X a axis =

3,284

⎛ Spacing ⎞ ⎛ 1 ⎞ × × Σ ( Mom o ent X product ) ⎝ 3 ⎠ ⎝ 2⎠

Moment o of area about the X a axis =

⎛ 4 ⎞ ⎛ 1⎞ × × 3, 284 = 2,189 1 .33 33 m3 ⎝ 3 ⎠ ⎝ 2⎠

The centre of area from the x axis can be found using: Centre of area from the X axis x =

Moment of area about the X a axxis 2,189.33 = = 7 60 m Area u under de the graph 288.00

The moment of area relative to the y axis can be found using: Ordinate

Offset

Lever Simpson’s Multipliers

MomentY product

0

0

0

1

0

4

8

1

4

32

8

15

2

2

60

12

20

3

4

240

16

17

4

2

136

20

10

5

4

200

24

0

6

1

0

Total

Moment of area about the Y axi a s=

9781408176122_Ch14_5_Rev_txt_prf.indd 571

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

668

m Y (Moment

product )

11/16/2013 2:42:08 AM

572 • Ship Stability, Powering and Resistance

Moment of area about the Y axi a s=

⎛ 4⎞ × 4 × 688 = 3, 532.67 ⎝ 3⎠

The centre of area from the y axis can be found using: Centre o of area from o the Y axis x =

Moment o of area about the Y a axxis 3, 532.67 = = 12.37 m Area u under de the graph 288.00

The inertia at the x axis, relative to the x axis can be found using: Ordinate

Offset

Offset3

Simpson’s Multipliers InertiaX product

0

0

0

1

0

4

8

512

4

2,048

8

15

3,375

2

6,750

12

20

8,000

4

32,000

16

17

4,913

2

9,826

20

10

1,000

4

4,000

24

0

0

1

0

Total

Inertia about the X axis x =

54,624

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 4 ⎞ ⎛ 1⎞ × × 54 , 624 = 24 , 277.33 3m4 ⎝ 3 ⎠ ⎝ 3⎠

The inertia at the y axis, relative to the y axis can be found using: Offset

Lever2

0

0

0

1

0

4

8

1

4

32

8

15

4

2

120

12

20

9

4

720

16

17

16

2

544

20

10

25

4

1,000

24

0

36

1

0

Ordinate

Simpson’s Multipliers InertiaY product

Total

9781408176122_Ch14_5_Rev_txt_prf.indd 572

2,416

11/16/2013 2:42:10 AM

Solutions to Questions • 573

Inertia about the Y axis x =

⎛ Spacing ⎞ × Spacing2 ⎝ 3 ⎠

Inertia about the Y axis x =

(InertiaY

product )

⎛ 4⎞ × 4 2 × 2, 416 = 51, 541.33 m4 ⎝ 3⎠

The inertia at the centre, relative to the x axis can be found using:

(

InertiaCENTRE + Area Distance E

InertiaREMOTE

(

24 , 277.33 = InertiaCENTRE + 288.00 × 7.602 E

)

)

InertiaCENTRE = 7, 642.45 m4 E The inertia at the centre, relative to the y axis can be found using:

(

)

InertiaCENTRE + Area Dstance E

InertiaREMOTE

(

51, 541.33 = InertiaCENTRE + 288.00 × 12.372 E

)

InertiaCENTRE = 7, 472.46 m4 E Q8.7 Ordinate

Offset

Simpson’s Multiplier

Area product 2.5

0

5

0.5

1

5.6

2

2

6

1.5

4

5.25

4

6

4

1.5

6

7

3

2

6

8

2

0.5

1

Total

56.7

Area =

9 21

⎛ Spacing ⎞ × Σ( Area p product d ) ⎝ 3 ⎠

Area =

9781408176122_Ch14_5_Rev_txt_prf.indd 573

11.2

⎛ 2⎞ × 56.7 = 37 37.8 units2 ⎝ 3⎠

11/16/2013 2:42:11 AM

574 • Ship Stability, Powering and Resistance Q8.8 Ordinate

Offset

Simpson’s Multipliers

Area product

0

0

0.5

0

2

4

1

4

4

8

1.5

12

8

15

4

60

12

20

2

40

16

17

4

68

20

10

1.5

15

22

5

2

10

24

0

0.5

0

Total

Area =

209

⎛ Spacing ⎞ × Σ( Area p product d ) ⎝ 3 ⎠

Area =

⎛ 4⎞ × 209 = 278.67 67 m2 ⎝ 3⎠

Q8.9 The area can be found using: X

Y

Simpson’s Multiplier

Area product

0

0

0.5

0

2

1

2

2

4

2

1.5

3

8

3

4

12

12

4

2

8

16

5

4

20

20

5

2

10

24

4

4

16

28

2

1.5

3

30

1

2

2

32

0

0.5

0

Total

76

9781408176122_Ch14_5_Rev_txt_prf.indd 574

11/16/2013 2:42:13 AM

Solutions to Questions • 575

Area =

⎛ Spacing ⎞ × Σ ( Area product ) ⎝ 3 ⎠

Area =

⎛ 4⎞ × 76 = 101.33 m2 ⎝ 3⎠

The moment of area relative to the x axis can be found using: X

Y

Y2

0

0

0

0.5

0

2

1

1

2

2

4

2

4

1.5

6

8

3

9

4

36

12

4

16

2

32

16

5

25

4

100

20

5

25

2

50

24

4

16

4

64

28

2

4

1.5

6

30

1

1

2

2

32

0

0

0.5

0

Simpson’s Multiplier MomentX product

Total

Moment of area about the X a axis =

298

⎛ Spacing ⎞ ⎛ 1 ⎞ × × Σ ( Mom o ent X product ) ⎝ 3 ⎠ ⎝ 2⎠

Moment of area about the X a axis =

⎛ 4 ⎞ ⎛ 1⎞ × × 298 = 198.67 6 m3 ⎝ 3 ⎠ ⎝ 2⎠

The centre of area from the x axis can be found using: Centre o of area from o the X axis x =

9781408176122_Ch14_5_Rev_txt_prf.indd 575

Moment of area about the X a axxis 198.67 = = 1.96 m Area under the graph 101.33

11/16/2013 2:42:14 AM

576 • Ship Stability, Powering and Resistance The moment of area relative to the y axis can be found using: X

Y

Lever

Simpson’s Multiplier

MomentY product

0

0

0

0.5

0

2

1

0.5

2

1

4

2

1

1.5

3

8

3

2

4

24

12

4

3

2

24

16

5

4

4

80

20

5

5

2

50

24

4

6

4

96

28

2

7

1.5

21

30

1

7.5

2

15

32

0

8

0.5

0

Total

Moment o of area about the Y axi a s=

314

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

Moment o of area about the Y axi a s=

m Y (Moment

product )

⎛ 4⎞ × 4 × 314 = 1, 674.67 ⎝ 3⎠

The centre of area from the y axis can be found using: Centre of area from the Y axis x =

Moment o of area about the Y a axxis 1, 674.67 = = 16.53 m Area u under de the graph 101.33

The inertia at the x axis, relative to the x axis can be found using: X

Y

Y3

0

0

0

0.5

0

2

1

1

2

2

4

2

8

1.5

8

3

27

4

108

12

4

64

2

128

16

5

125

4

500

20

5

125

2

250

9781408176122_Ch14_5_Rev_txt_prf.indd 576

Simpson’s Multiplier InertiaX product

12

11/16/2013 2:42:15 AM

Solutions to Questions • 577

Y3

X

Y

Simpson’s Multiplier InertiaX product

24

4

64

4

28

2

8

1.5

30

1

1

2

2

32

0

0

0.5

0

Total

Inertia about the X axis x =

256 12

1,270

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 4 ⎞ ⎛ 1⎞ × × 1, 270 = 564.44 m4 ⎝ 3 ⎠ ⎝ 3⎠

The inertia at the y axis, relative to the y axis can be found using: X

Y

Lever2

Simpson’s Multiplier

0

0

0

0.5

0

2

1

0.25

2

0.5

4

2

1

1.5

3

8

3

4

4

48

12

4

9

2

72

16

5

16

4

320

20

5

25

2

250

24

4

36

4

576

28

2

49

1.5

147

30

1

56.25

2

112.5

32

0

64

0.5

Inertia about the Y axis x =

0

Total

1,529

⎛ Spacing ⎞ × Spacing2 ⎝ 3 ⎠

(InertiaY

Inertia about the Y axis x =

9781408176122_Ch14_5_Rev_txt_prf.indd 577

InertiaY product

product )

⎛ 4⎞ × 4 2 × 1, 529 = 32, 618.67 m4 ⎝ 3⎠

11/16/2013 2:42:16 AM

578 • Ship Stability, Powering and Resistance The inertia at the centre, relative to the x axis can be found using:

(

InertiaCENTRE + Area Distance E

InertiaREMOTE 564.44 44 = I

ti

CENTRE E

(

+ 101.33 × 1.962

)

)

InertiaCENTRE = 175.17 m4 E The inertia at the centre, relative to the y axis can be found using: InertiaREMOTE

(

InertiaCENTRE + Area Distance E

(

32, 618.67 = InertiaCENTRE + 101.33 × 16.532 E

)

)

InertiaCENTRE = 4 , 931.17 m4 E Q8.10 The area can be found using: X

Y

Simpson’s Multiplier

Area product

0

4

0.5

2

3

5

2

6

6

1.5

12

7

4

28

18

7

2

14

24

6

4

24

30

5

2

10

36

4

4

16

42

3

1.5

4.5

45

2

2

4

48

0

0.5

0

Total

121.5

Area =

9781408176122_Ch14_5_Rev_txt_prf.indd 578

10 9

⎛ Spacing ⎞ × Σ ( Area product ) ⎝ 3 ⎠

11/16/2013 2:42:18 AM

Solutions to Questions • 579

Area =

⎛ 6⎞ × 121.5 = 243 m2 ⎝ 3⎠

The moment of area relative to the x axis can be found using: X

Y

Y2

0

4

16

0.5

3

5

25

2

50

6

6

36

1.5

54

12

7

49

4

196

18

7

49

2

98

24

6

36

4

144

30

5

25

2

50

36

4

16

4

64

42

3

9

1.5

13.5

45

2

4

2

8

48

0

0

0.5

0

Simpson’s Multiplier

MomentX product 8

Total

Moment o of area about the X a axis =

685.5

⎛ Spacing ⎞ ⎛ 1 ⎞ × × Σ ( Mom o ent X product ) ⎝ 3 ⎠ ⎝ 2⎠

Moment o of area about the X a axis =

⎛ 6 ⎞ ⎛ 1⎞ × × 685.5 = 6 68 85 5.5 5 m3 ⎝ 3⎠ ⎝ 2⎠

The centre of area from the x axis can be found using: Centre of area from the X axis x =

Moment o of area about the X a axxis 685.5 = = 2 82 m Area under the graph 243

The moment of area relative to the y axis can be found using: X

Y

0

4

0

0.5

0

3

5

0.5

2

5

6

6

1

1.5

9

12

7

2

4

9781408176122_Ch14_5_Rev_txt_prf.indd 579

Lever

Simpson’s Multiplier MomentY product

56

11/16/2013 2:42:19 AM

580 • Ship Stability, Powering and Resistance

X

Y

Lever

Simpson’s Multiplier MomentY product

18

7

3

2

42

24

6

4

4

96

30

5

5

2

50

36

4

6

4

96

42

3

7

1.5

31.5

45

2

7.5

2

30

48

0

8

0.5

0

Total

415.5

Moment o of area about the Y axi a s=

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

Moment o of area about the Y axi a s=

m Y (Moment

product )

⎛ 6⎞ × 6 × 415.5 = 4 , 986 ⎝ 3⎠

The centre of area from the y axis can be found using: Centre of area from the Y axis x =

Moment o of area about the Y a axxis 4 , 986 = = 20.52 m Area u under de the graph 243

The inertia at the x axis, relative to the x axis can be found using: X

Y

Y3

0

4

64

3

5

125

2

250

6

6

216

1.5

324

12

7

343

4

1,372

18

7

343

2

686

24

6

216

4

864

30

5

125

2

250

36

4

64

4

256

42

3

27

1.5

40.5

45

2

8

2

16

48

0

0

0.5

Simpson’s Multiplier InertiaX product 0.5

Total

9781408176122_Ch14_5_Rev_txt_prf.indd 580

32

0 4,090.5

11/16/2013 2:42:20 AM

Solutions to Questions • 581

Inertia about the X axis x =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 6 ⎞ ⎛ 1⎞ × × 4 , 090.5 = 2, 727 m4 ⎝ 3 ⎠ ⎝ 3⎠

The inertia at the y axis, relative to the y axis can be found using: X

Y

Lever2

0

4

0

0.5

0

3

5

0.25

2

2.5

6

6

1

1.5

9

12

7

4

4

112

18

7

9

2

126

24

6

16

4

384

30

5

25

2

250

36

4

36

4

576

42

3

49

1.5

220.5

45

2

56.25

2

225

48

0

64

0.5

Simpson’s Multiplier InertiaY product

0

Total

Inertia about the Y axis x =

1,905

⎛ Spacing ⎞ × Spacing2 ⎝ ⎠ 3

Inertia about the Y axis x =

(InertiaY

product )

⎛ 6⎞ × 62 × 1, 905 = 137,160 m4 ⎝ 3⎠

The inertia at the centre, relative to the x axis can be found using: InertiaREMOTE

( + (243 × 2.82 )

InertiaCENTRE + Area Distance E

2,727 727 = I

ti

)

2

CENTRE E

InertiaCENTRE = 794.57 m4 E The inertia at the centre, relative to the y axis can be found using: InertiaREMOTE

9781408176122_Ch14_5_Rev_txt_prf.indd 581

(

InertiaCENTRE + Area Distance E

)

11/16/2013 2:42:21 AM

582 • Ship Stability, Powering and Resistance 160 = I 137,160

ti

CENTRE E

(

+ 243 × 20.522

)

InertiaCENTRE = 34 , 839.89 m4 E Q8.11 Simpson’s Rule can be used to determine the area under the graph: Ordinate

Offset

Simpson’s Multiplier Area product

0

0

1

0

1

0.115

4

0.460

2

0.392

2

0.784

3

0.702

4

2.808

4

0.933

2

1.866

5

1.008

4

4.032

6

0.904

2

1.808

7

0.66

4

2.640

8

0.353

2

0.706

9

0.097

4

0.388

10

0

1

0

Total

15.492

The vessel has a waterline length of 9.414 m, and a total of 11 measuring points (and therefore ten ‘spacings’ between the points). This allows the spacing to be found: Spacing =

9.414 = 0.941 941 m 10

This allows the area to be found: Area =

Spacing p g × Σ(Totall a area ea p product d ) 3

Area =

9781408176122_Ch14_5_Rev_txt_prf.indd 582

0.914 × 15.492 = 4.861 m3 3

11/16/2013 2:42:24 AM

Solutions to Questions • 583 Note that as the units of the x axis are metres, and the units of the y axis are metres2, therefore the overall units of the area of the graph are these multiplied together, giving units of metres3, or volume. Therefore, the underwater volume of the lifeboat at the design draught is 4.861 m3. This can then be used to determine the displacement of the lifeboat: Δ = ∇ × ρ = 4.861× 1.025 = 4.983 tonnes Q8.12 The area of the section area curve can be found using Simpson’s Rule: Position (m FOAP) Immersed area (m2) Simpson’s Multiplier Area product 0

0.0

1

0.00

12

25.6

4

102.40

24

65.9

2

131.84

36

79.4

4

317.44

48

80.0

2

160.00

60

80.0

4

320.00

72

80.0

2

160.00

84

79.4

4

317.44

96

74.2

2

148.48

108

42.9

4

171.52

120

0.0

1

0.00

Total

Area under the graph =

1,829.12

⎛ Spacing ⎞ ⎛ 12 ⎞ × Σ ( Area product ) = ×1, 829.12 ⎝ ⎠ ⎝ 3⎠ 3

Area u under de the g graph ap = 7, 316.48 m3 ∴∇ = ,

9781408176122_Ch14_5_Rev_txt_prf.indd 583

3

∇ × ρ = Δ ∴ 7, 316.48 × 1.025 = 7, 499..392 392 tonnes

11/16/2013 2:42:25 AM

584 • Ship Stability, Powering and Resistance Q8.13 Simpson’s Rule can be used to determine the area under the graph: Draught (m)

Waterplane area (m2)

Simpson’s Multiplier

Area product

0

0

1

0

0.063

1.797

4

7.188

0.125

3.544

2

7.088

0.188

5.199

4

20.796

0.250

6.754

2

13.508

0.313

8.208

4

32.832

0.375

9.559

2

19.118

0.438

10.807

4

43.228

0.500

11.951

2

23.902

0.563

12.989

4

51.956

0.625

13.92

1

13.92

Total

233.536

The vessel has a draught of 0.625 m, and a total of 11 measuring points (and therefore ten ‘spacings’ between the points). This allows the spacing to be found: Spacing =

0.625 = 0.063 m 10

This allows the area to be found: Area =

Spacing p g × Σ(Totall a area ea p product d ) 3

Area =

0.063 × 233.536 = 4.904 m3 3

Note that as with the section area curve, the units of the x axis are metres, and the units of the y axis are metres2, therefore the overall units of the area of the graph are these multiplied together, giving units of metres3, or volume. The moment of area about the y axis can now be found, which can then be used to find the centre of area of the graph from the y axis:

9781408176122_Ch14_6_Rev_txt_prf.indd 584

11/16/2013 2:40:05 AM

Solutions to Questions • 585

Draught (m)

Waterplane area (m2)

Lever

Simpson’s Multiplier MomentY product

0

0

0

1

0

0.063

1.797

1

4

7.188

0.125

3.544

2

2

14.176

0.188

5.199

3

4

62.388

0.250

6.754

4

2

54.032

0.313

8.208

5

4

164.16

0.375

9.559

6

2

114.708

0.438

10.807

7

4

302.596

0.500

11.951

8

2

191.216

0.563

12.989

9

4

467.604

0.625

13.92

10

1

139.2

Total

1,517.268

Moment of area about the Y axi a s=

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

Moment of area about the Y axi a s=

m Y (Moment

product )

⎛ 0.063 ⎞ × 0.063 × 1, 517.268 26 = 2.007 m4 ⎝ 3 ⎠

Once the moment of area about the y axis has been found, the centre of the graph from the y axis, which is the KB value, can be found using: Centre o of area from o the Y axis x = ∴ KB =

Moment of area about the Y ax axis Area under the graph

2.007 = 0.409 m 4.904

Q8.14 The moment of area of the draught waterplane area curve can be found about the y axis: Draught (m)

Waterplane area (m)2

Lever

Simpson’s Multiplier MomentY product

0.00

0.0

0

1

0.90

1,807.2

1

4

7,228.8

1.80

2,620.8

2

2

10,483.2

2.70

2,808.0

3

4

33,696

9781408176122_Ch14_6_Rev_txt_prf.indd 585

0

11/16/2013 2:40:06 AM

586 • Ship Stability, Powering and Resistance

Draught (m)

Waterplane area (m)2

Lever

Simpson’s Multiplier MomentY product

3.60

2,880.0

4

2

23,040

4.50

2,952.0

5

4

59,040

5.40

3,024.0

6

2

36,288

6.30

3,096.0

7

4

86,688

7.20

3,168.0

8

2

50,688

8.10

3,240.0

9

4

116,640

9.00

3,312.0

10

1

33,120

Total

456,912

Moment of area about the Y axi a s=

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

Moment of area about the Y axi a s=

m Y (Moment

product )

⎛ 0 90 ⎞ × 0.90 × 456 , 912 = 123 2 , 366.24 m3 ⎝ 3 ⎠

This allows the centre of area from the y axis to be found, which is the KB value: Centroid from the Y axis x = KB =

Moment o of area about the Y axi a s Area

To use this formula, we need to know the area of the graph. We know that the area of the draught waterplane area curve is also equal to the underwater volume, which is given in the question as 24,693.12 m3. Therefore: Centroid from the Y axis x = KB =

123, 366.24 = 5 00 m 24 , 693.12

Q8.15 The underwater volume can be found from the displacement: ∇ × ρ = Δ ∴∇ =

7, 000 = 6 , 829.27 m3 1.025

This can be used in Morrish’s formula: KB =

⎞ ⎞ 1 ⎛ 5 × 7 ⎛ 6 , 829.27 ⎞ ⎞ 1 ⎛ 5D ⎛ ∇ ×⎜ −⎜ = × −⎜ 32 m ⎟ = 4.32 ⎝ 1, 500 3 ⎝ 2 ⎝ Waterplane area ⎟⎠ ⎟⎠ 3 ⎜⎝ 2 0 ⎠ ⎟⎠

9781408176122_Ch14_6_Rev_txt_prf.indd 586

11/16/2013 2:40:07 AM

Solutions to Questions • 587 Q8.16 Simpson’s Rule can be used to determine the inertia of the waterline half beam curve, relative to the x axis: Ordinate

Half beam

Half beam3

Simpson’s Multiplier InertiaX product

0

0

0

1

0

1

0.366

0.049

4

0.196

2

0.712

0.361

2

0.722

3

0.976

0.93

4

3.719

4

1.129

1.439

2

2.878

5

1.179

1.639

4

6.555

6

1.108

1.36

2

2.721

7

0.935

0.817

4

3.27

8

0.661

0.289

2

0.578

9

0.305

0.028

4

0.113

10

0

0

1

0

Total

20.752

The vessel has a waterline length of 9.414 m, and a total of 11 measuring points (and therefore ten ‘spacings’ between the points). This allows the spacing to be found: Spacing =

9.414 = 0.941 941 m 10

This allows the inertia about the x axis to be found: Inertia about the X axis x =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 0.941⎞ ⎛ 1⎞ × × 20.752 = 2.170 m4 ⎝ 3 ⎠ ⎝ 3⎠

This value is based on the half beams, so it needs to be doubled to take into account the entire waterplane: Transverse inertia e t a = 2.170 × 2 = 4.340 m4 This can be used with the volume to determine BM: BM =

9781408176122_Ch14_6_Rev_txt_prf.indd 587

Inertia 4.340 = = 0.885 885 m ∇ 4.904

11/16/2013 2:40:09 AM

588 • Ship Stability, Powering and Resistance Q8.17 The transverse inertia of the waterplane area can be found from the waterline half beam curve: Position (m FOAP)

Waterplane half beam (m)

Offset3 (m3)

Simpson’s Multiplier

InertiaX product

0

0.00

0

1

0

23

9.45

843.91

4

3,375.64

45

12.90

2,146.69

2

4,293.38

68

13.80

2,628.07

4

10,512.28

90

13.80

2,628.07

2

5,256.14

113

13.80

2,628.07

4

10,512.28

135

13.80

2,628.07

2

5,256.14

158

13.80

2,628.07

4

10,512.28

180

12.75

2,072.67

2

4,145.34

203

8.85

693.15

4

2,772.6

225

0.00

0

1 Total

0 56,636.08

Inertia about the X axis x =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 23 ⎞ ⎛ 1⎞ × × 56 , 636.08 = 141,5 590.2 2 m4 ⎝ 3 ⎠ ⎝ 3⎠

This value is for the half beams, so should be doubled to take into account the entire waterplane area: Transverse waterplane ate p a e inertia = 141, 590.2 × 2 = 283,180.4 m4 This can be divided by the volume (which is given in the question) to give the BM value: BM =

9781408176122_Ch14_6_Rev_txt_prf.indd 588

I 283,180.4 = = 5.87 m ∇ 48 , 228.75

11/16/2013 2:40:10 AM

Solutions to Questions • 589 Q8.18 The area of the section area curve can be found using Simpson’s Rule: Position (m FOAP)

Immersed area (m2)

Simpson’s Multiplier

0

0.0

1

0

15

40.0

4

160

30

103.0

2

206

45

124.0

4

496

60

125.0

2

250

75

125.0

4

500

90

125.0

2

250

105

124.0

4

496

120

116.0

2

232

135

67.0

4

268

150

0.0

1

0

Total

Area u under de the g graph ap = ∴∇ =

Area product

2,858

⎛ Spacing ⎞ ⎛ 15 ⎞ × Σ ( Area ea product ) = × 2, 858 = 14 , 290 m3 ⎝ ⎠ ⎝ 3⎠ 3 3

,

∇ × ρ = Δ ∴14 , 290 90 × 1.0 025 5 = 14 , 6 647.25 5 tonnes

The moment of area of the draught waterplane area curve can be found about the y axis: Draught (m)

Waterplane area (m2)

Lever

Simpson’s Multiplier

0.00

0.0

0

1

0

0.75

1,255.0

1

4

5,020

1.50

1,820.0

2

2

7,280

2.25

1,950.0

3

4

23,400

3.00

2,000.0

4

2

16,000

3.75

2,050.0

5

4

41,000

4.50

2,100.0

6

2

25,200

5.25

2,150.0

7

4

60,200

9781408176122_Ch14_6_Rev_txt_prf.indd 589

MomentY product

11/16/2013 2:40:11 AM

590 • Ship Stability, Powering and Resistance

Draught (m)

Waterplane area (m2)

Lever

Simpson’s Multiplier

MomentY product

6.00

2,200.0

8

2

35,200

6.75

2,250.0

9

4

81,000

7.50

2,300.0

10

1

23,000

Total

317,300

Moment of area about the Y axi a s=

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

Moment of area about the Y axi a s=

m Y (Moment

product )

⎛ 0 75 ⎞ × 0.75 × 317, 300 = 59 , 493.75 m3 ⎝ 3 ⎠

This allows the centre of area from the y axis to be found, which is the KB value: Centroid from the Y axis x = KB =

Moment o of area about the Y axi a s Area

To use this formula, we need to know the area of the graph. This can be found via Simpson’s Rule, however we know that the area of the draught waterplane area curve is also equal to the underwater volume, which we have previously found from the section area curve to be 14,290 m3. Therefore: Centroid from the Y axis x = KB =

59 , 493.75 = 4.16 m 14 , 290

The transverse inertia of the waterplane area can be found from the waterline half beam curve: Waterplane half beam (m)

Offset3 (m3)

Simpson’s Multiplier

0

0.00

0

1

15

6.30

250.05

4

1,000.2

30

8.60

636.06

2

1,272.12

45

9.20

778.69

4

3,114.76

60

9.20

778.69

2

1,557.38

75

9.20

778.69

4

3,114.76

90

9.20

778.69

2

1,557.38

Position (m FOAP)

9781408176122_Ch14_6_Rev_txt_prf.indd 590

InertiaX product 0

11/16/2013 2:40:12 AM

Solutions to Questions • 591

Waterplane half beam (m)

Offset3 (m3)

Simpson’s Multiplier

InertiaX product

105

9.20

778.69

4

3,114.76

120

8.50

614.13

2

1,228.26

135

5.90

205.38

4

821.52

150

0.00

0

1

0

Position (m FOAP)

Total

16,781.14

Inertia about the X axis x =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 15 ⎞ ⎛ 1⎞ × × 16 , 781.14 = 27, 968 96 .57 m4 ⎝ 3 ⎠ ⎝ 3⎠

This value is for the half beams, so should be doubled to take into account the entire waterplane area: Transverse waterplane inertia = 27, 968.57 × 2 = 55, 937.14 m4 This can be divided by the volume (which we have already found) to give the BM value: BM =

I 55, 937.14 = = 3.91 91 m ∇ 14 , 290

Finally, GM can be found: GM = KB + BM − KG = 4.16 + 3.91 − 7.07 = 1.00 m Q8.19 The area of the section area curve can be found using Simpson’s Rule: Position (m FOAP)

Immersed area (m2)

Simpson’s Multiplier

Area product

0

0.0

1

0

18

57.6

4

230.4

36

148.3

2

296.64

54

178.6

4

714.24

72

180.0

2

360

9781408176122_Ch14_6_Rev_txt_prf.indd 591

11/16/2013 2:40:13 AM

592 • Ship Stability, Powering and Resistance

Position (m FOAP)

Immersed area (m2)

Simpson’s Multiplier

90

180.0

4

720

108

180.0

2

360

126

178.6

4

714.24

144

167.0

2

334.08

162

96.5

4

385.92

180

0.0

1

0

Total

4,115.52

Area under the graph =

∴∇ =

Area product

⎛ Spacing ⎞ ⎛ 18 ⎞ × Σ ( Area product ) = × 4 ,115.52 = 24 , 693.12 m3 ⎝ ⎝ 3⎠ 3 ⎠

,

3

, ∇ × ρ = Δ ∴ 24 , 693.12 ×1 × 1.025 = 25, 310.448 tonnes

The moment of area of the draught waterplane area curve can be found about the y axis: Draught (m)

Waterplane Lever area (m2)

Simpson’s Multiplier

MomentY product

0.00

0.0

0

1

0.90

1,807.2

1

4

7,228.8

1.80

2,620.8

2

2

10,483.2

2.70

2,808.0

3

4

33,696

3.60

2,880.0

4

2

23,040

4.50

2,952.0

5

4

59,040

5.40

3,024.0

6

2

36,288

6.30

3,096.0

7

4

86,688

7.20

3,168.0

8

2

50,688

8.10

3,240.0

9

4

116,640

9.00

3,312.0

10

1

33,120

Total

456,912

Moment o of area about the Y axi a s=

9781408176122_Ch14_6_Rev_txt_prf.indd 592

⎛ Spacing ⎞ × Spacing ⎝ 3 ⎠

0

m Y (Moment

product )

11/16/2013 2:40:14 AM

Solutions to Questions • 593

Moment of area about the Y a axis =

⎛ 0 90 ⎞ × 0.90 × 456 , 912 = 123 2 , 366.24 m3 ⎝ 3 ⎠

This allows the centre of area from the y axis to be found, which is the KB value: Centroid from the Y axis x = KB =

Moment o of area about the Y axi a s Area

To use this formula, we need to know the area of the graph. This can be found via Simpson’s Rule, however we know that the area of the draught waterplane area curve is also equal to the underwater volume, which we have previously found from the section area curve to be 24,693.12 m3. Therefore: Centroid from o the Y axis x = KB =

123, 366.24 = 5 00 m 24 , 693.12

The transverse inertia of the waterplane area can be found from the waterplane half beam curve: Position (m FOAP)

Waterplane half beam (m)

Offset3 (m3)

Simpson’s InertiaX product Multiplier

0

0.00

0

1

0

18

7.56

432.08

4

1,728.32

36

10.32

1,099.1

2

2,198.2

54

11.04

1,345.57

4

5,382.28

72

11.04

1,345.57

2

2,691.14

90

11.04

1,345.57

4

5,382.28

108

11.04

1,345.57

2

2,691.14

126

11.04

1,345.57

4

5,382.28

144

10.20

1,061.21

2

2,122.42

162

7.08

354.89

4

1,419.56

180

0.00

0

1

0

Total

28,997.62

Inertia about the X axis x =

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 18 ⎞ ⎛ 1⎞ × × 28 , 997.62 = 57, 995.24 m4 ⎝ 3 ⎠ ⎝ 3⎠

9781408176122_Ch14_6_Rev_txt_prf.indd 593

11/16/2013 2:40:16 AM

594 • Ship Stability, Powering and Resistance This value is for the half beams, so should be doubled to take into account the entire waterplane area: Transverse waterplane inertia = 57, 995.24 × 2 = 115, 990.48 m4 This can be divided by the volume (which we have already found) to give the BM value: BM =

I 115, 990.48 = = 4.70 70 m ∇ 24 , 693.12

Finally, GM can be found: GM = KB + BM − KG = 5.00 + 4.70 − 9.55 = 0.15 m Q8.20 The waterplane area can be determined: Ordinate

Half beam (m)

Simpson’s Multiplier

Area product

0 (AP)

0

1

0

1

0.366

4

1.464

2

0.712

2

1.424

3

0.976

4

3.904

4

1.129

2

2.258

5

1.179

4

4.716

6

1.108

2

2.216

7

0.935

4

3.74

8

0.661

2

1.322

9

0.305

4

1.22

10 (FP)

0

1

0

Total

Area =

Spacing p g × Σ Area product 3

Area =

0.941 × 22.264 = 6.983 m2 3

22.264

This must be doubled to account for the entire waterplane area: Waterplane area a ea = 6.983 × 2 = 13 13.996 m2

9781408176122_Ch14_6_Rev_txt_prf.indd 594

11/16/2013 2:40:17 AM

Solutions to Questions • 595 This allows the TPC to be found: W rplane a Water area ea × 0.01× 01× ρ = 13 13.996 × 0.01× 1.025 = 0.143 t/cm

TPC Q8.21

The waterplane area can be determined: Position (m FOAP)

Waterplane half beam (m)

Simpson’s Multiplier

Area product

0

0

1

0

15

6.3

4

25.2

30

8.6

2

17.2

45

9.2

4

36.8

60

9.2

2

18.4

75

9.2

4

36.8

90

9.2

2

18.4

105

9.2

4

36.8

120

8.5

2

17

135

5.9

4

23.6

150

0

1

0

Total

230.2

Area =

Spacing p g × Σ Area product 3

Area =

15 × 230.20 = 1 1151 , .00 m2 3

This must be doubled to account for the entire waterplane area: Waterplane area = 1,151.00 × 2 = 2, 302.00 m2 This allows the TPC to be found: TPC

W rplane a Water area ea × 0.01× 01× ρ = 2, 302.00 × 0.01× 1.025 = 23.60 t/cm

9781408176122_Ch14_6_Rev_txt_prf.indd 595

11/16/2013 2:40:18 AM

596 • Ship Stability, Powering and Resistance Q8.22 The first stage is to plot the shape of the tank on an x–y graph. The 3 m wide aft transverse bulkhead is plotted on the y axis, and the 5 m long longitudinal bulkhead is plotted on the x axis. The 0.10 m wide forward transverse bulkhead forms the right hand vertical side of the graph, and the hull plating shape forms the curve of the graph: 3

Y

2

1

0 0

1

2

3

4

5

X

The first stage is to determine the area. This is calculated using Simpson’s Rule as before: Ordinate

Offset

Simpson’s Multiplier

Area product

0

3

1

3

1.25

2.6

4

10.4

2.5

1.9

2

3.8

3.75

1.1

4

4.4

5

0.1

1

0.1

Total

21.7

Area =

⎛ Spacing ⎞ ⎛ 1.25 ⎞ × Σ ( Area product ) = × 21.7 = 9.04 m2 ⎝ ⎝ 3 ⎠ 3 ⎠

9781408176122_Ch14_6_Rev_txt_prf.indd 596

11/16/2013 2:40:20 AM

Solutions to Questions • 597 The next stage is to determine the first moment of area about the x axis: Ordinate

Offset2 Simpson’s Multiplier MomentX product

0

9

1

1.25

6.76

4

27.04

2.5

3.61

2

7.22

3.75

1.21

4

4.84

5

0.01

1

0.01

Total

48.11

Moment o of area about the X a axis =

9

⎛ Spacing ⎞ ⎛ 1 ⎞ × × Σ ( Mom o ent X product ) ⎝ 3 ⎠ ⎝ 2⎠

Moment of area about the X a axis =

⎛ 1.25 ⎞ ⎛ 1 ⎞ × × 48.11 = 10.02 m3 ⎝ 3 ⎠ ⎝ 2⎠

This allows the centre of area from the x axis to be found: Centre of area from the X axis x =

Moment o of area about the X a axxis 10.02 = = 1.11 m Area under the graph 9 04

The inertia about the x axis can be found: Ordinate

Offset3 Simpson’s Multiplier

0

27

1

27

1.25

17.58

4

70.32

2.5

6.86

2

13.72

3.75

1.33

4

5.32

5

0

1

0

Total

116.36

Inertia about the X axis x =

InertiaX product

⎛ Spacing ⎞ ⎛ 1⎞ × × Σ (InertiaX product ) ⎝ 3 ⎠ ⎝ 3⎠

Inertia about the X axis x =

⎛ 1.25 ⎞ ⎛ 1⎞ × × 116.36 = 16.16 m4 ⎝ 3 ⎠ ⎝ 3⎠

Finally, the parallel axes theory can be used to determine the inertia through an axis parallel to the x axis but through the centroid: InertiaREMOTE

