Kinematic Synthesis of Mechanisms. Using Excel and Geogebra 9783031309540, 9783031309557

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Mechanisms and Machine Science Volume 131

Series Editor Marco Ceccarelli , Department of Industrial Engineering, University of Rome Tor Vergata, Roma, Italy Advisory Editors Sunil K. Agrawal, Department of Mechanical Engineering, Columbia University, New York, NY, USA Burkhard Corves, RWTH Aachen University, Aachen, Germany Victor Glazunov, Mechanical Engineering Research Institute, Moscow, Russia Alfonso Hernández, University of the Basque Country, Bilbao, Spain Tian Huang, Tianjin University, Tianjin, China Juan Carlos Jauregui Correa , Universidad Autonoma de Queretaro, Queretaro, Mexico Yukio Takeda, Tokyo Institute of Technology, Tokyo, Japan

This book series establishes a well-defined forum for monographs, edited Books, and proceedings on mechanical engineering with particular emphasis on MMS (Mechanism and Machine Science). The final goal is the publication of research that shows the development of mechanical engineering and particularly MMS in all technical aspects, even in very recent assessments. Published works share an approach by which technical details and formulation are discussed, and discuss modern formalisms with the aim to circulate research and technical achievements for use in professional, research, academic, and teaching activities. This technical approach is an essential characteristic of the series. By discussing technical details and formulations in terms of modern formalisms, the possibility is created not only to show technical developments but also to explain achievements for technical teaching and research activity today and for the future. The book series is intended to collect technical views on developments of the broad field of MMS in a unique frame that can be seen in its totality as an Encyclopaedia of MMS but with the additional purpose of archiving and teaching MMS achievements. Therefore, the book series will be of use not only for researchers and teachers in Mechanical Engineering but also for professionals and students for their formation and future work. The series is promoted under the auspices of International Federation for the Promotion of Mechanism and Machine Science (IFToMM). Prospective authors and editors can contact Mr. Pierpaolo Riva (publishing editor, Springer) at: [email protected] Indexed by SCOPUS and Google Scholar.

Eres Söylemez

Kinematic Synthesis of Mechanisms Using Excel® and Geogebra

Eres Söylemez Orta Do˘gu Teknik Üniversitesi Ankara, Türkiye

ISSN 2211-0984 ISSN 2211-0992 (electronic) Mechanisms and Machine Science ISBN 978-3-031-30954-0 ISBN 978-3-031-30955-7 (eBook) https://doi.org/10.1007/978-3-031-30955-7 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

In memory of Professor Ferdinand Freudenstein “f 2 ”

Preface

Historically, drawing has been the main source of understanding mechanical systems. Starting with Burmester, for the kinematic synthesis of planar mechanisms graphical techniques for position, path generation with prescribed timing and the correlation of crank angles have been used and developed for industrial applications. Although graphical methods give a big insight about the problem, solving these problems on the drawing board using basic drawing tools was time consuming and usually inaccurate. Alongside the advance of computers, new and more efficient methods have been sought in the synthesis of mechanisms and analytical methods suitable for computer implementation were developed. Some of these efforts led to commercial programs or tools in widely used commercial programs. The aim of this book is to explain the basics involved in kinematic synthesis, discuss the old and the new approaches for mechanism design and to help the reader to implement the theories involved in two readily available tools: Geogebra and Excel® . Graphical methods are very simple to implement in Geogebra, and the solutions obtained are as accurate as any other commercially available CAD program. When the difficulties faced during drawing board applications are eliminated, classical graphical methods can be reintroduced in the synthesis of mechanisms course. New developments in analytical synthesis are as important as the classical geometrical methods. Analytical methods can be effectively implemented using small function routines written in Excel® . Otherwise, the instructor has to use an expansive commercial program so that students can solve problems on synthesis. In teaching, this approach makes the user to be a software operator. Cost-effectiveness is also another important factor especially for students or engineers working in small companies who cannot afford special commercial programs for mechanism design. This is more critical especially in the developing countries. In this book, all examples are either solved in Geogebra or in Excel® . Users may write their own programs or they can download the VBA files used in the examples from http://makted.org.tr/kinematic-synthesis-of-mechanisms. One has to know and understand each of the function subroutines used and, whenever necessary, must be

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able to write their own subroutine for mechanism design. This approach also helps the students to use the same approach in other fields of engineering. In Chap. 1, brief history on how the machine science has developed throughout history is discussed from the viewpoint of mechanism design. Chapters 2–5 explain in steps two, three, four and five position synthesis of mechanisms in detail, respectively. Geometrical and analytical methods of guiding a rigid body between the given positions, path generation with prescribed timing and correlation of crank angles are covered. Analytical methods are explained using complex numbers. Using relative motion concept, the position synthesis of six link mechanisms is also taken into account. In Chap. 6 Roberts–Chebyshev theorem is stated and proved. Use of the theorem in practice is shown. Converting a continuous rotary motion into an oscillating or reciprocating motion has been the main task starting with the windmill or water wheel. Design of such mechanisms is explained in Chap. 7. Chapter 8 is on the analytical methods developed for the correlation of crank angles and function generation. Freudenstein’s equation for three, four and five precision points and least squares method for function generation using Freudenstein’s equation are explained. Chapter 9 is on the developments made in Russian school on mechanism synthesis starting with Chebyshev. Although little known in the West, their remarkable work on function generation is noteworthy. Other topics such as infinitesimally close position synthesis or multiply separated position synthesis, curvature theory and approximate position synthesis are not included in this text. Optimization is an important topic in every field of engineering and is usually treated as a separate course or as a part of numerical methods course. In Chap. 10, application of optimization in mechanism design is shown with examples. No attempt is made in the explanation of the methods. Solver tool as an add-in in Excel® is used. This tool provides a simple fast and easy-to-use platform for the optimization of mechanisms. This book assumes that the reader has already taken a basic course about mechanisms. The reader must be familiar with mechanism analysis using both graphical and analytical methods. Throughout the book, several examples are provided to make the reader understand how the methods explained can be applied in practice. Excel® and Geogebra files for the selected examples for each chapter can be downloaded from the website http://makted.org.tr/kinematic-synthesis-of-mechanisms. This book can be used as a textbook for a senior or graduate course or as a selflearning text for practicing engineers who are involved in mechanical design. Main aim of this text is to familiarize the reader with the kinematic synthesis tools and methods to be used in mechanism design. This book is dedicated to Prof. Ferdinand Freudenstein, Father of Modern Kinematics, to whom the author is indebted on what he has learned from him.

Preface

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The author is grateful to Prof. Gökhan Kiper of ˙Izmir Institute of Technology for proofreading and giving valuable comments about everything in this book. He has given his long time in correcting the mistakes author has made. Ankara, Turkey January 2023

Eres Söylemez

Contents

1

Introductıon History of Mechanism Design . . . . . . . . . . . . . . . . . . . . . . 1.1 Early Ages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Archimedes, Levers and Gears and Screws (300 BC–) . . . . . . . . . 1.3 Al–Jazari: Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Renaissance Period, Leonardo Da Vinci, Printing Press (1400–) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Early Industrial Developments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Industrial Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Birth of Machine Theory: Franz Realeaux . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 7 9 12 18 26 30

2

Two Positions of a Moving Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.1 Two Finitely Separated Positions of a Plane . . . . . . . . . . . . . . . . . . 31 2.2 Graphical Synthesis Methods for Two Positions . . . . . . . . . . . . . . 40 2.2.1 Performing Two Position Synthesis Using Solidworks® . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.2.2 Performing Two Position Synthesis Using Geogebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.3 Analytical Synthesis Methods for Two Positions . . . . . . . . . . . . . . 50 2.4 Numerical Solution of Analytical Methods Using Excel® . . . . . . 59 2.5 Principle of Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 2.6 Inverse Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 2.7 Relative Motion Between Two Moving Planes . . . . . . . . . . . . . . . . 71 2.8 Correlation of Crank Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 2.9 Correlation of Slider Displacement with Crank Rotation . . . . . . . 82 2.10 Design of Six Link Mechanisms for Two Positions . . . . . . . . . . . . 84 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

3

Three Positions of a Moving Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 3.1 Three Finitely Separated Positions of a Plane . . . . . . . . . . . . . . . . . 113 3.2 The Pole Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

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3.3

Graphical Methods for Three Position Synthesis . . . . . . . . . . . . . . 3.3.1 Performing Three Position Synthesis Using Solidworks® . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Performing Three Position Synthesis Using GeoGebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Analytical Synthesis Methods for Three Positions . . . . . . . . . . . . . 3.4.1 Direct Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Dyad Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Three Homologous Points on a Straight Line . . . . . . . . . . . . . . . . . 3.6 Principle of Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Relative Motion Between Two Moving Planes . . . . . . . . . . . . . . . . 3.8 Correlation of Crank Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.1 Geometrical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.2 Analytical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Path Generation with Prescribed Timing . . . . . . . . . . . . . . . . . . . . . 3.10 Design of Six Link Mechanisms for Three Positions . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Four Positions of a Moving Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Four Finitely Separated Positions of a Plane: Burmester Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Analytical Method for Four-Position Synthesis- Dyad Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Path Generation with Prescribed Timing . . . . . . . . . . . . . . . . . . . . . 4.4 Correlation of Crank Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Four Homologous Points on a Straight Line: Ball’s Point . . . . . . . 4.6 Four Position Synthesis Using Excel® . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Five Positions of a Moving Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Five Finitely Separated Positions of a Plane . . . . . . . . . . . . . . . . . . 5.2 Graphical Synthesis Method for Five Positions . . . . . . . . . . . . . . . 5.3 Analytical Synthesis Method for Five Positions . . . . . . . . . . . . . . . 5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

243 243 244 247 258 264

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Four-Bar and Slider-Crank Cognates . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Roberts-Chebychev Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Slider-Crank Mechanism Cognate . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Design of a Six-Link Mechanism with a Link in Rectilinear Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

265 265 272

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122 125 128 128 130 144 152 154 156 156 158 161 170 185

187 201 219 223 227 231 242

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Contents

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Four-Link Crank Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Type of Four-Link Crank Mechanisms . . . . . . . . . . . . . . . . . . . . . . . 7.2 Crank-Rocker Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Dead Centers of Crank-Rocker Mechanism . . . . . . . . . . . 7.2.2 Transmission Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3 The Classical Transmission Angle Problem . . . . . . . . . . . 7.3 Slider-Crank Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Quick Return Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Scotch-Yoke Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Continuous Rotation of the Output . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Design of Drag-Link Mechanisms with Optimum Transmission Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.2 Inverted Slider-Crank Mechanism . . . . . . . . . . . . . . . . . . . 7.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Correlation of Input and Output Motion; Function Generation; Freudenstein’s Equation in Mechanism Design . . . . . . . 8.1 Derivation of Freudenstein’s Equation . . . . . . . . . . . . . . . . . . . . . . . 8.2 Function Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Three Precision Point Function Approximation . . . . . . . . 8.2.2 Four or Five Precission Point Function Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.3 Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Order and Mixed Order Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Correlation of Two Crank Angles with Optimum Transmission Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Designing Six Link Mechanisms for Long Oscillation or for Long Dwell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chebyshev Methods in Kinematic Synthesis: Russian School . . . . . . 9.1 Chebyshev Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Chebyshev Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Chebyshev Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Application of Chebyshev Theorem to Straight-Line Motion Generation: Straight-Line Motion Generation by a Swinging Block Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Application of Chebyshev Theorem to Straight-Line Motion: Straight-Line Motion Generation by a Slider-Crank Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Application of Chebyshev Theorem to Obtain Constant Velocity for a Four-Bar Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Extension of Chebyshev Theorem to Function Generation . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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287 287 288 289 291 294 309 317 320 321 321 327 329 331 333 333 337 337 345 358 363 366 371 377 399 401 401 402 403

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10 Optimization Methods in the Synthesis of Mechanisms Using Excel® . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Lifting a Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Function Generation Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Path Generation Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.1 Ackermann Steering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

451 451 453 463 465 479 492

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495

Chapter 1

Introductıon History of Mechanism Design

“The farther backward you can look, the farther forward you can see.” Sir. W.Churchill

Abstract In this Chapter a brief history on how the machine science has developed through out history will be discussed from the viewpoint of mechanism design. Starting with a lever and a wheel from the early ages, human kind tried to improve their living by discovering new mechanical systems. After the invention of the steam engine, what was initially the problem of the blacksmith, mechanics or technicians at that time, design and construction of mechanical devices caught the attention of the academicians of the time and what we now call “Machine Science” was born. Keywords History of science · History of kinematics · History of machines

1.1 Early Ages It seems that the need for a mechanical motion has started in the very early stages of human life. Humans used their muscle to move things, to hunt for food and to build shelter. When muscle power was not enough, they first used stone and wood, and later any metal available at that age [1]1 . One of the oldest inventions is assumed to be the “potters wheel”. Historians think that the rotary motion was used to make fire, to drill a hole, grind wheat, etc., before the use of motion as a wheel in transportation (Fig. 1.12 ,3 ,4 ). Archeological remains also show that the civilization started in the region called the “green crescent” or “fertile crescent” (Fig. 1.2). 1

History of mechanisms is a broad topic. For those interested, the book series “History of Machines and Machine Science”, by Springer under the Editorship of Marco Ceccarelli is an important resource. This is a brief review of the events from mechanism design viewpoint. 2 http://ceramicstoday.glazy.org/articles/potters_wheel2.html. 3 https://www.pinterest.com/pin/758082549753815922/. 4 https://www.dekopasaj.com/kagni-tekeri-22. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_1

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1 Introductıon History of Mechanism Design

(2)

(3)

(4)

Fig. 1.1 Rotary motion as potters wheel, wheat grinding and Wheel of an ox driven cart

Fig. 1.2 Fertile crescent5

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https://commons.wikimedia.org/wiki/File:Fertile_crescent_Neolithic_B_circa_7500_BC.jpg.

1.2 Archimedes, Levers and Gears and Screws (300 BC–)

3

In later years one of the biggest problems that was faced by humans was to move huge blocks of stone as early as in the tenth century BC. There is no clear explanation on how the stones of Göbeklitepe were brought and erected. They must have used some kind of levers. We know that in the ancient Egypt around fourth century BC, they have moved stone blocks on logs and used ropes and pulleys. They also used triangular pieces of stone or wood (wedge) to separate or move stones. Since we do not have any written document we can only make prediction. However in the early ages we know that they were able to erect big T shaped stone pillars and were able to make use of hand tools.6

1.2 Archimedes, Levers and Gears and Screws (300 BC–) It was Archimedes (287–212 BC), who amongst other basic scientific findings, has defined a lever and pulley system to lift heavy objects [2, 3]. His statement, “Give me a place to stand on, and I will move the Earth” is well known throughout the ages (Fig. 1.3). The importance of Archimedes is that his work was documented in writing. In his book “The Method of Mechanical Theorems”,7 lever is decribed. Although considered to be a simple device, it is the first mechanical device with which the mankind has achieved mechanical advantage (Fig. 1.4). Archimedes Ellipsograph or Archimedes Trammel is the first recorded closed loop mechanism (Fig. 1.58 ). Screw, pulley and wedge were the devices they have found to amplify the force to lift heavy loads, apply force in squeezing oil from olive fruit, etc. Simple lever is the start of the mechanical movement with which we transmit force and motion to do useful work. Very probably this principle has been used much before Archimedes in cases such as raising water from the wells or rivers (Fig. 1.6) around the Mediterranean region and in China [1]. Lever was also used for weighing goods and as a catapult to throw stones. Archimedes screw has been used through out the ages to raise water (Fig. 1.7). It is still used today in sewage systems. His works were translated to Arabic and then to Latin. His work is the basis of scientific thinking up to renaissance period. Perhaps the rotary motion and pulley system is the one that was more understood by the craftsman. Even today, primary and high school students are thought about lever and pulley in science classes but not on mechanisms topics. The main reason is that the governing equations for pulley and gear are linear and for the lever by making small motion assumption we model lever using linear equations!

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https://en.wikipedia.org/wiki/G%C3%B6bekli_Tepe. https://en.wikipedia.org/wiki/The_Method_of_Mechanical_Theorems. 8 By National Museum of American History, Kenneth E. Behring Center, Gift of Wesleyan University—Elliptic Trammel http://americanhistory.si.edu/collections/search/object/ nmah_694104, Public Domain, https://commons.wikimedia.org/w/index.php?curid=57868477. 7

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1 Introductıon History of Mechanism Design

Fig. 1.3 “Give me one firm spot on which to stand, and I shall move the earth” (see Footnote 8)

Fig. 1.4 Simple lever

Fig. 1.5 Archimedes Ellipsograph (see Footnote 8)

1.2 Archimedes, Levers and Gears and Screws (300 BC–)

Fig. 1.6 Use of a lever with a counterweight to raise water (see Footnote 8)

Fig. 1.7 Arcimedes screw9

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https://commons.wikimedia.org/wiki/File:Chambers_1908_Archimedean_Screw.png.

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1 Introductıon History of Mechanism Design

Fig. 1.8 Wooden bevel gear from seventeenth century10

Gears were either made using wood or metal (usually brass) Wooden gears were made by placing equally spaced pins along the circumference (Fig. 1.8). Metal gears had triangular tooth around the circumference. A very famous finding is the mechanism with gears found in 1901 in the Aegean sea dated 200 BC (Fig. 1.911 ). It is believed that it was used to predict the motion of celestial bodies.

10

Böckler, Georg Andreas, “Theatrum Machinarum Novum”, 1661, Available from https://eco mmons.cornell.edu/handle/1813/57664 11 No machine-readable source provided. Own work assumed (based on copyright claims), CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=469865.

1.3 Al–Jazari: Automata

7

Fig. 1.9 Antikythera mechanism (see Footnote 11)

1.3 Al–Jazari: Automata One of the earliest documented mechanism we see is in Ibn al-Razzaz al-Jazari’s (1136–1206) Book “The Book of knowledge of Ingenious Mechanical Devices” [4] (Figs. 1.10 and 1.11). His work consists of clocks, fountains and perpetual flutes, pitchers for washing hands, moving humans serving drinks, machines for raising water and locks. In China, use of water powered machines during 10-fourteenth centuries in grinding wheels, tilt hammer, blast furnace, spinning wool and clocks has been recorded [5].

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1 Introductıon History of Mechanism Design

Fig. 1.10 Figure of a man sitting on an elephant and hands are in motion [4] (see Footnote 17)

Fig. 1.11 Raising water using a paddle wheel [4] (see Footnote 17)

1.4 Renaissance Period, Leonardo Da Vinci, Printing Press (1400–)

9

Fig. 1.12 Slider crank mechanism (Leonardo) (see Footnote 12)

1.4 Renaissance Period, Leonardo Da Vinci, Printing Press (1400–) In the works of Leonardo da Vinci (1452–1519) we see several different mechanisms. The examples shown are from his note book known as “Codec Madrid”.12 Apart from his paintings, in his life time he kept his work secret and collected them in several handwritten page known as codex, which were discovered after a few centuries. Some of his work is not original therefore his writing in some way is the state of art of mechanical systems in fifteenth century Renaissance period Italy in Figs. 1.12, 1.13, 1.14, 1.15, 1.16 and 1.17. One cannot say that his work has been influential in his time (we can state the same thing on Al Jazari’s work. Maybe these two ingenious people were ahead of their time.). In the construction of mechanical devices (except clocks) wood was mainly used as the construction material. The main reason was its availability and its ease to shape. In certain parts bronze and cast iron was utilized. Starting from the mids of fourteenth century, tower clocks operated by weight are seen. Starting from Italy, it spread through out Europe although the escapement used and the first clock dates to eleventh century China. In eighteenth century watch making was a big industry in Switzerland. Introduction of printing press in 1436 by Gütenberg had a tremendous effect in the spread of knowledge through out Europe. From this time on we see printed books on machinery.

12

https://archive.org/details/codex-madrid/.

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1 Introductıon History of Mechanism Design

Fig. 1.13 Paralelogram mechanism (Leonardo) (see Footnote 12)

Fig. 1.14 Cam mevhanism (Leonardo) (see Footnote 12)

1.4 Renaissance Period, Leonardo Da Vinci, Printing Press (1400–)

Fig. 1.15 A four-bar (Leonardo) (see Footnote 18)

Fig. 1.16 Worm gear and a slidercan mechanism (Leonardo) (see Footnote 12)

Fig. 1.17 Gearing (Leonardo) (see Footnote 12)

11

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1 Introductıon History of Mechanism Design

Fig. 1.18 Single-person-operated device for drawing water from a well” (Ramelli) (see Footnote 14)

1.5 Early Industrial Developments With the rise of the renaissance period, starting in Italy, the development of mechanical systems spread throughout Europe. There was a need for mechanical devices in production. Mechanical equipment were developed for agriculture (irrigation, grinding of wheat, sawing wood, etc.), warfare (guns, catapults, etc.), construction works (i.e. canals, cranes for lifting), mining and transportation of goods and humans. Main prime mower for these mechanical devices besides human power was animal power, wind and water. We can see these developments in Branca Giovanni’s book “Le Machine”13 (1629) in Italy, Ramelli’s (1515–1572) book “Le diverse et artificiose machine”14 (Figs. 1.18 and 1.19) in France and Georg Andreas Böckler’s (1617–1687) book “Theatrum Machinarum Novum”15 (Fig. 1.20) in Germany. For example, in Ramelli’s drawings apart from gears and pulleys, we see several different link mechanisms used to transmit force and motion. From these publications we see a gradual but considerable increase of the application of machines in all different fields. Unlike the writings of Al Jazari or Leonardo Da Vinci, the machines described were mainly those used in practice. These were the state of art of their time. We can easily state that the mechanization started in sixteenth century.

13

https://en.wikipedia.org/wiki/Giovanni_Branca#Le_Machine. Ramelli, Agostino. “Le Diverse Et Artificiose Machine Del Capitano Agostino Ramelli.” Paris, France, 1588. https://digital.sciencehistory.org/works/4b29b614k. 15 Bockler, Georg Andreas, “Theatrum Machinarum Novum”, 1661. Available from Cornell University Library: https://ecommons.cornell.edu/handle/1813/57664. 14

1.5 Early Industrial Developments

Fig. 1.19 Raising water using water and wind power (Ramelli) (see Footnote 14)

Fig. 1.20 Grinding wheat using water and muscle power (Andreas Böckler) (see Footnote 15)

13

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1 Introductıon History of Mechanism Design

Fig. 1.21 Windmill from Encyclopedie [6]

Historians divide the scientific discoveries to be of two types. One is the homogeneous developments, which occur gradually. A certain device gradually changes during time by the artisan who made the device or by someone who is using the device. These changes may be due to change in the use of the device or sometimes, due to the developments in some other field. Another type of change is known as heterogeneous development which results in a step change in technology. A discovery which is a very small change on a known system, may have a big impact on technology. In 1752 the Encyclopedie [5] written by D’Alambert and Diderot compiled the state of art achieved up to the industrial revolution (Fig. 1.21). The aim was to “to change the way people think” and” to incorporate all of the worlds knowledge”. These 17 volumes of articles and 11 volumes of illustrations contained, religıous and political controversies, politics and society and science and technology. In effect, this work may be considered as the eighteenth century Wikipedia.

1.5 Early Industrial Developments

15

Steam power was well known in ancient times but was not implemented in great scale. Through out the middle ages and during renaissance period mainly hydraulic power was used where available, otherwise human or animal power was used. It seems wind power was used in the very early ages along the Eageean costs. At the start of seventeenth century we see some applications which rotate a paddle wheel by steam instead of water. Figure 1.22 is from the book by Giovanni Bianca 16 (1629). Although Newton conceived a cart driven by steam in 1680, it was never realized [7] (Fig. 1.23). The first commercial steam engine was due to Newcommen. In 1705 he patented his steam engine (Fig. 1.24 [7]) which made 6–8 strokes per minute. The engine was used to replace hydraulic or wind driven machines which drive pumps to raise water. Pumps were used for the drainage of mines and for the supply water in towns. The valves were initially manually opened or closed and sealing between the piston and the cylinder was achieved by pouring water onto the piston (later, the valves were opened and closed by latches whose motion is imparted from the motion of the lever). It is a simple lever that converts a linear motion of the piston to a rotation and then to a linear motion of the pump. Continuous rotation was not achieved. Towards the end of seventeenth and eighteenth centuries there was a huge demand for power. With the availabity of cotton from its colonies, in England textile mills were able to produce cloth which was much cheaper than wool. There was a very high demand for textile goods produced in England. However, the textile mills had to be erected near the rivers and there was not enough hydraulic power. Another industry that required power was mining. The coal mines were below the sea level. To excavate, they had to pump out the water. Newcommen engine was invented to replace horses which drive pumps that were driven by horses to drain water from the mines (In fact the HorsePower unit for engines was due to the number of horses a steam engine can replace!!). Although very inefficient, Newcommen engine was very succesful especially in mining industry and in public water supply. Throughout the years it was improved and increased in size. For example in 1775 an engine with cylinder diameter of 180 cm and a stroke of 290 cm was making 9 strokes per minute. In places where hydraulic power was not available or scarce, water was raised into a reservoir using the Newcommen engine and textile mills were operated using this hydraulic power stored. The efficiency of the Newcommen engine was quite poor since the steam was condensed in the same cylinder, cylinder had to be be cooled by pouring water [7].

16

https://de.m.wikipedia.org/wiki/Datei:Dampfmaschine_Bianca_1629.png (Copyright expired, Public Domain).

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1 Introductıon History of Mechanism Design

Fig. 1.22 Steam power used in pulverizing mineral17

17

https://de.m.wikipedia.org/wiki/Datei:Dampfmaschine_Branca_1629.png (Copyright expired, Public Domain).

1.5 Early Industrial Developments

Fig. 1.23 Newton’s Steam engine [7]

Fig. 1.24 Newcommen steam engine [7]

17

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1 Introductıon History of Mechanism Design

1.6 Industrial Revolution It was james Watt who found the drawbacks of the Newcommen Engine. His first steam engine in 1774 was an improvement of Newcommen engine where the steam was condensed in a separate cylinder and thus keeping the main cylinder warm at all times (Fig. 1.25 [7]). (The first engine had a cylinder diameter of 460 mm. It was patented in 1769). Compared to Newcommen Engine this system performed the same task using less fuel. This was a big improvement which is believed to kindle the industrial revolution. This engine still had an oscillating output to drive a pump. For continous rotary output one had to wait for another 12 years. In 1781 James Watt and his partner issued another patent on the steam engine (Fig. 1.26 [7]). This engine had five improvements each of which is a breakthrough in his time: 1. The sun and planet arrangement to impart rotary motion. 2. A flywheel to keep the rotation of the shaft when the mechanism is at or near dead centers. 3. A centrifugal governor to regulate the speed by controlling the steam inlet valve to the cylinder. 4. A mechanism to move the piston rod vertically. 5. The engine was double acting. Unlike the previous single acting pumping engines, steam was injected alternatingly from both sides of the cylinder thus pulling and pushing the piston. Use of sun and planet arrangement was due to a patent of a crank obtained previously. In 1781, a crank was used (Fig. 1.26 [7]). The mechanism is basically a crank-rocker mechanism with rocker as the input. Continuous motion of the crank is due to the flywheel. Let us discuss the fourth invention, since it is the first document we have in history that shows the design of a mechanism to trace a certain curve. (Flywheel was invented by Wasborough, Watt and Boulton paid for royalty rights). In his letter to his son, James Watt describes his invention as [8]: The idea originated in this manner. On finding double chains, or racks and sectors, very inconvenient for communicating the motion of the piston-rod to the angulat motion of the working-beam, I set to work to try if I could not contrieve some means of performing the same form motions turning upon centres, and after some time it occured to me that AB, CD, being two equal radii revolving on the centres B and C, and connected together by a rod AD, in moving through arches of certain length, the variations from straight line would be nearly equal and opposite, and that the point E would describe a line nearly straight, and that if for convenice the radius CD was only half of AB, by moving the point E nearer to D, the same would take place; and from this the construction, afterwards called the parallel motion was derived. …. Though I am not over anxious after fame, yet I am more proud of parallel motion than any other inventions I have ever made.

Figure 1.26 is the hand sketch of James Watt from the letter [8], This mechanism we now call “Watt’s Straight line motion mechanism”.

1.6 Industrial Revolution

19

Fig. 1.25 James Watt’s first engine (1769) [7]

The importance of straight line motion was that during 1780s they were not able to manufacture cylinder that would guide the piston as in case of slider-crank mechanism used in internal combustion engines, both in terms of manufacture and material. Guiding the piston along a straight line and converting a reciprocating

20

1 Introductıon History of Mechanism Design

Fig. 1.26 James Watt’s Engine imparting continuous rotation (1781). Please note a flywheel, b centrifugal governor, c Watt’s straight line motion mechanism, d planetary gear train for continuous rataion, e double acting cylinder [7]

motion to a rotary motion had an important effect on the steam engine since it minimized the side force exerted by the piston onto the cylinder. Watt’s straight line motion describes an approximate straight line within a certain part of motion. Straight line motion drew the interest of mathematicians and those working in manufacture of steam engine. For example, James White patented what we now call “cardanic motion” for steam engines. Although it describes an exact

1.6 Industrial Revolution

21

Fig. 1.27 James Watt’s steam engine with a crank-rocker mechanism replacing the planetary gears [7]

straight line, it was not commercially succesful (Fig. 1.29 [7]). Another straight line motion mechanism (again Cardanic motion), patented by Fremantle in 1803, is commonly known as “Scott-Russell linkage” or “Evan’s Grasshopper” was used by an American Evans in 1805. This is basically an in line slider crank mechanism forming an isoceles triangle (Fig. 1.30 [7]). What was initially the problem of the blacksmith, mechanics or technicians at that time, the straight line motion caught the attention of the academicians of the time. The academics of the time were mainly mathematicians, phsicists or chemists. Especially the mathematicians of the period contributed a great deal. This interest eventually

22

1 Introductıon History of Mechanism Design

a(7)

b(6)

Fig. 1.28 a James Watt’s hand drawn sketch in explaining his straight line mechanism invention, b construction of mechanism [7, 8]

Fig. 1.29 Planetary gear mechanism to generate a straight line (Cardan motion)

ended up in a new field “kinematic analysis and synthesis of mechanisms”. It was Watt’s straight line motion that trigerred the study of mechanisms and his steam engine that created the mechanical engineering (up to that time only Military and Civil Engineering were the only two fields). We can rightfully state the field that we

1.6 Industrial Revolution

23

Fig. 1.30 Isoceles ˙Inline slider crank mechanism to generate exact straight line [7]

now call as “Theory of Machines” (including mechanism kinematic analysis and synthesis and dynamic analysis) started with Watts steam engine of 1781. After the breakthrough made by Watt, several inventions were made by him and others especially in opening and closing the valves. A typical example is shown in Fig. 1.31 [7]. Although the system looks like a Rube Goldberg Design,18 it was used to control the opening and closing of the valves of a steam engine. Watt has opened pandora’s box for mathematicians, scientests and all inventors. It was Peacullier who found an exact straight line motion using revolute joints only in 1864 (Fig. 1.32 [7]). The mathematicians Sylvester and Chebyshev devoted most of their time to mechanisms. It is interesting that in 1873 the Russian Mathematician Chebyshev wrote to his British colleague Sylvester “Take to kinematics. It will repay you. It is more fecund (productive) than geometry; it adds a fourth dimension to it” [7]. As a byproduct several ingenious mechanisms were found and implemented on the steam engine, textile machines, etc. People found out that they can automate different activities using mechanisms. The straight line problem had several byproducts. Chebyshev (1821–1894), found other mechanisms which also approximated a straight path. Although he is very well known as a mathematician, most of his work is from the study of mechanisms. He developed a theory to minimize the coupler curve path deviation from a straight line, which we now call “mini-max method”. This was the first problem of “optimization” in mechanisms. He also found that there are three different four-bars (cognates) that describe the same coupler curve (ındependently the English mathematician Roberts also found the same result), which we now call “Robert-Chebyshev theorem”.

18

Rube Goldberg was an American Cartonist, engineer and inventor. His drawings show complex gadgets intended to solve simple or ridiculous tasks. https://en.wikipedia.org/wiki/Rube_Goldberg.

24

1 Introductıon History of Mechanism Design

Fig. 1.31 Valve drive [7]

By the turn of the eighteenth century, steam engine was well developed. In the first stage steam engines were mainly used to raise water and to drain mines. With the rotary output the textile mills did not need running stream. Steam engines producing a few horsepower were replaced by those of several horsepowers, a few revolutions per minute rose to several hundred revolutions per minute. It was first applied in trains than to ships, opening up a new era in manufacturing and in transportation. With the advances in steel production, wood was replaced by steel. These developments also created new inovations involving mechanisms. In the mean time, interest in mechanisms in the academia also florished. For example It was the first time mechanical engineering and mechanisms was thought in Ecole Politechnik, France. The book published by Hachette in 1808 named Traité élémentaire des machines19 19

M. Hachette, “Traite Elementaire Des Machines”, Saint Petersburg, 1811 (MDCCCXI) https:/ /books.google.com.tr/books/about/Trait%C3%A9_%C3%A9l%C3%A9mentaire_des_machines. html?id=7JXmAAAAMAAJ&redir_esc=y.

1.6 Industrial Revolution

25

Fig. 1.32 Peacullier exact straight line motion [7]

Fig. 1.33 A page from Hachette’s book “Traité élémentaire des machines” (see Footnote 18)

(Fig. 1.33) was the first book on these topics thus starting formal education. Instead of learning in the machine shop, these topics were thought in class. Nineteenth century is the foundation of what we now call “Machine Theory”. In Great Britain the developments made were mainly in shop floors. These were mainly inventions and improvements without any theoretical background. James Watt, although his approach is very scientific, had no formal education at that time. Main characteristics of the people were that they were free thinkers, without any dogma to follow. French people were very much influenced by the “Encyclopedie” [5] of Diderot and d’Alambert. Military engineer Lazarre Carnot who first published “Essai sur les

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1 Introductıon History of Mechanism Design

Machines en General”20 in 1803, pioneered for the establishment of Ecole Polytechnique in Paris. From The Military school Gaspard Monge came to Ecole Polytechnique and started teaching descriptive geometry in 1794. Later Pierre Hachette was in charge of the course. In order to analyze and classify machines of his time he made a chart as shown in Fig. 1.33 (see Footnote 18). He classified the mechanisms by considering the conversion of motion. His basic motion types were circular, oscillating, continuous rectilinear, oscillating rectilinear. His book “Traite Elementaire des Machines” (see Footnote 18) (1811) was the reference for mechanisms for the next fifty years. Later Betancourt extended Hachette’s system. Ecole Polytechnique became the pioneer of kinematics. It was Gaspard-Gustave Coriolis who published his book in 1829 and is well known with the accelaration term in his name. Michael Chasles whose name is remembered from “Chasles’ Theorem” is also from the same school whose theories we use both in planar and special kinematics. Theoretical basis was founded at Ecole polytechnique, France. In fact the word “Kinematic” or “Cinematique” which was derived from the Greek word for movement is due Andre-Marie Ampere who is known from his works in physics. However, it was first Euler in 1775 who stated: “The investigation of the motion of a rigid body may be conveniently separated into two parts, the one geometrical, the other mechanical”. The works done in Ecole Polytechnique, drew the attention of other schools in the continent. In Great Britain, it was Robert Willis from Cambridge University who wrote “Principle of Mechanisms”21 in 1841. His work is mainly the analysis of the existing mechanisms. His classification is also based on the type of motion.

1.7 Birth of Machine Theory: Franz Realeaux Germany was very much influenced by the industrial revolution in Britain. Early in nineteenth century, industry through out Europe had to buy British made steam engines to run their mills. German scholars were following very closely the theoretical developments in France. Early in nineteenth century they started teaching mechanical engineering in line with the developments of Ecole polytechnic in their polytechic schools. It was Professor Redtenbacher (1809–1863) in the polytechnic school in Karlsruhe who enthusiasticaly seeked to find a unified system to base the study of mechanisms. Although he was not succesful, his student Franz Reuleaux (1829–1905) shaped the way we look on mechanisms today. Unlike his predecessors, he showed that the main charecteristics of a mechanism is due to the kinematic pairs (joints in between the rigid links and the general shape of a rigid body is immaterial in terms of kinematics. He defined “kinematic chain and considered the ground as any other rigid body. What was called “three-bar” by Willis was referred to as a 20

https://books.google.com.tr/books/about/Essai_sur_les_machines_en_g%C3%A9n%C3% A9ral.html?id=I8k2AAAAMAAJ&redir_esc=y. 21 https://books.google.com.tr/books/about/Principles_of_Mechanism.html?id=1CCEKSqQa qcC&redir_esc=y.

1.7 Birth of Machine Theory: Franz Realeaux

27

“four-bar” (actually in German it is “viergelenk”-four hinged, referring to four revolute joints). His definitions are still used in modern mechanism education. His book published in 1875 “Theoretische Kinematik” [10] was a milestone. It was translated to French, Italian and English in a very short time. His approach to mechanisms paved the way to new developments. Due to all his work rightfully he is called “Father of Kinematics”. He invented new mechanisms, such as the “Reuleaux triangle” used in constant breadth cams. What was left to the ingeniuty of the people can now be mathematically treated. His work is mainly based on the classification and identification of mechanisms. One other contribution of Reuleaux is the preparation of more than 800 models of mechanisms (Fig. 1.34). It is the first time a mechanism is viewed as separate identity instead of being described as a machine part. Some 350 models were marketed and bought by universities in Japan, Russia and USA. Most of the original models in Berlin were unfortunately destroyed during the wars. Cornell University has one of the best collections (220 models).22

Fig. 1.34 Reuleaux’s models as shown in Voigt catalogue23 22

Reuleaux Kinematic Mechanisms Collection | Cornell University Library Digital Collections. Voigt, Gustav , ““Kinematische Model nach Professor Reuleaux”, Berlin, Voigt 1907. https://eco mmons.cornell.edu/handle/1813/58764. 23

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One other important fact about Franz Reuleaux is that people working on mechanisms before his time were either craftsman or technician in the machine shop or scientist, mathematician in the universities. Now we see a new figure who we may call “Engineer–Scientist”, While working in the university on theoretical topics he also worked as a consultant and expert in the industry. Through Reuleaux’s efforts especially in polytechnical schools, both teory and practice were employed. The relation in between the two was established. A person working in a machine shop had a very good knowledge about geometry and a person working on the drawing board knew everything about the machine shop. The combination of theory with practice, university with factory or drawing board with machine shop had a great impact on German economy. For example after 1855 Paris World Fair, Sir William Fairbaine wrote24 “Although we are producing more (meaning the British industry), from what I have seen French and Germans are in advance of us in theoretical knowledge of the principles of higher branches of industrial art; and I think it arisis from the greater facilities afforded by the institutions of those countries for instruction afforded by the institutions of those countries for instruction in chemical and mechanical science”. In fact, Encyclopedia Americana states “During 1870–1895 as the German exports increased by 42%, British exports had an increase of 13%. The main reason being the education”. Franz Reuleaux was very much influential in this respect. Reuleaux also showed that although kinematics is based on the geometry of motion, it cannot be separated from the main goal: “Machine Design”, which encompasses kinematic synthesis and analysis, dynamics, strength and all other basic engineering topics in addition to the practical experience, intiution and aptitude. One must even be aware of social and economic conditions. Reuleaux also believed that for the invention of a new machine one need not be a genious but he must have a good understanding of the previous machines. As his contemporary Edison (1847–1931) states “Invention comes to the prepared minds” and “invention is 1% inspıration 99% perspiration”. One person who was very much influenced by Reuleaux work was Burmester (1840–1927) [11]. His formal education is on mathematics and geometry and he worked as a secondary school teacher. However he was appointed professor of descriptive and stynthetic Geometry in Dresden. Burmester had written several books on geometry. For example in 1871 he wrote a book named “Theory and Representation of well Defined Surfaces”. Under the influence of Professor Ritterhaus he got involved in kinematics—geometry in motion. Influenced by Reuleaux’s classification Burmester started to develop new theories. In 1888 his book “Lehrbook der Kinematik” [12] is his pioneering work on planar mechanism synthesis. What we now call “Burmester Theory” was a great achievement on mechanism synthesis. Burmester’s work was mainly based on synthetic geometry and geometrical construction was used in the synthesis of mechanisms. His work and approach found 24

https://books.google.com.tr/books?id=LPIDAAAAMAAJ&printsec=frontcover&dq=The+ Life+of+Sir+William+Fairbairn,+Bart&redir_esc=y#v=onepage&q=The%20Life%20of%20Sir% 20William%20Fairbairn%2C%20Bart&f=false.

1.7 Birth of Machine Theory: Franz Realeaux

29

several followers especially in Germany. It was M.F. Grübler (1851–1935) H. Alt (1889–1954), R. Beyer (1892–1960), Lichtenheldt (1901–1980), Hain (1908–1995) and many others who developed and tought in German schools. Three different but closely related problems were treated. These were the correlation of crank angles (function generation), path generation (for a point on the coupler link to describe a certain algebraic curve) and position synthesis (to move a rigid body through a set of positions) The work of Reuleaux and his followers had a big impact on mechanical engineering education and on machine industry. However, in English and French speaking countries these works did not find followers. Although Reuleaux’s book was translated into English and French in a few years after its publication, Burmester’s work was not translated. For example, During the second world war, the need for computing linkages in USA was solved by a technique devoloped by Svoboda [13] while working in Radiation Laboratory at MIT, had no relation with the methods developed in Germany. There was a great need for new mechanical devices. It was Hrones and Nelson’s Atlas [14] that compiled possible coupler curves of a four bar mechanism in a certain order that was used by the machine designers in USA for many years. The textbooks till 1970s printed in USA mainly contained graphical mechanism analysis only. Geometrical synthesis procedures developed in Germany was time consuming and required large skill during the construction. Russian school developed analytical methods following the works of Chebyshev. In this respect the works of N. I. Levitskii in function approximation was an important application of Chebyshev theorem to mechanisms [15]. In early 50s revitalization, application and development of Chebyshev Methods by S. Sch. Bloch [6] shows the importance of Chebyshev theorem in mechanisms design. Ivan Ivanovich Artobolevski was the most leading figure during the Soviet era. In addition to his several books, his editorship to seven volume collection of mechanisms is a lasting contribution [16]. It was F. Freudenstein’s (1920–2006) paper in 1954 [17] “An analytical Approach to the design of Four-Link mechanism” which opened a new era on kinematic synthesis in the west. Rightfully he is called “the father of modern kinematics”. In the following years we see a vast development on mechanism design using analytical methods, solved in now emerging computers. Initially Burmester Theory was analytically formulated and solved using numerical techniques. Analytical alternatives to the geometrical methods were first developed. Later new methods and numerical techniques were found. Through Freudenstein’s work, a new approach to mechanism synthesis and analysis was triggered especially in the United States. In later years optimization techniques in mechanism design was introduced. With the spread of personal computers these analytical methods gained more popularity. It was Erdman and Sandor’s Book [18] a milestone for analytical methods. Other books followed. For example the aplication of the analytical methods were implemented in Matlab by Russell, Shen and Sodhi [19]. However, in recent years with the availability of integrated geometry softwares, graphical methods can also be implemented easily by other freely available softwares.

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1 Introductıon History of Mechanism Design

In the light of all these developments, the aim of this book is to discuss the kinematic synthesis of mechanisms following the paths of Burmester, Chebyshev and Freudenstein. Both geometrical and analytical methods will be explained and examples will be shown in Geogebra and Excel® .

References 1. S.A. Paipetis, M. Ceccarelli, The genius of Archimedes—23 centuries of influence on mathematics, science and engineering, in Proceedings of an International conference held at Syracuse, Italy, 8–10 June 2010 (Springer, 2010) 2. M. Ceccarelli, in Distinguished Figures in Mechanisms and Machine Science. Their Contributions and legacies, Part 1 (Springer, 2007) 3. I.R. Al-Jazari, in The Book of Knowledge of Ingenious Mechanical Devices (Kitab fi ma ‘rifat al-hiyal al-handasiyya), Translated by Doland R. Hill (D. Reidel Publishing Co. Boston, 1974) 4. Z. Lu, X. Gao, The development of water-powered machines of China in 10–14th century, in International Symposium on History of Machines and Mechanisms—Proceedings HMM2000 (Springer, 2000) 5. F.M. Ricci editore in Parma, in Planches scelte dalla Encyclopedie di Diderot e D’Alambert (Paris 1751–1772) (Realized for Fiat-Divisione Mare, 1971) 6. S. Sch Bloch, Angenaherte Synthese von Mechanismen (Verlag Technik, Berlin, 1951) 7. R.H. Thurston, A History of the Growth of the Steam-Engine (Page 149, Fig.42 and Page 59, Fig.19) (D. Appleton and Company, 1902) 8. F. Reuleaux, Kinematics of Machinery, Outlines of a Theory of Machines (Translated by A.B.W. Kennedy) (MacMillan and Co. 1876, London) 9. E.S. Ferguson, Kinematics of Mechanisms from the Time of Watt, United national Museum Bulletin 228 (1962). Available from: https://www.gutenberg.org/files/s7106/27106-h/27106h.htm 10. F. Reuleaux, Kinematics of Machinery—Outlines of a Theory of Machines (translation of “Theoretiche Kinematic” by A.B. W. Kennedy) (Macmillan and Company, 1876) 11. M. Ceccarelli, T. Koetsier, Burmester and Allievi: a theory and its application for mechanism design at the end of 19th century. ASME J. Mech. Des. 130 (2008) 12. L. Burmester, Lehrbook der Kinematik, Für Studirende der Machinentechnik, Mathematik und Physik Geometrisch Dargestellt (Verlag Von Arthur Felix, Leipzig, 1888) 13. A. Svoboda, Computing Mechanisms and Linkages (Dover Publication, 1965) (first Printing Mc Graw Hill, 1948) 14. J.A. Hrones, G.L. Nelson, Analysis of the Four Bar linkage: Its Application to the Synthesis of Mechanisms (The Massachusetts Institute of Technology and John Wiley&Sons, New York, 1951). https://www.scribd.com/doc/214820022/Hrones-and-Nelson-Atlas 15. N.I. Levitskii, Desigrt of Plane Mechanisms with Lower Pairs (AH CCCP Izdatelstvo Akademii Nauk, Moscow-Leningrad, 1950) 16. I.I. Artoboleyevski, Mechanisms in Modern Engineering Design (Mir Publishers, 1975) (First Published in Russian in 1970, also published in French and German) 17. F. Freudenstein, An analytical approach to the design of four-link mechanisms. Trans. ASME, 76, 483–492 (1954) 18. A. Erdman, G. Sandor, Mechanism Design, Analysis and Synthesis, vol. 1 and 2 (Prentice Hall, 1984) 19. K. Russell, Q. Shen, R.S. Sodhi, Mechanism Design, Visual and Programmable Approaches (CRC Press, 2014)

Chapter 2

Two Positions of a Moving Plane

Abstract In this chapter design of mechanisms to move a rigid body in plane between two positions is discussed. This topic is based on Chasles’ theorem. In plane motion the pole and the rotation about the pole are the two invariants of motion between two positions. The motion of a moving plane can be with respect to a fixed frame or relative to another moving plane. Realization of motion between two positions with a four-bar, slider crank and double slider mechanism or, when necessary, design of six link mechanisms for two position synthesis are explained. The principle of superposition is discussed. Solutions using graphical and analytical methods are shown using Geogebra and Excel® . Keywords Motion generation · Two position synthesis · Burmester theory · Chasles theorem

2.1 Two Finitely Separated Positions of a Plane Consider a moving rigid plane E. This plane can be represented by a directed line segment AB (points A and B being any two arbitrary points on the plane E) or a point O’ as the origin of a moving frame and angle that defines orientation of the moving coordinate frame (Fig. 2.1). The moving planes are assumed to extend to infinity in both directions. For geometrical treatment of the problem a directed line AB is more useful whereas for analytical treatment polar or cartesian representation is preferred. The plane E is moved from a position #1 (E1 or A1 B1 ) to another position #2 (E2 or A2 B2 ) with respect to a fixed reference frame F (Fig. 2.2). A1 , A2 are the locations of the point A on the fixed frame for these two positions. They are the reflections of point A to the fixed frame at two different instances in time (i.e., using a camera, if we shoot the position of E1 and E2 onto the same frame of the film we shall have two points A1 , A2 for point A on the film). A1 , A2 are known as “homologous points”. The plane motion of the plane E from position #1 to position #2 can be most simply obtained by a rotation. This was first stated by Chasles (1830) as:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_2

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2 Two Positions of a Moving Plane

E X

E

x

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B (x ,y ) b b

y

l φ

O'(X, Y) A (x ,ya ) a

r

O

Y

Geometric Representation

O

θ

Y

Polar (r,θ,φ) or Cartesian (X,Y,φ) Representation

Fig. 2.1 Geometric and Cartesian representations of a moving plane Fig. 2.2 Two finite positions of a moving plane and the location of the pole at the intersection of the perpendicular bisectors of two pairs of homologous points A1 , A2 and B1 , B2

Chasles’ Theorem1 ,2 : The translation of a rigid body in a plane motion occurs most simply by a rotation about the pole P12 which is located at the point of intersection of the perpendicular bisectors of two pairs of homologous points. In order to prove the theorem, one must show that the triangle A1 P12 B1 has moved to the second position in an undeformed fashion when rotated about P12 , i.e., triangle A1 P12 B1 and A2 P12 B2 are congruent. This will show that the plane E has moved in between the two positions by rotating about P12 . |A1 B1 | = |A2 B2 | and both are equal to |AB| on the moving plane (homologous points). |A1 P12 | = |A2 P12 | since P12 is on 1

This theorem was first stated by Mozzi (1763). However it is mainly known as Chasles’ Theorem in literature. Also the original theorem is for general motion of a rigid body, stated as: “The most general rigid body displacement can be produced by a translation along a line (called the screw axis) followed by a rotation about an axis parallel to that line”. In plane motion the translation along the screw axis does not exist1 . 2 https://en.wikipedia.org/wiki/Chasles%27_theorem_(kinematics).

2.1 Two Finitely Separated Positions of a Plane

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Fig. 2.3 ΔA1 B1 P12 = ΔA2 B2 P12 . Hence the motion in between two positions is realized by a pure rotation

the perpendicular bisector to the line segment and A1 A2 (a12 ). Similarly, |B1 P12 | = |B2 P12 | since P12 is also on the perpendicular bisector to the line segment B1 B2 (b12 ) (Fig. 2.3). Note that P12 is a unique point on the moving plane, which has no displacement when the moving plane moves in between these two positions. It is the only coincident homologous points (there is no other point C on the moving plane where C1 and C2 are coincident, whereas P12 (1) and P12 (2) are coincident). All other points on the moving plane will move on concentric arcs. Another important point is that numbers and letters do not exist in real life. What one calls points A, B, C, etc. one may as well call Point 1, Point 2, etc. and what we call position #1 or #2, one may as well call position A, B, etc. In the real world, you will never see numbers or letters at points. In fact, you will not see points; it will be just a body moving from one position to another. These letters and numbers are what we use to communicate. If the two positions and the pole P12 is suitably located, the simplest way to move the rigid body in between these two positions is to attach a revolute joint between the moving and the fixed planes at P12 . In many applications this is the simplest mechanism you can obtain. Unfortunately, in some design tasks, attaching a revolute joint may not be possible when the pole is located too far away or if the path of the moving body interferes by some other bodies in the system when the body is rotating about a fixed axis (Figs. 2.4 and 2.5). When rotation about the pole is not feasible, one can realize the motion in between the two positions by means of a four-bar as follows: Let us select a point A0 on the perpendicular bisector of A1 A2 (a12 ) and select another point B0 on the perpendicular bisector of B1 B2 (b12 ). Now, let us attach two links A1 A0 and B1 B0 and join these links with the rigid body AB and the fixed link A0 B0 by revolute joints (Fig. 2.6). A and B on the moving plane will describe circular arcs with centers A0 and B0 respectively. |A0 A1 | = |A0 A2 | and |B0 B1 | = |B0 B2 | since A0 and B0 are on the perpendicular bisectors a12 and b12 respectively. Also, |A1 B1 | = |A2 B2 | and A0 B0 is fixed. The

34 Fig. 2.4 P12 is located far from the object to be moved in between two positions

Fig. 2.5 Rotation about P12 cannot be realized due to obstacle

2 Two Positions of a Moving Plane

2.1 Two Finitely Separated Positions of a Plane

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Fig. 2.6 Moving Plane AB in between two positions using a four-bar. A0 and B0 are selected on the perpendicular bisectors a12 and b12 respectively

four-bar will exist at the two defined positions. However, we cannot guarantee that the mechanism will be movable in between the two given positions. When we select a point A or B on the moving frame to attach a revolute joint, this point in the final stage will describe a circle and A0 or B0 will be the center of the circle. Points such as A, B etc. we shall call “circle point” or “moving pivot” and points A0 , B0 , etc. we shall call “center point” or “fixed pivot”. One can also attach a slider to the moving plane by selecting A0 and B0 at infinity (a circle with infinite radius is a straight line) (Fig. 2.7). Hence any plane motion between two positions can be attained by lower kinematic pairs with the following simple mechanisms: (a) Wheel (or crank): Plane being joined to the fixed plane at the pole P12 (Fig. 2.3). (b) Four-bar: Moving pivots can be attached at any two points on the moving plane and the fixed pivot is any point on the perpendicular bisector of the homologous points (Fig. 2.6). (c) Slider-Crank. Moving pivots can be attached at any point on the moving plane. While one of the fixed pivots is at a finite location on the perpendicular bisector of one homologous point, the other is at infinity for the slider (Fig. 2.7). (d) Double slider: Moving pivots can be attached at any point on the rigid body. Fixed pivot is at infinity on the perpendicular bisectors of the homologous points. In other words, the slider axis is the line joining the homologous points (Fig. 2.8). It must be noted that the mechanism thus obtained will pass through any two specified design positions. However, the motion in between the two design positions

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a12

a12

A0 b12

b12 P12 B0

P12 B2

B1 A1

B2

B1

A2

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A1

Fig. 2.7 Moving the plane AB in between two positions using slider-crank mechanism Fig. 2.8 Moving the plane AB in between two positions using double slider

is completely unspecified. The mechanism may not be movable in between these two positions or the transmission angle may not be suitable for the practical use of the mechanism. For example, for the two positions given in Fig. 2.3., while the mechanism shown in Fig. 2.6 is movable in between the two positions, the mechanism shown in Fig. 2.9 is immovable (You must disconnect one of the joints and reassemble for the second position). The type of the four-bar or slider-crank mechanism thus obtained (e.g., double rocker, crank-rocker, or double crank) is not known a priori. However, the number of possible solutions is infinite. If we want to realize the motion between two positions by a pure rotation, we have a unique solution. Whereas in the case of a four-bar, we can select the moving pivot points anywhere on the rigid body (A and B can be any two points on the rigid body) and the corresponding fixed pivots will be selected on the perpendicular bisectors of

2.1 Two Finitely Separated Positions of a Plane

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Fig. 2.9 The four-bar mechanism realized cannot be moved in between the two positions (it must be dismantled)

the homologous points. We thus say there are ∞6 possible solutions, each infinity referring to one parameter that can be changed in infinitely many ways. The number of possible solutions for the slider-crank and double slider mechanisms are ∞5 and ∞4 respectively. Let us derive certain conclusions for the four-bar mechanism that will synthesize two positions of a plane. This will be of use at the later stages of synthesis (Fig. 2.10). Referring to Fig. 2.10: Let α = ∠A1 P12 B1 = ∠A2 P12 B2 φ12 = ∠A1 P12 A2 = ∠B1 P12 B2 Then ∠A1 P12 B2 = ∠A1 P12 A2 + ∠A2 P12 B2 = φ12 + α. Also: ∠A1 P12 B2 = ∠A1 P12 N + ∠NP12 M + ∠MP12 B2 . Since Points M and N are on the perpendicular bisectors: ∠A1 P12 N = 1/2∠A1 P12 A2 = 1/2φ12

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2 Two Positions of a Moving Plane

M

N a12

α

b12

φ12/2 1

A1

φ12

B1

α

φ12 /2

A2

α

φ12

B2

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P12

Fig. 2.10 ∠A1 P12 B1 = ∠A2 P12 B2 = ∠NP12 M = ∠A0 P12 B0 = α, ∠A1 P12 N = ∠A1 P12 A0 = ∠B1 P12 N = ∠B1 P12 B0 = 1/2φ12

and: ∠MP12 B2 = 1/2 ∠B1 P12 B2 = 1/2φ12 Hence: ∠A1 P12 B2 = φ12 + α = φ12 + ∠NP12 M Therefore: ∠NP12 M = α We can derive the following conclusions: 1. Cranks subtend the same angle at the pole. This angle is one half the rotation of the plane about the pole: ½ φ12 or (½ φ12 + π). 2. Coupler and the fixed link subtend the same angle at the pole. This angle is α or α + π. α does not represent the motion property (i.e., if different moving pivot points are selected α will be different. Whereas φ12 does not depend on the selection of moving pivot). 3. P12 and φ12 completely define the displacement from position #1 to #2. The case where the cranks subtend angles differing by π is shown in Fig. 2.11. This case occurs if A0 and B0 are in opposite sides of the pole P12 .

2.1 Two Finitely Separated Positions of a Plane

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Fig. 2.11 ∠A1 P12 B1 = ∠A2 P12 B2 = ∠NP12 M = α, ∠A0 P12 B0 = α + π, ∠A1 P12 N = ∠A1 P12 A0 = ∠B1 P12 N = 1/2φ12 , ∠B1 P12 B0 = 1/2φ12 + π

Special cases: (a) Two parallel link positions In this case P12 is at infinity and the angular rotation is measured in terms of the distance between homologous points A1 A2 , B1 B2 or C1 C2 distance, s12 . Plane motion is a translation (Fig. 2.12). One can design a four-bar mechanism in the usual way: (b) Two positions symmetric about a line In this case the perpendicular bisectors of the pair of homologous points will coincide. P12 is on a12 and b12 such that < B1 P12 B2 = < A1 P12 A2 = φ12 . Hence P12 is located at the intersection of the perpendicular bisectors with the line A1 B1 and A2 B2 (one can locate P12 by selecting a point C on the moving frame which is not symmetric. The perpendicular bisector c12 to C1 C2 will locate the pole P12 (see Fig. 2.13). (c) Two Infinitesimally close plane positions The arcs A1 A2 and B1 B2 will become tangents to the paths of respective points. The perpendicular bisectors are the normal to the paths. In this case, the pole P12 is the “instantaneous center of rotation” for the rigid body and is usually denoted

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2 Two Positions of a Moving Plane

Fig. 2.12 Two finitely separated positions parallel. Pole, P12 is at infinity

Fig. 2.13 a Two finitely separated positions are symmetric about a line. Pole P12 is at the intersection of the two lines. b you can eliminate symmetry by selecting any other point, C on the moving plane to eliminate symmetry

by I (Fig. 2.14). This point is now called “instant center”, “instantaneous center”, “centro” or “rotation pole”. One can realize this motion by means of a pure rotation (about I), a four-bar (A0 and B0 selected on the respective normal), a slider-crank mechanism or a double slide (slider axes along the tangents), as before.

2.2 Graphical Synthesis Methods for Two Positions Nowadays we apply graphical techniques using a CAD program. In general, one can use any one of the programs available. Here two different packages will be used. One is a professional software SolidWorks® which has similar characteristics with other available CAD programs such as Siemens NX® , Catia® , etc. and the other is an opensource software known as Geogebra.

2.2 Graphical Synthesis Methods for Two Positions

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Fig. 2.14 Two infinitesimally close positions. “Instant Center”

2.2.1 Performing Two Position Synthesis Using Solidworks® The procedure will be explained in steps. You must have SolidWorks® program installed and ready. The procedure is not unique. Some tasks can be performed differently. Step 1. Open a new part file. On the front plane set up a drawing sheet. On this sheet draw the two given positions of the rigid body (in Fig. 2.15 it is shown as a rectangle but any other shape may be used). The two shapes must be exactly the same in different orientation. Therefore, define the required constraints to make similar sides equal to each other. Fix these two positions. These are the required two positions of the moving plane. Of course, depending on the problem you can use a drawing sheet on the top plane or any other plane of your choice. On this drawing sheet you can also draw some other rigid parts which may be important for the motion of the final mechanism. Step 2. Pick two points such C and D on the rigid body. These are your reference points (In Fig. 2.15, these two points are selected at two corners of the rectangle. However, any other two points on the rigid body can be used). These two points will be at C1 , D1 in position 1 and C2 , D2 in position 2. Step 3. Close the drawing sheet and open another drawing sheet on the same plane (frontal plane in this case). A point A that will be the axis of a revolute joint on the moving rigid body (moving plane) must be equidistant to both C and D. Hence draw lines C1 A1 D1 and C2 A2 D2 and define relations |C1 A1 = |C2 A2 | and |D1 A1 | = |D2 A2 |. In this way, when you hold point A1 with the cursor and move to another location, A2 will also move such that the distance |AC| and |AD| always remains the same for the two positions. Step 4. Select another point B as the axis of another revolute joint on the moving body and apply the procedure used for point A (|C1 B1 | = |C2 B2 | and |D1 B1 | = |D2 B2 |). You can move A1 , B1 or A2 , B2 anywhere in the plane and the other

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2 Two Positions of a Moving Plane

Fig. 2.15 Two position synthesis using SolidWorks® . Sheet 1 contains the moving body in two positions (rectangles and points CD). On Sheet 2 you select the location of two revolute joints A, B arbitrarily, locate the corresponding perpendicular bisectors of the homologous points and select corresponding center points A0 , B0 on these perpendicular bisectors

homologous point (A2 , B2 or A1 , B1 ) will automatically move. This means that you are free to select your moving revolute joint axes anywhere in the moving plane easily. Step 5. Join A1 A2 and B1 B2 and draw the perpendicular bisectors. (Define the relation that A1 A2 and a12 , B1 B2 and b12 are at right angles). You can now select a point on a12 as A0 and a point on b12 as B0 , to be the fixed revolute joint axes. Draw line segments A0 A1 , A0 A2 , B0 B1 , B0 B2 , which correspond to the two positions of the cranks A0 A and B0 B. Step 6. Draw circular arcs A1 A2 and B1 B2 with centers A0 and B0 . As you change the positions of A and/or B or A0 , B0 along the perpendicular bisectors, you will obtain different four-bar proportions. You can perform several trials in a very short time. No synthesis problem is complete without analysis. You must analyze the motion of the mechanism that you have selected as a design candidate. You can perform the analysis in SolidWorks easily if you follow the following steps.

2.2 Graphical Synthesis Methods for Two Positions

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Fig. 2.16 Initial four-bar A0 ABB0

Step 7. Exit from the drawing sheet. Open a third drawing sheet on the same plane that you used for the other sheets (i.e. frontal plane). Now you have another transparent sheet on the previous drawing sheets. Step 8. On this new sheet draw line segments A0 A, AB and B0 B. Select A0 and B0 , to coincide with A0 and B0 of the previous sheet (do not select A and B coincident to A1 and B1 or A2 and B2 ) (Fig. 2.16). Step 9. Select A0 A on this sheet and equate to A0 A1 on the previous sheet. Repeat for B0 B (B0 B1 ) and AB (A1 B1 ). Also fix A0 and B0 (Fig. 2.17). Step 10. You can add the rectangle by moving the four-bar to the first position and drawing the rectangle that represents the object in the first drawing sheet. The mechanism can be moved by first clicking on the link which is to be used as the input (This link can be A0 A, AB or B0 B depending on the application). This link has a certain length and an angle. You can change this angle by clicking on the up or down arrows or you can write any value of this angle. Step 11. Make the line AB and the rectangle drawn as a block (tools-Block menu). You can now move the rectangle (or any other shape drawn) from one position to another. After analyzing this mechanism, you can exit and save this file. Next, you open the second sheet and change the locations of A, B, A0 and/or B0 to obtain another set of link length proportions and then analyze this new mechanism in the third drawing sheet (when you change A, B, A0 , B0 on the second sheet, notice that the four-bar

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2 Two Positions of a Moving Plane

Fig. 2.17 A0 ABB0 after Step 9

mechanism on the third drawing sheet changes accordingly). The first sheet is your problem definition, the second is the design sheet. The third drawing sheet is for the analysis of the mechanism designed. The method described is not unique. One can find other means in SolidWorks® or in any other CAD program that is available to solve the same problem.

2.2.2 Performing Two Position Synthesis Using Geogebra Created by Markus Hohenewarter in 2001, Geogebra represents a digital learning environment that was designed to combine geometry, algebra and calculus in a single dynamic user interface.3

Geogebra is a mathematical environment that allows you to work with graphing, dynamic 2D and 3D geometry, spreadsheets, complex numbers, differential equations, curve fitting, etc. It has been written for high school students (grades 10–12) so it must not be hard for a senior mechanical engineering student to use. While the computer does the necessary computations and graphing, you have to translate 3

Hohenwarter et al. [1].

2.2 Graphical Synthesis Methods for Two Positions

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Fig. 2.18 Geogebra interface

your problem into a language with which the computer is able to solve your problem efficiently.4 You can download Geogebra from different web sites. There are lots of videos in YouTube and you can download the manual and some tutorials prepared for Geogebra from the internet as well. In this section, the use of Geogebra for two position synthesis will be explained. Of course, the procedure described is not unique and one can find other ways to solve the same problem in Geogebra. After you have installed and run the program, your computer screen will be as shown in Fig. 2.18. The first menu on the tool bar is “Move” tool, with which you can move the screen or mark an object. The second menu is for “Point” tools. Selecting the point and clicking on the graph view will define a point on the graph. Another way to define a point is in the input bar type A = (x, y). This will define a point named A with coordinates x and y. Instead, you can type A = (r; θ) to locate a point using polar coordinates (r is the distance from origin and θ is the angle made with respect to positive x axis, measured in counterclockwise direction as positive). With “PointIntersect” tool, you can select the point at the intersection of two curves. When you select a point using “Point on object”, the point will only move on the object you have selected. This object can be a line, circle, surface, etc. With “line” tools you can define lines and vectors of different types. With “special line tools” for example you can determine the perpendicular bisector line. With polygon Tool different polygons can be drawn. In circle, Arc menu, a circle a circle or an arc can be drawn using different parameters. With slide tool you can define a parameter whose value can be changed. Let us consider the two positions given as shown in Fig. 2.19. The box is selected as 50 × 20 mm in size. The construction will be explained in steps. First, we must define the positions to the program and then apply the rules we have found to determine the four-bar that 4

Iriarte et al. [2].

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2 Two Positions of a Moving Plane

Fig. 2.19 Two positions of a box #1 100 50

#1

50°

20°

50

100

150

200

satisfies the given two positions. Again, let us emphasize the fact that the procedure described is not unique. Step 1. Plot two points CD on the moving plane in positions 1 by typing to the input line: Line CD represents the bottom of the box. Since two points on the moving plane define the moving body, box is not drawn (if necessary, box can be easily drawn).5 C_1 = (0, 50) C_2 = (200, 100) D_1 = C_1 + (50∗ cos(−20◦ ), 50∗ sin(−20◦ )or D_1 = C_1 + (50; −20◦ )  ◦ D_2 = C_2 + (50∗ cos 50 , 50∗ sin(50◦ )or D_2 = C_2 + (50; 50◦ ) Note that C_1 will be seen as C1 on the screen. The screen will be as shown in Fig. 2.20. This is the given two positions of the moving plane. Step 2. Go to the slider menu and create two sliders. First slider will be the distance, ra , of a point A on the moving frame from point C and the next is the angle θa , AC makes with respect to the line CD. For these two variables select ranges as: 0 ≤ ra ≤ 100 and for increments select 5 mm, 0° ≤ θa ≤ 360° and use increments as 1°. Type A_1 = C_1 + (r_a; θ_a-20°) and A_2 = C_2 + (r_a; θ_a + 50°) to locate the homologous points (Fig. 2.21). You can change θa between 0° and 360° to sweep the whole plane. This procedure can locate a point A anywhere on the moving plane. Repeat the process to define another point B on the moving plane (first select two sliders, rb and θb , and repeat). Select the “polygon” command to form a moving plane ABCD in two positions. Note that with this construction the motion of ABCD from the first position to the second is a rigid body motion no matter where you have selected A and B. (A1 , A2 ) and (B1 , B2 ) are two pairs of homologous points which can be used to attach revolute joints. Their location can be changed by the four slides controlling the values of ra , θa , rb , θb as shown in Fig. 2.22. Normally points A and B 5

To type degree sign °, type Alt + o.

2.2 Graphical Synthesis Methods for Two Positions

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Fig. 2.20 Two positions of the plane represented by C1 D1 and C2 D2 drawn on Geogebra graphics

Fig. 2.21 Determining the location of a point A at two positions (A1 and A2 )

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2 Two Positions of a Moving Plane

Fig. 2.22 Rigid body in two positions. Location of A, B arbitrarily selected by the parameters ra , θa , rb , θb

should be selected close to the box. However selection of the location of these points will depend on the particular application in which the mechanism is to be used. Step 3. Using the perpendicular bisector tool draw the perpendicular bisectors of the homologous points (A1 , A2 ) and (B1 , B2 ). These lines are denoted by a12 (perpendicular bisector of A1 A2 ) and b12 (perpendicular bisector of B1 B2 ). The point of intersection is P12 . Let us select two other slides (a and b) with range 0 ≤ a ≤ 200 and 0 ≤ b ≤ 200 and increment by 5 mm. (You can always change the limits and the increment size by right click on the slide and select “object properties”). Now draw two circles with center P12 and radii a and b. Determine the intersection of the perpendicular bisector a12 with the circle of radius a (A0 ) and the intersection of the perpendicular bisector b12 with the circle of radius b (B0 ). A0 , B0 are arbitrarily selected fixed pivot points (you can change their location by changing the values of a and b using the slide). Hide the circles. Now, you have 6 free scalar parameters defined by the slides for the four-bar mechanism which will realize the required motion in between the two given positions. Instead of selecting a and b as the parameters which show the distances of the fixed pivots A0 and B0 from the pole P12 , we can select a and b as the length of two links A0 A and B0 B to be attached between the moving and fixed planes. In such a case, A0 and B0 will be determined as the intersections of circles drawn from A1 and B1 (or A2 and B2 ) of radii a and b with the perpendicular bisectors a12 and b12 respectively. You can change the parameters freely to obtain different four-bar proportions Change the names of labels for line A0 A1 to a2 , A1 B1 to a3 and B0 B1

2.2 Graphical Synthesis Methods for Two Positions

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to a4 (Fig. 2.23). The coordinates of A0 and B0 and the link lengths can be seen on the algebra screen. Of course, you must check your result. Step 4. For the analysis, define a new slide (this slide is for the motion, it is not a design parameter). The parameter for the slide must be angle, θ and let 0° < θ < 180° in 1° increments. Select the command “angle with given size” then click on A0 A1 and on the pop-up set the angle equal to θ- slide value and select counterclockwise rotation. You will see a point A where < AA0 A1 = α =θ. Draw line A0 A. Draw a circle with center B0 and passing through B1 (the circle will automatically pass through B2 ). Draw another circle with center A and radius equal to A1 B1 = a3 . Draw a line segment from AB and B0 B thus forming the four-bar at a general position (when θ = 0, AB will be on A1 B1 ) (Fig. 2.24). You can construct ABCD quadrilateral by translating the polygon A1 B1 C1 D1 by −−→ a vector A1 A using “Translate by Vector” command. Next, we measure the angle between A1 B1 and AB lines and use “Rotate around Point” command to rotate the translated polygon by that amount to its final position ABCD. You must than hide the intermediate constructions. The result is as shown in Fig. 2.25. By changing angle θ you can move the mechanism in between the two positions and you can simulate the motion. By changing the values in slides, ra , θa , rb , θb , a and b you can obtain different four-bar proportions which are possible solutions for the problem. Three possible solutions are shown in Fig. 2.26. In Fig. 2.26a the revolute joints are selected at the

Fig. 2.23 Fixed pivots A0 , B0 selected on the perpendicular bisector to A1 A2 and B1 B2 respectively

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2 Two Positions of a Moving Plane

Fig. 2.24 Construction of the four-bar for simulation

bottom corners of the box and in Fig. 2.26b the revolute joints are located below the box. In Fig. 2.26c the revolute joints are on one side of the box. Depending on how the box is to be handled, i.e., the type of gripper and possible obstacles, any one of these solutions (or some other) may be preferrable. However, the mechanism shown in Fig. 2.26d cannot be used since it will lock in between the two positions. The result is given in TwoPosSynthesis.ggb file posted in One can solve the same problem using different methods in Geogebra.

2.3 Analytical Synthesis Methods for Two Positions Although showing the moving plane with any two points is advantageous for graphical methods, for analytical methods we usually prefer a fixed and moving coordinate frames as shown in Fig. 2.27. We have a fixed plane X, Y and a moving plane at a jth position (for example j = 1 or 2) x, y. The position of the moving frame at a position j is determined by the locating its origin with respect to the fixed reference frame using a vector expressed in complex numbers in the form: aj = aj + ibj , and the angular orientation, φj with respect to the positive x axis of the reference frame. It is left to the student that one can determine these three parameters (aj , bj and φj ) if the position of two points A, B with respect to the fixed frame at position j (Aj (xAj , yAj ), Bj (xBj , yBj ) is given.

2.3 Analytical Synthesis Methods for Two Positions

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Fig. 2.25 Possible Four-bar mechanism that satisfies the motion in between the two given positions

(a)

(b)

(c)

(d)

Fig. 2.26 Four different Four-bar proportions (a), (b), (c) and (d). (Four bar shown in (d) locks in between positions) determined by six design parameters: ra , θa , rb , θb , a and b

52 Fig. 2.27 Moving plane at jth position defined by aj and fj. Aj is the location of a point A on the moving frame at jth position relative to the fixed frame

2 Two Positions of a Moving Plane

Y

Aj

y

#j

z O'j

x

φj

Zj aj O

X

The position of a point A on the moving plane is given by a vector shown in complex numbers z = x + iy with respect to the moving frame. Note that once this point A is selected the z vector is fixed and it does not change by the motion of the moving plane. When the moving plane is at position j, the projection of the point A on the fixed plane is Aj . Note that Aj will depend on z vector and the position of the moving frame determined by the parameters aj and φj . The position vector of Aj , with respect to the fixed frame is Zj = Xj + iYj and given by the equation: Z j = a j + zeiϕ j The above equation is valid for any point A on the moving plane and for any position j. Now let us consider two positions j and k. The homologous points Aj and Ak will be given by: Z j = a j + zeiϕ j Z k = a k + zeiϕk In general, Aj and Ak will not be coincident and they will be two different points on the fixed frame. However, there is one and only one point on the moving plane whose two homologous points coincide for two positions. This point is the pole Pjk , since the moving frame will be rotating about this point when moving from position j to k. Hence Zj = Zk (Aj is coincident with Ak ) if and only if Zj = Zk = Pjk . In such a case the point A on the moving frame will be the location of the pole with respect to the moving frame: z = pjk : Z j = Z k = P j k and z = p j k a j + p j k eiϕ j = a k + p j k eiϕk or

2.3 Analytical Synthesis Methods for Two Positions

53

  p j k eiϕ j − eiϕk = a k − a j p jk =

ak i e ϕj

− aj = the location of the pole with respect to the moving frame. − ei ϕk

When we substitute pjk into Pjk :  ak − a j ei ϕ j = a j + iϕ j e − eiϕk 

Z j = Z k = P j k =a j + p j k e P jk

iϕ j

    a j eiϕ j − ei ϕk + a k − a j eiϕ j a j ei ϕ j − a j eiϕk + a k eiϕ j − a j eiϕ j = = i ϕ iϕ e j −e k eiϕ j − eiϕk

P jk =

a k eiϕ j − a j ei ϕk = the location of the pole with respect to the fixed frame eiϕ j − eiϕk

Example 2.1 Two positions of a moving plane AB are given in Fig. 2.28. Determine the pole. From the figure, if we select the positive x axis of the moving frame along the line AB, then we have: ◦

a1 = 0, a2 = 2 + 1i, φ1 = 0 , φ2 = π/2 ei ϕ1 = eio = 1 and ei ϕ2 = eiπ/2 = i hence: p12 =

a2 − a1 2 + 1i − 0 (1 + i)(2 + i) = = ei ϕ1 − eiϕ2 1−i (1 + i)(1 − i)

Fig. 2.28 Two positions of a moving plane

Y

2

B2

1

A2 B1

A1

0

1

2

X

54

2 Two Positions of a Moving Plane

p12 =

1 3 1 + 3i = + i 2 2 2

and: 2+i a2 eiϕ1 − a1 ei ϕ2 (2 + i ) ∗ 1 − 0 ∗ i = = iϕ iϕ 1 2 e −e 1−i 1−i 1 3 = + i = p12 2 2

P 12 = P 12

Also φ12 = φ2 − φ1 = π/2 (=90°). Note that for this example P12 and p12 coincide, since at the first position the moving and fixed frames are coincident (a1 = 0 and φ1 = 0°) (Fig. 2.29). Let us consider another example (Fig. 2.30). In this case: a1 = 2i, a2 = 1, φ1 = −π/2, φ2 = 0eiϕ1 = e−iπ/2 = −i and eiϕ2 = eio = 1   ◦  eio = 1 φ12 = π/2 = 90 p12 =

a2 iϕ e 1

− a1 1 − 2i (1 − i)(1 − 2i) =− = i ϕ 2 −e −i − 1 (1 + i)(1 − i) p12 =

1 3 1 + 3i = + i 2 2 2

and: 3i a2 eiϕ1 − a1 ei ϕ2 1 ∗ −i − 2i ∗ 1 = = iϕ i ϕ 1 2 e −e −i − 1 1+i 3 3 = + i 2 2

P 12 = P 12

Note that for this second example P12 and p12 vectors are not the same, since the moving and fixed coordinate systems do not coincide (Fig. 2.31). Fig. 2.29 Pole P12 for the two positions given in Fig. 2.28

Y a 12

2 b 12

x B2

P12=p12

1

y y

A2

B1 x

A1

0

1

2

X

2.3 Analytical Synthesis Methods for Two Positions Fig. 2.30 Two positions of a moving plane with first position not coincident with the fixed frame coordinates

55

Y

2

y A1 B1

1 x

y B2

A2

0 Fig. 2.31 Pole for the two positions given in Fig. 2.30

x

2

1

X

Y b12

y

2

p

A1

12

a12 P12

B1

1 x

y P12

0

A2

1

B2

x

2

X

An alternative way of determining the pole is as shown in Fig. 2.32. The two positions of the rigid body, j and k defined by (aj , φj ) and (ak , φk ) is shown. The angular rotation of the rigid body is φjk = φk –φj . There exists a point Pjk , such that the rigid body moves from position j to k by rotating about Pjk by an angle φjk . If we denote O, j Pjk = Pjk then Pjk O, k = −Pjk eiφjk . Noting δjk = ak − aj ,: Pjk − Pjk eiφjk = δjk → jk = P

δ jk 1 − eiϕ jk

As an example, referring to Fig. 2.29: δ12 = 2 + i, φ12 = π/2: → 12 = 2 + i = 1 (2 + i)(1 + i) P 1−i 2 1 3 → 12 = + i P 2 2

56

2 Two Positions of a Moving Plane

Fig. 2.32 Alternative representation of two −→ → positions. δ j k = a→k − − aj and φ jk = φk − φ j

or, referring to Fig. 2.31: a1 = 2i, a2 = 1 δ12 = a2 − a1 = 1-2i: φ12 = π/2: → 12 = 1 − 2i = 1 (1 − 2i)(1 + i) P 1−i 2 1 3 → 12 = − i P 2 2 Note that P12 = 3/2 − i/2 is measured from a reference frame whose origin is (0, 2) and x axis is horizontal. Another form of obtaining the pole is as shown in Fig. 2.32. The two positions of − → → a j , φj and a→k , φk . Let us define δ jk and φjk as: the moving plane are given by − − → → a j and φ jk = φk − φ j δ jk = a→k − − Now, the pole Pjk is located such that the moving plane is rotated about Pjk by an angle φjk to move the moving plane from position j to k. Let us show the location −−−→ −→ of the pole by a vector O ,j P jk = P jk . This vector is measured with respect to the fixed reference frame. When the moving plane moves from position j to k, the vector −→ −−−→ − P jk = P jk O ,j will rotate by an angle φjk about the pole Pjk . We have a vector loop −−,−→ −−−→, − → O j P jk + P jk O j = δ jk . Or: P j k − P j k eiφ jk = δ j k

2.3 Analytical Synthesis Methods for Two Positions

57

Solving for the vector Pjk : P jk =

δ jk 1 − ei φ jk

where Pjk , δ j k are vectors expressed in complex numbers. In certain cases, it is useful to use this form of representation. Let us consider two positions of a moving plane given by a1 , φ1 and a2 , φ2 . Next consider a point A on the moving plane is given (Fig. 2.33). Point A can be defined by a vector zA relative to the moving plane. The coordinates of A1 and A2 relative to the fixed frame will be given by the equations: A1 = a1 + z A eiφ1 A2 = a2 + z A eiφ2 If A1 is given, first determine zA = (A1 -a1 ) e −iφ 1 , then determine A2 from the second equation. Another procedure is to determine the pole P12 and then rotate the vector (A1 -P12 ) about P12 by an angle φ12 = φ2 − φ1 . Hence A2 = P12 + (A1 -P12 ) eiφ 12 . If point A is to be on a circle point, one can select the corresponding center point of the circle A0 , as any point on the perpendicular bisector to A1 A2 . To locate A0 one must determine the perpendicular bisector. Midpoint M of the line A1 A2 will be a point on the perpendicular bisector. The coordinates of the midpoint M relative to the fixed coordinate frame will be given by: A M = ( A1 + A2 )/2 Fig. 2.33 Two positions of a moving frame; selection of a circle point A on the moving frame and determining its corresponding center point A0

A0

s A1

Y

M

y #1

zA

O

a1

#2

y

x

φ1

O'1 A1

A2

φa

a2 φm

A2

X

a12

zA O'2

x φ2

58

2 Two Positions of a Moving Plane

or in Real numbers: AXM = (A1X + A2X )/2, AYM = (A1Y + A2Y )/2 The angle the line A1 A2 makes with respect to the positive X axis of the fixed frame is: φa = arg (A2 -A1 ) or in real numbers: φa = tan−1 [(A2X − A1X )/(A2Y − A1Y )]. or the slope of A1 A2 is. ma = (A2Y − A1Y )/(A2X − A1X ) (tan−1 (X; Y) means that you must use double argument inverse tangent, ATAN2(X; Y) function rather than single argument ATAN(Y/X) function). Since the bisector is perpendicular to A1 A2 , the angle this line makes with the positive X axis is φb = φa + π/2, and its slope will be given by: mb = tan(φa + π/2) = −1/ tan(φa ) = −(A2X − A1X )/(A2Y − A1Y ) Hence, we have defined perpendicular bisector as a line passing through point M and having a slope mb . Point A0 can be any point on the perpendicular bisector. Let us define this point by a length s (−∞ < s < + ∞). Then the coordinate of A0 will be given by: XA0 = (A1X + A2X )/2 + s cos(φb ); YA0 = (A1Y + A2Y )/2 + s sin(φb ) ornn XA0 = AMX + s cos(φb ), YA0 = AMY s sin(φb ) or in complex numbers: ZA0 = (Al + A2 )/2 ± seiφb = AM ± seiφb = AM ± iseiφa Instead of using the midpoint M, note that the pole is also on the perpendicular bisectors. We can also define the perpendicular bisector as line passing through P12 and having a slope mb . To determine a point A0 at a distance S from P12 on the perpendicular bisector can be determined from the equation: ZA0 = P12 ± iSeiφa Note that both s and S are measured as distances (real numbers) on the perpendicular bisector. Whereas s = 0 is when A0 is at the midpoint, M and S = 0 is when A0 is at the pole. We have 3 parameters to determine a crank AA0 . The two coordinates of A1 or A2 (A1X , A1Y or A2X , A2Y ) or the coordinate of A on the moving frame (xA , yA ) and the parameter s (or S) to determine the location of A0 . The same applies for the other crank B0 B. If s or S for A0 or B0 is selected at infinity, then we have a slider-crank mechanism. If both A0 and B0 are at infinity, a double slider result. The above procedure can be implemented in Matlab® , MathCad® or Excel® in complex numbers (or in real numbers).

2.4 Numerical Solution of Analytical Methods Using Excel®

59

2.4 Numerical Solution of Analytical Methods Using Excel® Example 2.2 Parts on a rack must be picked one by one and placed on a conveyor as shown in Fig. 2.34. In Excel® we first enter the input values into cells as: Given the two positions Q1

0

0

φ1

0

0

Q2

−650

500

φ2

50

0.872664626

x and y coordinates of the moving coordinate frame for two positions and the angular orientation are the inputs. Using these values as the input we first determine the pole using a user defined function routine. The function routine Pole (A1, Fi1, A2, Fi2) is written in Visual Basic language to solve the equation: Pjk =

ak eiφj − aj eiφk eiφj − eiφk

A1 and A2 are two element arrays, representing the coordinates of the origin of → → a1 and − a2 vectors in Fig. 2.33) and Fi1, Fi2 are the moving plane for two positions (− the angles φ1 and φ2 . We first perform the required multiplications and additions and determine the real and imaginary parts of the numerator and denominator. Next, we divide the numerator with the denominator to obtain the x and y coordinates of the pole. Written in Basic, the function routine is:

Fig. 2.34 Parts to picked from a rack and placed horizontally on a conveyor

50°

500 Y

X O

650

60

2 Two Positions of a Moving Plane

Function Pole (A1, Fi1, A2, Fi2). Dim Num(2), Denom(2), P(2), Del Num(0) = A1(1) * Cos(Fi2) − A1(2) * Sin(Fi2) − (A2(1) * Cos(Fi1) − A2(2) * Sin(Fi1)) Num(1) = A1(1) * Sin(Fi2) + A1(2) * Cos(Fi2) − (A2(1) * Sin(Fi1) + A2(2) * Cos(Fi1)) Denom(0) = Cos(Fi2) − Cos(Fi1) Denom(1) = Sin(Fi2) − Sin(Fi1) Del = Denom(0) ^ 2 + Denom(1) ^ 2 P(0) = (Num(0) * Denom(0) + Num(1) * Denom(1))/Del P(1) = (Num(1) * Denom(0) − Num(0) * Denom(1))/Del Pole = P End Function When this function is available,6 we select two adjacent cells and select this function from user defined functions and determine the pole directly (φ12 = φ2 – φ1 ). P12

−861.1267301

−446.9647492

φ12

50

0.872664626

Now, we arbitrarily select two moving pivot points A1 and B1 on the moving plane. First, we determine the distance A1 P12 (B1 P12 ) and then rotate A1 P12 about P12 by an angle φ12 to obtain A2 (B2 ) (You can only select A1 and B1 ). Select

113 13 250 150 _ Moving Pivot 1 #2 611.0019833 _ Moving Pivot 2 #2 454.0650532 Moving Pivot 1 #1 Moving Pivot 2 #1

60 _ 40 _ 46 54 484.2470734 661.9428807

Now we determine the angle made by A1 A2 , ηa (also B1 B2 , ηb ) with respect to x axis of the fixed frame and draw a line of a certain arbitrary length from P12 making an angle ηa − π/2 (ηb − π/2). These lines are the perpendicular bisectors of the homologous points. We now select the crank lengths AA0 and BB0 . These lengths must be long enough to intersect the perpendicular bisectors when drawn from the first or second positions. We determine the distance MA0 (M, B0 ). using the triangle A1 MA0 (B1 M, B0 ) (Fig. 2.36). There will be two intersections and you must select one. Using the analytical procedure explained, A0 and B0 are determined. The result is shown in Fig. 2.35. 6

When the above function is written in Visual Basic module, which is available from the developer menu, it can be exported and imported whenever needed. This function is available from Formula menu in “user defined functions” when the program is imported to the file being used. You can download these modules and Excel example files from http://makted.org.tr/kinematic-synthesis-ofmechanisms.

2.4 Numerical Solution of Analytical Methods Using Excel®

61

Fig. 2.35 Excel Chart showing the two given positions and a four-bar mechanism that realizes the motion in between the positions

Next, we determine the link lengths A0 B0 and AB and determine the angles made with respect to reference frame by A0 A1 , A0 B0 . Using FourBar2(Crank, Coupler, Rocker, Fixed, Config, Theta) function written in basic the mechanism can be simulated in between the positions to make sure that the mechanism is movable in between the positions (TwoposSynthesis Example.xlsm7 ). Another important problem is the analysis of the mechanism A0 ABB0 that has been synthesized by the analytical procedures described (Fig. 2.35). At the end of the synthesis procedure, with the selection of the three design parameters for each crank, we shall end up with four vectors: A1 , ZA0 , B1 and ZB0 (A1 is the position vector for A and B1 is the position vector for B. relative to the fixed reference frame in the first position. zA and zB are the position vectors for A and B relative to the moving frame (Fig. 2.36).

7

*.xlsm extension is for Excel files with macro. When TwoPosSynthesisExample.xlsm file is opened from the Developer menu you can open VBA editor. You will see two modules containing small function routines written in basic. BASIC module functions are for the analysis of simple mechanisms such as four-bar, slider-crank, etc. and BASIC1 module functions are for the solution of problems in complex numbers. You can export these modules to your computer and then use it in any Excel file by importing the files written in basic (with.bas extension) to your Excel file.

62

2 Two Positions of a Moving Plane

Fig. 2.36 Selection of moving and fixed pivots

In the first position, the crank will be represented by a vector A0 A1 = A1 − ZA0 . The link length a2 = |A1 − ZA0 |. Similarly, B0 B1 = B1 − ZB0 and a4 = |B1 − ZB0 |. The coupler link and the fixed link in the first position is A1 B1 = B1 − A1 , A0 B0 = ZB0 − ZA0 and a3 = |B1 − A1 |. The arguments of these complex numbers will give us the angle these vectors make with respect to the positive X axis of the reference (1) (1) (1) is frame: ψ1 (1) = Arg (A0 B0 ) and ψ2 (1) = Arg (A0 A1 ) and θ(1) 12 = ψ2 − ψ1 . θ12 the angle Link A0 A makes with the fixed link in the first position. In a similar fashion (2) (1) (2) = Arg (A0 A2) = Arg (A2 -ZA0 ). one can determine θ(2) 12 = ψ2 − ψ1 where ψ2 (1) Now increment θ12 n times between θ12 (2) and θ12 (1) by Δθ12 = θ(2) 12 − θ12 /n and determine θ12 for these n positions. Determine θ14 and θ13 by performing motion analysis of the four-bar with the known dimensions. One can select B0 B as the crank, increment θ14 and determine θ13 and θ12 . Another method is since the moving link (coupler- link 3) rotates by an angle φ12 = φ2 − φ1 , (1) we can determine θ13 (1) = Arg(A1 B1 ), θ(2) 13 = θ13 + φ12 . Use θ13 as the input and determine θ12 and θ14 . If the input link is not obvious, we must animate in different ways in order to decide the link to be used as an input.

2.5 Principle of Superposition “In physics and systems theory superposition principle, also known as superposition property, states that for all linear systems, the net response caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually” (from Wikipedia). This principle has been used in force analysis,

2.5 Principle of Superposition

63

strength calculations. In kinematic synthesis, the principle of superposition has a slightly different meaning. In certain cases, one may be required to synthesize a certain motion by an already existing mechanism or the mechanism may be too complex to yield a simple synthesis procedure. In such cases, the method of superposition can be used. The main concept is that any given plane motion is defined by the pole P12 and the amount of angular rotation about the pole φ12 . If two different plane motions (one motion is the required plane motion and the other is the motion of a link in a given mechanism) have the same angular rotation between their two positions, the required motion may be superimposed onto the plane of the link in the given mechanism. This is the principle of superposition in mechanism design. Example 2.3 In Fig. 2.37a, the two positions of a moving plane are given. We want to realize this motion by a gear and rack mechanism as shown in Fig. 2.37b. For the required motion we determine the pole P12 and the angular rotation about the pole φ12 as shown in Fig. 2.38. In order to realize this motion on a plane attached to the pinion, we must first determine the two positions of the pinion such that the angular rotation of the pinion is that of the required motion, φ12 . When the pinion rotates by φ12 (CCW) it also translates on the rack by an amount s12 = rφ12 towards left. We can draw the pinion at two potions as shown in Fig. 2.39. The next thing is to superimpose the two motions. If the figures are drawn in a computer using a CAD program, simply copy Fig. 2.38 onto Fig. 2.39, such that the poles of the two motions coincide (Fig. 2.40). Since the two planes have the same angular rotation in between the positions, the two moving planes can be rigidly fixed to each other by any means convenient. The rack and pinion will move the rigid body AB from the first position to the second, when it translates by an amount s = rφ12 (Fig. 2.41). Y

s

2

B2

1

r

θ

A2 B1

A1

0

X

1

(a)

2

(b)

Fig. 2.37 Given two positions of a plane (a), the gear and rack (b) to realize the motion in between the two positions

64 Fig. 2.38 Determination of the pole P12 and the angular rotation φ12 for two positions given in Fig. 2.39a

2 Two Positions of a Moving Plane

Y a 12

2 b 12

x

1

φ12

y

A2

y

B1 x

A1

0 Fig. 2.39 Determination of the two positions of the gear pinion such that the angular rotation in between the two positions is the same as that found for the given motion, φ12

B2

P12=p12

X

2

1

s=r φ12

d12

c12 P12

D1

Fig. 2.40 The two positions of the given motion and the pinion superimposed

Y

C2

D2

C1

s=r φ12

d12x

c12

2

B2

P12

D1

1 C2

D2

C1

B1

A1

0

A2

2.5 Principle of Superposition

65

s=r φ12 B2

A2 A1

B1

0

Second Position

First Position Fig. 2.41 The two positions of the pinion with plane AB rigidly attached

Example 2.4 We would like to realize the two positions of the moving plane given in Fig. 2.42a by a planetary gear train shown in Fig. 2.42b. 1. Determine the pole and the angular rotation about the pole for the plane AB (φ12 = 120° CCW) (Fig. 2.43). 2. Determine two positions of the planetary gear train such that the amount of angular rotation of the planet is equal to the amount of angular rotation φ12 of the plane AB between two positions. The planet must rotate by an angle φ12 =120°, which is the rotation of the plane AB in between two positions. For the planet:

30T B2 3 A1

B

12

1

0°

45 °

A2 20T 2 1

(a)

(b)

Fig. 2.42 Given the two positions of a plane AB (a) and the planetary gear mechanism to realize the motion in between the two positions (b)

66

2 Two Positions of a Moving Plane

Fig. 2.43 Determination of the pole P12 and the angular rotation φ12 for two positions given in Fig. 2.42a

a 12 b12 P12 A1

B2

B1

A2

φ =120° 12 ω planet − ωar m Tsun 20 2 =− =− =− ωsun − ωar m T planet 30 3 Since ωsun = 0, ωplanet = (1 + 2/3) ωarm = 5/3ωarm or ωarm = 3/5ωplanet . The result is: Δθarm = 3/5Δθplanet and Δθplanet = φ12 . In such a case Δθarm = 3/5*120° = 72°. For the planet to rotate 120° CCW, the arm must rotate by an angle 72° CCW. Hence, we determine the second position of the planet (Fig. 2.44). Now superimpose the motion of the plane AB onto the planet such that P12 for the two planes coincide (in Fig. 2.45, for clarity the construction lines are not shown). In Fig. 2.46 one possible construction is shown for the two positions.

2.6 Inverse Motion The motion of a body is always measured with respect to some reference that we select. This reference need not be the fixed plane. In some other cases we may want to observe the motion from the moving body itself. In that case, the reference frame, which is assumed as fixed, will now be moving. As a very simple example, if a car is moving from left to right on a straight line, an observer on the fixed plane will see that the car is traveling from left to right and the displacement s that he measures is from left to right (Fig. 2.47). Whereas an observer on the car will see a fixed post (or any other object) on the ground, moving from right to left. If the car has moved by an amount s towards right within a time Δt, the observer in the car will see the post as if it has moved by an amount x = –s (negative sign means that the motion is towards left). Consider a moving plane represented by two points AB. The two positions of the moving plane are as shown in Fig. 2.48 and for an observer sitting on the fixed frame the moving plane has rotated about P12 by a counterclockwise angle φ12 .

2.6 Inverse Motion

67

c 12 D1

d 12 C2

3 72°

C1 D2

P12 2

1

Fig. 2.44 Determination of the two positions of the planetary gear train such that the angular rotation of the planet in between the two positions is the same as that found for the given motion, f12

D1

d 12 C2

3 C1

D2

B2

P12 A1

B1

2

1

A2

=

Fig. 2.45 The two positions of the given motion and the planet superimposed

68

2 Two Positions of a Moving Plane

D1 3 C2

3 C

A1

1

D2

P12

B2

B1 2

2

1

A2

1 POSITION #1

POSITION #2

Fig. 2.46 The two positions of the planet with AB rigidly attached

Fig. 2.47 Change in the position of the observer: Observer on the ground (distance s) on the translating body (distance −s) Fig. 2.48 Two positions of the moving frame AB

2.6 Inverse Motion

69

If the observer is sitting on plane AB, he will not observe the motion of the moving body since he is sitting on the body. However, he will see that the fixed frame is moving relative to the plane AB. If the relative motion between the moving and fixed frames are the same, then the observer will not move and the fixed frame will move such that the relative positions are the same. We can determine the relative position of the fixed frame by moving and rotating the quadrilateral A2 B2 CD such that A2 B2 coincides with A1 B1 (Fig. 2.49). AB is the original position of the moving plane which is now considered as fixed). The two positions of the fixed frame relative to the moving frame is C1 D1 and C2 D2 , while the moving frame is at the first position A1 B1 . This is the inverse motion of the fixed frame relative to the moving frame. If we determine the pole and the angular rotation about the pole, we will see that the pole P12 of the original motion and the inverse motion are the same. Although the angular rotation magnitudes are the same, the angular rotation for the inverse motion is opposite to the angular rotation of the original motion: –φ12 (Fig. 2.50). Fig. 2.49 Inverse motion: moving frame fixed in the first position, fixed frame moved (keeping the relative positions the same)

C2

D2

B1 A1 C

D

1

Fig. 2.50 The pole and the rotation around the pole for the inverse motion

1

C2

−φ12

D2

c 12

P12

B1 d 12

A1 C

1

D

1

70

2 Two Positions of a Moving Plane

Analytically, the jth position of the moving frame was defined with a displacement δj and an angular rotation αj with respect to the first position (Figs. 2.32 and 2.51). In the inverse motion we move the jth position of the moving plane to the first position and rotate so that the two positions overlap each other. In this case the fixed frame will rotate by an angle –αj and it will be displaced by a vector –δj e−iαj (Fig. 2.52). Consider a four-bar mechanism (Fig. 2.53a). In the original motion, revolute joints will be on circular arcs and A0 , B0 , will be the centers of the circular arcs. Due to this characteristics revolute joint axes A, B are called “circle point” and A0 , B0 are called “center point” (Points A and B are also called “moving pivot” and A0 , B0 are called “fixed pivot”). If we invert the motion such that we fix the coupler link AB and release the fixed link. The revolute joints A0 and B0 will be on circular arcs

x y

#j

O'j

Y

αj φ1

y #1

δj

φ1

x X

O O'

j

Fig. 2.51 Motion from the first to the jth position: Displacement by δj , rotation by αj = φj -φ1

Fig. 2.52 Inverse motion: Displacement of the plane XY by −δj , rotation by −αj = −(φj -φ1 )

2.7 Relative Motion Between Two Moving Planes

71

B A

B A

A0

A0 B0

(a)

B0

(b)

Fig. 2.53 Four bar mechanism a A0 B0 fixed, A, B describe circles, b Invert the motion: AB fixed, A0 , B0 , describe circles

and A, B will be center of these arcs. What was the circle point is now the center point and what was the center point is now the circle point (Fig. 2.53b). In the previous sections, for two position synthesis, one first selected two points arbitrarily on the moving plane as the moving pivot. The fixed pivot was then selected on the perpendicular bisectors of the homologous points. With the kinematic inversion we now have another possibility. One can select the fixed pivots (E0 , F0 , say) anywhere in the fixed plane. Consider the inverse motion and determine the two positions (E0 F0 and E, 0 F, 0 ). Draw the perpendicular bisectors. You can now select E1 and F1 on the corresponding perpendicular bisectors (Fig. 2.54). A resulting mechanism E0 EFF0 with AB attached to the coupler link EF is shown in Fig. 2.55. One may use the original motion and select the moving pivots first and then determine the fixed pivot or invert the motion to select the fixed pivots first and then determine the moving pivots, depending on the constraints imposed for that particular problem.

2.7 Relative Motion Between Two Moving Planes Consider two planes moving relative to a third frame that is assumed as fixed. The motion of the two planes relative to the fixed frame can be determined as described before (Fig. 2.56). For two positions the plane E will be rotating about point P12 by an angle φ12 and the plane E, will be rotating about point P’12 by an angle ψ12 .8 Now, how will an observer sitting on plane E see the motion of plane E, . Similar to kinematic inversion, we can “subtract” the motion of plane E from the motion of E, . This can be achieved by fixing planes E and E, to each other in the second position and moving the two planes together such that plane E is at its first 8

In order to follow, the angles are selected as φ12 = 80° and ψ12 = 45° counterclockwise.