9781408176122_Ch14_6_Rev_txt_prf.indd 597

InertiaCENTRE + ( Area Distance 2 ) E

11/16/2013 2:40:20 AM

598 • Ship Stability, Powering and Resistance 16 = I 16.16

(

)

tiaCENTRE + 9.04 1.11 112 ∴ E

CENTRE E

= 5.02 m4

This value is the free surface moment of the tank, and is used in the same way as the values in the hydrostatic databook. Q8.23 The plan area of the tank can be found: Ordinate

Offset Simpson’s Multiplier

Area product

0.00

2.50

1

2.50

2.50

2.40

4

9.60

5.00

2.30

2

4.60

7.50

2.20

4

8.80

10.00

2.00

1

2.00

Total

27.50

Area =

Spacing p g 2 50 × Σ Area p product oduct = × 27.50 = 22.9 92 m2 3 3

The plan area can be multiplied by the depth of the fluid to determine the volume of fluid, which can be multiplied by the density of the fluid to determine the mass of ballast: Volume = Plan a area ea × Dept e h = 22.92 × 3.00 = 68 76 m3 Volume × ρ = Mass ∴ 68.76 × 1.025 = 70.48 tonnes The KG of the ballast can be found: KG = Distance from keel to centre of mass = 2 00 +

3 00 = 3 50 m 2

The centre of area of the tank from the longitudinal bulkhead can be found: Offset

Offset2

Simpson’s Multiplier

0

2.50

6.30

1

6.30

2.5

2.40

5.80

4

23.00

5

2.30

5.30

2

10.60

7.5

2.20

4.80

4

19.40

10

2.00

8.00

1

8.00

Total

67.20

Ordinate

9781408176122_Ch14_6_Rev_txt_prf.indd 598

MomentX product

11/16/2013 2:40:22 AM

Solutions to Questions • 599

Moment about the X a axis =

⎛ Spacing ⎞ 1 × × Σ Moment X product ⎝ 3 ⎠ 2

Moment about the X a axis = Centre o of Area ffro om the X a axis =

⎛ 2 5⎞ 1 × × 67.20 = 28.01 m3 ⎝ 3 ⎠ 2

Moment about the X a axis 28.01 = = 1.22 22 m Area 22.92

Therefore, the centre of the tank from the centreline of the ship can be found: Tank centre from centreline = 7.00 + 1.22 = 8 22 m The inertia of the area about the x axis can be found: Offset

Offset3

Simpson’s Multiplier

InertiaX product

0

2.50

15.60

1

15.60

2.5

2.40

13.80

4

55.30

5

2.30

12.20

2

24.30

7.5

2.20

10.60

4

42.60

10

2.00

8.00

1

8.00

Ordinate

Total

Inertia about the X axis x =

145.80

⎛ Spacing ⎞ 1 × × Σ (InertiaX product c ) ⎝ 3 ⎠ 3

Inertia about the X axis x =

⎛ 2 5⎞ 1 × × 145.80 = 40.51 m4 ⎝ 3 ⎠ 3

The parallel axes formula can then be used to find the inertia at the centre of the planform area: InertiaREMOTE 40.51 51 = I

InertiaCENTRE + ( Area Distance 2 ) E

(

tiaCENTRE + 22.92 × 1.22 222 E

)

CENTRE E

6.40 m4

This is the free surface moment of the tank. It can be multiplied by the ballast density, and used in the loading table to account for the free surface effect: Corrected FSM FS = 6.40 × 1.025 = 6 56 tonne metres

9781408176122_Ch14_6_Rev_txt_prf.indd 599

11/16/2013 2:40:24 AM

600 • Ship Stability, Powering and Resistance A loading table can then be used to find the new fluid KG: Item

Mass (tonnes)

Ship

KG (m)

5,000

Ballast

70.48

Ballast FSM

0

Total

Moment (tonne metres)

7.00

35,000.00

3.50

246.68

0

6.56

5,070.48

KG =

35,253.24

Moment 35, 253.24 = = 6.95 m Mass 5, 070.48

This allows GM to be found: GM = KM − KG = 8.00 − 6.95 = 1.05 m A loading table can be used to determine the TCG (note the TCG is negative as the tank is to starboard of the centreline): Item

Mass (tonnes) TCG (m)

Ship

5,000

Ballast

70.48

Total

Moment (tonne metres)

0.00

0.00

–8.22

–579.35

5,070.48

TCG =

579.35

Moment −579.35 = = −0.11 11 m Mass 5, 070.48

Finally, this allows the list to be found: tanθ

TCG ⎛ 0.11⎞ θ tan−1 − ∴θ = −6.0 degrees (starboard) d ⎝ 1.05 ⎠ GM

Q8.24 The remote inertia through the parallel can be found: InertiaRemote

InertiaCentroid + ( Area Distance 2 )

InertiaRemote = 8 , 000 +

9781408176122_Ch14_6_Rev_txt_prf.indd 600

(

×

) = 88, 000 m

4

11/16/2013 2:40:26 AM

Solutions to Questions • 601 Q8.25 The area can be found: t Beam = 60 × 20 = 1,200 200 m2 Area Length The transverse inertia through the centre line can be found: Inertia =

LB 3 60 × 203 = = 40 , 000 m4 12 12

The remote inertia through axis XX can be found: InertiaCentroid + ( Area Distance 2 )

InertiaRemote

InertiaRemote = 40 , 000 +

(

,120 , 000 m ) = 1120

×

4

Q8.26 The KB of one hull can be found: KB =

D 3 = = 1.50 m 2 2

Note that as described previously, the overall KB of both hulls will be the same. The underwater volume of one hull can be found: ∇ = L × B × D = 40 × 4 × 3 = 480 m3 The waterplane area of one hull can be found: Area L × B = 40 × 4 = 160 m2 The transverse inertia of the waterplane area of one hull about the centreline of the hull can be found: I=

LB 3 40 × 4 3 = = 213.34 34 m4 12 12

This can be converted to be the transverse inertia of the waterplane area of one hull about the centreline of the ship: InertiaREMOTE

InertiaREMOTE = 213.3 34 +

9781408176122_Ch14_6_Rev_txt_prf.indd 601

(

)

×

) = 5, 973.34 m

InertiaCENTRO Distance E ID + Area

(

4

11/16/2013 2:40:27 AM

602 • Ship Stability, Powering and Resistance This can be used to determine the BM: BM =

Inertia 5, 973.34 × 2 = = 12.45 m Volume 480 × 2

Note that we have doubled the transverse inertia of the waterplane area of one hull about the centreline of the ship to obtain the overall value for both hulls, and doubled the underwater volume of one hull to obtain the overall underwater volume. Finally, GM can be found: GM = KB + BM − KG = 1.50 + 12.45 − 5.00 = 8.95 m Q8.27 The waterplane area can be found: Station

Waterline half beam (m)

Simpson’s Multiplier

Area product

0

0.1

1

0.1

1

1.0

4

4

2

1.7

2

3.4

3

2.0

4

8

4

1.5

2

3

5

0.7

4

2.8

6

0.0

1

0

Total

21.3

⎛ 60 ⎞ Spacing p g ⎜ ⎟ Area = × Σ ( Area product ) = ⎜ 6 ⎟ × 21.3 = 71 71.0 m2 3 ⎜⎝ 3 ⎟⎠ This value is for half of the waterplane area of one hull, therefore the total waterplane area of one hull can be found: Total w water aterp pla anee area o of one hull u = 71.0 × 2 = 142.0 m2

9781408176122_Ch14_6_Rev_txt_prf.indd 602

11/16/2013 2:40:30 AM

Solutions to Questions • 603 The transverse inertia about the hull centreline can be found: Waterline half beam3

Simpson’s Multiplier

InertiaX product

0

0.001

1

0.001

1

1

4

4

2

4.913

2

9.826

3

8

4

4

3.375

2

6.75

5

0.343

4

1.372

6

0

1

0

Station

32

Total

InertiaX =

53.95

Spacing p g 1 × × Σ (InertiaX product ) 3 3

60 1 InertiaX = 6 × × 53.95 = 59.94 m4 3 3 This value is for half of the transverse inertia of the waterplane area of one hull (measured about the hull centreline), therefore the total transverse inertia of the waterplane area of one hull (measured about the hull centreline) can be found: Total transverse w waterrplane area of one hull about the hu ull centreline = 59.94 × 2 = 119.89 m2 The transverse inertia of the waterplane area of one hull about the ship centreline can be found: InertiaREMOTE

(

InertiaCENTRO Distance E ID + Area

)

2 ⎛ ⎛ 10 ⎞ ⎞ InertiaREMOTE = 119.89 + ⎜ 142.0 × = 3, 669.89 m4 ⎝ 2 ⎠ ⎟⎠ ⎝

This value is for one hull (about the centreline of the ship), so it must be doubled to account for both hulls: Total transverse waterpl r ane area of two hulls about the h sship p centreline = 3, 669.89 × 2 = 7, 339.78 m2

9781408176122_Ch14_6_Rev_txt_prf.indd 603

11/16/2013 2:40:31 AM

604 • Ship Stability, Powering and Resistance This allows BM to be found: BM =

Inertia 7, 339.78 = = 13.85 85 m ∇ 530

This allows GM to be found: GM = KB + BM − KG = 1+ 13.85 − 6 = 8.85 m Q8.28 The waterplane area can be determined for each sponson: Waterplane a area ea = l b = 15 × 1.5 = 22.5 m2 The centre of the waterplane area from the centreline of the ship can be found: Sponson w water aterp pla anee centre =

⎛ 10 ⎞ ⎛ 1 5 ⎞ + = 5.75 m from the centreline ⎝ 2⎠ ⎝ 2⎠

The additional underwater volume for each sponson can be found: ∇ SPONSON = Sponson waterpl aterp a anee area × Sponson drau draught r ght = 22.5 × 2 = 45 45 m3 The total underwater volume including the sponsons can be found: ∇ = ∇HULL + ∇ SPONSONS = (50 × 10 × 2) + 45 + 45 = 1,090 090 m3 The transverse inertia of each sponson waterplane, measured about the centreline of the sponson, can be found: I=

LB 3 = 12

(

×

) = 4.22 m

4

12

This can be converted (using the parallel axes formula) to be about the centreline of the ship – note that for the sponson waterplane, this will be the remote inertia, as it does not go through the centre of the sponson waterplane area:

(

)

IREMOTE = ICENTRO Distance E ID + Area IREMOTE = 4.22 2+

(

×

) = 748.13 m

4

The transverse inertia of the hull waterplane measured about the centreline can be found: I=

9781408176122_Ch14_6_Rev_txt_prf.indd 604

LB 3 = 12

(

× 12

) = 4 ,166.67 m

4

11/16/2013 2:40:32 AM

Solutions to Questions • 605 The total transverse inertia of the main hull and the two sponsons can be found by adding up the transverse inertia of the components: Total transverse inertia = Hull transverse inertia Spo S nsons transverse t inertia Total ttransverse a s e se inertia = 4 ,166 66.67 + 748.13 + 748 8.13 = 5, 662.92 m4 This allows the overall BM to be found: BM =

I 5, 662.92 = = 5.20 20 m ∇ 1, 090

The KB can be found: KB =

D 2 00 = = 1.00 m 2 2

Therefore, GM can be found: GM = KB + BM − KG = 1.00 + 5.20 0 − 5.00 = 1.20 20 m Q8.29 The first stage in determining the effect of the sponsons on BM is to determine the waterplane area of each sponson. This can be found using Simpson’s Rule: Ordinate

Offset

Simpson’s Multiplier

Area product

0

0.00

1

0.00

1

1.00

4

4.00

2

1.50

2

3.00

3

1.00

4

4.00

4

0.00

1

0.00

Total

11.00

Area =

Spacing p g 1 × Σ ( Area product ) = × 11.00 .00 = 3.6 67 m2 3 3

As the sponson can be assumed to be a constant planform, the underwater volume of the sponson can be found (the question states that the vessel floats with the waterline at half of the depth of the sponson): ∇ Sponson = 3 67 ×

9781408176122_Ch14_6_Rev_txt_prf.indd 605

⎛ 2.10 ⎞ = 3.85 m3 ⎝ 2 ⎠

11/16/2013 2:40:35 AM

606 • Ship Stability, Powering and Resistance The next stage is to determine the centre of area of the sponson waterplane from the side of the ship. This can be found by determining the moment of the waterplane area about the x axis of the waterplane area: Ordinate

Offset2 Simpson’s Multiplier MomentX product

0

0.00

1

0.00

1

1.00

4

4.00

2

2.25

2

4.50

3

1.00

4

4.00

4

0.00

1

0.00

Total

12.50

Moment about the X a axis =

Spacing p g 1 1 1 × × Σ ( Moment o e t X product ) = × ×12 12.50 = 2.08 m3 3 2 3 2

This allows the centre of the sponson waterplane from the x axis to be found: Centre o of area from o the X axis x =

Moment about the X a axis 2 08 = = 0.57 57 m Area 3 67

The inertia of the waterplane about the x axis can be found: Ordinate

Offset3 Simpson’s Multiplier

InertiaX product

0

0.00

1

0.00

1

1.00

4

4.00

2

3.38

2

6.75

3

1.00

4

4.00

4

0.00

1

0.00

Total

14.75

Inertia about the x axis x =

Spacing p g 1 × × Σ (InertiaX product ) 3 3

Inertia about the x axis x =

1 1 × × 14.75 = 1.64 m4 3 3

This is the remote value measured through the x axis. The parallel axes formula can be used to determine the inertia through an axis parallel to the x axis but running through the centre of the shape:

9781408176122_Ch14_6_Rev_txt_prf.indd 606

11/16/2013 2:40:37 AM

Solutions to Questions • 607 InertiaREMOTE 1.64

CENTRO E ID

InertiaCENTRO Distance 2 ) E ID + ( Area (3.67 × 0.572 )

CENTRO E ID

0.46 m4

As the vessel will roll about the centreline of the ship, the inertia of the waterplane area must be found relative to the centreline of the ship. This is a remote axis, away from the centre of the sponson waterplane area: InertiaREMOTE

InertiaCENTRO Distance 2 ) E ID + ( Area

2 ⎛ ⎛7 ⎞ ⎞ InertiaREMOTE = 0.46 6 + ⎜ 3.67 × + 0.57 ⎟ = 61.24 m4 ⎝2 ⎠ ⎠ ⎝

The new total inertia and total underwater volume can be used to determine the new BM: BM =

Inertia 857.43 + 61.24 + 62.14 = = 1.54 m 645.75 ∇ + (3 85 × 2) 1.025

Q8.30 The first stage in determining the effect of the sponsons on BM is to determine the original waterplane inertia of the vessel. This can be found from the displacement and the original BM: BM =

Inertia Inertia ∴11.39 = ∴ Inertia = 33, 336.59 m4 3, 000 ∇ 1.025

The next stage is to calculate waterplane area of each sponson. This can be found using Simpson’s Rule: Ordinate

Offset

Simpson’s Multiplier

Area product

0

0.00

1

0.00

1

2.00

4

8.00

2

3.00

2

6.00

3

2.00

4

8.00

4

0.00

1

0.00

Total

22.00

9781408176122_Ch14_6_Rev_txt_prf.indd 607

11/16/2013 2:40:38 AM

608 • Ship Stability, Powering and Resistance

Area =

Spacing p g 4 × Σ ( Area p product oduct ) = × 22.00 = 29 9.33 m2 3 3

As the sponson can be assumed to be a constant planform, the underwater volume of the sponson can be found (the question states that the vessel floats with the waterline at half of the depth of the sponson): ∇ Sponson = 29.33 ×

⎛ 4⎞ = 58.66 m3 ⎝ 2⎠

The next stage is to determine the centre of area of the sponson waterplane from the side of the ship. This can be found by determining the moment of the waterplane area about the x axis of the waterplane area: Offset2

Simpson’s Multiplier

MomentX product

0

0.00

1

0.00

1

4.00

4

16.00

2

9.00

2

18.00

3

4.00

4

16.00

4

0.00

1

0.00

Total

50.00

Ordinate

Spacing p g 1 × × Σ ( Moment X product ) 3 2 4 1 = × × 50.00 = 33.33 m3 3 2

Moment about the X a axis =

This allows the centre of the sponson waterplane from the x axis to be found: Centre of area from the X axis x =

Moment about the X a axis 33.33 = = 1.14 m Area 29.33

The inertia of the waterplane about the x axis can be found: Offset3

Simpson’s Multiplier

InertiaX product

0

0.00

1

0.00

1

8.00

4

32.00

2

27.00

2

54.00

3

8.00

4

32.00

4

0.00

1

0.00

Ordinate

Total

9781408176122_Ch14_6_Rev_txt_prf.indd 608

118.00

11/16/2013 2:40:40 AM

Solutions to Questions • 609

Inertia about the x axis x =

Spacing p g 1 × × Σ (InertiaX product ) 3 3

Inertia about the x axis x =

4 1 × × 118.00 = 52.44 m4 3 3

This is the remote value measured through the x axis. The parallel axes formula can be used to determine the inertia through an axis parallel to the x axis but running through the centre of the shape: InertiaCENTRO Distance 2 ) E ID + ( Area

InertiaREMOTE 52.44

CENTRO E ID

(29.33 ×1 1.14 2 )

CENTRO E ID

14.32 m4

As the vessel will roll about the centreline of the ship, the inertia of the waterplane area must be found relative to the centreline of the ship. This is a remote axis away from the centre of the sponson waterplane area: InertiaREMOTE

InertiaCENTRO Distance 2 ) E ID + ( Area

2 ⎛ ⎛ 20 ⎞ ⎞ InertiaREMOTE = 14.3 32 + ⎜ 29.33 × + 1.14 ⎟ = 1,120.05 05 m4 ⎝ 2 ⎠ ⎠ ⎝

The new total inertia and total underwater volume can be used to determine the new BM: BM =

Inertia 33, 336.59 + 1120 ,120.05 + 1120 ,120.05 = = 11.69 m 3, 000 ∇ + (58.66 × 2) 1.025

Q8.31 The spacing between the stations can be determined: Spacing =

60 = 10.00 00 m 6

The section area data can be used to determine the underwater volume: Station

Section area (m2)

0

1.57

Simpson’s Multiplier 0.5

Area product 0.785

0.5

25.13

2

50.26

1

56.55

1.5

84.825

2

76.97

4

307.88

3

76.97

2

153.94

9781408176122_Ch14_6_Rev_txt_prf.indd 609

11/16/2013 2:40:41 AM

610 • Ship Stability, Powering and Resistance

Section area (m2)

Station

Simpson’s Multiplier

Area product

4

56.55

4

226.2

5

25.13

1.5

37.695 12.56

5.5

6.28

2

6

0.00

0.5

0

Total

873.36

Area =

S ⎛ 10.00 ⎞ × Σ ( Area product ) = × 873.36 = 2, 911.2 m3 ⎝ 3 ⎠ 3

Therefore, the underwater volume is 2,911.2 m3. The section area data can be used to determine the LCB by finding the moment of area of the section area curve about the Y axis, and therefore the centre of area from the Y axis: Station 0

Section area (m2) 1.57

Lever

Simpson’s Multiplier

MomentY product

0

0.5

0

0.5

25.13

0.5

2

25.13

1

56.55

1

1.5

84.825

2

76.97

2

4

615.76

3

76.97

3

2

461.82

4

56.55

4

4

904.8

5

25.13

5

1.5

188.475

5.5

6.28

5.5

2

6

0.00

6

0.5

0

Total

2,349.89

Moment about Y axis x = Moment about Y a axis xs=

S ×S 3

69.08

( MomentY p product d )

10.00 × 10.00 × 2, 349.89 = 78 , 329.67 m4 3

The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCB: LCB =

Moment about Y axis x 78 , 329.67 = = 26.91 m FOAP Area 2, 911.2

9781408176122_Ch14_6_Rev_txt_prf.indd 610

11/16/2013 2:40:43 AM

Solutions to Questions • 611 Q8.32 The station spacing can be determined: Spacing =

120 = 20 m 6

Simpson’s Rule can be used to determine the area under the curve of waterline half beams: Station

Waterline half beam

Simpson’s Multiplier

Area product

0

0

1

0

1

6

4

24

2

12

2

24

3

12

4

48

4

12

2

24

5

3

4

12

6

0

1

0

Total

Area =

132

Spacing p g 20 × Σ (Total a area ea product ) = × 132 = 880 m2 3 3

This is derived from the half beams, so is half of the waterplane area. Therefore, this value needs to be doubled to determine the overall waterplane area: Waterplane area a ea = 880 × 2 = 1, 760 m2 Simpson’s Rule can be used to determine the moment of area under the curve about the y axis: Station

Waterline half beam

Lever

Simpson’s Multiplier

MomentY product

0

0

0

1

0

1

6

1

4

24

2

12

2

2

48

3

12

3

4

144

4

12

4

2

96

5

3

5

4

60

6

0

6

1

0

Total

9781408176122_Ch14_6_Rev_txt_prf.indd 611

372

11/16/2013 2:40:44 AM

612 • Ship Stability, Powering and Resistance Spacing p g × Spacing T Moment o e tY prod o uct ) (Total 3 20 = × 20 × 372 = 49 , 600 m3 3

Moment about the Y a axis =

This is derived from the half beams, so is half of the total moment of area of the waterplane about the y axis. Therefore, this value needs to be doubled to determine the overall moment of area of the waterplane about the y axis: w rplane about the Y axis x = 49 , 600 × 2 = 99 ,200 200 m2 Moment of area of the water The LCF is the centre of area of the waterplane, which can be found by dividing the moment of area of the waterplane by the waterplane area: LCF =

Moment o of area o of the w water aterp pla anee about tthee Y axis x 99 , 200 = = 56.36 m FOAP Waterrpl p ane area 1, 760

Q8.33 The station spacing can be determined: Spacing =

90 = 15 m 6

Simpson’s Rule can be used to determine the area under the curve of waterline half beams: Station

Waterline half beam

Simpson’s Multiplier

Area product

0

0

0.5

0

0.5

4

2

8

1

8

1.5

12

2

10

4

40

3

10

2

20

4

9

4

36

5

6

1.5

9

5.5

3

2

6

6

0

0.5

0

Total

9781408176122_Ch14_7_Rev_txt_prf.indd 612

131

11/16/2013 6:52:39 PM

Solutions to Questions • 613

Area =

Spacing p g 15 × Σ (Total area product ) = × 131 = 665 m2 3 3

This is derived from the half beams, so is half of the waterplane area. Therefore, this value needs to be doubled to determine the overall waterplane area: Waterplane area a ea = 665 × 2 = 1, 310 m2 Simpson’s Rule can be used to determine the moment of area under the curve about the y axis: Station

Waterline half beam

Lever

Simpson’s Multiplier MomentY product

0

0

0

0.5

0

0.5

4

0.5

2

4

1

8

1

1.5

12

2

10

2

4

80

3

10

3

2

60

4

9

4

4

144

5

6

5

1.5

45

5.5

3

5.5

2

33

6

0

6

0.5 Total

0 378

Spacing p g × Spacing o e tY prod o uct ) (TTotal Moment 3 15 = × 15 × 378 = 28 ,350 350 m3 3

Moment about the Y axi a s=

This is derived from the half beams, so is half of the total moment of area of the waterplane about the y axis. Therefore, this value needs to be doubled to determine the overall moment of area of the waterplane about the y axis: Moment of area of the waterpl r ane about the Y axis x = 28 , 350 × 2 = 56 , 700 m2 The LCF is the centre of area of the waterplane, which can be found by dividing the moment of area of the waterplane by the waterplane area: LCF =

Moment o of area o of the waterpl aterp a anee about tthee Y axis x 56 , 700 = = 43.28 m FOAP Waterrpl p ane area 1, 310

9781408176122_Ch14_7_Rev_txt_prf.indd 613

11/16/2013 6:52:41 PM

614 • Ship Stability, Powering and Resistance Q8.34 The waterline half beam values can be used to draw the waterline half beam curve. Simpson’s Rule can then be used to determine the inertia, or second moment of area, of the area under the curve about the y axis: Waterline half beam

Lever2

Simpson’s Multiplier

0

0

0

1

0

1

10

1

4

40

2

20

4

2

160

3

20

9

4

720

4

16

16

2

512

5

8

25

4

800

6

0

36

1

0

Station

Total

Inertia about the Y axis x =

⎛ Spacing ⎞ × Spacing2 ⎝ 3 ⎠

Inertia about the Y axis x =

InertiaY product

2,232

(InertiaY

product )

⎛ 30 ⎞ × 302 × 2, 232 = 20 , 088 , 000 m4 ⎝ 3⎠

As the curve is based on the waterline half beams, this value is for half of the waterplane, and so must be doubled to give the overall inertia of the waterplane area about the y axis. This gives a value for the longitudinal inertia of the waterplane measured through the y axis of 40,176,000 m4. As this value is measured about the y axis, and not the centre of the waterplane, it is a remote value. To determine the longitudinal inertia of the waterplane measured through the centre (the LCF), the parallel axes formula needs to be used: REMOTE

= ICENTRO Distance 2 ) E ID + ( Area

(

2 40 ,176 , 000 = ICENTRO E ID + 4 , 480 × 86.79

)

4 ICENTRO E ID = 6 , 430 , 381.63 m

Finally, BML can be found: BML =

9781408176122_Ch14_7_Rev_txt_prf.indd 614

IL 6 , 430 , 381.63 = = 214.35 m ∇ 30 , 000

11/16/2013 6:52:42 PM

Solutions to Questions • 615 Q8.35 The spacing between the stations can be determined: Spacing =

90 = 15.00 00 m 6

The section area data can be used to determine the underwater volume: Station

Section area (m2)

Simpson’s Multiplier

1.57

0.5

0

Area product 0.785

0.5

14.14

2

28.28

1

39.27

1.5

58.905

2

76.97

4

307.88

3

100.53

2

201.06

4

76.97

4

307.88

5

39.27

1.5

58.905

5.5

14.14

2

28.28

6

0.00

Area =

0.5

0

Total

991.19

S ⎛ 15.00 ⎞ × Σ ( Area p product oduct ) = × 99 991.19 9 = 4 , 955.95 m3 ⎝ 3 ⎠ 3

Therefore, the underwater volume is 4,955.95 m3. This can be used to determine the displacement: Δ = ∇ × ρ = 4 , 955.95 × 1.025 = 5, 079.85 tonnes The section area data can be used to determine the LCB by finding the moment of area of the section area curve about the Y axis, and therefore the centre of area from the Y axis: Station 0

Section area (m2)

Lever

1.57

0

0.5

Simpson’s Multiplier MomentY product 0

0.5

14.14

0.5

2

14.14

1

39.27

1

1.5

58.905

2

76.97

2

4

615.76

3

100.53

3

2

603.18

9781408176122_Ch14_7_Rev_txt_prf.indd 615

11/16/2013 6:52:43 PM

616 • Ship Stability, Powering and Resistance

Section area (m2)

Lever

4

76.97

4

4

5

39.27

5

1.5

294.525

5.5

14.14

5.5

2

155.54

6

0.5

0

Total

2,973.57

Station

6

0.00

Moment about Y axis x = Moment about Y a axis xs=

Simpson’s Multiplier MomentY product

S ×S 3

1,231.52

( MomentY p product d )

15.00 × 15.00 × 2, 973.57 = 223, 017.75 m4 3

The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCB: LCB =

Moment about Y axis x 223, 017.75 = = 45.00 m FOA O P Area 4 , 955.95

The waterline half beam values can be used to determine the transverse inertia of the waterplane area: Station

Waterline half beam (m)

Waterline half beam3

Simpson’s Multiplier

0

1.00

1.00

0.5

0.5

3.00

27.00

1

5.00

125.00

1.5

2

7.00

343.00

4

1,372.00

3

8.00

512.00

2

1,024.00

4

7.00

343.00

4

1,372.00

5

5.00

125.00

1.5

5.5

3.00

27.00

6

0.00

0.00

9781408176122_Ch14_7_Rev_txt_prf.indd 616

0.50

2

54.00 187.50

187.50

2

54.00

0.5 Total

Inertia about the X axis x =

InertiaX product

0.00 4,251.50

S 1 × × Σ(InertiaX p product d ) 3 3

11/16/2013 6:52:44 PM

Solutions to Questions • 617

Inertia about the X axis x =

15.00 1 × × 4 , 251.50 = 7, 085.83 m4 3 3

This uses the half beams, so the value must be doubled for the overall waterplane inertia about the centreline: 6 m4 Waterplane inertia about the centreline = 7, 085.83 × 2 = 14 ,171.66 This, along with the underwater volume allows BM to be found: BM =

I 14 ,171.66 = = 2.86 m ∇ 4 , 955.95

The waterline half beam values can also be used to determine the area of the waterplane: Station

Waterline half beam (m) Simpson’s Multiplier

Area product

0

1.00

0.5

0.50

0.5

3.00

2

6.00

1

5.00

1.5

7.50

2

7.00

4

28.00

3

8.00

2

16.00

4

7.00

4

28.00

5

5.00

1.5

7.50

5.5

3.00

2

6.00

6

0.00

0.5

0.00

Total

99.50

a= Area =

S × Σ( Area product p d ) 3

15.00 × 99.50 = 497.5 m2 3

This uses the half beams, so the value must be doubled for the overall waterplane area: Waterplane area a ea = 497.5 × 2 = 995 m2

9781408176122_Ch14_7_Rev_txt_prf.indd 617

11/16/2013 6:52:45 PM

618 • Ship Stability, Powering and Resistance The waterline half beam values can be used to determine the moment of area of the waterplane about the y axis: Station

Waterline half beam (m)

Lever

Simpson’s Multiplier

MomentY product

0

1.00

0

0.5

0.00

0.5

3.00

0.5

2

3.00

1

5.00

1

1.5

7.50

2

7.00

2

4

56.00

3

8.00

3

2

48.00

4

7.00

4

4

112.00

5

5.00

5

1.5

37.50

5.5

3.00

5.5

2

33.00

6

0.00

6

0.5

0.00

Total

Moment about Y axis x = Moment about Y axis x =

S ×S 3

297.00

( MomentY p product d )

15.00 × 15.00 × 297.00 = 22, 275 m4 3

This uses the half beams, so the value must be doubled for the overall moment of waterplane area about the y axis: Moment of waterplane area about the y a axis = 22, 275 × 2 = 44 44 , 550 m3 The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCF: LCF =

Moment about Y axis x 44 , 550 = = 44.77 m FOAP Area 995

The waterline half beam values can be used to determine the inertia of area of the waterplane about the y axis: Waterline half beam (m)

Lever2

0

1.00

0

0.5

0.00

0.5

3.00

0.25

2

1.50

1

5.00

1

1.5

7.50

Station

9781408176122_Ch14_7_Rev_txt_prf.indd 618

Simpson’s Multiplier

InertiaY product

11/16/2013 6:52:47 PM

Solutions to Questions • 619

Waterline half beam (m)

Lever2

2

7.00

4

4

112.00

3

8.00

9

2

144.00

4

7.00

16

4

448.00

5

5.00

25

1.5

187.50

5.5

3.00

30.25

2

181.50

6

0.00

36

0.5

Station

Simpson’s Multiplier

Total

Inertia about the Y axis x = Inertia about the Y axis x =

S × S2 3

InertiaY product

0.00 1,082.00

(Total T l inertiaY p product d )

15.00 × 15.002 × 1, 082.00 = 1, 217, 250 m4 3

This uses the half beams, so the value must be doubled for the overall inertia of the waterplane area about the y axis: Inertia o of w waterrplane a area ea about tthee y axis x = 1, 217, 250 × 2 = 2,4 434 ,500 500 m4 This is about the y axis, which is a remote axis. To use to determine the longitudinal BM, this must be converted to be about the LCF: IREMOTE AXIS = ICENTRO Distance 2 ) E ID AXIS + ( Area 2, 434 , 500 = ICENTRO 44.772 ) E ID AXIS + ( 995 4 ICENTRO E ID AXIS = 440 ,168.86 m

This, along with the underwater volume allows BML to be found: BML =

IL 440 ,168.86 = = 88.82 m ∇ 4 , 955.95

Q8.36 The spacing between the stations can be determined: Spacing =

9781408176122_Ch14_7_Rev_txt_prf.indd 619

120 = 20.00 m 6

11/16/2013 6:52:48 PM

620 • Ship Stability, Powering and Resistance The section area data can be used to determine the underwater volume: Station

Section area (m2)

Simpson’s Multiplier

Area product

0.00

0.5

0

0 0.5

25.13

2

1

76.97

1.5

115.455

2

157.08

4

628.32

3

190.07

2

380.14

4

157.08

4

628.32

5

56.55

1.5

84.825

5.5

14.14

2

28.28

6

0.00

Area =

50.26

0.5

0

Total

1,915.60

S ⎛ 20.00 ⎞ × Σ ( Area product ) = × 1, 915.60 = 12, 770.67 m3 ⎝ 3 ⎠ 3

Therefore, the underwater volume is 12,770.67 m3. This can be used to determine the displacement: Δ = ∇ × ρ = 12, 770.67 × 1.025 = 13, 089.94 94 tonnes The section area data can be used to determine the LCB by finding the moment of area of the section area curve about the Y axis, and therefore the centre of area from the Y axis: Station 0

Section area (m2) 0.00

Lever

Simpson’s Multiplier MomentY product

0

0.5

0

0.5

25.13

0.5

2

1

76.97

1

1.5

2

157.08

2

4

1,256.64

3

190.07

3

2

1,140.42

4

157.08

4

4

2,513.28

5

56.55

5

1.5

424.125

5.5

14.14

5.5

2

155.54

6

0.5

0

Total

5,630.59

6

0.00

9781408176122_Ch14_7_Rev_txt_prf.indd 620

25.13 115.455

11/16/2013 6:52:50 PM

Solutions to Questions • 621

Moment about Y axis x = Moment about Y a axis xs=

S ×S 3

( MomentY p product d )

20.00 × 20.00 × 5, 630.59 = 750 , 745.33 m4 3

The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCB: LCB =

Moment about Y axis x 750 , 745.33 = = 58.79 m FO OAP A Area 12, 770.67

The waterline half beam values can be used to determine the transverse inertia of the waterplane area: Station

Waterline half beam (m)

Waterline half beam3

0

0.00

0.00

0.5

4.00

64.00

1

7.00

343.00

2

10.00

1,000.00

4

4,000.00

3

11.00

1,331.00

2

2,662.00

4

10.00

1,000.00

4

4,000.00

5

6.00

216.00

5.5

3.00

27.00

6

0.00

0.00

Simpson’s Multiplier

InertiaX product

0.5 2

128.00

1.5

514.50

1.5 2

Inertia about the X axis x =

324.00 54.00

0.5 Total

Inertia about the X axis x =

0.00

0.00 11,682.50

S 1 × × Σ(InertiaX p product d ) 3 3

20.00 1 × × 11, 682.50 = 25, 961.11 m4 3 3

This uses the half beams, so the value must be doubled for the overall waterplane inertia about the centreline: 922. 2 m4 Waterplane inertia about the centreline = 25, 961.11× 2 = 51, 922.2

9781408176122_Ch14_7_Rev_txt_prf.indd 621

11/16/2013 6:52:51 PM

622 • Ship Stability, Powering and Resistance This, along with the underwater volume allows BM to be found: BM =

I 51, 922.22 = = 4.07 m ∇ 12, 770.67

The waterline half beam values can also be used to determine the area of the waterplane: Station

Waterline half beam (m)

Simpson’s Multiplier

Area product

0

0.00

0.5

0.00

0.5

4.00

2

8.00

1

7.00

1.5

10.50

2

10.00

4

40.00

3

11.00

2

22.00

4

10.00

4

40.00

5

6.00

1.5

9.00

5.5

3.00

2

6.00

6

0.00

0.5

0.00

Total

a= Area =

135.50

S × Σ( Area p product d ) 3

20.00 × 135.50 = 903.33 m2 3

This uses the half beams, so the value must be doubled for the overall waterplane area: Waterplane area = 903.33 × 2 = 1, 806.66 m2 The waterline half beam values can be used to determine the moment of area of the waterplane about the y axis: Station