72 Fig. 2.54 In inverse motion select the fixed pivots and determine the corresponding moving pivots

Fig. 2.55 A four-bar mechanism realized using kinematic inversion shown in Fig. 2.56

2 Two Positions of a Moving Plane

2.7 Relative Motion Between Two Moving Planes

73

Fig. 2.56 Two moving frames. When E is at E1 , E, is at E, 1 and when E is at E2, E, is at E, 2

position. This motion can be performed by rotating planes E2 and E, 2 about P12 by an angle –φ12 . A person sitting on plane E will see the plane E, has moved from position E, 1 to E,, 2 . (Fig. 2.57). Relative to plane E (when E is at E1 ), E, now has two positions E, 1 and E,, 2 . According to Chasles’ theorem the motion from E, 1 to E,, 2 most simply occurs by a rotation about the pole by an angle γ12 (Fig. 2.58). This pole can again be determined at the intersection of the perpendicular bisectors of two pairs of homologous points. This time we shall call this pole “Relative pole” and denote it as R12 1 at position 1 (motion of E, relative to the moving plane E). Unlike P12 , which is the same point for Fig. 2.57 Fix E2 and E, 2 to each other and rotate such that E2 moves to its first position E1 . “Subtract” the motion of the plane E from E, . Motion from E, 1 to E,, 2 is the “relative motion” of E, with respect to E for two positions

74

2 Two Positions of a Moving Plane

Fig. 2.58 Determine the pole (R12 1 ) and the angular rotation about the pole (γ12 ) for the relative motion E, 1 to E,, 2

positions 1 and 2, the position of R12 1 is not the same on the fixed plane for the two positions. In other words, When the moving plane E is at position E2 , the relative pole R12 1 will rotate about the pole P12 and will be at a position R12 2 . We can move planes E and E, from position 1 to position 2 by rotating the planes about P12 and P’12 by an angle φ12 and ψ12 or we can first rotate plane E’1 about R12 (1) by an angle γ12 (This will bring E, to E,, 2 ) and then fix the two planes to each other and rotate both planes about P12 by an angle φ12 . This will bring the two planes to their second positions. Since rotation angles are additive in plane motion (not in spatial motion), the total rotation of the plane must not change. Hence γ12 +φ12 = ψ12 or the relative rotation angle γ12 = ψ12 − φ12 (for the case shown γ12 = 45◦ − 80◦ = −35◦ (negative sign is due to clockwise angle) (Fig. 2.59). From Fig. 2.59. | | | | |P12 R1 | = |P12 R2 | 12 12 | | , 1 | | |P R | = |P, R2 | Therefore Triangles P12 R1 P, and P12 R2 P, are equal 12 12 12 12 12 12 12 12 (Side-Side-Side) and

2.7 Relative Motion Between Two Moving Planes

75

Fig. 2.59 To move from E, 1 to E, 2 a Rotate about P, 12 by ψ12 , b Rotate from E, 1 to E,, 2 about P12 by φ12 and then rotate E, about R12 1 by γ12

∠R112 P12 R212 = φ12 ∠R112 P,12 R212 = ψ12 Hence ∠XP12 R112 = −φ12 /2 ∠XP,12 R112 = −ψ12 /2 where X is in the direction of the line P12 P,12 . Without going into a long construction, we can now determine the relative pole R12 1 as follows (Fig. 2.60): (a) Determine the poles and the angular rotation angles of the planes E and E, using the usual procedure. (b) Draw a line that makes an angle –φ12 /2 from P12 X from P12 . Draw another line that makes an angle –ψ12 /2 from P’12 X.

76

2 Two Positions of a Moving Plane

Fig. 2.60 Determining the relative pole using the poles and the angle of rotation about the poles of the two moving planes E and E,

Fig. 2.61 Determination of relative pole R12 at the second position of the moving planes (R12 2 )

(c) Point of intersection of these two lines is the position of the relative pole in the fixed plane in position 1, R12 1 . R12 2 can be determined by taking the mirror image of R12 1 with respect to P12 P,12 . You can determine the location of R12 2 in the following three different ways: 1. Rotate P12 R12 1 about P12 by an angle φ12 . 2. Rotate P, 12 R12 1 about P, 12 by an angle ψ12 . 3. Determine the image of R12 1 about the line P12 P, 12 (Drop a perpendicular and let R12 1 n = R12 2 n (Fig. 2.61).

2.8 Correlation of Crank Angles Consider two links 2 and 4 rotating about A0 and B0 . When link 2 rotates by an angle φ12 , we would like to have link 4 rotate by ψ12 . These are directed angles and are considered positive when measured counterclockwise, negative when clockwise.

2.8 Correlation of Crank Angles Fig. 2.62 Two links in a fixed axis rotation. As link to rotates by an angle φ12 , link 4 is to rotate by an angle ψ12

77

ψ

φ12

12

2 A0 (P12)

4 B 0 (P'12 )

For the two positions of these moving planes, the rotation poles are obviously A0 (P12 ) and B0 (P, 12 ) (Fig. 2.62). We want to design a four-bar mechanism to satisfy this task. We can solve this problem using two different methods, and for small angles, we can use linearization approach. Method I. Consider the relative motion between the two planes. It was shown that the motion of link 4 relative to link 2 for two positions is a rotation about the relative pole R1 12 , in the first position by an angle ψ12 − φ12 . We also know how to locate R1 12 relative to P12 P, 12 . Furthermore, whatever stated for two positions of a moving plane in Sect. 1.1. is valid for the motion of link 4 relative to link 2. First determine (R12 (1) ) by drawing a line from A0 making an angle – φ12 /2 and another line from B0 making an angle – ψ12 /2 with respect to the directed line A0 B0 . If its length is not given, A0 B0 length can be taken arbitrarily (when correlating crank angles, it’s the ratio of the links that is important). These two lines intersect at R12 (1) . This is the relative pole for the motion of the plane A0 A (or B0 B) relative to the plane B0 B (or A0 A) (A and B are to be determined). For example, if we were sitting on the plane A0 A we would see the motion of B0 B as a rotation about R12 (1) . by an angle (ψ12 - φ12 ). In this relative motion A0 B0 and AB (not known) are the two cranks (B0 , B1 are the two moving pivots in the first position and A0 A1 are the fixed pivots). In Fig. 2.63 the angles are selected as φ12 = 120◦ and ψ12 = 50◦ counterclockwise. Of course, the construction can be performed for any angle. Referring to Sect. 2.1 (see Fig. 2.10), it was shown that the cranks subtend equal angles at the pole (and this angle is half the rotation angle between the two positions). The same is true for the relative motion. When we fix A0 A (link 2), AB and A0 B0 are Fig. 2.63 Determine the relative pole R12 1

78

2 Two Positions of a Moving Plane

the two cranks and B0 B (link4) is the moving link. Hence the following construction results (Fig. 2.64): 1. From R12 (1) draw a line (any line) L1 . 2. Draw another line, L2 from R12 (1) such that ∠L1 R12 (1) L2 = ∠A0 R12 (1) B0 = α = (φ12 ψ12 ). 3. Select A1 anywhere on L1 and B1 anywhere on L2 . In the resulting four-bar mechanism when A0 A1 is rotated by −φ12 , B0 B1 will rotate by ψ12 . 4. Check your result (Fig. 2.65). This construction is because “the cranks subtend the same angle at the pole”, which was proven for two positions of a moving plane (see Sect. 2.1). Although it is guaranteed that the two positions exist, there is no information for the motion in between the two positions. Therefore, the result must always be checked (the mechanism may lock or the transmission angle may be unacceptable in between the two positions). There are infinitely many ways to select L1 . When L1 is selected, L2 line is fixed. However, there are infinitely many ways to select A1 and B1 on L1 and L2 . Hence, there are 3 scalar parameters that can be selected freely. Also, note that you can change the order of construction (i.e. first select L2 , A1 or B1 ) as long as you make sure < A1 R12 (1) B1 = < A0 R12 (1) B0 = α = (φ12 − ψ12 ).

Fig. 2.64 To locate A1 and B1 draw two lines such that ∠A0 R12 1 B0 = ∠L1 R12 1 L2

Fig. 2.65 Select A1 and B1 on lines L1 and L2 arbitrarily

2.8 Correlation of Crank Angles

79

ψ12

Fig. 2.66 Select B1 arbitrarily and determine B2 (∠B1 B0 B2 = φ12 )

B2

A

B1

B0

0

Method II. In the second method we can invert the motion directly, i.e., keep A0 A or B0 B fixed and release A0 B0 . If we keep A0 A as fixed, then A0 B0 must rotate about A0 by an angle –φ12 . Hence, we have the following procedure: 1. Select B1 arbitrarily. Rotate B0 B1 by ψ12 to determine B0 B2 . These are the two positions for link B0 B relative to the fixed link (Fig. 2.66). 2. Fix A0 B0 and B0 B2 to each other and rotate these two planes about A0 by an angle −φ12 . You have “subtracted” or “eliminated” the motion of the plane A0 A from that of B0 B. The motion of the plane B0 B from B0 B1 to B, 0 B, 2 is the relative motion of B0 B with respect to A0 A for two positions (Fig. 2.67). 3. Draw the perpendicular bisector to B1 B, 2 . A1 can be selected as any point on the perpendicular bisector (Fig. 2.68). One possible solution is shown in Fig. 2.69. Again note that there is no guarantee for the motion in between the positions. In fact, the solution shown in Fig. 2.69 may not be a good solution since the transmission angle in the first position is rather small and for the motion between two positions, link B0 B will first rotate clockwise and then rotate counterclockwise, which may not be desirable.

B1

Fig. 2.67 Rotate A0 B0 B2 by an angle −φ12 about A0 . A0 B, 0 B, 2 is the rotated position

A

0

B0 −φ

12

B'0

B'2

80

2 Two Positions of a Moving Plane

Fig. 2.68 Draw the perpendicular bisector (b12 ) to B1 B, 2 . A1 can be selected anywhere on b12

Fig. 2.69 Four-bar A0 A1 B1 B0 is a possible solution. When A0 A rotates by φ12 , B0 B will rotate by ψ12

Example 2.5 We would like to have an output link to rotate 60° CCW corresponding to an input crank rotation of 160° CCW. In this case let us invert the motion such that the output link is fixed. Following the steps explained for method 2, first we select A0 A1 and rotate it by angle φ12 = 160° CCW to determine A2 . Next, we rotate B0 A0 A2 about B0 by angle –ψ12 = 60° CW. To determine A, 2 , we can select B1 anywhere on the perpendicular bisector a, 12 . However, if we select B1 such that B0 B1 is perpendicular to a, 12 , then we can make the deviation of the transmission angle from 90° at the two given positions nearly equal and transmission angle is 90° somewhere in between the two positions (Fig. 2.70).

2.8 Correlation of Crank Angles

81

Fig. 2.70 Determine the second position of A0 A relative to B0 B in the first position. B1 can be selected anywhere on the perpendicular bisector to A1 A, 2 (a12 ). For better transmission angle characteristics B1 is selected such that B0 B1 is perpendicular to a12

Fig. 2.71 A possible four-bar mechanism solution

The mechanism is shown in Fig. 2.71 at its two positions. While satisfying the positions the transmission angle is quite satisfactory. Linearization approach: Perhaps one of the milestones of human civilization is the invention of the lever by Archimedes (200 BC). As shown in Fig. 2.72, the basic idea is that, neglecting energy loss, force times displacement (work) is constant. The displacement at A and B when the lever is horizontal, is linearly related by the ratio: Δθ |A0 B| ∼ = s B and Δθ |A0 A| ∼ = sA

Fig. 2.72 Lever

82

2 Two Positions of a Moving Plane

Fig. 2.73 Four-bar mechanism where the input and output rotations are approximately linearly related for small rotations

Therefore: s A ∼ |A0 A| = |A0 B| sB Let us now extend this idea. Referring to Fig. 2.73a, consider a four-bar mechanism with the cranks parallel to each other. At the instant shown, the velocity of A and B will be equal (vA = vB ) or considering small displacements ΔsA ~ ΔsB . Also, |A0 A|Δφ∼ = ΔsA and |B0 B|Δψ∼ = ΔsB . We have: (If A0 and B0 are selected on the same side, the links will rotate in the same direction (Fig. 2.73b). If the input and output is to rotate in opposite directions, A0 and B0 must be on opposite sides of AB (Fig. 2.73c). Δψ B0 B ψ12 ≈ = Δφ φ12 A0 A I believe especially in the early years of civilization (and even today in most machine shops) the artisans constructed simple machines using this concept. In this construction, note that the transmission angle is also at or near optimum, which ever link is driving.

2.9 Correlation of Slider Displacement with Crank Rotation While one link is connected to the fixed link by a revolute joint, the other link is connected by a slider. As the crank rotates by an angle φ12 , we would like to have the slider to be displaced by an amount s12 along the slider axis (Fig. 2.74). For the crank, A0 is the rotation pole P12 . Whereas for the slide the pole is at infinity in the direction perpendicular to the slider axis. Note that if the two links were rotating at a pole located at a finite space, the relative pole was found by drawing two lines making angles –φ12 /2 and –ψ12 /2 with respect to P12 P, 12 . Now instead of ψ12 we have s12 —a linear displacement. Which is the arc length when the pole is at infinity. Hence the relative pole is determined at the intersection of a line making

2.9 Correlation of Slider Displacement with Crank Rotation

83

Fig. 2.74 Two positions of a rotating and a sliding link to be correlated

an angle—φ12 /2 with respect to P12 P, 12 (P, 12 is at infinity perpendicular to the slider axis) and a line parallel to P12 P, 12 at a distance—s12 /2 (Fig. 2.75). One can use either method 1 or method 2 as described in Sect. 2.6. The only difference is instead of angle ψ12 , one must use the displacement s12 . Method 1 is shown in Fig. 2.75 and Method 2 is shown in Fig. 2.76 (with slider assumed fixed for the relative motion). Fig. 2.75 Correlation of crank angle with slider displacement using Method I

Fig. 2.76 Correlation of crank angle with slider displacement using Method II

84

2 Two Positions of a Moving Plane

2.10 Design of Six Link Mechanisms for Two Positions Relative motion can occur between any two moving bodies and one can synthesize mechanisms with higher number of links. In cases where simple four-link mechanisms do not yield a feasible solution, one may seek mechanisms with higher number of links. For example, if the planar mechanism with one degree-of- freedom and with revolute or prismatic joints are to be used, in the order of complexity after the fourlink mechanisms the next mechanism group is six link mechanisms (of Watt or Stephenson type). Example 2.6 Consider designing a lamp post for a table. We want to move the lamp horizontally in between two positions. The moving links must be within the trapezoidal area shown (Fig. 2.77). Also, we can only place the fixed pivots on the table on the dashed rectangular area as shown. Although there are infinitely many solutions, one cannot possibly design a four-bar mechanism that will move the lamp in between the two given positions and satisfy the constraints. Therefore, we shall seek a six-link mechanism with revolute joints. Whenever the number of links increase, the number of design parameters also increase. We are free to select some link lengths beforehand. Consider attaching two links (called dyad) as shown in Fig. 2.78. (We can select A1 and B1 freely). We have an open chain with three degrees of freedom (This is like a robot arm: by attaching 3 motors for each joint, you can move this rigid body from its first position to the second!!). We want to reduce the degree of freedom to one. Consider the motion of the lamp relative to the moving plane A0 A. Select another moving pivot on the lamp (C1 in the first position). The relative motion is obtained when we “subtract” or “eliminate” the motion of A0 A from the motion of the lamp. Therefore, fix the three links (A0 A, AB and BC- lamp) at position two and rotate about A0 such that A2 coincides with A1 . The result is shown in Fig. 2.79. The motion from B1 C1 to B, 2 C, 2 is the relative motion of the lamp with respect to A0 A. One can now draw the perpendicular bisector to C1 C, 2 and select a point D1 on link A0 A. Now the system has two degrees of freedom. To move the lamp from the first position to the second first Rotate A1 B1 about A1 to A1 B, 2 (this will automatically Fig. 2.77 Two positions of a lamp. Fixed pivots can only be placed on the rectangular area on the left

2.10 Design of Six Link Mechanisms for Two Positions

85

Fig. 2.78 Construction of a three degrees of freedom open chain arbitrarily

move C1 to C, 2 ), then fix A0 A1 , A1 B, 2 to each other and rotate the links about A0 to move A1 to A2 . In order to have a constrained motion, the motion of AB or CD must result from the motion of A0 A. Let us select a point E (moving pivot) on link CD (not necessarily on the line CD, on the plane of CD. However, it is preferable to have this point close to the line CD You can as well select a pivot point on AB). E is located at E1 and E2 for two positions. Draw the perpendicular bisector e12 . Now you can Select E0 on e12 (Fig. 2.80). For a four-bar mechanism we have seen that there were 6 scalar design parameters (∞6 ). However, as it can be seen in this problem, no solution was obtained. Now in this case there are 10 scalar design parameters[selection of A, B, C, E (4*2 = 8), Fig. 2.79 Relative motion of the lamp with respect to link A0 A (in position 1) and construction of link DC to reduce the degree of freedom by one

86

2 Two Positions of a Moving Plane

Fig. 2.80 Considering two positions of link AB with respect to the fixed link, selecting a point E1 on link AB as moving pivot and determining a corresponding fixed pivot (E0 ). As a result a six link mechanism is designed with one dof

selection of D, E0 on their respective perpendicular bisector lines (2*1 = 2)]. Final sketch of the mechanism constructed at the given two positions is shown in Fig. 2.81. Example 2.7 Two positions of a garage door are as shown in Fig. 2.82. You can design a slider crank (Fig. 2.83) or a double slide mechanism to move the door in between the given two positions (such designs do exist in practice). However, when you try to

1 1

A0

Fig. 2.81 Mechanism in the given two positions of the lamp

87

7°

900 mm

300 mm

2.10 Design of Six Link Mechanisms for Two Positions

60°

2m

#1

#2

Fig. 2.82 Two positions of a garage door

find a four-bar mechanism for the synthesis of this motion, you realize that the door interferes with the ceiling in between the two positions. A sliding joint is usually an element that must be avoided due to high friction and misalignment properties (one must use two identical mechanisms at the two sides of the door and the two sides is usually greater than 2–3 m. At such a distance a slight misalignment of the two slides at the opposite sides of the wall or deflection of the parts due to their weight will lock the mechanism. An alternative would be to design a six-link mechanism to synthesize this motion. In such a case, first let us select a moving pivot A behind the door close to the bottom (but not exactly at the edge) and select corresponding fixed pivot A0 on the perpendicular bisector and on the vertical column that will be fixed to the wall at the two sides of the door. The angular rotation of the crank A0 A in between positions 1 and 2 is measured as ∠A1 A0 A2 = 145° (CW). Select a moving pivot B (B1 ) on this crank and select a fixed pivot C0 anywhere in the fixed plane (Again you would prefer to select C0 on the vertical column). Assume the crank C0 C is to rotate 52° in between the two positions (you can select another angle as well). Now, design a four-bar mechanism A0 BCC0 , in which as A0 B rotates by 124° (CW), C0 C rotates by 52° (CCW). Locate the relative pole (R12 (1) ) by drawing lines that make angles 72.5° (CCW) from A0 and 26° (CW) from C0 with respect to the line A0 C0 . Then draw line Lb and Lc such that Lb passes from the arbitrarily selected point B1 and ∠Lb RLc = ∠A0 RC0 . Select C1 anywhere on Lc (Fig. 2.84).

2 Two Positions of a Moving Plane

300 mm

88

7°

900 mm

2m

#1

60°

#2

Fig. 2.83 Realization of two positions by a slider crank mechanism

26°

Lc

C1

C0

A1

α

Lb

α

B1 ° .50 72

60°

145

°

2m

R112 A0 B2 A2

Fig. 2.84 Select fixed and moving pivots A and A0 for a crank. Construct a four-bar A0 BCC0 to correlate the CW angular rotation of the crank A0 A to another to an arbitrary CCW of C0 C

2.10 Design of Six Link Mechanisms for Two Positions

Q1

89

c'12 D1

Q2

C1

A1

C0 B1

52

C'2 C2

60°

145

°

2m

R112 A0 B2 A2

Fig. 2.85 Determine C1 and C2 . Invert the motion such that the door is fixed in the first position. Determine C, 2 , for the inverse motion. Select a point D1 on the door located at the perpendicular bisector of C1 C, 2

Now, we have a point C (C1 and C2 at the two positions). If we attach a link between point C and the door (say point D) by revolute joints, we shall obtain a six link mechanism that will guide the door in between the given two positions (Fig. 2.85). To find the point D where this link must be attached by a revolute joint to the door, note that point C will describe a circle relative to the door. To find this relative motion move the triangle A2 C2 Q2 such that A2 Q2 coincides with A1 Q1 . The new position of C2 is C2 , . Point D1 can be selected as any point on the perpendicular bisector to C1 C2 , . Hence, we have completed the design as shown in Fig. 2.86. Of course, one may come up with a better solution. There are 10 free parameters. Hence there are plenty of possibilities to come up with a good design. You must always check your result. If you cannot satisfy yourself with the software simulation, then you must make a model of the mechanism using cardboard, wood or the like, to check your result. Example 2.8 Figure 2.87 is from CN104139841 issued on November 12 2014 for a steering column of an earth moving vehicle such as a grader which can be for sitting and standing positions of the operator. The mechanism used is basically a four-bar mechanism. A similar patent US4664221 issued on May 12, 1987 “Function control linkage for a vehicle” describe a similar design. The steering wheel is used at two positions of the driver. The two positions of the steering wheel are as shown in Fig. 2.88. In one case the driver is in a seated position. In the other case the driver stands up and drives the vehicle.

90

2 Two Positions of a Moving Plane

Q1

D1

Q2 C1

A1

C0

C0

B1 C2

D2

60°

2m

R112

A0 B2 A2

Fig. 2.86 One degree of freedom six link door mechanism in two positions

Fig. 2.87 Steering wheel support for a grader. A Four bar mechanism

Kinematically we can state the requirement as: when the steering column rotates by an angle 50° in counterclockwise direction, the steering wheel must rotate 50° in clockwise direction. A company wanted to obtain a similar motion using a mechanism other than the four-bar. One possible solution is to use a planetary chain drive with fixed sprocket twice the size of the moving sprocket as shown in Fig. 2.89. When the arm is rotated 50° CCW, the planet (to which the steering wheel is attached) rotates 50° CW if the gear

2.10 Design of Six Link Mechanisms for Two Positions

91

Fig. 2.88 Solid model of the steering wheel support for two positions Fig. 2.89 Cardanic motion utilizing planetary chain drive

Moving sprocker Radius=r

Fixed sprocker Radius=2r

ratio is 2:1. This is known as “Cardioid motion”, which is the inverse of Cardanic motion.9 Another alternative way of eliminating the patent is to use a six link mechanism for the same task. Consider the two positions of the steering column (A0 A) and the steering wheel (AE) as shown in Fig. 2.90. Assume A0 A1 E1 and φ12 , ψ12 given. Let us now select a point B1 on the column (link 2) and attach a link (3) B1 C1 . When link 2 rotates by an angle φ12 (CCW) select an angular rotation for link B1 C1 as η12 (CCW in the figure). Determine B2 C2 . Draw the perpendicular bisector to C1 C2 and select C0 on the perpendicular bisector (Fig. 2.91). We thus have a fourbar mechanism in which as the crank A0 B rotates by an angle φ12 , the coupler link BC rotates by an angle η12 . Now consider the motion of link 6 (AE) relative to link 3 (BC) (Fig. 2.92). We eliminate the motion of BC from AE by moving the quadrilateral B2 C2 A2 E2 such that 9

Zwikker [3].

92

2 Two Positions of a Moving Plane

Fig. 2.90 Steering column A0 A and the steering wheel platform AE in two positions

ψ

12

E2

A2

φ

E1

12

6 A1

B1

B2

2

A0 Fig. 2.91 Construct a four-bar A0 BCC0 such that BC rotates by an angle η12 in between the two positions

E2

A2

E1

C2 B2

6

12

c 12

A1

B1

η

3

C0

2

C1 A0

B2 C2 coincides with B1 C1 . The relative second position of AE is A, 2 E, 2 . Determine the perpendicular bisector to E1 E, 2 as D1 on link 3. We have designed a six-link mechanism. The final mechanism at two positions is shown in Fig. 2.93 and its solid model representation is given in Fig. 2.94.

2.10 Design of Six Link Mechanisms for Two Positions

93

E2

A2

E1

C2 e'12

6 A1

B1

B2

2

D1 3

C0

C1 A0

A'2

E'2 Fig. 2.92 Invert the motion by keeping BC fixed Determine D1 as the centerpoint for E in this relative motion between two positions

E2

A2

E1

C2

6 5 B1

B2 C0

C0

4

D1 3 C1

A0

A0

Fig. 2.93 Six Link mechanism for the steering column

2

A1

94

2 Two Positions of a Moving Plane

Fig. 2.94 Solid model of the steering column

Of course, some effort is required to obtain a reasonably optimum solution. Note that space requirements play a very important role. Problems

20

#1

36

43°

53

Fig. 2.95 Two positiond of a rigid body to be moved by rotation only

10

1. The rigid body shown in the Fig. 2.95 is to be guided from position #1 to position #2 by a rotation about a single fixed pivot. Locate the pivot.

30 #2

2.10 Design of Six Link Mechanisms for Two Positions

95

Fig. 2.96 Open and closed position of a missile wing

B2

A1

A2

β

α α

B1 C

2. Figure 2.96 shows one of the four folding wings of a missile. The curved wing of radius R subtends an angle β. When the wing is at its open position, A2 B2 and C (the axis of the missile) must be collinear. Since there are four wings: 2α + β = π/2 Determine the location of the hinge about which the wing can be rotated in terms of R and α in its most simple form. What is the amount of rotation of the wing? (Hint: Select your reference axis such that the pole is in one of the principal directions). 3. In Sect. 2.2.2 when describing a procedure for two position synthesis using Geogebra, in step 2, for the selection of moving points (A or B) the distance from two reference points were selected (a,b for Ai and c,d, for Bi ). Instead, Show that one can also use (ra ,θa ) to locate Ai and (rb ,θb ) to locate Bi . 4. Write a Matlab m file such that given the two positions, the file determines the rotation φ12 between the two positions and the location of the pole P12 . 5. You are to prepare a computation file in a computer enviroment of your choice [i.e. mathematical packages such as Matcad® , or Matlab® or a spread program (such as Excel® ) or write a program in one of the computer languages (C, C+, Pascal, Visual Basic, Fortran, etc.)]. The user will input the two positions of a moving plane. Then the user will select the coordinates of two points A, B on the moving body to be used as moving pivots (circle points) or two positions on the frame (A0 , B0 ) to be used as fixed pivots (center points). The computer will determine the perpendicular bisectors and using a scalar parameter the user will select the fixed pivot (center point) or the moving pivot (circle point).

96

2 Two Positions of a Moving Plane

Visual effects and the ease of use will make the program more effective. The solution of the following problems (problems 5 to 16) can be performed graphically or analytically. 6. You are to design a “Hinge” for a ventilation roof window (or attic window). The window is shown at its closed position in Fig. 2.97. Some overlap is required to make the window waterproof. The open position is not shown, but it must be such that it is away from the opening (so that you can climb the roof). The hinge must be inside the opening so that nobody can open the door from outside and the hinge is away from the extreme outside wheather conditions). Design this “Hinge” (as you can expect, this hinge must be of a four-bar type). 7. The back of the railway carriage seat is to be reversible as shown so that the passangers can sit in either way with respect to the direction of travel. Design a four-bar mechanism for this purpose. In the actual construction two identical four-bars will be placed on each side of the seat. You must first think where can you place the moving and fixed pivots of the four-bar. After the design you must check the movability and how you can construct such a four-bar. Also note that the passangers are going to sit from one side of the seat. The links should not block the sides at these two positions (Fig. 2.98). 8. Figure 2.99 shows two positions of a folding seat used in aisles of busses to accommodate extra seated passengers. In the open position the extra seat is at the aisle and is in a stable position so that a person can sit. There are two different designs. In (a) when folded, the seat is below the passenger seats. In design (b) the folded position of the seat is the arm rest for the passenger seat. Design a four-bar mechanism to support the seat so that it will lock in the open position and fold to a closing position as shown. Note that in the closed position there can be no link inside the aisle and you cannot place a fixed pivot outside the bus!. 9. In an automatic packaging machinery, the boxes coming from a conveyor are to be turned right-side-up and placed onto another conveyor as shown. Design a four-bar mechanism for this purpose. Also think how you can pick, hold and place the boxes during this transfer (Fig. 2.100). 10. Figure 2.101 shows the door of a bus in open and closed positions. You are to design a four-bar mechanism to synthesize this motion. 11. Consider the baggage compartment in cars. In Fig. 2.102a and b the open and closed positions of the hood is shown. You are to formulate this problem as a 1000

Fig. 2.97 Ventilation window in closed position

950

25

2.10 Design of Six Link Mechanisms for Two Positions

97

Fig. 2.98 Railway carriage seat which can face either side

Fig. 2.99 Two different folding seat arrangements to be used for extra passanger. When folded extra seat is a) below the normal seat, b) an arm rest for the normal seat

two position synthesis problem. Since the edge of the hood is of curved shape, the hood must translate forwards as shown in Fig. 2.102c and rotate for the open position. Design a four-bar mechanism for the solution of this problem. 12. In Fig. 2.103 the open and closed positions of a engine hood is shown. Design a four-bar mechanism to guide the car hood considering all the practical limitations. Also investigate the car hoods in existing cars. There are cars that use a

98

2 Two Positions of a Moving Plane

Fig. 2.100 Turning a package 90° when moved from one conveyor to the other

Fig. 2.101 Door of a bus in open an closed positions

simple hinge, a four-bar and a six-link mechanism. Can you use a simple hinge? Why? 13. You are to design collapsible seat for city bu sor metro trains. The open and closed positions are as shown in Fig. 2.104. Dimensions are in centimeters. When closed all links must be within 10 cm to the wall. Design a four-bar mechanism for this task. 14. We would like to guide a door between two positions as shown in Fig. 2.105a. Figure 2.105b shows a possible construction (not to scale) using a four-bar mechanism (note that the two cranks are made from steel strips). Design a four-bar mechanism and show how it can be constructed.

2.10 Design of Six Link Mechanisms for Two Positions

Fig. 2.102 Baggage hood of a car Fig. 2.103 Engine hood of a car

99

100

2 Two Positions of a Moving Plane

35 25

40

5

10

18

2

3

Fig. 2.104 Collapsible bus or metro seat

(a)

(b)

Fig. 2.105 Cabinet door hinge

15. Kitchen Cabinet doors shown in Fig. 2.106 must be hinged to the side walls of the cabinet wall such that when closed, the side wall is hidden from sight and it can be opened with no interference with the adjacent door or the cabinet wall. The two positions of the two adjacent doors are as shown. You are asked to design a standard hinge with such characteristics. (use the two positions of the door on the right. The motion of the other door is the image of the first door.) You may or may not find a feasible solution using a four-bar. In which case a six link mechanism solution may be considered. See how this has been realized in practice.

2.10 Design of Six Link Mechanisms for Two Positions

101

Side Wall

20

18

8 Door

Door #1 8

#1

#1 #2 Door

Door Closed

Door #2

#1 #2

#2

Door Open

Fig. 2.106 Another type of cabinet door in open and closed positions

15. In Fig. 2.107a shows another kitchen cabinet door which is opened upwards. Because of the other cabinet doors, the two positions are as shown in the drawing (b). Design a suitable four-bar mechanism. Why is spring used? Think how this spring will be placed. 16. Figure 2.108 shows a kitchen cabinet door in closed, midpoint and fully open positions. Consider a 1000 mm high cabinet (say width is 800 mm). The two halves of the door will be 500 mm high. Design a suitable four-bar mechanism. (You can decide on the open and closed positions of the door yourself). Problems on the principle of superposition 17. Two positions of a plane AB is given in Fig. 2.109 (A1 B1 and A2 B2 ). We would like to realize this motion by a double slider mechanism schematically as shown. In the existing mechanism axes of sliders are perpendicular and length CD = l = 80 mm. Superimpose the required motion of the plane AB on the existing double slider mechanism. (Hint: First choose an appropriate initial configuration for the double slider). 18. Two positions of a plane are defined as (Fig. 2.110). A1 B1 : a1 = 0, φ1 = 0

◦

A2 B2 : a2 = 2 + i, φ2 = 90

◦

102

2 Two Positions of a Moving Plane Upper Door Cabinet

20

10

Door Open

30

P12

20 Cabinet

70

°

Door Closed

(a)

(b)

Fig. 2.107 Cabinet door opened upwards

Fig. 2.108 Another Cabinet door. Two shelves are opened at the same time

AB = 1 unit (let 1 unit = 30 mm). Synthesize this motion by means of: (a) A planetary gear train with gear ratio = –2 (planet being the small gear)

2.10 Design of Six Link Mechanisms for Two Positions

103

Fig. 2.109 Two positions of a plane and the mechanism to realize this motion Fig. 2.110 Representation of a moving plane by vector aj and angle φj

Y y #j

Aj

x

φj

aj O

X

(e.g. if R = radius of the sun gear and r = radius of the planet: R = 2r) (b) By a gear and rack. Relative motion: Correlation of crank angles 19. Given θ12 and φ12 . How should A1 be chosen so that B1 can be anywhere in the plane of the drawing? 20. If in a four-bar mechanism θ12 , φ1 , φ2 , Ao A, Bo B are prescribed, determine a simple construction for the remaining mechanism proportions. Hence, synthesize a four-bar mechanism such that: θ12 = 90◦ CW, φ1 = 46◦ , φ2 = 120◦

Ao A = 1 unit Ao Bo = 4 units

21. Find the proportions of a four-bar mechanism with A0 B0 = 80 mm; driving crank (A0 A) = 26 mm; θ12 = 180° CCW; φ12 = 90° CCW; driven crank (B0 B) = 40 mm (Fig. 2.111). Draw the mechanism in two positions and state whether it is possible to go from position #1 to position #2 without temporary disconnection.

104

2 Two Positions of a Moving Plane

Fig. 2.111 Cranks A0 A = 26 mm and B0 B = 40 mm and A0 B0 = 80 mm to be connected by a link to form a four-bar, satisfying correlated crank rotations

Hint: since θ12 = 180°, we must have a crank-rocker mechanism proportions. Use Grashof’s rule and also note that for crank-rocker proportions, the rocker (B0 B) must not pass the fixed line of centers (A0 B0 ). 22. Two rigid bodies A and B have the following displacements (Fig. 2.112): A1 (0, 0), φ1 = 0

◦

A2 (1, 0), φ2 = 40

B1 (1, 1), φ1 = 30 ◦

◦

B2 (−1, 0), φ2 = 90

◦

Synthesize the proportions of a planetary gear train with A as the arm and B as the planet. What is the gear ratio? What will happen if B is the arm and A is the planet? 23. Design a four-bar mechanism to convert a clockwise rotation of 80° into: (a) A clockwise rotation of 90° (b) A counterclockwise rotation of 90°. Also the maximum deviation of the transmission angle from 90° must not be greater then 45°. 24. Synthesize a slider-crank mechanism for a slider displacement of 40 mm (S12 ) from left to right and a corresponding crank rotation of (a) 90° CW: (b) 90° CCW. 25. Design a four-bar mechanism with the following crank rotations φ12 = 60° CW; φ12 = 90° CW. Due to the design requirement, we would like to have doublecrank (drag-link) proportions. Take Ao A = 4 units Ao Bo = 1 unit and Bo B = 3 units. Fig. 2.112 Two moving planes to be correlated for two positions by a planetary gear train

2.10 Design of Six Link Mechanisms for Two Positions

105

26. Design a four-bar mechanism such that 60° CW rotation of one crank results in 35° rotation of the other crank. Solve the problem twice by taking 35° crank rotation clockwise and counterclockwise (also, solve the two problems in two different methods discussed in the text. Also consider the force transmission characteristics). 27. Design a slider crank mechanism such that when the crank is rotated 48° CW the slider is displaced 100 mm towards right in horizontal direction. Repeat with crank rotating 480 CCW. Also, we would like to have optimum force transmission characteristics when a) crank is the input, slide is the output b) Slide is the input, crank is the output. 28. You are to design a four-bar mechanism such that a 40° rotation of one crank results in 30° rotation of the other crank. Solve the problem using the relative pole and using linearization approach. 29. Figure 2.113 shows a 6-link mechanism to convert a slider motion to a large link oscillation. The reason why a 6-link is used instead of a slider crank mechanism is that the deviations of the transmission angles involved from 90° must be less than 45°, which would be impossible with a single slider-crank mechanism. Note that the 6-link can be broken up into a slider-crank and a four-bar mechanism. Links 1,2,3 and 4 form the slider-crank mechanism in which link 2 is the input and link 4 is the output (what is the transmission angle?) Links 1, 4, 5 and 6 form the four-bar mechanism which is driven by the slider-crank mechanism. s12 = 250, D0 D = 125, θ12 = 100◦ CCW, φ12 = 160◦ CCW Hint: You have to synthesize each mechanism separately for the given motion. e.g. For the given s12 and θ12 design the slider-crank mechanism and check that the transmission angle deviation is less than 45° for this mechanism. (the minimum and the maximum values of the transmission angle will occur at Fig. 2.113 Slider crank and four-bar mechanism in series (forming a six link) to convert a slider displacement to a large angular oscillation

106

2 Two Positions of a Moving Plane

Fig. 2.114 Garbage can with a lid to be opened by a foot pedal (not shown)

the two extreme positions). Next design the four-bar mechanism for the given θ12 and φ12 and check the transmisston angle. 30. Design a garbage can with a lid (Fig. 2.114). The lid is to be opened by a foot pedal (not shown). 31. You are to design a furniture in which when the top of the cabinet is opened the front of the cabinet is also opened up thus resulting as a nice table (Fig. 2.115). Such a furniture may be used as liquor storage and service table or as a small writing table. (The mechanism must be inside the cabinet when closed. Remember the cabinet when closed, forms a closed volume.) Hint: think of two four-bars in series. Relative motion, general 32. Figure 2.116 shows a portable chair in the open position such that when closed, all the rigid bodies involved collapse onto link 1. The global dimensions are as shown. Determine the link length dimensions for the chair. 33. In certain design problems, we may have to design a mechanism with two degrees of freedom. Figure 2.117 shows a mechanism encountered in the design of a textile machine. When φ12 and y12 are given inputs for the two positions of links 2 and 5 respectively, link 4 must move by an amount x12 relative to link 5. Determine a method for the synthesis of this mechanism. Design a mechanism for which φ12 = 45°, y12 = 100 mm and x12 = 500 mm 34. Figure 2.118 shows a trailer which carries bulk load such as coal, beat, etc. (Such systems are also used in railway cars). In order to dump the load to the side rather than rotating the whole system by 90°, the side of the trailer is made as two seperate pieces as shown (Planes E and F). When the trailer (shown as plane G) is rotated by 45°, Plane F rotates so that it is level with the bottom of the trailer, Plane E rotates by an angle 40° relative to the trailer so that it does

2.10 Design of Six Link Mechanisms for Two Positions

107

not create any obstacle for the falling material. Design a mechanism to realize this motion. 35. When generating waves in a laboratory, a flap is placed inside a channel as shown. The wave magnitude generated is proportional to the amount of water displaced. A wave generator in A hydraulics lab is as shown in Fig. 2.119. The upper portion of the flap is attached to sliders on both sides of the channel wall. Link 2 is driven by a four-bar mechanism (not shown). You are to design this six-link mechanism fort he two positions of the flap shown in Fig. 2.120 while the crank (link 2) oscillates by φ12 = 60° 36. Let us reconsider the attic door problem. The ventilation door is to be rotated by 110° and translated as shown in Fig. 2.121. In such a case a feasible solution using a four bar mechanism was not possible. Also, we would like to place the moving pivots in the rectangular region labeled as A under the door and the fixed pivots in a rectangular region B near the side wall. Design a six-link mechanism to realize this motion. 37. In kitchens when there are corners it is hard to use the space. In this particular example there are two shelves inside the cupboard as shown in Fig. 2.122.10 The motion for the two shelves are as follows. i. Shelf#1 slides outwards by approximately 380 mm (door is fixed to Shelf#1). There is a slide on link 2 which is hinged to the frame at A0 . During this time

Fig. 2.115 A cabinet forming a closed container when closed and a table when open

10

https://www.hafele.com/us/en/info/residential/kitchen-solutions/corner-solutions/6995/.

108

2 Two Positions of a Moving Plane

Front foot and the backrest (one piece) (1)

1 Arm rest (6)

5 Seat

3

Back foot

2

(4)

(2)

Fig. 2.116 A folding arm chair Fig. 2.117 A two degree of freedom five link mechanism to be designed for two positions

link 2 cannot move (it cannot rotate relative to the frame. Why?) and shelf #2 is stationary. The positions of the shelves are as shown in Fig. 2.123 at the end. ii. When Shelf#1 slides to the end of the joint, it is fixed to link 2 and can rotate about A0 . Clockwise angular rotation of link 2 (say 60°) imparts a translation of Shelf#2 from right to left so that one can reach to the items in that shelf. The final position of the two shelves are as shown in Fig. 2.124. You are to design a mechanism to perform this task. You may use any additional links as necessary. Note: In practice you can see a solution as shown in Fig. 2.125. The door is hinged at A0 and there is a single shelf with base as a half circle that can be rotated CCW for open position. There is some space lost.

2.10 Design of Six Link Mechanisms for Two Positions

Fig. 2.118 A trailer whose side opens when it is tilted sideways

Fig. 2.119 A wave generator mechanism

109

110

Fig. 2.120 Two required positions of the flap

Fig. 2.121 Ventilation door and its two given positions

2 Two Positions of a Moving Plane

2.10 Design of Six Link Mechanisms for Two Positions

Fig. 2.122 A kitchen cupboard for a corner Fig. 2.123 Shelf #1 is moved out

Fig. 2.124 Shelf#1 rotates and Shelf#2 moves to the first position of Shelf #1

111

112

2 Two Positions of a Moving Plane

Fig. 2.125 An alternative used for the corner cupboards

A0

References 1. M. Hohenwarter, D. Jarvis, Z. Lavicza, Linking geometry, algebra and mathematics teachers: Geogebra software and the establishment of the Geogebra Institute. Int. J. Technol. Math. Educ. 16(2) (2009) 2. X. Iriarte, J. Aginaga, J. Ros, Teaching mechanism and machine theory with GeoGebra, in New Trends in Educational Activity in the Field of Mechanism and Machine Theory, Mechanisms and Machine Science, vol. 19 (Springer International Publishing Switzerland 2014) 3. C. Zwikker, The Advanced Geometry of Plane Curves and Their Applications (Dover Publication Inc., 1963) (First Published in 1950)

Chapter 3

Three Positions of a Moving Plane

Abstract In this Chapter the realization of motion in between three given positions is discussed. In plane motion the pole triangle is the invariant of motion between three positions. Synthesis of a four-bar mechanism using graphical and analytical methods are shown. The Locus of points on the moving plane whose three homologous positions lie on a straight line is derived and design for slider-crank or double slider discussed. Correlation of crank angles, path generation with prescribed timing and design of six-link mechanisms for three positions are explained. Keywords Motion generation · Three position synthesis · Burmester theory

3.1 Three Finitely Separated Positions of a Plane Consider the motion of a moving body relative to a fixed reference frame given for three positions as shown in Fig. 3.1. For three positions of a moving plane, whatever we have stated for two positions of a moving plane is still valid, since we can always neglect one of the given positions and consider the other two. Consider positions 1 and 2. If you are to attach a crank from arbitrary points A and B on the moving body, then A0 and B0 must be selected on the perpendicular bisectors a12 and b12 respectively, as it was shown in Chap. 2. Now, if you also consider positions 1 and 3 (or 2 and 3), then the A0 B0 must also lie on the perpendicular bisectors a13 and b13 . In order to satisfy all the three positions then A0 and Bo must be selected at the intersection of the perpendicular bisectors a12 , a13 and b12 , b13 respectively (in Fig. 3.1 if a23 and b23 were used, the intersection points will be exactly the same.). You thus have a four-bar mechanism that will move your rigid body in between the three given positions (Fig. 3.2). What is stated in the above paragraph is the construction of a circle passing through any three points. In our case the three points are the homologous points A1 , A2 , A3 and B1 , B2 , B3 . Note that you are free to select A and B as your moving pivots. Once the moving pivots are selected, the fixed pivots A0 , B0 are fixed. We thus have ∞2 possible solutions (actually, when selecting A and B one must select four scalar parameters, therefore you have ∞4 possible solutions). Of course, the construction © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_3

113

114

3 Three Positions of a Moving Plane

A2

A2

B2

B2 A3

A3 A1

B1

A1

B3

B3

B1

B0 A0 F

F

a13

a12

b12

b13

Fig. 3.1 Three positions of a moving frame. Two arbitrarily selected circle points A and B, determination of the corresponding center points A0 and B0 Fig. 3.2 Construction of a four-bar using two arbitrarily selected circle points

A2

A1

B2 A3

B1

B3

B0 A0 F

does not guarantee the movability of the mechanism in between the three positions. The mechanism may lock in between the positions and the three positions may not be on the same branch of the four-bar. The resulting mechanism must always be analyzed.

3.2 The Pole Triangle The solution explained for three positions of a moving plane does not give us an insight about the motion. For example, for two positions of a moving plane the pole and the angular rotation about the pole were the two invariants of motion between

3.2 The Pole Triangle

115

two positions. Let us try to find such characteristics for the motion in between three positions. For three positions, considering two positions at a time, we can determine three poles as P12 , P13 and P23 as shown in Fig. 3.3. These three poles will form a triangle, which is known as “pole triangle”. These poles on the moving plane will remain stationary for the positions shown by their indices. For example, P12 will be the same point for position 1 and 2. When the plane moves to position 3, P12 will be at a different position on the fixed frame which will be denoted by P12 (3) . P12 (3) is known as the “image” of P12 or “image pole”. Similarly, there will be image poles P13 (2) , P23 (1) . Consider the plane represented as a line CD. Initially the plane is at C1 D1 (Fig. 3.4). Let us rotate the moving plane about C1 by an angle φ12 . C1 = C2 = P12 . D1 will move to D2 . Now, let us rotate C2 D2 about D2 by an angle φ23 . D2 = D3 = P23 . C2 will move to C3 . The position C3 D3 can also be accomplished by a rotation about the pole P13 which is located at the intersection of the perpendicular bisectors C1 C3 and D1 D3 . Since the perpendicular bisectors bisect the angles φ23 and φ13, ∠P13 P12 P23 = φ12 /2, ∠P12 P23 P13 = φ23 /2. Let us label the sides of the pole triangle with the common index of the poles of the adjacent vertices i.e. The side P12 P13 has common index 1, call that side as 1. Hence, we have the “theorem for the pole triangle” [1]: Theorem of the Pole Triangle 1. φ12 + φ23 = φ13 and φ12 + φ23 + φ31 = 0 or 2π . The internal angles of the pole triangle for three finite positions of the moving plane are equal to the halves of the corresponding angles of rotation (measured as directed angles. i.e., ∠P12 P23 P13 =

1 φ23 .(φ23 = −φ32 ). 2

and: 2. The pole triangle completely defines the motion between three positions of a moving plane (invariant). Fig. 3.3 Pole triangle for three positions

P23

P13 P12 B2 E1 A1 F

B1

E2 A2

B3

E3 A3

116

3 Three Positions of a Moving Plane

Fig. 3.4 Three positions of a moving plane: C1 D1 , C2 D2 , C3 D3 with pole triangle P12 P13 P23 and the theorem of the pole triangle

Consider the pole P23 . It is a common point for positions 2 and 3. To determine its location in position 1, one can rotate P23 P12 line about P12 by an angle −φ12 , or rotate P23 P13 line about P13 by an angle −φ13 (Fig. 3.5) in either case, P23 will move to a unique location P23 (1) . (image pole). Another Way of obtaining P23 (1) will be by dropping a perpendicular from P23 to side 1 of the pole triangle, and determining P23 (1) as the “mirror image” of P23 with respect to side 1. i.e., make P23 n = P23 (1) n. The triangle P12 P13 P23 (1) is the “image pole triangle” at position 1. There are two other pole triangles for positions 2 and 3 (P12 P13 (2) P23 , P12 (3) P13 P23 respectively). Pole triangle completely defines the motion between the three positions of a moving plane. Although initially the motion described seems special, referring to three general positions as shown in Fig. 3.3, If C2 is selected as P12 and D2 is selected as P23 as the location of two points on the moving frame at position 2, we shall obtain Fig. 3.4. Fig. 3.5 Pole triangle and the image pole triangle at the first position

3.2 The Pole Triangle

117

Hence, the motion is general, however the points selected on the moving body is special. Consider a point C1 on the moving plane AB. One can determine the homologous points C2 and C3 by rotating C1 P12 about P12 by φ12 (for C2 ) and rotating C1 P13 about P13 by φ13 (for C3 ). Another way of determining the homologous points is taking the image of C1 with respect to side 1. This image is known as the ground point or base point and is usually denoted by subscript g or 123 (Cg or C123. ). If we now take the image of C123 with respect to side 2 or 3, we shall determine the location of the homologous points C2 and C3 respectively (the proof is left to the reader) (Fig. 3.6. Most of the construction lines are not shown). If point C is to be used as a circle point, then the corresponding center point, C0 can be obtained at the intersection of the perpendicular bisectors of the homologous points. Let us now join the center point C0 and the ground point C123 to one of the poles (say P12 ). Since P12 C0 is the perpendicular bisector of C1 C2 , we have the following angular relations (Fig. 3.7): 1 φ12 2 1 ∠P13 P12 P23 = φ12 2 ∠C0 P12 P23 = ∠C0 P12 C2 − ∠P23 P12 C2 ∠C0 P12 C2 =

Fig. 3.6 Ground point C123 and the corresponding homologous points C1 , C2 and C3 as the image of C123 with respect to the corresponding sides of the pole triangle

118

3 Three Positions of a Moving Plane

Since ∠P23 P12 C2 = ∠C123 P12 P23 and ∠C0 P12 C2 = ∠P13 P12 P23 =

1 φ12 , 2

∠C0 P12 P23 = ∠P13 P12 P23 − ∠C123 P12 P23 = ∠P13 P12 C123 = −∠C123 P12 P13 Similarly ∠C0 P13 P23 = −∠C123 P13 P12 and ∠C0 P23 P13 = −∠C123 P23 P12 Hence, we have the following theorem: Theorem The angle formed at a pole by one adjacent side of the pole triangle and a line drawn from the ground point or the center point is equal and opposite to the angle formed by the other adjacent side and a line drawn to the center point or the ground point. Now, using this theorem, rather than selecting circle points on the moving frame and then determining the center points, one can select center points and then determine the corresponding circle points.

Fig. 3.7 Use of image pole triangle P12 P13 P23 1 to determine the circle point C1 when the center point C0 is selected

3.2 The Pole Triangle

119

Another method for selecting the center point and then determining the circle point is to perform kinematic inversion. When we invert the motion the rotation poles P12 and P13 will be that of the original motion. However, the angular rotation will be negative of the original motion. Since the internal angles are half the angular rotations, when we invert the motion, the image pole triangle P12 P13 P23 (1) is the pole triangle of the original motion. We can state this as a theorem: Theorem When the motion is inverted the pole triangle and the image pole triangle change their role. What is the image pole triangle for the original motion, is the pole triangle of the inverse motion. As shown in Fig. 3.7, One can select a center point C0 on the fixed plane and determine its three corresponding positions as C0 (2) and C0 (3) . The center of the circle passing through these three points will be C1 on the moving plane. Example 3.1 In Fig. 3.8 three positions of a moving body and, due to space restriction, the location of a fixed pivot is given. We would like to determine the corresponding circle point. This problem can be solved by performing kinematic inversion. If we keep the moving plane fixed and determine the motion of the fixed frame relative to the moving frame, A0 will be the circle point. We can then determine the second and third positions (A0 (2) and A0 (3)) and then determine A1 as the center of the circle passing through the homologous points A0 (Fig. 3.9). Another way of obtaining A1 is first we determine the poles and pole triangle by intersecting the perpendicular bisectors of any two pairs of homologous points. Next, we can determine the ground point A123 by constructing angles ∠A0 P23 P12 =–∠A123 P23 P13 and ∠A0 P12 P23 = – ∠A123 P12 P13 as shown in Fig. 3.10 (from the two adjacent sides of the pole triangle the angles for A0 and A123 are equal but in opposite direction.) Fig. 3.8 Three positions of a moving plane and the center point A0

120

3 Three Positions of a Moving Plane

Fig. 3.9 Determine image pole triangle at position 1 and locate A0 (2) , A0 (3) for the inverse motion. A1 is at the intersection of the perpendicular bisectors of the homologous points

A(3)0

P23

2

A1 (2)

0

P23

3

P13 A A0

#1

(1)

3

2

1

P12

A2

A0123

#3 #2

Fig. 3.10 For a given center point determine the ground point using the angular relation: ∠ P23 P12 A0 = − ∠P13 P12 A123 , ∠P12 P23 A0 = −∠P13 P23 A123

P23 3

15° 80

15°

°

A0 P 12

1

P13 80°

2

A

123

#1

#3 #2

If we take the image of A123 with respect to side 1 of the pole triangle, we shall obtain the location of the circle point Ai (Fig. 3.11) We can thus attach a crank between A and A0 (Fig. 3.12).

3.3 Graphical Methods for Three Position Synthesis

121

Fig. 3.11 Determine circle point using the pole triangle and the ground point A123

Fig. 3.12 Crank A0 A in three positions

A1 A0 #1

A2 A3 #3 #2

3.3 Graphical Methods for Three Position Synthesis The Graphical method will be explained using Solidworks® and Geogebra. One can easily extend these methods for any commercial CAD program they use.

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3 Three Positions of a Moving Plane

3.3.1 Performing Three Position Synthesis Using Solidworks® The procedure will be explained in steps. You must have Solidworks® program installed and ready. The procedure is not unique. Some tasks can be performed differently. Step 1. Open a part file. On the frontal plane set up a drawing sheet. On this sheet draw the three given positions of the rigid body (in the figure, it is shown as a rectangle but any other shape may be used) (Fig. 3.13). Fix these three positions. These are the required positions. Of course, depending on the problem you can use a drawing sheet on the top plane or any other plane of your choice. On this drawing sheet you can also draw some other rigid parts which may be important for the motion of the final mechanism. Step 2. Pick two points such C and D on the rigid body. These are your reference points (In the figure, these two points are selected at the two corners of the rectangle. However any other two point on the rigid body can be used).

Fig. 3.13 Three given positions of the moving frame drawn on the input sheet

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Fig. 3.14 On the construction sheet select A and B on the moving plane, Determine A0 and B0 at the center of the circle defined by three homologous points, A1 , A2 , A3 and B1 , B2 , B3

Step 3. Close the drawing sheet and open another drawing sheet on the same plane (Fig. 3.14) (frontal plane in this case). A point A that will be the axis of the revolute joint (moving pivot- circle point) on the rigid body (moving pivot must be equidistant to both C and D. Hence draw lines C1 A1 D1 and C2 A2 D2 and insert relations C1 A1 = C2 A2 = C3 A3 and D1 A1 = D2 A2 , D1 A1 = D3 A3 . In this way, when you hold point A1 with the cursor and move to another location, A2 and A3 will also move such that the distance AC and AD is always the same for the three positions. Step 4. Select another point B as the axis of another revolute joint on the moving body and apply the procedure used for point A (C1 B1 = C2 B2 = C3 B3 and D1 B1 = D2 B2 = D3 B3 ). You can move A1 , B1 or A2 , B2 ; A3 , B3 anywhere in the plane and the other homologous point (A2 , B2 or A1 , B1 or A3 , B3 ) will automatically move. This means that you are free to select your moving revolute joint axes anywhere on the moving plane easily.

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3 Three Positions of a Moving Plane

Fig. 3.15 On the analysis sheet, draw the four-bar mechanism such that the link lengths and points A0 , B0 coincide with that of the construction sheet

Step 5. Join A1 A2 , B1 B2 and A1 A3 and B2 B2 or A2 A3 and B2 B3 and draw the perpendicular bisectors. (insert the relation that A1 A2 , A2 A3 and a12 , a23 , B1 B2 , B2 B3 and b12 , b23 are at right angles). Determine the intersections of a12 and a23 (or a13 ) and b12 and b23 (or b13 ). These intersection points will give the location of the fixed pivot points A0 and B0 . A1 A0 , B1 B0 are the two cranks of the four-bar constructed (Fig. 3.14). One can also determine A0 and B0 using “Arc through three points” command instead of drawing the perpendicular bisectors. Step 6. Draw arcs A1 A3 and B1 B3 with centers A0 and B0 . As you change the positions of A and/or B or A0 , B0 along the perpendicular bisectors, you will obtain different four-bar proportions. You can perform several trials in a very short time (Fig. 3.15). Step 7. Close the drawing sheet and open another sheet on the same drawing plane. Note that the first drawing sheet is the problem definition, the second drawing sheet is your construction sheet and the third sheet will be the analysis sheet to check whether your construction works and satisfies your requirements. You can animate the mechanism and make a solid model if required.

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One can of course perform all the above on one drawing sheet. In such a case, the drawing becomes rather crowded and it is hard to debug the drawing. Instead of selecting a circle point on the moving plane and determining the corresponding center point, one can as well select a center point and then determine the corresponding circle point. The procedure is the same up to Step 5. Instead of Step 5 now select the center points B0 and A0 and fix these points. Draw 3 lines from A0 to A1 , A2 and A3 and 3 lines from B0 to B1 , B2 and B3 . Normally, B0 Bi and A0 Ai (i = 1, 2, 3) will not be equal. However, if you select these three lines and make equal, Ai and Bi will be the circle points for the corresponding center points A0 and B0 .

3.3.2 Performing Three Position Synthesis Using GeoGebra Let us consider the three positions given in Figs. 3.7 and 3.11. The coordinates of points Ci , Di for three positions are1 ◦

C1 (0, 0), D1 (100, 0); C2 (30, −30), D2 (30 + 100 ∗ cos(50 ), ◦

− 30 + 100 ∗ sin(50 ); C3 (90, −20)and D3 (90, 80).

Fig. 3.16 Three positions of the moving plane and selection of a circle point in position 1 (A1 ) using two design parameters In GeoGebra you will type: C1 = (0, 0), C2 = (30, −30), C3 = (90,–20), D1 = (100, 0), D2 = C2 + (100;50°), D3 = C3 + (100;90°). 1

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3 Three Positions of a Moving Plane

These values are typed into the input and line segments are drawn to show the three positions (Fig. 3.16). Next, we set up two slides a and b. We draw circles from C1 of radius a and from D1 of radius b. Use one of the intersections of the circles A1 or A1 ’as a circle point. When we repeat the procedure using C2 , D2 and C3 , D3 as the centers for the circles of radius a and b respectively, we determine A2 and A3 . As you move the slides, you will select different points A on the moving plane (Fig. 3.17). Once you determine the intersection points, you can hide the circles by clicking on the marker to the left of the equation shown in algebra screen. Of course, one can select Ai by using distance CA = ra and the angle ∠DCA = αa as two free parameters. Next, similar procedure is used to determine another moving pivot B and its three positions using the two slides c and d (Fig. 3.17). Determine the centers of the circles A0 and B0 defined by three homologous points A1 , A2 , A3 and B1 , B2 , B3 . Note that once A and B are defined by the four parameters a, b, c, d or ra , αa , rb , αb , center points A0 and B0 are fixed. A0 A1 B1 B0 is the position of the four-bar mechanism at the first position of the moving plane. You can determine different four-bar proportions

Fig. 3.17 Selection of circle points using four design parameters, determination of the center points and the construction of a four-bar to move plane CD in between three positions

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satisfying the given three positions by varying the value of the four design variables a, b, c and d or ra , αa , rb , αb . Of course, some of the solutions found will not be movable, may have bad transmission angle or the link lengths may be out of proportion or there may be an obstacle to be avoided. As always, the mechanism must be analyzed. For the analysis let us select another slide with variable β as an angle for the crank A0 A measured from the first position A0 A1 . Draw Line A0 A’ of length a2 and making an angle β with respect to A0 A1 . From A’ draw a circle with radius equal to A1 B1 and draw another circle from B0 with radius a4 = B0 B1 . Determine the intersection of these circles, B’ (There will be two points of intersection. Select the correct one). Draw the lines B0 B’ and AB’. Next, determine points C and D (for example C will be at a distance C1 A1 and C1 B1 from A’ and B’. Draw two circles with radii A1 C1 and B1 C1 from A’ and B’ to determine C do the same for D). You may animate the mechanism obtained to make sure that it moves in between the positions and passes through the positions (Fig. 3.18). You can find different solutions by changing the values of a, b, c, d using the slides.

Fig. 3.18 Simulation of the four-bar mechanism whose coupler link pass through the given three positions (ThreePositions.ggb file)

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3 Three Positions of a Moving Plane

3.4 Analytical Synthesis Methods for Three Positions 3.4.1 Direct Formulation In direct formulation, what has been explained in geometrical construction will be applied into analytical equations for three-position synthesis. Let us consider three positions of a moving plane given by a1 ,φ1 , a2 , φ2 and a3 , φ3 . Now consider a point A on the moving plane as given. Point A can be defined by a vector z relative to the moving plane. In such a case the coordinates of A1 and A2 relative to the fixed frame will be given by the equations: A1 = a1 + z A eiφ1 A2 = a2 + z A eiφ2 A3 = a3 + z A eiφ3 If A1 is given, first determine zA = (A1 –a1 )ei1 , then determine A2 and A3 from the second and third equations. Another procedure is to determine the pole P12 and P13 and then rotate the vector (A1 –P12 ) about P12 by an angle φ12 = φ2 − φ1 and iφ about P13 by an angle φ13 = φ3 − φ1 . Hence, A2 = P12 + (A1 –P12 ) e12 and A3 = iφ P13 + (A1 –P13 ) e13 . Once Point A on the moving plane is selected, the location of the corresponding center point A0 is fixed. It is the center of the circle defined by the three homologous points. There are several different methods to determine the center of the circle defined by three points. One method is the use of perpendicular bisectors to the two pairs of homologous points A1 A2 , A1 A3 , A2 A3 . These three perpendicular bisectors will intersect at one point, A0 . Hence one can determine A0 by using any two of the three perpendicular bisectors. We must determine the midpoints M12 and M23 of the lines A1 A2 and A2 A3 (one could as well use A1 A3 to locate the midpoint M13 ) and the slope of the perpendicular bisectors. The coordinates of the midpoint M12 and M23 relative to the fixed coordinate frame will be given by: M12 = (A1 + A2 )/2 M23 = (A2 + A3 )/2 or in Real numbers: M12x = (A1x + A2x )/2, M12Y = (A1Y + A2Y )/2 and M23x = (A2x + A3x )/2, M23Y = (A2Y + A3Y )/2 The angles the lines A1 A2 and A2 A3 makes with respect to the positive X axis of the fixed frame are:

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φ12 = arg (A2 − A1 )or in real numbers : φ12 = tan−1 [(A2x − A1x )/(A2Y − A1Y )] and2 φ23 = arg (A3 − A2 ) or in real numbers : φ23 = tan−1 [(A3x − A2x )/(A3Y − A2Y )] or the slope of A1 A2 and A2 A3 is: m12 = (A2Y − A1Y )/A2X − A1X ), m23 = (A3Y − A3Y )/(A3X − A2X ) (tan−1 (X; Y) means that you must use double argument inverse tangent, ATAN2(X; Y) function rather than single argument ATAN(Y/X) function). Since the bisectors are perpendicular to A1 A2 and A2 A3 , the angle these lines make with the positive X axis is φbi = φai + π/2, and their slope will be given by: m,12 = tan(φ12 + π/2) = −1/tan(φ12 ) m,12 = −(A2x − A1x )/A2Y − A1Y ); m,23 = −(A3x − A2x )/A3Y − A2Y ) Point A0 will be at the intersection of the two perpendicular bisectors. This point will have the coordinates: ] [ 1 ) m ,12 M12x − m ,23 M23x − M12y + M23y X A0 = ( , , m 12 − m 23 and Y A0 = m ,12 (X A0 − M12x ) + M12y We have 2 parameters to determine a crank AA0 . In this formulation these are the coordinates of point A on the moving frame. Using kinematic inversion, one can develop an algorithm so that the coordinates of A0 are the two free parameters (i.e. if we use a→ ,j = a→ j e−i j and φ ,j = −φ j , we have inverted the motion and one can select the coordinates of A0 as free parameters). This procedure can be implemented in MATLAB® , Mathcad® or Excel® in complex numbers (or in real numbers).

2

”arg” is the argument of a complex number. This is the angle formed by the vector with respect to positive real axis, measured counterclockwise as positive.

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3 Three Positions of a Moving Plane

3.4.2 Dyad Formulation 3.4.2.1

Basic Form

Instead of the above approach, Sandor and Erdman give another approach utilizing complex numbers [2, 3]. Consider a plane moving from the first position to another j th position. We want to realize this motion by attaching two cranks to the moving body at points A and B (Fig. 3.19). Points A and B will move on circular arcs (circle points) and A0 and B0 will be the centers (center points). The mechanism can be said to be formed by two vector pairs called “dyads”. For example, in Fig. 3.19 A0A and AO’ form a vector pair (dyad) and B0 B and BO’ form any vector pair (dyad). Once the magnitudes of these vectors are known, one can determine the dimensions of the mechanism. The positions of the moving body are given by the vector ai and φi . The moving plane will rotate by an angle αj = φj − φ1 , when it moves from position 1 to position j, hence any vector attached to the moving frame will rotate by αj during this motion. One vector pair is shown in Fig. 3.20. In the first position let A0 A1 = W and A1 O’1 = Z. Note that neither W nor Z are known initially. Let us assume that the crank A0 A will rotate by an angle βj , when the moving frame is moved from the first position

Fig. 3.19 A four-bar mechanism made up of two dyads: A0 AO’ and B0 BO’. Where O’ is the origin of the moving coordinate frame on the coupler link

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Fig. 3.20 Dyad A0 AO’ represented as two vectors W and z in the first position Weiβj and Zeiαj in jth position

to the j th position. At the j th position vector W will be rotated by βj and vector Z will be rotated by an angle αj . These vectors at j th position will be Weiβj and Zeiαj . One can write a vector loop equation: −−→ −−→, −−,−→, −−,−→ −−−→ A0 A + AO j + O j O1 + O1 A1 + A1 A0 = 0 or: W eiβ j + Z eiα j − δ j − Z − W = 0

(3.1)

which can be written as: ( ) ( ) W eiβ j − 1 + Z ei α j − 1 = δ j

(3.2)

where δj = aj − a1 . Note that the fixed reference frame O is arbitrary. The equation is independent of the reference frame selected. Therefore, one can select the origin of the reference frame to coincide with the origin of the moving frame in the first position. For 3 positions of the moving frame a1 , a2 , a3 and φ1 , φ2 , φ3 are given. One can easily determine δ2 = a2 − a1 , δ3 = a3 − a1 and α2 = φ2 − φ1 , α3 = φ3 − φ1 . When Eq. (3.2) is written for positions 2 and 3: ( ) ( ) W eiβ2 − 1 + Z eiα2 − 1 = δ2

(3.3)

( ) ( ) W eiβ3 − 1 + Z ei α3 − 1 = δ3

(3.4)

In these two complex number equations δ2 , δ3 , α2 , α3 are known parameters. W, Z and β2 , β3 are the variable parameters. If β2 and β3 are assumed beforehand, Eqs. (3.3) and (3.4) forms a linear set and can be solved for W and Z as:

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3 Three Positions of a Moving Plane

| | | δ2 eiα2 − 1 | | | | δ3 eiα3 − 1 |

( ( ) ) δ2 eiα3 − 1 − δ3 eiα2 − 1 )( ) ( )( ) − 1 eiα3 − 1 − eiβ3 − 1 eiα2 − 1

W = | iβ | = ( iβ | e 2 − 1 eiα2 − 1 | e 2 | | | eiβ3 − 1 eiα3 − 1 | | iβ | | e 2 − 1 δ2 | | | ( ( ) ) | eiβ3 − 1 δ3 | δ3 eiβ2 − 1 − δ2 eiβ3 − 1 )( ) ( )( ) Z = | iβ | = ( iβ | e 2 − 1 eiα2 − 1 | e 2 − 1 eiα3 − 1 − eiβ3 − 1 eiα2 − 1 | | | eiβ3 − 1 eiα3 − 1 |

(3.5)

(3.6)

Noting that O’1 A1 = –Z and O’1 A0 = –Z–W = –R1 , we have determined the location of a circle point (A1 ) and its corresponding center point (A0 ) when we solve for W and Z. If any other set of β2 and β3 are assumed, then the other dyad (B1 , B0 ) will be determined. Note that there are two free parameters to determine a circle point and its corresponding center point (crank angles β2 , β3 ). Hence, you have 4 free design parameters to construct a four-bar mechanism. Equation (3.2) is known as the “standard form” or “basic form” of the dyad equation.

3.4.2.2

Numerical Solution of Basic Form Using Excel

Example 3.2 Consider the problem given in Fig. 3.8. (Repeated in Fig. 3.21). For the given three positions: δ2 = 30 − 30i, α2 − 50◦ δ3 = 90 − 20i, α3 − 90◦ For the first dyad let us use β2 = 40°,β3 = 110o for the crank rotation and use subscript A (i.e. WA and ZA ) To identify the dyad O’AA0 . Solving Eqs. 3.4 and 3.5 yield: WA = −40.182 − 20.664 i ZA = −13.315 − 3.342 i Relative to the reference coordinate frame, A1 will be: O, A = −ZA = 13.315 + 3.342 i and A0 will be at: O, A0 = −ZA − WA = 53.497 + 24.006 i

3.4 Analytical Synthesis Methods for Three Positions Fig. 3.21 Three positions of a moving plane (Example 3.2)

133

Y

50°

#2

#3

20

30

#1 X

30 90

For the second dyad (O’BB0 ) let us selectβ2 = 300 ,β3 = 900 for the rotation of the crank BoB and use subscript B. Again, using Eqs. 3.4 and 3.5: WB = −29.645 − 36.98 i ZB = −25.355 + 1.987 i Relative to the reference coordinate frame B1 will be at: O, B = −ZB = 25.355 − 1.987 i and B0 will be at: O, B0 = −ZB − WB = 42.961 + 40.33 i Mechanism is shown in Fig. 3.22. Of course, the result must be analyzed to make sure that it is movable in between the positions and satisfies all other possible constraints. Mathematical packages like MATLAB® and Mathcad® can perform complex number calculations. In excel, you must first go to “Options- Add-ins” menu and select Excel Add-In. On the pop-up menu you must click onto the “Analysis Tool Pack”. Once this add-in is loaded you can first define numbers as complex using COMPLEX (real number, imaginary number, i). With the analysis tool pack you will be able to use functions starting with IM. (like IMPRODUCT, IMDIV, etc.) in the Engineering Functions subgroup. A better and easier way of solving this problem is to identify complex numbers as an array with two elements and write real and imaginary part of a complex number

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3 Three Positions of a Moving Plane

Fig. 3.22 Synthesis of a Four-bar mechanism for three positions using crank angle specification

in two adjacent cells in a row. We can then use these variables in a small function program that we write in Visual Basic for three positions. As the input we define αj andδj that define the three positions and the crank angles βj (j = 2,3). The function program in BASIC is written as: Function ThreePosition_Basic(Alfa2, Alfa3, delta2, delta3, Beta2, Beta3) ‘This function returns both W and Z vectors when two positions and the crank angles beta2 and beta3 are given. In the output the real and imaginary parts of W and Z are given in order. Dim A, B, C, D, E, F, Del As Double Dim W(2), Z(2), Result(4)As Double A = delta2(1) ∗ (Cos(Alfa3) − 1) − delta2(2) ∗ Sin(Alfa3) − delta3(1) ∗ (Cos(Alfa2) − 1) + delta3(2) ∗ Sin(Alfa2) B = delta 2(1) ∗ Sin(Alfa 3) + delta 2(2) ∗ (Cos(Alfa 3) − 1) − delta 3(1) ∗ Sin(Alfa 2)

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− delta 3(2) ∗ (Cos(Alfa 2) − 1) C = Cos(Beta 2 + Alfa 3)−Cos(Beta 3 + Alfa 2) −Cos(Beta 2)−Cos(Alfa 3) + Cos(Beta 3) + Cos(Alfa 2) D = Sin(Beta 2 + Alfa 3)−Sin(Beta 3 + Alfa 2) −Sin(Beta 2) − Sin(Alfa 3) + Sin(Beta 3) + Sin(Alfa 2) E = delta 3(1) ∗ (Cos(Beta 2) − 1) − delta 3(2) ∗ Sin(Beta 2) − delta 2(1) ∗ (Cos(Beta 3) − 1) + delta 2(2) ∗ Sin(Beta 3) F = delta 3(2) ∗ (Cos(Beta 2) − 1) + delta 3(1) ∗ Sin(Beta 2) − delta 2(2) ∗ (Cos(Beta 3) − 1) − delta 2(1) ∗ Sin(Beta 3) Del = C 2 + D 2 W(1) = (A ∗ C + D ∗ B)/Del W(2) = (B ∗ C − A ∗ D)/Del Z(1) = (E ∗ C + F ∗ D)/Del Z(2) = (F ∗ C − E ∗ D)/Del Result(0) = W(1) Result(1) = W(2) Result(2) = Z(1) Result(3) = Z (2) ThreePosition_Basic = Result End Function In order to use this program in an Excel Sheet. you must first mark four adjacent cells in a row. You may than go to Formulas-Insert function and choose “user defined functions” select “ThreePositions” Basic function. A screen shown in Fig. 3.23 will appear. Enter the required arguments (keep in mind that Deltaj arguments require two inputs and Alfaj must be in radians (It is advisable to write each of the arguments of the function in cells and use these cell addresses as the input to the function. In such a case, the function output will change automatically when you change the value of any one of the arguments) After completion of the input arguments, press Ctrl and Shift keys first and then press Enter. The first two cells written will be W vector (real and imaginary parts written in sequence) and the third and fourth cells will be the Z vector. Similar to the above function routine, the following function routines are also available.

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Fig. 3.23 Function menu for ThreePosition_Basic function

ThreePosition_FixedPivot_Beta(Alfa2, alfa3, Delta2, Delta3, R) This function returns Beta2 and Beta3 for given 3 positions and fixed pivot. Function ThreePosition2_FixedPivot(Alfa2, alfa3, Delta2, Delta3, R). This function returns –Z vector (moving pivot location) when R (fixed pivot location) is specified. Function ThreePosition3_MovingPivot(Alfa2, alfa3, Delta2, Delta3, Zp). This function returns both R vector (fixed pivot location) when –Z (= Zp) (moving pivot location) is specified. Function ThreePosition_Basic2(Alfa2, alfa3, Delta2, Delta3, Beta2, Beta3). This function returns both –Z and R vectors (location of moving and fixed pivot points when two positions and the crank angles Beta2 and Beta3 are given, Depending on the problem, any one of these function routines can be easily used. Example 3.3 Consider the problem considered in Example 3.2. We write the input values for δj and αj in an ordered fashion as shown in Fig. 3.24. Then the angular rotations of the two cranks are selected. The angles are converted into radians. Then we mark cells E6 to H6 and enter the function Threeposition_Basic as: = ThreePosition_Basic(F2, F3, B2 : C2, B3 : C3, C6, C8)

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Fig. 3.24 Excel sheet for three position synthesis with crank angle specification. W and Z vector components obtained using ThreePositions_Basic function

Fig. 3.25 Excel sheet for three position synthesis with crank angle specification. –Z and R vector components obtained using ThreePositions_Basic2 function. Mechanism is drawn in the first position

While pressing Ctrl and Shift keys with your left fingers, press enter to obtain the result for W and Z written in cells E6 to H6. Repeat the same procedure using another set of β2 and β3 .

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3 Three Positions of a Moving Plane

Instead of ThreePosition_Basic function if you use ThreePosition_Basic2 function, you obtain –Z and R vectors directly and draw the mechanism at the first position as shown in Fig. 3.25. You can easily change β values and visualize different possibilities.

3.4.2.3

Circle or Center Point Specification

By writing the vector loop equation formed in another form one can use the coordinates of the moving or fixed pivots as free parameters instead of the crank angles. If we redraw Fig. 3.20 by adding new vectors Ri as shown in Fig. 3.26, for three finitely separated positions we have: W + Z = R1 W eiβ2 + Z ei α2 = R2 = δ2 + R1 W eiβ3 + Z eiα3 = R3 = δ3 + R1

(3.1)

Instead of specifying the crank angles β2 , and β3 , if we now specify O’1 A0 = R = –R1 , which is the location of the fixed pivot (center point), this set of 3 complex equations (corresponding to 6 scalar equations), there are two complex unknowns (W and Z) and two scalar unknowns (β2 , and β3 ) to be solved. (for three positions δ2 , δ3 , α2 , α3 are known parameters). The equations nonlinear in terms of the parameters β2 , and β3 . Let us assume that we have solved for β2 , and β3 , in some way (say by trial and error). Then these three complex equations yield 3 linear equations in two complex unknown W, and Z. From linear algebra, we know that the only way to have a solution for W and Z from these three equations is that these equations must be linearly dependent (must be of rank 2). For linear dependence, the determinant of the augmented matrix of the coefficients must be identically zero: Fig. 3.26 Dyad A0 AO’ in the first and jth positions W + Z = R1

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139

| | | 1 1 R1 | | | | eiβ2 eiα2 R2 | = 0 | | | eiβ3 eiα3 R | 3

(3.2)

Equation 3.2 is a complex equation in two unknowns β2 , and β3 . When these unknowns are solved W and Z vectors can be solved using any two of the complex equations in (3.1). If we expand the determinant with respect to the first column: | iα | | | | | | e 2 R2 | | | | | | | − eiβ2 | 1 R1 | + eiβ3 | 1 R1 | = 0 | eiα3 R3 | | eiα3 R3 | | eiα2 R2 | (

) ( ) ( ) R3 eiα2 − R2 eiα3 + eiβ2 R1 eiα3 − R3 + eiβ3 R2 − R1 ei α2 = 0

(3.3)

Let: ) ( Δ1 = ( R3 eiα2 − R2 )eiα3 Δ2 = ( R1 eiα3 − R3 ) Δ3 = R2 − R1 eiα2 For three positions, when R1 is specified, one can calculate the complex numbers Δi (i = 1, 2, 3) and one can write Eq. 3.3 and its complex conjugate as: Δ1 + Δ2 eiβ2 + Δ3 eiβ3 = 0 Δ1 + Δ2 e−iβ2 + Δ3 e−iβ3 = 0

(3.4)

If we want to solve for β2 or β3 , we must eliminate β3 or β2. Let us eliminate β3 and solve for β2 . We leave the term containing β2 on one side of the equations: Δ3 eiβ3 = −( Δ1 + Δ2 eiβ2 ) Δ3 eiβ3 = −( Δ1 + Δ2 eiβ2 )

(3.5)

Multiplying these two equations side by side: Δ3 Δ3 = Δ1 Δ1 + Δ2 Δ2 + Δ1 Δ2 e−iβ2 + Δ1 Δ2 eiβ2 Let d = Δ1 Δ1 + Δ2 Δ2 − Δ3 Δ3 ,a = Δ1 Δ2 and t = eiβ2 e−iβ2 =

1 eiβ2

=

1 t

Equation 3.6 can now be written as: at + a/t + d = 0

(3.6)

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3 Three Positions of a Moving Plane

Fig. 3.27 Three positions of a moving plane and center points A0 , B0 specified

Y

B0 A0

50°

#2

#3

20

30

#1 X

30 90

or at 2 + dt + a

(3.7)

This is a quadratic equation in terms of t. Note that t is a unit vector (|t| = 1) and there are two solutions for t (t1 , t2 , say). β2i = atan2(Re(ti ),Im(ti )). β21 or β22 will be equal to α2 . This is a trivial solution. One must use the nontrivial solution as β2 . β3 can now be solved using Eq. 3.4 and W and Z can be solved using any two of the Eq. 3.1. Actually, one need not solve for β3 . If Eq. (3.3) is solved for β2 , then one can use the first two equations of (1) to solve for –Z to locate the coordinates of the moving pivot. Example 3.4 Three-position problem solved in the previous examples will be solved. In this case the coordinates of the fixed pivots A0 and B0 are given as (Fig. 3.27): RA = 51.48 + 20.44i and RB = 55.56 + 29.26i and the three positions are given by: δ2 = 30 − 30i, α2 = 50◦ δ3 = 90 − 20i, α3 = 90◦ Using R1 = –51.48–20.44i we determine Δi as: Δ1 = 5.299 + 24.994 i Δ2 = −18.08 − 11.04 i Δ3 = −4.047 + 2.135 i

3.4 Analytical Synthesis Methods for Three Positions

141

Calculate d and a as: d = 1081 a = −371.737 + 2393.385 i the solution to the quadratic equation yields: t1 = 0.643 + 0.766 i t2 = 0.728 + 0.685 i note that the magnitudes of these complex numbers are unity and the arguments of these complex numbers yield β2 as: β21 = 500 , β22 = 43.241◦ Note that β21 = α2 , trivial solution. Therefore, we select β2 = β22 = 39.966°. We determine β3 : β3 = 121.675◦ Using the first two equations in (3.1) we obtain the coordinates of the moving pivot in position 1 as: −ZA = 14.092 + 10.021 Repeating the same calculation with R1 = −55.56–29.26i we determine β21 = − 50 o , β22 = 25.393° Eliminating the trivial solution, (β3 = 106.06°) we determine the coordinates of the moving pivot B1 as: −ZB = 34.051 + 9.522 i The result is shown in Fig. 3.28. If one specifies circle point coordinates O’A1 = – Z instead of specifying the center point coordinates R = −R1 , Eq. (3.1) can be written as: W + R = −Z = Z 1 W eiβ2 + R = δ2 − Z eiα2 = Z 2 W eiβ3 + R = δ3 − −Z ei α3 = Z 3

(3.8)

Noting that the terms on the right-hand side of the equations are known when the circle point (−Z) and the three positions are specified, we again have 3 complex equations in 2 complex (W and R) and two real (β2 , and β3 ) unknowns. Again, the condition for obtaining a solution for W and R requires the determinant of the augmented matrix of the coefficients must be zero:

142

3 Three Positions of a Moving Plane

Fig. 3.28 Four-Bar mechanism obtained using center point specification

Y

B0 A0

50°

#2

#3

20

30

#1 X

30 90

| | | 1 1 Z1 | | | | eiβ2 1 Z 2 | = 0 | | | eiβ3 1 Z | 3 Expanding the determinant about the first column: | | | | | | | 1 Z2 | | | | | | | − eiβ2 | 1 Z 1 | + eiβ3 | 1 Z 1 | = 0 | 1 Z3 | | 1 Z3 | | 1 Z2 | (Z 3 − Z 2 ) + eiβ2 (Z 1 − Z 3 ) + eiβ3 (Z 2 − Z 1 ) = 0

(3.9)

Δ1 = (Z 3 − Z 2 ) Δ2 = (Z 1 − Z 3 ) Δ3 = (Z 2 − Z 1 )

(3.10)

If we specify:

Solution for β2 or β3 will follow in the same way as described for the center point specification. One can then determine the coordinates of the center point (R) using any two of Eq. (3.8). Another way of solving circle point specification is to use the center point algorithm and to apply kinematic inversion for the specified three positions. For example, for three positions of a plane, when δ2 , δ3 and α2 , α3 are given if we have a package program that will determine the location of the circle point (–Z) when the location of the center point (R) is given, we can use the same program to determine the location of center point (R) when the location of the circle point (–Z) is given by defining: δ,2 = −δ2 e2−iα2 , δ,3 = −δ3 e3−iα3 , α,2 = −α2 , α,3 = −α3 . Now, due to the inverse motion, we can specify the center point of the inverse motion R’as the circle point (–Z) of the original motion and obtain –Z’ as a result which will be the center point (R) of the original motion.

3.4 Analytical Synthesis Methods for Three Positions

143

Example 3.5 Three positions specified in the previous examples will be used. In this example we would like to have the moving pivot points A and B to be at A1 (15,10) and B1 (35,10) as shown in Fig. 3.29. To determine A0 , we first determine Zi : Z1 = 15 + 10 i Z2 = 31.981 − 12.081 i Z3 = 90 − 5 i Determine Δi : Δ1 = 48.019 + 7.081 i Δ2 = −65 + 15 i Δ3 = 16.981 − 22.081 i When we solve for β2 we again obtain two solutions (0 and 42.767). 0 is the trivial solution. We obtain β2 = 42.767° and β3 = 121.655°. Using these values for the location of the fixed pivot point we obtain: RA = 51.687 + 20.643 i Applying the same procedure for B, we obtain: RB = 55.616 + 29.46 i

Fig. 3.29 Three positions of a plane and circle points A, B specified at position 1

Y

A1

B1

50 °

#2 30 90

#3

20

30

#1 X

144

3 Three Positions of a Moving Plane

Fig. 3.30 Four-Bar mechanism obtained using circle point specification

Y

B0 A0

A1 B1

B2 A3

30

#2

50°

A2

#3

20

#1

B3

X

30 90

The result is shown in Fig. 3.30. Depending on the problem to be solved, one can use center or circle point as the input or use the crank angles. If the region for the fixed pivot location is restricted then center point specification may be more appropriate. If the moving pivots can only be placed at certain regions of a moving body, one must use circle point specification. If the correlation of the moving body positions with a crank angle is required, crank angle specification may be preferable. For each dyad of the four-bar, one can use any three of these parameter specifications. Note that in all these methods discussed for each dyad one has two free scalar parameters (β2 , β3 , A0 (x0 , y0 ) or A1 (x1 , y1 )). If one has prepared a routine to determine the circle point when center point is given, the same program can be easily used to determine the center point when a circle point is given by simply inverting the motion, i.e. keep the moving frame fixed in the first position and move the fixed frame. In such a case we replace δj and αj by δ,j = –δj e−iαj and α,j = −αj .

3.5 Three Homologous Points on a Straight Line For three positions of a moving plane, any point selected on the moving plane will have three homologous coordinated points (A1 , A2 , A3 ). We know that these three points will always define a circle. Only in a particular case the radius of this circle will be at infinity (A0 at infinity) and the homologous points will lie on a straight line. If we can find these points than we can attach slider as a constraint, rather than a crank.

3.5 Three Homologous Points on a Straight Line

145

Fig. 3.31 Moving plane in the first and j th position. Homologous points Aj lie on a straight line

Consider the moving plane in first and j th position. We are looking for a point A1 on the moving plane such that the homologous points Aj lie on a straight line (Fig. 3.31). The slider is displaced by an amount sj when the moving plane moves from the first position to the jth position. The straight line makes an angle γ with respect to the reference axis. The vector loop A1 Aj O’j O’A1 can be written as: s j eiγ + Z eiα j − δ j − Z = 0 Or Z (eiα j − 1) + s j ei γ = δ j For three positions of a moving plane, we can write two equations in the form: Z (ei α2 − 1) + s2 eiγ = δ2 Z (ei α3 − 1) + s3 eiγ = δ3 δ2 , δ3 and α2 , α3 are given. The unknown parameters are Z, s2 , s3 and γ. Hence the system yields 4 scalar equations in 1 complex and 3 scalar unknowns. Let us define two new variables as: S = s2 eiγ and λ =

s3 , so that s3 eiγ = λS s2

146

3 Three Positions of a Moving Plane

There are 2 complex and one scalar unknowns. If λ is assumed, the equations yield two complex equations linear in terms of two complex unknowns Z and S: ( ) Z( eiα2 − 1) + S = δ2 Z ei α3 − 1 + λS = δ3 Solving for –Z: δ3 − λδ2 δ3 − λδ2 ) ( ) ) ( )=( −Z = ( i α 1 − eiα3 − λ( 1 − eiα2 λ e 2 − 1 − eiα3 − 1 Hence there are infinite number of possible points on the moving plane whose 3 homologous points will lie on a straight line. These points are determined by changing λ within –∞ < λ < + ∞. –Z is going to describe a locus. In complex number algebra an equation in the form: z=

a + λb c + λd

where a, b, c and d are complex numbers and λ is a real number –∞ < λ< + ∞, the locus is a circle [4, 5]. This locus passes through a/c (λ = 0) and b/d (λ = ∞). In our case when λ = 0: δ3 ) = P13 −Z = ( 1 − eiα3 When λ = ∞: δ2 ) = P12 −Z = ( 1 − eiα2 And when λ = 1: δ3 − δ2 (1) ) = P23 −Z = ( iα e 2 − ei α3 We can state this result as a theorem: Theorem The locus of points on the moving plane whose three homologous positions lie on a straight line is the circumcircle of the image pole triangle P12 P13 P23 (1) . The straight line passes through the altitude H 3 of the pole triangle P12 P13 P23 . 3

Altitude of a triangle is determined by dropping perpendiculars from the corners of a triangle to the opposite side. All three perpendiculars will intersect at one point known as the “altitude” of the triangle. Altitude of the pole triangle is on the circumcircle of the image pole triangle (of course the reverse is also true- altitude of the image pole triagle is on the circumcircle of pole triangle).

3.5 Three Homologous Points on a Straight Line

147

The designer is now free to design a four-bar, slider-crank or double slider to realize the motion between the three given positions of a moving plane (of course, the designer must check his result for movability in between the positions). To determine the angle the slider makes with respect to the reference frame selected, one must solve for S: ( ) ( ) δ2 1 − eiα3 − δ3 1 − ei α2 ) ( ) S= ( 1 − eiα3 − λ 1 − ei α2 The argument of this complex number S will yield the angle γ. Let us consider kinematic inversion. When the moving plane is fixed and the fixed plane is moved, in case of revolute joints the circle and center points change their role. In case of a slider, the inverse motion results with a straight line passing through a fixed point. Now, for the inverse motion if we determine a point whose three homologous points lie on a straight line, this point in the original motion will be a point on the fixed frame at which three corresponding points of a line on the moving frame will be concurrent. Since for the inverse motion the pole triangle and the image pole triangle change their role, we have another theorem: Theorem The locus of points on the fixed frame where three corresponding positions of a line on the moving plane are concurrent is the circumcircle of the pole triangle P12 P13 P23 . The line on the moving plane passes through the altitude of the image pole triangle P12 P13 P23 (1) , H’. One can use the equation for –Z derived for the determination of a point in the first position whose three homologous points lies on a straight line to the determine the point at which three homologous lines are concurrent, by simply using δ,j = −δj e−iαj and α,j = −αj instead of δj and αj . Example 3.6 In Fig. 3.32, for three positions of the moving plane the pole triangle is found and the image pole triangle is determined. A point A1 on the circumcircle of the image pole triangle (P12 P13 P23 1 ) is selected. Note that this line passes through H, the altitude of the pole triangle. The homologous points A2 and A3 lie on the line defined by A1 H. Also note that the ground point for A1 (whose three homologous points lie on a straight line), A123 , lies on the circumcircle of the pole triangle. For the same three positions, a point A0 , on the circumcircle of the pole triangle is selected (Fig. 3.33). Line L1 on the moving plane in the first position passes through A0 H’ (H’ is the altitude of the image pole triangle). The corresponding positions of the line L on the moving plane (L2 and L3 ) pass through A0 . Hence, a swinging block is designed (this is also called “inverted slider”). Example 3.7 Consider three positions of a moving plane CD in three positions as shown in Fig. 3.34a. We would like to realize by a four-link RPRP mechanism known as “Conchoidal Motion Mechanism” shown in Fig. 3.34.b.

148

3 Three Positions of a Moving Plane

Fig. 3.32 Point A1 selected on the circumcircle of the image pole triangle P12 P13 P23 1 . Three homologous points A2 , A3 lie on a straight line which passes through the altitude of the pole triangle

Fig. 3.33 Point A0 selected on the circumcircle of the pole triangle P12 P13 P23 . The straight line on the moving plane in the first position pass through the altitude of the image pole triangle P12 P13 P23 1

Fig. 3.34 a Three positions of a moving Plane CD to be synthesized by a conchoidal motion mechanism shown in (b)

3.5 Three Homologous Points on a Straight Line

149

Using Geogebra we first draw the given three positions determine the poles and draw the pole triangle and the image pole triangle. We then draw the circumcircles of the pole and image pole triangles and determine the orthocenters H and H’ (Fig. 3.35). Any point of the on the circumcircle of the pole triangle can be selected as A0 , point where three homologous positions of Line1 will be concurrent and point on the circumcircle of the image pole triangle can be selected as B1 , where the three homologous points of B will lie on line B1 H. The resulting mechanism is shown in Fig. 3.36. The same problem can be solved analytically using Excel. Three Positions of the moving plane will be defined as: Degrees

Radians

δ2

0

0

α2

30

0.5236

δ3

0.36603

−0.366

α3

60

1.0472

Fig. 3.35 Construction of the pole and image pole triangles and their circumcircle. Determine the orthocenter of the pole triangle (H) and image pole triangle (H’). A0 and B1 on the corresponding circle loci are selected arbitrarily

150

3 Three Positions of a Moving Plane

Fig. 3.36 Conchoidal motion mechanism whose coupler link passes through the three design positions

To obtain the locus of points which lie on a straight line for 3 homologous points we must change the free parameter λ= s3 /s2 in between −h ≤ λ ≤ −h One way of obtaining the values of l, we enter angles in between −90° ≤ γ≤ + 90°. if we let λ= tanγ, we obtain 180 values of λ from −h to + h. Now using Threeposition_Slider (φ12 , φ13 , δ2 , δ3 , λ) user defined function we obtain a set of values which are on the the circumcircle of the image pole triangle. To obtain the locus of points where three homologous lines on the moving frame are concurrent we must invert the motion. For the inverted motion δ ,j = −δ j e−iα j and a,j = −aj s. Three positions will now be defined as: δ,2 δ,3

0 0.13397

Degrees

Radians

0

α,2

−30

−0.524

0.5

α,3

−60

−1.047

, , Again we use Threeposition_Slider (φ12 , φ13 , δ2, , δ3, , λ) to obtain the locus which is the circumcircle of the pole triangle. This function routine will return –Z vector, which is the location of points in position 1 (with respect to the first position C1 ) whose three homologous points lie on a straight line. Next, we can select A0 and B1

3.5 Three Homologous Points on a Straight Line

151

on the corresponding loci to realize the conchoidal motion which will move the plane in between the three positions as shown in Fig. 3.37 (for A0 , γ= 77° or λ = 2.7475 and for B1 γ = 66° or λ = 2.2460 is selected. If horizontal slider axis is required, one must select γ= 67.0887° (λ= 2.3660). Instead of using λ as a free parameter and then determining the point A1 whose three homologous points lie on a straight line, it may be preferable to use the angle γ, the angle slider axis makes with respect to the fixed reference frame. Considering the two vector equations obtained for three positions in complex numbers: Z (ei α2 − 1) + s2 eiγ = δ2 Z (ei α3 − 1) + s3 eiγ = δ3 Now, we must eliminate s2 and s3 and obtain –Z as a function of γ. Leaving the term containing si on one side: s2 eiγ = δ2 − Z(eiα2 − 1) s3 eiγ = δ3 − Z(ei α3 − 1)

Fig. 3.37 Solution of conchoidal motion mechanism using excel

152

3 Three Positions of a Moving Plane

Dividing the two equations by their complex conjugate: s2 ei γ s2 e−i γ s3 ei γ s3 e−i γ

= e2iγ = = e2iγ =

δ2 −Z(ei α2 −1) δ 2 −Z(eiα2 −1) δ3 −Z(ei α3 −1) δ 3 −Z(ei α3 −1)

The above equations can be written as: Z (eiα2 − 1) − Z(e−iα2 − 1)e2iγ = δ2 − δ 2 e2iγ Z (eiα3 − 1) − Z(e−iα3 − 1)e2iγ = δ3 − δ 3 e2iγ In complex plane, these equations constitute two linear equations in terms of two unknowns Z and Z. Solving for –Z results: −Z =

(δ2 − δ 2 e2iγ )(e−iα3 − 1) − (δ3 − δ 3 e2iγ )(e−iα3 − 1) 2i[sin(α3 − α2 ) − sin(α3 ) + sin(α2 )]

or [ ] ] [ δ2 (e−iα3 − 1) − δ3 (e−iα2 − 1) + e2i γ −δ 2 (e−i α3 − 1) + δ 3 (e−iα2 − 1) −Z = 2i[sin(α3 − α2 ) − sin(α3 ) + sin(α2 )] Again, this is the equation of a circle, i.e. as γ changes from 0° to 360°, –Z will describe a circle. Depending on the problem to be solved one can use λ or γ as the free parameter and determine points whose three homologous positions lie on a straight line.

3.6 Principle of Superposition For two positions of a moving plane the invariant of motion was the pole and the angle of rotation about the pole. In case of three positions of a moving plane the invariant of motion is the pole triangle. Two different plane motions for three positions will be the same provided that their pole triangles are equal. If the motion of a rigid body in plane motion is given for three positions, we must first determine its pole triangle. Next, if we are to superimpose this motion onto the one of the links of an existing mechanism, we must determine three positions of this link such that the angular rotations in between the three positions are equal to the angular rotations of the given motion. In such a case the two pole triangles will be similar (since the internal angles will be equal). The two plane motions will only be different from each other by what is called “Stretch-rotation” e.g. by changing the orientation of the pole triangles and/or by changing the scale of both or one of the pole triangles, the two triangles can be superimposed on each other and thus the plane motion can be realized by a preselected mechanism.

3.6 Principle of Superposition

153

Example 3.8 Consider three positions of a plane AB as shown in Fig. 3.38a We would like to realize this motion by a planetary gear train of gear ratio −1. We determine the pole triangle for the given motion as shown in Fig. 3.38b. Next, we draw the mechanism which will synthesize this motion at its three positions such that the angular rotation of the plane that is to be used for superposition has the same angular rotation between the positions as the angles of rotation of the original required motion. In the example shown φ12 = 120° (CW) and φ23 = 120° (CW). For the planetary gear train with gear ratio –1 speed ratio is given by the equation: ω planet − ωar m = −1 ωsun − ωar m Since ωsum = 0(fixed sun gear)ω planet = 2ωar m . Integrating and letting the integration constant to be zero: θ planet = 2θar m For the planet to have the same rotations in between the positions of the original motion, the arm must rotate by angle 60° clockwise in between the positions 1–2 and 2–3. Hence, we draw the mechanism for these three positions and determine the corresponding pole triangle (Fig. 3.39). The pole triangle obtained is similar to the pole triangle of the required motion given in Fig. 3.39. In order to make the pole triangles equal one must rotate the mechanism shown in Fig. 3.39 by 60° clockwise and scale the mechanism dimensions such that the sides of the pole triangles are equal (or scale the original motion). The result is shown in Fig. 3.40. The principle of superposition can be applied using any mechanism as long as one can find a link on the mechanism, which rotate by the required amounts defined by the given three positions. The final mechanism is obtained by a rotation and scaling.

Fig. 3.38 a Three positions of a moving plane. b The pole triangle

154

3 Three Positions of a Moving Plane

Fig. 3.39 Planetary gear train in three positions where the planet performs the same angular rotations between the positions as that of the given three positions

Fig. 3.40 a Superposition of the required motion onto the planet motion. b Mechanism in the first position with moving plane AB superimposed

3.7 Relative Motion Between Two Moving Planes Referring to Sect. 2.6, when there are two moving planes, for two positions of a moving plane there is a relative pole between these two planes and an angular rotation about this relative pole equal to the difference of the rotations of these planes relative to the fixed plane. Hence when there are three positions of the moving planes, there will be three relative poles which will form a relative pole triangle. In Fig. 3.41 three positions of planes E and E’ are given. One can determine the pole triangles for the

3.7 Relative Motion Between Two Moving Planes

155

planes E and E’ as explained (construction lines are removed for clarity). Relative pole R12 (1) (or R13 (1) ) can be determined as explained in Sect. 2.6 using P12 and P’12 (or P13 and P’13 ) and drawing lines that make angles –φ12 /2 and −ψ12 /2 (or –φ13 /2 and −ψ13 /2) with respect to poles P12 P’12 (P13 P’13 ) line. R12 (1) and R13 (1) are the relative poles for the plane E’ when the moving plane is at position 1. To determine R23 (1) the simplest method is to draw lines that make angles γ12 = (ψ12 − φ12 )/2 and γ13 = (ψ13 − φ13 )/2. R23 (1) is the relative pole of the plane E’ when moving from the second to third position relative to E1 . It is a common point for positions 2 and 3 of the plane E’. One can treat the relative pole triangle R12 (1) R13 (1) 23 (1) as the pole triangle P12 P13 P23 . In this case the motion of the plane E’ is defined relative to the moving plane E in position 1. (sometimes it may be simpler to invert the motion directly, i.e. determine the three positions of E’ relative to E1 initially by “subtracting the motion of plane E from the motion of the plane E’ as explained in Sect. 2.6)

Fig. 3.41 Three positions of two moving planes and the Relative pole triangle R12 1 R13 1 R23 1 for the three positions of plane E’ relative to the first position of plane E

156

3 Three Positions of a Moving Plane

3.8 Correlation of Crank Angles 3.8.1 Geometrical Methods Consider two links 2 and 4 rotating about A0 and B0 (Fig. 3.42). When link 2 rotates by an angle φ12 , link 4 is to rotate by ψ12 and when link 2 rotates by an angle φ13 we want link 4 to rotate by an angle ψ13 . A0 B0 length can be taken as any value if it is not specified by the problem statement. Since A0 and B0 are fixed pivot points, these are the permanent poles for links 2 and 4. Geometrically this problem will be solved using two different methods. Method 1: Since everything stated for two positions is valid for three positions, consider satisfying positions 1 and 2 and positions 1 and 3 simultaneously. The following construction can be performed (In the figures φ12 = 700 ψ12 = 400 andφ13 = 1200 ψ13 = 600 are used. Of course, any angle can be used. 1. Determine R12 (1) and R13 (1) using A0 as P1j and B0 as P’1j . 2. Select a point A1 on link 2 (or B1 on link 4). Draw lines L1 and L’1 from R12 (1) and R13 (1) . 3. Draw Lines L2 and L’2 such that < L1 R12 (1) L2 = < A0 R12 (1) B0 and < L’1 R13 (1) L’2 = < A0 R13 (1) B0 . 4. The intersection of L2 and L’2 is B1 . The mechanism A0 A1 B1 B0 will satisfy the angular rotations between the positions (Fig. 3.43). The validity of this method stems from the conclusion stated for two positions: “Cranks subtend the same angle at the pole. This angle is one half the rotation of the plane about the pole: ½ φ12 or (½ φ12 + π)” stated in Sect. 2.1. In relative motion the angular rotation of the plane is ½ (ψ12 − φ12 ) or ½ (ψ12 − φ12 ) + π. It must be obvious that the correlation of slider displacement with crank rotation will follow. R12 (1) and R13 (1) will be determined as described in Sect. 2.8 (P’12 and P’13 is at infinity) and the method described above will be used (select A1 or B1 , determine the corresponding B1 or A1 ). Method 2. In this method, we shall perform direct kinematic inversion by keeping link 2 or 4 and releasing the fixed link. First, we can assume the length and the first position of link 2 or 4 by selecting A1 or B1 at any appropriate location. One can then determine B2 and B3 as shown in Fig. 3.44. Fig. 3.42 Crank angles φ12 , φ13 of link 2 to be correlated with the crank angles ψ12 , ψ13 of link 4

3.8 Correlation of Crank Angles

157

Fig. 3.43 Use of relative poles R12 1 and R13 1 to locate B1 for a given A1 for the correlation of three crank angles Fig. 3.44 Selection of B1 . Determine B2 , B3 using ψ12 and ψ13 given

Fig. 3.45 Relative positions of B0 B with respect to link 2 in position 1 and determination of A1

Now we invert the motion by keeping link 2 fixed and keeping the relative position between links 1 and 2 the same. This is performed by rotating A0 B0 B2 and A0 B0 B3 fixed to each other and rotating about A0 by angles –φ12 and –φ13 as shown in Fig. 3.45. The center of the circle defined by the three points B1 , B’2 and B’3 will be

158

3 Three Positions of a Moving Plane

the center point for the inverse motion (Fig. 3.45). A point A1 on link 2. Of course, one can select any appropriate point A1 and determine the corresponding B1 in the same way. The resulting mechanism is A0 A1 B1 B0 as shown. One must always check the result for movability and other constraints which may be imposed according to the application. Note that in the correlation of crank angles we select A1 or B1 . We have two scalar design parameters. The same method can be used for the correlation of slider displacement with crank rotation.