Waterline half beam (m)

Lever

Simpson’s Multiplier

MomentY product

0

0.00

0

0.5

0.00

0.5

4.00

0.5

2

4.00

1

7.00

1

1.5

10.50

2

10.00

2

4

80.00

3

11.00

3

2

66.00

9781408176122_Ch14_7_Rev_txt_prf.indd 622

11/16/2013 6:52:52 PM

Solutions to Questions • 623

Station

Waterline half beam (m)

Lever

Simpson’s Multiplier

MomentY product

4

10.00

4

4

5

6.00

5

1.5

45.00

5.5

3.00

5.5

2

33.00

6

0.00

6

0.5 Total

Moment about Y a axis xs= Moment about Y axis x =

S ×S 3

160.00

0.00 398.50

( Moment o e tY p product d )

20.00 × 20.00 × 398.50 = 53,133.33 m4 3

This uses the half beams, so the value must be doubled for the overall moment of waterplane area about the y axis: Moment of waterplane area about the y a axis = 53,133.33 × 2 = 106 , 266.66 m3 The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCF: LCF =

Moment about Y axis x 106 , 266.66 = = 58.82 m FOA O P Area 1, 806.66

The waterline half beam values can be used to determine the inertia of area of the waterplane about the y axis: Station

Waterline half beam (m)

Lever2

Simpson’s Multiplier

InertiaY product

0

0.00

0

0.5

0.00

0.5

4.00

0.25

2

2.00

1

7.00

1

1.5

2

10.00

4

4

160.00

3

11.00

9

2

198.00

4

10.00

16

4

640.00

5

6.00

25

1.5

225.00

5.5

3.00

30.25

2

181.50

6

0.00

36

0.5 Total

9781408176122_Ch14_7_Rev_txt_prf.indd 623

10.50

0.00 1,417.00

11/16/2013 6:52:53 PM

624 • Ship Stability, Powering and Resistance

Inertia about the Y axis x = x = Inertia about the Y axis

S × S2 3

(Total T l inertiaY p product d )

20.00 × 20.002 × 1, 417.00 = 3, 778 , 666.67 6 m4 3

This uses the half beams, so the value must be doubled for the overall inertia of the waterplane area about the y axis: Inertia o of w waterrplane area about the y axis x = 3, 778 , 666.67 × 2 = 7, 557, 333.34 m4 This is about the y axis, which is a remote axis. To use to determine the longitudinal BM, this must be converted to be about the LCF: IREMOTE AXISS = ICENTRO Distance 2 ) CE OID D AXIS + ( Area 7, 557, 333.34

CENTRO E ID AXIS

(1, 806.66 × 58.822 )

4 ICENTRO E ID AXIS = 1, 306 , 664.8 m

This, along with the underwater volume allows BML to be found: BML =

IL 1, 306 , 664.8 = = 102.32 m ∇ 12, 770.67

Q8.37 The spacing between the stations can be determined: Spacing =

105 = 17.50 m 6

The section area data can be used to determine the underwater volume: Station

Section area (m2)

Simpson’s Multiplier

Area product 7.07

0

14.14

0.5

0.5

56.55

2

113.1

1

100.53

1.5

150.795

2

127.23

4

508.92

3

127.23

2

254.46

4

100.53

4

402.12

5

39.27

5.5

14.14

6

9781408176122_Ch14_7_Rev_txt_prf.indd 624

0.00

1.5

58.905

2

28.28

0.5

0

Total

1,516.58

11/16/2013 6:52:54 PM

Solutions to Questions • 625

Area =

S ⎛ 17.50 ⎞ × Σ ( Area product ) = × 1, 516.58 = 8 , 846.72 m3 ⎝ 3 ⎠ 3

Therefore, the underwater volume is 8,846.72 m3. This can be used to determine the displacement: Δ = ∇ × ρ = 8 , 846.72 × 1.025 = 9 , 067.89 89 tonnes The waterline half beam values can be used to determine the transverse inertia of the waterplane area: Station

Waterline half beam (m)

Waterline half beam3

Simpson’s Multiplier

0

3.00

27.00

0.5

0.5

6.00

216.00

2

432.00

1

8.00

512.00

1.5

768.00

2

9.00

729.00

4

2,916.00

3

9.00

729.00

2

1,458.00

4

8.00

512.00

4

2,048.00

5

5.00

125.00

1.5

5.5

3.00

27.00

6

0.00

0.00

Inertia about the X axis x =

13.50

187.50

2

54.00

0.5 Total

Inertia about the X axis x =

InertiaX product

0.00 7,877.00

S 1 × × Σ(InertiaX p product d ) 3 3

17.50 1 × × 7, 877.00 = 15, 316.39 m4 3 3

This uses the half beams, so the value must be doubled for the overall waterplane inertia about the centreline: Waterplane inertia e t a about tthee centreline = 15, 316.39 × 2 = 30 , 632.7 632. 8 m4 This, along with the underwater volume allows BM to be found: BM =

9781408176122_Ch14_7_Rev_txt_prf.indd 625

I 30 , 632.78 = = 3.46 m ∇ 8 , 846.72

11/16/2013 6:52:56 PM

626 • Ship Stability, Powering and Resistance The waterline half beam values can also be used to determine the area of the waterplane: Station

Waterline half beam (m) Simpson’s Multiplier Area product

0

3.00

0.5

0.5

6.00

2

12.00

1

8.00

1.5

12.00

2

9.00

4

36.00

3

9.00

2

18.00

4

8.00

4

32.00

5

5.00

1.5

7.50

5.5

3.00

2

6.00

6

0.00

0.5

0.00

Total

a= Area =

1.50

125.00

S × Σ( Area p product d ) 3

17.50 × 125.00 = 729.17 m2 3

This uses the half beams, so the value must be doubled for the overall waterplane area: Waterplane area = 729.17 × 2 = 1, 458.34 m2 The waterplane area and volume can be used with Morrish’s method to determine KB: KB =

⎞ ⎞ 1 ⎛ 5 × 9 ⎛ 8 , 846.72 ⎞ ⎞ 1 ⎛ 5D ⎛ ∇ ×⎜ −⎜ = × −⎜ ⎟ = 5 48 m ⎝ 1, 458 3 ⎝ 2 ⎝ Waterplane area ⎟⎠ ⎟⎠ 3 ⎜⎝ 2 5 .34 ⎠ ⎟⎠

This allows GM to be found: GM = KB + BM − KG = 5.48 + 3.46 − 7.00 = 1.94 m The waterline half beam values can be used to determine the moment of area of the waterplane about the y axis:

9781408176122_Ch14_7_Rev_txt_prf.indd 626

11/16/2013 6:52:58 PM

Solutions to Questions • 627

Station

Waterline half beam (m)

Lever

Simpson’s Multiplier MomentY product

0

3.00

0

0.5

0.00

0.5

6.00

0.5

2

6.00

1

8.00

1

1.5

12.00

2

9.00

2

4

72.00

3

9.00

3

2

54.00

4

8.00

4

4

128.00

5

5.00

5

1.5

37.50

5.5

3.00

5.5

2

33.00

6

0.00

6

0.5 Total

about Y axis x = Moment about Y a axis xs=

S ×S 3

0.00 342.50

( MomentY p product d )

17.50 × 17.50 × 342.50 = 34 , 963.54 m4 3

This uses the half beams, so the value must be doubled for the overall moment of waterplane area about the y axis: Moment o of waterplane a area ea about tthee y a axis = 34 , 963.54 × 2 = 69 ,927 9 .08 m3 The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCF: LCF =

Moment about Y axis x 69 , 927.08 = = 47.95 m FOAP Area 1, 458.34

The waterline half beam values can be used to determine the inertia of area of the waterplane about the y axis: Waterline half beam (m)

Lever2

0

3.00

0

0.5

0.00

0.5

6.00

0.25

2

3.00

1

8.00

1

1.5

2

9.00

4

4

144.00

3

9.00

9

2

162.00

Station

9781408176122_Ch14_7_Rev_txt_prf.indd 627

Simpson’s Multiplier InertiaY product

12.00

11/16/2013 6:52:59 PM

628 • Ship Stability, Powering and Resistance

Station

Waterline half beam (m)

Lever2

Simpson’s Multiplier InertiaY product

4

8.00

16

4

512.00

5

5.00

25

1.5

187.50

5.5

3.00

30.25

2

181.50

6

0.00

36

0.5 Total

Inertia about the Y axis x = Inertia about the Y axis x =

S × S2 3

0.00 1,202.00

(Total T l inertiaY p product d )

17.50 × 17.502 × 1, 202.00 = 2,147, 322.92 9 m4 3

This uses the half beams, so the value must be doubled for the overall inertia of the waterplane area about the y axis: Inertia o of w waterrplane area about the y axis x = 2,147, 322.92 × 2 = 4 , 294 , 645.84 m4 This is about the y axis, which is a remote axis. To use to determine the longitudinal BM, this must be converted to be about the LCF: IREMOTE AXISS = ICENTRO Distance 2 ) CE OID D AXIS + ( Area 4 , 294 , 645.84

CENTRO E ID AXIS

(1, 458.34 × 47.952 )

4 ICENTRO E ID AXIS = 941, 626.87 m

This, along with the underwater volume allows BML to be found: BML =

IL 941, 626.87 = = 106.44 44 m ∇ 8 , 846.72

This allows GML to be found: GML = KB B + BML − KG = 5.48 + 106 06.44 − 7.00 = 104.92 m This can be used to determine the MCTC: MCTC =

ΔGML 9 , 067.89 × 104.92 = = 90.61 61 tonne metres 100LBP 100 × 105

9781408176122_Ch14_7_Rev_txt_prf.indd 628

11/16/2013 6:53:01 PM

Solutions to Questions • 629 The section area data can be used to determine the LCB by finding the moment of area of the section area curve about the Y axis, and therefore the centre of area from the Y axis: Section area (m2)

Station

Lever

Simpson’s Multiplier

MomentY product 0

0

14.14

0

0.5

0.5

56.55

0.5

2

56.55

1

100.53

1

1.5

2

127.23

2

4

1,017.84

3

127.23

3

2

763.38

4

100.53

4

4

1,608.48

5

39.27

5

1.5

294.525

5.5

14.14

5.5

2

155.54

6

0.5

0

Total

4,047.11

6

0.00

Moment about Y a axis xs= Moment about Y a axis xs=

S ×S 3

150.795

( Moment o e tY p product d )

17.50 × 17.50 × 4 , 047.11 = 413,142.48 m4 3

The moment about the y axis can be divided by the area of the graph to give the centroid from the y axis, which is the LCB: LCB =

Moment about Y axis x 413,142.48 = = 46.70 m FOA O P Area 8 , 846.72

This can be used to determine the trim: Trim =

LCG ) Δ ( 46.70 − 45.00 ) 9 , 067.89 = = 170 cm by the h stern MCTC 90.61

(LCB

Q9.1 Ship speed , v = 10 × 0.514 = 5.14 m/s 825 Frictional resistance = fSv 1.825 = 1.392 × 4 , 000 × 5.141.825 = 110 , 460 6 N = 110.5 kN

9781408176122_Ch14_7_Rev_txt_prf.indd 629

11/16/2013 6:53:03 PM

630 • Ship Stability, Powering and Resistance Q9.2 Although both the coefficient of friction and wetted surface area are unknown in this question, it is important to know that neither property will vary with the speed of the vessel. This information is used to solve the problem. Frictional resistance = fSv 1.825 f (10 × 0.514 )1.825

134 , 044 ∴ fS =

134 , 044 = 6 , 756.8 (10 × 0.514 )1.825

The frictional resistance of the vessel at any speed can now be found: 825 Frictional resistance = fSv 1.825 = 6 , 756.8 × (11× 0.514 )1.825 = 159 , 511 N

Q9.3 The first stage is to determine the speed of the ship at the corresponding speed to the test (the Froude number of the ship and model should be the same): vM gLM 2.2 9.81 4 vS =

=

=

vS gLS vS 9.81× 120

2.2 9.81× 120 9 81× 4

= 12.05 m/s

The next stage is to determine the frictional resistance of the model, RFM: RFM

fM sM v Mn = 1.609 × 4.5 × 2.21.825 = 30.527 N

Now the frictional resistance has been found, the residual resistance of the model, RRM, can be found. This is the difference between the total model resistance, RTM, and the model friction resistance, RFM: RRM

RTM − RFM = 70 − 30.527 = 39.473 N

The next step is to scale up the residual resistance of the model, RRM, to find the residual resistance of the ship, RRS. The scale factor for residual resistance is the displacement: RRS

ΔS ΔM

RRM =

9781408176122_Ch14_7_Rev_txt_prf.indd 630

4 , 959 × 39.473 = 1, 092.404 N = 1, 092.404 kN 0.1792

11/16/2013 6:53:04 PM

Solutions to Questions • 631 The next stage is to determine the frictional resistance of the ship, RFS: RFS

fS sS v Sn = 1.483 × 4 , 050 × 12.051.825 = 564.145 N = 564.145 kN

The frictional resistance of the ship, RFS, can now be added to the residual resistance of the ship, RRS, to find the total resistance of the ship, RTS: RTS = RFS + RRS = 564.145 5 + 1, 092.404 = 1, 656.549 549 kN Q9.4 LSHIP L 150 = λ ⇒ LMODEL = SHIP = = 7.5 m LMODEL λ 20 SSHIP S 6 , 500 = λ 2 ⇒ SMODEL = SHIP = = 16.25 m2 SMODEL λ2 202 Δ SHIP ρ = λ 3 × SEAWATER ⇒ Δ MODEL = Δ MODEL ρFFRESHWATER

Δ SHIP 9 , 800 = = 1.195 tonnes ρ 1.025 3 3 SEAWATTTEEER 20 × λ × 1.000 ρFRESHWATER

Q9.5 The first step is to find the ship properties from the model from the length scale: LSHIP = λ ⇒ LSHIP λ LMODEL OD = 20 × 5 = 100 m LMODEL SSHIP 2 2 = λ 2 ⇒ SSHIP λ 2 SMODEL OD = 20 × 5.6 = 2 , 240 m SMODEL The next stage is to determine the speed of the ship at the corresponding speed to the test (the Froude number of the ship and model should be the same): vM gLM 1.4 9.81 5 vS =

=

=

vS gLS vS 9.81× 100

1.4 9.81× 100 9 81× 5

= 6.261 261 m/s

The next stage is to determine the frictional resistance of the model, RFM: RFM

9781408176122_Ch14_7_Rev_txt_prf.indd 631

fM sM v Mn = 1.655 × 5.6 × 1.41.825 = 17.127 N

11/16/2013 6:53:07 PM

632 • Ship Stability, Powering and Resistance The next stage is to determine the residual resistance of the model, RRM. This is the difference between the total model resistance, RTM, and the model friction resistance, RFM: RTM − RFM = 25 − 17.127 = 7.873 N

RRM

The next stage is to scale up the residual resistance of the model, RRM, to find the residual resistance of the ship, RRS: RRS

ΔS ΔM

= 203 ×

3

⎛ L ⎞ ρ = ⎜ SHIP ⎟ × SSEAWATER × RRM ρFFRESHWATER ⎝ LMODEL ⎠

RRM

1.025 × 7.873 = 64 , 562N = 64 6 .562 56 kN 1.000

The next stage is to determine the frictional resistance of the ship, RFS: RFS

fS sS v Sn = 1.401× 2, 240 × 6.2611.825 = 89 , 240 N = 89.24 kN

The frictional resistance of the ship, RFS, can now be added to the residual resistance of the ship, RRS, to find the total resistance of the ship, RTS: RTS = RFS + RRS = 89.240 + 64.562 = 153.802 kN Q9.6 Rn =

Lv 10 × v = ν 1.004 × 10 −6

CF =

0.075

(

RF Speed (m/s) 0.1

Rn −

)2

SSv 2 C F

Reynolds number

CF

Friction (N)

996,016

0.0047

0.235

1

9,960,159

0.0030

15.000

10

99,601,594

0.0021

1,050.000

Q9.7 Rn =

9781408176122_Ch14_7_Rev_txt_prf.indd 632

Lv 100 × (15 × 0.5144 ) = = 734 , 857,142 ν 1 05 × 10 −6

11/16/2013 6:53:09 PM

Solutions to Questions • 633 0.075

CF =

(

Rn −

=

)

2

0.075

(

−

= 0.0021

)2

C F (1+ k ) = 0.0021× (1+ 0.1) = 0.0023

CV

SSv 2 CV = 0.5 × 1, 025 × 4 , 000 × (15 × 0.5144 ) × 0.0023 = 280 , 715.3 N 2

RV

RV = 280.7 kN Q9.8 Rn = CF =

Lv 2×8 = = 15, 238 , 095 ν 1 05 × 10 −6

0.075

(

Rn −

(1

RV

)

2

=

0.075

(

k ) = 1+ 2

− ⎛t⎞ ⎝ c⎠

60

⎛t⎞ ⎝ c⎠

)2

= 0.0028

4

4

4⎞ ⎛ 0.4 ⎛ 0.4 ⎞ + 60 = 1.496 ⎝ 2 ⎠ ⎝ 2 ⎠

(1

k ) = 1+ 2

CV

C F (1+ k ) = 0.0028 × 1.496 = 0.0042

SSv 2 CV = 0.5 × 1, 025 × (2 × 5 × 2.24 ) × 82 × 0.0042 = 3, 085.8 N

Q9.9 The geometry of the full scale ship can be determined: LS = λ ∴ LS LM SS = λ 2 ∴ SS SM

LM × λ = 5 × 50 = 250 m

SM × λ 2 = 6.40 × 502 = 16 16 , 000 m2

The total resistance coefficient of the model can be determined: CTM =

RT M 25 = = 6.457 × 10 −3 0.5ρSM v M2 0.5 × 1, 000 × 6.40 × 1.102

The Reynolds number of the model can be determined: RnM =

9781408176122_Ch14_7_Rev_txt_prf.indd 633

Lv 5 × 1.10 = = 5, 500 , 000 ν 1 00 × 10 −6

11/16/2013 6:53:12 PM

634 • Ship Stability, Powering and Resistance The frictional resistance coefficient of the model can be determined: CF M =

0.075

(

)

2

Rn −

=

0.075

(

)

2

−

= 3.338 × 10 −3

The viscous resistance coefficient of the model can be determined: CV M

C F M (1+ k ) = 3.338 × 10 −3 × (1.15) = 3.838 × 10 −3

The wave resistance coefficient of the model can be determined: CT M − CV M = 6.457 × 10 −3 − 3.838 × 10 −3 = 2.618 × 10 −3

CW M

This will be the same for the ship. The speed of the ship can be determined from the Froude number: Fn =

vM

(g

LM )

=

vS

(g

∴

LS )

1.10

( g × 5)

=

vS

∴ v S = 7.775 m/s

(g × 250 )

The Reynolds number of the ship can be determined: RnS =

Lv 250 × 7.775 = = 1, 833, 726 , 415 υ 1 06 × 10 −6

The frictional resistance coefficient of the ship can be determined: CF S =

0.075

(

Rn −

)

2

=

0.075

(

−

)

2

= 1.422 × 10 0 −3

The viscous resistance coefficient of the ship can be determined: CV S

C F S (1+ k ) = 1.422 × 10 −3 × (1.15) = 1.635 × 10 −3

The total resistance coefficient of the ship can be determined: CT S

CV S + CW S = 1.635 × 10 −3 + 2.618 × 10 −3 = 4.254 × 10 −3

The total resistance of the ship can be found: Total sship ip resistance

ρSS v S2 CTS

Total sship ip resistance = 0.5 × 1, 025 × 16 , 000 × 7.7752 × 4.254 × 10 −3 = 2, 2,108 , 481.4 N ∴ Total sship ip resistance = 2,108.5 kN Q9.10 Ship correlation factor ( SCF SC ) Weather Weather allowance Appendage allowanc a e

9781408176122_Ch14_7_Rev_txt_prf.indd 634

11/16/2013 6:53:15 PM

Solutions to Questions • 635 SCF = 1.05 × 1.12 = 1.176 RT × v = 850 × 10 = 8 , 500 kW

PEN

PEN × SCF = 8 , 500 × 1.176 = 9 , 996 kW

PE Q10.1

RT × v = 900 × 11 = 9 , 900 kW

PEN PE

PEN × SCF = 9 , 900 × 1.15 = 11,385 385 kW

ηH =

PE P 11, 385 ⇒ PT = E = = 11, 617.45 kW PT ηH 0 98

ηp =

PT P 11, 617.35 ⇒ PD = T = = 20 , 381.3 kW PD ηp 0 57

ηT =

PD P 20 , 381.3 ⇒ PS = D = = 21, 011.7 kW PS ηT 0 97

ηM =

PS P 21, 011.7 ⇒ PI = S = = 26 , 264.6 kW PI ηM 08

Q10.2 2

AC =

2

Δ 3 v 3 90 , 000 3 × 153 = = 339 PS 20 , 000

2

PS =

2

Δ 3 v 3 90 , 000 3 × 173 = = 29 ,105 105 kW AC 339

Q10.3 2

2

2 Δ 3 v 3 Δ 3 × 83 AC = = = 0.0512 Δ 3 PS 10 , 000 2

PS =

2

Δ 3v 3 Δ 3 × 73 = = 6 , 699 kW 2 AC 0.0512 Δ 3

Q10.4 For the original vessel: 2

AC =

9781408176122_Ch14_7_Rev_txt_prf.indd 635

2

Δ 3 v 3 7, 000 3 × 14 3 = = 125.5 PS 8 , 000

11/16/2013 6:53:18 PM

636 • Ship Stability, Powering and Resistance V

Froude number u be , Fn =

gL

=

14 × 0.514 9.81× 100

= 0.23

For the scaled up vessel: V

Fn =

gL

⇒V

FFn gL

V = 0.23 9.81× 110 = 7.55 2

AC =

PS

( =

/s = 14 14.7 knots 2

Δ 3v 3 Δ 3v 3 ⇒ PS = PS AC

)

×

2

125.5

3

× 14.73

= 11,207 207 kW

Q10.5 2

Fuel coefficient =

2

Δ 3v 3 90 , 000 3 × 153 = = 193, 657 Daily fuel consumption 35

At the new speed of 15 + 2 17 knots 2

2

Δ 3v 3 90 , 000 3 × 173 Daily fuel ue consumption = = = 50.9 tonnes/day Fuel coefficient 193 9 , 657 Q10.6 2

Daily fuel consumption =

2

Δ 3v 3 80 , 000 3163 = = 58.50 tonnes/day Fuel coefficient 130 ,000

Voyage fuel consumption = Daily fuel consumption × = 58.50 ×

Voyage distance a 24 × Ship speed

2, 000 = 304.7 tonnes 24 × 16

Q10.7 2

Daily fuel ue consumption =

2

ν 3 v3 80 , 000 314 3 = = 39.19 tonnes/day Fuel coefficient 130 ,000

Voyage fuel consumption = Daily fuel consumption ×

9781408176122_Ch14_7_Rev_txt_prf.indd 636

Voyage distance a 24 × Ship speed

11/16/2013 6:53:22 PM

Solutions to Questions • 637

= 39.19 ×

2, 000 = 233.3 tonnes 24 × 14

Reduction in ffuel co consumption su pt o = 304.7 − 233.3 = 71.4 tonnes Reduction in ffuel consumption c mpti (%) =

71.4 × 100 = 23. % 304.7

Q11.1 Pn = 5 × 2.5 = 12 12.5 m/s

vT Q11.2 SA =

vT v 12.6 − 12 = = 0.048 = 4.8% vT 12.6

Q11.3

ω = 0.5C 5 b − 0.05 ω = 0.5Cb − 0.05 = 0.5 × 0.6 − 0.05 = 0.25 v (1 −ω ) = 13 (1 − 0.25) = 9.75 m/s

vA

v = v A + vW ⇒ vW = v v A vW = 13 − 9.75 = 3.25 m/s Q11.4 vA

10 − 7 = 0.3 = 30% 10

SR =

vT

vT

Pn = 4.5 × 2.55 = 11.475 475 m/s

vT

=

Q11.5

SA =

vT v 11.475 − 11 = = 0.041 = 4.1% vT 11.475

ω = 0.5Cb − 0.05 = 0.5 × 0.66 − 0.05 = 0.28 vA vW

9781408176122_Ch14_7_Rev_txt_prf.indd 637

v (1 − ω ) = 11(1 − 0.28 ) = 7.92 m/s v − v A = 11 − 7.92 = 3.08 m/s

11/16/2013 6:53:25 PM

638 • Ship Stability, Powering and Resistance

SR =

vT

vA

=

vT

11.475 − 7.92 = 0.31 = 31% 11.475

Q12.1 First convert the boat speed to metres per second: ν s = 60 ×

1, 852 = 30.87 87 m/s 3, 600

Now the jet velocity can be found from the jet efficiency: η jet =

2ν s 2ν 2 × 30 87 ⇒ν j = s − ν s = − 30.87 = 72.02 m/s ν j +ν s η jet 0 60

The required thrust is approximately equal to the resistance, T = 50kN, allowing the flowrate to be found: ρ Aν j ν abs b = ρ Aν j

T ⇒ fl

t = Aν j =

ρ

(

T

)

−

=

(

j

−

s

)

50 , 000 = 1.185 185 m3 s 1, 025 (72.02 − 30.87)

Now the cross-sectional area and diameter of the jet are calculated: Aν j A=

⇒ A=

π D2 4A ⇒D = 4 π

1.185 1.185 = = 0.01646 m2 νj 72.02

4 × 0.1646 = 0.1448 m = 144.8 mm π

Q12.2 First convert the boat speed to metres per second: v s = 60 ×

1, 852 = 30.87 87 m/s 3, 600

Next the boat resistance is found from the effective power: PE

9781408176122_Ch14_7_Rev_txt_prf.indd 638

RT v s

RT =

PE 1, 600 = = 51.84 kN v s 30.87

11/16/2013 6:53:28 PM

Solutions to Questions • 639 Using the thrust deduction factor the total thrust requirement is obtained: T (1 − t ) ⇒ T =

RT

RT 51.84 = = 56.34 kN 1 − t 1 − 0.08

The thrust requirement for each waterjet unit is 28.17 kN and the thrust power requirement is: PT

Tv s = 28.17 × 30.87 = 869.7 kW Tv

Next the jet system power can be found from the engine shaft power and the efficiencies of the pump, inlet and duct: Ps × 0.87 × (1 − 0.07) = 1, 300 × 0.87 × (1 − 0.07) = 1, 051.8 kW

PJs

The jet efficiency must therefore be: η jet =

PT 869.7 = = 0.827 PJJs 1, 051.8

Now the jet velocity can be found from the jet efficiency: η jet =

2v s 2v 2 × 30 87 ⇒ v j = s − vs = − 30.87 = 43.79 m/s v j + vs η jet 0.827

Now the cross-sectional area and diameter of the jet are calculated:

(

T ρ Av A j v aabs bs = ρ Av j v j b ⇒ A=

A=

ρvj

(

T −

)

π D2 ⇒D= 4

=

vs

)

28.17 × 103 = 0.04860 04860 m2 1, 025 × 43.79 ( 43.79 − 30.87)

4A = π

4 × 0.04860 = 0.249 m = 249 mm π

Q13.1 FN cosa = 20 × cos15 = 19.32 kN

Transverse force Drag force o ce

FN sina s a = 20 × sin s 15 = 5 18 kN

Q13.2 FN FN

9781408176122_Ch14_7_Rev_txt_prf.indd 639

(

560 A × V 2 .

)2

sina

sin = 76 , 584 N = 76.58 kN sin15

11/16/2013 6:53:30 PM

640 • Ship Stability, Powering and Resistance Q13.3 FN

500 A × V 2

sina 500 × 10 × (10 × 0.514 )2 × sin (10 ) = 22939 N

T

FN × Torque lever = 22, 939 × 1.00 = 22, 939 Nm

M

FN × Bendinglever = 22, 939 × 4.00 = 91, 754 Nm

Q13.4 The normal force can be determined: FN

480 A × V 2

sina

480 × 15 × (10 × 0.514 )2 × sin25 sin = 80 , 391 N

The torque can be found: T

FN × Torque lever = 80 , 391× 0.5 = 40 ,195 Nm

The bending moment can be found: M

FN × Bendinglever = 80 , 391× 2.60 = 209 ,017 017 Nm

These can be combined to give the equivalent torque: TE = M + M 2 + T 2 = 209 , 017 + 209 , 0172 + 40 ,1952 = 421, 864 Nm Q13.5 The normal force can be determined: FN

505 A × V 2

sina 505 × 12 × (13 × 0.514 )2 × sin 35 = 155,195 N

The torque can be found: T

FN × Torque lever = 155,195 × 0.60 = 93,117 117 Nm

The bending moment can be found: M

FN × Bendinglever = 155,195 × 2.30 = 356 ,949 949 Nm

These can be combined to give the equivalent torque: TE = M + M 2 + T 2 = 356 , 949 + 356 , 9492 + 93,1172 = 725, 844 Nm This allows the stress to be found: q=

16TE 16 × 725, 844 = = 29 , 573, 545 Nm = 28. 28.57 57 MNm π D3 π × 0 53

9781408176122_Ch14_7_Rev_txt_prf.indd 640

11/16/2013 6:53:33 PM

Solutions to Questions • 641 Q13.6 The normal force can be determined: 18.01AV 2α = 18.01× 6 × (18 × 0.154 )2 × 35 = 323, 746 N

FN

The torque can be found: T

FN × Torque lever = 323, 746 × 0.15 = 48 , 562 Nm

The bending moment can be found: M

FN × Bendinglever = 323, 746 × 1.65 = 534 ,180 Nm

These can be combined to give the equivalent torque: TE = M + M 2 + T 2 = 534 ,180 + 53 534 ,1802 + 48 , 5622 = 1, 070 , 563 Nm The limiting stress can be used to find the allowable diameter: q=

16TE 16TE ∴3 =D 3 πD q ×π

D=

3

16 × 1, 070 , 563 = 0.414 m 77 × 10 6 × π

Q13.7 From the stock diameter and limiting stress, we can find the maximum allowable equivalent torque: q=

16TE q D 3 ∴ = TE π D3 16

TE =

77 × 10 6 × π × 0.4003 = 967, 611 Nm 16

The bending moment and torque can be written in terms of the normal force: T

0 5 FN

M

65 FN

These two equations can be substituted into the equivalent torque formula: TE = M + M 2 TE

.

FN + (1.65FN )2

T2 FN )2

This can be simplified: TE TE

9781408176122_Ch14_7_Rev_txt_prf.indd 641

FN + 1.652 FN2

FN + 2.723FN2

2

FN2

FN2 = 1.65FFN

FN2

11/16/2013 6:53:37 PM

642 • Ship Stability, Powering and Resistance TE

FN + 1.657FN

FN

As the maximum equivalent torque is known, this can be rewritten in terms of the formula for normal force, and appropriate values substituted so that the limiting speed can be found: TE FN = 3.307 × (18.01AV 2α ) = 59.559 AV 2α ∴ 967, 611 = 59.559 AV 2α ∴ 967, 6 611 = 59.559 × 6 × V 2 × 35 ∴V =

967, 611 = 8 80 m/s 59.559 × 6 × 35

8 80 = 17.12 knots 0.514

9781408176122_Ch14_7_Rev_txt_prf.indd 642

11/16/2013 6:53:40 PM

APPENDIX 1 MV REED – SAMPLE STABILITY DATA BOOKLET Principal Dimensions Length between perpendiculars = 100.00 m Lightship displacement = 2,615.0 tonnes Lightship KG = 7.000 m Lightship TCG = 0.000 m Lightship LCG = 44.000 m FOAP Tropical displacement = 7,530.9 tonnes Summer displacement = 7,329.0 tonnes Winter displacement = 7,129.4 tonnes Tropical draught fresh water allowance = 135 mm Summer draught fresh water allowance = 133 mm Winter draught fresh water allowance = 131 mm Tropical draught = 7.146 m Summer draught = 7.000 m Winter draught = 6.854 m

9781408176122_App01_1_Rev_txt_prf.indd 643

12/3/2013 5:32:59 PM

644 • Ship Stability, Powering and Resistance Upright Hydrostatics Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (tonnes/ cm)

MCTC (tonne metres)

8.00

8,750

44.607

40.272

4.451

7.581

14.600

101.238

7.90

8,604

44.681

40.299

4.392

7.548

14.530

100.002

7.80

8,460

44.755

40.335

4.333

7.515

14.458

98.735

7.70

8,315

44.832

40.381

4.273

7.482

14.383

97.429

7.60

8,172

44.909

40.436

4.214

7.450

14.306

96.081

7.50

8,029

44.988

40.502

4.155

7.418

14.225

94.683

7.40

7,887

45.068

40.578

4.096

7.385

14.141

93.239

7.30

7,746

45.149

40.667

4.036

7.354

14.054

91.742

7.20

7,606

45.231

40.770

3.977

7.322

13.962

90.191

7.10

7,467

45.313

40.888

3.918

7.291

13.868

88.601

7.00

7,329

45.395

41.023

3.859

7.260

13.770

86.961

6.90

7,192

45.477

41.174

3.800

7.230

13.667

85.254

6.80

7,056

45.559

41.341

3.741

7.200

13.560

83.475

6.70

6,921

45.639

41.526

3.682

7.172

13.449

81.624

6.60

6,787

45.718

41.732

3.624

7.144

13.333

79.698

6.50

6,654

45.796

41.953

3.566

7.118

13.214

77.725

6.40

6,522

45.871

42.175

3.507

7.092

13.095

75.799

6.30

6,392

45.944

42.395

3.449

7.068

12.978

73.925

6.20

6,263

46.015

42.614

3.392

7.046

12.861

72.102

6.10

6,135

46.084

42.830

3.334

7.025

12.747

70.337

6.00

6,008

46.150

43.043

3.277

7.006

12.634

68.639

5.90

5,882

46.215

43.254

3.219

6.988

12.524

67.005

5.80

5,757

46.276

43.460

3.162

6.973

12.415

65.433

5.70

5,634

46.336

43.662

3.106

6.959

12.309

63.924

5.60

5,511

46.393

43.858

3.049

6.948

12.205

62.479

5.50

5,390

46.448

44.049

2.993

6.939

12.104

61.104

5.40

5,269

46.501

44.233

2.937

6.931

12.005

59.800

5.30

5,150

46.552

44.409

2.880

6.927

11.910

58.573

5.20

5,031

46.600

44.578

2.825

6.925

11.818

57.407

5.10

4,913

46.647

44.738

2.769

6.925

11.728

56.315

5.00

4,796

46.691

44.889

2.713

6.928

11.643

55.305

9781408176122_App01_1_Rev_txt_prf.indd 644

12/3/2013 5:32:59 PM

Appendix 1 • 645

Draught (m)

Displacement LCB LCF KB (m) KM (m) TPC (tonnes) (m FOAP) (m FOAP) (tonnes/ cm)

MCTC (tonne metres)

4.90

4,680

46.734

45.035

2.658

6.934

11.559

54.340

4.80

4,565

46.775

45.174

2.603

6.942

11.477

53.425

4.70

4,451

46.815

45.309

2.547

6.954

11.397

52.558

4.60

4,337

46.853

45.446

2.492

6.969

11.319

51.751

4.50

4,224

46.888

45.579

2.437

6.987

11.244

51.004

4.40

4,112

46.922

45.706

2.382

7.009

11.168

50.265

4.30

4,001

46.954

45.828

2.328

7.034

11.095

49.574

4.20

3,890

46.985

45.947

2.273

7.063

11.023

48.913

4.10

3,781

47.013

46.061

2.219

7.097

10.951

48.286

4.00

3,671

47.040

46.173

2.164

7.134

10.882

47.704

3.90

3,563

47.064

46.292

2.110

7.177

10.816

47.229

3.80

3,455

47.087

46.400

2.055

7.224

10.749

46.737

3.70

3,348

47.107

46.505

2.001

7.276

10.684

46.273

3.60

3,241

47.125

46.605

1.947

7.335

10.619

45.838

3.50

3,136

47.141

46.713

1.893

7.399

10.557

45.498

3.40

3,030

47.154

46.811

1.839

7.470

10.496

45.155

3.30

2,926

47.165

46.905

1.785

7.549

10.434

44.831

3.20

2,822

47.173

46.992

1.731

7.635

10.372

44.517

3.10

2,718

47.179

47.064

1.677

7.730

10.309

44.155

3.00

2,615

47.182

47.141

1.623

7.834

10.247

43.868

2.90

2,513

47.181

47.212

1.569

7.949

10.185

43.587

2.80

2,412

47.179

47.273

1.515

8.076

10.122

43.291

2.70

2,311

47.173

47.327

1.461

8.215

10.057

42.992

2.60

2,211

47.165

47.375

1.407

8.368

9.992

42.696

2.50

2,111

47.154

47.416

1.353

8.536

9.925

42.394

2.40

2,012

47.140

47.45

1.299

8.722

9.856

42.083

2.30

1,914

47.123

47.479

1.245

8.928

9.786

41.767

2.20

1,816

47.104

47.499

1.191

9.155

9.713

41.431

2.10

1,720

47.082

47.511

1.137

9.407

9.638

41.089

2.00

1,624

47.057

47.506

1.083

9.688

9.557

40.659

These hydrostatic values have been calculated with the vessel floating on an even keel.