3.8.2 Analytical Method In Chap, 6 another way of treating the correlation of crank angles will be given. In this chapter this problem is treated as a position synthesis problem using kinematic inversion. We can also use the dyad formulation for the correlation of crank angles. Let us select B0 B1 arbitrarily. It will be denoted by a vector ZB . Since we want B0 B to rotate by angles ψ12 and ψ13 , we can determine B2 and B3 . δ2 and δ3 will be given by (Fig. 3.46): δ2 = Z B eiψ12 − Z B = Z B (eiψ12 − 1) δ3 = Z B eiψ13 − Z B = Z B (eiψ13 − 1) Referring to Fig. 3.47 we have a vector loop in the form: W eiφ1 j + Z − δ j − z − W = 0 or ( ) ( ) W ei φ1 j − 1 + Z eiα j − 1 = δ j Where αj is the rotation of the coupler link from the first to the j th position. For three positions: Fig. 3.46 Selection of B1 . Determine δ2 , δ3 vectors using ψ12 and ψ13 given

3.8 Correlation of Crank Angles

159

Fig. 3.47 Vector Dyad A0 AB at first and jth positions for the correlation of crank angles

( ) ( ) W eiφ12 − 1 + Z ei α2 − 1 = δ2 ( ) ( ) W eiφ13 − 1 + Z ei α3 − 1 = δ3 When B1 is selected, δ2 , δ3 and φ12 , φ13 are known. We have two complex equations in terms of two complex (W and Z) and two real (α2 and α3 ) unknowns. If we assume α2 and α3, we can solve for W and Z. Example 3.9 We would like to design a four-bar mechanism such that: φ12 = 70o ψ12 = 40o and φ13 = 120o ψ13 = 60o where φ and ψ are the rotation angles of the two links connected to the fixed link. Let B0 B1 = 45 mm < 60° and α2 = 15°, α2 = 25o Using the given values, we calculate δ2 = −28.925 + 10.528i and δ3 = −44.316 + 7.814i. Using the dyad equations, we solve for W and z: W = 4.239 + 10.425 i Z = 59.075 + 55.356 i The resulting mechanism and its three positions are shown in Fig. 3.48. (A0 A = 11.254 mm, AB = 80.957 mm, A0 B0 = 46.506 mm and A0 B0 makes 42.316° with the horizontal reference frame.). If required, one can rotate the whole mechanism 42.316° CW to make A0 B0 horizontal and scale the links such that A0 B0 is of a certain required length. Note that in this case there are four free design parameters (B0 B1 vector and α2 , α3 ). (You can rotate the whole mechanism such that A0 B0 is horizontal). In case of the correlation of slider displacement with crank angle, the same method can be easily used, since the resulting dyad equation will be the same for the correlation of crank angles (with δj given as slider displacement. Figure 3.49)

160

3 Three Positions of a Moving Plane

Fig. 3.48 Resulting Four-bar mechanism for Example 3.8 (arbitrarily selected coupler link angles)

B3

B2 B1

B0 A2 A3

A1 A0

Fig. 3.49 Vector Dyad A0 AB at first and jth positions for the correlation of crank angle with slider displacement

We can rewrite the dyad equation and specify the location of A0 instead of using α2 and α3 as free parameters, by defining a vector R = B1 A0 . For three positions now we can write three equations as: W + Z = −R = R1 W eiφ12 + Z eiα2 = −R + δ2 = R2 W eiφ13 + Z eiα3 = −R + δ3 = R3 If A0 is specified, R is known and one can calculate R1 , R2 and R3 . This is a set of three complex equations in 2 complex (W and z) and two real (α2 and α3 ) unknowns. In order to have a solution for W and z the determinant of the augmented matrix must be equal to zero. The solution method is the same as described in Sect. 4.2.2 for center or circle point specification.

3.9 Path Generation with Prescribed Timing

161

Fig. 3.50 Resulting four-bar mechanism for Example 3.9 (B0 A0 arbitrary)

Example 3.10 In addition to the angles to be correlated in the previous example, we would like to have A0 to be located 100 mm to the left of B0 . We calculate –R = 128.925 + 34.472i and: R1 = 128.925 + 34.472i R2 = 100 + 45i R3 = 84.609 + 42.286i When we expand the determinant of the augmented matrix and solve for α2 , we obtain two solutions α2,1 = 70° and α2,2 = 4.307°. The first solution is equal to φ12 and therefore is a trivial solution. Using α2 = 4.307° we determine α3 = 11.587°. Using any two of the three complex equations we can solve for W and Z: W = 15.153 + 18.558i z = 113.775 + 15.918i The resulting mechanism and its three positions are as shown in Fig. 3.50 (A0 A = 23.958 mm, AB = 114.883 mm, A0 B0 = 100 mm).

3.9 Path Generation with Prescribed Timing Another type of problem that we see in certain applications is to correlate the path of a point on the moving plane with the angular rotation of one of the cranks. If we assume that the crank is rotating at a constant speed, the amount of angular rotation will correspond to the time it takes to move in between the prescribed positions. Therefore, this problem is commonly known as “path generation with prescribed timing” (Fig. 3.51). For geometrical synthesis, first let us assume that this is a position synthesis problem where a point on the moving plane is displaced about the given path and rotated by the given crank angles (Fig. 3.52). We can easily select a point on the

162

3 Three Positions of a Moving Plane

Fig. 3.51 Path with prescribed timing (φ12 , φ13 and δ2 , δ3 given)

moving plane (B1 ), identify the homologous points B2 , B3 and determine the center of the circle (B0 ) that passes through the three homologous points or invert the motion and select center point B0 and determine B1 . The result is a dyad B0 BP, in which the moving plane PB will move through the given specified positions. However, rather than φ12 and φ13 being the angular rotation of the moving plane, we want a crank to rotate by these angles when point P moves from P1 to P2 and to P3 . In order to achieve this, we utilize a simple parallelogram construction as shown in Fig. 3.53. Consider line B0 A1 parallel to B1 P1 and line A1 P1 parallel to B0 B1 . In such a case, since the sides of a parallelogram will always be parallel, the angular rotation BP, will be equal to the angular rotation of A0 A. When point P1 moves to P2 , A1 will move to A2 , rotating by an angle φ12 . For the second dyad the moving plane at its three positions are A1 P1 , A2 P2 , A3 P3 . We can select a circle point C on this moving plane and determine its corresponding

Fig. 3.52 Three positions of the coupler link, displaced by δj and rotated by φ1j (j = 2,3). Select B1 (or B0 ) and determine B0 (or B1 )

3.9 Path Generation with Prescribed Timing

163

center point, C0 . Hence a four-bar mechanism A0 ACC0 is designed in which as the crank A0 A1 rotates by angles φ12 , φ13 and in the meantime point P on the coupler link moves from P1 to P2 and P3 . (Fig. 3.54). Fig. 3.53 Construction of a paralelogram B0 A1 P1 B1 . The angular rotation of B0B1 in between the positions are φ12 and φ13

Fig. 3.54 Determination of the second dyad for path generation with prescribed timing: Select C1 (or C0 ) and determine C0 (or C1 ) for three positions of the coupler A1 P1 , A2 P2 and A3 P3

164

3 Three Positions of a Moving Plane

Consider the dyad formulation for the analytical treatment of the path generation with prescribed timing (Fig. 3.55). Writing the basic form of the dyad equation: W (eiβ j − 1) + Z (ei α j − 1) = δ j For three position path generation with prescribed timing, δ2 , δ3 and β2, β3 will be specified. In this case one can assume the coupler angle α2 , α3 and solve these two linear complex equations for two complex unknowns WA and ZA to determine the circle and center point A1 and A0 . For the second dyad one must use the angles α2 , α3 which were assumed for the first dyad as given values and assume the crank angles β2, β3 and solve for WB , ZB to determine the circle and center points B and B0 . Instead of using the basic form, one can use circle or center point specification using the dyad equation in the form: W + Z = R1 W eiβ2 + Z ei α2 = R2 = δ2 + R1 W eiβ3 + Z eiα3 = R3 = δ3 + R1 where R1 = −R = O’1 A0 , the location of the center point. For path generation with prescribed timing if we now specify the location of the center point in the above equations δ2 , δ3 and β2, β3 will be given and one can compute R1 , R2 , R3 . The unknowns are W, Z and the coupler link angles α2 , α3 . Equating the determinant of the augmented matrix to zero, one will obtain a complex equation in two unknowns α2 , α3 , which will yield a quadratic equation for α2 or α3 . one of the solutions will be trivial (α2 = β2 and α3 = β3 ). The other solution is the result. Using the given δ2 , δ3 and the computed coupler link angles from the first dyad (α2 , α3 ), we can select another center point or circle point for the second dyad and determine the corresponding circle point or center point as described in Sect. 3.4.2.2. Instead of preparing a new routine for this problem, one can easily use an algorithm prepared for three position synthesis. If we replace α2 and α3 with the values of β2 Fig. 3.55 Dyad A0 AO’ in the first and j th positions

3.9 Path Generation with Prescribed Timing

165

Fig. 3.56 In path generation with prescribed timing W and Z vectors change their role. Final solution is obtained through parallelogram construction

and β3 in path generation with prescribed timing, from the above equations W and Z change their role (Fig. 3.56). What was the z vector is now W and W vector is now z. after solving for W and z, we form a parallelogram loop as shown to reverse the roles W and z. As a solution instead of A,1 , the circle point will be A1 . Example 3.11 A straight-line path generating four-bar mechanism is to be designed. The coupler point has to pass through the precision points C1 (0,0), C2 (250,0) and C3 ((440,0). During this displacement of the coupler point crank A0 A, A0 located at (330.7, −66), has to rotate φ12 = 16o and φ13 = 28° (CCW) when C1 moves to C2 and C3 respectively. B0 must be located at (134.2, −208). We assume the angular rotation of the crank is the angular rotation of the coupler plane. Hence, we construct three positions of a moving body C1 D1 , C2 D2 , C3 D3 as shown in Fig. 3.57. Please note that CD is not the coupler plane motion since the crank angle is assumed to be the angular rotation of this plane. For three positions of plane CD, we first locate the poles at the intersection of a pair of the perpendicular bisectors of the corresponding homologous points, construct the pole and image pole triangles. Considering the motion as inverted (i.e., moving plane fixed, fixed plane moving), we consider A0 as a circle point and take the image of A0 with respect to side 1 of the image pole triangle to determine the ground point A0G , and take the image of A0G with respect to side 2 and 3 of the image pole triangle to determine A02 and A03 . A’1 is the center of the circle defined by these points. A’ is a point on the moving plane CD which moves on a circle with center A0 and the plane ACD rotates by angles φ12 and φ13 and the crank A0 A’ will rotate by α2 and α3 in between the positions. To convert the angular rotation of the coupler to the angular rotation of the crank we construct a paralelogram with AO A’1 and A’1 C1 as the to

166

3 Three Positions of a Moving Plane

Fig. 3.57 Path generation with prescribed timing. 3 points on the coupler path (C1 , C2 , C3 ), the angular rotation of crank A0 A (φ12 , φ13 ) and the fixed pivot point locations A0 , B0 are given

sides and determine A1 . Now A0 A1 will rotate by angles φ12 and φ13 and the coupler will rotate by α2 and α3 in between the positions (Fig. 3.58). Now the three positions of the moving plane is C1 A1 , C2 A2 , C3 A3 . We can construct the pole triangle as shown in Fig. 3.59. Since the center point B0 is known one can determine B1 , by inverting the motion and determine B02 , B03 as described for the previous dyad. Instead, one can use the theorem “The angle formed at a pole Fig. 3.58 Construction of the pole and image pole triangles. Using the image pole determine A’1 as the center of the circle defined by A0 j. Construction of a parallelogram to convert the angular rotation of the coupler CD to crank rotation A0 A

3.9 Path Generation with Prescribed Timing

167

Fig. 3.59 Construction of the pole triangle for three positions of the plane CA (C1 A1 , C2 A2 , C3 A3 ) and determining the circle point for a given center point

by one adjacent side of the pole triangle and a line drawn from the ground point or the center point is equal and opposite to the angle formed by the other adjacent side and a line drawn to the center point or the ground point” to determine the Ground point BG and take the image of BG with respect to side 1 of the pole triangle to determine B1 (Fig. 3.59). The four-bar mechanism is shown in Fig. 3.60. When the crank A0 A rotates by φ12 and φ13 , coupler point C will be displaced from C1 to C2 and C3 respectively. When solving the same problem numerically using Excel®, as input again we use the crank angles φ12 and φ13 as the angular rotation of the coupler link. For three positions we use: First Dyad

X

Y

Degrees

Radians

δ2

250

0

φ12

16

0.279253

δ3

440

0

φ13

28

0.488692

and

168

3 Three Positions of a Moving Plane

Fig. 3.60 Four-bar mechanism synthesized geometrically using Geogebra. As point C moves from C1 to C2 and C3 , A0 A rotates by φ12 and φ13 . Note that the solution is unique when A0 and B0 are specified

A0 = R

X

Y

330.7

−66

We now utilize the function the function Pivot(φ12 , φ13 , δ2 , δ3 , R) to obtain –Z vector:

– Z

ThreePositions2_Fixed

X

Y

159.552

413.2436

−−−→ −−−→ This vector is C1 A’1 . It is not the vector C1 A1 . To obtain the vector C1 A1 . we again consider the parallelogram: −−−→ C1 A1 = R − (−Z) = (330.7 − 66i) − (159.5521 + 413.2436i) = 171.1478−479.2436i To locate A2 and A3 with respect to the origin we must rotate A0 A1 (=−(−Z)) by angle φ1j (j = 2, 3) and R vector. We obtain:

A1

X

Y

171.1479

−479.244 (continued)

3.9 Path Generation with Prescribed Timing

169

(continued) X

Y

A2

291.2341

−507.214

A3

383.83

−505.778

For the second dyad we must determine three positions of the moving plane CA. To determine the angular positions we use Atan2() function as: )) ( ( φk = a tan 2 (Akx − Ckx ), Aky − Cky . where k = 1, 2, 3 and Akx , Aky are the x and y coordinates of point Ak . Radians

Degrees

φ1

−1.22779

−70.3473

φ2

−1.48968

−85.3523

φ3

−1.6814

−96.3371

Again we use ThreePosition2_FixedPivot function with input data as: X

Y

Radians

Degrees

δ2

250

0

α2 = φ2 − φ1

6.021298

344.9949

δ3

440

0

α3 = φ3 − φ1

5.829577

334.0102

and

B0 = R

X

Y

134.2

−208

X

Y

77.85723

−553.286

The result is:

C1 B1 = −Z

To determine the coordinates of B2 and B3 , we rotate the Z vector by αj and add δj vector (C1 Bj = δj + Zeiαj ). The coordinates are: X

Y

B1

77.85723

−553.286

B2

181.9545

−554.578

B3

267.5275

−531.451

170

3 Three Positions of a Moving Plane

Knowing the coordinates of all the joints, one can then construct the four-bar mechanism in three positions. If we are to animate the mechanism, the link lengths and their angular orientation in the first position must be determined (use mag(x,y) and atan2(x,y) functions to determine the length and orientation of vectors). Radians

Degrees

A0 B0

a1

242.43814

γ1

−2.51583

−144.146

A0 A1

a2

442.97537

γ2

−1.93926

−111.111

AB

a3

119.10242

γ3

−2.47072

−141.562

B0 B

a4

349.85247

γ4

−1.73255

−99.2677

CB

558.7368

CA

508.8871

< BAC

1.898662

We use θ as the angular rotation of A0 A from its first position and use FourBar2 function as Fourbar2(a2 , a3 , a4 , a1 , 1,θ + γ2 − γ1 ). The output are the angles θ13 and θ14 measured from the fixed link A0 B0 . To determine the orientation of the links with respect to the positive x axis we must add γ1 to these angles. θ

θ13

θ14

3

−0.01093718

0.834770046

Relative to A0 B0 (in radians)

−2.52676613

−1.681058913

Relative to x axis (in radians)

−144.773035

−96.31758081

Relative to x axis (in degrees)

The mechanism can now be sketched at three positions and it can be animated, as shown in Fig. 3.61 (Excel file available as PathGeneration.xls).

3.10 Design of Six Link Mechanisms for Three Positions Using the relative motion concept, similar to two position synthesis we can synthesize mechanisms with higher number of links for three positions. Since there are more design parameters, certain constraints may be easier to achieve using a six-link mechanism. The first example is the lamp post problem. Since we would like to have the lamp move horizontal above the table, we define three positions of the lamp as shown in Fig. 3.62. We want the lamp supports to be within the dashed area on one side of the table.

3.10 Design of Six Link Mechanisms for Three Positions

171

Fig. 3.61 Numerically synthesized four-bar mechanism. As point C moves from C1 to C2 and C3 , A0 A rotates byφ12 and φ13

#1

#2

#3

Lamp 750 1.500 Table

Fig. 3.62 Three positions of a lamp. The fixed pivots must be located within the rectangular area (dashed lines)

Let us first select a dyad A0 AB where A0 is within the region specified for the fixed pivots and B on the lamp (Fig. 3.63). Next, let us invert the motion and determine the motion of the lamp relative to link A0 A. If we select a point C on the lamp and determine its three positions relative

172

3 Three Positions of a Moving Plane

Fig. 3.63 Select dyad A0 AB

A1 A2 A3

B1

B2

#1

#2

B

#33

Lamp

A

750

0

1.500 Table

to link A0 A as shown, we can pass a circle through C1 , C’2 , C’3 . The center of the circle is D1 on link A0 A (Fig. 3.64) (one can also select select D1 and Locate C1 ). Now consider the motion of link AB or CD relative to the fixed plane (in Fig. 3.65 link AB is selected). If we select a point E on one of these planes Its three homologues points will be E1 , E2 , E3 , Through these three points we can draw a circle with center E0 . Of course, you must make some trials to make sure that E0 is within the specified region or, you can select E0 first and invert the motion (keep plane AB fixed and determine the motion of the fixed frame in this inverted motion) and determine E1 as the center point of the circle that passes through E0 , E’02 , E’03 . The final mechanism is shown in first and third positions in Fig. 3.66. Fig. 3.64 Relative motion of the lamp with respect to A0 A for three positions. Select C1 (or D1) and determine D1 (or C1)

D1

A1 B' C'33 B'2 C'2 B1 #1 C1 Lamp

A0

Table

3.10 Design of Six Link Mechanisms for Three Positions Fig. 3.65 6 Link mechanism synthesized for three positions

A1E

1

A2 E 2

173

A3 E3

B1

B2

#1

#2

B

#33

Lamp 750

A0

1.500

E

0

Table

#3

Fig. 3.66 Lamp post mechanism in the first and third positions

Problems In the solution of the following problems one can use Geogebra or Excel. When using Excel use of function routines will make solution simple. One can either write these function routines or download from the site. 1. A rigid body in plane motion is shown in Fig. 3.67 in three positions. (a) Locate the poles and draw the pole triangle. (b) Locate the point on the moving system which has its three positions on a circular arc about D0 (2,-7) as the center (Locate D1 )

174

3 Three Positions of a Moving Plane

(c) Locate the locus of points on the moving system having their three corresponding positions on a straight line. 2. 3 positions of a link AB is given in Fig. 3.8 Determine the proportions of the following mechanisms which are to synthesize this motion and draw the mechanisms in position # 1. (a) Link AB is guided by two pins fitting into straight guides. (b) Link AB is guided on the planet gear of an epicyclic gear train, in which the sun and planet gears have the same number of teeth. (External mesh) (c) Link AB is attached to the coupler link of a 4-bar mechanism (Fig. 3.68). 3. Three positions of a body in plane motion is specified. Positions 2 and 3 coincide. For two points A and B, velocity directions are given in position 2 (or 3). Can the pole triangle be drawn? Can you construct a four-bar mechanism to realize this motion. How would you formulate this problem by utilizing the dyad formulation? 4. A four-link, planar RPRP mechanism is to be designed to realize the motion shown in Fig. 3.69. (This mechanism is known as conchoidal motion mechanism). Synthesize the proportions of this mechanism. Fig. 3.67 Three positions of a moving plane for Problem 1

B1(0,5)

A (0,0)

B2 (7,0) B3

A1(2,0)

1

Fig. 3.68 Three positions of a moving plane for Problem 1

A3(12,0)

B 3

A (0,0) 1

A 2(1,0) B 1

B2

A (2,0) 3

3.10 Design of Six Link Mechanisms for Three Positions

175

5. Figure 3.69. 6. A mechanically reproduced scooping motion incorporated in a truck designed principally for handling lead ore is shown in Fig. 3.70. Three positions of the crank A0 A and coupler AC are given. It is required to determine the follower position (B1 B0 ). Preferably, we would like to have B0 withing the region enclosed in dotted lines and B on the line AC. A0 A = 550 mm AC = 1550 mm. 7. We would like to replace the slider guidance of the slider crank mechanism by a follower (BB0 ) attached to the coupler for the 3 positions shown in Fig. 3.71. A0 A is to be the crank for this 4-bar mechanism, and we would like to have crank-rocker proportions. Determine a follower (A0 A = 50, AB = 200). 8. Figure 3.72 shows the thread feed of a central spool sewing machine. The eye of the thread lever has to be on the lines l1 , l 2 , l 3 when the crank rotates by angles

Fig. 3.69 Three positions of plane CD to be realized by a chonchoidal motion (RPRP) mechanism

C 3 A1 32°

15 50

° 60

400

C 2

550

47°

550

60 °

10°

12°

A2

C 1

A3 400

Fig. 3.70 Loader mechanism. Three positions of the crank and the coupler

176

3 Three Positions of a Moving Plane A1

° 60

B3 B 2

B 1

60

°

A2

A3

Fig. 3.71 Three positions of a slider-crank mechanism to be replaced by a four-bar

θ12 and θ23 . Originally, the motion was achieved by a disk cam. A four-bar mechanism was found to be more advantageous. Since the first and the third positions must be at the dead center positions of the crank (the eye must not be above l1 and below 13 ) a crank AA0 was selected and a certain length AC for the coupler was assumed. Due to space limitations we have to select B0 , the center of the other crank, on the machine head. An appropriate position is shown on the figure. Determine the corresponding circle point at position # 1 (B1 ) so that the coupler point C on the four-bar A0 ABB0 satisfies the given motion requirements. Check whether the four-bar thus obtained is of crank-rocker proportions. If you want, you can select B0 at another position on the machine head.

34 0

C1

309

80

150

500

B0

A1 47°

1° A2 12

80

55

A0 C2 300

Fig. 3.72 Three positions of a thread feed for a central spool sewing machine

3.10 Design of Six Link Mechanisms for Three Positions

177

9. Figure 3.73. shows an existing in-line slider-crank mechanism in a part of a machine (A0 A = 50, AB = 200). We would like to obtain a dwell motion from this slider-crank mechanism. Link 6 is required to be in an approximate dwell while the crank (link 2) rotates by angle φ = θ13 = θ12 + θ23 = 120◦ as shown. A possible solution would be to find a point C on the coupler link C (C1 , C2 , C3 ) whose 3 corresponding positions lie on a straight line. The shape of the mechanism thus obtained is shown in Fig. 3.74. If the path of point C corresponds with the prismatic joint C deviates from this line link 6 will translate. Synthesize such a mechanism. 10. Figure 3.75. shows an existing 8-link mechanism. Link 2 is the driven crank and link 8 is the oscillating output. However, due to bad transmission angle of the slider at position # 1 the machine will not function properly. Instead of a slider (link 6), we would like to attach a swinging block mechanism (Fig. 3.76), while keeping the characteristics of the original motion. Hence when link 2 rotates, link 4 has its limit positions at B1 and B3 We determine B2 at the midpoint of the swing. Our aim is to determine a line on link 5 whose 3 corresponding Fig. 3.73 Three positions of a slider-crank mechanism to be used for a dwell motion

A1 B1 B3 A0

B2

A2 A3

Fig. 3.74 A possible dwell mechanism when the coupler point C of the slider-crank mechanism describes an approximate straight line

Path of c ci

6 ck

A1 2 A0

5

C

3 4 B1

178

3 Three Positions of a Moving Plane

positions are concurrent at a point while the plane of link 5 is at positions B1 C1 , B2 C2 , B3 C3 . Determine the mechanism thus obtained. Compare the motion of point 8 for both mechanisms. 11. Figure 3.77 shows the principle of a lift truck in which the fork is guided by a linkage. The fork must move up and down in nearly rectilinear translation. Determine the dimensions of the required linkage. Note that A0 ABB0 is a fourbar linkage, and that point C should have an approximate straight-line path of 1800 mm length. Link DE maintains the fork in a motion of translation: to determine this link, use three positions of accuracy for which the fork isto be exactly horizontal at ground level, halfway up and all the way up. Points A0 and B0 may be chosen in any convenient location within 1250 × 2500 mm rectangle. Keep in mind that the truck must have maneuverability in close quarters where the headroom is low. There are several methods to solve this problem one possible procedure is: (a) First omit link DE and the fork CD. Choose a certain link length for A0 A and AC (keeping in mind the place limitations, etc.) (b) Point C on link AC must move on a vertical line. Select 3 corresponding points of C at equal distance (or use Chebyshev spacing which is explained in Chap. 8) and determine the three positions of link AC. (c) Draw the pole triangle for the 3 positions of AC. Select a certain point B, and determine its center-point B0 , or, you can select a certain center-point B0 and use the theorem “The coupler and the fixed link subtend the same angle at the pole”. We would like to have B on the line AC, but you can omit this requirement if necessary. (d) Now that you have selected A0 ABB0 , at the three positions of C draw a vertical (or horizontal) line from point C. This line represents the required positions of the fork. To determine the link DE we are interested in the D B3

B2 8

7

B0

5 B1

3 C

A2

6

BB0=125 AA0=70 CB =160 AB =325 CD=280 DD0=300

Fig. 3.75 An existing eight-link mechanism

A3

2 A0 A1

30

4

D0

3.10 Design of Six Link Mechanisms for Three Positions

179

E0 6

D

B 8

7

B0

5

C BB0=125 AA0=70 CB =160 AB =325 CD=280 DD0=300

3 A 2 A0

30

4

D0

Fig. 3.76 Mechanism to replace The eight link mechanism of Fig. 3.75

Fig. 3.77 Mechanism for a vertical lift

relative motion between the fork plane and the crank BB0 . To investigate this relative motion rotate C2 B2 B0 , C3 B3 B0 (and the vertical line you have drawn). About B0 such that B2 and B3 will coincide with B1 . The new

180

3 Three Positions of a Moving Plane ,

,

,

positions of C(C1 , C2 , C3 , say) and the fork plane is the motion relative to the crank BB0 . Draw the pole triangle for this relative motion, select a point D1 and determine its corresponding center point (E1 ). 12. Figure 3.78 shows the principle of the linkage guidance of a mixer motor. The motor must move up and down in nearly rectilinear translation approximately 250 mm. Determine the dimensions of the required linkage for 3 positions of the motor, lowest midway and highest (or use Chebyshev spacing (refer to Chap. 8)). Keep space limitations in mind when choosing your links. (Look at the method discussed for the fork-lift. Later we shall see another method for the synthesis of this motion). 13. Design an in-line slider crank mechanism so that 90o clockwise rotation of link A0 A results in a 300 mm rectilinear translation of point B from left to right along a horizontal line. Displacement of B along the line must be approximately proportional to the corresponding rotations of A0 A within the 90° interval. Use three accuracy points. Check your result. 14. Consider the in line slider crank mechanism you have designed in problem 12. Find a point C on the coupler link AB, whose three homologus points C1 , C2 and C3 lie on a circle with center C0 as shown in Fig. 3.79. Thus you can design a four-bar mechanism A0 ACC0 with point B as the coupler point whose three positions lie on a straight line. Fig. 3.78 Mixer mechanism for guiding the motor approximately on a vertical path

E D C A

B

250

A0

B0

3.10 Design of Six Link Mechanisms for Three Positions Fig. 3.79 Location A0 and C0 for problems 12 and 13

181

C0

Path of point B A0

15. We would like to correlate a crank displacement with a slider displacement as shown in Fig. 3.80a). In order to achieve this, double slide mechanism shown in Fig. 3.80b) is considered (the moving and fixed slides may be selected as perpendicular or at an angle with each other). Derive a procedure for the synthesis of such a mechanism and using this procedure, synthesize a mechanism for which: ◦

◦

◦

θ1 = 30 s1 = 100 mm θ2 = 40 s2 = 80 mm θ3 = 50 s3 = 60 mm Hint: Consider kinematic inversion. What is the locus of A1 for this inverted motion? 16. As a part of automation process, a four-bar mechanism must be designed to remove boxes from one conveyor belt and deposit to an upper conveyor belt as

(a)

(b)

Fig. 3.80 a Correlation of slider displacement si with crank rotation θi b Double slide mechanism to realize the three positions defined in (a)

182

3 Three Positions of a Moving Plane

shown in Fig. 3.81. Both the moving and fixed pivots must be located between the upper and lower conveyor belts. (a) Design a four-bar mechanism using graphical method. (b) Design a four-bar mechanism by using the basic form of the dyad equation (specify the crank rotations arbitrarily) (c) Design a four-bar mechanism by specifying the fixed pivot points (or moving pivot points) 17. A crank-rocker path generating four-bar is required to advance film in a camera as shown in Fig. 3.82. Given C1 , C2 , C3 and φ2 , φ3 and selecting A0 = 0–75i; B0 = –56–119i. Determine the four-bar mechanism proportions. 720 40°

UP

432

576

UP

UP

144

Fig. 3.81 Three positions of a box when transported from one conveyor to the other Fig. 3.82 Three positions of a coupler point and the corresponding crank rotations. Location of fixed pivots are given

3.10 Design of Six Link Mechanisms for Three Positions

183

C1 = 0 C2 = −0 + 26 i C3 = 31 + 10 i φ12 = 800 , φ23 = 1100 18. A Small autoclave is to be used to sterilize medical instruments. The door must be stored on the inside of the autoclave when it is open. The door must be closed by a mechanism from the inside to form a seal with a gasket that allows the steam pressure to reach 15 psi on the inside of the vessel, forcing the door to stay closed (as in the case of a steam cooker). Synthesize a four-bar mechanism for the autoclave door using the three positions given in Fig. 3.83. (If you cannot find a satisfactory solution using the given three positions, you can change the location of the mid position and its angle) 19. Consider the overhead garage door problem solved for two positions in Chap. 2. Let us consider the third position in between the open and closed positions of the door (Fig. 3.84). Design a six-link mechanism with the same kinematic structure. Q1 = 0 + 0 i, Q2 = 750 − 100 i, Q3 = 1400 + 0 i 20. The structure shown in Fig. 3.85a is commonly known as “level luffing jib crane” which is commonly used at ports ship docks, etc. The hook of the crane is above 28 m and must have an outreach of 37 m (minimum 9.5 m). After the load is lifted to a certain height, it must remain level to the ground when moved inwards. Point C is a coupler point of the four bar which must describe an approximate straight line within the working range of the crane. One way of synthesizing this mechanism is to select A0 A (at the extreme position you can locate point C and select an appropriate distance AC). Select three precision points C1 , C2 , C3 at x1 = 35 m, x2 = 23.25 m and x3 = 11.5 m at a height of 28 m from point A0 . Select a moving pivot point B (or select a fixed pivot point Fig. 3.83 Three positions of an autoclave door

184

3 Three Positions of a Moving Plane

Fig. 3.84 Three positions of a garage door

Q1= 0+0i, Q2=750 -100i, Q3=1400+0i

B0 ) and determine the corresponding fixed pivot B0 (or corresponding moving pivot B) that point C will be on the straight line for the given three positions. 21. A small truck is to be used as a dump truck. To minimize the force acting on the piston when raising or lowering, the center of gravity is to move on a line making a small angle with the horizontal. The three positions of the dump are given by C1 (0,0), φ1 = −5◦ , C2 (920, 290), φ2 = −50◦ , C3 (1860, 160), φ3 = −80◦ (Fig. 3.86). You are to design a four-bar mechanism to realize this motion. Note

A

C

C2

40 m C2

C2

xi 28 m

9.50 m

9.50 m 37.00m

37.00m

A0

0.8 m

Fig. 3.85 a Level luffing jib crane, b A0 AC dyad selected. Designer must determine B0 B

References

185

Fig. 3.86 Dump truck

that you can place the fixed pivots in the dark region shown. Moving pivot can be placed on the side of the dump.

References 1. R. Beyer, Kinematısche Getriebesynthese (Springer, Heidelberg, 1953). (Translated into English by H. Kuenzel as: “The Kinematic Synthesis of Mechanisms”, Chapman and Hall, London, 1963) 2. A.G. Erdman, Three and four precision point kinematic synthesis of planar linkages. Mech. Mach. Theory 16, 227–245 (1981) 3. G. N. Sandor, A. G. Erdman, Mechanism Design- Analysis and Synthesis, Vol. 1& 2 (Prentice Hall, 1984) 4. C. ZWIKKER, The Advanced Geometry of Plane Curves and Their Applications. Dover Publication Inc., (1963) (First Published in 1950) 5. R. DEAUX, Introduction to the Geometry of Complex Numbers. Dover Publication Inc (1956) (First published in French, 1893, Translation by H. Eves

Chapter 4

Four Positions of a Moving Plane

Abstract Design for four positions of a moving plane is based on Burmester Theory. The invariants of motion are the complementary pole quadrilaterals. In order to attach a crank to the moving plane to move through four prescribed positions one must determine points on the moving plane whose four homologous positions lie on a circle. Using this condition the analytical equation for the circle point curve (by inverting the motion the center point curve) is derived and the geometric construction of circle and center point curves are explained. The design concept for four positions is generalized for the path generation with prescribed timing and correlation of crank angles. Keywords Motion generation · Four position synthesis · Burmester curves · Ball’s point · Burmester theory

Design for four positions of a moving plane is based on Burmester Theory. The invariants of motion are the complementary pole quadrilaterals. In order to attach a crank to the moving plane to move through four prescribed positions one must determine points on the moving plane whose four homologous positions lie on a circle. Using this condition the analytical equation for the circle point curve (by inverting the motion the center point curve) is derived and the geometric construction of circle and center point curves are explained. The design concept for four positions is generalized for the path generation with prescribed timing and correlation of crank angles.

4.1 Four Finitely Separated Positions of a Plane: Burmester Curves Consider four positions of a moving body relative to a fixed reference frame as shown in Fig. 4.1. For four positions of a moving plane, whatever we have stated for two and three positions of a moving plane is still valid, since we can always neglect one or two of the given positions. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_4

187

188

4 Four Positions of a Moving Plane

Fig. 4.1 Four finitely separated positions of a moving plane

B2

B1 A2

A1

A3

B3

A4

B4

F

In case of two position synthesis, there were 6 design parameters and we were able to select circle points A and B arbitrarily then select the center points on the corresponding perpendicular bisectors. In case of three position synthesis, we were able to select circle point A and B (or the center points A0 and B0 , if we consider kinematic inversion), the corresponding center point (or circle point) was a unique point. Hence, the number of free design parameters has reduced to 4. For four positions, you cannot select the circle points A and B arbitrarily. The moving pivots must be on a curve on the moving plane (circle point curve) and the fixed pivots, which are corresponding center points, will be on a curve in the fixed plane (center point curve). These curves are also known as “Burmester Curves”,1 in memory of the person who first found these curves. Our first aim is to determine these two curves when four positions are given. For four position synthesis the number of free design parameters has reduced to two (selection of A and B or A0 and B0 on a defined locus). If we consider two positions at a time, we are going to have 6 poles (P12 , P13 , P14 , P23 , P24 and P34 ). If we consider three positions at a time, we are going to have four pole triangles (ΔP12 P13 P23 , ΔP12 P14 P24 , ΔP13 P14 P34 and ΔP23 P24 P34 ). Also, there will be image pole triangles and image poles. In the geometric treatment of four position synthesis problem, we first define “opposite poles” which are also called “complementary poles”, as the two poles with different indices. (P12 , P34 ); (P14 , P23 ); (P13 , P24 ) are the opposite poles. Using these opposite poles we can form three quadrilaterals (Four sided figures) known as “opposite pole quadrilateral” or “complementary pole quadrilateral”: (P12 , P34 , P14 , P23 ); (P12 , P34 , P13 , P34 ) and (P14 , P23 , P13 , P24 ). In these quadrilaterals, complementary poles are the diagonals. In Fig. 4.2 two different pole quadrilaterals are shown. Dashed lines connect the complimentary pole pairs. We label the sides of the quadrilateral with the common subscript of the vertices. Furthermore, these sides are directed lines. For example, consider two opposite sides 1 and 3. When we consider 1

Ludwig Burmester was a German Kinematician and geometer who lived in the years 1840–1927. He has served as a Professor in Dresden University. He is the pioneeer in kinematic synthesis of mechanisms.

4.1 Four Finitely Separated Positions of a Plane: Burmester Curves

189

vector direction on side 1 as P12 P14 , then the direction for side 3 must be P23 P34 .Hence all sides are directed lines. There will also be “Complementary image pole quadrilaterals”. For example, in the first position There will be (P12 , P34 1 , P14 , P23 1 ); (P12 , P34 1 , P13 , P24 1 ) and (P14 , P23 1 , P13 , P24 1 ) as image pole quadrilaterals. Using the pole quadrilateral and the image pole quadrilaterals, one can geometrically determine circle and center point curves pointwise. In Fig. 4.3. For the given four positions the poles and image poles are determined using the method explained in the second chapter (construction lines are not shown). The three complimentary pole quadrilaterals and image pole quadrilaterals (in position 1) are shown in Figs. 4.4 and 4.5. respectively. One can always draw a circle through any three points. For four points to lie on a circle certain condition must be satisfied. Consider 4 points on a circle as shown in Fig. 4.6. In complex plane the position of these four points are given by complex numbers zk (k = 1, 2, 3, 4). Consider the angles ∠A2 A3 A1 and ∠A1 A4 A2 . Since P23

3

P23

P34

P34

3 2

4

P14

2

P

12

4 1

1

P14

P

12

Fig. 4.2 Opposite (or complementary) pole quadrilateral. Dashed line shows the diagonals, which join complementary poles

P12 B1

B2

(1)

P23 A1 (1)

(1)

P34

P13

P 14

P24

A3 P23

A2

P34

P24 B3

A4 B4

Fig. 4.3 Four Positions of a moving plane. Poles (P12 , P13 , P14 , P23 , P24 , P34 ) and image poles (P23 1 , P24 1 , P34 1 ),in the first position

190

4 Four Positions of a Moving Plane P12 B1

P12

P12 B1

B2

B1

B2

B2 (1)

(1)

P23

(1)

P34

A1

P 14

A3

P13

(1)

P23

P24

A1

P24

P34

P23

(1)

P23

A2

B3

(1) P34

P13

(1)

A3

P 14

P24

A4

P34

B4

P23

(1)

P34

A1

A2 P24

P13

(1) P24

B3

P 14

P23 A 2 A3 P34

A4

P24 B3

A4

B4

B4

Fig. 4.4 Three different complementary pole quadrilaterals for four positions shown in Fig. 4.3

P12 B1

B1

B2

B1

(1)

P23 A1 (1)

P12

P12

(1)

P34

P 14 P13

A3

P23

P34

P24

A1

P24 B3

A4

(1) P23

(1)

P23

A2

(1)

(1)

P34

P13

P 14

P24

B4

B2

B2

A3

P34

P23

A1

A2 P24

(1) P24

B3

(1)

P34

P 14

P13

P23 A2 A3 P34

A4

P24 B3

A4

B4

B4

Fig. 4.5 Three different complementary image pole quadrilaterals for four positions shown in Fig. 4.3

these circumferential angles subtend the same arc length, they must be equal, say β. zk –zl are vectors Al Ak , in complex numbers. These vectors can also be represented as rkl ei γkl : rkl eiγkl = z k − zl = Al Ak β = γ13 − γ23 = γ14 − γ24 Now, the angle β is the difference of the angles γ as: These angles are nothing but the arguments of the corresponding vectors. Therefore: Arg(z1 − z3 ) − Arg(z2 − z3 ) = Arg(z1 − z4 ) − Arg(z2 − z4 ) = β Noting θ2 − θ1 = arg(a1 ei θ2 ) − arg(a2 ei θ1 ) = arg Consider ∠A1 A3 A2 . we have:

(

a 1 e i θ2 a 2 e i θ1

)

( ) a1 i(θ2 −θ1 ) = arg e a2

4.1 Four Finitely Separated Positions of a Plane: Burmester Curves

191

Fig. 4.6 Condition for four points to lie on a circle. ∠A2 A3 A1 = ∠A2 A4 A1

(

z1 − z3 Arg(z 1 − z 3 ) − Arg(z 2 − z 3 ) = Arg z2 − z3

) =β

also note that: aeiθ z = −i θ = e2iθ z ae Hence: ( (

z 1 −z 3 z 2 −z 3 z 1 −z 3 z 2 −z 3

) )=

(z 1 − z 3 ) (z 2 − z 3 ) = e2iβ (z 1 − z 3 ) (z 2 − z 3 )

Similar result will be obtained for 1. One can use the initial crank

7.2 Crank-Rocker Mechanism

297

Fig. 7.8 Locus of points, Ae , Be , location of the moving pivots at extended dead center position satisfying the given swing angle ψ and the corresponding crank rotation φ

angle β as the free parameter rather than λ. In such a case the link lengths are given by the following equations: ( 1 cos 21 φ a2 = −sin ψ [ 1 2 sin 2 (φ ( sin 21 φ 1 [ a3 = + sin ψ 2 cos 21 (φ

+ β

)

] − ψ) ) + β ] − ψ)

a24 = (a2 + a3 )2 + 1 − 2(a2 + a3 )cosβ a1 = 1

(7.14)

(7.15)

(7.16)

One can either use λ or β as a free parameter and obtain different four-bar mechanism proportions. The mechanism proportions that are of crank rocker type will all have the given swing angle and corresponding crank rotation. In order to obtain crank rocker proportions we have limits on swing angle and corresponding crank rotation as: ◦

0 < ψ < 180◦ 1 1 90◦ + ψ < ϑ < 270◦ + ψ 2 2 If we let: ( t = tan

) 1 φ , 2

298

7 Four-Link Crank Mechanisms

] 1 (φ − ψ) , 2 ) ( 1 v = tan ψ 2 [

u = tan

The link lengths can be expressed as: a21

( ( ) ) u2 + λ1 2 1 2 2 1 = = sin (φ − ψ) + λ cos (φ − ψ) 1 + u2 2 2 ( ) v2 2 1 a22 = = sin ψ 1 + v2 2 ( ) λ2 v2 2 2 2 1 ψ a3 = = λ sin 1 + v2 2 ( ) ( ) t2 + λ1 2 1 2 2 1 φ + λ φ a24 = = sin cos 1 + t2 2 2

(7.17)

(7.18)

(7.19)

(7.20)

Example 7.2 Determine the proportions of a four-bar mechanism of 120 mm fixed link length and with ψ = 40° and φ = 160°. (a) Let us select the initial crank angle β = 60◦ .. Using Eqs. (7.15)–(7.17): cos(80◦ + 60◦ ) = 0.30254 sin(80◦ − 20◦ ) sin(80◦ + 60◦ ) = 0.43969 a3 = sin(20◦ ) cos(80◦ − 20◦ ) a2 = −sin(20◦ )

and a24 = (0.43969 + 0.30254)2 + 1 − 2 ∗ (0.43969 + 0.30254)cos60◦ = 0.80867 a4 = 0.89926 For a1 = 120 mm, a2 = 36.30 mm, a3 = 52.76 mm and a4 = 107.91 mm. The maximum deviation of the transmission angle from 90° will be (Eq. (7.4)): cos μ

min max

=

120 ∗ 36.30 107.912 + 52.762 − 1202 − 36.302 ± 2 ∗ 107.91 ∗ 52.76 107.91 ∗ 52.76

= − 0.113247 ± 0.765106

7.2 Crank-Rocker Mechanism

299

From which we obtain: μmax = 151.44◦ (Δ1 = 61.44◦ ) and μmin = 49.32◦ (Δ2 = 40.68◦ ). Since μmax deviates by 61.44° from right angle, μmax is the critical transmission angle (or μmin = 180 − 151.44 = 28.56◦ is the critical transmission angle). (b) If we select λ = 1.4, Using Eqs. (7.18)–(7.21): t = tan(80◦ ) = 5.671282, u = tan(60◦ ) = 1.732051, v = tan(20◦ ) = 0.36397 a21 a22 a23 a24

+ 1.4 = 1.732051 = 1.24 1 + 1.7320512 0.3639702 = 0.3639702 + 1 = 0.116978 = 1.42 ∗ 0.116978 = 0.229276 2 + 1.42 = 5.671282 = 1.028948 1 + 5.6712822 2

2

a1 a2 a3 a4

= = = =

1.113553 0.342020 0.478828 1.014371

120 = 36.86 mm, a3 = 51.60 mm, a4 = When a1 = 120 mm: a2 = 0.342020 1.113553 109.31 mm. The critical transmission angles will be: cos μ

min max

=

109.312 + 51.602 − 1202 − 36.862 120 ∗ 36.86 ± 2 ∗ 109.31 ∗ 51.60 109.31 ∗ 51.60

= − 0.101715 ± 0.784200 From which we obtain: μmax = 152.36◦ (Δ = 62.36◦ ) and μmin = 49.96◦ (Δ = 43.04◦ ). Since μmax deviates by 62.36° from right angle, μmax (or μmin = 180 − 152.36 = 27.64◦ ) is the critical transmission angle. For each value of λ one will obtain a set of linkage proportions. After the mechanism is checked for crank rocker proportions, one can determine the minimum value of the transmission angle for the four bar proportions obtained for each value of λ. Then we can plot a curve similar to the one shown in Fig. 2.7. for φ = 160◦ and ψ = 60◦ . There is a certain range λmin < λ < λmax within which we have crank rocker proportions. These limits depend on φ and ψ.. Within this range every λ value gives a mechanism with a different minimum transmission angle as shown. In order to optimize the force transmission characteristics, we would prefer to use the proportions whose minimum transmission angle deviates least from 90°. This is the maximum of the minimum transmission angle, max(μmin ), corresponding to λopt .. For a given swing angle and corresponding crank rotation this problem can be solved by determining μmin for every crank rocker mechanism corresponding to a value of λ.. We obtain a curve as shown in Fig. 7.9. The value of λ, λopt , for which μmin min is a maximum (max(μmin )), results with a crank rocker mechanism with the best

300

7 Four-Link Crank Mechanisms

Fig. 7.9 Change of minimum transmission angle, μ/ min , as a function of λ = a3 a2 . For a given φ and ψ

35 30

max(μmin)

μmin

25 20 15 10 5 0

0

1

2

3

4

5

6

7

8

λ force transmission characteristics. One can solve this problem as a one parameter constraint optimization problem, which will be shown in Example 7.4. An elegant solution found by Freudenstein and Primrose 1 will be explained. Aim is to find λ for a given swing angle and crank rotation, such that the resulting mechanism has the highest minimum transmission angle (one must select the one for which the maximum deviation of the transmission angle from 90° is a minimum). Substituting the link lengths expressed in terms of a single parameter (λ), into the equation for minimum (or maximum) transmission angle (Eq. (7.21)): cos μ

min max

=

a24 + a23 − a21 − a22 a1 a2 ± 2a3 a4 a3 a4

(7.21)

Equation ( ) (7.21) will be a function of λ only. One can determine the value of λ λopt which minimises the maximum deviation of the transmission angle by setting the derivative of Eq. (7.21) with respect to λ to zero. The value of λopt is then obtained from the solution of the cubic equation: ( ) t 2 1 + t2 Q + 2Q − t Q − = 0 u2 3

2

2

2

(7.22)

where Q = λt2 . The positive real root between u12 < Q < t2 yields opt the four-bar mechanism with crank rocker proportions for which the maximum deviation of the transmission angle is a minimum and which satisfies the given swing angle and corresponding crank rotation. This solution due to Freudenstein and Primrose [2].

1

Freudenstein, F. Primrose, EJF [2].

7.2 Crank-Rocker Mechanism

301

One can use the root finding routines available in MathCad® or MatLab® and Goal Seek and Solver tools in Excel® to determine the correct root of this cubic. A simple numerical solution for the root of the cubic equation can be obtained by applying Newton–Raphson method. Making an initial guess for the root within the feasible region (the midpoint of the interval u12 < Q < t2 is a good choice ( ) i.e., Q0 = 21 u12 + t2 and refining the estimate on the root Q by the recursion formula: ) 2 ( 2Q2k (Qk + 1) + ut 2 1 + t2 Qk + 1 = (7.23) Qk (3Qk + 4) − t2 Q

−Q

The iteration can be stopped when k Q+ 1 k is sufficiently small number k+1 (say 10–6 ). The solution for optimum transmission angle for given swing angle corresponding crank rotation is shown in a chart form in Chart 7.1 (this chart is known as Alt chart), This Chart is from VDI 2130,2 a publication of the German Engineering Association. From these charts the value of β and maxμmin can be obtained when the swing angle and corresponding crank rotation is given. The link lengths can also be found using the equations which express the link lengths in terms of β. (Besides this chart, charts prepared for the ratio of the link lengths are also included in VDI2130). Example 7.3 Determine the proportions of a four-bar mechanism of 120 mm fixed link length and with ψ = 40◦ and φ = 160◦ and for which the maximum deviation of the transmission angle from 90° is a minimum. Noting that t = 5.671282 and u = 1.732051, 0.333333 < Q < 32.163437. Let us assume Q0 = 16.248. The values for seven iterations are: i

0

1

2

3

4

5

6

7

Qi

16.24839 11.4723 8.905767 7.982081 7.857866 7.855707 7.855706 7.855706

After seven iterations, Q = 7.855706 is correct to at least six significant digits. Alternatively, one can enter the initial guess value of Q to a cell in Excel sheet and calculate the value of Eq. (7.22) in another cell using this value of Q, which will not be zero. Now one can use “Goal seek” tool to set the function value to zero by changing the value of Q you have typed. / / Using λopt = t2 Q, λopt = 2.023432. Corresponding link lengths are:

2

VDI Richtlinie 2130: “Ebene Kurbelgetriebe—Konstruktion von Kurbelschwingen zur Unwandlung einer umlaufenden Bewegung in eine Schwingbewegung”, VDI/AWF Handbuch Getriebetechnik, 1959, VDI-Verlag, Dusseldorf.DI 2130.

302

7 Four-Link Crank Mechanisms

Chart 7.1.Optimum transmission angle maxμ μmin and initial crank angle for a given swing angle ψ and corresponding crank rotation φ

Chart. 7.1 Optimum transmission angle maxμmin and initial crank angle for a given swing angle ψ and corresponding crank rotation φ

7.2 Crank-Rocker Mechanism

303

Fig. 7.10 Crank-rocker mechanism with 40° swing angle and 160° corresponding crank rotation and the minimum transmission angle is optimum. a1 = 1, a2 = 0.2565, a3 = 0.5201 and a4 = 0.784

a21 a22 a23 a24

+ 2.023432 = 1.732051 = 1.773569 1 + 1.7320512 0.3639702 = 0.3639702 + 1 = 0.116978 = 2.0234322 ∗ 0.116978 = 0.478939 2 + 2.0234322 = 5.671282 = 1.093304 1 + 5.6712822 2

2

a1 a2 a3 a4

= = = =

1.331754 0.342020 0.692054 1.045612

For a1 = 120 mm: a2 = 0.342020 120 = 30.82 mm a3 = 62.36 mm, a4 = 94.22 mm. 1.331754 The maximum deviation of the transmission angle from 900 will be: cosμ

min max

=

120 ∗ 30.82 94.222 + 62.362 − 1202 − 30.822 ± 2 ∗ 94.22 ∗ 62.36 94.22 ∗ 62.36

= 0.219938 ± 0.629455 From which we obtain: μmax = 114.17◦ (Δ1 = 24.17◦ ) and μmin = 31.85◦ (Δ = 58.15◦ ). Since μmin deviates most, μmin is the critical transmission angle. If we refer to Chart 7.1, we see that for ψ = 40◦ and φ = 160◦ , max(μmin ) ∼ = 50.5◦ . Using the formula for the link lengths: = 32◦ and β ∼ a1 = 1, a2 = 0.2565, a3 = 0.5201 and a4 = 0.784 for a1 = 120 mm; a2 = 30.78 mm, a3 = 62.41 mm and a4 = 94.08 mm. The final mechanism is shown in Fig. 7.10. Note that as the mechanism deviates from the centric configuration for a higher time ratio, the transmission angle reaches unacceptable values. Looking at Chart 7.1 we can also say that as the swing angle increases, the maximum value of the transmission angle that is possible decreases. For example, if a minimum transmission angle of 40° is necessary, the possible swing angle and corresponding crank rotation that one can specify must be within the region shown in Fig. 7.11. Another point is to note that when the crank rotation from the extended to folded dead centers (φ) is greater than 180°, we obtain a better transmission angle compared to 360◦ − φ angle, which results with the same time ratio. In the solution of the transmission angle problem, there are two special cases. One special case occurs when φ = 180◦ (centric four-bar), for which the maximum and minimum values of the transmission angle is:

304

7 Four-Link Crank Mechanisms

Fig. 7.11 Region in Chart 7.1 where the transmission angles of the possible four bar solutions are greater than 40°

cosμ

min max

= ±

a1 a2 a3 a4

If the maximum deviation of the transmission angle is to be minimized, then for optimum a2 = 0 (since a1 = 1), a4 = 0 and a3 = 1, which is physically impossible four-bar mechanism proportion. In this case, we search for a crank-rocker mechanism for which the transmission angle deviation is of reasonable magnitude while the ratio of the link lengths is of acceptable proportions (the transmission angle will improve as λ or β increases). Using Eqs. (7.15)–(7.17) we obtain the link lengths in terms of initial crank angle and the swing angle (φ = 180◦ ) as: ( / ) a2 = tan ψ 2 sinβ a1 a3 = cosβ a1 a4 sinβ ( / ) = a1 cos ψ 2 Substituting the link ratios into the expression for cos μmin : cosμmin =

( / ) tan ψ 2 sinβ sinβ cosβ cos(ψ / 2)

=

( / ) sin ψ 2 cosβ

Now, the link lengths ratios can be expressed in terms of minimum transmission angle μmin as:

7.2 Crank-Rocker Mechanism

cosβ = a2 = a1 a3 = a1 a4 = a1

305

( / ) sin ψ 2 cosμmin ( / ) ( / )) tan ψ 2 /( 2 cos μmin − sin2 ψ 2 cosμmin ( / ) sin ψ 2 cosμmin /( ( / )) cos2 μmin − sin2 ψ 2 ( / ) cos ψ 2 cosμmin

In Fig. 7.12, the change of the link length ratios a4 /a1 and a2 /a1 as a function of the swing angle is shown for different minimum transmission angle values. For example, for ψ = 90◦ the best minimum transmission angle (with a4 /a1 = 0 and a2 /a1 = 0) is 45°. By reducing the transmission angle to 40° from the chart a4/a1 ≈ 0.54 and a2 /a1 ≈ 0.38. Using the equations for μmin = 40◦ , ψ = 90◦ yield (to five-digit accuracy): a4 /a1 = 0.54398, a3 /a1 = 0.92306 and a2 /a1 = 0.38463. If the transmission angle is 43° then a4 /a1 = 0.25970, a3 /a1 = 0.98299 and a2 /a1 = 0.18363. A higher value of the transmission angle will result with a smaller crank and rocker and the coupler and the/fixed link/ratio close to unity. Note that for any crank-rocker mechanism μmin < π 2 − ψ 2 i.e., for 90° swing angle the transmission angle cannot be greater than 45° (this statement is true for any corresponding crank angle, φ. For a given swing angle centric four-bar will have a better force transmission characteristics). For the centric four-bar mechanism a very simple design procedure can also be used (Fig. 7.13). Select an arbitrary length for the rocker and draw the two extreme 1

1

a4/a1

0,9

a2/a1

0,8

0,8

0,7

0,7

0,6

0,6

0,5

0,5

0,4

0,4

0,3

0 0 70 0 65 60 800 75

0,2

0

55

0

500 450 400 350 300 250 200 150

0 μ=5

0,3

100

0,2

0,1 0

0,1 0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

ψ

Fig. 7.12 Link length ratios as function of swing angle for centric four-bar

160

170

180

0

a4/a1

a2/a1

0,9

306

7 Four-Link Crank Mechanisms

a2 Be A

f

Bf

a2 A0

Ae

β a1

Q ψ/2 ψ

a4

B0

Fig. 7.13 Geometrical construction of a centric four-bar for a given swing angle

positions (B0 Be and B0 Bf ) such that ∠Be B0 Bf = ∠Be B0 Bf = ψ, specified swing angle. Draw a line passing through Be Bf . Drop a perpendicular ( / ) to this line from B0 . It intersects at Q and a2 = Be Q = Be Bf /2 = a4 sin ψ 2 . Select a point A0 on line Be Bf and draw a circle with a2 as the radius. The intersection of the circle is the location of the extended positions of the crank Ae and Af (A0 Q = Af Bf = Ae Be = a3 ). You must check Grashof’s rule for crank rocker proportions. Note that there are infinite solutions, each having different motion and transmission angle characteristics. Theoretical optimum transmission angle will be obtained when A0 is selected at infinity or when a2 /a1 ratio is zero, which is practically impossible. The previous equations given for the centric crank mechanism can be derived from the geometry given in Fig. 7.13. ◦ Another special case is when /[ φ − ψ ]= 180 . The transmission angle deviation is optimum when λopt = 1 + sin11 ϕ . The link lengths are: 2

1 a2 = sin ψ 2 /[(

) ] 1 1 a3 = sin ψ + 1 sin ψ 2 2 /( ) 1 1 + sin ψ a4 = 2 a1 = 1

As a rule of thumb, we can say: (a) If the corresponding crank angle is not important, for a certain swing angle ψ, centric four-bar mechanism (φ = 180◦ ) is the best choice for minimum transmission angle. At the same time, acceleration characteristics is also favorable.

7.2 Crank-Rocker Mechanism

307

(b) Crank rocker mechanisms with φ − ψ = 180◦ have the optimum transmission angle for a given φ (or the highest crank angle for a given μmin ). Example 7.4 We require a crank-rocker mechanism of 40° swing angle and s time ratio of 1.25. Determine the dimensions of the mechanism with optimum transmission angle and the ratio of maximum to minimum link length is not to exceed three. When time ratio is specified, there are two possibilities for the angle measured from extended to folded dead center positions in counterclockwise direction: TR =

φ 360◦ − φ or TR = 360◦ − φ φ

For TR = 1.4 we can have φ1 = 150◦ or φ2 = 210◦ . When we synthesize for ψ = 40◦ and φ = 150◦ with optimum transmission, λopt = 1.676 and max maxμmin = 21.9◦ . Maximum to minimum link length ratio is 3.69 which is greater than 3. When we synthesize for ψ = 40◦ and φ = 210◦ with optimum transmission, λopt = 2.179 and maxμmin = 35.65◦ which is better than the first solution. However, maximum to minimum link length ratio is 3.27 which is also greater than 3. Therefore, the global optimum lies outside the feasible range. We can either treat the problem as a constrained optimization and use solver tool in Excel to determine the value of λ which optimizes the transmission angle and subject i) < 3, since the optimum will be when max(ai )/min(ai ) = 3, to constraint max(a min(ai ) we can use Goal seek tool and determine the value of l which makes the maximum to minimum link length ratio equal to 3. Since the maximum transmission angle is greater, we synthesize for φ = 210◦ and ψ = 40◦ and obtain: λopt = 1.337 and maxμmin = 28.47◦ and the link lengths: a1 = 1, a2 = 0.3420, a3 = 0.4574, a4 = i) = 3. 1.0030 and max(a min(ai ) Besides the transmission angle, for high speeds the maximum acceleration is also critical. Note that when swing angle and the corresponding crank rotation is specified, we have one parameter solution for the determination of the four-bar mechanism proportion (λ or β). Instead of optimizing for the transmission angle deviation, one can minimize the maximum acceleration of the output link. Example 7.5 We would like to design a four-bar mechanism with 80° swing angle and 220° corresponding crank rotation. If we are to optimize for the transmission angle, we solve the cubic equation for Q and then solve λ. It turns maxμmin = 24.298◦ when λ = 1.7495. Link lengths are: a1 = a4 = 1, a2 = 0.577, a3 = 1.009 units. Next let us perform motion, velocity and acceleration analysis for the whole cycle for every 1o increment. (ω12 = 1 s−1 , constant). The analytical equations for the velocity and acceleration are given as:

308

7 Four-Link Crank Mechanisms

a2 sin(θ12 a3 sin(θ14 a2 sin(θ12 = a4 sin(θ14

ω13 = ω14

α14

− − − −

θ14 ) ω12 θ13 ) θ13 ) ω12 θ13 )

⎡

⎤ cos(θ12 − θ13 ) ⎢ sin(θ − θ )(ω − ω ) ⎥ 1 a2 14 13 12 13 ⎢ ⎥ = + ω12 2 ⎢ ⎥ ⎦ a4 −cos(θ14 − θ13 ) sin (θ14 − θ13 ) ⎣ sin(θ12 − θ13 )(ω14 − ω13 )

The result is shown in graphic form in Fig. 7.14. The maximum deviation of the transmission angle from 900 is 65.70° (μmin = 24.298◦ ) and the maximum acceleration of the rocker is (assuming ω12 = 1 unit, constant) 2.627 s−2 . If it is a high-speed application, instead of optimizing for the transmission angle, we can find the value λ which makes the maximum value of α14 a minimum. Again, using the Solver tool in Excel, we obtain λ = 2.629 and max(α14 ) = 2.412 s−2 . The minimum transmission angle for this case is less than the optimum: μmin = 20.61◦ . Of course, the result will be more accurate if the analysis for the whole cycle is made for every 0.1° instead of 1° . For the case with minimum acceleration link lengths are a1 = a4 = 1, a2 = 0.494, a3 = 1.299. The result is shown in Fig. 7.15.

Fig. 7.14 Displacement and acceleration of the output link and the transmission angle of a four-bar mechanism with ψ = 80◦ , φ = 220◦ and λ = 1.7495 for optimum transmission angle

7.3 Slider-Crank Mechanism

309

Fig. 7.15 Displacement and acceleration of the output link and the transmission angle of a four-bar mechanism with ψ = 80◦ , φ = 220◦ and λ = 2.6285 for minimum acceleration of the output link

7.3 Slider-Crank Mechanism Another mechanism that has a very wide range of use in machine design is the slider-crank mechanism. It is mainly used to convert rotary motion to a reciprocating motion. In order to convert reciprocating motion to a rotary motion, one must only impart motion only for a certain part of motion and let the inertia of the rotating output link sustain the motion for the rest of the cycle. In Fig. 7.16, The slidercrank mechanism is shown and the parameters that are used to define the angles and the link lengths are given. As in the four-bar mechanism, the extended and folded dead center positions are when the crank and the coupler are collinear (coupler link is commonly called connecting rod in slider-crank mechanisms). Full rotation of the crank is possible if the eccentricity, c, is less than the difference between the connecting rod and the crank lengths and the crank length is less than the connecting rod length (e.g., c < (a3 − a2 ) and a3 > a2 ) (Fig. 7.17). The case where a full rotation is not possible is shown in Fig. 7.18. Using the right-angled triangles formed at the dead center positions: ]1 2 [ se = (a2 + a3 )2 − c2 / ] [ 1 2 sf = (a2 − a3 )2 − c2 / Noting s = se − / sf = stroke = / the distance slider travels between dead-centers. If we let λ = a2 a3 and ε = c a3 , the stroke will be given by:

310

7 Four-Link Crank Mechanisms y μ

A a2

a3

θ13

3

θ12

B

4

c

0

x

s14

Fig. 7.16 Slider crank mechanism A1

Fig. 7.17 Dead centers of a slider-crank mechanism

φ

Af

φe

c

Ae Bf

B1

sf

Be

s se

A

3 B

4

2

A0

Fig. 7.18 A slider-crank mechanism where a full rotation of the crank is not possible

[ ]1 2 [ ]1 2 s = a3 (1 + λ)2 − ε2 / − a3 (1 − λ)2 − ε2 / If the eccentricity, c (or a1 ), is zero (c = 0) the slider crank mechanism is called an in-line slider-crank and the stroke is twice the crank length (s = 2a2 ). If the eccentricity is not zero (c /= 0), it is usually called an offset slider-crank mechanism. The transmission angle can be determined from the equation (see Fig. 7.16): a3 cosμ = a2 sinθ12 − c

(7.24)

7.3 Slider-Crank Mechanism

311

μ max

A1

B1 c

A0

B2 μ min

A2 Fig. 7.19 Positions of a slider-crank mechanism where transmission angle is critical

Maximum deviation of the transmission angle occurs when the derivative of μ with respect to θ12 is zero. Hence differentiating Eq. (7.24) with respect to θ12 : dμ −cosθ12 = = 0 dθ12 sinμ

(7.25)

Maximum or minimum deviation occurs when θ12 is 90° or 270° (Fig. 7.19) and the value of the maximum or minimum transmission angle is given by: cosμ

max min

=

− c ± a2 a3

(7.26)

If c is positive as shown in Fig. 7.19, transmission angle is critical when θ12 = 270◦ . If c is negative, then the most critical transmission angle is at θ12 = 90◦ . (the angle that deviates most from 90° is the critical transmission angle). If the eccentricity, c, is zero, maximum value of the transmission angle is: cosμ

max min

=

a2 a3

(7.27)

In reciprocating pumps, the crank-to-connecting rod ratio is kept less than 1/4, which corresponds to 14.48° maximum deviation of the transmission angle from 90°. Since the crank length is fixed by the required stroke (a2 = s/2) one must increase the connecting-rod length for better transmission angles. However, this will increase the size of the mechanism. In internal combustion engines, although this angle is not the transmission angle (slide is driving), it is still critical, since a small angle μ will result with a high normal force component between the cylinder and the piston.

312

7 Four-Link Crank Mechanisms

Similar to the transmission angle problem in the four-bar mechanisms, the transmission angle problem in slider-crank mechanisms can be stated as follows: Determine the slider-crank proportions with a given stroke, s, and corresponding crank rotation between dead-centers, φ, such that the maximum deviation of the transmission angle from 90° is a minimum.

This Problem was first studied by Meyer zur Capellen [3], later by Volmer [4] and published by VDI Guideline VDI2132.3 The analytical solution was given by E. Söylemez [5]. The problem can again be considered in two parts. The first part is the determination of slider crank mechanisms with a given stroke and corresponding crank rotation. The second part is the determination of one particular slider-crank mechanism with optimum transmission angle variation. For the first part of the problem note that the stroke s is a function of the link length ratios, i.e. if we double the length of the links, the stroke will be doubled. Therefore, without loss of generality, let s = 1 (the link lengths thus found will be multiplied by the stroke to give the actual values). Referring to Fig. 7.17, the vector loop equations at the dead centers are: A0 Be + Be Ae + Ae A0 = 0

(7.28)

A0 Bf + Bf Af + Af A0 = 0

(7.29)

− ic + se + (a3 + a2 )eiφe = 0

(7.30)

− ic + sf + (a3 − a2 )ei(φe + φ − π ) = 0

(7.31)

or in complex numbers:

Subtracting Eq. (7.31) from Eq. (7.30) and noting se − sf = s = 1: 1 + (a3 + a2 )eiφ1 + (a3 − a2 )ei(φ1 +φ) = 0

(7.32)

/ If we let Z = a3 eiφ1 and λ = a2 a3 Eq. (7.31) can be rewritten in the form: Z(1 + λ) + Z(1 − λ)eiφ + 1 = 0

(7.33)

For a full rotation of the crank a necessary (but not sufficient) condition is |λ| < 1. Eq. (7.33) can be solved for Z to yield:

3

VDI Richtlinie 2132 “Ebene Kurbelngetriebe-Konstruktion von Schubkurbeln”. VDI/AWF Handbuch Getriebetechnik, 1959, VDI-Verlag, Düsseldorf.

7.3 Slider-Crank Mechanism

313

Fig. 7.20 Locus of all possible moving (ka ) and fixed (k0 ) pivots of slider crank mechanisms having a unit stroke and a corresponding crank rotation φ

Z =

−1 ) ( ) ( iφ + 1 + eiφ λ 1 − e

(7.34)

If λ is taken as a free parameter, as it varies, the tip of Z given by (7.30), will generate a circle which is the locus of all possible moving pivots for the crank when the crank and the coupler are at extended position (ka circle). The locus of all possible fixed pivots is another circle (ko circle) which is given by Z(1 + λ) (the origin for both vectors is Be with the real axis parallel to the slider axis). Any line drawn from Be intersects these circles at Ae and Ao respectively, yielding the slider-crank mechanism in extended dead center position. In Fig. 7.20 these circles are shown for φ = 160◦ . The eccentricity, c can be obtained as the imaginary component of the vector Be A0 = Be Ae + Ae A0 which can be written as: 2ic = (a3 + a2 )eiφ1 − (a3 + a2 )e−iφ1

(7.35)

2ic = Z(1 + λ) − Z(1 + λ)

(7.36)

or using Z and λ:

and substituting the value of Z: ) ( 1 − λ2 sinφ 1 [( ) ( ) ] c = 2 1 + λ2 + 1 − λ2 cosφ The link lengths can now be expressed as:

(7.37)

314

7 Four-Link Crank Mechanisms

1 1 [( ) ( ) ] 2 2 1 + λ + 1 − λ2 cos(φ) 1 1 = [ ] 4 cos2 φ2 + λ2 sin2 φ2

a23 =

a22 = a23 λ2 = =

(7.38)

λ2 1 ] [( ) ( ) 2 1 + λ2 + 1 − λ2 cos(φ) λ2 1 [ ] φ 4 cos2 2 + λ2 sin2 φ2

(7.39)

Equations (7.37)–(7.39) yield a singly infinite set of solutions for the slider-crank mechanisms satisfying a given crank rotation (stroke = 1 unit). One can also use the eccentricity, crank or coupler link length as a free parameter to determine the other link lengths instead of λ. Example 7.6 Determine the link lengths of the slider crank mechanism with a stoke s = 120 mm, corresponding crank rotation φ = 160◦ and the crank to coupler link ratio λ = 0.5. Using unit stroke, from Eqs. (7.37), (7.38) and (7.39) the link lengths are: a2 = 0.47881, a3 = 0.95762 and c = 0.23523. For s = 120 : a2 = 114.91 mm, a3 = 57.46 mm and c = 28.23 mm. The minimum transmission angle for this mechanism is μmin = 41.79◦ . Example 7.7 Determine the link lengths of the slider-crank mechanism having the same stoke and corresponding crank rotation as in Example 7.6 but instead of crank to coupler link ratio specified, the eccentricity is specified as c = 20 mm. For unit stroke c = 20/120 = 0.16667. Solving Eq. (7.33) for λ yields: / 1 − 2c cot φ 2 / λ = 1 + 2c tan φ 2 2

(7.40)

For c = 0.16667, λ2 = 0.325635. Substituting into Eqs. (7.34) and (7.35), a2 = 0.48508 and a3 = 0.85006. For s = 120 mm, c = 20 mm, a2 = 58.21 mm and a3 = 102.01 mm. The minimum transmission angle for this mechanism is μmin = 39.94◦ . Note that a similar procedure can be carried out if crank or coupler link length is specified. / The minimum transmission angle is when θ = π 2 : μmin = cos− 1

[

c + a2 a3

] (7.41)

For a full rotation of the crank, c + a2 < a3 or c < a3 − a2 . At the limit position (c = a3 − a2 ), μmin = 0. Using Eq. (7.33), (7.34) and (7.35) this condition

7.3 Slider-Crank Mechanism

315

yields the limits of φ for a full rotation of the crank: ( ) π − 1 −1 ≤ φ ≤ tan 2 c and cot2

(φ) 2

(7.42)

< λ < 1

Expressing μmin in terms of λ and φ: cosμmin

) ( 1 − λ2 sin(φ) 1 = λ + √ [( ) ( ) ]1 2 1 + λ2 + 1 − λ2 cos(φ) 2

(7.43)

since λ is a free design parameter, the necessary condition for the minimum transmission angle to be a maximum is dμdλmin = 0. If the value of λ which makes the derivative equal to zero is λ = λopt , differentiating Eq. (7.43) and setting dμdλmin = 0 yields. ) ) ( ( ) ( t2 Q3 + 1 − t2 Q2 − t4 + t2 + 1 Q + 1 + t2 = 0

(7.44)

( / ) where Q = λ2opt t2 and t = tan φ 2 . The roots of Eq. (7.44) are: / ) 1 ( 1 + 5 + 4t2 2 2 / ) 1 ( 1 Q2 = − − 5 + 4t2 2 2 1 Q3 = 2 t Q1 = −

(7.45)

/ Since Q must be positive, Q2 is not a solution. Corresponding to Q3 , λ = 1 t2 , the deviation of the minimum transmission angle 90° is ( maximized / ) (cosμmin = 1). The root Q1 yields the value of λopt within the range 1 t2 , 1 , and this value satisfies the necessary and sufficient condition for slider-crank mechanism with optimum transmission angle characteristics. Therefore: λ2opt

1 = 2 2t

[/ ] ( ) 2 5 + 4t − 1

(7.46)

is the unique optimum solution. Example 7.8 For slider stroke s = 120 mm and corresponding crank rotation φ = 160◦ , determine the slider crank mechanism with optimum force transmission characteristics.

316

7 Four-Link Crank Mechanisms

Fig. 7.21 Slider crank mechanism with 120 mm stroke 160° corresponding crank rotation and the force transmission is optimum. A0 A = 55.86 mm; AB = 137.88 mm; eccentricity = 45.29 mm

Af

160 0 A0 Ae Bf

Be 120 mm

From Eq. (7.46), λopt = 0, 405185. Utilizing Eqs. (7.41), (7.42) and (7.43) for unit stroke the link lengths are: a2 = 0.465542, a3 = 1.14896, c = 0.377378 and for 120 mm stroke: a2 = 55.86 mm; a3 = 137.88 mm; c = 45.29 mm The minimum transmission angle for the mechanism is μmin = 42.81◦ . The mechanism is shown in Fig. 7.21. The results are given in Chart 7.2. The slider-crank link lengths (a2 , a3 , c) and λopt values and the minimum transmission angle μmin as function of crank rotation between dead centers is given. In Chart 7.3, [5] all the possible solutions and their

140

1.4

120

1.2

a2/s

1

λ

80

0.8

maxμmin 60

0.6

40

0 180

0.4

a2/s c/s

20

a3/s, a2/s, c/s

maxμmin

100

0.2

0 170

160

150

140

130

120

110

100

90

φ Chart. 7.2 Slider-crank proportions with optimum transmission angle variation for a given crank rotation between dead centers, φ

7.4 Quick Return Mechanism

317

Chart. 7.3 Transmission angle variation corresponding to crank rotation φ and eccentricity c of a slider-crank mechanism [5]

minimum transmission angle values are given (note that the horizontal axis is not to a linear scale).

7.4 Quick Return Mechanism As can be seen, neither a four-bar nor a slider-crank mechanism is a good choice if one requires a time ratio much different than 1. For example, for φ < 140◦ or φ > 220◦ you cannot expect a reasonable transmission angle. A mechanism which

318

7 Four-Link Crank Mechanisms

4

B 4

3 A0

2

A

θ12 A2

a1

4

2

A0

s43

B0

φ

ψ

φ/2

A

3 A0

A

A1

3

2

A1

φ A2

ψ/2

ψ

B

B

0

0

θ14

c

Fig. 7.22 Centric and eccentric quick return mechanism

is more appropriate in terms of time ratio is an inverted slider-crank mechanism. Due to its characteristics, is commonly known as “Quick-return” mechanism (Fig. 7.22). For the oscillation of the output link (link 4) corresponding to a continuous crank rotation a2 + ecc < a1 . The dead center occurs when the slider axis is tangent to the crank circle (A1 and A2 ). As the crank rotates from A2 to A1 by an angle 360◦ − φ counter clockwise, the output link swings by an angle ψ(CCW). Whether it is centric or eccentric, in both cases the swing angle and the corresponding crank rotation add up to 180◦ (φ + ψ = 180/◦ ). For the( centric / ) crank (for/a )given swing angle the unique design variable is a2 a1 = sin ψ 2 = cos φ 2 . For the eccentric: φ = φ1 + φ2 = cos− 1

(

a2 + c a1

)

+ cos− 1

(

a2 − c a1

)

Due to the sliding joint, force transmitted to link 4 will be perpendicular to the slider axis. In case of centric inverted slider-crank this force will tend to drive the output link whereas in case of eccentric inverted slider-crank mechanism, only the force component perpendicular to the line B0 A will tend to drive and there will be a force component that will tend to apply bearing pressure only. Because of this characteristics, centric inverted slider crank mechanisms are usually preferred (Fig. 7.23). However, If a larger swing angle for the same crank to fixed link length ratio is required, eccentricity will be a used. In Fig. 7.24. the motion curve for ψ = 60◦ is shown. For 240° counter clockwise input crank rotation, the output link rotates counterclockwise by 60°. For the remaining 120° crank rotation output link rotates clockwise 60°. The curve is a continuous smooth curve where, during a big portion of forward stroke, the change in velocity is very small. For most tasks such as metal cutting, this approximate constant velocity characteristics is very important (this is why quick return mechanism is used in shapers). One drawback is that you cannot select the corresponding crank rotation (φ) independent of the swing( angle. i.e./ if ψ = )60◦ , φ = 120◦ (TR = 2) and when ψ = 30◦ , φ = 150◦ TR = 7 5 = 1.4 and is fixed.

7.4 Quick Return Mechanism

319

Fig. 7.23 Force transmission

130

0.4

120

0.2

ω14

110

0 -0.2

θ14

90

-0.4

80

-0.6

70

-0.8

60

-1

50

-1.2

40

ω14

θ 14

100

-1.4 -30

0

30

60

90

120

150

180

210

240

270

300

330

θ 12 Fig. 7.24 Motion curve of a quick return mechanism with output oscillation ψ = 60◦

Although different in construction, mechanism called as “swinging block” mechanism is another inverted slider-crank construction governed by the same equations with the “quick return” mechanism that we have discussed. Although they are kinematically equivalent (i.e. for the same link dimensions they have the same motion characteristics) their constructions are different and are used for different purpose (Fig. 7.25).

320

7 Four-Link Crank Mechanisms

A

A 2

A0

2

A0

A1 A2

3

A1

3 A2

B

0

B 4

0

4

Fig. 7.25 Swinging Block Mechanism

7.5 Scotch-Yoke Mechanism Another mechanism that is used to transform a rotary motion to reciprocating motion is Scotch-Yoke mechanism (This mechanism is sometimes called “Slotted link“) as shown in Fig. 7.26, consisting of two prismatic and two revolute joints. There are two design parameters: The crank length A0 A = r and the angle between the two prismatic joint axes, α. For any value of r and α, the crank rotation between Fig. 7.26 Scotch yoke mechanism α

θ

A r

Af A0

Ae

s0/2 s0 s

7.6 Continuous Rotation of the Output

321

the two limiting positions of the stroke is 180°. In most applications α is 90° and the resulting mechanism has a stroke s = 2r. If required, the stroke can be increased by changing angle α. In such a case, the stroke is: s0 =

2r sinα

Slider displacement as a function of crank rotation is given by: ) ( sinθ + c s = r cosθ − tanα When α = 90◦ , s = rcosθ + c. Because of this characteristic, the mechanism is also called “harmonic motion mechanism”.

7.6 Continuous Rotation of the Output 7.6.1 Design of Drag-Link Mechanisms with Optimum Transmission Angle Drag-link mechanisms are used to convert a uniform continuous rotation into a nonuniform continuous rotary motion. They can be used in series with crank-rocker or slider-crank mechanism to obtain a different motion characteristic. One of the applications is to obtain quick-return characteristics without the drawbacks of the inverted slider crank mechanism. A graphical method for the design of drag link mechanism is given by Kurt Hain [6] Analytical solution is due to Lung Wen Tsai [7] In Fig. 7.27, the four-bar mechanism with drag-link proportions is shown in positions where the deviation of the transmission angle from 90° is maximum (the input crank and the fixed link are collinear). It develops that the transmission angle is optimum when the two maximum deviations are equal (μmax − 90◦ = 90◦ − μmin ). Corresponding to 180° input crank rotation, the output link will be required to rotate by ψ (Note that ψ < 180◦ , since the output link will rotate 360◦ − ψ for the next 180° rotation of the crank). Let us also assume a minimum value of the transmission angle μmin . The two extreme positions will be as shown in Fig. 7.27. For the best transmission angle characteristics, at these two extreme positions if we equate the transmission angle deviations from 90◦ (μmax − 90◦ = 90◦ − μmin ), we obtain the loop equations as: − (a2 + a1 ) + a3 ei(ψ1 − π + μmin ) = a4 eiψ1

(7.47)

(a2 − a1 ) + a3 ei(ψ1 + ψ − μmin ) = a4 ei(ψ1 + ψ)

(7.48)

322

7 Four-Link Crank Mechanisms A

Fig. 7.27 Drag link mechanism

3 2 ψ1

B2

B a

3

μ max

A2

4

a4

a2

A1

a1 A

0

B 0

ψ

μ min

B1

Writing the cosine theorem for these two positions: (a2 − a1 )2 = a23 + a24 − 2a4 a3 cosμmin

(7.49)

(a2 + a1 )2 = a23 + a24 + 2a4 a3 cosμmin

(7.50)

Adding these two equations: a21 + a22 = a23 + a24

(7.51)

In drag link mechanism / for a certain portion of the cycle the output will rotate faster than the input link (dθ14 dθ12 > 1 input link will rotate/by an angle ψ < π ) and in the other portion of the cycle it will move slower (dθ14/ dθ12 < 1 input link will will be when dθ14 dθ12 = 1.. For a four-bar rotate by 2π − ψ / > π ). Transition / mechanism dθ14 dθ12 = ω12 ω14 is given by: ω12 a2 sin(θ12 − θ13 ) = ω14 a4 sin(θ14 − θ13 )

(7.52)

/ When dθ14 dθ12 = 1, a2 sin(θ12 − θ13 ) = a4 sin(θ14 − θ13 ). Referring to Fig. 7.28, this condition can only be satisfied if a2 sinθ13 = 0 or when θ13 = 0. There / are two positions of the drag link mechanism where θ13 = 0.θ and dθ14 dθ12 = 1. These two positions are as shown in Fig. 7.29. (Positions A0 A1 B1 B0 and A0 A2 B2 B0 ). Let us draw lines parallel to B0 B1 (A0 C1 ) and B0 B2

7.6 Continuous Rotation of the Output Fig. 7.28 Figure showing a2 sin(θ12 − θ13 ) = a4 sin(θ14 − θ13 ) can only be satisfied if a1 sin(θ13 ) = 0 or θ13 = 0 or π

323

B θ13

A

a sin( θ - θ13 ) 4

14

θ14 θ13 θ12 θ13

a sin( θ - θ13 ) 2

θ14

θ12

12

B0

A0 a sin( θ ) 13 1

Fig. / 7.29 Positions at which dθ14 dθ12 = 1

A1

C1

B1

α1 β1

ψ ψ

B0

A0

β2

C2

1

ψ2

α2

B

A2

(A0 C2 ) and write the cosine theorems for the triangles A0 A1 C1 and A0 A2 C2 to determine the angles αi and βi as:

a24 = a22 + (a3 − a1 )2 − 2a2 ∗ (a3 − a1 )sinα1

(7.53a)

a24 = a22 + (a3 + a1 )2 − 2a2 ∗ (a3 + a1 )sinα2

(7.53b)

a22 = a24 + (a3 − a1 )2 − 2a4 ∗ (a3 − a1 )sinβ1

(7.53c)

a22 = a24 + (a3 + a1 )2 − 2a4 ∗ (a3 + a1 )sinβ2

(7.53d)

324

7 Four-Link Crank Mechanisms

In Eq. (7.49), it was shown that to obtain equal deviations of the transmission angle from 90° at the two extreme positions, the condition a21 + a22 = a23 + a24 must be satisfied. Substituting this condition into Eqs. (7.53): ( / ) α1 = α2 = cos− 1 a3 a2

(7.54)

( / ) β2 = π − β1 = cos− 1 a1 a4

(7.55)

and

Also, from Fig. 7.29 φ1 = π − α1 , φ2 = 2π − α2 , ψ1 = β1 , ψ2 = π + β2 , φ = φ2 − φ1 = π, ψ = ψ2 − ψ1 = 2β2 .ψ = ψ2 − ψ1 = 2β2 . Now, from Eq. (7.55): a4 1 ( / ) = a1 cos ψ 2

(7.56)

If we subtract Eq. (7.49) from Eq. (7.50), we obtain: a1 a2 − a3 a4 cosμmin = 0

(7.57)

Solving for a2 : ( a2 = a3

) a4 cosμmin a1

Substituting into equation a21 + a22 = a23 + a24 yield: [( a23 (

a3 a1

a4 a1

)2

cos μmin − 1 2

( )2

= [( )2 a4 a1

]

)2

a4 a1

− 1

cos2 μmin

− 1

= a24 − a21

( / ) sin2 ψ 2 ( / ) ] = cos2 μmin − cos2 ψ 2

(7.58)

Lastly, using Eq. (7.57) we have: (

a2 a1

)

( =

a3 a1

)(

) a4 cosμmin a1

(7.59)

Equations (7.56), (7.58) and (7.59) can be used to design a drag link mechanism when ψ and minimum transmission angle μminmin are given.

7.6 Continuous Rotation of the Output

325

Example 7.9 Determine the four-bar mechanism with drag link proportions for which within half revolution of the input link, output link rotates by 120°. We would like to have a transmission angle of greater than 45°. Taking a1 = 1 unit with ψ = 120◦ and μmin = 50◦ we have: a4 =

1 = 2.00 cos(60◦ ) sin2 (60◦ ) cos2 (45◦ ) − cos2 (60◦ )

a23 = = 3 a3 = 1.7321 a2 = 2.1439 ∗ 2.00 ∗ cos(45◦ ) = 2.4495 If we take a1 = 100 mm, a2 = 245 mm, a3 = 173 mm and a4 = 200 mm. The result is ( shown ( / )) in Fig. 7.30. When input link is at 135◦ = π − cos− 1 a3 a2 the lag cycle starts and output link is at 120°. When the input link rotates 180°, the output link rotates by 140°, in the remaining 180° rotation of the input, output link rotates by 220°. In the first half of the cycle output link angular velocity is less than 1. Let us now change the role of input and output links. In such a case, for the first 220° crank rotation, the output link will rotate by 180° and for the remaining 140° it will rotate by 180° (μmin used in the design of the drag-link mechanism is no longer the transmission angle). Let us attach an in-line slider-crank mechanism in series such that the slider crank mechanism is at the top dead center (i.e., B0 B’ is in line with the slider axis at the start of lag cycle as shown in Fig. 7.31a (A0 A = 200 mm, B0 B = 276 mm)). For the first 220° crank rotation the slide will move from left to right (when the slide is vertical this motion is from top to bottom dead center). For the 1.8

300 270

1.6

240

θ14

210 180

1.4

ω14/ω12

1.2

120

μ

90

1 0.8

60 0.6

30 0 -30

0.4 0

30

60

90

120 150 180 210 240 270 300 330 360 0.2

-60 -90

θ12

Fig. 7.30 Characteristics of the drag-link mechanism in Example 7.9

0

ω14/ω12

θ14, μ

150

326

7 Four-Link Crank Mechanisms

remaining 140° crank rotation slide returns to its original position. A time ratio of 2 is obtained. The minimum transmission angle for the drag-link mechanism is 19.5°. The motion of the slide is as shown in Fig. 7.31b. (Such a mechanism construction is used for certain type of press. Displacement of s is taken as decreasing when moving from top to bottom positions). In drag-link mechanisms, since both links connected to the fixed link make a complete rotation, there is the problem of interference of the links. Either the input and output planes must be in two different planes and in between we have the coupler plane, or the input link must be made as a ring (usually driven by a pinion gear from the outside). The two cases are as shown in Fig. 7.32.

120

A0 4

10

100

B0

8

A

s

2

80

6

60

4

3 5

2

40 B

0 20

-2

6 C

ds/d

B'

12

0

-4 0

60

120

180

240

300

360

Fig. 7.31 a 6-Link mechanism with quick return properties; b Displacement and velocity of the slide (link 6)

Fig. 7.32 Construction of drag link mechanism to eliminate interference between the links

7.6 Continuous Rotation of the Output

327

7.6.2 Inverted Slider-Crank Mechanism Another mechanism which converts a uniform continuous rotation into a nonuniform continuous rotary motion is the inverted slider-crank mechanism. If crank length is greater than the fixed link length, the output link will have a full rotation. Referring to Fig. 7.33, there is a single design parameter (a2 /a1 > 1). When there is no eccentricity, the mechanism is symmetric about the line A0 B0 . When the input crank rotates 180° in between positions marked 1 and 2 CCW (θ12 = 0◦ and 180◦ respectively), the output link will rotate by an angle ψ which is given by: ( / ) ψ = 2 ∗ tan−1 a2 a1 . When the input link rotates from 2 to 1 (CCW) the angular rotation of the output link will be 360◦ − ψ.. When the output link rotates 180° between positions ii to i (CCW) the input link rotates by an angle φ given by: ( / ) φ = 2 ∗ cos−1 a1 a2 . Note that at positions i and ii, the input and output angular velocities / will be equal ◦ ω12 > 1 and 360 − φ from ii to i ω and when the output link rotates CCW 14 / when it rotates from i to ii CCW ω14 ω12 < 1 by an angle φ.. There is / only one design parameter. Therefore, one can select either ψ, φ or a2 a1 ratio and determine the value of the other parameters. In Fig. 7.34 for Fig. 7.33 Inverted slider-crank mechanism with a2 /a1 > 1

ψ

ψ/2

3 a2 A0

2 i

θ12

A

4

1

a1

ii

B0 φ φ/2

θ14

328

7 Four-Link Crank Mechanisms

390 360 330 300 270 240 210 180 150 120 90 60 30 0

3.5 3 2.5 2 1.5

ω14/ω12

θ14

a2 /a1 = 1.5 the displacement and velocity of the output link is shown (input velocity ◦ is assumed constant). θ12 = 221.8◦ and θ12 = 318.19 / are the input crank angles at positions i and ii respectively (at these positions ω14 ω12 = 1). If we attach another inverted slider crank mechanism for which the output is in oscillation by selecting a2 /a1 < 1 (say a2 /a1 = 0.5 ( < 1), a six-link mechanism as shown in Fig. 7.35a or b can be formed (you can as well have point C on link 4 on the other side of B0 . Since link 4 makes a full rotation, link 2 is usually constructed as a ring driven by a pinion as shown in Fig. 7.32. When the mechanism is constructed as in Fig. 7.35b the input and output shafts are collinear). Displacement and velocity

1 0.5 0

30

60

90

120 150 180 210 240 270 300 330 360

0

θ12

Fig. 7.34 Motion characteristics of an inverted slider-crank mechanism with a2 /a1 = 1.5

A 5 C

B

0

6

2

3

4

θ12

A0

θ14

3 θ16

2 A0 6

A

θ12

4

θ14 B

θ16

0

C 5

D0

(a)

(b)

Fig. 7.35 Six-link mechanism with two inverted slider-crank mechanisms in series

7.7 Problems

329

140

1

120

0 -1

100

θ16

-3 60

ω16/ω12

-2 80

-4 40

-5

20

-6

0

-7 0

60

120

180

240

300

360

θ12

Fig. 7.36 Displacement and the velocity of the output link for the mechanism with two inverted slider-cranks in series (a2 /a1 = 1.5 for the first and a2 /a1 = 0.5 for the second inverted slider-crank mechanism)

of the output as a function of input crank rotation is as shown in Fig. 7.36 (For the construction shown in Fig. 7.35a. Note that for 300° crank rotation the output is in a forward stroke and rotates approximately 60° with almost constant velocity and returns to the original position in 60° crank rotation. Of course, one must be concerned with the high accelerations that will occur during the return stroke. These four link mechanisms coupled with another four-link mechanism can be designed to obtain criteria like increasing the time ratio, for long oscillations or for long approximate dwells while keeping the transmission angle within reasonable limits. These will be discussed in Chap. 8.

7.7 Problems 1. For all the following problems the output link oscillates 30° in between dead centers. (a) Design a centric four-bar. a2 /a1 ratio cannot be less than 0.1 and force transmission is optimum. What is the minimum transmission angle? (b) Design a four-bar mechanism such that the input crank rotates 160° in between dead centers and the transmission angle is optimum. What is the minimum transmission angle? (c) Design a four-bar mechanism such that the input crank rotates 200° between dead centers and the transmission angle is optimum. What is the minimum transmission angle?

330

7 Four-Link Crank Mechanisms

(d) Design a four-bar mechanism such that the input crank rotates 200° between dead centers and the maximum to minimum link length ratio is less than 4. (e) Design a four-bar mechanism such that the input crank rotates 200° between dead centers and the maximum acceleration of the output link when the input crank rotates at a constant speed is a minimum. (f) Comment about the results you have obtained. 2. (a) Design a slider-crank mechanism for with 100 mm stroke and 160° corresponding crank rotation between dead centers and the force transmission is optimum. What is the minimum transmission angle? (b) Design a slider-crank mechanism with 100 mm stroke and 160° corresponding crank rotation between dead centers with optimum force transmission characteristics and the eccentricity cannot be greater than 20 mm. (c) Design a slider-crank mechanism with 100 mm stroke and 160° corresponding crank rotation between dead centers and the maximum acceleration of the slide is a minimum. 3. (a) Design a drag link mechanism for which within half revolution of the input link, output link rotates by 100°. We would like to have a transmission angle of greater than 45° . (b) Using a centric crank rocker mechanism in series with the drag link mechanism designed in part (a), design a six-link mechanism in which as the input crank rotates by 100° the output rotates 60° from extended dead center to folded dead center and oscillates back slowly in the remaining 260°. What is the minimum transmission angle for this mechanism? Compare your result with a crank-rocker mechanism which has a 60° swing angle and 100° (or 260°, whichever gives a better result in terms of transmission angle) crank rotation. How would you connect the drag link with the crank rocker mechanism if you were to design the resulting six link mechanism so that for 100° crank rotation the output rotates from folded dead center to extended dead center? (c) For drag link mechanisms comment what happens when the minimum transmission angle is increased or decreased. 4. In Example 7.9 when coupling the drag-link mechanism with an inline slidercrank mechanism, the input and output of the drag link mechanism was reversed, i.e. link 4 was used as input and link 2 was the output. A similar six-link mechanism without reversing the input and output links is possible for quick-return application such as used for a press. Design this six-link mechanism. Compare this mechanism with the result of Example 7.9. In terms of input–output motion curve and its derivatives. Keep in mind that the drag link must be coupled with the slider-crank mechanism such that during the downward stroke of the slide, the drag link output must rotate slower than the upward stroke.

References

331

References 1. H.M. Alt, Über die totlagen des gelenkviereck (about the dead points of the four-bar linkage). Z. Angew. Math. Mech. 5(4), 337–346 (1925) 2. F. Freudenstein, E.J.F. Primrose, “The classical transmission-angle problem”. The Institution of Mechanical Engineers, C96/ 72, Mechanisms, London (1972) 3. W.M.Z. Capellen, Die totlagen des ebenen gelenkvierecks in analyticher darstellung. Forsch. Ing-Wes. Bd 22/Heft 2, 42–50 (1956) 4. J. Volmer, Die kontruktion von schubkurbeln mit hilfe von kurventafeln. Machinenbautechnik H 12, 120–125 (1957) 5. E. Söylemez, Classical transmission-angle problem for slider-crank mechanisms. Mech. Mach. Theor. 37(4), 419–425 (2002) 6. K. Hain, Applied kinematics (Mc Graw Hill, New York, 1961), pp.358–360 7. L.-W. Tsai, Design of drag-link mechanisms with optimum transmission angle. Trans ASME 105(2), 254–258 (1983)

Chapter 8

Correlation of Input and Output Motion; Function Generation; Freudenstein’s Equation in Mechanism Design

Abstract Analytical function synthesis methods has become popular with Freudenstein’s equation that gives the designer a simple relation between the input and output link rotation angles of a four-bar. Correlation of crank angles for three, four and five positions are addressed. Using Chebyshev spacing and relating the crank angles with the dependent and independent variables of a function, the procedure is extended to function generation. The procedure is extended to least squares minimization, order and mixed order synthesis, correlation of two crank angles with optimum transmission angle and to the correlation of crank angle with the slider-displacement of a slider crank mechanism. Keywords Correlation of crank angles · Function generation · Precision point synthesis

8.1 Derivation of Freudenstein’s Equation Considering the four-bar mechanism (Fig. 8.1), the loop closure equation in vectorial form is: A0 A + AB = A0 B0 + B0 B or in complex numbers: a2 eiθ12 + a3 eiθ13 = a1 + a4 eiθ14

(8.1)

If we equate the real and imaginary parts of this equation separately, we shall obtain two scalar equations in three position variables (θ12 , θ13 and θ14 ). If one of the position variables is the input variable whose value is given, then we shall be able to solve for the values of the other two variables. In complex plane, when we have an equation in complex numbers, the complex conjugate of the equation is also true. The complex conjugate yields vectors which are the mirror image of the original vectors with respect to the real axis (x-axis); © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_8

333

334

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.1 Four-bar mechanism

e.g., in case of a mechanism, if we place a mirror about the real axis, as we move the original mechanism its image will also move and corresponding to the original closed loop, the loop formed on the mirror image will also be closed at every position. Hence, we obtain another loop closure equation in terms of complex numbers as: a2 e−iθ12 + a3 e−iθ13 = a1 + a4 e−iθ14

(8.2)

The original Eq. (8.1) and its complex conjugate (8.2) are the two independent equations in the complex plane (if we equate the real and imaginary parts of these two equations, they will yield the same two scalar equations in the real plane). Using the Eqs. (8.1) and (8.2), if we are to find θ14 as a function of θ12 , we must eliminate θ 13 from the above equations. We can write the loop closure equations in the form: a3 eiθ13 = a1 + a4 eiθ14 − a2 eiθ12

(8.3)

a3 e−iθ13 = a1 + a4 e−iθ14 − a2 e−iθ12

(8.4)

Multiplying Eqs. (8.3) and (8.4):    a23 ei(θ13 −θ13 ) = a1 + a4 eiθ14 − a2 eiθ12 a1 + a4 e−iθ14 − a2 e−iθ12

(8.5)

Noting ei(θ−θ) = ei0 = 1; 2a1 a4 cos θ14 − 2a1 a2 cos θ12 − 2a2 a4 cos(θ14 − θ12 ) + a12 + a22 − a32 + a42 =0

(8.6)

Since cos θ = (eiθ + e−iθ )/2, Eq. (8.6) reduces to the form: 2a1 a4 cos θ14 − 2a1 a2 cos θ12 − 2a2 a4 cos(θ14 − θ12 ) + a12 + a22 − a32 + a42 =0 or, dividing every term by 2a2 a4 :

8.1 Derivation of Freudenstein’s Equation

335

a1 a1 a 2 + a22 − a32 + a42 cos θ14 − cos θ12 + 1 = cos(θ14 − θ12 ) a2 a4 2a2 a4 or: K 1 cos θ14 − K 2 cos θ12 + K 3 = cos(θ14 − θ12 )

(8.7)

where   2 a1 + a22 − a23 + a24 a1 a1 K1 = , K2 = , K3 = a2 a4 2a4 a2 Equation (8.7) is called “Freudenstein’s Equation” which can be used for analysis of four-bar mechanisms [1]. It gives an implicit relation between the position variables θ14 and θ12 . We can also use this equation for synthesis since it gives us a relation between the input and output links with three link-length parameters. We must note that Freudenstein’s equation is an important milestone in modern kinematics. It is the first analytical approach to a classical problem for which till that period only geometrical methods were known in the west. Up to now, we used subscript for the angles (i.e., θ12 ) to mean we are measuring the position of a link (link 2) relative to another link (link 1). Since we shall use the subscript for different positions of a link, let θ = θ 12 and φ = θ 14 . Equation (8.7) can be written as: K 1 cos φ − K 2 cos θ + K 3 = cos(φ − θ )

(8.7)

Consider a case when link 2 is at an angle θ1 , link 4 is at angle φ1 , when link 2 is at θ2 link 4 is at φ2 and when link 2 is at angle θ3 , link 4 is at angle φ3 (Fig. 8.2). If these three positions of the links have to be satisfied by a four-bar mechanism, then Eq. (8.7) must be satisfied for all the three positions: K 1 cos φ1 − K 2 cos θ1 + K 3 = cos(φ1 − θ1 ) K 1 cos φ2 − K 2 cos θ2 + K 3 = cos(φ2 − θ2 ) K 1 cos φ3 − K 2 cos θ3 + K 3 = cos(φ3 − θ3 )

(8.8)

In synthesis problems the angles are given and the link lengths are to be determined. We want to satisfy the angular relationship between the links given by Eq. (8.1). Since the angles are specified, in the set of Eqs. (8.8), K1 , K2 , K3 are Fig. 8.2 Three crank angles to be correlated

336

8 Correlation of Input and Output Motion; Function Generation …

unknowns. These unknowns are related to the link lengths which we want to determine. Equation (8.8) forms a set of linear equations in terms of K1 , K2 , K3 . When Kj are solved, one can determine the link lengths as: a2 =

a1 a1 a4 = and a3 = K1 K2

/  a12 + a22 + a42 − 2K 3 a4 a2

If a1 is not specified, we can let a1 = 1 (for the correlation of crank angles it is the ratio of the link lengths that is important). Numerical solution of these equations can be performed using any mathematical package such as MatLab® , MathCad® , Excel® or Geogebra. Here we shall use Excel® . Example 8.1 Design a four-bar mechanism to correlate the following three design positions (angles in degrees): i

θi

φi

1

20

45

2

60

50

3

90

60

We write the three equations and obtain (In Excel you can write each term into a cell and thus solve 3 × 3 group for the coefficient matrix and 1 × 3 vector for the right-hand side): ⎡

⎤⎡ ⎤ ⎡ ⎤ 0.707107 −0.93969 1 0.906308 K1 ⎣ 0.642788 −0.5 1 ⎦⎣ K 2 ⎦ = ⎣ 0.984808 ⎦ K3 0.5 −6.1E − 17 1 0.866025 In Excel the screen will look as shown in Fig. 8.3 (of course, the cell rows and columns may be different). Now mark 3 cells in a row (C5; C7) in the figure and type “ = MMULT (MINVERSE(B1:D3),(F1:F3))”,1 and while pressing Ctrl and Shift keys press Enter key. You will obtain the thee values of K in cells C5 to C7 as: K1 = 2.98722, K2 = 0.61551 and K3 = −0.62758 Solving for link lengths: a1 = 1, a2 = 0.33476, a3 = 2.10577 and a4 = 1.62467 If the crank is equal to 200 mm (a2 = 200), than a1 = 597.4, a4 = 970.6 and a3 = 1258.1 mm. Figure 8.4 shows the resulting mechanism. 1

In Excel If you are using the decimal separator as coma, use semicolon between the two arrays rather than coma. i.e. type: “= MMULT (MINVERSE(B1:D3);(F1:F3)).