9781408176122_App01_1_Rev_txt_prf.indd 645

12/3/2013 5:33:00 PM

646 • Ship Stability, Powering and Resistance No. 1 Double Bottom Port

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

1.880

0.000

100.000

13.483

84.877

1.413

1.348

0.000

1.858

0.023

97.900

13.199

84.871

1.411

1.334

1.938

1.800

0.081

92.600

12.486

84.853

1.404

1.299

1.842

1.700

0.181

83.700

11.284

84.816

1.392

1.239

1.678

1.600

0.281

75.100

10.119

84.771

1.379

1.178

1.516

1.500

0.381

66.700

8.994

84.717

1.366

1.116

1.359

1.400

0.481

58.700

7.913

84.650

1.353

1.054

1.206

1.300

0.581

51.000

6.879

84.569

1.339

0.992

1.058

1.200

0.681

43.700

5.897

84.470

1.324

0.929

0.915

1.100

0.781

36.900

4.970

84.349

1.309

0.865

0.779

1.000

0.881

30.400

4.105

84.199

1.294

0.801

0.651

0.900

0.981

24.500

3.306

84.015

1.277

0.736

0.529

0.800

1.081

19.100

2.581

83.789

1.260

0.670

0.418

0.700

1.181

14.400

1.936

83.516

1.243

0.603

0.314

0.600

1.281

10.200

1.382

83.209

1.223

0.536

0.223

0.500

1.381

6.800

0.921

82.873

1.202

0.468

0.146

0.400

1.481

4.100

0.555

82.509

1.177

0.401

0.086

0.300

1.581

2.100

0.283

82.092

1.147

0.332

0.042

0.220

1.661

1.000

0.135

81.710

1.120

0.277

0.019

0.200

1.681

0.800

0.106

81.596

1.112

0.263

0.015

0.100

1.781

0.100

0.018

80.977

1.067

0.193

0.002

0.000

1.881

0.000

0.000

80.977

1.067

0.193

0.000

9781408176122_App01_1_Rev_txt_prf.indd 646

12/3/2013 5:33:01 PM

Appendix 1 • 647 No. 1 Double Bottom Starboard

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

1.880

0.000

100.000

13.483

84.877

–1.413

1.348

0.000

1.858

0.023

97.900

13.199

84.871

–1.411

1.334

1.938

1.800

0.081

92.600

12.486

84.853

–1.404

1.299

1.842

1.700

0.181

83.700

11.284

84.816

–1.392

1.239

1.678

1.600

0.281

75.100

10.119

84.771

–1.379

1.178

1.516

1.500

0.381

66.700

8.994

84.717

–1.366

1.116

1.359

1.400

0.481

58.700

7.913

84.650

–1.353

1.054

1.206

1.300

0.581

51.000

6.879

84.569

–1.339

0.992

1.058

1.200

0.681

43.700

5.897

84.470

–1.324

0.929

0.915

1.100

0.781

36.900

4.970

84.349

–1.309

0.865

0.779

1.000

0.881

30.400

4.105

84.199

–1.294

0.801

0.651

0.900

0.981

24.500

3.306

84.015

–1.277

0.736

0.529

0.800

1.081

19.100

2.581

83.789

–1.260

0.670

0.418

0.700

1.181

14.400

1.936

83.516

–1.243

0.603

0.314

0.600

1.281

10.200

1.382

83.209

–1.223

0.536

0.223

0.500

1.381

6.800

0.921

82.873

–1.202

0.468

0.146

0.400

1.481

4.100

0.555

82.509

–1.177

0.401

0.086

0.300

1.581

2.100

0.283

82.092

–1.147

0.332

0.042

0.220

1.661

1.000

0.135

81.710

–1.120

0.277

0.019

0.200

1.681

0.800

0.106

81.596

–1.112

0.263

0.015

0.100

1.781

0.100

0.018

80.977

–1.067

0.193

0.002

0.000

1.881

0.000

0.000

80.977

–1.067

0.193

0.000

9781408176122_App01_1_Rev_txt_prf.indd 647

12/3/2013 5:33:02 PM

648 • Ship Stability, Powering and Resistance No. 2 Double Bottom Port

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

100.860

67.769

2.579

1.172

0.000

1.965

0.030

97.900

98.740

67.759

2.572

1.154

113.437

1.900

0.096

93.400

94.170

67.738

2.555

1.116

109.882

1.800

0.196

86.600

87.307

67.704

2.529

1.058

104.425

1.700

0.296

79.900

80.591

67.667

2.502

1.000

98.922

1.600

0.396

73.400

74.030

67.628

2.474

0.942

93.413

1.500

0.496

67.100

67.628

67.585

2.445

0.884

87.899

1.400

0.596

60.900

61.393

67.539

2.415

0.826

82.398

1.300

0.696

54.900

55.330

67.488

2.384

0.768

76.927

1.200

0.796

49.000

49.448

67.432

2.351

0.710

71.471

1.100

0.896

43.400

43.755

67.370

2.317

0.652

66.035

1.000

0.996

37.900

38.260

67.300

2.282

0.595

60.632

0.900

1.096

32.700

32.975

67.221

2.244

0.537

55.241

0.800

1.196

27.700

27.911

67.130

2.204

0.479

49.859

0.700

1.296

22.900

23.084

67.022

2.161

0.422

44.493

0.600

1.396

18.400

18.513

66.891

2.113

0.364

39.131

0.500

1.496

14.100

14.219

66.726

2.061

0.306

33.736

0.400

1.596

10.200

10.238

66.505

1.999

0.248

28.245

0.300

1.696

6.600

6.626

66.188

1.922

0.190

22.406

0.200

1.796

3.500

3.492

65.670

1.817

0.130

15.538

0.100

1.896

1.100

1.075

64.617

1.645

0.069

6.986

0.096

1.899

1.000

1.004

64.557

1.636

0.067

6.638

0.000

1.996

0.000

0.000

64.557

1.636

0.067

0.000

9781408176122_App01_1_Rev_txt_prf.indd 648

12/3/2013 5:33:03 PM

Appendix 1 • 649 No. 2 Double Bottom Starboard

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

100.860

67.769

–2.579

1.172

0.000

1.965

0.030

97.900

98.740

67.759

–2.572

1.154

113.437

1.900

0.096

93.400

94.170

67.738

–2.555

1.116

109.882

1.800

0.196

86.600

87.307

67.704

–2.529

1.058

104.425

1.700

0.296

79.900

80.591

67.667

–2.502

1.000

98.922

1.600

0.396

73.400

74.030

67.628

–2.474

0.942

93.413

1.500

0.496

67.100

67.628

67.585

–2.445

0.884

87.899

1.400

0.596

60.900

61.393

67.539

–2.415

0.826

82.398

1.300

0.696

54.900

55.330

67.488

–2.384

0.768

76.927

1.200

0.796

49.000

49.448

67.432

–2.351

0.710

71.471

1.100

0.896

43.400

43.755

67.370

–2.317

0.652

66.035

1.000

0.996

37.900

38.260

67.300

–2.282

0.595

60.632

0.900

1.096

32.700

32.975

67.221

–2.244

0.537

55.241

0.800

1.196

27.700

27.911

67.130

–2.204

0.479

49.859

0.700

1.296

22.900

23.084

67.022

–2.161

0.422

44.493

0.600

1.396

18.400

18.513

66.891

–2.113

0.364

39.131

0.500

1.496

14.100

14.219

66.726

–2.061

0.306

33.736

0.400

1.596

10.200

10.238

66.505

–1.999

0.248

28.245

0.300

1.696

6.600

6.626

66.188

–1.922

0.190

22.406

0.200

1.796

3.500

3.492

65.670

–1.817

0.130

15.538

0.100

1.896

1.100

1.075

64.617

–1.645

0.069

6.986

0.096

1.899

1.000

1.004

64.557

–1.636

0.067

6.638

0.000

1.996

0.000

0.000

64.557

–1.636

0.067

0.000

9781408176122_App01_1_Rev_txt_prf.indd 649

12/3/2013 5:33:04 PM

650 • Ship Stability, Powering and Resistance No. 3 Double Bottom Port

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

245.966

49.375

4.159

1.057

0.000

1.961

0.038

97.900

240.800

49.369

4.153

1.037

539.733

1.900

0.098

94.600

232.574

49.360

4.144

1.005

536.830

1.800

0.198

89.000

219.031

49.345

4.128

0.953

531.593

1.700

0.298

83.600

205.546

49.330

4.111

0.900

525.930

1.600

0.398

78.100

192.126

49.313

4.092

0.848

519.630

1.500

0.498

72.700

178.778

49.296

4.072

0.795

512.618

1.400

0.598

67.300

165.511

49.278

4.050

0.743

504.900

1.300

0.698

61.900

152.334

49.260

4.026

0.690

496.119

1.200

0.798

56.600

139.259

49.240

4.001

0.637

486.390

1.100

0.898

51.300

126.299

49.220

3.972

0.584

475.297

1.000

0.998

46.100

113.469

49.198

3.941

0.532

462.694

0.900

1.098

41.000

100.786

49.175

3.907

0.479

448.363

0.800

1.198

35.900

88.271

49.150

3.870

0.426

431.958

0.700

1.298

30.900

75.949

49.122

3.828

0.373

413.302

0.600

1.398

26.000

63.846

49.089

3.781

0.320

392.560

0.500

1.498

21.100

51.992

49.048

3.727

0.268

369.749

0.400

1.598

16.400

40.423

48.994

3.664

0.215

344.387

0.300

1.698

11.900

29.191

48.917

3.585

0.162

315.354

0.200

1.798

7.500

18.382

48.787

3.475

0.109

280.723

0.100

1.898

3.300

8.194

48.497

3.282

0.056

232.330

0.037

1.961

1.000

2.459

47.978

2.963

0.023

165.470

0.000

1.998

0.000

0.000

47.978

2.963

0.023

0.000

9781408176122_App01_1_Rev_txt_prf.indd 650

12/3/2013 5:33:04 PM

Appendix 1 • 651 No. 3 Double Bottom Starboard

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

245.966

49.375

–4.159

1.057

0.000

1.961

0.038

97.900

240.800

49.369

–4.153

1.037

539.733

1.900

0.098

94.600

232.574

49.360

–4.144

1.005

536.830

1.800

0.198

89.000

219.031

49.345

–4.128

0.953

531.593

1.700

0.298

83.600

205.546

49.330

–4.111

0.900

525.930

1.600

0.398

78.100

192.126

49.313

–4.092

0.848

519.630

1.500

0.498

72.700

178.778

49.296

–4.072

0.795

512.618

1.400

0.598

67.300

165.511

49.278

–4.050

0.743

504.900

1.300

0.698

61.900

152.334

49.260

–4.026

0.690

496.119

1.200

0.798

56.600

139.259

49.240

–4.001

0.637

486.390

1.100

0.898

51.300

126.299

49.220

–3.972

0.584

475.297

1.000

0.998

46.100

113.469

49.198

–3.941

0.532

462.694

0.900

1.098

41.000

100.786

49.175

–3.907

0.479

448.363

0.800

1.198

35.900

88.271

49.150

–3.870

0.426

431.958

0.700

1.298

30.900

75.949

49.122

–3.828

0.373

413.302

0.600

1.398

26.000

63.846

49.089

–3.781

0.320

392.560

0.500

1.498

21.100

51.992

49.048

–3.727

0.268

369.749

0.400

1.598

16.400

40.423

48.994

–3.664

0.215

344.387

0.300

1.698

11.900

29.191

48.917

–3.585

0.162

315.354

0.200

1.798

7.500

18.382

48.787

–3.475

0.109

280.723

0.100

1.898

3.300

8.194

48.497

–3.282

0.056

232.330

0.037

1.961

1.000

2.459

47.978

–2.963

0.023

165.470

0.000

1.998

0.000

0.000

47.978

–2.963

0.023

0.000

9781408176122_App01_1_Rev_txt_prf.indd 651

12/3/2013 5:33:05 PM

652 • Ship Stability, Powering and Resistance No. 4 Double Bottom Port

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

209.601

31.529

3.893

1.088

0.000

1.963

0.035

97.900

205.198

31.542

3.886

1.069

471.414

1.900

0.098

94.100

197.302

31.566

3.873

1.034

466.058

1.800

0.198

88.200

184.953

31.605

3.852

0.979

457.376

1.700

0.298

82.400

172.750

31.644

3.829

0.925

448.425

1.600

0.398

76.700

160.698

31.685

3.806

0.870

439.064

1.500

0.498

71.000

148.802

31.726

3.782

0.816

429.341

1.400

0.598

65.400

137.068

31.769

3.756

0.762

419.155

1.300

0.698

59.900

125.504

31.813

3.729

0.707

408.404

1.200

0.798

54.400

114.117

31.859

3.700

0.653

396.988

1.100

0.898

49.100

102.916

31.907

3.670

0.599

384.795

1.000

0.998

43.900

91.912

31.959

3.637

0.544

371.561

0.900

1.098

38.700

81.118

32.015

3.602

0.490

357.127

0.800

1.198

33.700

70.552

32.077

3.564

0.436

341.253

0.700

1.298

28.700

60.233

32.147

3.522

0.382

323.809

0.600

1.398

23.900

50.186

32.230

3.477

0.328

304.900

0.500

1.498

19.300

40.440

32.332

3.426

0.274

284.519

0.400

1.598

14.800

31.036

32.463

3.369

0.220

262.250

0.300

1.698

10.500

22.036

32.645

3.301

0.166

237.208

0.200

1.798

6.500

13.548

32.928

3.214

0.112

207.886

0.100

1.898

2.800

5.808

33.476

3.076

0.057

167.918

0.045

1.954

1.000

2.095

34.169

2.899

0.027

125.991

0.000

1.998

0.000

0.000

39.786

1.950

0.002

0.000

9781408176122_App01_1_Rev_txt_prf.indd 652

12/3/2013 5:33:06 PM

Appendix 1 • 653 No. 4 Double Bottom Starboard

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

209.601

31.529

–3.893

1.088

0.000

1.963

0.035

97.900

205.198

31.542

–3.886

1.069

471.414

1.900

0.098

94.100

197.302

31.566

–3.873

1.034

466.058

1.800

0.198

88.200

184.953

31.605

–3.852

0.979

457.376

1.700

0.298

82.400

172.750

31.644

–3.829

0.925

448.425

1.600

0.398

76.700

160.698

31.685

–3.806

0.870

439.064

1.500

0.498

71.000

148.802

31.726

–3.782

0.816

429.341

1.400

0.598

65.400

137.068

31.769

–3.756

0.762

419.155

1.300

0.698

59.900

125.504

31.813

–3.729

0.707

408.404

1.200

0.798

54.400

114.117

31.859

–3.700

0.653

396.988

1.100

0.898

49.100

102.916

31.907

–3.670

0.599

384.795

1.000

0.998

43.900

91.912

31.959

–3.637

0.544

371.561

0.900

1.098

38.700

81.118

32.015

–3.602

0.490

357.127

0.800

1.198

33.700

70.552

32.077

–3.564

0.436

341.253

0.700

1.298

28.700

60.233

32.147

–3.522

0.382

323.809

0.600

1.398

23.900

50.186

32.230

–3.477

0.328

304.900

0.500

1.498

19.300

40.440

32.332

–3.426

0.274

284.519

0.400

1.598

14.800

31.036

32.463

–3.369

0.220

262.250

0.300

1.698

10.500

22.036

32.645

–3.301

0.166

237.208

0.200

1.798

6.500

13.548

32.928

–3.214

0.112

207.886

0.100

1.898

2.800

5.808

33.476

–3.076

0.057

167.918

0.045

1.954

1.000

2.095

34.169

–2.899

0.027

125.991

0.000

1.998

0.000

0.000

39.786

–1.950

0.002

0.000

9781408176122_App01_1_Rev_txt_prf.indd 653

12/3/2013 5:33:07 PM

654 • Ship Stability, Powering and Resistance No. 5 Double Bottom Port

Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

27.847

15.490

1.793

1.213

0.000

1.936

0.029

97.900

27.261

15.486

1.787

1.196

12.820

1.900

0.064

95.300

26.543

15.482

1.781

1.176

12.352

1.800

0.164

88.200

24.562

15.471

1.762

1.118

11.145

1.700

0.264

81.200

22.625

15.464

1.744

1.061

10.076

1.600

0.364

74.500

20.733

15.460

1.725

1.004

9.124

1.500

0.464

67.800

18.883

15.461

1.707

0.947

8.269

1.400

0.564

61.300

17.075

15.466

1.689

0.890

7.495

1.300

0.664

55.000

15.309

15.477

1.670

0.833

6.786

1.200

0.764

48.800

13.588

15.495

1.651

0.776

6.127

1.100

0.864

42.800

11.915

15.521

1.630

0.718

5.511

1.000

0.964

37.000

10.293

15.557

1.608

0.660

4.924

0.900

1.064

31.300

8.729

15.607

1.585

0.602

4.353

0.800

1.164

26.000

7.232

15.672

1.559

0.543

3.791

0.700

1.264

20.900

5.813

15.759

1.530

0.483

3.230

0.600

1.364

16.100

4.484

15.875

1.498

0.423

2.659

0.500

1.464

11.700

3.265

16.032

1.460

0.362

2.086

0.400

1.564

7.800

2.180

16.256

1.416

0.301

1.512

0.300

1.664

4.500

1.261

16.594

1.363

0.237

0.944

0.200

1.764

2.000

0.556

17.101

1.293

0.173

0.448

0.144

1.820

1.000

0.276

17.523

1.244

0.136

0.233

0.100

1.864

0.400

0.121

18.011

1.197

0.106

0.098

0.000

1.964

0.000

0.000

18.011

1.197

0.106

0.000

9781408176122_App01_1_Rev_txt_prf.indd 654

12/3/2013 5:33:08 PM

Appendix 1 • 655 No. 5 Double Bottom Starboard Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m3)

(m)

(m)

(m)

(m4)

2.000

0.000

100.000

27.847

15.490

–1.793

1.213

0.000

1.936

0.029

97.900

27.261

15.486

–1.787

1.196

12.820

1.900

0.064

95.300

26.543

15.482

–1.781

1.176

12.352

1.800

0.164

88.200

24.562

15.471

–1.762

1.118

11.145

1.700

0.264

81.200

22.625

15.464

–1.744

1.061

10.076

1.600

0.364

74.500

20.733

15.460

–1.725

1.004

9.124

1.500

0.464

67.800

18.883

15.461

–1.707

0.947

8.269

1.400

0.564

61.300

17.075

15.466

–1.689

0.890

7.495

1.300

0.664

55.000

15.309

15.477

–1.670

0.833

6.786

1.200

0.764

48.800

13.588

15.495

–1.651

0.776

6.127

1.100

0.864

42.800

11.915

15.521

–1.630

0.718

5.511

1.000

0.964

37.000

10.293

15.557

–1.608

0.660

4.924

0.900

1.064

31.300

8.729

15.607

–1.585

0.602

4.353

0.800

1.164

26.000

7.232

15.672

–1.559

0.543

3.791

0.700

1.264

20.900

5.813

15.759

–1.530

0.483

3.230

0.600

1.364

16.100

4.484

15.875

–1.498

0.423

2.659

0.500

1.464

11.700

3.265

16.032

–1.460

0.362

2.086

0.400

1.564

7.800

2.180

16.256

–1.416

0.301

1.512

0.300

1.664

4.500

1.261

16.594

–1.363

0.237

0.944

0.200

1.764

2.000

0.556

17.101

–1.293

0.173

0.448

0.144

1.820

1.000

0.276

17.523

–1.244

0.136

0.233

0.100

1.864

0.400

0.121

18.011

–1.197

0.106

0.098

0.000

1.964

0.000

0.000

18.011

–1.197

0.106

0.000

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656 • Ship Stability, Powering and Resistance No. 1 Wing Tank Port Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m )

(m)

(m)

(m)

(m4)

4.600

0.000

100.000

32.219

85.533

3.587

8.904

0.000

4.569

0.032

97.900

31.542

85.509

3.586

8.881

18.861

4.400

0.201

87.200

28.090

85.376

3.582

8.757

17.263

4.200

0.401

75.400

24.307

85.204

3.582

8.610

15.462

4.000

0.601

64.700

20.841

85.010

3.586

8.462

13.760

3.800

0.801

54.900

17.683

84.791

3.595

8.312

12.127

3.600

1.001

46.000

14.831

84.546

3.610

8.161

10.516

3.400

1.201

38.100

12.286

84.274

3.631

8.007

8.906

3.200

1.401

31.200

10.043

83.975

3.658

7.852

7.380

3.000

1.601

25.100

8.088

83.647

3.691

7.695

5.861

2.800

1.801

19.900

6.413

83.302

3.728

7.537

4.289

2.600

2.001

15.600

5.010

82.972

3.762

7.379

2.844

2.400

2.201

12.000

3.855

82.673

3.792

7.221

1.844

2.200

2.401

9.000

2.914

82.399

3.817

7.066

1.198

2.000

2.601

6.700

2.153

82.141

3.840

6.912

0.758

1.800

2.801

4.800

1.546

81.896

3.861

6.758

0.465

1.600

3.001

3.300

1.071

81.663

3.880

6.604

0.273

1.400

3.201

2.200

0.709

81.442

3.898

6.452

0.152

1.200

3.401

1.400

0.444

81.234

3.913

6.300

0.080

1.079

3.521

1.000

0.322

81.110

3.923

6.209

0.052

1.000

3.601

0.800

0.256

81.028

3.929

6.149

0.037

0.800

3.801

0.400

0.131

80.825

3.945

5.998

0.016

0.600

4.001

0.200

0.055

80.624

3.961

5.846

0.005

0.400

4.201

0.100

0.017

80.435

3.977

5.695

0.001

0.200

4.401

0.000

0.002

80.247

3.995

5.544

0.000

0.000

4.601

0.000

0.000

80.247

3.995

5.544

0.000

9781408176122_App01_2_Rev_txt_prf.indd 656

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Appendix 1 • 657 No. 1 Wing Tank Starboard Sounding

Ullage

Fill

Capacity

LCG

TCG

KG

FSM

(m)

(m)

(%)

(m )

(m)

(m)

(m)

(m4)

4.600

0.000

100.000

32.219

85.533

–3.587

8.904

0.000

4.569

0.032

97.900

31.542

85.509

–3.586

8.881

18.861

4.400

0.201

87.200

28.090

85.376

–3.582

8.757

17.263

4.200

0.401

75.400

24.307

85.204

–3.582

8.610

15.462

4.000

0.601

64.700

20.841

85.010

–3.586

8.462

13.760

3.800

0.801

54.900

17.683

84.791

–3.595

8.312

12.127

3.600

1.001

46.000

14.831

84.546

–3.610

8.161

10.516

3.400

1.201

38.100

12.286

84.274

–3.631

8.007

8.906

3.200

1.401

31.200

10.043

83.975

–3.658

7.852

7.380

3.000

1.601

25.100

8.088

83.647

–3.691

7.695

5.861

2.800

1.801

19.900

6.413

83.302

–3.728

7.537

4.289

2.600

2.001

15.600

5.010

82.972

–3.762

7.379

2.844

2.400

2.201

12.000

3.855

82.673

–3.792

7.221

1.844

2.200

2.401

9.000

2.914

82.399

–3.817

7.066

1.198

2.000

2.601

6.700

2.153

82.141

–3.840

6.912

0.758

1.800

2.801

4.800

1.546

81.896

–3.861

6.758

0.465

1.600

3.001

3.300

1.071

81.663

–3.880

6.604

0.273

1.400

3.201

2.200

0.709

81.442

–3.898

6.452

0.152

1.200

3.401

1.400

0.444

81.234

–3.913

6.300

0.080

1.079

3.521

1.000

0.322

81.110

–3.923

6.209

0.052

1.000

3.601

0.800

0.256

81.028

–3.929

6.149

0.037

0.800

3.801

0.400

0.131

80.825

–3.945

5.998

0.016

0.600

4.001

0.200

0.055

80.624

–3.961

5.846

0.005

0.400

4.201

0.100

0.017

80.435

–3.977

5.695

0.001

0.200

4.401

0.000

0.002

80.247

–3.995

5.544

0.000

0.000

4.601

0.000

0.000

80.247

–3.995

5.544

0.000

9781408176122_App01_2_Rev_txt_prf.indd 657

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658 • Ship Stability, Powering and Resistance No. 2 Wing Tank Port Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

8.000

0.000

100.000

200.066

68.448

6.039

7.300

0.000

7.912

0.088

97.900

195.846

68.419

6.035

7.242

40.949

7.500

0.500

88.300

176.612

68.280

6.016

6.974

38.082

7.000

1.000

77.200

154.486

68.095

5.995

6.647

34.849

6.500

1.500

66.800

133.717

67.890

5.976

6.320

31.648

6.000

2.000

57.100

114.332

67.662

5.960

5.993

28.639

5.500

2.500

48.200

96.334

67.406

5.947

5.664

25.757

5.000

3.000

39.900

79.766

67.116

5.937

5.334

22.783

4.500

3.500

32.300

64.655

66.786

5.931

5.002

19.845

4.000

4.000

25.500

51.041

66.410

5.930

4.667

16.764

3.500

4.500

19.500

38.982

65.980

5.933

4.331

13.511

3.000

5.000

14.300

28.538

65.500

5.939

3.992

10.215

2.500

5.500

9.900

19.735

64.962

5.948

3.650

7.321

2.000

6.000

6.300

12.604

64.352

5.961

3.306

4.773

1.500

6.500

3.600

7.174

63.672

5.978

2.959

2.676

1.000

7.000

1.700

3.398

62.935

5.999

2.617

1.185

0.731

7.269

1.000

1.995

62.530

6.010

2.437

0.666

0.500

7.500

0.600

1.103

62.192

6.019

2.288

0.362

0.000

8.000

0.000

0.000

62.192

6.019

2.288

0.000

No. 2 Wing Tank Starboard Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

8.000

0.000

100.000

200.066

68.448

–6.039

7.300

0.000

7.912

0.088

97.900

195.846

68.419

–6.035

7.242

40.949

7.500

0.500

88.300

176.612

68.280

–6.016

6.974

38.082

7.000

1.000

77.200

154.486

68.095

–5.995

6.647

34.849

6.500

1.500

66.800

133.717

67.890

–5.976

6.320

31.648

9781408176122_App01_2_Rev_txt_prf.indd 658

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Appendix 1 • 659

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

6.000

2.000

57.100

114.332

67.662

–5.960

5.993

28.639

5.500

2.500

48.200

96.334

67.406

–5.947

5.664

25.757

5.000

3.000

39.900

79.766

67.116

–5.937

5.334

22.783

4.500

3.500

32.300

64.655

66.786

–5.931

5.002

19.845

4.000

4.000

25.500

51.041

66.410

–5.930

4.667

16.764

3.500

4.500

19.500

38.982

65.980

–5.933

4.331

13.511

3.000

5.000

14.300

28.538

65.500

–5.939

3.992

10.215

2.500

5.500

9.900

19.735

64.962

–5.948

3.650

7.321

2.000

6.000

6.300

12.604

64.352

–5.961

3.306

4.773

1.500

6.500

3.600

7.174

63.672

–5.978

2.959

2.676

1.000

7.000

1.700

3.398

62.935

–5.999

2.617

1.185

0.731

7.269

1.000

1.995

62.530

–6.010

2.437

0.666

0.500

7.500

0.600

1.103

62.192

–6.019

2.288

0.362

0.000

8.000

0.000

0.000

62.192

–6.019

2.288

0.000

No. 3 Wing Tank Port Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

8.000

0.000

100.000

337.599

49.589

7.068

6.103

0.000

7.848

0.152

98.000

330.846

49.581

7.067

6.025

0.000

7.841

0.159

97.900

330.509

49.581

7.067

6.022

18.331

7.500

0.500

93.400

315.380

49.564

7.065

5.847

18.201

7.000

1.000

86.900

293.245

49.537

7.062

5.590

18.016

6.500

1.500

80.300

271.197

49.509

7.058

5.333

17.817

6.000

2.000

73.800

249.251

49.478

7.055

5.076

17.605

5.500

2.500

67.400

227.409

49.446

7.051

4.819

17.404

5.000

3.000

60.900

205.675

49.410

7.047

4.562

17.201

4.500

3.500

54.500

184.065

49.371

7.043

4.305

16.968

4.000

4.000

48.200

162.591

49.328

7.038

4.048

16.744

3.500

4.500

41.800

141.265

49.280

7.033

3.791

16.502

9781408176122_App01_2_Rev_txt_prf.indd 659

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660 • Ship Stability, Powering and Resistance

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

3.000

5.000

35.600

120.112

49.226

7.027

3.534

16.235

2.500

5.500

29.400

99.159

49.163

7.021

3.277

15.949

2.000

6.000

23.200

78.448

49.089

7.014

3.020

15.617

1.500

6.500

17.200

58.043

49.001

7.006

2.763

15.232

1.000

7.000

11.300

38.047

48.894

6.996

2.507

14.743

0.500

7.500

5.500

18.621

48.764

6.984

2.252

14.088

0.092

7.908

1.000

3.368

48.635

6.973

2.046

13.337

0.000

8.000

0.000

0.000

48.635

6.973

2.046

0.000

No. 3 Wing Tank Starboard Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

8.000

0.000

100.000

337.599

49.589

–7.068

6.103

0.000

7.848

0.152

98.000

330.846

49.581

–7.067

6.025

0.000

7.841

0.159

97.900

330.509

49.581

–7.067

6.022

18.331

7.500

0.500

93.400

315.380

49.564

–7.065

5.847

18.201

7.000

1.000

86.900

293.245

49.537

–7.062

5.590

18.016

6.500

1.500

80.300

271.197

49.509

–7.058

5.333

17.817

6.000

2.000

73.800

249.251

49.478

–7.055

5.076

17.605

5.500

2.500

67.400

227.409

49.446

–7.051

4.819

17.404

5.000

3.000

60.900

205.675

49.410

–7.047

4.562

17.201

4.500

3.500

54.500

184.065

49.371

–7.043

4.305

16.968

4.000

4.000

48.200

162.591

49.328

–7.038

4.048

16.744

3.500

4.500

41.800

141.265

49.280

–7.033

3.791

16.502

3.000

5.000

35.600

120.112

49.226

–7.027

3.534

16.235

2.500

5.500

29.400

99.159

49.163

–7.021

3.277

15.949

2.000

6.000

23.200

78.448

49.089

–7.014

3.020

15.617

1.500

6.500

17.200

58.043

49.001

–7.006

2.763

15.232

9781408176122_App01_2_Rev_txt_prf.indd 660

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Appendix 1 • 661

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

1.000

7.000

11.300

38.047

48.894

–6.996

2.507

14.743

0.500

7.500

5.500

18.621

48.764

–6.984

2.252

14.088

0.092

7.908

1.000

3.368

48.635

–6.973

2.046

13.337

0.000

8.000

0.000

0.000

48.635

–6.973

2.046

0.000

No. 4 Wing Tank Port Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

8.000

0.000

100.000

334.633

30.455

7.079

6.189

0.000

7.843

0.157

97.900

327.606

30.464

7.079

6.109

18.538

7.500

0.500

93.300

312.312

30.486

7.077

5.935

18.536

7.000

1.000

86.700

289.992

30.521

7.074

5.680

18.530

6.500

1.500

80.000

267.685

30.563

7.070

5.424

18.471

6.000

2.000

73.300

245.405

30.610

7.066

5.167

18.406

5.500

2.500

66.700

223.153

30.667

7.062

4.910

18.326

5.000

3.000

60.100

200.952

30.733

7.056

4.651

18.180

4.500

3.500

53.400

178.822

30.813

7.050

4.391

18.007

4.000

4.000

46.900

156.789

30.910

7.043

4.130

17.757

3.500

4.500

40.300

134.894

31.031

7.034

3.867

17.431

3.000

5.000

33.800

113.203

31.185

7.024

3.602

16.989

2.500

5.500

27.400

91.818

31.384

7.012

3.334

16.405

2.000

6.000

21.200

70.923

31.646

6.999

3.064

15.656

1.500

6.500

15.200

50.832

31.988

6.986

2.792

14.774

1.000

7.000

9.600

32.011

32.412

6.976

2.521

13.588

0.500

7.500

4.500

14.987

32.867

6.966

2.256

11.929

0.118

7.882

1.000

3.346

33.212

6.958

2.059

10.552

0.000

8.000

0.000

0.000

33.212

6.958

2.059

0.000

9781408176122_App01_2_Rev_txt_prf.indd 661

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662 • Ship Stability, Powering and Resistance No. 4 Wing Tank Starboard Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

8.000

0.000

100.000

334.633

30.455

–7.079

6.189

0.000

7.843

0.157

97.900

327.606

30.464

–7.079

6.109

18.538

7.500

0.500

93.300

312.312

30.486

–7.077

5.935

18.536

7.000

1.000

86.700

289.992

30.521

–7.074

5.680

18.530

6.500

1.500

80.000

267.685

30.563

–7.070

5.424

18.471

6.000

2.000

73.300

245.405

30.610

–7.066

5.167

18.406

5.500

2.500

66.700

223.153

30.667

–7.062

4.910

18.326

5.000

3.000

60.100

200.952

30.733

–7.056

4.651

18.180

4.500

3.500

53.400

178.822

30.813

–7.050

4.391

18.007

4.000

4.000

46.900

156.789

30.910

–7.043

4.130

17.757

3.500

4.500

40.300

134.894

31.031

–7.034

3.867

17.431

3.000

5.000

33.800

113.203

31.185

–7.024

3.602

16.989

2.500

5.500

27.400

91.818

31.384

–7.012

3.334

16.405

2.000

6.000

21.200

70.923

31.646

–6.999

3.064

15.656

1.500

6.500

15.200

50.832

31.988

–6.986

2.792

14.774

1.000

7.000

9.600

32.011

32.412

–6.976

2.521

13.588

0.500

7.500

4.500

14.987

32.867

–6.966

2.256

11.929

0.118

7.882

1.000

3.346

33.212

–6.958

2.059

10.552

0.000

8.000

0.000

0.000

33.212

–6.958

2.059

0.000

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

7.000

0.000

100.000

122.670

13.668

6.928

7.807

0.000

6.956

0.080

97.900

121.245

13.681

6.926

7.782

12.825

6.500

0.536

86.000

106.472

13.832

6.905

7.517

12.348

6.000

1.036

73.200

90.613

14.036

6.878

7.220

11.473

Port Bunker

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Appendix 1 • 663

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

5.500

1.536

60.700

75.229

14.301

6.847

6.914

10.494

5.000

2.036

48.800

60.492

14.648

6.812

6.597

9.326

4.500

2.536

37.700

46.660

15.116

6.774

6.265

7.995

4.000

3.036

27.500

34.112

15.743

6.736

5.914

6.676

3.500

3.536

19.000

23.475

16.486

6.699

5.547

5.110

3.000

4.036

12.300

15.224

17.174

6.656

5.179

3.568

2.500

4.536

7.400

9.124

17.780

6.600

4.812

2.297

2.000

5.036

3.900

4.860

18.313

6.527

4.445

1.292

1.500

5.536

1.700

2.140

18.785

6.433

4.077

0.572

1.241

5.795

1.000

1.241

19.010

6.374

3.884

0.311

1.000

6.036

0.500

0.667

19.208

6.315

3.705

0.154

0.500

6.536

0.100

0.091

19.592

6.176

3.331

0.014

0.000

7.036

0.000

0.000

19.984

1.902

4.502

0.000

Starboard Bunker Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

7.000

0.000

100.000

122.670

13.668

–6.928

7.807

0.000

6.956

0.080

97.900

121.245

13.681

–6.926

7.782

12.825

6.500

0.536

86.000

106.472

13.832

–6.905

7.517

12.348

6.000

1.036

73.200

90.613

14.036

–6.878

7.220

11.473

5.500

1.536

60.700

75.229

14.301

–6.847

6.914

10.494

5.000

2.036

48.800

60.492

14.648

–6.812

6.597

9.326

4.500

2.536

37.700

46.660

15.116

–6.774

6.265

7.995

4.000

3.036

27.500

34.112

15.743

–6.736

5.914

6.676

3.500

3.536

19.000

23.475

16.486

–6.699

5.547

5.110

3.000

4.036

12.300

15.224

17.174

–6.656

5.179

3.568

2.500

4.536

7.400

9.124

17.780

–6.600

4.812

2.297

2.000

5.036

3.900

4.860

18.313

–6.527

4.445

1.292

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664 • Ship Stability, Powering and Resistance