8.2 Function Generation

337

Fig. 8.3 Typical excel sheet for the solution of three linear equations

B1

Fig. 8.4 Four-bar mechanism synthesized in Example 8.1

B2 B1

A3

A2 A1

A0

B0

Note that a3 must always be positive. If the term under the square root is negative, then there is no solution. If K1 , or K2 (or both) turns out to be negative, a2 or a4 (or both) will be negative. Link lengths has to be positive. −a2 means that the direction of link 2 must change. When driving the Freudenstein’s equation note that A0 A was represented by a2 eiθ . If the angle is θ + π than a2 ei(θ +π) = −a2 eiθ . Therefore, the negative sign means that the link length must be measured opposite to the direction defined by the angle θ.

8.2 Function Generation 8.2.1 Three Precision Point Function Approximation Usually rather than correlating three pairs of crank angles, one is faced with the problem of obtaining a certain functional relationship between the rotations of the two cranks in a certain range of motion. An example would be to correlate a pedal motion with the opening of a valve or the correlation of the front tire rotations when taking a curve.

338

8 Correlation of Input and Output Motion; Function Generation …

We would like to realize a function y = f(x) mechanically within the interval xin < x < xfin . We represent the independent parameter x with the input crank angle rotation, θ and the dependent parameter y, with the output crank angle φ. Hence, we introduce a scale factor for each of these variables: rx =

Δθ Δφ ry = Δx Δy

(8.9)

whereΔx = xfin − xin and Δy = ymax − ymin (If the function is monotonically increasing or decreasing within the interval, ymax and ymin will be the values of y at xfin and xin . Otherwise, they will be the maximum and/or the minimum values of y within the interval). Δφ, Δθ and θin , φin corresponding to xin and yin , can be arbitrarily selected (hence we have 4 free variables). With the above arbitrarily selected values, we can convert x and y to θ and φ using a linear relation as: θ = θin + r x (x − xin )φ = φin + r y (y − yin )

(8.10)

Or x = xin +

1 1 (θ − θin ) y = yin + (φ − φin ) rx ry

The function to be realized by the four-bar mechanism will be φ = g , (θ ). When we use the above relations to convert φ and θ to x and y, the function realized by the mechanism will be: y = g(x). This functional relation depends on the link lengths. In general, this equation is different than the function we want to realize: y = f(x). If we define error function as: ε(x) = f(x) − g(x) This error is known as “structural error” to separate from manufacturing errors. As we have shown previously, we can determine the four-bar proportions such that for certain defined positions we can correlate the crank angles. Hence, we can satisfy the function y = f(x) at certain positions within the interval. For example, we can select x1 , x2 , x3 within xin < x < xfin , and determine the corresponding y values, x1 , x2 , x3 from yj = f(xj ), using Eqs. 10 determine θ1 , θ2 , θ3 and corresponding φ1 , φ2 , φ3 . We can determine the four-bar mechanism proportions that satisfy these three positions. Hence the error between the generated and the required functions at these points will be zero. At any other point the error will not be zero. Error curve ε(x) will be as shown in Fig. 8.5. The points at which the error is zero will be called “precision points”). If there are more precision points the error will be less. However, the number of precision points will depend on the number of free design parameters (in case of a four-bar we now have 3 design parameters, hence we shall have three precision points). If we select x1 , x2 , x3 within xin < x < xfin , arbitrarily

8.2 Function Generation

339

Fig. 8.5 Error curve. x1 , x2 and x3 are the precision points

we shall have an error curve as shown in Fig. 8.5. Depending on where we select these precision points the error curve will change. The selection of precision points is arbitrary. In most engineering problems the aim is to minimize the area underneath the error curve. Chebyshev,2 has found out for a certain class of polynomials (known as “Chebyshev polynomials”) the maximum deviation of the error function from zero can be minimized if the absolute value of the maximums and minimums within the interval are made equal. Furthermore, for n precision points there should be n + 1 minimum or maximum (alternating) and the end points are minimum or maximum. For Chebyshev polynomials of n precession points the precision points are located at:   1  1 (2 j − 1) π j = 1,2, . . . n x j = x f in + xin − x f in − xin cos 2 2 2n for n = 3: xj =

  1  1 (2 j − 1) x f in + xin − x f in − xin cos π j = 1,2,3 2 2 6

Precision points can be shown geometrically as in Fig. 8.6. In general, a circle of diameter (x f in − xin ) and center at 21 (x f in + xin ) is drawn and on the circumference of the circle we determine 2n equally spaced points symmetrically located about the x axis. The projection of these points onto the x axis are the precision points of Chebyshev polynomials. If the error curve is a Chebyshev polynomial the local maximum and minimum values will be equal (in alternating sign) and the maximum error within the interval is minimized (Fig. 8.7) (this is the basis of Chebyshev theorem and its a theory that has been rigorously proven). Our error function is not a Chebyshev polynomial. Therefore, the local maximum and minimum magnitudes will not be equal. However, 2

Chebyshev Theorem and Chebyshev polynomials will be discussed more in detail in Chap. 9.

340

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.6 Chebyshev spacing of precision points

it has been found out that when Chebyshev spacing for the precision points is used the local maximum and minimum values are close and the theory is approximately satisfied. Rather than using equal spacing or any other spacing for the precision points, Chebyshev spacing has been found to be a good choice. Freudenstein recommends the following rules of thumb when using Freudenstein’s equation: 1. Avoid abrupt changing functions. Four-bar mechanisms are good in approximating smooth functions. 2. Avoid generation of symmetric functions, such as y = sinx within the range π/3 < x < 2π/3, or y = x2 within the range −2 < x < 2. (whereas approximating y = sinx within π/6 < x < 4π/9 or y = x2 within 0 < x < 4 is acceptable and can be approximated by a four-bar). 3. The total angular rotations of the cranks (Δθ and Δφ) must be less than 120°. Example 8.2 Determine the link length proportions of a four-bar mechanism that will realize the function y = lnx within the interval 1 ≤ x ≤ 2. Let us use Chebyshev spacing for the precision points. We obtain: Fig. 8.7 Error curve for a Chebyshev polynomial with three precision points. Magnitudes of the local minimum and maximum errors are equal and if there are n precision points there are n + 1 local extremums (in alternating sign)

8.2 Function Generation

341

x1 = 1.06699; x2 = 1.5; x3 = 1.93301. using yi = lnxi we obtain: y1 = 0.06484; y2 = 0.40546; y3 = 0.65908. Now, we are going to correlate x with θ and y with φ. We have 4 free parameters: θin , φin , Δθ and Δφ. You may need a number of trials to come up with a good solution. Let θin = 30° φin = 30° (corresponding to x = 1 and y = 0 respectively) Also let Δθ = 90° and Δφ = 60°. Then the crank angles that we want to correlate can be determined according to Chebyshev spacing. Hence, we can form a table as (angles in degrees): i

x

θi (o )

y

φi (o )

1

1.0670

0.0648

36.0289

35.6126

2

1.5000

0.4055

75.0000

65.0978

3

1.9330

0.6591

113.9711

87.0511

We form a system of linear equations: ⎤ ⎤⎡ ⎤ ⎡ 0.99997 0.81297 −0.80872 1 K1 ⎣ 0.42107 −0.25882 1 ⎦⎣ K 2 ⎦ = ⎣ 0.98510 ⎦ K3 0.89164 0.05145 0.40628 1 ⎡

The solution is: K1 = −0.72316, K2 = −0.54242, K3 = 1.149216 The Link lengths are: a1 = 1, a2 = −1.383, a3 = 0.672, a4 = −1.844 a1 = 1, a2 = −1.383, a3 = 0.672, a4 = −1.844. Note that both cranks have negative values. Hence the links are drawn opposite to the angles defined as shown in Fig. 8.8. The required and generated curves are plotted in Fig. 8.9. The error between the two curves is given in Fig. 8.10. These local maximum extreme values are: ε1 = − Fig. 8.8 Four-bar mechanism synthesized in Example 8.2. (θin = 30° φin = 30°)

2

1.8 1.6

A0

A

B

1.4 1.2

1

ln2

y =lnx ln1

x B0

342

8 Correlation of Input and Output Motion; Function Generation …

0.0128, ε2 = 0.0132, ε3 = −0.0199 and ε4 = 0.0403. Note that the magnitude of local maximum and minimum errors are not exactly equal. If the resulting error is considered to be unacceptable, one can change the initial crank angles θin and φin while keeping the ranges Δθ and Δφ the same. For example, in the problem considered if we select θin = 45°. φin = 30°, we obtain a1 = 1, a2 = −0.210, a3 = 0.866 and a4 = −0.327 with a maximum error of emax = 0.007 The mechanism and the error curve are shown in Figs. 8.11 and 8.12. Maximum error has decreased from 0.04 to 0.007. This is much better than the previous result. One can find a better solution by changing the input crank angles. A solution may not exist in certain cases and for some initial crank angles one may obtain a very low maximum error but the ratio between the longest link to the shortest link for the resulting mechanism may not be acceptable. Usually mechanisms with maximum to minimum link length ratio greater than 10 or 15 (of course, it will depend on the application) is considered undesirable in practical applications. For example, for the problem considered if we select θin = 48° φin = 28°, maximum error is εmax = Fig. 8.9 The function generated and the function realized (Example 8.2). (θin = 30° φin = 30°)

0.7

y generated

0.6

y required

0.5

y

0.4 0.3 0.2 0.1 0

1

1.2

1.4

x

1.6

1.8

Fig. 8.10 Resulting error curve ε = yrequired − ygenerated in Example 8.2. (θin = 30°, φin = 30°)

2

8.2 Function Generation

343

Fig. 8.11 Mechanism synthesized in Example 8.2 when θin = 45°. φin = 30o

2 1.8 1.6

1.4

1.2

ln2 y=lnx ln1 y

1

x

A0

B0

A

B

Fig. 8.12 Resulting error curve in Example 8.2 when θin = 45° φin = 30°

0.0048. However, a2 = −0.002 and a4 = −0.004 when a1 = 1 unit. It will be very hard, if not impossible, to construct such a mechanism. As seen in this example, selection of initial crank angles changes the result extensively. Also, in cases where Chebyshev spacing on x does not yield a good solution, another alternative is to apply Chebyshev spacing on y and/or take the average of the x and y spacings. Spacing on y rather than on x need not necessarily yield a better solution for all cases considered. However, when approximating certain types of functions different spacings may result in a better approximation. Example 8.3 Consider the same problem discussed in Example 8.2. Using the same range and the same initial crank angles (θin = 30° φin = 30° and Δθ = 90° Δφ = 60°) and applying Chebyshev spacing on y the resulting mechanism is not movable in between 115° < θ < 120°, corresponding to 1.95 < x < 2. Therefore, let θin = 35° and consider Chebyshev spacing on y. We have: i

x

y

1

1.04750

0.0464

2

1.41420

3

1.90930

θi (o )

φi (o )

39.2774

34.0192

0.3466

72.2792

60.0000

0.6467

116.8333

35.9808

344

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.13 Error curve for Example 8.3 (Chebyshev spacing on y), (θin = 35° φin = 30°)

When we solve the system of linear equations: K1 = −0.8929, K2 = −0.6649, K3 = 1.2212 and a1 = 1, a2 = −1.11993, a3 = 0.6343, a4 = −1.504 The error curve is shown in Fig. 8.13. Local extremum values are:ε1 = −0.0065, ε2 = 0.0086, ε3 = −0.0203 and ε4 = 0.0472. For this problem, although the first two extreme values are less when spacing on y is used, the error at x = 2 is larger than the value obtained when we performed Chebyshev spacing on x. If we take the average of the xi values obtained from Chebyshev spacings on x and y: than we have (x-I and x-II are the x values obtained when Chebyshev spacing on x and on y are made respectively. xaverage = (x-I + x-II)/2: i

x-I

x-II

xAverage

y

1

1.0475

1.0670

1.0573

0.0464

2

1.4142

1.5000

1.4571

3

1.9093

1.9330

1.9212

θi (o )

φi (o )

39.2774

34.0192

0.3466

72.2792

60.0000

0.6467

116.8333

85.9808

When we solve the system of linear equations: K1 = −0.6961, K2 = −0.5260, K3 = 1.1413 and a1 = 1, a2 = −1.4367, a3 = 0.6661,a4 = −1.9010

8.2 Function Generation

345

Fig. 8.14 Error curve for Example 8.3 (average of the x and y spacings), (θin = 35° φin = 30°)

The error curve is shown in Fig. 8.14. Local extremum values are: ε1 = −0.0109, ε2 = 0.0135, ε3 = −0.0218 and ε4 = 0.0498. For this particular problem, Chebyshev spacing on x yield the minimum maximum error (εmax = 0.0403). Depending on the function to be approximated and the value of the assumed parameters, spacing on y or average spacing may yield a better result. For this particular problem all three spacings yield relatively close results. One must always check the final result for movability, transmission angle and link length ratios. If the result is not feasible there are four parameters (θin , φin , Δθ and Δφ) which can be changed. Usually, we first select Δθ and Δφ of reasonable magnitude and change θin and φin . If we cannot achieve a good result, then we try to reduce the size of Δθ and Δφ In great number of problems some of these 4 parameters may be defined or the range for these parameters may be restricted.

8.2.2 Four or Five Precission Point Function Approximation The beauty in three precision point synthesis is that there are three linear equations to be solved. As design parameters, besides the three coefficients of Freudenstein’s equation (which are functions of the three link length ratios ratios: a2 /a1 , a3 /a1 , a4 / a1 ), one can measure the input and output angles from references other than the line joining the fixed pivots (Fig. 8.15). Using Freudenstein’s equation one can solve for four or five precision points [2, 3]. The orientation of these reference axes θin and φin will be the two new design parameters as shown. However, unlike the three-position synthesis problem, the equations to be solved are nonlinear in terms of θin and φin . In most cases higher number of precision points will yield a better result in terms of

346

8 Correlation of Input and Output Motion; Function Generation …

structural error for the same range. Increasing the number of precision points may in some cases limit the range of feasible solutions and may not yield a better result. Depending on the problem, rather than increasing the number of precision points these free parameters can be used such that a mechanism with better transmission angle characteristics or mechanisms with a more favourable link proportions can be obtained. Example 8.4 A four-bar mechanism that correlates the four crank angles given in Fig. 8.16 (angles in degrees) is to be designed. In addition to the link length ratios, we can use θin as the fourth design variable. Now Freudenstein’s equation can be written for four positions:     K1 cos φj − K2 cos θj +θs + K3 = cos θj − θj −θin j = 1, 2, 3, 4 The equations are linear in terms of K1 , K2 , K3 but nonlinear interms of θin . If we assume that we know the value of θin, then we have 4 linear equations in terms of three unknowns. In order this linear set of equations to have a solution the determinant of the augmented matrix must be zero. / / Cos(φ1 ) −Cos(θ1 + θin ) 1 / / Cos(φ2 ) −Cos(θ2 + θin ) 1 / / Cos(φ3 ) −Cos(θ3 + θin ) 1 / / Cos(φ ) −Cos(θ + θ ) 1 4 4 in

/ Cos(φ1− θ1− − θin ) // Cos(φ2− θ2− θin ) // =Δ=0 Cos(φ3− θ3− θin ) // Cos(φ4− θ4− θin ) /

Fig. 8.15 Correlation of more than three crank angles using θin and φin as design variables

Fig. 8.16 Correlation of crank angles for four precision points. a2 /a1 , a3 /a1 , a4 /a1 and θs are four design variables.

8.2 Function Generation

347

The aim is to determine the values of θin which makes Δ = 0. If we are to use Excel, let us assume a value for θin . The determinant will not necessarily be zero (unless you have hit the jackpot!!). Now use “Goal seek” or “Solver” tool in Excel to make the value of the determinant to be equal to zero by changing the value of θin . According to the initial guess, you will have different solutions for θin . In general, there are at most 4 solutions for θin which pairwise differ by 180°. To determine these solutions different initial guesses must be made. (Also make sure that you have increased the number of iterations and reduced maximum change in Excel (go to File-Options-Formula menu). Once you determine θin , you can solve the parameters K1 , K2 , K3 using any 3 of the four equations. For the values given, when we assume θin = 15°, Δ = 0.006726. Using solver algorithm (GRG nonlinear and unconstrained variables can be negative) to make the determinant zero, by changing value of θin , we end up with θin = 69.58566° (Δ = −3.5 × 10–9 ). If different guesses for θin is made, four different values of θin are found. These angles are 69.58566°, −110.414, 96.46821, −83.5318. Note that the first and second, third and fourth values of θin which make determinant zero refer to the same mechanism (difference between the first and second, third and fourth angles are180o ). Now, one can use any three of the four equations to determine K1 , K2 , K3 and then a2 , a3 , a4 . The results are shown in Fig. 8.17. As you can see from the table two different four-bar mechanism proportions satisfying four positions may be possible. Depending on the function to be approximated and the value of θin or φin selected, one may obtain only one solution or no solution. In this particular example, resulting two mechanisms are shown in Fig. 8.18. Instead of θin one can use φin as a free parameter and obtain a different result. Example 8.5 Consider the same problem discussed in Example 8.2 (to generate a function y = ln(x) 1 ≤ x ≤ 2). The same range and the same initial input crank angle are to be used (θin = 30° and Δθ = 90° Δφ = 60°). We consider the initial output angle φin as a design variable. Starting from a different arbitrary guesses for φin , using Solver (GRG nonlinear, unconstrained variables can be negative) we obtain four different values which result with two different four-bar mechanisms. Initial crank angles, link lengths mechanisms at their first position and the error curves are as shown in Figs. 8.19. When we consider θin as the design variable and keep φin = 30° no solution exists. However, when φin = 110° the determinant of the augmented matrix can be made

Fig. 8.17 Four possible solutions for θs and two four-bar proportions obtained in Example 8.4

348

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.18 Resulting two four-bar mechanisms correlating four crank angles (Example 8.4)

Fig. 8.19 Two different four-bar proportions and their error curves obtained for four precision points (φin as the design variable, θin = 30°) for function y = ln(x) 1 ≤ x ≤ 2). (Example 8.5)

zero when θin = 73.8277° and −16.7282°. Mechanisms and the resulting structural errors are as shown in Fig. 8.20 (εmax = 9.54 × 10–4 ) and Fig. 8.21 (εmax = 2.93 × 10–3 ). The same method can be applied for five precision points by considering both θin and φin as design variables. We can write Freudenstein’s equation for five precision points. The augmented matrix will have 5 rows and 4 columns. For the rank of the

8.2 Function Generation

349

Fig. 8.20 Four precision point synthesis result for function y = ln(x) 1 ≤ x ≤ 2). (Example 8.5).θin = −16.7282°, φin = 110°. A0 A = −4.40471, AB = 4.57331, B0 B = 1.99627 (A0 B0 = 1)

Fig. 8.21 Four precision point synthesis result: θin = 73.8227°, φin = 110°, a1 = 1, a2 = 3.29602, a3 = 1.03359, a4 = 3.24167

augmented matrix to be 3, the determinants of all the five 4 × 4 matrices formed from 5 × 4 augmented matrix must be zero. Using an initial guess for θin and φin , the values of the determinants can be determined (Δ1 , Δ2 , Δ3 , Δ4 , Δ5 ) The requirement for 5 determinants to be zero can be reduced to a single condition by summing the square of the determinants: ε=

/ Δ21 + Δ22 + Δ23 + Δ24 + Δ25

ε can only be zero, if and only if the values of all the determinants are equal to zero. Now we can use “solver” tool in Excel to make ε zero by changing the values of our initial guess for θin and φin . Note that a solution mayo r may not exist. Example 8.6 Consider the same problem discussed in Example 8.2 (to generate a function y = ln(x), 1 ≤ x ≤ 2). The same range is to be used (Δθ = 90° Δφ = 60°). Applying Chebyshev spacing for five precision points:

350

8 Correlation of Input and Output Motion; Function Generation …

 x j = 1.5 − 0.5 cos

(2 j − 1)π , 10

j = 1, 2, 3, 4, 5

Assuming θin = 30° and φin = 60° we have the following precision points: θj

φj

j

xj

yj

1

1.0245

0.0242

32.2025

62.0928

2

1.2061

0.1874

48.5497

76.2215

3

1.5000

0.4055

75.0000

95.0978

4

1.7939

0.5844

101.4503

110.5856

5

1.9755

0.6808

117.7975

118.9343

When we write Freudenstein’ sequation for these positions, assuming θin and φin as known, we have 5 linear equations in three unknowns K1 , K2 and K3 . The augmented matrix takes the form:

0.46804 -0.84617 0.23817 -0.66197 -0.08886 -0.25882 -0.35161 0.19852 -0.48381 0.46635

1 1 1 1 1

0.86698 0.88562 0.93911 0.98732 0.99980

In order to have a solution for K1 , K2 and K3 , the rank of this augmented matrix must be 3. Eliminating one row at a time we have five 4 × 4 determinants. The values of these determinants are: Δ1 = 0.000578, Δ2 = 0.001479, Δ3 = −0.001847, Δ4 = 0.001603 and Δ5 = 0.000657

/ and ε = Δ21 + Δ22 + Δ23 + Δ24 + Δ25 = 0.002989. In order to have a solution for Kj , φs and θs must be selected such that ε = 0. Using Excel Solver tool (GRG nonlinear) we search for values φin and θin to make the value of ε to be zero the result is: ε = 2.3 × 10−9 Δ1 = 2.04 × 10−10 , Δ2 = 6.56 × 10−10 , Δ3 = −1.22 × 10−9 Δ4 = 1.65 × 10−9 and Δ5 = 8.79 × 10−10 and θs = 63.26148°, φs = 105.4232°. The augmented matrix is:

8.2 Function Generation

-0.30097 -0.41527 -0.52465 -0.14244 -0.77186 0.31335 -0.91361 0.70354 -0.96296 0.87512

351

1 1 1 1 1

0.74254 0.76791 0.84564 0.93171 0.97319

One can now use any three of the above four equations to solve for Kj : K1 = 0.279466, K2 = 0.322115 and K3 = 0.960411 The link lengths are: a1 = 1, a2 = 3.57835, a3 = 1.4505 and a4 = 3.1045. The mechanism and the error curve is as shown in Fig. 8.22 (|εmax |=0.00026). Since numerical method is used, the results may change according to the set-up. The number of iterations and maximum change and the initial guess for values φin and θin play an important role. In certain cases one must try different initial guesses. As the number of precision points increase the maximum error decreases. However, mechanism design is usually not based on a single criterion. The space occupied, transmission angle link length ratios and constraints imposed by the particular problem to be solved must all be taken into account. In addition to increasing the number of precision points, there are other ways of minimising the error. Note that when Chebyshev spacing is used, the resulting error curve is not optimized according to Chebyshev theorem (absolute values of local maximums and minimums within the interval are not made equal). Freudenstein

Fig. 8.22 Four-bar mechanism and its error curve obtained using five precision point synthesis in Example 8.6

352

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.23 Error curve showing local extremum values and the intervals

[2] has shown that the error can be minimized in Chebyshev sense by successive respacing of the precision points such that the absolute maxima and minima are made equal. Intuitively, we can say the maximum local error that occurs within the interval is proportional to the size of the interval. Let Ei be the absolute maximum error within the interval Δi . (Fig. 8.23) we want to change the location of precision points so that the new values of intervals Δ’i. result with maximum errors that are all of the same value, Emin . We have two basic equations. The first equation is obtained from the assumption that the local errors are proportional to the size of the interval in the form: E min = Ei



Δi, Δi

m (8.1)

where power m is to be determined empirically (this is an assumption) and the sum of the intervals adds up to the total range of x: ∑Δi, = x f in − xin = Δ Solving for Δ’i : Δ, i =Δi 1/m

also E min

1/m

E min

1/m

Ei Δ = ∑ Δi hence Δi, = 1/m

Ei

ΔΔi ∑

1/m Ei

Δi 1/m Ei

(8.2)

With the new values Δi we determine the new precision points and determine the maximum errors. Since the proportionality of the maximum error to the size of the interval is an assumption and the value of m is an empirical value, the new values may not yield equal maximum values. The process can be repeated till equal maxima

8.2 Function Generation

353

are reached. The values of m in between 3 and 4 is recommended by Freudenstein (Eq. 8.2 is usually called as “Freudenstein’s respacing formula”). Example 8.7 Consider the problem discussed in Example 8.2. When Chebyshev spacing is used on x, the intervals and the maximum errors within the intervals are:

Select m = 3. Now, we can calculate Δ’i and the new precision points: With Chebyshev spacing, the maximum error at x = 2 was Emax = 0.040321. Using the new precision points, the maximum errors are not made equal but maximum error is still at x = 2 with a value of Emax = 0.019059. Which is a big decrease. When we repeat the procedure 5 times The maximum error is reduced to Emax = 0.0163 (at x = 2). The two error curves are as shown in Fig. 8.24. The link length dimensions of the final four bar mechanism are: a1 = 1, a2 = −1.186, a3 = 0.693, a4 = −1.615 When a1 = 120.0 mm a2 = 141; a3 = 83; a4 = 194 mm or when a2 = 100 mm a1 = 84; a3 = 58; a4 = 136 mm. 0.04

0.03

0.02

Respacing after 5th iteration

Chebyshev Spacing

0.01

0 1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

-0.01

-0.02

Fig. 8.24 Three precision point synthesis. The error curve using Chebyshev spacing and improvement made after fifth iteration using Freudenstein’s respacing formula

354

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.25 Initial crank angles, link lengths and the maximum error values for four-bar mechanisms approximating function in Example 8.8 using four equally spaced precision points

When the structural error is less than the manufacturing tolerances and clearances, one can easily stop searching for a better solution, since even if a better solution exists, it cannot be realized. If one considers these different possibilities in solving function generation, it is very probable that a good satisfactory solution can be reached after certain number of trials. The respacing shown for three precision points can easily be applied for four or five precision points. Example 8.8 A four-bar mechanism is to be designed to generate the function [4]: x y = − (x + 6) 8 in the range 0 ≤ x ≤ 6. Both cranks are to rotate 90° for this range of x. When we design the mechanism using four precision points and select φin = 0°, 15°, 30°, 45°, 60° and 75° and use θin as a design variable. In reference (8.6) precision points are selected as x = 0, 2, 4 and 6. The result of the solutions made for the 6 different output link initial angle is given in a tabular form in Fig. 8.25. In Fig. 8.26 the error curves when φin = 0°, 45°, and 75° are shown. Instead of equally spaced precision points if we use Chebyshev spacing, results shown in Fig. 8.27 is obtained. Four-bar mechanism solution when φin = 45° is shown in Fig. 8.28. as it can be seen, for all φin values there is a considerable decrease in maximum error involved when we use Chebyshev spacing instead of equally spaced precision points. Example 8.9 As another example, consider approximating y = tan x within the range 0° ≤ x ≤ 60°. Select Δθ = 90° and Δφ = 90°. We are free to select the initial crank angles. (a) Applying three position synthesis and using Chebyshev spacing on x and after several trials in initial crank angles,3 for θin = −51° and φin = −7°, we obtain An easy way to change θin and φin values in Excel will be to use “spin button” as a control unit and change these values. If you have prepared an Excel sheet showing the structural error, then with each click of the botton the structural error will change. Thus, a very fast scanning for a good solution can be performed and one can recognize the regions of the initial crank angles where a good solution is available or where there is no solution.

3

8.2 Function Generation

355

Fig. 8.26 Error curve when φin = 0°, 45°, and 75° for four-bar mechanisms approximating function in Example 8.8 using four equally spaced precision points

Fig. 8.27 Initial crank angles, link lengths and the maximum error values for four-bar mechanisms approximating function in Example 8.8 using four precision points according to Chebyshev spacing

a1 = 1, a2 = −1.889, a3 = 1.186, and a4 = −2.382 the error curve is as shown in Fig. 8.29. The maximum error is εmax = 0.0040. Note that we have 4 precision points. (b) Applying three position synthesis and using Chebyshev spacing on y and after several trials in initial crank angles, for θin = −52° and φin = −6°, we obtain a1 = 1, a2 = −1.844, a3 = 1.216, and a4 = −2.349 the error curve is as shown in Fig. 8.30 (εmax = 0.004). Please note the similarity of the two curves obtained in (a) Fig 8.29 and (b) (Fig. 8.30). (c) Using four precision points and selecting φin = −18°, and using Chebyshev spacing, we obtain θin = 60.8229° and the link lengths a2 = −1.4820, a3 = 2.5894, a4 = −3.1370 (a1 = 1). The resulting mechanism and the error curve is shown in Fig. 8.31 (εmax = 1.1 × 10–3 ). Solution for five position synthesis using Chebyshev spacing is shown in Fig. 8.32. we obtain θin = 82.1044°, θin = 149.449° and the link lengths a2 = −1.6509, a3 = 5.3870, a4 = 5.2151 (a1 = 1) and εmax = 6 × 10–3 . Note that there is an increase in

356

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.28 Four-bar mechanism obtained in Example 8.8 using four precision points according to Chebyshev spacing and φin = 45o Fig. 8.29 Structural error curve in approximating y = tan x, 0 ≤ x ≤ 60°. Three precision points using Chebyshev spacing on x. (Example 8.9)

Fig. 8.30 Structural error curve in approximating y = tanx, 0o ≤ x ≤ 60°. Three precision points using Chebyshev spacing on y

8.2 Function Generation

357

Fig. 8.31 Mechanism sketch and structural error curve in approximating y = tanx, 0 o ≤ x ≤ 60°. Four precision points using Chebyshev spacing on x

maximum error. In terms of maximum error and the ratio of maximum to minimum link lengths, solution obtained using four-precision points seems to be a better choice. Example 8.10 In this example a problem solved in reference [5] (Example 6.2 page 122). The aim is to design a four-bar dial to approximate the flow rate (Q) and pump head (h) readings, which is given by the equation h = 0.0017Q2 within 300 ≤ Q ≤ 800. Using 6 data points and selecting Δθ=Δφ = 60° the problem was solved. A0 A1 = 0.8614 − 0.27762i, A0 B0 = 1, B0 B1 = −0.9096 −1.7535i (the resulting link lengths are: a1 = 1, a2 = 2.9068, a3 = 1.2810 and a4 = 1.9753. Mechanism and the structural error results are as shown in Fig. 8.33. Maximum structural error is 1.0083. When we apply Chebyshev spacing on Q and select θin = 107.2381°, φin = 62.5775° (which is the result from the reference solution), Δθ=Δφ = 60° and solve the problem using three precision points we obtain four-bar mechanism proportions

Fig. 8.32 Mechanism sketch and structural error curve in approximating y = tanx, 0 o ≤ x ≤ 60°. Five precision points using Chebyshev spacing on x

358

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.33 Four-bar mechanism and its structural error curve in approximating h = 0.0017Q2 300 ≤ Q ≤ 800. Using 6 equally spaced data points for approximation [5] Fig. 8.34 Structural error curve in approximating h = 0.0017Q2 300 ≤ Q ≤ 800, using three precision points Chebyshev spacing and the result from References [5]. (θin = 107.2381°, φin = 62.5775°)

as: a1 = 1, a2 = −2.9278, a3 = 1.22965 and a4 = −1.9881. The structural error curve is as shown in Fig. 8.34 (εmax = 1.2807). The link lengths and the structural error is close to the solution given by the reference. If we simply change θin and φin , we end up with different results. For example, using spin button for changing initial crank angles θin and φin and observing the change in the structural error, when θin = 94° and φin = 62°, we end up with a solution a1 = 1, a2 = −2.9712, a3 = 0.8684 and a4 = −2.4633. The mechanism and its structural error are as shown in Fig. 8.35 (εmax = 0.5692). If we select θin and φin as design variables and use Chebyshev spacing (five precision points) we obtain: θin = 92.5303° and φin = 62.6093°, a1 = 1, a2 = −3.0334, a3 = 0.81904 and a4 = − 2.5589. The structural error is shown in Fig. 8.36 (εmax = 0.3654).

8.2.3 Method of Least Squares Rather than approximating a function using a number of precision points, one can consider the “method of least squares”. In this method the aim is to minimize the

8.2 Function Generation

359

Fig. 8.35 Four bar mechanism and its structural error curve in approximating h = 0.0017Q2 300 ≤ Q ≤ 800, using three precision points Chebyshev spacing and the result from reference (7 ). (θin = 94°, φin = 62°)

Fig. 8.36 Structural error curve in approximating h = 0.0017Q2 300 ≤ Q ≤ 800, using five precision points Chebyshev spacing. (θin = 92.5303°, φin = 62.6093°)

area underneath the error curve, instead of minimizing the maximum error. Due to alteration in sign, we square the error, and then take the sum of the square of the errors for n number of data points within the interval. Consider a case where we want to correlate θi with φi for n points (not necessarily precision points or design points and usually n is greater then the number of design parameters), If we substitute these values into Freudenstein’s Equation, Freudenstein’s equation will not be satisfied and there will be some error involved as [6]: K 1 cos φi − K 2 cos θi + K 3 − cos(φi − θi ) = εi (i = 1,2,..n)

360

8 Correlation of Input and Output Motion; Function Generation …

Let us square the error at each point and take the sum: S=

∑

εi2 =

∑

n

[K 1 cos φi − K 2 cos θi + K 3 − cos(φi − θi )]2

(8.1)

n

We want to minimize S. A necessary (but not sufficient) condition for S to be a minimum is if the partial derivatives of S with respect to design parameters is zero. Since there are three design parameters (K1 , K2 and K3 ), this results in three equations: ∑ ∂S =2 cos φi [K 1 cos φi − K 2 cos θi + K 3 − cos(φi − θi )] = 0 ∂ K1 n ∑ ∂S = −2 cos θi [K 1 cos φi − K 2 cos θi + K 3 − cos(φi − θi )] = 0 ∂ K2 n ∑ ∂S =2 (8.2) [K 1 cos φi − K 2 cos θi + K 3 − cos(φi − θi )] = 0 ∂ K2 n Expanding the terms: K1

∑

cos2 φi − K 2

∑

n

K1

∑

n

cos φi cos θi − K 2

n

K1

∑

cos φi − K 2

∑

n

∑

cos φi cos θi + K 3 ∑

cos φi =

n

cos2 θi + K 3

n

cos θi + n K 3 =

n

∑

∑

∑

cos φi cos(φi − θi )

n

cos θi =

n

∑

cos θi cos(φi − θi )

n

cos(φi − θi )

(8.3)

n

Again we obtain three linear equations in terms of three unknowns. Example 8.11 We shall again consider approximating the function y = lnx within 1 ≤ x ≤ 2. Let θin = 30°, φin = 30° (corresponding to x = 1 and y = 0 respectively), Δθ = 90° and Δφ = 60° as in the previous examples. Let us select xi with 0.1 intervals. We can form a table as shown in Fig. 8.37. We take the cosine of the respective angles and determine the sum to obtain the linear set of Eqs. (8.3) as: ⎡

∑

cos2 φi

⎢∑ n ⎢ cos φi cos θi ⎢ ⎣ n ∑ cos φi ⎡

n

−

∑

∑ ⎤ ⎤ ⎡∑ cos φi cos θi cos φi = ⎡ cos φi cos(φi − θi ) ⎤ n ∑ n n K ⎥ ⎥ ⎢∑ 1 ∑ ⎥ ⎢ − cos2 θi cos θi = ⎥ ⎥⎣ K 2 ⎦ = ⎢ cos θi cos(φi − θi ) ⎥ n n n ⎦ ⎦ ⎣ ∑ ∑ K3 − cos θi n cos(φi − θi ) n

⎤⎡

⎤

⎡

⎤

K1 2.860 −2.436 4.699 4.649 ⎣ 2.436 −2.766 2.508 ⎦⎣ K 2 ⎦ = ⎣ 2.615 ⎦ K3 4.699 −2.508 11 10.605

n

8.2 Function Generation

361

Fig. 8.37 10 equally spaced data points for y = lnx within 1 ≤ x ≤ 2 and their correlated input and output angles

When the three linear equations are solved: K1 = −0.690, K2 = −0.519 and K3 = 1.1407. The corresponding link lengths are (a1 = 1 unit): a2 = −1.448, a3 = 0.666, a4 = −1.927. Now we perform the kinematic analysis of the resulting mechanism, which is shown in Figs. 8.37 and 8.38. In Fig. 8.39.ε = yrequired − ygenerated (structural error) is plotted in terms of x. Note that there are 3 precision points. The maximum error is is εmax = 0.07 which is greater than the error obtained in three precision synthesis with the same initial crank angle (Fig. 8.10. εmax = 0.0403). Comparing the two approaches in terms of maximum error is not correct since in one case the aim is the minimization of the maximum (known as minimax method, in short) the other is the minimization of the sum of the squares of the errors within the range. Although they are related (both methods tend to reduce the area under the error curve), they are not the same. If a better solution is searched by changing the initial crank angles, the result shown in Fig. 8.40 is obtained when θin = 29°, φin = 14°. Note that there are four precision points and the maximum error has reduced to 0.005. The link lengths are:

Fig. 8.38 Kinematic analysis of the resulting mechanism in Example 8.11

362

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.39 Structural error curve fort the four-bar mechanism obtained using least squares minimization method in Example 8.11 (θin = 30°, φin = 30°)

(a1 = 1 unit): a2 = −0.220, a3 = 0.804, a4 = −0.400. Mechanism is shown in Fig. 8.41. There are other methods for function generation which will be discussed in the following two chapters. In an actual industrial problem, it is recommended to try different methods and compare the results in terms of all constraints particular to that problem. For example, in certain cases the force transmission effect may be more important than the structural error or the constraints on the link lengths may be more effective.

Fig. 8.40 Four-bar mechanism and its structural error curve obtained using least squares minimization method in Example 8.11 (θin = 29°, φin = 14°)

8.3 Order and Mixed Order Synthesis

363

A

45°

45° B0

A0

B Fig. 8.41 Four-bar mechanism approximating a gear pair with −1 gear ratio

8.3 Order and Mixed Order Synthesis For certain applications rather than approximating a function in a certain range, we would like to approximate the function in a small interval with high accuracy e.g. we would like to approximate the function and its derivatives. Freudensteins equation is: K 1 cos φ − K 2 cos θ + K 3 = cos(φ − θ )

(8.7)

Successive differentiation with respect to the input crank angle θ yields:    2 d 2φ dφ dφ − 1 sin(φ − θ )K 1 sin φ + cos φ dθ dθ 2 dθ 2  d 2φ dφ − 1 −K 2 cos θ = sin(φ − θ ) + cos(φ − θ ) dθ 2 dθ

dφ sin φ − K 2 sin θ = K1 dθ



If the crank angles and its derivatives are given, we again obtain three linear equations in three unknowns, K1 , K2 and K3 . Example 8.12 Motion is being transferred between two shafts by means of two equal gears (gear ratio = −1) for 20° of shaft rotations. Gear ratio need not be exact, therefore we would like to replace these gears by a four-bar mechanism. Let the crank angles be equal to θ = 45° and φ = 45°. In order to approximate a 2 = −1 and ddθφ2 = 0. Substituting these values gear with a four-bar we must have dφ dθ into the equations: K 1 cos 45◦ − K 2 cos 45◦ + K 3 = 1 −K 1 sin 45◦ − K 2 sin 45◦ = 0 K 1 cos 45◦ − K 2 cos 45◦ = 4

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.42 Structural error curve in approximating the gear ratio for the four-bar mechanism obtained in Example 8.12

1.5 1 0.5 0 -0.5

ε

364

10

30

θ

50

70

90

The result is: K1 = 2/0.707 = 2.8284 K2 = −K1 = −2.8284 K3 = −3 The link lengths are: a 1 = 1; a2 = 0.3535; a4 = −0.3535; a3 = 0.7071. If we let a1 = 120 mm a2 = 42.4, a4 = −42.4, a3 = 84.9 mm. The mechanism is shown in Fig. 8.41. And the error between the required and the generated functions is shown in Fig. 8.42. Note that the error within 35° ≤ θ ≤ 55° is quite negligible. One can also make mixed approximation such as correlating two pairs of crank angles and specifying the rate of change of output link angle with respect to the input. An example would be to specify two positions of the cranks and to have one of the positions to be a dead center position (dφ/dθ= 0). Example 8.13 Design a four-bar mechanism such that when θ1 = 110°, φ1 = 40° and when θ2 = 200°, φ2 = 140°. when θ2 = 200°, we would like the mechanism to be at a dead center position (dφ/dθ = 0). The three equations which must be satisfied are: K 1 cos40◦ − K 2 cos110◦ + K 3 = cos(−70◦ ) K 1 cos140◦ − K 2 cos 200◦ + K 3 = cos(−60◦ ) K 2 sin(200◦ ) = − sin(−60◦ ) We can solve for K2 from the third equation (K2 = 2.532089). Substituting into the first two equations result in: 

0.766044 1 −0.766044 1



K1 K3



 =

−0.524005 −1.879385



Solving for K1 and K3 : K1 = 0.884661 and K3 = −1.201695. Solving for the link lengths yield: a1 = 1, a2 = 1.130376, a3 = 1.872603 and a4 = 0.394931

8.3 Order and Mixed Order Synthesis

365

The mechanism, output angle φ and transmission angle μ as a function of input crank angle are shown in Fig. 8.43a–c. When θ = 200° the output angle curve is horizontal, meaning dφ/dθ = 0 (Note that a few iterations for the initial crank angles is made to come up with a reasonable solution with the link lengths and the transmission angle).

Fig. 8.43 Mechanism synthesized in Example 8.13 a mechanism sketch, b change of output angle φ and c transmission angle μ as a function of input crank angle θ

366

8 Correlation of Input and Output Motion; Function Generation …

8.4 Correlation of Two Crank Angles with Optimum Transmission Angle In great number of problems only the correlation of crank angles for two positions is required. If we use Freudenstein’s equation to satisfy these two positions we have two linear equations in terms of 3 unknowns. If θ1 , φ1 and θ2 , φ2 K1 cos φ1 − K2 cos θ1 + K3 = cos(φ1 − θ1 )

(8.1)

K1 cos φ2 − K2 cos θ2 + K3 = cos(φ2 − θ2 )

(8.2)

Hence we have one free parameter. What we can do is use this free parameter to optimize for the transmission angle. Equating the diagonal AB0 using the cosine theorem for the two triangles B0 A0 A and B0 AB, the transmission angle μ for a four-bar mechanism in terms of crank angle is obtained as (Fig. 8.44): a22 + a12 − 2a2 a1 cosθ = a32 + a42 − 2a3 a4 cosμ

(8.3)

The equation can be written in the form.   2 a1 + a22 − a32 + a42 a3 a4 a1 cos μ = − + + cos θ a2 2a2 a4 a2 a4 or in terms the parameters Ki : a3 K1 cos μ = −K 1 + + K 2 cos θ a2 K2 Fig. 8.44 Four-bar mechanism

(8.4)

8.4 Correlation of Two Crank Angles with Optimum Transmission Angle

367

Transmission angle is maximum or minimum when θ = 0 or π. When 0 < θ < π or π < θ < 2π the transmission angle is monotonically increasing or decreasing. When the crank angle changes in between θ1 and θ2 , if crank angles 0 and π is not in between θ1 and θ2 , the transmission will be maximum and minimum at these two positions. If we make the deviation of the transmission angle from 90° at the two positions equal, then we can make the transmission angle deviation within θ1 < θ < θ2 optimum. ◦ ◦  When θ =◦ θ1 let μ = ◦μ1 = 90 − δ and when θ = θ2 μ = μ2 = 90 + δ or μ1 = 90 + δ, μ2 = 90 − δ Then for the two positions: a3 K1 sin δ = −K 1 + + K 2 cos θ1 a2 K2

(8.5)

a3 K1 sin δ = K 1 − − K 2 cos θ2 a2 K2

(8.6)

When we equate the two deviations at θ1 and θ2 ; 2K 3 −

2K 1 − K 2 (cos θ1 + cos θ2 ) = 0 K2

(8.7)

We can solve for K1 and K2 in terms of K3 from Eqs. (8.1) and (8.2) as: K1 =

1 [P1 K 3 + Q 1 ] Δ

(8.8)

K2 =

1 [P2 K 3 + Q 2 ] Δ

(8.9)

where:

Δ = cos θ1 cos ∅1 − cos θ2 cos ∅2 P1 = (cos θ1 − cos θ2 ) Q 1 = cos θ1 cos(∅2 − θ2 ) − cos θ2 cos(∅1 − θ1 ) P2 = (cos ∅2 − cos ∅1 ) Q 1 = cos ∅1 cos(∅2 − θ2 ) − cos ∅2 cos(φ1 − θ1 ) If we substitute K1 and K2 into Eq. (8.7) and simplify we obtain a quadratic equation for K3 as: A1 K 32 + A2 K 3 + A3 = 0 where:

(8.10)

368

8 Correlation of Input and Output Motion; Function Generation …

  1 A1 = P2 Δ − P2 (cos θ2 + cos θ1 ) 2 A2 = (Q 2 − P1 )Δ − P2 Q 2 (cos θ2 + cos θ1 )   Q2 A3 = − Q 1 Δ + 2 (cos θ2 + cos θ1 ) 2 Now, we can solve for K3 from Eq. (8.10) and substitute into Eqs. (8.8) and (8.9) to solve for K1 and K2 and then determine the link lengths in the usual way. This will yield a four-bar mechanism that correlates two pairs of crank angles for which the transmission angle will be optimum. Example 8.14 We want to design a four-bar mechanism such that as one crank rotates 60°, the other crank will rotate 110o and the transmission angle is optimum. Let us assume θ1 = 25° and φ1 = 65°. Then θ2 = 85° and φ1 = 175o We determine Δ = −0.93969, P1 = −0.81915, P2 = −1.41881, Q1 = −0.06677 and Q2 = 0.763129 and calculate A1 = 0.33312, A2 = −0.4112, A3 = −0.35202. Solving the quadratic we obtain two solutions: K31 = 1.815423 and K32 = −0.58175 When K31 = 1.815423 is used K11 = 1.653596 and K21 = 1.928945 yielding a2 = 0.604743 a3 = 0.704391 and a4 = 0.518418 (a1 = 1). Transmission angle deviation is: δ = 42.7° When K32 = −0.58175 is used K12 = −0.43608 K22 = − 1.69047 yielding a2 = −2.29318, a3 = 2.861281, a4 = −0.59155. Transmission angle deviation is: δ = 33.7°. Second solution is selected. The resulting mechanism is as shown in Fig. 8.45. The transmission angle and the change of the output angle as a function of crank angle is as shown in Fig. 8.46. One can change the initial crank angles to obtain different solutions. These solutions will have different transmission angle characteristics (better or worse maximum deviation within the range) and different link lengths. Depending on the application one can use the one that best suits his needs. If the angular position correlation θ1 , φ1 is not important, and if only the angular rotation of the input Δθand corresponding output rotation Δφin between two positions is specified, then method of linearization discussed in Chap. 2 yields a much simpler solution when Δθ and Δφ is less than 180°. Example 8.15 Example 8.14 will be solved using “small angle assumption”. As the ◦ ◦ input crank rotates by Δφ = 60 the output must rotate by Δθ = 110 . Consider the input links as horizontal when they are at the middle of their angular rotations. Referring to Fig. 8.47, B1B2 = 2a4 sin (Δφ/2) and A1 A2 = 2a2 sin (Δθ/2). In order the vertical distance A1 A2 and B1 B2 to be equal, the relation between a2 and a4 is: a4 sin(Δθ/2) = a2 sin(Δφ/2)

8.4 Correlation of Two Crank Angles with Optimum Transmission Angle

369

Fig. 8.45 Four-bar mechanism in Example 8.14. Correlation of two positions with optimum force transmission characteristics

Let a2 = 1 unit and let a3 = 3 units (one can arbitrarily select these two lengths) ◦ ◦ when Δφ = 60 and Δθ = 110 a4 /a2 0.610 a4 = 0.610 and. a1 =

/ a32 + (a2 − a4 )2 = 3.03

The mechanism obtained will not satisfy Δφ = 110° when Δθ = 60° exactly (In this example corresponding to Δθ = 60°, Δφ110.0576°). In order to obtain 110° more accurately, we can use “solver” or “goal seek” tool (make Δφ110° by changing a3 ). When the coupler length a3 changes slightly (a3 = 3.003), a very close result is obtained. The Transmission angle and change of the output as a function of input crank angle are shown in Fig. 8.48 (θ = 0 is when links 2 and 4 are horizontal). Transmission angle is 90° whenθ = 0° (φ = 0o also) and at the end positions the deviations from 90° is almost the same (52.6° and 57.4°).