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

1.500

5.536

1.700

2.140

18.785

–6.433

4.077

0.572

1.241

5.795

1.000

1.241

19.010

–6.374

3.884

0.311

1.000

6.036

0.500

0.667

19.208

–6.315

3.705

0.154

0.500

6.536

0.100

0.091

19.592

–6.176

3.331

0.014

0.000

7.036

0.000

0.000

19.984

–1.902

4.502

0.000

Sounding (m)

Ullage (m)

Fill (%)

Capacity (m3)

LCG (m)

TCG (m)

KG (m)

FSM (m4)

7.000

0.000

100.000

69.778

98.044

0.000

2.957

0.000

6.788

0.212

97.900

68.313

98.064

0.000

2.873

1.259

6.500

0.500

95.200

66.442

98.091

0.000

2.767

1.043

6.000

1.000

90.900

63.418

98.138

0.000

2.600

0.815

5.500

1.500

86.800

60.534

98.186

0.000

2.450

0.738

5.000

2.000

82.600

57.619

98.236

0.000

2.309

0.779

4.500

2.500

78.100

54.481

98.290

0.000

2.168

0.951

4.000

3.000

72.800

50.788

98.343

0.000

2.018

1.382

3.500

3.500

66.100

46.142

98.391

0.000

1.844

2.337

3.000

4.000

57.600

40.202

98.418

0.000

1.638

3.830

2.500

4.500

47.600

33.192

98.426

0.000

1.404

5.255

2.000

5.000

36.600

25.547

98.420

0.000

1.151

6.199

1.500

5.500

25.300

17.654

98.395

0.000

0.884

6.237

1.000

6.000

14.400

10.017

98.333

0.000

0.602

4.986

0.500

6.500

5.000

3.512

98.166

0.000

0.306

2.263

0.180

6.820

1.000

0.697

97.905

0.000

0.111

0.493

0.000

7.000

0.000

0.000

97.905

0.000

0.111

0.000

Fore Peak

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Appendix 1 • 665 After Peak Sounding

Ullage

Fill (%)

LCG

TCG

KG

FSM

(m )

(m)

(m)

(m)

(m4)

Capacity 3

(m)

(m)

6.954

0.000

100.000

106.994

2.398

0.000

5.511

0.000

6.925

0.030

97.900

104.747

2.411

0.000

5.479

778.357

6.500

0.454

70.400

75.370

2.638

0.000

4.978

456.216

6.000

0.954

45.600

48.819

2.987

0.000

4.253

185.227

5.500

1.454

29.900

31.971

3.297

0.000

3.427

53.167

5.000

1.954

22.200

23.780

3.340

0.000

2.772

10.366

4.500

2.454

18.700

20.001

3.287

0.000

2.385

1.710

4.000

2.954

16.600

17.717

3.251

0.000

2.137

0.531

3.500

3.454

14.700

15.752

3.239

0.000

1.929

0.381

3.000

3.954

12.800

13.671

3.247

0.000

1.722

0.453

2.500

4.454

10.600

11.312

3.277

0.000

1.500

0.629

2.000

4.954

8.100

8.652

3.340

0.000

1.256

0.851

1.500

5.454

5.400

5.786

3.460

0.000

0.990

1.013

1.000

5.954

2.800

2.994

3.680

0.000

0.700

0.940

0.577

6.377

1.000

1.066

4.015

0.000

0.436

0.559

0.500

6.454

0.700

0.799

4.096

0.000

0.386

0.456

0.000

6.954

0.000

0.000

4.824

0.000

0.046

0.000

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666 • Ship Stability, Powering and Resistance Grain Hold Data

Sounding (m) Grain volume Grain KG (m3) (m)

Volumetric heeling Grain TCG moment (m4) (m)

Grain LCG (m FOAP)

8.00

1,200

6.00

102.00

0

50

7.50

1,125

5.75

327.60

0

50

7.00

1,050

5.50

590.40

0

50

6.50

975

5.25

789.60

0

50

6.00

900

5.00

925.20

0

50

5.50

825

4.75

1,006.80

0

50

5.00

750

4.50

1,006.80

0

50

4.50

675

4.25

1,006.80

0

50

4.00

600

4.00

1,006.80

0

50

3.50

525

3.75

1,006.80

0

50

3.00

450

3.50

1,006.80

0

50

2.50

375

3.25

804.00

0

50

2.00

300

3.00

705.00

0

50

1.50

225

2.75

608.00

0

50

1.00

150

2.50

506.00

0

50

0.50

75

2.25

360.00

0

50

0.00

0

2.00

0.00

0

50

Maximum Permissible Mass Grain Heeling Moments (tonne metres)

Displacement (tonnes)

Effective KG values (m) 5.00

5.20

5.40

5.60

5.80

6.00

8,750

4,824.1

4,460.2

4,096.4

3,732.5

3,368.7

3,004.8

8,604

4,685.6

4,327.8

3,970.0

3,612.3

3,254.5

2,896.7

8,460

4,550.2

4,198.4

3,846.6

3,494.8

3,143.0

2,791.2

8,315

4,416.2

4,070.4

3,724.7

3,378.9

3,033.1

2,687.4

8,172

4,286.9

3,947.1

3,607.3

3,267.5

2,927.6

2,587.8

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Appendix 1 • 667

Displacement (tonnes)

Effective KG values (m) 5.00

5.20

5.40

5.60

5.80

6.00

8,029

4,159.5

3,825.6

3,491.7

3,157.9

2,824.0

2,490.1

7,887

4,032.8

3,704.8

3,376.8

3,048.9

2,720.9

2,393.0

7,746

3,911.8

3,589.7

3,267.6

2,945.5

2,623.4

2,301.3

7,606

3,791.5

3,475.2

3,158.9

2,842.6

2,526.3

2,210.1

7,467

3,675.0

3,364.5

3,054.0

2,743.5

2,433.0

2,122.5

7,329

3,560.8

3,256.1

2,951.3

2,646.6

2,341.8

2,037.0

7,192

3,450.4

3,151.3

2,852.3

2,553.2

2,254.1

1,955.1

7,056

3,342.1

3,048.7

2,755.3

2,461.9

2,168.5

1,875.1

6,921

3,238.9

2,951.1

2,663.3

2,375.5

2,087.7

1,799.9

6,787

3,137.6

2,855.4

2,573.2

2,290.9

2,008.7

1,726.5

6,654

3,041.1

2,764.5

2,487.8

2,211.1

1,934.4

1,657.7

6,522

2,946.6

2,675.4

2,404.2

2,133.0

1,861.8

1,590.6

6,392

2,857.0

2,591.2

2,325.4

2,059.6

1,793.8

1,528.0

6,263

2,771.7

2,511.3

2,250.8

1,990.4

1,730.0

1,469.5

6,135

2,689.3

2,434.2

2,179.1

1,924.0

1,668.9

1,413.8

6,008

2,611.0

2,361.2

2,111.3

1,861.5

1,611.7

1,361.9

5,882

2,535.3

2,290.7

2,046.1

1,801.6

1,557.0

1,312.4

5,757

2,464.6

2,225.2

1,985.8

1,746.5

1,507.1

1,267.7

5,634

2,396.7

2,162.4

1,928.1

1,693.9

1,459.6

1,225.3

5,511

2,332.9

2,103.8

1,874.6

1,645.5

1,416.3

1,187.1

5,390

2,272.8

2,048.7

1,824.6

1,600.4

1,376.3

1,152.2

5,269

2,214.2

1,995.1

1,776.0

1,556.9

1,337.8

1,118.7

5,150

2,161.2

1,947.1

1,732.9

1,518.8

1,304.6

1,090.5

5,031

2,110.4

1,901.2

1,692.0

1,482.8

1,273.6

1,064.4

4,913

2,062.2

1,857.9

1,653.6

1,449.3

1,245.1

1,040.8

4,796

2,017.4

1,818.0

1,618.6

1,419.2

1,219.7

1,020.3

4,680

1,975.8

1,781.2

1,586.6

1,392.0

1,197.4

1,002.8

9781408176122_App01_3_Rev_txt_prf.indd 667

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668 • Ship Stability, Powering and Resistance Maximum Permissible Mass Grain Heeling Moments (tonne metres)

Displacement (tonnes)

Effective KG values (m) 6.00

6.20

6.40

6.60

6.80

7.00

8,750

3,004.8

2,641.0

2,277.1

1,913.3

1,549.4

1,185.6

8,604

2,896.7

2,538.9

2,181.2

1,823.4

1,465.6

1,107.8

8,460

2,791.2

2,439.4

2,087.6

1,735.9

1,384.1

1,032.3

8,315

2,687.4

2,341.6

1,995.9

1,650.1

1,304.4

958.6

8,172

2,587.8

2,248.0

1,908.2

1,568.4

1,228.6

888.8

8,029

2,490.1

2,156.3

1,822.4

1,488.6

1,154.7

820.8

7,887

2,393.0

2,065.0

1,737.0

1,409.1

1,081.1

753.2

7,746

2,301.3

1,979.2

1,657.1

1,335.0

1,012.9

690.8

7,606

2,210.1

1,893.8

1,577.5

1,261.2

945.0

628.7

7,467

2,122.5

1,812.0

1,501.6

1,191.1

880.6

570.1

7,329

2,037.0

1,732.3

1,427.5

1,122.8

818.0

513.3

7,192

1,955.1

1,656.0

1,357.0

1,057.9

758.8

459.8

7,056

1,875.1

1,581.7

1,288.3

994.8

701.4

408.0

6,921

1,799.9

1,512.1

1,224.3

936.5

648.7

360.9

6,787

1,726.5

1,444.3

1,162.1

879.8

597.6

315.4

6,654

1,657.7

1,381.0

1,104.3

827.6

550.9

274.3

6,522

1,590.6

1,319.4

1,048.2

777.0

505.8

234.6

6,392

1,528.0

1,262.2

996.4

730.6

464.8

199.0

6,263

1,469.5

1,209.1

948.7

688.2

427.8

167.4

6,135

1,413.8

1,158.7

903.6

648.5

393.4

138.2

6,008

1,361.9

1,112.0

862.2

612.4

362.5

112.7

5,882

1,312.4

1,067.8

823.2

578.6

334.0

89.4

5,757

1,267.7

1,028.3

788.9

549.5

310.1

70.7

5,634

1,225.3

991.0

756.8

522.5

288.2

53.9

5,511

1,187.1

958.0

728.8

499.7

270.5

41.3

9781408176122_App01_3_Rev_txt_prf.indd 668

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Appendix 1 • 669

Displacement (tonnes)

Effective KG values (m) 6.00

6.20

6.40

6.60

6.80

7.00

5,390

1,152.2

928.1

703.9

479.8

255.7

31.5

5,269

1,118.7

899.6

680.5

461.4

242.3

23.3

5,150

1,090.5

876.3

662.2

448.0

233.9

19.7

5,031

1,064.4

855.2

646.0

436.8

227.6

18.4

4,913

1,040.8

836.5

632.2

427.9

223.6

19.3

4,796

1,020.3

820.9

621.4

422.0

222.6

23.2

4,680

1,002.8

808.2

613.6

419.0

224.4

29.8

Maximum Permissible Mass Grain Heeling Moments (tonne metres)

Displacement (tonnes)

Effective KG values (m) 5.00

5.20

5.40

5.60

5.80

6.00

4,680

1,975.8

1,781.2

1,586.6

1,392.0

1,197.4

1,002.8

4,565

1,936.2

1,746.4

1,556.6

1,366.7

1,176.9

987.1

4,451

1,900.4

1,715.3

1,530.2

1,345.1

1,160.1

975.0

4,337

1,866.7

1,686.3

1,506.0

1,325.6

1,145.3

965.0

4,224

1,835.3

1,659.6

1,484.0

1,308.4

1,132.7

957.1

4,112

1,806.9

1,635.9

1,464.9

1,294.0

1,123.0

952.0

4,001

1,780.4

1,614.1

1,447.7

1,281.3

1,114.9

948.6

3,890

1,756.0

1,594.3

1,432.5

1,270.8

1,109.0

947.2

3,781

1,735.1

1,577.9

1,420.7

1,263.4

1,106.2

949.0

3,671

1,714.5

1,561.8

1,409.2

1,256.5

1,103.9

951.2

3,563

1,697.5

1,549.3

1,401.2

1,253.0

1,104.9

956.7

3,455

1,681.5

1,537.8

1,394.1

1,250.5

1,106.8

963.1

3,348

1,667.2

1,528.0

1,388.8

1,249.6

1,110.4

971.2

3,241

1,655.4

1,520.7

1,385.9

1,251.1

1,116.4

981.6

3,136

1,645.3

1,514.9

1,384.5

1,254.1

1,123.7

993.3

9781408176122_App01_3_Rev_txt_prf.indd 669

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670 • Ship Stability, Powering and Resistance

Displacement (tonnes)

Effective KG values (m) 5.00

5.20

5.40

5.60

5.80

6.00

3,030

1,636.2

1,510.2

1,384.2

1,258.2

1,132.2

1,006.2

2,926

1,629.9

1,508.2

1,386.6

1,264.9

1,143.2

1,021.5

2,822

1,624.3

1,506.9

1,389.6

1,272.2

1,154.9

1,037.6

2,718

1,620.0

1,507.0

1,394.0

1,280.9

1,167.9

1,054.9

2,615

1,617.1

1,508.4

1,399.6

1,290.9

1,182.1

1,073.4

2,513

1,616.1

1,511.6

1,407.1

1,302.6

1,198.1

1,093.6

2,412

1,616.9

1,516.6

1,416.3

1,316.0

1,215.7

1,115.4

2,311

1,618.1

1,522.0

1,425.9

1,329.8

1,233.7

1,137.6

2,211

1,620.5

1,528.6

1,436.7

1,344.7

1,252.8

1,160.8

2,111

1,623.2

1,535.4

1,447.6

1,359.8

1,272.1

1,184.3

2,012

1,627.1

1,543.5

1,459.8

1,376.1

1,292.5

1,208.8

1,914

1,632.2

1,552.6

1,473.0

1,393.4

1,313.8

1,234.2

1,816

1,636.7

1,561.2

1,485.7

1,410.2

1,334.7

1,259.2

1,720

1,642.8

1,571.3

1,499.7

1,428.2

1,356.7

1,285.2

1,624

1,648.5

1,581.0

1,513.5

1,445.9

1,378.4

1,310.9

Maximum Permissible Mass Grain Heeling Moments (tonne metres)

Displacement (tonnes)

Effective KG values (m) 6.00

6.20

6.40

6.60

6.80

4,680

1,002.8

808.2

613.6

419.0

224.4

29.8

4,565

987.1

797.3

607.5

417.6

227.8

38.0

4,451

975.0

789.9

604.8

419.7

234.6

49.6

4,337

965.0

784.6

604.3

423.9

243.6

63.2

4,224

957.1

781.4

605.8

430.1

254.5

78.9

4,112

952.0

781.0

610.0

439.0

268.0

97.1

4,001

948.6

782.2

615.8

449.5

283.1

116.7

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7.00

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Appendix 1 • 671

Displacement (tonnes)

Effective KG values (m) 6.00

6.20

6.40

6.60

6.80

7.00

3,890

947.2

785.5

623.7

462.0

300.2

138.5

3,781

949.0

791.8

634.5

477.3

320.1

162.9

3,671

951.2

798.6

645.9

493.3

340.6

188.0

3,563

956.7

808.5

660.4

512.2

364.1

215.9

3,455

963.1

819.5

675.8

532.1

388.5

244.8

3,348

971.2

831.9

692.7

553.5

414.3

275.1

3,241

981.6

846.8

712.1

577.3

442.5

307.8

3,136

993.3

862.9

732.5

602.1

471.7

341.3

3,030

1,006.2

880.2

754.2

628.2

502.2

376.2

2,926

1,021.5

899.9

778.2

656.5

534.9

413.2

2,822

1,037.6

920.2

802.9

685.5

568.2

450.8

2,718

1,054.9

941.9

828.9

715.8

602.8

489.8

2,615

1,073.4

964.7

855.9

747.2

638.5

529.7

2,513

1,093.6

989.1

884.6

780.1

675.6

571.1

2,412

1,115.4

1,015.1

914.8

814.5

714.2

613.9

2,311

1,137.6

1,041.5

945.4

849.3

753.2

657.1

2,211

1,160.8

1,068.9

977.0

885.0

793.1

701.1

2,111

1,184.3

1,096.5

1,008.7

920.9

833.2

745.4

2,012

1,208.8

1,125.1

1,041.5

957.8

874.2

790.5

1,914

1,234.2

1,154.7

1,075.1

995.5

915.9

836.3

1,816

1,259.2

1,183.6

1,108.1

1,032.6

957.1

881.6

1,720

1,285.2

1,213.7

1,142.1

1,070.6

999.1

927.6

1,624

1,310.9

1,243.4

1,175.8

1,108.3

1,040.8

973.2

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0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

2,000

2,200

2,400

2,600

2,800

3,000

3,200

3,400

3,600

3,800

4,000

4,200

4,400

4,600

4,800

0.00

0.00

1,800

5,000

0.00

0 (degrees)

1,600

Displacement (tonnes)

0.61

0.61

0.61

0.61

0.61

0.61

0.62

0.63

0.63

0.64

0.65

0.67

0.69

0.71

0.73

0.76

0.80

0.85

5 (degrees)

1.21

1.21

1.22

1.22

1.22

1.23

1.24

1.25

1.27

1.29

1.31

1.34

1.37

1.41

1.46

1.52

1.59

1.68

10 (degrees)

KN Data (Free to Trim, Based on Zero Trim)

1.46

1.46

1.47

1.47

1.48

1.48

1.50

1.51

1.53

1.55

1.58

1.61

1.65

1.70

1.75

1.82

1.91

2.01

12 (degrees)

1.83

1.83

1.83

1.83

1.84

1.85

1.87

1.88

1.91

2.45

2.45

2.45

2.46

2.47

2.48

2.50

2.52

2.54

2.57

2.61

1.96 1.93

2.66

2.71

2.77

2.85

2.93

3.03

3.15

20 (degrees)

2.00

2.05

2.11

2.17

2.25

2.35

2.47

15 (degrees)

3.08

3.08

3.08

3.09

3.09

3.11

3.12

3.14

3.17

3.20

3.24

3.28

3.33

3.39

3.45

3.53

3.62

3.72

25 (degrees)

4.34

4.33

3.71 3.71

4.32

4.32

4.32

4.32

4.32

4.32

4.33

4.34

4.36

4.38

4.40

4.44

4.48

4.53

4.59

4.66

35 (degrees)

3.71

3.71

3.72

3.72

3.74

3.75

3.77

3.79

3.82

3.85

3.89

3.94

3.99

4.05

4.13

4.22

30 (degrees)

4.93

4.93

4.92

4.90

4.89

4.88

4.87

4.87

4.86

4.86

4.87

4.87

4.89

4.91

4.94

4.97

5.01

5.07

40 (degrees)

672 • Ship Stability, Powering and Resistance

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0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

0.00

5,600

5,800

6,000

6,200

6,400

6,600

6,800

7,000

7,200

7,400

7,600

7,800

8,000

8,200

8,400

8,600

8,800

0.00

0.00

5,400

9,000

0.00

0 (degrees)

5,200

Displacement (tonnes)

0.67

0.66

0.66

0.66

0.65

0.65

0.64

0.64

0.64

0.63

0.63

0.62

0.62

0.62

0.61

0.61

0.61

0.61

0.61

0.61

5 (degrees)

1.33

1.33

1.32

1.31

1.30

1.29

1.29

1.28

1.27

1.26

1.26

1.25

1.24

1.24

1.23

1.23

1.22

1.22

1.22

1.21

10 (degrees)

1.61

1.60

1.59

1.58

1.58

1.57

1.56

1.55

1.54

1.52

1.52

1.51

1.50

1.49

1.48

1.47

1.47

1.47

1.46

1.46

12 (degrees)

2.00

1.99

1.98

1.96

1.95

1.94

1.93

1.92

1.91

1.90

1.89

2.66

2.65

2.64

2.62

2.61

2.59

2.58

2.56

2.55

2.54

2.52

2.51

2.50

1.87 1.88

2.49

2.48

2.47

2.46

2.45

2.45

2.45

20 (degrees)

1.86

1.85

1.84

1.84

1.83

1.83

1.83

15 (degrees)

3.25

3.25

3.25

3.25

3.25

3.24

3.23

3.21

3.20

3.18

3.17

3.76

3.78

4.23

4.25

4.27

4.29

3.80 3.79

4.31

4.33

4.34

4.36

4.37

4.38

4.39

4.39

4.39

4.39

4.39

4.39

4.38

4.37

4.36

4.35

35 (degrees)

3.81

3.82

3.82

3.82

3.82

3.82

3.81

3.80

3.79

3.14 3.15

3.78

3.76

3.75

3.74

3.73

3.72

3.71

30 (degrees)

3.13

3.11

3.10

3.09

3.09

3.08

3.08

25 (degrees)

4.64

4.67

4.69

4.72

4.75

4.77

4.80

4.82

4.84

4.86

4.88

4.89

4.91

4.92

4.93

4.93

4.94

4.94

4.94

4.94

40 (degrees)

Appendix 1 • 673

11/16/2013 2:37:04 AM

5.43

5.40

5.38

5.37

5.36

5.36

5.36

5.36

5.37

5.38

5.40

5.41

5.41

5.42

5.43

5.43

5.43

5.42

5.42

5.41

1,800

2,000

2,200

2,400

2,600

2,800

3,000

3,200

3,400

3,600

3,800

4,000

4,200

4,400

4,600

4,800

5,000

5,200

5,400

45 (degrees)

1,600

Displacement (tonnes)

9781408176122_App01_3_Rev_txt_prf.indd 674

5.78

5.79

5.81

5.82

5.83

5.84

5.85

5.86

5.86

5.86

5.86

5.86

5.85

5.84

5.82

5.81

5.79

5.78

5.76

5.75

50 (degrees)

6.06

6.08

6.10

6.12

6.14

6.16

6.18

6.20

6.21

6.22

6.23

6.24

6.24

6.24

6.23

6.21

6.19

6.16

6.12

6.07

55 (degrees)

6.26

6.29

6.31

6.34

6.36

6.39

6.41

6.44

6.46

6.39

6.42

6.45

6.48

6.51

6.53

6.56

6.58

6.61

6.63

6.65

6.49 6.48

6.67

6.68

6.70

6.70

6.71

6.71

6.69

6.67

6.63

65 (degrees)

6.51

6.52

6.52

6.52

6.51

6.50

6.48

6.44

6.39

60 (degrees)

6.46

6.49

6.52

6.54

6.57

6.60

6.62

6.65

6.67

6.70

6.72

6.74

6.76

6.78

6.79

6.81

6.81

6.81

6.80

6.76

70 (degrees)

6.47

6.49

6.52

6.54

6.57

6.59

6.62

6.64

6.66

6.69

6.71

6.73

6.76

6.78

6.80

6.82

6.83

6.83

6.83

6.81

75 (degrees)

6.41

6.43

6.46

6.48

6.50

6.52

6.54

6.57

6.59

6.61

6.64

6.66

6.68

6.71

6.73

6.75

6.76

6.77

6.77

6.76

80 (degrees)

6.14

6.15

6.17

6.18

6.20

6.21

6.23

6.24

6.26

6.28

6.30

6.32

6.34

6.35

6.37

6.39

6.41

6.42

6.44

6.47

90 (degrees)

674 • Ship Stability, Powering and Resistance

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5.39

5.38

5.36

5.35

5.33

5.31

5.29

5.26

5.24

5.21

5.18

5.16

5.13

5.10

5.07

5.03

5.00

5,800

6,000

6,200

6,400

6,600

6,800

7,000

7,200

7,400

7,600

7,800

8,000

8,200

8,400

8,600

8,800

9,000

5.32

5.35

5.38

5.42

5.45

5.48

5.51

5.54

5.56

5.59

5.61

5.64

5.66

5.68

5.70

5.72

5.74

5.76

50 (degrees)

5.58

5.61

5.64

5.68

5.71

5.74

5.76

5.79

5.82

5.85

5.87

5.90

5.92

5.94

5.97

5.99

6.01

6.04

55 (degrees)

5.79

5.82

5.85

5.88

5.91

5.93

5.96

5.95

5.97

6.00

6.02

6.05

6.07

6.09

6.12

6.14

6.01 5.98

6.17

6.19

6.21

6.24

6.26

6.29

6.31

6.34

6.37

65 (degrees)

6.04

6.06

6.09

6.11

6.13

6.16

6.18

6.21

6.24

60 (degrees)

These KN values have been calculated with the vessel on an even keel.

5.40

45 (degrees)

5,600

Displacement (tonnes)

6.05

6.07

6.09

6.11

6.13

6.15

6.17

6.19

6.22

6.24

6.26

6.28

6.31

6.33

6.36

6.38

6.41

6.43

70 (degrees)

6.10

6.12

6.14

6.15

6.17

6.18

6.20

6.22

6.24

6.26

6.28

6.30

6.32

6.34

6.37

6.39

6.42

6.44

75 (degrees)

6.11

6.12

6.13

6.14

6.15

6.16

6.18

6.19

6.21

6.22

6.24

6.26

6.28

6.30

6.32

6.34

6.37

6.39

80 (degrees)

5.97

5.97

5.97

5.98

5.98

5.98

5.99

6.00

6.01

6.02

6.03

6.04

6.06

6.07

6.08

6.10

6.11

6.12

90 (degrees)

Appendix 1 • 675

11/16/2013 2:37:05 AM

676 • Ship Stability, Powering and Resistance Down-Flooding Angle Data Draught (m)

Displacement (tonnes)

Down-flooding angle (degrees)

8.00

8,750

27.50

7.90

8,604

28.30

7.80

8,460

29.10

7.70

8,315

29.90

7.60

8,172

30.70

7.50

8,029

31.50

7.40

7,887

32.24

7.30

7,746

32.98

7.20

7,606

33.72

7.10

7,467

34.46

7.00

7,329

35.20

6.90

7,192

35.88

6.80

7,056

36.56

6.70

6,921

37.24

6.60

6,787

37.92

6.50

6,654

38.60

6.40

6,522

39.24

6.30

6,392

39.88

6.20

6,263

40.52

6.10

6,135

41.16

6.00

6,008

41.80

5.90

5,882

41.88

5.80

5,757

41.96

5.70

5,634

42.04

5.60

5,511

42.12

5.50

5,390

42.20

5.40

5,269

43.30

5.30

5,150

44.40

5.20

5,031

45.50

5.10

4,913

46.60

5.00

4,796

47.70

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Appendix 1 • 677

Draught (m)

Displacement (tonnes)

Down-flooding angle (degrees)

4.90

4,680

48.26

4.80

4,565

48.82

4.70

4,451

49.38

4.60

4,337

49.94

4.50

4,224

50.50

4.40

4,112

51.02

4.30

4,001

51.54

4.20

3,890

52.06

4.10

3,781

52.58

4.00

3,671

53.10

3.90

3,563

53.60

3.80

3,455

54.10

3.70

3,348

54.60

3.60

3,241

55.10

3.50

3,136

55.60

3.40

3,030

56.08

3.30

2,926

56.56

3.20

2,822

57.04

3.10

2,718

57.52

3.00

2,615

58.00

2.90

2,513

58.50

2.80

2,412

59.00

2.70

2,311

59.50

2.60

2,211

60.00

2.50

2,111

60.50

2.40

2,012

61.12

2.30

1,914

61.74

2.20

1,816

62.36

2.10

1,720

62.98

2.00

1,624

63.60

These down-flooding angle values have been calculated with the vessel on an even keel and assuming that the sea surface is flat. The angles may be reduced in the event of a swell or a trimmed vessel.

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678 • Ship Stability, Powering and Resistance Limiting KG Data

Displacement

Limiting KG (m) SOLAS criteria

Section

2,600

7.375

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 30 to 40

2,800

7.305

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 40

3,000

7.205

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

3,200

7.104

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

3,400

7.019

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

3,600

6.950

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

3,800

6.893

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

4,000

6.848

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

4,200

6.814

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

4,400

6.789

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

4,600

6.772

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

4,800

6.763

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

5,000

6.762

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

5,200

6.766

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

5,400

6.776

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

5,600

6.792

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

5,800

6.812

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

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Appendix 1 • 679

Displacement

Limiting KG (m) SOLAS criteria

Section

6,000

6.836

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

6,200

6.864

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

6,400

6.895

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

6,600

6.928

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

6,800

6.963

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

7,000

6.997

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

7,200

7.031

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

7,400

7.064

A.749(18) Ch3 – Design criteria applicable to all ships

3.1.2.1: Area 0 to 30

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11/16/2013 2:37:05 AM

APPENDIX 2 DERIVATION OF ARCHIMEDES’ PRINCIPLE FOR BOX SHAPED VESSELS When an object is immersed or partially immersed in a fluid it will experience ‘hydrostatic pressure’. This is a pressure which is the result of the weight of the static fluid pushing from above and from the sides of the object. For any depth of submergence in a fluid, the hydrostatic pressure, P, can be found from: P

ρ g×h

Increasing water depth

In this equation, P is in units of Newtons per metre squared, g is the acceleration due to gravity, which can be taken to be 9.81 m/s2, ρ is the fluid density in kilograms per metres cubed (taken to be 1025 kg/m3 for seawater, and 1000 kg/m3 for fresh water) and h is the depth of submersion in the fluid in metres. As g and ρ are constant for a given fluid, it can be seen that the hydrostatic pressure increases linearly with depth.

▲ Figure A2.1 Hydrostatic pressure on a box shaped vessel

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Appendix 2 • 681 Consider a simple box shaped vessel partially submerged in a fluid, as shown in Figure A2.1. The hydrostatic pressure (indicated by the arrows) will push at right angles against any surface of the vessel in contact with the fluid, and as the depth increases, the hydrostatic pressure increases. The hydrostatic pressure acting on the sides of the vessel acts to ‘crush’ the vessel inwards. In the case of a box shaped vessel, this produces no vertical force, but the hydrostatic pressure acting upwards on the keel plates of the vessel creates an upwards force – this is the buoyancy force. The force created as a result of pressure can be found from: Force

Pressure × Area

The buoyancy force is created by the hydrostatic pressure pushing upwards on the keel of the vessel, therefore in the context of buoyancy: Buoyancy force o ce

Hyd ostat c pressure Hydrostatic r × Keel area

Combining these equations gives an expression for the buoyancy force: Buoyancy force ρ × g × h K Keel eel area For the box shaped vessel shown in Figure A2.1, the keel area can be found from the length (in units of metres) and the beam (also in units of metres) of the vessel: Keel area = Length × Beam = L B Combining these equations gives an expression for the buoyancy force acting on the vessel: Buoyancy force ρ × g × h L × B The pressure is acting on the keel, so the depth at the point where the pressure is acting to create buoyancy is the same as the draught of the vessel, given the symbol D, and measured in metres. Therefore: Buoyancy force

ρ×g×D L×B

If we examine this equation, we can see that the last three terms are the draught, length and beam of the vessel. For a box shaped vessel, the underwater volume can be found by multiplying the length, beam and draught, therefore: Buoyancy force ρ × g × Underwater volume

9781408176122_App02_Rev_txt_prf.indd 681

11/16/2013 2:43:34 AM

682 • Ship Stability, Powering and Resistance The underwater volume of the object is given the symbol ∇, and is measured in units of m3. Using this notation: Buoyancy force

ρ×g×∇

The mass (in kilograms) of an object is found by multiplying the volume of an object (in metres cubed) by the density of the object (in kilograms per metres cubed). If we consider the amount of water displaced, or pushed aside by a vessel, it will therefore be equal to the underwater volume of the ship, multiplied by the density of water: Mass of fluid displaced = ∇ × ρ The weight (in Newtons) of an object is equal to the mass of an object multiplied by the acceleration due to gravity, therefore: Weight of fluid displaced = ρ × g × ∇ As can be seen: Buoyancy force o ce W Weigh e g t of ffluid u d displaced This confirms Archimedes’ Principle, which states that: When an object is immersed or partially immersed in a fluid, it experiences an upwards thrust equal to the weight of water that it pushes aside, or displaces.

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11/16/2013 2:43:35 AM

APPENDIX 3 DERIVATION OF ARCHIMEDES’ PRINCIPLE FOR SEMICIRCULAR SECTION VESSELS Consider a vessel with a length L, and a constant semi-circular cross-section, as shown in Figure A3.1.

θ =π

dθ r

θ =0 θ

dA

▲ Figure A3.1 Definitions for a semi-circular hull form

An element of surface area L dA will have an area given by Lr dθ. Therefore: = Lr dθ The vertical depth underwater of the element dA can be found by resolving the radius of the hull and the angle θ: Dept e h r sinθ

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11/16/2013 2:43:18 AM

684 • Ship Stability, Powering and Resistance The hydrostatic pressure acting on the element can be found from: P ρ g×h Therefore: P ρ g × r sinθ The normal force acting on the element can be found by multiplying the element area by the hydrostatic pressure, giving: Force ρ × g × r × sinθ dA The force can be resolved into a vertical component, which is buoyancy: Buoyancy =ρ × g × r × sin n2 θ dA d The element dA can be expressed in terms of the geometry: dA = Lr dθ This gives: Buoyancy =ρ × g × r 2 × L sin2 θ dθ The buoyancy can be integrated to determine the overall buoyancy force: π

L∫ssin n2 θ d θ

Total buoyanc o cy ρ × g × r 2

0

π

Total buoyanc o cy ρ × g × r 2 Total buoyanc o cy ρ × g × r 2

L×

⎡θ 1 ⎤ − sin θ ⎥ ⎣2 4 ⎦0

⎡⎛ π 1 ⎞ L×⎢ − sin2 π ⎝ ⎠ 2 4 ⎣

Total buoyanc o cy ρ × g × r 2

⎛0 ⎝2

1 ⎞⎤ sin0 ⎥ ⎠⎦ 4

⎡⎛ π ⎞ ⎤ L×⎢ − ( 0 )⎥ ⎝ ⎠ 2 ⎣ ⎦

Total buoyanc o cy ρ × g × r 2

L×

π 2

The underwater volume can be determined from the cross-sectional area and the length: ∇ = Underwater cross section area × Length

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Appendix 3 • 685

∇=

1 π × π × r2 × L = × r2 × L 2 2

The weight of displaced fluid can be found: Displaced ffluid = ∇ × ρ × g Displaced ffluid =

π 2 ×r 2

L×ρ×g

Archimedes tells us that the buoyancy force is equal to the displaced mass, which can be seen from these equations.

9781408176122_App03_Rev_txt_prf.indd 685

11/16/2013 2:43:21 AM

APPENDIX 4 LINEAR INTERPOLATION USING THE EQUATION OF A STRAIGHT LINE As an alternative method, the formula for a linear equation can be used to calculate the value. A straight line on a graph always follows the following equation: y = mx + c In this, y is the value up the y axis of a graph, m is the gradient or slope of the line, x is the value along the x axis and c is the point where the line will cross the y axis. Using the graph in Figure 1.5, and the data in Table 1.1, we can use this equation to create a formula linking the draught (on the x axis) and displacement (on the y axis). Using these values, this becomes: Displacement = m Draught + c The gradient, m, is found using: m=

Change in y vvalues Change in x vvalues

In this example, this gives: m=

3, 890 − 3, 781 = 1, 090 4.20 − 4.10

Using this, we can substitute in the value for m, and the draught and displacement from one row of the data, to find the value c: Displacement = m Draught + c

9781408176122_App04_Rev_txt_prf.indd 686

11/16/2013 2:41:35 AM

Appendix 4 • 687 3, 890 = 1, 090 × 4.20 + c 3, 890 = 4 , 578 + c c = −688 Therefore, this becomes: Displacement = 1, 090 × Draught − 688 This formula allows us to determine the displacement at any draught between 4.10 and 4.20 m. For example, at a draught of 4.12 m, the displacement will be: Displacement = 1, 090 × Draught − 688 Displacement = 1, 090 × 4.12 − 688 Displacement = 3, 802.8 tonnes Combining the equations, along with the above processes, gives us a formula for linear interpolation between two rows of data, labelled as shown in Table A4.1. Table A4.1 Naming convention for interpolation by formula Draught (m)

Displacement (tonnes)

Y1 (4.20)

X1 (3,890)

Y2 (4.10)

X2 (3,781)

Consider the equation: y = mx + c Substituting in the gradient gives: y=

Change g in y vvalues x+c Change in x vvalues ⎛ Y Y ⎞ y =⎜ 1 2 ⎟ x+c ⎝ X1 X 2 ⎠

9781408176122_App04_Rev_txt_prf.indd 687

11/16/2013 2:41:37 AM

688 • Ship Stability, Powering and Resistance Substituting in values Y1 and X1 with the gradient gives us c: ⎛ Y Y ⎞ Y1 = ⎜ 1 2 ⎟ X1 c ⎝ X1 X 2 ⎠ ⎛ Y Y ⎞ Y1 − ⎜ 1 2 ⎟ X1 ⎝ X1 X 2 ⎠

c

Putting this back gives: ⎛ ⎛ Y Y ⎞ ⎛ Y Y ⎞ ⎞ y = ⎜ 1 2 ⎟ x Y1 − ⎜ 1 2 ⎟ X1 ⎟ ⎝ X1 X 2 ⎠ ⎝ X1 X 2 ⎠ ⎠ ⎝ Therefore, using our example values to find the displacement at a draught of 4.12 m gives: Displacement =

⎛ ⎞ ⎛ 3, 890 − 3, 781⎞ ⎛ 3, 890 − 3, 781⎞ Draught + 3, 890 − 4.20⎟ ⎝ 4.20 − 4.10 ⎠ ⎝ 4.20 − 4.10 ⎠ ⎝ ⎠ Displacement = (1, 090 ) Draught gh + ( −688 )

Displacement = (1, 090 ) 4.12 + ( −688 ) = 3, 802.8 tonnes

9781408176122_App04_Rev_txt_prf.indd 688

11/16/2013 2:41:39 AM

APPENDIX 5 THE DERIVATION OF THE TPC FORMULA As mass is added to the vessel, she will sink lower into the water, and therefore increase her underwater volume. Previously, in Formula 1.1, we have seen the relationship between underwater volume and the mass of the ship: ∇×ρ = Δ If we assume that we load x tonnes to the vessel, so that the vessel sinks by 1 cm, causing the mass of the ship and the underwater volume to increase, then:

(∇ + Change in ∇) × ρ = Δ + Change in Δ In shorthand notation, the change in a value is given the symbol δ. Using this notation gives:

(∇ +δ ∇) × ρ = Δ +δ Δ If we are loading x tonnes to the vessel, then the change in mass of the vessel must be x, therefore:

(∇ +δ ∇) × ρ = Δ + x Multiplying out the left hand side gives:

(∇ + ) + (

∇+

)= Δ+ x

Substituting Formula 1.1 back gives: Δ + (δ ∇ × ρ ) = Δ + x

9781408176122_App05_Rev_txt_prf.indd 689

11/16/2013 2:47:01 AM

690 • Ship Stability, Powering and Resistance Transposing this for δ∇ gives: δ∇=

Δ+ x−Δ ρ

∴δ ∇ =

x ρ

Over a change in draught of 1 cm, the sides of the vessel can be assumed to be vertical. Therefore, the increase in underwater volume, in metres cubed, as a result of sinking 1 cm can be assumed to be given by the waterplane area multiplied by 1 cm (in units of metres): δ ∇ = Waterplane area × 0.01 From previously, we know that waterplane area, length and beam are related by the waterplane area coefficient: CW =

Waterplane area L B

This can be transposed to give the waterplane area: Waterplane area = CW

L×B

Substituting this gives: δ ∇ = CW × L × B × 0 01 Substituting this for δ∇ gives: x = CW ρ

L × B × 0 01

This can be transposed for x: x

CW × L B × 0 01× ρ

Using Formula 1.6, we can determine the relationship between sinkage and added mass: Sinkage =

9781408176122_App05_Rev_txt_prf.indd 690

Mass TPC

11/16/2013 2:47:02 AM

Appendix 5 • 691 If adding x tonnes causes 1 cm of sinkage, then these values can be substituted in Formula 1.1: 1=

x TPC

∴ x = TPC This can be substituted to give an expression for the TPC: TPC

9781408176122_App05_Rev_txt_prf.indd 691

CW × L B × 0 01× ρ

11/16/2013 2:47:04 AM

APPENDIX 6 THE DERIVATION OF THE FRESH WATER ALLOWANCE FORMULA The fresh water allowance formula, given in Formula 1.10, can be proved from first principles. If the mass of the vessel is assumed to remain constant, then as the vessel moves from sea water to fresh water, she will sink down (as a result of lost buoyancy) by a distance δD, and therefore increase her underwater volume, by an amount δ ∇. To differentiate between sea water and fresh water, the subscripts SW and FW will be used. Previously, in Formula 1.1, we have seen the relationship between underwater volume and the mass of the ship: ∇×ρ = Δ If we consider the vessel in sea water, then using the SW subscripts gives: ∇ SW × ρ SW = Δ If we consider the vessel in fresh water, then using the FW subscripts gives: ∇FW × ρFFW = Δ The underwater volume of the vessel increases as she moves into fresh water, by an amount δ∇, therefore: ∇ SW + δ ∇ = ∇FW Substituting this value gives:

(∇

9781408176122_App06_Rev_txt_prf.indd 692

+ ∇) × ρFFW = Δ

11/16/2013 2:46:44 AM

Appendix 6 • 693 As the mass of the vessel is assumed to be constant, we can equate these equations, which both give the mass of the vessel. Therefore:

(∇

+ ∇) × ρFFW = Δ SW × ρSSW

The left hand side can be multiplied out:

(∇

) + (δ ∇ × ρFW ) = Δ SW × ρSSW

×

This can be transposed to make δ ∇ the subject of the formula: ρFW ) = (∇

(δ

δ∇=

(∇

) (Δ

×ρ

) (Δ

×ρ

)

×ρ

)

×ρ

ρFW

The numerator can be factorised for the ∇SW: δ∇=

∇ SW ( ρ − ρ

)

ρFW

Formula 1.1 substituted with SW subscripts gives us the relationship between underwater volume, density and mass in sea water. Transposing this for underwater volume gives: ∇ SW =

Δ ρSW

This can be substituted back to give: δ ∇=

ρ Δ (ρ ρSW ρFW

) = Δ (ρ

ρ

)

ρSW × ρFW

This can be transposed to give: δ

ρSW =

Δ (ρ

ρ

)

ρFW

The left hand side gives the mass of water displacement by the additional volume as the vessel moves from fresh to sea water. Given that the change in draught is small, the vessel can be considered to be wall-sided.

9781408176122_App06_Rev_txt_prf.indd 693

11/16/2013 2:46:45 AM

694 • Ship Stability, Powering and Resistance Therefore, the mass of the additional water displaced as the vessel moves from fresh to sea water can be expressed in terms of the TPC: δ

ρSW =δδ D × TPC SW

Substituting gives: δ D TPC T SW =

Δ ( ρSW ρFW ) ρFW

This can be transposed to give the change in draught: δD=

Δ ( ρSW ρFW ) ρFW × TPC SW

The change in draught, δD, is the difference in draught between fresh water and sea water, and is therefore the ‘Fresh Water Allowance’. Substituting this, along with the constants for density, gives: FWA W =

Δ ( .0 5 − .000 ) = 1.000 × TPC SW

Δ (0.025) Δ = TTPC SW 40 × TPC SW

An analysis of the units that make up this formula show that the FWA here is in units of centimetres: ⎡ ⎛ t ⎞ ⎤ ⎡ ⎤ t× Δ (1.025 − 1.000 ) ⎢ ⎝ m3 ⎠ ⎥ ⎢ t ⎥ ⎡ cm t ⎤ ⎥=⎢ FWA W = =⎢ =[ ⎥= 1.000 × TPC SW ⎢ ⎛ t ⎞ × t ⎥ ⎢ t ⎥ ⎢⎣ t ⎥⎦ ⎢⎣ ⎝ m3 ⎠ cm ⎥⎦ ⎣ cm ⎦

]

Given the required standards of accuracy, the FWA is often calculated in units of millimetres. To convert centimetres to millimetres, the FWA formula is multiplied by ten: FWA W in mm =

9781408176122_App06_Rev_txt_prf.indd 694

Δ × 10 = 40 × TPC SW

0×Δ Δ = TTPC SW 40 × TPC SW

11/16/2013 2:46:46 AM

APPENDIX 7 THE DERIVATION OF THE CHANGE IN KG FORMULA The short-cut for determining the change in the position of the centre of gravity due to moving cargo can be proved algebraically. Assuming that a ship has a displacement Δ, with a vertical centre of gravity KG1, and a mass w removed from a position d1 and replaced at a position d2, causing the centre of gravity of the ship to shift to KG2, then the loading table would show: Table A7.1 Loading table for a change in KG Item

Mass (tonnes)

KG (m)

Vertical moment (tonne metres)

Ship

Δ

KG1

ΔKG1

Mass off

−w

d1

−wd2

Mass on

W

D2

−wd2

Total

Δ

ΔKG1−wd1+wd2

The final centre of gravity, KG2, of the ship can be found from Formula 2.1: Overall K KG G after loading = KG2 =

Total vertical e t ca moment Total mass

ΔKG1 − wd d1 + wd2 Δ

This can be factorised for the load w: KG2 =

9781408176122_App07_Rev_txt_prf.indd 695

ΔKG1 + w d2 − d1 ) Δ

11/16/2013 2:47:06 AM

696 • Ship Stability, Powering and Resistance Both sides can be multiplied by Δ to give: Δ ΔKG KG1 + w (d2 − d1 )

ΔKG2

Both sides can have ΔKG1 subtracted to give: ΔKG1 = w (d2 − d1 ) ΔKG

ΔKG2

This can be factorised for the displacement Δ: Δ(

2

1

) = w (d2 − d1 )

Dividing both sides by the displacement Δ gives:

(KG2 − KG1 ) =

w (d2 − d1 ) Δ

In the left hand side, KG2 − KG1 is the shift in the centre of gravity of the ship, and in the right hand side, d2 − d1 is the shift in the mass. Substituting in the change in G and the distance moved gives Formula 2.4. Note that this also works with the TCG and LCG, with appropriate values substituted.

9781408176122_App07_Rev_txt_prf.indd 696

11/16/2013 2:47:07 AM

APPENDIX 8 THE DERIVATION OF THE FORMULAE GIVING THE TRANSVERSE INERTIA OF A RECTANGULAR WATERPLANE AND A SHIP SHAPE WATERPLANE MEASURED THROUGH THE CENTRELINE The inertia (sometimes also referred to as the second moment of area) is defined as the area multiplied by the distance from a datum point squared. Consider a waterplane area of length L metres and beam B metres. An infinitely thin strip of area dy thick and L long will have an area dA, which can be found from L dy, with a centre y from the centreline of the vessel. Therefore, the area of the strip can be found from: dA = L d dy The inertia of the strip, relative to the centreline, can be found from: Inertia d dA y 2 = L y 2 dy

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698 • Ship Stability, Powering and Resistance L

B/2

dy

B

y

–B/2

▲ Figure A8.1 Inertia geometry

The overall inertia of the waterplane can be found by integrating the waterplane between the limits of B/2 and −B/2: B/2

B/2

Inertia L

∫

2

y d dy

−B / 2

⎛ ⎡ B 3 ⎤ ⎡ B 3 ⎤⎞ ⎡ y3 ⎤ ⎛ B 3 B 3 ⎞ LB 3 L⎢ ⎥ = L⎜ ⎢ 3 −⎢ 3 =L + = ⎥ ⎥ ⎟ ⎝ 24 24 ⎟⎠ 12 ⎝ ⎣ 2 × 3 ⎦ ⎣ 2 3 ⎦⎠ ⎣ 3 ⎦ −B / 2

For a box shaped vessel, the length and beam of the waterplane are the same as the length and beam of a vessel. For a real ship form, the beam varies along the length of the vessel making the transverse inertia of the waterplane about the centreline harder to determine. dx

y dA

dy y

x

0

L

▲ Figure A8.2 Waterplane area geometry

Consider the waterplane as shown in Figure A8.2. The beam at any point along the vessel is y metres. If the waterplane is broken down into elements with a length of dx and a width of dy then: a of element dA = dy dx

9781408176122_App08_Rev_txt_prf.indd 698

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Appendix 8 • 699 The definition of inertia is area multiplied by distance from the reference axis squared, therefore: of element d dA from ce centreline te e Inertia o

y 2 dA d = y 2d dy dx

Integrating this along the length of the waterplane gives: Ly

Total inertia e t a of half a waterplane from o centreline

2 ∫∫ y ddyy dx 00

L

1 3 y dx 3 ∫0

This value must be doubled to take account of both sides of the waterplane: Ly

Total inertia nertia of waterplane from centreline 2∫∫ y 2 dy dx = 00

9781408176122_App08_Rev_txt_prf.indd 699

L

2 3 y dx 3 ∫0

11/16/2013 2:40:49 AM

APPENDIX 9 THE DERIVATION OF THE RELATIONSHIP BETWEEN TRANSVERSE INERTIA, BM AND UNDERWATER VOLUME FOR A BOX SHAPED VESSEL Consider a box shaped vessel, length L, beam B, as shown in Figure A9.1, inclined through a very small angle θ (exaggerated for clarity). As the vessel inclines, a wedge of underwater volume will emerge on the high side of the vessel, while a wedge of an identical amount of underwater volume will be immersed on the low side of the vessel. The result of this is that the centre of buoyancy will move, both transversely, in a direction normal to the centreline, and vertically, in a direction normal to the keel. The volume of each of these emerged and immersed wedges can be determined from the length and beam of the vessel, as shown in Figure A9.2. Volume o of each wedge

L×

2 ⎛B ⎞ ⎛ B ⎞ ⎛ ⎞ LB × tanθ × × = tanθ ⎝2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ 8

An expression for the shift in the centre of buoyancy can be derived from a table of transverse moments of buoyancy, in a similar way to a loading table for TCG. Assuming that a ship has a volume ∇, with a initial centre of buoyancy TCB1, and a volume v removed from a position tcb1 and replaced at a position tcb2, causing the centre of buoyancy of the ship to shift to TCB2, can be represented as in Table A9.1.

9781408176122_App09_Rev_txt_prf.indd 700

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Appendix 9 • 701 Vessel centreline

Emerged wedge

θ

Old WL

Immersed wedge

WL

Original centre of buoyancy

Old WL New centre of buoyancy

Shift in centre of buoyancy

▲ Figure A9.1 The shift in B as a vessel rolls

Vessel centreline

Old WL

Centre of emerged wedge

θ Centre of immersed wedge

(1/3) × (B/2) x tan θ WL

(B/2) x tan θ

(B/2)

Old WL (B/2) x (2/3)

▲ Figure A9.2 Measuring the shift in B

The final centre of buoyancy, TCB2, of the ship can be found from: Overall ce centre tre after loading =

9781408176122_App09_Rev_txt_prf.indd 701

Total vertical ertica moment Total volume

11/16/2013 2:47:41 AM

702 • Ship Stability, Powering and Resistance Table A9.1 Transverse movement in B Volume (m3) TCB (m)

Item

Vertical moment (m4)

Ship

∇

TCB1

∇TCB1

Emerged wedge

−v

tcb1

−vtcb1

Immersed wedge

v

tcb2

vtcb2

Total

∇

TCB2 =

∇TCB1 − vtcb1 + vtcb2

∇TCB1 − vtcb1 vtcb2 ∇

This can be factorised for the wedge volume v: TCB2 =

∇TCB1 + v (tcb b2 ∇

tcb b1 )

Both sides can be multiplied by ∇ to give: ∇

2

=∇

+ v (tcb b2 − tcb b1 )

1

Both sides can have ∇TCB1 subtracted to give: ∇

2

−∇

1

= v (tcb b2 − tcb b1 )

This can be factorised for the volume ∇: ∇(

2

−

1

) = v(( b2 − b1 )

Dividing both sides by the volume ∇ gives: (

2

1

)=

v(( b2 ∇

b1 )

In the left hand side, TCB2− TCB1 is the shift in the centre of buoyancy of the ship, and in the right hand side, tcb2− tcb1 is the shift in the wedge centre. Therefore, the movement of the centre of buoyancy of the ship, when a edge of volume is transferred, can be found using: Change in B =

9781408176122_App09_Rev_txt_prf.indd 702

v Distance moved ∇

11/16/2013 2:47:41 AM

Appendix 9 • 703 The transverse movement of the centre of buoyancy can therefore be determined from: Transverse movement of B =

Volume moved o ed TTransvers a s e se distance mo m ved Total volume

2 LB 2 ⎛B ⎞ LB ⎛ B ⎞ LB 3 tanθ × × × 2 tanθ × × 2 tanθ ⎝2 3 ⎠ ⎝3 ⎠ 8 8 ∴ Movement of o B= = = 12 ∇ ∇ ∇

The vertical movement of the centre of buoyancy can be determined from: Vertical movement o e e t of B =

Volume moved V Vertical distance move v d Total volume

LB 2 ⎛1 B ⎞ LB 3 tanθ × × tanθ × 2 tan2 θ ⎝3 2 ⎠ 8 24 ∴ Movement of B = = ∇ ∇ As tan2 θ can be assumed to be zero as θ is a small angle, the vertical movement of B can be assumed to be zero. Consider the right angle triangle defined by the original centre of buoyancy, the new centre of buoyancy and the metacentre, as shown in Figure A9.3. Vessel centreline

θ Old WL

Metacentre BM

WL

Old WL

▲ Figure A9.3 The metacentre and the shift in B

9781408176122_App09_Rev_txt_prf.indd 703

11/16/2013 2:47:43 AM

704 • Ship Stability, Powering and Resistance

tanθ =

Transverse movement in B BM

This can be transposed to give BM: BM =

Transverse movement o e e t in B tanθ

Combining the equations gives: LB 3 tanθ LB 3 12 Inertia ∇ BM = = 12 = tanθ ∇ ∇

9781408176122_App09_Rev_txt_prf.indd 704

11/16/2013 2:47:44 AM

APPENDIX 10 THE DERIVATION OF THE RELATIONSHIP BETWEEN TRANSVERSE INERTIA, BM AND UNDERWATER VOLUME FOR A SHIP Consider a real ship form with local half beam y, as shown in Figures A10.1 and A10.2, inclined through a very small angle θ (exaggerated for clarity). As the vessel inclines, a wedge of underwater volume dx metres long will emerge on the high side of the vessel, while a wedge of underwater volume will be immersed on the low side of the vessel. The result of this is that the centre of buoyancy will move, both transversely, in a direction normal to the centreline, and vertically, in a direction normal to the keel. The volume of each of these emerged and immersed wedges can be determined from the length and beam of the vessel, as shown in Figure 9.5: L

Volume o of each wedge

∫ ( y × tan 0

) × ( y ) × ⎛⎝

L

1⎞ y2 d = ∫ tanθ dx dx 2⎠ 2 0

The transverse movement of the centre of buoyancy can be determined from: Transverse movement o e e t of B =

Volume moved o ed TTransvers a s e se distance mo m ved Total volume

y2 2 ⎞ ⎛ tanθ d dx × y × × 2 0 2 ⎝ 3 ⎠ ∴ Movement o of B = ∇

∫

9781408176122_App10_Rev_txt_prf.indd 705

L

11/16/2013 2:34:20 AM

706 • Ship Stability, Powering and Resistance Vessel centreline

Emerged wedge

θ

Old WL

Immersed wedge

WL

Original centre of buoyancy

Old WL New centre of buoyancy

Shift in centre of buoyancy

▲ Figure A10.1 The shift in B as a vessel rolls Vessel centreline

Old WL

θ

Centre of emerged wedge

Centre of immersed wedge

(1/3) × (y) × tan θ WL

(y) × tan θ

(y) (y) × (2/3)

Old WL

▲ Figure A10.2 Measuring the shift in B as a vessel rolls

y2 ⎛ 4y⎞ d × ∫ 0 2 tanθ dx ⎝ 3⎠ Movement o of B = ∇ L

4 y3 2 L 3 tan d dx tan nθ dx d ∫ 0y ta 0 6 Movement of o B= =3 ∇ ∇

∫

9781408176122_App10_Rev_txt_prf.indd 706

L

11/16/2013 2:34:21 AM

Appendix 10 • 707 The vertical movement of the centre of buoyancy can be determined from: Vertical movement o e e t of B =

Volume moved V Vertical distance move v d Total volume

1 L 2 ⎛1 ⎞ 1 L 3 2 y tanθ d dx × × y tanθ ×2 ⎝3 ⎠ 3 ∫ 0y tan θ dx 2∫0 ∴ Movement of B = = ∇ ∇ As tan2θ can be assumed to be zero as θ is a small angle, the vertical movement of B can be assumed to be zero. Consider the right angle triangle defined by the original centre of buoyancy, the new centre of buoyancy, and the metacentre, as shown in Figure A10.3. Vessel centreline

θ Old WL

Metacentre BM

WL

Old WL

▲ Figure A10.3 The shift in B and the metacentre

tanθ =

Transverse movement in B BM

This can be transposed to give BM: BM =

Transverse movement in B tanθ

Combining the equations gives: 2 L 3 y tanθ dx 2 L 3 3∫0 ∫ 0y ddx Inertia ∇ BM = =3 = tanθ ∇ ∇

9781408176122_App10_Rev_txt_prf.indd 707

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APPENDIX 11 MAXIMUM ANGLES OF LIST ON PASSENGER VESSELS DUE TO PASSENGER CROWDING As well as considering the metacentric height, officers on passenger vessels are required to determine the list angle that would be caused by all of the passengers crowding on one side of the vessel. This is detailed in chapter 3, section 3.1 of the 2008 IS Code (International Maritime Organisation, 2008). They are also required to determine the heel angle when turning – see Appendix 14: Passenger Vessel Heel in a Turn. Each passenger is assumed to have a minimum mass of 75 kg, with a centre of gravity 1 m above the deck when stood up-right, or 0.30 m above the deck if seated. Passengers are assumed to crowd around the vessel in whichever way is least favourable to the metacentric height and angle of list of the vessel, with an assumed crowding of four passengers per square metre. The resulting angle of list should not exceed 10 degrees.

9781408176122_App11_Rev_txt_prf.indd 708

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APPENDIX 12 THE DERIVATION OF THE FORMULA GIVING THE ANGLE OF LIST FOR A NEUTRALLY STABLE VESSEL Figure A12.1 shows the movement of G as a result of the movement of mass, and the vertical and horizontal movement of B as a result of the vessel inclining. As the vessel inclines, the transverse position of G, which is also M in the case of a neutrally stable vessel, moves in the opposite direction relative to B as a result of the two points being opposite sides of the point of rotation.

Original G BM sin θ

TCG

Final G Original B

Transverse shift of B

Final B Vertical shift of B

▲ Figure A12.1 Vertical and transverse movement of B and G

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710 • Ship Stability, Powering and Resistance The volume of each of these emerged and immersed wedges can be determined as previously shown: Volume o of each wedge

L×

2 ⎛B ⎞ ⎛ B ⎞ ⎛ ⎞ LB × tanθ × × = tanθ ⎝2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ 8

As previously shown, the transverse movement of the centre of buoyancy can be determined from: Transverse movement o e e t of B =

Volume moved TTransverse distance mo m ved Total volume

Substituting in the values for the volume and shift in volume (see equation and Figure A9.2) gives: 2 LB 2 ⎛B ⎞ LB ⎛ B ⎞ LB 3 tanθ × × × 2 tanθ × × 2 tanθ ⎝2 3 ⎠ ⎝3 ⎠ 8 8 ∴ Movement of o B= = = 12 ∇ ∇ ∇

The same process can be used to give the vertical movement of the centre of buoyancy, which can be determined from: Vertical movement o e e t of B =

Volume moved o ed Vertica Ve t cal distance move v d Total volume

LB 2 ⎛1 B ⎞ LB 3 tanθ × × tanθ × 2 tan2 θ ⎝3 2 ⎠ 8 24 ∴ Movement of o B= = ∇ ∇ The transverse and vertical shift in B can be resolved to their horizontal components: LB 3 tanθ Horizontally resolved transverse movement of B = 12 cos oθ ∇ LB 3 tan2 θ 24 Horizontally resolved vertical movement of B = sinθ ∇ Note that for real ship forms, this becomes: 2 L 3 a θ ∫ 0y ddx tan Horizontally resolved transverse movement of B = 3 cosθ ∇ 1 L 3 2 ∫ 0y ddx tan θ 3 Horizontally resolved vertical movement of B = sinθ ∇

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Appendix 12 • 711 Therefore, the total horizontal movement of B can be found from: LB 3 LB 3 tanθ tan2 θ 12 24 Resolved horizontal o o ta movement o of B = cosθ + sinθ ∇ ∇ Note that for real ships forms, this becomes: 2 L 3 1 L 3 2 y d dx tanθ ∫ ∫ 0y ddx tan θ 0 3 3 Resolved horizontal movement of B = cosθ + sinθ ∇ ∇ The inertia divided by the volume gives BM: Resolved horizontal movement of B

BM tanθ tanθ cosθ

1 BM tan2 θ sinθ 2

The resolved horizontal movement of G can be found: Resolved horizontal o o ta movement o of G TTCG cosθ Note that relative to the original B, G has moved a transverse distance sideways given by: BM sinθ As described previously, the resolved horizontal shift in G must equate to the resolved horizontal shift in B: 1 BMtan2 θ sinθ 2

TCG cosθ BM sinθ BMtanθ cosθ Dividing through by cos θ gives: TCG BMtanθ BMtan = BMtanθ

1 BMtan3 θ 2

Collecting like terms for θ gives: TCG

1 BMtanθ BMta + BMtan3 θ BMtanθ 2

This simplifies to: TCG

9781408176122_App12_Rev_txt_prf.indd 711

1 BMtan3 θ 2

11/16/2013 2:35:58 AM

712 • Ship Stability, Powering and Resistance This can be transposed to give an expression for θ: 2 × TCG = BMtan3 θ 2 × TCG = tan3 θ BM 3

2 × TCG = tanθ BM

As seen in Formula 2.4, the movement of G as a result of a transverse shift in a mass can be expressed in terms of the mass moved and the distance moved by: Movement of G =

w d Δ

Therefore, the list angle of a vessel with neutral stability can be found from:

3

9781408176122_App12_Rev_txt_prf.indd 712

2×

w ×d Δ = tanθ BM

11/16/2013 2:36:00 AM

APPENDIX 13 THE DERIVATION OF THE FORMULAE DESCRIBING THE EFFECT OF FREE SURFACES ON METACENTRIC HEIGHT Consider a cross-section of a rectangular planform tank with a length l metres and beam b metres, containing a fluid with a density ρ. As the vessel heels through an angle θ, the free surface will move through an angle θ relative to the original free surface, as shown in Figure A13.1. This will cause a wedge of fluid to move across the tank, resulting in a transverse movement of the centre of gravity of the fluid, from g1 to g2. b

Original free surface

θ g1

g2

b tan θ 2

2 b x 3 2

New free surface

▲ Figure A13.1 The geometry of moving fluids

The volume of the transferred wedge of fluid can be found from the geometry of the wedge: Wedge volume

9781408176122_App13_Rev_txt_prf.indd 713

l×

2 ⎛ b⎞ ⎛ b ⎞ 1 lb × tanθ × = tanθ ⎝ 2⎠ ⎝ 2 ⎠ 2 8

11/16/2013 2:43:24 AM

714 • Ship Stability, Powering and Resistance The mass of the transferred wedge of fluid can be found from the volume of the wedge: Wedge mass Wedge W volume × ρ = ρ

lb2 tanθ 8

As proven in Formula 2.4, the shift in the centre of gravity of a vessel can be found from: Shift f in in G =

W Distance moved Δ

Therefore, the overall transverse shift in the centre of gravity of the vessel can be found by:

Shift f in G =

The term

ρ

lb2 2 b lb3 tanθ × × × 2 ρ tanθ 8 3 2 = 12 Δ Δ

lb3 is the transverse inertia of the free surface, written as IFS: 12

Shift f in G =

ρ

lb2 2 b tanθ × × × 2 ρI tanθ 8 3 2 = FS = TCG Δ Δ

Now consider the vessel heeling through an angle θ. The centre of gravity will move transversely, as shown in Figure A13.1. This will move the line of action of gravity closer to the line of action of buoyancy, so that it crosses the centreline at a position known as the virtual centre of gravity, GV, and therefore gives a reduction in the torque trying to right the vessel. Effectively, the initial metacentric height has been reduced by the movement of the fluid. This reduction is termed the ‘virtual’ reduction, as it only occurs as the vessel inclines, and will not exist when the vessel is perfectly upright and still (a condition that the vessel will never reach in reality). The line of action of gravity, the transverse shift in G, and the centreline from the original centre of gravity to GV form a right angle triangle. This can be used to determine the distance from G to GV, which is the ‘virtual’ loss in metacentric height. Tanθ = tanθ =

9781408176122_App13_Rev_txt_prf.indd 714

opp adj d

TCG G to GV

11/16/2013 2:43:24 AM

Appendix 13 • 715

M Gv

G TCG

B

▲ Figure A13.2 The effective, fluid or virtual centre of gravity

G to GV = Loss in GM =

TCG tanθ

Combining gives: G to GV = Loss in GM =

ρ IFS tanθ ρ I tanθ ρ IFS ÷ tanθ = FS = = FSC Δ Δ tanθ Δ

Therefore, the loss in GM, also known as the free surface correction, or FSC, can be directly determined from the inertia of the free surface, the density of the fluid and the displacement of the ship. The final effective, or fluid GM, can be found from: GM = KM − KG − FSC Although this method may seem different to the loading table method initially shown, as the FSC includes the displacement as a denominator, it can be combined in the loading table.

Table A13.1 Sample loading table for KG with FSM Item

Mass (tonnes)

KG (m)

Vertical moment (tonne metres)

Ship

Δ1

KG1

Δ1KG1

Fluid cargo

ΔL

KGL

ΔLKGL

FSE

0

0

FSM

Total

Δ

9781408176122_App13_Rev_txt_prf.indd 715

Δ1KG1+ ΔLKGL + FSM

11/16/2013 2:43:26 AM

716 • Ship Stability, Powering and Resistance Consider a vessel with a displacement Δ1 and KG of KG1, loading a liquid cargo with a mass of ΔL, at a KG of KGL, with a free surface moment of FSM tonne metres. The loading table and subsequent effective KG would be given by: Effective v KG after loading =

Total vertical ertica moment (including FSM ) Total mass

Effective v KG KG after loading =

Δ1KG G1 + Δ L KG GL + FSM Δ

The overall effective GM would be given by: GM = KM − KG = KM −

Δ1KG1 + Δ L KGL + FSM Δ

The KG term can be expanded to: GM = KM − KG = KM −

Δ1KG G1 + Δ L KGL FSM − Δ Δ

This is the same form as GM = KM − KG − FSC, therefore either method should give the same final answer.

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APPENDIX 14 PASSENGER VESSEL HEEL IN A TURN Officers on passenger vessels are required to determine the heel angle when turning, as described in the 2008 IS Code (International Maritime Organisation, 2008). As with the passenger list angle, the resulting angle of heel should not exceed 10 degrees. Rather than using the approach described previously for heel angles in a turn, a different formula is used: MR = 0.200 ×

v 02 d⎞ ⎛ × Δ × KG − ⎝ LWL 2⎠

In this equation, MR is the heeling moment, in units of kilo-Newton metres, v0 is the service speed in units of metres per second, LWL is the waterline length of the vessel in units of metres, Δ is the displacement of the vessel in units of tonnes, d is the mean draught in units of metres and KG is the height of the centre of gravity above the keel in units of metres. Once this has been determined, the angle of heel can be found using: sinθ =

9781408176122_App14_Rev_txt_prf.indd 717

MR Δ × GM

11/16/2013 2:39:58 AM

APPENDIX 15 DERIVATION OF THE INCREASE IN DRAUGHT WHEN HEELING FORMULA Figure 2.28 shows the increase in the draught of a vessel as a result of heeling. If the vessel is assumed to be rectangular in cross-section (which is a reasonable assumption for a modern ship) then the increase in draught can be calculated from the cross-section geometry of the vessel. Consider the inclined and upright vessel as shown in Figure A15. On the submerged side, consider the geometry of the original and new waterlines, as shown in detail in Figure A15.2. It can be seen that the vertical submergence of the intersection of the original waterline B and the hull side is given by sinθ . 2

θ

WL

WL

▲ Figure A15.1 A box shaped vessel heeled

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11/16/2013 2:42:29 AM

Appendix 15 • 719

WL 0.5B sin θ

θ

WL

0.5B

▲ Figure A15.2 Submerging the original waterline

The original draught (D) can also be resolved vertically, as shown in Figure A15.3.

WL 0.5B sin θ

θ

WL

D cos θ

0.5B

▲ Figure A15.3 Resolved vertical draught

Combining the two components, as shown in Figure A15.3, gives the equation for the heeled draught: Heeled drau r ght

9781408176122_App15_Rev_txt_prf.indd 719

B D cosθ cos + sinθ 2

11/16/2013 2:42:30 AM

APPENDIX 16 THE MATHEMATICAL PROOF OF THE DETERMINATION OF GM FROM THE GZ CURVE This process of determining GM from the GZ curve is based on small angle approximations. We have already seen that at small angles, Formula 3.2 gives: GZ ≈ GM × sinθ If θ is in radians, then at small angles, sin θ ≈ θ, therefore: GZ ≈ GM θ In Figure 3.14, the straight line drawn through the initial slope has the equation y = mx + c, where y is the GZ, m is the gradient and x is the angle θ. As the line goes through the origin of the graph, c, the intercept is zero. Therefore, equating this with the equation of a straight line, the equation of the line in Figure 3.14 is: y

mx equates to : GZ = GM θ

Equation GZ ≈ GM θ can be differentiated to find the slope: d (GM G ) = GM dθ Therefore, the gradient of the equation GZ = GM θ is the initial GM of the vessel. If dθ is 1, then the resulting dGZ value must be equal to GM. Therefore, reading the value of the line at one radian (where θ = 1) gives us the value of GM.

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APPENDIX 17 THE DERIVATION OF THE ANGLE OF HEEL IN A TURN FORMULA It is this branch of stability theory that lets us calculate the angle of heel in a turn, as used previously. Centripetal force is given by: FC =

mv 2 r

Substituting in the mass of the vessel in tonnes, and dividing by g to keep the units correct, gives the centripetal force in units of tonnes force: FC =

Δv 2 gr

In this formula, the centripetal force FC is in units of Newtons, with the speed v in metres per second and the turn radius r in metres. The mass, m, is in units of kilograms. The centripetal force acts through the centre of buoyancy, with the lever of the force being the resolved vertical distance from the centre of gravity to the centre of buoyancy, as shown in Figure A17.1.

M G BG cos θ Fc B

▲ Figure A17.1 Forces when turning

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722 • Ship Stability, Powering and Resistance This force and lever create a heeling moment acting on the vessel: Heeling moment

Force × Lever =

Δv 2 (KG − KB ) cosθ gr

The righting moment created by the movement of the centre of buoyancy is equal to the buoyancy force multiplied by the righting lever: Righting moment = Buoyancy × GZ At small angles, GZ can be approximated: GZ = GM sinθ Therefore: Righting moment = Buoyancy × GM sinθ For the vessel to be in equilibrium, the buoyancy force (in units of tonnes force) must be equal to the displacement, therefore: Righting moment = Δ × GM sinθ When the vessel reaches a steady angle of heel, the forces and moment are in equilibrium, therefore the righting moment must equal the heeling moment, therefore: Righting moment = Heeling moment Δ × GM sinθ =

Δv 2 (KG − KB ) cosθ gr

This can be transposed for θ to give: Δv 2 (KG − KB ) sinθ gr = cosθ Δ × GM From trigonometric identities, we know that: sinθ = tanθ cosθ Therefore: Δv 2 (KG − KB ) gr tanθ = Δ × GM This simplifies to: tanθ =

9781408176122_App17_Rev_txt_prf.indd 722

v 2 (K KG KB ) g × r × GM

11/16/2013 2:41:06 AM

APPENDIX 18 THE MATHEMATICAL PROOF OF SIMPSON’S RULE Simpson’s Rule is a method of integrating, and work as follows. If a parabolic curve intersects a box at the curve apex, as shown in Figure A18.1, then the area of the box 2 is always split into a section which is of the area (the shaded section in Figure A18.1), 3 1 and a section which is of the total box area: 3

▲ Figure A18.1 Parabola and containing box

This can be proved with calculus: Consider a curve, as shown in Figure A18.2, with the equation y = ax2. The shaded area under the curve between the origin and x can be directly determined by integration: x

Area u under de the cu curvee = ∫a ax 2 dx 0

x

⎡ ax 3 ⎤ ⎡ ax 3 ⎤ ⎡ a03 ⎤ 1 3 Area u under de the curve = ⎢ =⎢ a ⎥ ⎥−⎢ ⎥ = ax ⎣ 3 ⎦0 ⎣ 3 ⎦ ⎣ 3 ⎦ 3 The area of the box itself can be determined by the width and depth of the box: Area Width W × Dept e h

9781408176122_App18_Rev_txt_prf.indd 723

x × ax 2 = ax 3

11/16/2013 2:39:36 AM

724 • Ship Stability, Powering and Resistance y

y = ax 2

ax 2

x

▲ Figure A18.2 Area under a parabola

Therefore, it can be seen that the shaded area under the curve is

1 of the total box area, 3

2 of the total box area. This also works if 3 the box is ‘skewed’, as shown below in Figure A18.3. The ‘box’ area of the trapezoid can still be found by length x depth: therefore the area above the curve must be

2 box area 3

▲ Figure A18.3 Skewed parabola

Now consider the curve shown in Figure A18.4. The curve is a smooth, parabolic curve that runs through three points, with coordinates x1,y1 and x2,y2 and x3,y3. The spacing between the x values, S, is constant: In Figure A18.4, the area of the solid shaded section can be found from trapezoidal integration: ⎛ y + y3 ⎞ Solid shaded section area = ⎜ 1 × 2S ⎝ 2 ⎟⎠

9781408176122_App18_Rev_txt_prf.indd 724

11/16/2013 2:39:38 AM

Appendix 18 • 725

d2 d2 d2

y3

y2

d1 y1

x1

S

x2

S

x3

▲ Figure A18.4 Area under a parabola and trapezoid

The depth of the ‘box’ surrounding the curve is ‘d2’. The distance between the x axis and the dashed line linking the base of the parabola curve, at the mid-point of the line, is ‘d1’. Distance ‘d1’ can be found from the mean of values y1 and y3: d1 =

y1 + y 3 2

It can be seen from Figure 9.20 that the distance d2 can be found using: d1

y 2 − d1

Substituting in the equation for d1 gives: d2 d2

y 2 − d1

⎛ y + y3 ⎞ y2 − ⎜ 1 ⎝ 2 ⎟⎠

2 As previously seen, the area of the hatched section of Figure A18.4 is of the area of the 3 ‘box’ surrounding the parabola. The area of the box surrounding the parabola can be found by multiplying the depth of the parabola box (d2) by the length of the parabola box (2S): Area under the parabola

2 Box length × Box o depth 3

Area u under de the parabola

9781408176122_App18_Rev_txt_prf.indd 725

2 3

S × ( d2 )

11/16/2013 2:39:38 AM

726 • Ship Stability, Powering and Resistance

Area under the parabola

⎛ ⎛ y + y3 ⎞ ⎞ S × y2 − ⎜ 1 ⎝ 2 ⎟⎠ ⎟⎠ ⎝

2 3

This can be transposed to give: 2 3

Area under the parabola

y y ⎞ ⎛ S × y2 − 1 − 3 ⎟ ⎝ 2 2⎠

These values can be used to determine the total area under the curve: Total a area ea

Solid So d shaded a area ea + Hatched atc ed shaded area

⎛ ⎛ y + y3 ⎞ ⎞ Total area = ⎜ ⎜ 1 × 2S ⎟ ⎝⎝ 2 ⎠ ⎠

⎛2 ⎝3

y y ⎞⎞ ⎛ S × y2 − 1 − 3 ⎟ ⎟ ⎝ 2 2 ⎠⎠

⎛ ⎛ y + y3 ⎞ ⎞ Total area = ⎜ ⎜ 1 ⎟ × 2S ⎝⎝ 2 ⎠ ⎠

⎛2 ⎝3

y y ⎞⎞ ⎛ S × y2 − 1 − 3 ⎟ ⎟ ⎝ 2 2 ⎠⎠

Total area

(( y

+y

)

S) +

y y ⎞⎞ ⎛4 ⎛ ×S y − 1− 3 ⎝ 2 2 2 ⎟⎠ ⎟⎠ ⎝3

Total area =

S S ( y + y )+ ( y − y − y 3 3

)

Total area =

S (( y + y ) + ( y − y − y 3

))

Total area =

S ( y+ y + y − y− y 3

)

Total area =

S ( y 3

)

Total area =

y + y + y − y

S ( − 3

Total area =

y + y +

−

S (y+ y + y 3

)

y

)

It can be seen that this gives the same final equation as Simpson’s Rule for three equally spaced coordinates. If there are an odd number of equally spaced ordinates (Figure A18.5), then the process simply repeats itself: Total area =

S S S ( y + y + y )+ ( y + y + y )+ ( y + y + y 3 3 3

9781408176122_App18_Rev_txt_prf.indd 726

)

11/16/2013 2:39:40 AM

Appendix 18 • 727

y3

y2

y1

y4

y5

y6 y7

S

S

S

S

S

S

▲ Figure A18.5 Simpson’s Rule notation

S Total area = (1y1 + 4 y 2 + 1y 3 + 1y 3 + 4 y 4 + 1y 5 + 1y 5 + 4 y 6 + 1y7 ) 3 S Total area = (1y1 + 4 y 2 + 2 y 3 + 4 y 4 + 2 y 5 + 4 y 6 + 1y7 ) 3 It can be seen that this gives the same final equation as Simpson’s Rule for an odd number of equally spaced coordinates.

9781408176122_App18_Rev_txt_prf.indd 727

11/16/2013 2:39:44 AM

APPENDIX 19 ALTERNATIVE CRITERIA FOR LARGE ANGLE STABILITY TIMBER DECK CARGO Under the 2008 IS Code, for vessels carrying a timber deck cargo in accordance with the loading described in the section on metacentric height requirements, alternative large angle stability criteria may be followed.

Requirement

Details

Peak GZ value

Must be greater than 0.25 m

Area under the graph between 0 and 40* degrees

Must be greater than 0.08 m radians

Initial GM

0.10 m

* If the angle of down-flooding (the angle of heel at which the vessel will progressively flood through weather-tight fittings) is less than 40 degrees, then that angle should be used instead.

9781408176122_App19_Rev_txt_prf.indd 728

11/16/2013 2:40:59 AM

APPENDIX 20 THE DIRECT CALCULATION OF THE AREA UNDER A GENERIC GZ CURVE As the GZ curve in the example used can be modelled by a mathematical function, integration can be used to determine the actual area under the curve between 0 and 40 degrees (with the angles converted to radians): θ

Area 1∫ssinθ cossθ dθ 0

If x

ssinθ then

dx = cosθ dθ ∴ dx = cosθ dθ θ

θ ⎡ x2 ⎤ ∴ Area = 1∫ xdx = ⎢ ⎥ ⎣ 2 ⎦0 0

However , x = sinθ 0.6981

⎡ ( sin )2 ⎤ ∴ Area = ⎢ ⎥ ⎣ 2 ⎦0

=

( sin .6981)2 ( sin 0 )2 − = 0.2066 m radians 2 2

It can be seen that Simpson’s Rule slightly underestimates the area under a typical GZ curve shape, and therefore approximates to the safe side of the actual value.

9781408176122_App20_Rev_txt_prf.indd 729

11/16/2013 2:37:55 AM

APPENDIX 21 THE DERIVATION OF THE WALLSIDED FORMULA FOR APPROXIMATING THE RIGHTING LEVER The wall-sided formula can be derived from the geometry of the vessel before either keel emersion or deck edge immersion. As the vessel heels, a wedge of underwater volume is transferred from the emerged to the immersed side of the vessel. This causes the centre of buoyancy to move transversely and vertically in relation to the centreline of the vessel, as shown in Figure A21.1. The volume of each of these emerged and immersed wedges can be determined as previously shown, from the length and beam of the vessel, as shown in Figure 9.2: Volume of each wedge

L×

Original B

2 ⎛B ⎞ ⎛ B ⎞ ⎛ ⎞ LB × tanθ × × = tanθ ⎝2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ 8

Final B

Transverse shift of B

Vertical shift of B

▲ Figure A21.1 Vertical and transverse movement of B

9781408176122_App21_Rev_txt_prf.indd 730

11/16/2013 2:41:10 AM

Appendix 21 • 731 The transverse movement of the centre of buoyancy can be determined from: Transverse movement o e e t of B =

Volume moved TTransverse distance mo m ved Total volume

2 LB 2 ⎛B ⎞ LB ⎛ B ⎞ LB 3 tanθ × × ×2 tanθ × ×2 tanθ ⎝2 3 ⎠ ⎝3 ⎠ 8 8 ∴ Movement of B = = = 12 ∇ ∇ ∇

As the wall-sided formula is concerned with large angles, tan2 θ cannot be assumed to be zero, and so must be accounted for. The vertical movement of the centre of buoyancy can be determined from: Vertical movement of B =

Volume moved o ed V Vertica e t cal distance move v d Total volume

LB 2 ⎛1 B ⎞ LB 3 tanθ × × tanθ ×2 tan2 θ ⎝ ⎠ 8 3 2 24 ∴ Movement of B = = ∇ ∇ These values for the shift in the centre of buoyancy can be resolved along the original waterline of the vessel, as can the original KB value. The sum of these resolved distances is equal to KN. Therefore: KN = KBsinθ KBsi T Transvers e B shift hf

θ Vertica V l B shift f sinθ

As shown previously, the waterplane inertia can be substituted into these equations. In addition, the inertia divided by the volume gives BM: ⎛ LB 3 ⎞ ⎜⎝ 12 tanθ ⎟⎠ I Transverse B shift = = tanθ BM tanθ ∇ ∇ ⎛ LB 3 2 ⎞ ⎜⎝ 24 tan θ ⎟⎠ I BM Vertical B shift = = tan2θ = tan2θ ∇ 2∇ 2 Substituting these values gives: KN = KB sinθ BMtanθ × cosθ +

9781408176122_App21_Rev_txt_prf.indd 731

BM tan2 θ × sinθ 2

11/16/2013 2:41:11 AM

732 • Ship Stability, Powering and Resistance From trigonometric identities, it is known that: tanθ =

sinθ cosθ

Therefore: KN = KBsinθ BM

sinθ BM × cosθ + tan2 θ × sinθ cosθ 2

This can be simplified to: KN = KB sinθ BM sinθ +

BM tan2 θ sinθ 2

We know from Formula 3.1: GZ = KN − KG sinθ Substituting this gives: GZ = KB sinθ BM sinθ +

BM ta 2 θ sin tan si θ − KG sinθ 2

This can be factorised for sin θ: BM ⎛ ⎞ GZ = sin si θ K KB BM + ta 2 θ K tan KG ⎝ ⎠ 2 Rearranging this to collect the terms not associated with θ gives: BM ⎛ ⎞ GZ = sin si θ K KB BM KG K + tan2 θ ⎝ ⎠ 2 However, from Formula 2.6 we know that: GM = KB + BM − KG Substituting this gives: BM ⎛ ⎞ GZ = sinθ GM + tan2 θ ⎝ ⎠ 2 For a real ship hull form, the beam will again be in terms of the local beam, y, with the volume based on a length of ship dx metres long. The volume of each of these emerged

9781408176122_App21_Rev_txt_prf.indd 732

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Appendix 21 • 733 and immersed wedges can be determined as previously shown, from the length and beam of the vessel, as shown in Figure A9.2: L

Volume o of each wedge

⎛ ∫ ( y × tan ) × ( y ) × ⎝ 0

L

1⎞ y2 d = ∫ tanθ dx dx 2⎠ 2 0

The transverse movement of the centre of buoyancy can be determined from: Transverse movement o e e t of B =

Volume moved TTransverse distance mo m ved Total volume

y2 2 ⎞ 2 L 3 ⎛ d × y × ×2 y tanθ dx ∫ 0 2 tanθ dx ⎝ 3 ⎠ 3∫0 ∴ Movement of B = = ∇ ∇ L

As the wall-sided formula is concerned with large angles, tan2 θ cannot be assumed to be zero, and so must be accounted for. The vertical movement of the centre of buoyancy can be determined from: Vertical movement o e e t of B =

Volume moved V Vertical distance move v d Total volume

1 L 2 ⎛1 ⎞ 1 L 3 2 y tanθ d dx × × y tanθ ×2 ∫ 0 ⎝ ⎠ 3 ∫ 0y tan θ dx 2 3 ∴ Movement of o B= = ∇ ∇ 1 L ∫0 ∴ Movement o of B = 3

∇

1 2 L 3 2 × ∫ y tan θ dx 0 =2 3 ∇

These values for the shift in the centre of buoyancy can be resolved along the original waterline of the vessel, as can the original KB value. The sum of these resolved distances is equal to KN. Therefore: KN = KB sin si θ Transvers T e B shift h f cos cosθ + Vertical B shift × sinθ As shown previously, the waterplane inertia can also be substituted. In addition, the inertia divided by the volume gives BM: 2 L 3 ∫ 0y tanθ dx I Transverse B shift = 3 = tanθ BM tanθ ∇ ∇ 1 2 L 3 2 × ∫ y tan θ dx I BM 2 3 0 Vertical B shift = = tan2 θ = tan2 θ ∇ 2∇ 2

9781408176122_App21_Rev_txt_prf.indd 733

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734 • Ship Stability, Powering and Resistance Substituting these values gives: KN = KB sinθ BMtanθ × cosθ +

BM tan2 θ × sinθ 2

From trigonometric identities, it is known that: tanθ =

sinθ cosθ

Therefore: KN = KB sinθ BM

sinθ BM × cosθ + tan2 θ × sinθ cosθ 2

This can be simplified to: KN = KB sinθ BM sinθ +

BM tan2 θ sinθ 2

We know from Formula 3.1: GZ = KN − KG sinθ Substituting this gives: GZ = KB sinθ BM sinθ +

BM tan2 θ sin si θ − KG sinθ 2

This can be factorised for sin θ: BM ⎛ ⎞ GZ = sin si θ K KB BM + ta 2 θ K tan KG ⎝ ⎠ 2 Rearranging this to collect the terms not associated with θ gives: BM ⎛ ⎞ GZ = sin si θ K KB BM KG K + tan2 θ ⎝ ⎠ 2 However, from Formula 2.6 we know that: GM = KB + BM − KG Substituting this gives: BM ⎛ ⎞ GZ = sinθ GM + tan2 θ ⎝ ⎠ 2

9781408176122_App21_Rev_txt_prf.indd 734

11/16/2013 2:41:18 AM

APPENDIX 22 THE DERIVATION OF THE FORMULA GIVING THE EFFECTIVE METACENTRIC HEIGHT AT AN ANGLE OF LOLL The metacentric height can be found from the slope of the GZ curve. Therefore, the metacentric height of the vessel at the angle of loll can be found by determining the slope of the GZ curve at the angle of loll. Assuming that the GZ can be modelled by the wall-sided formula, the slope of the GZ curve can be found by differentiating the wall-sided formula with respect to the angle. BM ⎛ ⎞ GZ = sinθ GMI + tan2 θ ⎝ ⎠ 2 Note that for clarity, in this formula GM has been written as GM1 to indicate that it is the initial metacentric height of the vessel. As the wall-sided formula is composed of two multiplied sections which each contain the variable used for differentiation, the product rule of differentiation must be used. This states that for a function of the form: y

uv

The differentiated function can be found from: dy dv du = u +v dx dx dx Using this method, it can be expressed as: GZ = u × v

9781408176122_App22_Rev_txt_prf.indd 735

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736 • Ship Stability, Powering and Resistance Where: u

sinθ sin v

GM +

BM tan2 θ 2

Differentiating these two components with respect to the angle gives: du = cosθ dx dv BM = 2tanθ sec 2 θ = BMtanθ sec 2 θ dx 2 The differentiation of tan2 θ will be shown at the end of this proof. Therefore, using the product rule, the gradient of the GZ curve can be found from: dy BM ⎛ ⎞ = sinθ BM tanθ sec 2 θ GM G + tan2 θ cosθ ⎝ I ⎠ dx 2 At the angle of loll, it has previously been shown that: GM +

BM ta 2 θ = 0 tan 2

Substituting this gives: dy = sinθ BM tanθ sec 2 θ + ( )cosθ = sinθ BM tanθ sec 2 θ dx From trigonometric identities, it is known that: 1 sec θ = cosθ Substituting gives: dy 1 = sinθ BM tanθ × dx cos2 θ This can be simplified to: dy BM tanθ sinθ = dx cos2 θ From trigonometric identities, it is known that: sinθ tanθ = cosθ

9781408176122_App22_Rev_txt_prf.indd 736

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Appendix 22 • 737 Substituting gives: dy BM sinθ sinθ = dx cosθ cos 2 θ This can be simplified to: BM sin2 θ dy = dx cosθ cos2 θ Substituting this gives: dy BM tan2 θ = dx cosθ We know that the angle of loll can be expressed in terms of the initial metacentric height and BM: ⎛ −2GMI ⎞ tan2 θ = ⎜ ⎝ BM ⎟⎠ Substituting this gives: dy = dx

BM

⎛ −2GMI ⎞ ⎝ BM ⎠ cosθ

This can be simplified to: dy −2GMI = = GM at the a angle g e of loll dx cosθ Proof of the differentiation of tan2 θ: If: tan2 θ tanθ tanθ Then it can be expressed in terms of the product rule: y uv Using the product rule: dy dv du = u +v dx dx dx Therefore: dy = tanθ sec 2 θ + tanθ sec 2 θ = 2tanθ sec 2 θ dx

9781408176122_App22_Rev_txt_prf.indd 737

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APPENDIX 23 THE DERIVATION OF THE FORMULA GIVING THE LONGITUDINAL INERTIA OF A RECTANGLE The longitudinal inertia (sometimes also referred to as the longitudinal second moment of area) is defined as the area multiplied by the distance from a datum point squared. Consider a waterplane area of length L metres and beam B metres. An infinitely thin strip of area dx thick and L long will have an area dA, which can be found from B dx, with a centre x from the longitudinal centre of area: L dx B x

–L/2

L/2

▲ Figure A23.1 Longitudinal inertia geometry

Therefore, the area of the strip can be found from: dA = B d dx

9781408176122_App23_Rev_txt_prf.indd 738

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Appendix 23 • 739 The longitudinal inertia of the strip, relative to the longitudinal centre of area, can be found from: InertiaL

dA x 2 = B x 2 dx d

The overall longitudinal inertia of the waterplane can be found by integrating the waterplane between the limits of L/2 and −L/2: L/2

L/2

InertiaL

B

∫

x2 d dx

−L / 2

9781408176122_App23_Rev_txt_prf.indd 739

⎛ ⎡ L3 ⎤ ⎡ L3 ⎤⎞ ⎡ x3 ⎤ ⎛ L3 L3 ⎞ BL3 B⎢ ⎥ = B⎜ ⎢ 3 − = B + = ⎥ ⎢ 3 ⎥⎟ ⎝ 24 24 ⎟⎠ 12 ⎝ ⎣ 2 × 3 ⎦ ⎣ 2 3 ⎦⎠ ⎣ 3 ⎦ −L / 2

11/16/2013 2:47:37 AM

APPENDIX 24 THE DERIVATION OF THE FORMULA GIVING THE LONGITUDINAL BM FOR A BOX SHAPED VESSEL Consider a box shaped vessel, length L, beam B, as shown in Figure A24.1, pitched through a very small angle θ (exaggerated for clarity). As the vessel pitches, a wedge of underwater volume will emerge on the high end of the vessel, while a wedge of an identical amount of underwater volume will be immersed on the low end of the vessel. The result of this is that the centre of buoyancy will move, both longitudinally, in a direction normal to a vertical line through amidships, and vertically, in a direction normal to the keel. The volume of each of these emerged and immersed wedges can be determined from the length and beam of the vessel, as shown in Figure A24.2. Emerged wedge

θ

Amidships

Old WL

Immersed wedge

WL Original centre of buoyancy

Old WL Shift in centre of buoyancy

New centre of buoyancy

▲ Figure A24.1 The shift in B as a vessel pitches

9781408176122_App24_Rev_txt_prf.indd 740

11/16/2013 2:47:48 AM

Appendix 24 • 741 (1/3) × (L/2) × tan θ

Old WL

θ

Centre of emerged wedge

Amidships

Centre of immersed wedge

WL (L/2) × tan θ (L/2)

(L/2) × (2/3)

Old WL

▲ Figure A24.2 Measuring the shift in B

Volume o of each wedge

B×

2 ⎛L ⎞ ⎛ L ⎞ ⎛ ⎞ BLL × tanθ × × = tanθ ⎝2 ⎠ ⎝ 2⎠ ⎝ 2⎠ 8

An expression for the shift in the longitudinal centre of buoyancy can be derived from a table of longitudinal moments of buoyancy, in a similar way to a loading table for LCG. Assuming that a ship has a volume ∇, with a initial centre of buoyancy LCB1, and a volume v removed from a position lcb1 and replaced at a position lcb2, causing the centre of buoyancy of the ship to shift to LCB2, then the table would show: Table A24.1 Transverse movement in B Item

Volume (m3)

LCB (m) Vertical moment (m4)

Ship

∇

LCB1

∇LCB1

Emerged wedge

−v

lcb1

−vlcb1

Immersed wedge

v

lcb2

vlcb2

Total

∇

∇LCB1 − vlcb1 + vlcb2

The final centre of buoyancy, LCB2, of the ship can be found from: Overall centre r after loading = LCB2 =

9781408176122_App24_Rev_txt_prf.indd 741

Total vertical e t ca moment Total volume

∇LCB1 − vlc v b1 vvlcb2 ∇

11/16/2013 2:47:48 AM

742 • Ship Stability, Powering and Resistance This can be factorised for the wedge volume v: LCB2 =

∇LCB1 + v (lcb l b2 ∇

l b1 ) lcb

Both sides can be multiplied by ∇ to give: ∇

=∇

2

1

+ v (lcb l b2 − lcb l b1 )

Both sides can have ∇TCB1 subtracted to give: ∇

−∇

2

1

= v (lcb l b2 − lcb l b1 )

This can be factorised for the volume ∇: ∇(

2

−

1

) = v (l b2 − l b1 )

Dividing both sides by the volume ∇ gives: (

2

1

)=

v (l b2 l b1 ) ∇

In the left hand side, LCB2 − LCB1 is the shift in the centre of buoyancy of the ship, and in the right hand side, lcb2 − lcb1 is the shift in the wedge centre. Therefore, the movement of the centre of buoyancy of the ship, when a edge of volume is transferred, can be found using: Change in B =

v Distance moved ∇

The longitudinal movement of the centre of buoyancy can therefore be determined from: Longitudinal movement of B =

Volume moved Longitudinal distanc n e moved Total volume

2 BLL2 ⎛L ⎞ BL ⎛ L ⎞ BLL3 tanθ × × × 2 tanθ × × 2 ⎝2 3 ⎠ ⎝ 3 ⎠ 12 tanθ 8 8 ∴ Movement of B = = = ∇ ∇ ∇

The vertical movement of the centre of buoyancy can be determined from: Vertical movement o e e t of B =

Volume moved Vertica V l distance move v d Total volume

BLL2 ⎛1 L ⎞ BLL3 2 tanθ × × tanθ × 2 ⎝3 2 ⎠ 24 tan θ 8 ∴ Movement of B = = ∇ ∇

9781408176122_App24_Rev_txt_prf.indd 742

11/16/2013 2:47:49 AM

Appendix 24 • 743 As tan2 θ can be assumed to be zero as θ is a small angle, the vertical movement of B can be assumed to be zero. Consider the right angle triangle defined by the original centre of buoyancy, the new centre of buoyancy, and the metacentre, as shown in Figure A24.3.

θ

Amidships

Old WL BML

WL

Longitudinal metacentre

Old WL

▲ Figure A24.3 The longitudinal metacentre and the shift in B

tanθ =

Longitudinal movement in B BML

This can be transposed to give BML: BML =

Longitudinal movement o e e t in B tanθ

Combining the equations gives: BLL3 tanθ BLL3 12 InertiaL ∇ BML = = 12 = tanθ ∇ ∇

9781408176122_App24_Rev_txt_prf.indd 743

11/16/2013 2:47:52 AM

APPENDIX 25 THE DERIVATION OF THE MOMENT TO CHANGE TRIM BY 1 CM FORMULA Assume that a trimming moment is applied to the vessel, so that the trim changes. As the vessel trims, B moves, and a righting lever, GZL, is created in the longitudinal direction, as shown in Figure A25.1.

ML

GML

ZL

G

New B

▲ Figure A25.1 Trim geometry

As the centre of buoyancy moves aft, a restoring moment will be generated, which will oppose the trimming moment. The restoring moment will be given by the displacement multiplied by the restoring lever, in a similar manner to a righting moment in the transverse case: Restoring moment

Δ G GZ L

For small angles of trim (which is realistic for trim angles) the restoring lever, GZL, can be found in terms of the longitudinal metacentric height, GML: GZ L = GML × sinθ

9781408176122_App25_Rev_txt_prf.indd 744

11/16/2013 2:46:39 AM

Appendix 25 • 745 Combining these gives: Δ GML

Restoring moment

sinθ

For the vessel to be in equilibrium, the restoring moment must be equal to the trimming moment: Trimming moment o e t = Restoring esto g moment Combining these gives: Trimming moment = Δ × GML × sinθ The definition of trim is the difference in draught of the ship measured at the forward perpendicular (abbreviated to DF) and the draught measured at the aft perpendicular (abbreviated to DA). If the length between perpendiculars (the LBP) is known, the trim angle can be determined in terms of the trim and the LBP, as shown in Figure A25.2 and the equation: Vertical movement of B =

Volume moved o ed V Vertica e t cal distance move v d Total volume

FP θ

AP

LBP DA

DF

▲ Figure A25.2 Trim angle, LBP and draughts

Sinθ =

DA DF LBP

If the vessel trim as a result of the trimming moment is 1 m, then this becomes: Sinθ =

9781408176122_App25_Rev_txt_prf.indd 745

1 LBP

11/16/2013 2:46:40 AM

746 • Ship Stability, Powering and Resistance If the vessel trim as a result of the trimming moment is 1 m, then this becomes: Moment to change trim 1 m = Δ × GML × sinθ Combining these equations gives: Moment to change ttrim 1 m = Δ × GML ×

Δ × GML 1 = LBP LBP

To determine the moment to change the trim by 1 cm, this can be divided by 100: MCTC =

9781408176122_App25_Rev_txt_prf.indd 746

ΔGML 100 LBP

11/16/2013 2:46:41 AM

APPENDIX 26 THE DERIVATION OF THE TRIM EQUATION As seen in Appendix 25: The Derivation of the Moment to Change Trim by 1 cm Formula, the MCTC can be derived from the geometry of the vessel. If the loading of the vessel changes so that the centre of gravity moves forward or aft, then it will be misaligned with the longitudinal centre of buoyancy. This creates a longitudinal GZ, as shown in Figure A25.1. The misalignment of the longitudinal centre of gravity and the longitudinal centre of buoyancy create a trimming moment: Trimming moment = GZ L × Δ The GZL can be expressed in terms of the LCG and the LCB: Trimming moment = (LCB LCG ) × Δ The definition of the MCTC is the trimming moment required to change the trim by 1 cm. Dividing the trimming moment by the moment to change trim by 1 cm will give the units of the trim in centimetres:

Trim =

tm Trimming moment = ⎛ t ⎞ = cm MCTC ⎝ cm ⎠

Therefore, the trim can be found from: Trim =

9781408176122_App26_Rev_txt_prf.indd 747

LCG ) Δ MCTC

(LCB

11/16/2013 2:42:32 AM

APPENDIX 27 THE DERIVATION OF THE FORMULA LINKING CARGO DENSITY, STOWAGE FACTOR AND COMPARTMENT PERMEABILITY The stowage factor (abbreviated to SF) tells us the volume required to stow 1 tonne of cargo. This is a practical measure which includes air gaps between the items of cargo, and is used to determine the mass of cargo could be stowed in a hold of a given volume: Mass of cargo =

Hold volume SF

The density of the cargo (the units are t/m3) tells us the mass of cargo that occupies one metre cubed of volume of matter, therefore it does not include any air gaps in the cargo as a result of inefficient stowage: Actual volume occupied by cargo =

Mass of cargo Density t of carg r o

The SF and cargo density (ρcargo) can be used to determine the permeability (μ) of a compartment which is bilged, assuming all of the cargo remains in the compartment: If a hold is completely filled with cargo, subtracting the actual volume occupied by the cargo from the volume of the hold tells us the amount of the hold which is taken by air, and is effectively floodable:

9781408176122_App27_Rev_txt_prf.indd 748

11/16/2013 2:38:11 AM

Appendix 27 • 749 Floodable volume = Hold volume − Actual volume occupied by carrgo g Substituting gives: Floodable volume = Hold volume −

Mass of cargo Density t of cargo r

We have previously defined the percentage or decimal measure of the amount of a hold that could flood in the event of bilging as the permeability: Permeability t =μ=

Floodable volume Hold volume

Substituting gives: Mass o of cargo Density t of of cargo r Hold volume

Hold volume − μ=

The volume of a hold with the mass of cargo and the stowage factor are linked in the Hold volume . This can be transposed to give the volume of SF the hold in terms of the stowage factor and mass of the cargo: equation Mass of cargo =

Hold volume o u e = Mass ass of cargo SF Substituting this gives:

μ=

(Mass of

Mass of cargo Density t of cargo r Mass of carg c o SF

cargo × SF ) −

The numerator can be factorised for the mass of cargo: ⎛ ⎞ 1 Mass o of cargo ⎜ SF − ⎝ Density t of of cargo r ⎟⎠ μ= Mass o of cargo × SF Finally, the mass of cargo can be cancelled from the numerator and denominator to give the permeability of the compartment in terms of the cargo SF and density: ⎛ ⎞ 1 ⎜⎝ SF − Density t of of cargo r ⎟⎠ μ= SF

9781408176122_App27_Rev_txt_prf.indd 749

11/16/2013 2:38:12 AM

APPENDIX 28 DERIVATION OF THE FORMULA GIVING THE TRANSVERSE INERTIA OF A RECTANGULAR WATERPLANE MEASURED FROM THE LONG EDGE The inertia (sometimes also referred to as the second moment of area) is defined as the area multiplied by the distance from a datum point squared. Consider a waterplane area of length L metres and beam B metres. An infinitely thin strip of area dy thick and L long will have an area dA, which can be found from L dy, with a centre y from the edge of the vessel. B

dy

y

0 L

▲ Figure A28.1 Inertia geometry when measured from the edge

9781408176122_App28_Rev_txt_prf.indd 750

11/16/2013 2:39:46 AM

Appendix 28 • 751 Therefore, the area of the strip can be found from: dA = L d dy The inertia of the strip, relative to the edge, can be found from: Inertia d dA y 2 = Ly 2 dy The overall inertia of the waterplane can be found by integrating the waterplane between the limits of B and zero: B

Inertia L ∫ y 2 dy d 0

9781408176122_App28_Rev_txt_prf.indd 751

B

⎛ ⎡ B 3 ⎤ ⎡ 03 ⎤⎞ ⎡ y3 ⎤ ⎛ B3 ⎞ LB LB 3 L ⎢ ⎥ = L ⎜ ⎢ ⎥ − ⎢ ⎥ ⎟ = L ⎜ + 0⎟ = 3 ⎝ 3 ⎠ ⎝ ⎣ 3 ⎦ ⎣ 3 ⎦⎠ ⎣ 3 ⎦0

11/16/2013 2:39:46 AM

APPENDIX 29 DERIVATION OF THE PARALLEL AXES THEOREM HUYGENSSTEINER THEOREM Consider a waterplane area of length L metres and beam B metres, as shown in Figure A29.1. An infinitely thin strip of area dy thick and L long will have an area dA, which can be found from Ldy, with a centre y from the centreline of the vessel. L

dy y

B

InertiaGG h

X

X InertiaXX

▲ Figure A29.1 Parallel axes geometry

The inertia of the strip about the axis XX can be found:

(h + y )2 ddA (h + y )2 L dy

InertiaXX

The inertia of the shape about XX can be found by the summation of the strips: B

InertiaXX

∫ (h + y ) 0

9781408176122_App29_Rev_txt_prf.indd 752

B

d dA

∫ (h + y )

2

L dy

0

11/16/2013 2:47:55 AM

Appendix 29 • 753 B

B

2 hy + h hyy + y 2 )L d dy ∫ (h + y )((h + y ) L dy = ∫(h + hy

InertiaXX

0

0

B

InertiaXX

∫ 0

B

h + hy hy + y L dy d = ∫LLh2 + 2Lhyy LLy 2 dy 0

B

B

∫Lh ddy ∫

InertiaXX

0

0

B

Lh2 ∫dy d

InertiaXX

B

Lhydy + ∫LLy 2 dy 0

B

B

0

0

Lh∫ yydy + L∫ y 2 dy

0

B

However, as can be seen below, the final term, L∫ y 2 dy is also the same as the inertia 0

measured at the centre: B

L∫ y dy 2

0

B 2

B

0

L∫ y dy + L ∫ 2

0

−

B 2

0

⎡ y3 ⎤ 2 ⎡ − y3 ⎤ y dy = L ⎢ ⎥ + L ⎢ ⎥ ⎣ 3 ⎦0 ⎣ 3 ⎦−B 2

2

Solving for the right hand section gives: B 2

0

B ⎡ y3 ⎤ ⎡ − y3 ⎤ L∫ y 2 dy = L ⎢ ⎥ + L ⎢ ⎥ ⎣ 3 ⎦0 ⎣ 3 ⎦− B 0

2

⎛ ⎡⎛ B⎞3 ⎤ ⎞ ⎛ ⎡ ⎛ B ⎞ 3 ⎤⎞ 3 ⎢ ⎥ ⎜ ⎝ ⎠ ⎟ ⎜ ⎡ (0 ) ⎤ ⎢ − ⎝ ⎠ ⎥ ⎟ 2 ⎥ ⎡ 03 ⎤⎟ 2 ⎥⎟ ⎢ ⎜ =L − + L⎜ ⎢ ⎥−⎢ ⎜ ⎢ 3 ⎥ ⎢⎣ 3 ⎥⎦⎟ ⎜ ⎢⎣ 3 ⎥⎦ ⎢ 3 ⎥⎟ ⎥ ⎢ ⎥⎟ ⎜⎢ ⎟ ⎜ ⎝⎣ ⎠ ⎝ ⎦ ⎣ ⎦⎠

Simplifying gives: B ⎛ ⎡ (0 )3 ⎤ ⎡ −B 3 ⎤⎞ ⎛ ⎡ B 3 ⎤ ⎡ 03 ⎤⎞ L∫ y 2 dy = L ⎜ ⎢ ⎥ − ⎢ ⎥⎟ + L ⎜ ⎢ ⎥−⎢ ⎥⎟ ⎝ ⎣ 24 ⎦ ⎣ 3 ⎦⎠ 0 ⎝ ⎢⎣ 3 ⎦ ⎣ 24 ⎦⎠ B ⎛ ⎡ B 3 ⎤⎞ ⎛ ⎡ −B 3 ⎤⎞ L∫ y 2 dy = L ⎜ ⎢ ⎥⎟ + L 0 − ⎢ ⎥⎟ ⎝ ⎣ 24 ⎦⎠ ⎝ ⎣ 24 ⎦⎠ 0 B

L∫ y 2 dy = 0

B

L∫ y 2 dy = 0

⎡ −B 3 ⎤⎞ LB 3 ⎛ + ⎜ (0 × L) L ⎢ ⎥⎟ 24 ⎝ ⎣ 24 ⎦⎠

LB 3 ⎡ −LB 3 ⎤ LB 3 ⎡ LB 3 ⎤ LB 3 − = + = 24 ⎢⎣ 24 ⎥⎦ 24 ⎢⎣ 24 ⎥⎦ 12

This is the inertia of the shape about the geometric centre, therefore the equation becomes: B

InertiaXX

Lh2 ∫dy d 0

9781408176122_App29_Rev_txt_prf.indd 753

B

Lh∫ yydy + IGG 0

11/16/2013 2:47:55 AM

754 • Ship Stability, Powering and Resistance B

The term Lh2 ∫d dy can be evaluated: 0

(

B

Lh2 [ y ] − [ y ]

Lh2 ∫d dy = Lh2 [ y ]B0 0

B

0

)

Lh2 (B − 0 ) = BLh2

However, the term LB is the area of the shape, therefore: B

Lh2 ∫d dy = Area h2 0

Substituting this gives: B

InertiaXX = Area h2 + 2Lh∫ yydy d IGG 0

The middle term can be evaluated: B

2Lh∫ ydy d 0

B

2Lh∫ yydy d 0

B 2

B

0

2Lh∫ yydy + 2Lh ∫ 0

−

B 2

ydy d

0

⎡ y2 ⎤ 2 ⎡ − y2 ⎤ 2Lh ⎢ ⎥ + 2Lh ⎢ ⎥ ⎣ 2 ⎦0 ⎣ 2 ⎦−B

2

⎡⎛ ⎛ B ⎞ 2 ⎞ ⎤ ⎡ ⎛ ⎛ B⎞2 ⎞ ⎤ ⎢⎜ ⎥ ⎢ 2 ⎛ − (0 ) ⎞ ⎜ − ⎝ 2 ⎠ ⎟ ⎥ ⎝ 2 ⎠ ⎟ ⎛ 02 ⎞ ⎥ ⎢ ⎢ ⎜ ⎟ ⎟⎥ 2Lh − + 2Lh ⎜ −⎜ ⎢⎜ 2 ⎟ ⎜⎝ 2 ⎟⎠ ⎥ ⎢⎝ 2 ⎟⎠ ⎜ 2 ⎟ ⎥ ⎢⎜ ⎥ ⎢ ⎟ ⎜ ⎟⎥ ⎠ ⎝ ⎠ ⎥⎦ ⎢⎣⎝ ⎥⎦ ⎢⎣ B

2Lh∫ yydy d 0

⎡⎛ ⎛ B ⎞ 2 ⎞ ⎤ ⎡ ⎛ ⎛ B⎞2 ⎞ ⎤ ⎢⎜ ⎥ ⎢ ⎜− ⎥ ⎟ ⎝ 2⎠ ⎥ ⎝ ⎠ ⎟ ⎟ + 2Lh ⎢0 − ⎜ 2 ⎟ ⎥ 2Lh ⎢⎜ ⎢⎜ 2 ⎟ ⎥ ⎢ ⎜ 2 ⎟⎥ ⎢⎜ ⎢ ⎜ ⎟⎥ ⎟⎥ ⎠ ⎥⎦ ⎠ ⎥⎦ ⎢⎣⎝ ⎢⎣ ⎝ B

2Lh∫ yydy d = 0

LhB 2 Lh( B )2 − =0 4 4

Substituting this gives: InertiaXX = Area h2 + IGG This can be transposed to give: InertiaXX − Area h2 = IGG

9781408176122_App29_Rev_txt_prf.indd 754

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APPENDIX 30 DIRECT CALCULATION OF SHEAR FORCE AND BENDING MOMENT The process of calculating shear force and bending moments can be speeded up by expressing the shear force as a function of the position from the AP. The resulting load function can then be integrated to determine the shear force and bending moment. The load can be expressed as a function of a position along the length of the vessel, and the function integrated twice to determine the bending moment at any point along the length of the vessel. For example, consider Question 7.1. The load in each hold can be found using: Load =

Δ LOADED Δ Mass o of hold cargo O − LIGHTSHIP − L L Hold length

Load (Hold 1, 0 m FOAP to 20 m FOAP ) =

322.26 + 600 322.26 300 3 − − = −5 t/m 60 60 20

Load (Hold 2, 20 m FOAP to 40 m FOAP ) = Load (Hold o d 3, 40 m FOAP to 60 m FOAP ) =

322.26 + 600 322.26 0 − − = 10 t/m 60 60 20

322.26 + 600 322.26 300 − − = −5 t/m 60 60 20

If the position along the vessel from the AP is x metres, the load values can be expressed as a function of x: Load (0

x

20 )

5 t/m

Load (20 ≤ x < 40 ) = 10 t/m Load ( 40 ≤ x < 60 ) = −5 t/m The shear force can therefore be expressed (using a Maccaulay step function) as: SF ( x )

9781408176122_App30_Rev_txt_prf.indd 755

5 x + 5[ x − 20] + 10[ x 20] 10[ x − 40] − 5[ x − 40]

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756 • Ship Stability, Powering and Resistance This can be integrated to give the bending moment function: x

BM = ∫SSF ( x ) d dx 0

x

BM = ∫ − 5 x + 5[ x − 20] + 10[ x 20] 10[ x − 40] − 5[ x 4 ]d dx 0

BM = −

5x2 5 + [x − 2 2

10 2

10 2

]2 + [ x − ]2 − [ x

]2

5 [x − 2

]2

Shear force (tonnes), Bending moment (tonne metres)

Note that as the bending moment must be zero at x = 0, the constant of integration must also be zero. The resulting shear force and bending moment diagrams are shown in Figure A30.1. It can be seen that these agree with the original solution.

200 0

10

20

30

40

50

60

–200 –400 –600 –800 –1,000 –1,200 Shear force Bending moment

–1,400 –1,600 Position (m FOAP)

▲ Figure A30.1 Shear force and bending moment diagram

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APPENDIX 31 PROPERTIES OF WATER Fresh water

Sea water (salinity 3.5%)

Temperature, t (°C)

Density, ρ (kg m–3)

Kinematic viscosity, ν (×10–6 m2 s–1)

Density, ρ (kg m–3)

Kinematic viscosity, ν (×10–6 m2 s–1)

0

999.8

1.78667

1,028.0

1.82844

1

999.8

1.72701

1,027.9

1.76915

2

999.9

1.67040

1,027.8

1.71306

3

999.9

1.61665

1,027.8

1.65988

4

999.9

1.56557

1,027.7

1.60940

5

999.9

1.51698

1,027.6

1.56142

6

999.9

1.47070

1,027.4

1.51584

7

999.8

1.42667

1,027.3

1.47242

8

999.8

1.38471

1,027.1

1.43102

9

999.7

1.34463

1,027.1

1.39152

10

999.6

1.30641

1,026.9

1.35383

11

999.5

1.26988

1,026.7

1.31773

12

999.4

1.23495

1,026.6

1.28324

13

999.3

1.20159

1,026.3

1.25028

14

999.1

1.16964

1,026.1

1.21862

15

999.0

1.13902

1,025.9

1.18831

16

998.9

1.10966

1,025.7

1.15916

17

998.7

1.08155

1,025.4

1.13125

18

998.5

1.05456

1,025.2

1.10438

19

998.3

1.02865

1,025.0

1.07854

20

998.1

1.00374

1,024.7

1.05372

21

997.9

0.97984

1,024.4

1.02981

22

997.7

0.95682

1,024.1

1.00678

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758 • Ship Stability, Powering and Resistance

Fresh water

Sea water (salinity 3.5%)

Temperature, t (°C)

Density, ρ (kg m–3)

Kinematic viscosity, ν (×10–6 m2 s–1)

Density, ρ (kg m–3)

Kinematic viscosity, ν (×10–6 m2 s–1)

23

997.4

0.93471

1,023.8

0.98457

24

997.2

0.91340

1,023.5

0.96315

25

996.9

0.89292

1,023.2

0.94252

26

996.7

0.87313

1,022.9

0.92255

27

996.5

0.85409

1,022.6

0.90331

28

996.2

0.83572

1,022.2

0.88470

29

995.9

0.81798

1,022.0

0.86671

30

995.6

0.80091

1,021.7

0.84931

9781408176122_App31_Rev_txt_prf.indd 758

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WORKS CITED Davies, J. (2004, May). ww2ships.com. Retrieved 31 January 2010, from ww2ships.com: http:// ww2ships.com/acrobat/us-os-001-f-r00.pdf Felkins, K., Leighly, H., & Jankovic, A. (1998). The Royal Mail Ship Titanic: Did a Metallurgical Failure Cause a Night to Remember? JOM, The Minerals, Metals and Materials Society, 12–18. International Maritime Organisation (1991). International Code for the Safe Carriage of Grain in Bulk. London: IMO. — (2002). Code on Intact Stability 2002. London: IMO. — (2008). Code in Intact Stability 2008. London: IMO. International Towing Tank Conference (2006). International Towing Tank Conference. Retrieved 31 January 2010, from International Towing Tank Conference – Recommended Procedures and Guidelines: http://ittc.sname.org/2006_recomm_proc/7.5–02–07–04.3.pdf Maritime and Coastguard Agency (1998). MSN 1698 M The Merchant Shipping (Passenger Ship Construction: Ships of Classes I, II and II(A)) Regulations 1998). Southampton: Maritime and Coastguard Agency. — (2000). MSN 1752(M) The Merchant Shipping (Load Line) Regulations 1998, as Amended by the Merchant Shipping (Load Line), (Amendment) Regulations 2000. Southampton: Maritime and Coastguard Agency. — (2007). Maritime and Coastguard Agency Guidelines for Surveyors. Retrieved 31 January 2010, from Maritime and Coastguard Agency: www.mcga.gov.uk/c4mca/loli_pt6.pdf — (n.d.). Guidance for Surveyors, Loadline, Part 8. Retrieved 23 January 2010, from Maritime and Coastguard Agency: www.mcga.gov.uk/c4mca/mcga07-home/shipsandcargoes/mcgashipsregsandguidance/mcga-dqs-ss_guidance_to_surveyors/dqs-ssb_load_line/mcgagr_gos_loadline-chapter8.htm Merchant Shipping (Load Line) Regulations 1998 (n.d.). Statutory Instruments 1998 No. 2241. Wright, D. (2005). Fracture – Some Maritime Examples. Retrieved 31 January 2010, from University of Western Australia School of Mechanical Engineering: http://school.mech.uwa. edu.au/~dwright/DANotes/fracture/maritime/maritime.html

9781408176122_Ref_Rev_txt_prf.indd 759

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INDEX abandonment 265 absorption 54, 323 acceleration 3–4, 81, 383 admiralty coefficient 359–60 advance 366–70, 372, 374–5 aeration 384 aerofoil 391 aligned 104–5, 140, 207, 270, 383 allowance 3, 25–7, 354, 400, 402 apparent slip 366, 369 apparent wind 148 appendage 325, 332, 350–2, 354, 380 Archimedes 3–5, 257, 272, 396 astern 357, 363–4, 383, 391–3 asymmetric 59, 91, 118, 148–9, 172 balance 174, 319 ballast 65–6, 68–71, 73–4, 78, 83, 118–9, 168, 194, 196, 267, 307 ballasting 73–4 baseline 267, 380 beaching 380 blade area 363–4, 372, 375–6, 378–9, 381–2, 384 blades 364, 366, 371, 372–5, 380, 382–3 block 2, 7–8, 18, 33, 35, 325, 364, 367 blockage 331 BM 38, 55–8, 64–5, 76, 78, 84–5, 87, 138, 140, 146, 150, 175, 177, 187, 198, 204, 223, 239, 243–5, 250–4, 260, 269, 297–9, 301, 303, 307–12, 315–16, 412–14 BML 175–7, 251, 269, 315 boundary layer 321, 332, 336–42, 348, 372, 389 breadth 31 bulb 324 bulbous 323–25 bulbous bows 323–25 bulkhead 30–1, 74, 244, 252–6, 258–9, 262–5, 267, 276, 305–6 bunkers 66, 74, 97, 119, 168–9, 184, 196 buoyancy 2–35, 37–40, 46–7, 55, 59, 64, 77, 80, 85, 86, 89, 91–2, 94–5, 104–6, 112, 115, 120, 140, 142–4, 146, 152, 170, 175, 178, 223–4, 239–40, 248, 253–4, 256–7, 260, 270–2, 312, 402 BWL 7–9 caissons 207 canal 331 cancellation 345, 347 capsize 78, 94, 176, 209 capsizing 38, 53, 77, 86, 89, 157–8, 170, 264 catamaran 309 cavitates 371, 375, 377, 379, 382–3 centreline 39, 42–3, 46–9, 55–6, 59, 62–5, 71, 74, 79, 104, 106, 112, 117–19, 122–3, 128, 139, 149, 159, 201–2, 242, 244–5, 248–9, 257, 260, 292, 297, 306–11, 392, 402–4 centrifugal 80 centripetal 80 centroid 245–7, 249–50, 280, 282, 286–7, 307–8, 313, 315

9781408176122_Index_Rev_txt_prf.indd 760

CFD 340, 369 characteristic 344–5 charts 186, 350, 372–4 collision 237, 257 correlation 319, 352, 354, 358 counter-ballasted 118 counter-clockwise 156–7 crane 52, 112, 201 criteria 53, 129–30, 148, 163 criterion 261 cross-section 67, 146, 271, 293 CSA (cross section area) 386 damage 33–4, 54, 222–69, 271, 375, 379 datum 261 davit 263 de-ballasted 75 deadweight 2, 6, 14–15 deceleration 383 deck cargo 29, 53–4, 62, 73, 129, 143–5 deck edge 78, 90, 119, 155, 266 deck edge immersion 90, 119 deck structure 31, 34 decks 31, 147, 264–5 deduction factor 370–1, 387, 389–90 delta 373 density 2–5, 21, 23–5, 27–8, 35–6, 66, 72–6, 84, 97, 119, 168, 184–93, 196, 200, 210, 212–13, 216–17, 219–20, 228, 293, 335, 338, 340, 386 derrick 44–5, 52, 112 deterministic 262–3, 266–7 diameter 81, 83, 364–5, 371–4, 378–80, 386, 388, 390, 395 discharge 28, 385 displacement 2, 4–6, 10–18, 22, 25–8, 35, 40–1, 43–6, 52–3, 61, 63–4, 66, 73–6, 79, 85, 89–90, 95–7, 101–4, 111, 115–16, 119–21, 124, 130, 139, 144–5, 147–8, 150, 155–7, 162, 166, 168–70, 178, 180–204, 210, 212–21, 225, 253, 257, 271, 276, 279, 293–4, 297, 301, 303, 306, 311–12, 316–18, 328, 333–5, 359–61, 397–8, 401, 408, 410–11, 413 divergent 323, 343 doors 30–1, 264 draft 189, 200–1, 206, 275, 378–80 drag 311, 319–20, 322, 325, 328, 336, 338–42, 347–52, 354, 379–80, 391–93, 395 draught 2–23, 26–30, 35–6, 39–40, 52, 55, 57–8, 61, 76, 79, 81–5, 88, 112, 139, 145, 155, 161, 174, 178–228, 236, 238–9, 241–4, 248, 250–9, 267, 269, 271, 279, 292–303, 309–13, 315–6, 318, 397–8, 401–2, 406–9, 413 dredgers 21 driveline 381, 383 drives 378, 380–1, 384 dry cargo 32, 266–8 dry-dock 181, 186, 206–13, 221 DSummer (summer draught) 399, 401 duct 386–9 DWA (dock water allowance) 3, 27, 36, 400–2

12/3/2013 5:22:43 PM

Index • 761 DWT (deadweight) 6 dynamic resistance 307 dynamic stability 90, 128, 159, 171–2, 281, 288 dynamic viscosity 338 dynamometer 331 economic 18, 213, 329 eddies 320, 325, 350 effective power 355–6, 387 efficiency 355–7, 375, 386–7 emerged 82, 105, 138, 155 energy 125–6, 160, 163, 323, 386–90 engine 101, 265, 353–6, 360, 378, 381, 384–5, 387, 390 entrained 366–7, 379 equilibrium 3–4, 64, 77, 104, 122, 161–2, 224, 263, 266, 276, 319, 337, 389 equivalent drag 348 equivalent propeller 386 equivalent thrust 390 equivalent torque 394–5 equivalent volume 165 exemptions 21 exhaust 101, 325, 350, 384 exposed 18, 31–2, 147, 161–2, 380 failure 271 ferries 155, 329 fixed systems 381 fixed trim 153–4 floodable 262 flotation 2–35, 313 flow 321, 336–7, 339, 341, 367, 391–2 fluid 47, 65–71, 152, 320 force 3–5, 39–40, 46–7, 49, 54, 77, 79–80, 91–2, 94, 112, 115, 123, 126, 161–3, 175, 178, 207–8, 224, 257, 265, 270–7, 319, 321–2, 339, 357, 380, 386, 391–5 forecastle 29 forepeak 66, 228 form drag 320, 322, 325 form factor 340–3, 348–9, 351 freeboard 18, 20–1, 26, 29–35, 91, 139, 143–4, 147–9, 153, 155, 172, 260–1, 265 freeboard deck 18, 20, 31, 33–4, 139, 155, 261 freeze 147 fresh water allowance 3, 25–7, 36, 400–2 fresh water berth 189 fresh water mark 25 friction 319–27, 330, 333–4, 336, 338–9, 340–2, 348–9, 350, 352, 378, 380, 387–8 friction calculation 340 friction coefficient 327 friction drag 338, 341, 350 friction line 339, 342, 348, 350 Froude numbers 325, 328–30, 336, 344, 346–7, 349, 351, 359–60 Froude, Robert Edmund 325 FSC 39, 72–4, 88 FSE 38–9, 67, 72–5, 88, 91, 101, 118, 151, 172, 198 FSM 38, 65, 72–6, 88, 151–2 fuel coefficient 360 fuel consumption 196, 311, 353, 360–1 fuel oil 196

9781408176122_Index_Rev_txt_prf.indd 761

fuel required 360 fuel saved 361 gangways 207 gases 350 gasketed 32 gaskets 31 gates 207 gearbox 356, 374, 378, 383, 385 geometrically similar 330, 333–5, 353, 361 GM (metacentric height) 38–9, 48–64, 67, 71–91, 97–102, 107, 112–18, 129, 131, 137–50, 160, 165, 168, 170, 172, 175–79, 191, 197–8, 200, 202–3, 207, 209–12, 223, 239, 243, 250, 253–4, 256–7, 260, 264, 266, 269, 279, 299, 301, 307–11, 318, 407–16 GML (longitudinal metacentric height) 175–6, 223, 269 gradient 118–20, 143–4, 146, 148–9, 151, 202, 276 grain 53, 91, 101, 129, 164–9, 172–3, 332 gravitational acceleration 3–4, 37–8, 40–1, 44, 46–7, 49, 59, 64–5, 67, 71–2, 76–7, 80–1, 86–8, 91–2, 94, 101, 104–6, 111–2, 118–20, 140–2, 147, 149, 175, 197, 200, 224, 254, 256, 260, 270–1 ground 207–8 ground reaction force 207–8 grounding 186, 206–7, 221 GZ approximation 98, 137, 159 GZ change 91, 94, 101, 103, 172 GZ convention 157 GZ curve 89–91, 94–5, 97–9, 101–2, 106–7, 112, 115–21, 124–6, 129–32, 137, 139, 143–6, 148, 150–7, 159–60, 163–4, 166–73, 223, 260, 263, 266, 269 half beams 313 half station 278, 288–9, 317 handed propeller 364 hatch 32, 198 heel angle 33, 49, 67, 80–2, 91, 94, 97, 101, 119–20, 123, 162–3, 168, 172, 200, 206, 263–4 heeling angles 264 heeling arm 90, 166–7, 171 heeling moment 62, 80, 90–1, 122–3, 142, 161–9, 171, 173, 198, 202–3, 263 heeling moments 168 heeling force 123 heeling lever 162 high-speed 347, 379–83, 385, 387, 389–90 hog 194–5, 213, 219–20, 271 hollow 329–30, 347 hollows 329, 346–7 hull centreline 308–9 curvature 343 designs 384 efficiency 356–7 form 7, 9, 48, 51, 137, 150, 279, 291, 344, 351, 359, 367 length 344 penetration 267 plating 306 resistance 350, 389 sheer 18 speed 328–9 hydrodynamic drag 311, 391, 393

12/3/2013 5:22:56 PM

762 • Index hydrodynamic resistance 307 hydrostatic curves 16, 95 hydrostatic data 11, 15–16, 29, 37, 38, 51, 72, 86–7, 95, 101, 174, 180, 184–6, 188, 220, 279, 305, 411 hydrostatic parameters 18 hydrostatic pressure 379 hydrostatic tables 10, 16, 22, 24, 180, 182, 187–91 hydrostatic value 11, 22, 178, 185–8, 191, 204, 220–1, 279 ice accretion 91, 147–9, 172 immersed area 294, 300–2 immersed section 293–4, 313, 315–16 inclining mass 201–4 inclining test 186–204, 221 inclinometer 200 inertia 38, 51, 56–7, 72, 75–6, 84, 87, 150, 177, 222–3, 244–50, 252–3, 260, 269, 278–9, 284–8, 290–1, 297, 305, 307–11, 315, 317–18, 412 inflection 120 instability 77 interaction 344–5, 389 interpolate 11–12, 14, 17, 29, 52, 76, 83, 178, 183, 186, 191, 200, 212, 216–17, 268, 402, 408 intersection 17–18, 71, 94–5, 99–100, 112, 118, 121–2, 124, 162–3, 168, 175, 373 iterative 371, 374 ITTC 319, 332, 339–40, 342, 348, 350, 352 jet 377, 384–90 keel clearance 82–3, 189, 196 keel emergence 82, 105, 138 keel laying 266 Kelvin wave pattern 323 kinematic viscosity 338, 342–3 kinetic energy 387, 389–90 KM 15–17, 51–2, 55, 57–8, 65, 76, 78, 175, 187–8, 198, 204, 209–13, 306, 407–11, 415–16 knot 81, 83, 161–2, 319, 321, 326, 328, 342, 359–61, 382, 385, 388, 390, 393–5 KQ 374 KWh 360 laminar flow 339 laminar sublayer 337 lateral area 147, 161–2 layer correction 185, 190–2, 194, 197, 213, 220–1 LBP 178, 183, 190–1, 214, 218, 254, 304, 313, 315–16, 332, 345–7 LCB 15–17, 37, 39–40, 85, 175, 178–80, 187, 192, 200, 206, 222, 241–2, 254, 268, 279, 308, 312–13, 316, 318, 403 LCF 10–11, 15–17, 174, 181–4, 186–7, 190–1, 206, 208, 210, 214, 216–18, 244, 252–3, 279, 313–16, 318 LCG 37, 40–3, 65–6, 72, 86, 175, 178–9, 180, 184–5, 192, 197–8, 200, 213, 221, 254, 317, 403–4 leading edge 338, 340, 363, 381–2 leeway 391 length factor 262 lifeboat 265, 267, 293–6, 298, 304 lightship condition 6, 76, 79

9781408176122_Index_Rev_txt_prf.indd 762

lightship displacement 200 lightship KG 221 linear interpolation 11–12, 14, 29, 52, 200, 216, 268 linear thrust 363 list angle 38, 59, 62–4, 71, 80, 87, 90–1, 111, 121, 123–4, 159, 168–9, 171, 173, 197, 200 listing arm 124–5 listing moment 59, 61, 67, 122–4, 178, 202 lists 1, 41, 67, 122, 127, 165, 280, 283–5 load diagram 273, 275–6 load displacement 6, 150 load draught 313, 315–16 load line 18, 25, 27, 36, 53 load waterline 163 loading table 40–3, 45, 51–2, 61–2, 72–3, 75–6, 85, 95, 147, 165, 180, 197–8, 209, 212, 239, 408–10, 412–15 longitudinal bulkhead 74, 259, 265, 305–6 longitudinal inertia 177, 250, 252, 279, 315, 318 longitudinal metacentre 175–6 longitudinal metacentric height 174, 178–9, 184 longitudinal stability 175, 177–9, 181–3, 253 longitudinal waterplane 177, 223, 250, 252, 269 lumber 3, 29–30, 36, 53 lumber load lines 3, 29–30, 36 LWL 7–8 manoeuvre 25, 196, 264, 379–80 margin line 261–3 Maritime and Coastguard Agency 1, 53–4, 94, 129, 147, 163, 198, 261 MCTC 15–17, 174, 178–81, 184, 187, 192, 205–6, 208, 210, 214, 217–18, 250, 254–6 mean draughts 10–11, 79, 112, 174, 180, 182–6, 189–94, 197, 201, 204–5, 210, 212–13, 216–17, 220–1, 228, 243 mean GM 198 mechanical efficiency 355–6 mechanical energy 386 mechanical power 388 metacentre 38, 47–50, 59, 86, 91, 97–8, 107, 141–2, 175–6, 284 metacentric diagram 58 metacentric height 33, 38–9, 49, 52–4, 58, 61–3, 67, 72, 74, 83, 86–8, 90, 129, 139, 142–3, 147, 164, 170, 174–5, 178–9, 184, 186, 210, 213, 221, 257–9, 263, 279, 317–18, 413 metacentric stability 37–87, 91 model resistance 333, 354 model scale 332 model ship 351 model speed 333 model test 319, 333, 348, 352 mono-hull 307–8 Morrish’s method 317 multihull 279, 307–8, 318 naked power 353–4 negative GM 77, 100, 112–14, 160 negative GZ 89, 94, 149, 157–60, 260 negative metacentric height 87, 143

12/3/2013 5:22:58 PM

Index • 763 negative righting moment 157 neutral axis 245 neutral stability 64, 115 non-dimensional ratio 364 non-dimensionalised 339, 373 nonlinear 344 nozzle 385–9 observed trim 214, 218 opening 18, 31, 33, 165, 325, 385 outboard 71, 77, 105, 143, 146, 378 outdrives 378 overhang 149, 152, 204, 383–4 parallel axes 245–6, 249, 252–3, 278, 286, 305, 307–8, 310–11, 315, 317 parallel sinkage 223–38, 260–1, 406 parametric roll 91, 172 parasitic drag 379–80 passenger vessels 21, 54, 64, 81, 200, 223, 261, 263–9, 325 passive 384 peak GZ 129, 166, 263, 267 peak tanks 66 pendulum 197–8, 200–4 permeability 32, 222–3, 227–8, 232, 234–5, 237–44, 249– 50, 252–3, 255–60, 262, 264, 266–9 permissible length 262 pivot point 10, 123, 181, 252 planform 306, 311–12 planing 384 plank 325–6 plating 31, 34, 74, 306, 336, 338–42, 348, 383, 385 plimsoll mark 31 pressed 66–7, 75, 97, 118–19, 168–9, 196, 200 pressure 5, 161–2, 263, 267, 322, 340–1, 344, 357, 364, 366, 375–6, 379, 381, 383, 393 probabilistic methods 264 Prohaska method 349 propeller blade 363–4, 366, 371, 378–80 damage 379 data 373 design 369, 372, 378–9 diameter 364–5, 371, 378–80 disc 365 efficiency 356–7 induced side force 380 pitch 364 section 357 slip 379 submergence 378, 384 thrust 369 tips 371 torque 369 types 381 quasi-propulsive coefficient 358 railing 30, 32, 363 rake 363, 382

9781408176122_Index_Rev_txt_prf.indd 763

rectangular compartments 39, 88 rectangular sponsons 310 rectangular tank 72, 76, 279, 305, 318 rectangular waterplane 56, 247–9, 308 residual area 166–9 residual resistance 322–3, 333–4, 343 residual stability 33, 91, 173 Reynolds number 338–40, 351 Ro-Ro 67, 264, 309 roughness 337 rudder 39, 325, 332, 336, 342–3, 378, 380–1, 385, 391–5 sagging 194–5, 213, 271, 273 screw propeller/s 362–76, 377 shafts 325, 378, 380–1, 385 shear force 270–1, 273–7 sheer 18, 34 Simpson’s Rule 126–7, 130–3, 168, 279, 288, 293, 295, 305, 311, 317 single-screw 372, 376, 380 sinkage 2–3, 22, 27–9, 33, 36, 83–4, 222–38, 260–1, 268, 399–402, 406 sister ship 200 skeg 380 skew 363 slack 75, 200 slamming 264 slip 366, 368–9, 376, 379, 384 slipstream 383, 386 sloshing 67 slow-turning propellers 378 SOLAS 148, 200, 266 sponson 279, 309–12, 318 squat 83, 328, 331 stability arm 94 stability computer 200 stability criteria 53, 130 stability data 10–11, 41, 52, 65, 130, 168, 186, 403 statical stability 89–90, 94, 120, 170–1 stimulators 332 Stockholm agreement 264–5 streamline 311, 322, 325 strength 31–2 stress 271, 391, 394–5 subdivision 30–2, 39, 74, 88, 200, 262–3, 265, 267 sublayer 337 submergence 105, 378, 380, 382, 384 suction 379 summer displacement 26–8, 139, 148, 189, 194, 301, 303 summer draught 20, 21, 23, 26–8, 30, 35, 145, 189, 193, 294, 296, 300–3, 402 summer Load Line 21, 23, 28, 31–2, 35, 402 summer mark 20–1, 25–6, 28, 30, 189, 399, 401 summer waterline 6, 39, 294, 299, 301 summer zone 23, 26, 28–9, 189 supercavitating 379 superposition 90, 122–3, 125, 163, 171, 323, 344–7 superstructure 31, 33–4, 54, 147–8, 350 superventilating 379 survey 186, 200, 213–14, 217–21, 329

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764 • Index survival 261, 263–7 swimming pool 72, 79 symmetrical ice 91, 147–9, 172 synchronous 91, 172

underwater volume 2, 5, 7, 8, 10, 35, 39, 46, 51, 55–7, 76–7, 84, 175, 186–7, 224–6, 239, 254, 256, 293–6, 298–9, 301, 303, 308–10, 313, 316, 396, 402 up-thrust 207–12

tank 30, 65–7, 71–2, 74–6, 83, 118, 139, 151–2, 165, 168, 200, 260–1, 279, 305–7, 318, 325–6, 331–4, 348, 351 Taylor wake fraction 366–7, 369–71, 374 TCB 37, 39–40, 85, 222, 242–3, 268, 403 TCG 37, 40, 42–3, 59–66, 71–2, 74, 86, 90, 104, 106–11, 122–5, 159, 170–1, 197–8, 403–4, 414–16 temperature 375 tender 54 theoretical speed 365–6, 368–9 thickness 321, 336, 338, 340–3 thrust blocks 356 thrust coefficient 373–4 thrust deduction factor 370–1, 387–90 thrust power 355–7, 369–70, 374, 387 thrust pressures 379 timber deck cargo 29, 53–4, 129, 143–5 timber load lines 29, 53 tips 371, 380 tolerance 332, 380, 383 tonnage 2, 6, 35 tonnes per centimetre immersion (Also see TPC) 22 torque 47, 49, 54, 59, 92, 120, 126, 363, 369, 373–4, 384, 393–5 TPC 2–3, 15–17, 22–9, 36, 83, 187–9, 204–6, 214, 216, 218, 279, 303, 318, 399–402 transition 338–9, 384 transmission efficiency 356, 387 transom 377, 383–5, 389 transverse axis 315 transverse bulkhead 258–9, 265, 276, 305–6 transverse GM 176 transverse inertia 38, 51, 56–7, 72, 75, 87, 223, 245, 247–9, 269, 279, 297, 307–11, 318 transverse stability 186, 206–7, 221, 253, 257 transverse waterplane 150, 222, 247, 260, 269 transverse wave 323, 343–4 trim 10–11, 82–3, 91, 149–54, 172, 174, 178, 179, 180–86, 190–3, 196, 200, 204–19, 221, 223, 228, 241, 250–6, 264, 267, 269–70, 317, 378, 380–1 tropical and winter lines 21 tropical fresh water allowance 20, 25, 30 tropical to summer 399, 401 tropical zone 21, 23, 25 trough 328, 344 turbulence 332, 337–9, 371–2, 384 twin-screw 376

vacuum 379 vanishing stability 90, 94, 101–2, 106, 119, 125, 144, 146, 148–9, 153, 170 vapour 375 ventilation 350, 379 vertical damage 267 vertical forces 273 vertical side 138 vibration 325, 371, 379, 383 virtual centre of gravity 39, 71, 88 viscosity 321–2, 338, 340–3, 348–50 volumetric heeling moment 165–6 voyage fuel consumption 196, 353, 361

ullage 65, 74, 139, 165 under-keel clearance 82–3, 189, 196 underwater noise 379 underwater section 294, 299, 301 underwater surface 39, 340

9781408176122_Index_Rev_txt_prf.indd 764

wake 322, 349, 366–7, 369–72, 374, 384 wake fraction 366–7, 369–71, 374 wake speed 367, 369 wake turbulence 384 wall-sided 78, 91, 138–41, 159, 172, 204 water-tight 237 waterjet 319, 377, 380, 383–6, 388–90 waterline 6–9, 18, 21–4, 27–8, 33, 39, 83, 123, 163, 189, 219, 222, 236, 241–2, 265, 268, 270, 279, 292–4, 297–9, 301–4, 309–10, 312–18, 344, 377, 384, 402 waterplane 2, 8, 9, 23–4, 28, 35, 38, 51, 56, 83–5, 87, 149–50, 177, 181, 204, 222–3, 226–36, 238, 242–5, 247–53, 260, 269, 279, 292, 295–7, 299–305, 307–11, 313–16, 318, 397 area 35, 177, 181, 204, 295–6, 300–2, 308, 310 centre 245, 248 curve 295 half beam 299, 301–2, 304–5, 315–16 inertia 56, 84, 150, 177, 222–3, 247, 250, 252, 260, 269, 311 watertight 18, 30–1, 33, 78, 101, 222, 225, 228, 236, 241–2, 244, 251–9, 262, 264–5, 267–8, 309 bulkheads 262, 267 compartment 31, 252–3, 265 decks 31 division 264 flats 241 volume 33, 309 wave interference 344–5 wave making resistance 322–3, 328, 343–4, 346–50, 354 wave pattern 323, 330, 343–7 wave speed 344, 346 wave system 323, 328–9, 344–5 wavelength 328, 330, 345 weather-tight 30–1, 129 windage 163, 264 yachts 325, 329

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