370

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.46 The transmission angle and the change of the output angle as a function of crank angle in Example 8.14 Fig. 8.47 Four-bar mechanism in Example 8.15

8.5 Designing Six Link Mechanisms for Long Oscillation or for Long Dwell

371

Fig. 8.48 Output link angle φ and transmission angle μ as a function of input crank angle θ obtained in Example 8.15

8.5 Designing Six Link Mechanisms for Long Oscillation or for Long Dwell One can use the above solutions for a four-bar mechanism as a building brick to design six-link mechanisms which satisfy different motion requirements such as large output swing angle or long dwell. Using a four-bar mechanism of crank-rocker proportions theoretically one can obtain at most 180° swing angle (for 120° the best minimum transmission angle that one can archive is less than 30°). If a long swing angle is required or due to high force transmission, a mechanism which has better force transmission characteristics then a four-bar is required, two four-bars in series can be used which results in a six-link mechanism. Another practical design requirement that cannot be achieved with a four-bar mechanism is that especially in production machinery, we would like the output to remain stationary for a certain portion of the input rotation. This is known as “dwell”. If the output is to remain completely stationary a cam mechanism is used. If small oscillations during the dwell period is permissible a six-link mechanism can be effectively used. This will result in what is called as “approximate dwell”. Again, this problem can be solved by designing two four-bar mechanisms and joining them in series to obtain a six-link mechanism. These solutions will be explained by solving different examples.

372

8 Correlation of Input and Output Motion; Function Generation …

Example 8.16 We would like to oscillate a shaft with an amplitude of 200°. One can not obtain 200° swing using a four-bar mechanism only. Let us first convert the continuous rotation of input link to 100° rotation of a rocker (this angle is arbitrary). We can then convert 100° rotation of this rocker oscillation to 200° rotation of the output link by another four-bar in series. It is a good idea to design this four-bar with optimum transmission angle. i.e. make the deviation of the transmission angle equal at the two extreme positions. Let us assume θ1 = 10° and φ1 = 35°. Then θ2 = 110° and φ1 = 235o We determine Δ = −0.2847, P1 = −1.32683, P2 = −1.39273, Q1 = −0.25489 and Q2 = 0.04999 and calculate A1 = −0.2269, A2 = −0.34722, A3 = −0.07337. Solving the quadratic equation, we obtain two solutions: K31 = −1.27709 and K32 = −0.25319 When K32 = −0.25319 is used, K12 = −0.28472 and K22 = −1.41421 yielding a2 = −3.51226 a3 = 3.885046 and a4 = −070, 711 (a1 = 1). Transmission angle deviation is: δ = 58.01° (μmin ≈ 32°) (You can try K31 ). If we want to reduce the maximum transmission angle deviation, we can change the initial crank angles. Increasing φ1 and decreasing θ1 gives us a better result. However, the ratio between the maximum link length to the minimum link length increases. If we limit max to min link length ratio to 6, we can increase φ1 = 36.5°, decreasing δ to 56.95° (μmin ≈33°). Of course, this deviation can be decreased at the cost of link length ratio. Now we can design a centric four-bar mechanism. With 100° swing angle which is the rotation of the input crank of the four-bar mechanism we have designed for two positions (with optimum transmission angle. If required, you can design noncentric four-bar for a fast return). Selecting β = 14°, ψ = 100°, (φ = 180°), we obtain the link lengths as (refer to centric four-bar design): a1 = 2 (selected), a2 = 0.576623, a3 = 1.94059 and a4 = 0.752727. If we connect the two four-bar mechanisms we have designed a six-link mechanism to obtain 220° swing angle. Its two extreme positions will be as shown in Fig. 8.49. We can thus obtain large swing angles with reasonable deviation of the transmission angle. Note that the fixed revolute joint centers A0 , B0 , C0 need not be collinear as shown. You can couple the two four-bars and change the location of C0 by changing the angle Be B0 B1 as long as when the first four-bar is at the extended dead center position, the second four-bar is at position 1. Also, one can increase or decrease the size of the two four-bar chains separately (i.e. increase the link lengths of the first four bar by a factor of 2), without effecting the output motion. In Fig. 8.50, the transmission angle of the first crank-rocker (dashed line) and the second four-bar (solid line) mechanism are shown. If the maximum deviation is be kept at a minimum while satisfying he maximum to minimum link length ratios, one can change the intermediate swing angle (100°). For example, increasing this swing angle will increase the maximum deviation of the transmission angle for the crank

8.5 Designing Six Link Mechanisms for Long Oscillation or for Long Dwell

373 C2

Be Bf

A Af

B'

Ae 0

B0

C0

C1

1

B'2

Fig. 8.49 Six link mechanism at two extreme positions synthesized for 200° output oscillation as the input crank rotates by 180° in Example 8.16

rocker mechanism, while decreasing the maximum deviation of the transmission angle for the second four-bar. Note that the solution is not unique. There are four free parameters: initial crank angle β, intermediate swing angle of the crank-rocker mechanism and initial crank angles of the second four-bar. One can treat the problem as a constraint optimization problem (optimize for the transmission angle). The constraint will be the maximum link length ratio. If a large swing angle using a linear actuator is required, then one can design an inverted slider-crank mechanism with a certain slider displacement and corresponding swing angle and than using a four-bar mechanism (and using two position synthesis with optimum transmission angle characteristics) one can convert this swing angle to a large swing angle. An example for the same 200° swing is shown in Fig. 8.51 (A0 A2 is the closed length and A0 A1 is the extended length of the piston-cylinder. Example 8.17 We would like to have an output link oscillation of 120° for a heavyduty application. Also, the maximum to minimum link length ratio cannot be greater than 5.

374

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.50 Transmission angles as a function of input crank rotation for the mechanism shown in Fig. 8.49

Fig. 8.51 Six link mechanism with a piston-cylinder as the input at two extreme positions synthesized for 200° output oscillation with a piston displacement A1 A2

8.5 Designing Six Link Mechanisms for Long Oscillation or for Long Dwell

375

If we are to design a four-bar mechanism which satisfies 120° swing angle with the given constraint the best transmission angle that can be achieved will be with a centric four bar (μmin = 29.32°). Instead, let us use a centric crank-rocker for 65° swing angle and then use another four-bar in series to convert this 65° swing to 120°. We can design the second four-bar for optimum transmission while satisfying the maximum to minimum link length ratio. A centric four bar mechanism with 68o swing angle and maximum-to-minimum link length ratio less than five results with μmin = 54.16°; the link lengths: a1 = 1, a2 = 0.200, a3 = 0.955, a4 = 0.358 and initial crank angle β = 17.248°. When we are correlating the 68° to 120° crank rotations with optimum transmission angle, we perform several trials for initial crank angles θin = 96o and φin = 3° seems a good choice with μmin = 54.09°. The link lengths of the first four-bar is: a1 = 1; a2 = 295, a3 = 2.895 and a4 = 0.579. For the realization, let us select the fixed link length for the centric crank-rocker as 500 and the second four-bar as 175 units. The resulting mechanism is as shown in Fig. 8.52 (in first position). In this mechanism A0 A = 100, AB = 478, B0 B = 179 A0 B0 = 500, B0 C = 402, CD = 507, DD0 = 101, B0 D0 = 175. If A0 B0 and D0 are collinear, < CB0 B = 96–73.25 = 22.75° (if you rotate A0 B0 by 22.75° CCW, B0 BC will be collinear) Maximum to minimum link length ratio is close to 5. Example 8.18 In production machinery, we would like to have the output of a mechanism to remain stationary while performing an operation on the product. This momentary stop while the input is rotating is an “approximate dwell”. A four-bar mechanism is at a momentary dwell when the mechanism is at a dead out = 0). However, the duration of dwell will be short (but finite due to center ( dω dωin clearances, etc.). If we get two four-bar mechanisms and have their dead centers coincide, then the duration of dwell can be increased.

Fig. 8.52 Six-link mechanism with 120° output link oscillation in Example 8.17

376

8 Correlation of Input and Output Motion; Function Generation …

We want a link to oscillate in one direction 100° and and dwell at one of its dead centers for 90° crank rotation and return to its original position within the remaining portion of the cycle. Consider the four-bar mechanism designed in example 8.4 for two positions and a dead center. Corresponding to 90° input link rotation the output link rotates by 100°. Let us design a centric four-bar mechanism corresponding to 90° swing angle and let us select the initial crank angle β = 17° (so that the link length ratios are acceptable. If β is close to 0° than the ratios will get large). Selecting a1 = 2, a2 = 0.584743, a3 = 1.91261 and a4 = 0.826952. If we join these two four-bars in series at their folded dead center positions, i.e., the output link of the crank-rocker is joined rigidly with the input link of the second four-bar (B1 B0 B’1 is one link. The six-link mechanism will be as shown in Fig. 8.53 (when folded mechanism is A0 A1 B1 B0 B’1 C1 C0 , and A0 A2 B2 B0 B’2 C2 C0 when extended). The motion curve is as shown in Fig. 8.54. The motion of the output within 175o < θ < 220° is less than 0.1° as seen in Fig. 8.55. Hence, we have an approximate dwell for 45° crank rotation. We can increase the dwell period provided that we permit some amount of angular oscillation of the output during the dwell period. The quality of dwell will decrease while its period increases. Consider having a phase angle such that the second four-bar mechanism is at a dead center before the first four-bar (which is of crank rocker proportions) is at a dwell. Let us say that the second four bar is at a dwell when the crank of the first four-bar is 30° away from folded dead center position. (i.e. when the input crank is at 197° the first four-bar is at a dead center. Consider the second four bar is at dead center when the input crank is at 197–30 = 167°. Now, the output will have zero velocity when the input crank is at 167°, 197° and 232°. Within 155o ≤ θ ≤ 248° of crank rotation (approximately 93° of crank rotation) the output link will remain

Fig. 8.53 Two four-bar mechanisms connected in series when both mechanisms are at dead center. A six-link mechanism with dwell characteristics is obtained. (Example 8.18)

Output Swing Angle

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

150 135 120 105 90 75 60 45 30 15 0

377

Output vs. Input

10

60

110

160 210 260 input crank angle

310

360

Fig. 8.54 Angular rotation of the output φ, as a function of input crank angle θ for the mechanism obtained in Example 8.18

Fig. 8.55 Output link angleφ as a function of input crank angle θ within 175o < θ < 220o

almost stationary (fluctuating within ± 0.2°) (Figs. 8.56 and 8.57). Of course, the phase angle can be increased thus increasing the dwell period if one permits higher fluctuations at the output link within the dwell period (There will also be a small amount of decrease on the oscillation angle of the output, which can be modified if required).

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank Mechanism Considering the slider-crank mechanism (Fig. 8.58.), the loop closure equation in vectorial form is:

8 Correlation of Input and Output Motion; Function Generation …

Output Swing Angle

378

150 135 120 105 90 75 60 45 30 15 0

Output vs. Input

10

60

110

160 210 260 input crank angle

310

360

Fig. 8.56 Output link angle φ as a function of input crank angle θ. Two four-bars connected in series with 30° phase angle in between the dead centers

Fig. 8.57 Output link angle φ as a function of input crank angle θ within 155o < θ < 255° (Deviation from 139.8°)

A0 A = A0 B + BA or in complex numbers a2 eiθ12 = s14 + ic + a3 ei θ13

(8.1)

If we equate the real and imaginary parts of this equation separately, we shall obtain two scalar equations in three position variables (θ12 , θ13 and s14 ). If one of the position variables is the input variable whose value is given, then we shall be able to solve for the values of the other two variables. In complex plane, when we have an equation in complex numbers, the complex conjugate of the equation is also true. The complex conjugate yields vectors which are the mirror image of the original vectors with respect to the real axis (x-axis);

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

379

Fig. 8.58 Slider-crank mechanism

e.g. in case of a mechanism, if we place a mirror about the real axis, as we move the original mechanism its image will also move and corresponding to the original closed loop, the loop formed on the mirror image will also be closed at every position. Hence, we obtain another loop closure equation in terms of complex numbers as: a2 e−iθ12 = s14 − ic + a3 e−iθ13

(8.2)

The original Eq. (8.1) and its complex conjugate (8.2) are the two independent equations in the complex plane (if we equate the real and imaginary parts of these equations they will yield the same two scalar equations in the real plane). Using the Eqs. (8.1) and (8.2), if we are to find θ14 as a function of θ12 , we have to eliminate θ13 from the above equations. We can write the loop closure equations in the form: a3 eiθ13 = a2 eiθ12 − s14 − ic

(8.3)

a3 e−iθ13 = a2 e−i θ12 − s14 + ic

(8.4)

Multiplying Eqs. (8.3) and (8.4):    a32 ei(θ13 −θ13 ) = a2 eiθ12 − s14 − ic a2 e−iθ12 − s14 + ic

(8.5)

Noting ei(θ−θ) = ei0 = 1; 2 a32 = a22 + s14 + c2 − a2 s14 (eiθ12 + e−iθ12 ) + ia2 c(ei θ12 − e−i θ12 )

(8.6)

Since cos θ = (eiθ + e−iθ )/2, sinθ = (eiθ − e−i θ )/2i Eq. (8.6) reduces to the form: 2 2a2 s14 cos θ12 + 2a2 c sin θ12 − (a22 + c2 − a32 ) = s14

or:

380

8 Correlation of Input and Output Motion; Function Generation … 2 K 1 s14 cos θ12 + K 2 sin θ12 − K 3 = s14

(8.7)

where K 1 = 2a2 K2 = 2a2 c K3 = a22 + c2 − a32 Equation (8.7) is also “Freudenstein’s Equation” for slider-crank mechanisms, which can be used for analysis of four-bar mechanisms. It gives an implicit relation between the position variable s and θ12 . We can also use this equation for synthesis since it gives us a relation between the input and output links with three link-length parameters. Let us also let θ = θ12 s = s14 . Equation (8.7) can be written as: K 1 s cos θ + K 2 sin θ − K 3 = s 2

(8.7)

Now consider a case when link 2 is at an angle θ1 link 4 is a position s1 , when link 2 is at θ2 link 4 is at s2 and when link 2 is at angle θ3 , link 4 is at s3 (Fig. 8.59). If these three positions of the links have to be satisfied by a slider-crank mechanism, then Eq. (8.7) must be satisfied for all the three positions: K 1 s1 cos θ1 + K 2 sin θ1 − K 3 = s12 K 1 s2 cos θ2 + K 2 sin θ2 − K 3 = s22 K 1 s3 cos θ3 + K 2 sin θ3 − K 3 = s32

(8.8)

In this equation the crank angles and the slider positions are given (we want to satisfy the relationship between the links). K1 , K2 , K3 are unknowns. These unknowns are related to the link lengths which we want to determine. Equation (8.8) forms a set of linear equations in terms of K1 , K2 , K3. We can determine the link lengths in terms of K’s as:

Fig. 8.59 Correlation of slider displacement with crank rotation

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

a2 =

K1 K2 a3 = , c= 2 K1

381

/  a22 + c2 − K 3

Example 8.19 Design a slider-crank mechanism to correlate the given three design positions (angles in degrees): j

θj (o )

sj (mm)

1

20

100

2

80

50

3

150

30

The three equations for the given three positions can be written in matrix form as: ⎤ ⎤⎡ ⎤ ⎡ 10000 93.96926 0.34202 −1 K1 ⎣ 8.682409 0.984808 −1 ⎦⎣ K 2 ⎦ = ⎣ 2500 ⎦ 900 −25.9808 0.5 −1 K3 ⎡

When solved: K1 = 73.3083, K2 = −1941.18 and K3 = −3775.19 Solving for link lengths: a2 = 36.65 mm, a3 = 76.29 mm and c = −26.48 mm Figure 8.60 shows the resulting mechanism. Note that a3 must always be positive (if the term under the square root is negative, then there is no solution. If K1 , or K2 (or both) turns out to be negative, a2 or c (or both) may be negative. 40 30 20 10 0 -40

-30

-20

-10

0

10

20

30

40

50

60

70

80

90

100

110

120

130

-10 -20 -30 -40

Fig. 8.60 Slider crank mechanism synthesized for the correlation of three crank angles with slider displacements (Example 8.19)

382

8 Correlation of Input and Output Motion; Function Generation …

Links lengths has to be positive. −a2 means that the direction of link 2 must change. When deriving the Freudenstein’s equation note that A0 A was represented by a2 eiθ . If the angle was θ + π than a2 ei (θ +π) = −a2 eiθ the negative sign means that the link length must be measured opposite to the direction defined by the angle. One can perform function generation to generate a function y = f(x) by relating x or y with θand y or x with s. What has been discussed for a four-bar mechanism will apply. One can perform 3 precision point synthesis, order approximation or mixed order approximation (i.e., correlating two crank positions with two slider displacement and defining the rate of change of s with respect to θ Thus one can design a six-link mechanism with slider as the output for which the slider dwells for a certain portion of the cycle. These are left as exercise. Example 8.20 Determine the link length proportions of a slider-crank mechanism that will realize the function y = tanx within the interval 0o ≤ x ≤ 60°. ◦

Consider using 3 design parameters. Δx = 60 and Δy = 1.73205. Let x corre◦ spond to θ(Δx = Δθ) and θin = 15 , Δs = 100 units (Rx = 1, Ry = Δs/Δy = 57.73503) and sin = 20 units. Using Chebyshev spacing for 3 precision points. Precission points on x, y and θ, s according to Chebyshev spacing will be:

Writing Eq. (8.8) for three positions: ⎞ ⎞⎛ ⎞ ⎛ 578.72515 22.74343 0.32589 −1 K1 ⎝ 37.71236 0.70711 −1 ⎠⎝ K2 ⎠ = ⎝ 2844.44444 ⎠ K3 11137.36903 34.39193 0.94541 −1 ⎛

Solution is K1 = −542.42449, K2 = 27,242.01570 and K3 = −4037.53902. The link lengths are: a2 = −271.212, c = 50.223 and a3 = 283.048. Mechanism is shown in Fig. 8.61 and the error curve ε = ygen − yreq is shown in Fig. 8.62 (|e|max = 0.041). Instead of using three design parameters, let sin be the fourth design parameter. Assuming sin = 20 as an initial guess, using Chebyshev spacing for 4 positions we have:

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

383

Fig. 8.61 Slider-crank mechanism to realize y = tanx within 0o ≤ x ≤ 60° using three design parameters (Example 8.20)

Fig. 8.62 Structural error curve for the slider crank mechanism realizing function y = tanx within 0o ≤ x ≤ 60° (Example 8.20)

384

8 Correlation of Input and Output Motion; Function Generation …

Wring Eq. (8.8) for four positions, the determinant of the augmented matrix formed is: / / / 21.29530 0.29710 −1 497.39438 / / / / 32.79744 0.55222 −1 1547.61343 / / / / 39.23231 0.83370 −1 5047.33845 / = −4020.47793 / / / 33.09287 0.95485 −1 12406.75973 / When we use “Goal Seek” and try to make the value of the determinant zero while changing sin we end up with sin = −10.2128 and the value of the determinant is 1.15 × 10–11 which can be considered as zero. Using any 3 of four equations, we solve for Ki and determine the resulting link lengths (K1 = −872.1965, K2 = 5198.0976, K3 = 2190.3544 and a2 = −416.098, c = −59.525 and a3 = 414.457). Mechanism and the resulting error curves are shown in Figs. 8.63 (|ε|max = 0.00563). Of course, in addition to sin using θin as a new variable, one can design using five design parameters. Note that in these problems we were not concerned with the force transmission characteristics of the resulting mechanisms. If we are to design a slider crank mechanism (with crank as the input) for two positions than we have two equations in three unknowns. There is one free parameter. One can use this free parameter to optimize for the transmission angle. However, for angles ◦ Δφ < 180 , it turns out that for best transmission angle characteristics symmetry is the best choice. Instead of taking θ1 , θ2 and s1 , s2 as the given values, let us assume that we want the output slide to be displaced by Δs (= s2 − s1 ) as the input crank rotates by Δθ(= θ2 − θ1 ). If we make the crank symmetric about a line perpendicular to the slider axis (Fig. 8.64): θ1 = (π − Δφ)/2 and θ2 = (π + Δφ)/2. a2 =

Δs 2

sin(Δφ/2) 1 c = a2 (1 + cos(Δφ/2)) 2

Now, let us select a2 and the eccentricity c as: These values of a2 and c make the transmission angle deviation from 90° at the three extreme positions equal (These positions are the two limit positions θ1 , θ2 and the position at which the transmission angle is minimum (θ = 0)). The transmission angle which deviates most from 90° is μmin = cos

−1



a2 − c a3



In this equation we see that as the length of the coupler link a3 increases, the transmission angle is close to 90°. a3 = ∞ is the theoretical optimum. a3 is usually determined by the available size and the condition a3 > a2 − c must be satisfied. When Δφ = 180°, a2 = Δs/2 and c = 0, i.e. centric slider-crank is obtained. For Δφ

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

385

a)

b)

Fig. 8.63 Slider-crank mechanism a and its structural error b to generate function y = tanx within 0o ≤ x ≤ 60° using four design parameters (Example 8.20)

> 180°, symmetry can not be achieved. For Δφ > 180° we may require one or both end conditions to be a dead center position. If both end positions are dead centers (Fig. 8.65), Then the equations derived for the dead center positions of a slider crank mechanism is valid. We have a one parameter solution with λ = a2 /a3 (s = 1) as:   1 − λ2 sin φ 1     c =  2 1 + λ2 + 1 − λ2 cos φ

(8.14)

386

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.64 Slider-crank mechanism for two positions with optimum transmission angle

1 1 1 1     =   (8.15) 2 2 φ 2 1 + λ + 1 − λ cos(φ) 4 cos2 + λ2 sin2 φ 2 2 ⎡ λ2 λ2 1 1    =   (8.16) a22 = a23 λ2 = ⎣  2 2 2 4 cos2 φ + λ2 sin2 φ 1 + λ + 1 − λ cos(φ) a23 =

2

2

In these equations we must select φas φ= 360 − Δφand select a value of λ The transmission angle is critical when the crank is perpendicular to the slider axis and at the end point 2. Using one parameter optimization (λ as the parameter), we can minimize the maximum deviation of the transmission angle from 90°. It turns out the optimum is when μ2 = μmin and μ1 > μmin . A different solution to this problem has been put into a chart form in VDI 2126 [7]. Example 8.21 We would like to design a slider-crank mechanism for which when the input crank rotates by 100°, the slider is displaced by 100 mm. Crank can be made symmetric about a line perpendicular to the slider axis. We have:

Fig. 8.65 Two position synthesis with optimum transmission angle when Δφ > 180°

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

a2 =

Δs 2

=

387

50 = 65.3 mm sin(50◦ )

sin(Δφ/2) 1 c = a2 (1 + cos(Δφ/2)) = 65.3(1 + cos(50◦ )/2 2 = 53.6 mm

The coupler length a3 is selected as 100 mm. The transmission angle deviation is shown in Fig. 8.66. Note that the maximum transmission angle deviations are equal at the three positions of the crank and there are two positions where the transmission angle is 90°. If the transmission angle deviation from 90° is to be decreased, one must increase the coupler length a3 (the mechanism will occupy more space). If we want 220° rotation of the crank to displace the slider by 100 mm. Then we can select an arbitrary value for λ(say λ = 0.5) and calculate the slider crank mechanism satisfying 360°–120° = 140° between dead centers. The result is a2 = 0.4302, a3 = 0.3569, c = 0.3569. The minimum transmission angle is μ1 = 33.95° is the critical transmission angle. If we maximize the minimum transmission angle by using the parameter λ(i.e. use “solver” tool in Excel), we obtain: a2 = 0.4863, a3 = 0.5935, c = 0.0744 with μmin = μ1 = 46.05°. The transmission angle deviation is shown in Fig. 8.67 for 100 mm stroke a2 = 48.6 mm, a3 = 59.4 mm and c = 7.4 mm. In certain applications of the slider crank mechanism slide is the input. Mechanism can be driven by the use of a hydraulic or pneumatic cylinder as the slide or a screw mechanism for the linear motion of the slide can be used. This mechanism is used to convert linear motion to an oscillation of the link 2. Although the motion is not continuous, there is a vast number of applications that can be seen in practice. Equation 7 that relates the crank angular rotation (θ) to the displacement of the slide (s) can be used for the correlation of the slider displacement with the crank rotation. If we are to design a slider-crank mechanism such that for a certain amount of slider displacement Δs, we would like to have a certain amount of crank rotation, ψ, we have several free parameters. For example, we can arbitrarily select θ1 and s1 , the initial position of the crank and the slider. Than we can write:

Fig. 8.66 Transmission angle of the slider crank mechanism as a function of input crank angle (Example 8.21)

388

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.67 Transmission angle deviation for slider crank mechanism for which 220° crank rotation results with 100 mm slider displacement and the transmission angle is optimum

K 1 s1 cos θ1 + K 2 sin θ1 − K 3 = s12 K 1 (s1 + s) cos(θ1 + ψ) + K 2 sin(θ1 + ψ) − K 3 = (s1 + s)2 where: K 1 = 2a2 K2 = 2a2 ck K3 = a22 + c2 − a32 Now, we are free to select one additional parameter, which can be one of the link lengths. There are three design parameters. Again, one can optimize for the transmission angle. However, a more straightforward solution has been found if we consider the transmission angle deviation in detail. In Fig. 8.68 three positions where the transmission angle is the most critical is shown. At these positions the transmission angles are given by the equations: a22 + a32 − (Δs − s , )2 − c2 2a2 a3 , a22 + a32 − s 2 − c2 cos μ2 = 2a2 a3 a22 + a32 − c2 cos μ3 = 2a2 a3 cos μ1 =

If we require μ1 = μ2 then Δs–s’ = s’ and s’ =Δs/2. As in the case of crank as the input, symmetry improves the force transmission characteristics (Fig. 8.69). Also note that ψ = ∠A1 A0 A2 = ∠B1 A0 B2 . Using The right-angle triangle B1 B3 A0 we can determine the eccentricity c and the length A0 B1 = A0 B2 as:

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

389

Fig. 8.68 Slider-crank mechanism with slide as the input at three critical positions

c=

Δs 2 tan ψ/2

A0 B1 =

Δs 2 sin ψ/2

(8.1) (8.2)

In order to have the deviation of μ3 , μ1 and μ2 from 90° equal we must have: μ1 = μ2 and μ3 = π − μ3 . It also turns out that a2 = a3 yields the best transmission angle (maxμmin ). Using the triangles A0 A1 B1 and A0 A3 B3 we can write: = 2a2 sin(μ3 /2) = c

(8.3)

and for triangle A0 A1 B1 : 2a2 sin(μ1 /2) =

Fig. 8.69 Slider-crank mechanism with slide as the input. For a displacement Δs, crank rotates by an angle ψ. B1 and B2 Symmetric about A0 B3 . a2 = a3

√ c2 + (Δs/2)2

(8.4)

390

8 Correlation of Input and Output Motion; Function Generation …

In order to make the deviation of the transmission angles μ1 and μ3 from 90° equal, μ1 = π − μ3 . 2a2 sin((π − μ3 )/2) =

√

c2 + (Δs/2)2 = 2a2 cos(μ3 /2)

And 

Δs 2

2 = 2a32 + 2a22 − 2c2

(8.5)

a2 = a3 results in √ √  2 Δs 1 √ a2 = a3 = √ 2c2 + 2 2

(8.6)

The result for different output angular rotation is shown in Fig. 8.70. Example 8.22 In a certain application for 100 mm stroke we would like to have 60° rotation of the crank. From Fig. 8.70, μmin > 80°. And a2 /Δs = a3 /Δs = 0.6 and c/Δs = 0.8. If we use the equations, for 100 mm stroke c = 86.6 mm, a2 = a3 = 66.1 mm and μmin = 81.8°. The motion curve is shown in Fig. 8.71. In certain cases, instead of using a2 = a3, one can arbitrarily select a2 or a3 while still keeping μ1 = μ2 . For example, if we assume a3 , using Eq. (8.5), a2 will be given by:

Fig. 8.70 a2 /Δs = a3 /Δs, c/Δs and μmin as a function of output oscillation ψ for optimum force ◦ transmission for slider crank mechanisms. (Slide as the input, output oscillation less than 76.35 )

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

391

Fig. 8.71 Output link and transmission angle as a function of slider displacement for a slider crank mechanism with optimum transmission angle (Example 8.22). c = 86.6 mm, a2 = a3 = 66.1 mm and μmin = 81.8o

√ ! √  2 √ 1 Δs a2 = √ + c2 − a32 2 2 If we take a3 = 30 mm, a2 = 88.6 mm (c = 86.6 mm, as before). The motion curve and the transmission angle variation will be as shown in Fig. 8.72. μmin = 64.1° is still an acceptable transmission angle. When we compare the two motion curves, we see that when a3 = 30 mm, the angular rotation changes more linearly (i.e. dθ/ds is almost constant). This may be an advantage for certain applications.

Fig. 8.72 Output link and transmission angle as a function of slider displacement for a slider crank mechanism with optimum transmission angle when a2 /= a3 (Example 8.22). c = 86.6 mm, a3 = 30 mm, a2 = 88.6and μmin = 64.1°

392

8 Correlation of Input and Output Motion; Function Generation …

 ◦ It turns out that when tanψ = cosψ ψ∗ = 76.35 there is a transition. For ◦ ψ < 76.35 a2 = a3 results with the optimum transmission angle and the three extreme positions of the transmission angle deviation from 90° can be made equal [8–11]. When ψ= ψ* = 76.35° the coupler BA is perpendicular to the slider axis at the second position and if the required second position is beyond this position, there is reversal of motion i.e. it is a dead center position. This position can be determined by taking the derivative of θ with respect to the slider displacement s and setting dθ/ ds = 0. Taking the derivative of Eq. (8.7) with respect to s yields: K 1 cos θ − K 1 s sin θ

dθ dθ + K 2 cos θ = 2s ds ds

Usually we require the crank to rotate in one direction only as the slide is displaced in one direction. Now for the second position, dθ/ds = 0 condition must also be satisfied (Fig. 8.73). Hence we have: K 1 (Δs − s , ) cos θ1 + K 2 sin θ1 − K 3 = (Δs − s , )2 ,

− K 1 s , cos(θ1 + ψ) + K 2 sin(θ1 + ψ) − K 3 = s 2 K 1 cos(θ1 + ψ) = −2s ,

(8.1)

There are two free design parameters: θ1 and s, . When these two parameters are specified, one can solve for K1 , K2 and K3 and hence the link lengths. A twoparameter optimization can be performed. It turns out that the optimum solution is reached when the deviations of μ1 and μ3 from 90° are equal (μ2 is not the critical transmission angle). However, the optimum point is “weak”, i.e., when a numerical two parameter optimization method is performed, the optimum is very much dependent on the initial conditions selected and the solution will differ. Another point is that in certain applications the dead center position at the end may not be Fig. 8.73 Three positions of slider crank mechanism with slider as the input

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

393

wanted. In Fig. 8.74 the link length dimensions, initial crank angle as a function of output rotation is given when the transmission angle is optimum. Example 8.23 If we are to obtain 150° rotation when the slider is displaced by 100 mm, from Fig. 8.69 we estimate θ1 ~ 53°, a3 /Δs ~ 0.52, a2 /Δs ~ 0.32, c/Δs ~ 0.38, s, /Δs = 0.32 and maxμmin ~ 41°. Using s’ = 0.3 and θ1 = 53°, one can use the “Solver” tool in Excel. The result is: a2 = 35.1 mm, a3 = 46.0 mm, c = 32.8 mm s’ = 32.6 mm θ1 = 51.9° andμmin = 45.4° (at positions 1 and 3), The transmission angle and the motion curve of the resulting mechanism is shown in Fig. 8.75.

Fig. 8.74 a2 /Δs = a3 /Δs, c/Δs, μmin , initial slider and output link positions (s1, q1) as a function of output oscillation ψfor optimum force transmission for slider crank mechanisms. (slider as the input)

Fig. 8.75 Slider crank mechanism (slider as the input) with optimum transmission angle for 100 mm slider displacement 150° output oscillation (Example 8.23). a2 = 35.1 mm, a3 = 46.0 mm, c = 32.8 mm s, = 32.6 mm θ1 = 51.9° andμmin = 45.4o

394

8 Correlation of Input and Output Motion; Function Generation …

Fig. 8.76 The motion curve and the transmission angle variation of slider-crank mechanism. When the slide is displaced 100 mm output link rotates 150°. a2 = 31.4 mm, a3 = 60 mm (specified), c = 43.9 mm, s, = 27 mm μmin = 45.1°

One can also specify certain links to be of a certain size. For example, if we want the coupler length a3 to be of 60 mm length instead of 46 mm for optimum transmission angle, we can use the length of a3 as a constraint in the optimization and obtain a solution as: a2 = 31.4 mm, a3 = 60 mm c = 43.9 mm, s, = 27 mm for which μ1 = μ2 = 45.1° as it can be seen a change in a3 has a very small effect on μmin . The motion curve and the transmission angle variation is shown in Fig. 8.76. Example 8.24 Large diameter butterfly valves are driven by an electric motor and in emergency conditions, by hand (Fig. 8.77). There must be a large reduction ratio. The valve has to rotate 90° between the two positions. Using a screw drive, we can impart linear motion to the slide with a high reduction ratio (i.e. a few millimetres of translation for 360° crank rotation). For ψ = 90° rotation we refer to Fig. 8.74. Exact reading of the chart is not possible. Approximately we have: s, /Δs ∼ 0.42 ◦

80 < θ1 < 90

◦

Now, you can pick any value within the above limits for the two parameters. As ◦ an initial guess let us assume Δs = 1 and let s, = 0.42 and θ1 = 85 (s2 = − ◦ 0.42, θ2 = 175 ). With these values we solve the three Eqs. (8.1) for K1 , K2 , K3 we have: K1 = 0.8432, K2 = 0.517, K3 = 0.2215. We then determine the dimensions:

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

395

Fig. 8.77 Butterfly valve

a2 = 0.4216, c = 0.6134 and a3 = 0.5767. Now we can determine the two critical transmission angles μ1 and μ3 (since ψ76.35°, μ2 is not critical) using the equations: cos μ1 =

a22 + a32 − c2 − s12 2a2 a3

cos μ2 =

a22 + a32 − c2 2a2 a3

Transmission angle using the calculated dimensions results as:  ◦ ◦ ◦ ◦ δ = 16 μ1 = 114.60 δ = 24.6 , μ3 = 74.00 The two values are not equal but they are close. If you use these link lengths the minimum transmission angle for this mechanism will be 65.4° (= 90°−24.6°). You can optimise within the vicinity either by trial–error or use a genetic algorithm that is available in most mathematical packages. For example, if you prepare your calculations on an Excel sheet, you can take the difference of the absolute cosines of the two transmission angles and try to make this difference go to zero using the “Solver” tool. The result is s, = 0.434, θ1 = 86.73° and a2 = 0.434, c = 0.588, a3 = 0.563 with μ1 = μ2 = μmin = 70.87°. (Δ = 19.13°). Please note that the optimum is weak. If one starts with slightly different initial condition for s, and θ1 , one may obtain a different result although the minimum transmission angle value changes very little (for example a solution s, = 0.474, θ1 = 78.01° and a2 = 0.485, c = 0.530, a3 = 0.429 with μmin = 70. 56°. If we set a3 = 0.6, we may have s, = 0.421, θ1 = 88.84°, a2 = 0.422 c = 0.608 and μmin = 70.67° as a solution depending on the initial conditions.) Practically there is negligible difference between these three solutions. Final mechanism is as shown in Fig. 8.78.

396

8 Correlation of Input and Output Motion; Function Generation …

90°

Fig. 8.78 Slider-crank mechanism using screw as the slider input to transmit slider displacement to 90° rotation of the output link. Δs = 1unit, a2 = 0.434, c = 0.588, a3 = 0.563 μmin = 70.87°

Problems (1) Design a four-bar mechanism using Freudenstein’s equation to coordinate the following crank angles. Check your result. (a) i

θj (o )

φj (o )

1

15

40

2

45

60

3

75

90

i

θj (o )

φj (o )

1

−30

70

2

−60

100

3

−90

120

(b)

(2) Design a four-bar mechanism to generate the function y = x 3/2 within 1 ≤ x ≤ 3. Correlate x with θ and y with φ and let Δθ = 60° and Δφ = 90°. You are free to select θin and φin . Apply Chebyshev spacing on x. Check your result and make sure that the mechanism is movable. Draw the structural error within the range 1 ≤ x ≤ 3 and try to minimize the maximum error by changing θin and φin . (3) Consider the solution obtained in problem 2. Use Freudenstein’s respacing formula to change the location of precision points to minimize maximum structural error. Letθin = 40°, φin = 60°.

8.6 Correlation of Slider Displacement with Crank Angle in a Slider-Crank …

397

Fig. 8.79 Four Precission point synthesis problem

(4) Design a four-bar mechanism to generate the function y = x 3/2 at x = 2 using Freudenstein’s equation and its derivatives. You are free to select the crank angle at the design position. Determine the structural error within the range 1 ≤ x ≤ 3 and compare the result obtained for problem 2. (5) Design a four-bar mechanism to generate the function y = x 3/2 within 1 ≤ x ≤ 3 using Freudenstein’s equation. Correlate x with θand y with φand let Δθ60° and Δφ = 90°. You are free to select θin and φin . Use least square approximation. Select 10 equally spaced x values within 1 < x < 3. (6) Design a four-bar mechanism to coordinate the following four crank angles. Use K1 , K2 , K3 and θs as four free design parameters. You select different initial guess for θs and determine determine 4 values of θs which pairwise differ by 180o (Fig. 8.79). (Make sure that you have increased the number of iterations and reduced maximum change in Excel, Options, Formula)) You can solve the parameters K1 , K2 , K3 using any 3 of the four equations and then determine the link lengths. (7) Generate function 1. y = ex 0 ≤ x ≤ 1 2. y = 1/ × 1 ≤ x ≤ 2 3. y = x1.5 0 ≤ x ≤ 1 Use: (a) Three, four or five design variables. (b) Chebyshev Spacing for the precision points As expected, there are several different ways for the formulation of the problem. Also, you can arbitrarily select the range for the crank angles. Try different cases and compare the results for each of the functions. (8) Design a four-bar mechanism to convert 90° input rotation to 210° rotation with best force transmission characteristics. You can select the initial crank angles. Also, make sure that the maximum to minimum link length ratio is less than 5. What is the minimum transmission angle? (9) Design a four-bar mechanism of crank-rocker proportions with 90° swing angle and 170° corresponding crank rotation. Now join this four-bar with the fourbar mechanism obtained in problem 7, thus obtain a six-link mechanism with

398

8 Correlation of Input and Output Motion; Function Generation …

210° swing angle. When moving from extended to folded position the crank rotates 170° and 190°, when the first four-bar rotates from folded to extended position. Also, design another four-bar with 190° corresponding crank rotation and use this four-bar to drive the four-bar obtained in problem 7. Compare the two results. (10) Design a six-link mechanism such that the output has a swing angle of 80° and dwells at the end of the stroke by approximately 30°. Make sure that the angular motion of the output during dwell period is less than ± 0.02° while the input has a continuous rotation. (11) Design a slider crank mechanism using Freudenstein’s equation to correlate the slider displacement with crank rotation, satisfying the following three positions. Check your result and determine the minimum transmission angle. j

θj (o)

sj (cm)

1

10

20

2

50

50

3

100

100

(12) You are to generate a function y = logx 1 < x < 10 using a slider-crank mechanism by correlating x with θ (crank rotation) and y with s (slider displacement). Let Δθ = 80° and Δs = 100 mm. You are free to select the initial positions θin and sin . Design the mechanism and check your result. Draw the structural error within the range. Determine the minimum transmission angle and try to improve the transmission angle and the structural error by changing the initial positions (if necessary, you may select θin = 45° and sin = 10 mm). (13) Design a slider-crank mechanism for which a 90° crank rotation results in 200 mm stroke and the transmission angle is optimum. (14) Design a slider-crank mechanism for which a 210° crank rotation results in 200 mm stroke and the transmission angle is optimum. (15) Design a six-link mechanism such that the output slide translates 100 mm and dwells at the end of the stroke by approximately 30o of crank rotation. Make sure that the angular motion of the output during dwell period is less than ± 0.1 mm while the input has a continuous rotation. (16) Design a slider-crank mechanism with slide as the input such that 100 mm displacement of the slide results in 70° crank rotation and force transmission characteristics is an optimum. (17) Design a slider-crank mechanism with slide as the input such that 100 mm displacement of the slide results in 120° crank rotation and force transmission characteristics is an optimum. (18) For a slider crank mechanism with slide as the input, as it can be seen in Fig. 8.74, transmission angle decreases as you increase the output swing angle.

References

399

Therefore, design a six link mechanism the first loop is a slider-crank mechanism with slide as the input. The output link is connected to a four-bar loop. Design the mechanism such that 100 mm displacement of the slide results in 180° oscillation and the transmission angles are optimum.

References 1. F. Freudenstein, An analytical approach to the design of link mechanisms. Trans. ASME 76, 483–492 (1954) 2. F. Freudenstein, Structural error analysis in Plane Kinematic synthesis. ASME J. Eng. Ind. (Feb 1959) 3. K.H. Sieker, VDI 2128 Ebene Kurbelgetriebe- Berechnung von Gelenkvierecken für gegebene Winkellagen in endlischem Abstand (Aug 1959) 4. A.S. Hall, Kinematics and Linkage Design. Balt Publishers (1966) 5. K. Russell, Q. Shen, R.S. Sodhi, Mechanism Design: Visual and Programmable Approaches. (CRC Press, 2014) 6. AH Soni, Mechanisms Synthesis and Analysis. (McGraw Hill Book Co., 1974/1981) 7. VDI 2126 K. Hain, Krafte und Bewegungen in Krafthebergetrieben, vol. 6. (Grundlagen der Landtechnik, Heft, 1955) 8. K. Hain, Krafte und Bewegungen in Krafthebergetrieben, vol. 6. (Grundlagen der Landtechnik, Heft, 1955); Or K. Hain, Applied Kinematics (English transl.). (McGraw Hill, 1961) 9. K. Hain, Applied Kinematics (English transl.). (McGrawHill, 1961) 10. VDI 2125, Ebene Gelenkgetriabe, Ubertragungsgünstigste Umwandlung einer Sc- hubschwigin eine Drehschwingbewegung: (Planar mechanisms: transfer of slider motion into a rocker motion with regard to optimum transmission angle) (2014) 11. A.J.K. Breteler, On the conversion of translational into rotational motion with slider rocker mechanism, regarding transfer of quality. http://resolver.tudelft.nl/uuid:3ab9c5ac-a2e2-4859bf0e-2e6dc13b6da7,2010

Chapter 9

Chebyshev Methods in Kinematic Synthesis: Russian School

Abstract Although developed for mechanism synthesis, The works of Chebyshev have been very influential in approximation of functions in several fields of engineering. When using Freudenstein’s equation, determination of precission points as the roots of Chebyshev polynomials has been commonly accepted. In this Chapter Chebyshev polynomials and their application in the design of certain mechanisms will be explained. Extension of Chebyshev theory for function generation will be discussed. Keywords Four-bar approximate function generators · Chebyshev theorem on approximation · Least-squares minimization

9.1 Chebyshev Approximation The deviation between ideal and generated motions due to a finite number of adjustable design parameters (link lengths) is known as structural error. It must be distinguished from other errors such as the dimensional tolerances, clearances, elastic deformation of the links, which are other sources of error during the realization of the mechanism. However, if the structural error can be reduced so that it is comparable with other errors, the theoretical phase of the mechanism design can be considered to be successful. The prediction, analysis and minimization of the structural error and the mathematics developed during the 19th Century by Chebyshev (Pafnuti Lvovich Chebyshev1 [1821–1894]) and the following developments in Russian school of kinematics will be discussed.

The name П. Л. ЧEБЫШEBA has been translated to latin alphabet in different ways throughout the years. For example in the literature you can encounter Tchebychef, Tchebycheff, Tchebyshef, Tchebichev, Chebyshev, etc. In English literature Chebyshev is more common. This spelling will be used.

1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_9

401

402

9 Chebyshev Methods in Kinematic Synthesis: Russian School

9.2 Chebyshev Theorem [1] Let: P(t) = p0 φ0 (t) + p1 φ1 (t) + p2 φ2 (t) + . . . + pn−1 φn−1 (t) where: φi (t) ith function of t (i = 0, 1, 2, …, n − 1) Linearly independent continuous functions defined within the range considered. ith parameter. These are adjustable real numbers, not all of which is zero. pi n number of parameters in P(t). β)

t1 ≤ t ≤ tn+1 Range of P(t) under consideration (α ≤ t ≤ β where t1 = α, tn+1 =

P(t) is said to constitute a linear system if its value is equal to zero not more than n-1 times within the interval, regardless of the value of the parameters pi . For example, a a polynomial of any order P(t) = p0 + p1 t + p2 t2 + … + pn−1 tn − 1 is linear in any real interval defined, whereas P(t) = p0 + p1 cos(t) may be linear or not depending on the interval (i.e. if 0 ≤ t < π, then this system is linear regardles of po and p1 and nonlinear if −2 π ≤ t ≤ + 2 π). Now, let f(t) be an arbitrary function of defined within the same interval as P(t). Let: L = | f (t) − P(t)|max Chebyshev Theorem: The maximum absolute difference between an arbitrary function f (t) and an approximating linear system P(t) is said to be minimized within the interval t1 ≤ t ≤ tn+1 if and only if, the values of the parameters pi in P(t) are selected such that the difference F(t) = f (t) − P(t) assumes values ± L at least n + 1 times within the interval with successive alterations in sign. F(t) is the general error function and L is the maximum value of F(t). If P(t) is a linear function, the number of zero points must be one less than the number of parameters pi . The number of extreme values (minimums and maximums) of the error function F(t) is one more than the number of parameters. The end points of the interval are generally the points where the error function F(t) is a maximum or a minimum. In mathematical terms, Chebyshev theorem can be stated as: Let t = ti [I = 1, 2, …, n + 1] be the values of t at which F(t) has an extreme value. Then: F(ti ) = ±L

9.2 Chebyshev Theorem [1]

403

F , (ti )(ti − t1 )(ti − tn+1 ) = 0

9.2.1 Chebyshev Polynomials For a particular case when f(t) = tn and P(t) =

∑n−1 k=0

pk t k ;

F(t) = Tn (t) = t n + pn−1 t n−1 + pn−2 t n−2 + . . . + p1 t + p0 These polynomials are known as the Chebyshev polynomials of the of the first kind. If pi are chosen so as to minimize the maximum absolute value of Tn (t), Tn (t) represents a polynomial of nth degree with the leading coefficient unity, that deviates least from zero within the given interval. These polynomials are known as Chebyshev polynomials and are well known in mathematics field [2]. The Chebyshev polynomials defined within the interval −1 ≤ t ≤ + 1 up to order 4 and the maximum deviations of these polynomials from zero are given as: T1 (t) = t T2 (t) = t 2 − 21 T3 (t) = t 3 − 43 t T4 (t) = t 4 − t 2 +

1 8

L1 L2 L3 L4

= ±1 = ± 21 = ± 41 = ± 18

In Fig. 9.1 the plot of these polynomials is shown. One can obtain the Chebyshev polynomial of order n + 1 using polynomials of order n and n-1 using the recursive formula (for n > 2): 4Tn+1 (t) = 4t Tn (t) − Tn−1 (t) For example, for n + 1 = 5: 4T5 (t) = 4t T4 (t) − T3 (t) Or   1 3 5 4T5 (t) = 4 t 5 − t 3 + t − t 3 + t = 4t 5 − 5t 3 + t 8 4 4 5 5 T5 (t) = t 5 − t 3 + t 4 16 It is interesting to note that Tn (t) can be expressed as:

404

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Fig. 9.1 Chebyshev polynomials T1 , T2 , T3 and T4 within −1 ≤ t ≤ + 1

Tn (t) =

  1 cos n cos−1 (t) 2n−1

if we let cosθ = t, then Tn (t) =

1 cos[nθ ] 2n−1

The Chebyshev polynomials can be considered as the projection of a cosine curve drawn on the surface of a cylinder to the diametral plane. If cosine function is drawn on the surface of a cylinder and projected on the diametral plane, then the zeros and extreme values on this projection are the zeros and extreme values of the polynomial Tn (t) (function is zero when nθ= π/2(2j−1) or θ = (π/2n)(2j−1), j = 1, 2, 3, … n and extreme values when θ = kπ/n, k = 0, 1, 2, 3, … n). Using the trigonometric identity: cos[nθ ] =

 1 (cos θ + i sin θ )n + (cos θ − i sin θ )n 2

results Tn (t) =

1 2n−1

cos[nθ ] =

n 

n  √ √ 1  t + t2 − 1 + t − t2 − 1 n 2

For example, for n = 2:

2 

2 √ √ 1 1  cos[2θ ] = 2 t + t 2 − 1 + t − t 2 − 1 2 2



 √ √    1  2  2 t + t − 1 + 2t t 2 − 1 + t 2 + t 2 − 1 − 2t t 2 − 1 = 4

T2 (t) =

9.2 Chebyshev Theorem [1]

=

405

 1 2 1 4t − 2 = t 2 − 4 2

1 For the range (−1, +1) the maximum error is given as L n = ± 2n−1 . If we change n variables so that the range is α ≤ t ≤ β, pk will be replaced by pk h nand the maximum and the zero and error will be Ln hn , where h = ½ (β−α)For example: L n = ± (β−α) 22n−1 extreme positions will occur when:

t ∗j =

1 1 (t1 + tn+1 ) − (tn+1 − t1 ) cos α j 2 2

ro points: α j = (2 j − 1)π/2n j = 1, 2, . . . n Extreme values: tj =

1 1 (t1 + tn+1 ) − (tn+1 − t1 )cosα j 2 2 α j = j π/n j = 0, 1, 2, . . . n

Note that in all the functions, the coefficients pi (and the maximum error L) are functions of αand β, the end points. When n > 2 certain relations must exist between the coefficients. These are known as the compatibility conditions. Example 9.1 Consider T3 (t) = t3 + p2 t2 + p1 t + p0 within the range α ≤ t ≤ β Let us determine the relation between the coefficients and the end points α and β. When t =α or β the function will reach an extremum value ± L. We can write the following four equations: T3 (β) = β 3 + p2 β 2 + p1 β + p0 = L

(9.1)

T3 (α) = α 3 + p2 α 2 + p1 α + p0 = −L

(9.2)

T3, (β) = 3β 2 + 2 p2 β + p1 = 0

(9.3)

T3, (α) = 3α 2 + 2 p2 α + p1 = 0

(9.4)

If we subtract Eq. (9.4) from Eq. (9.3), we can eliminate p1 and solve for p2 as: 3 p2 = − (α + β) 2

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9 Chebyshev Methods in Kinematic Synthesis: Russian School

Substituting into Eq. (9.3) or (9.4), we obtain p1 as: p1 =

3 (α + 3β)(β + 3α) 16

Using Eqs. (9.1) and (9.2), we can solve for P3 and L: p3 = L=

  1 (α + β) α 2 + β 2 + 14αβ 32

   1 3 α − β 3 + p2 α 2 − β 2 + p1 (α − β) 2

Since p0 , p1 and p3 are functions of α and β only, this results with a compatibility condition that must be satisfied between the coefficients:   27 p0 = p2 9 p1 − 2 p22 We use a similar approach to determine the coefficients of Chebyshev polynomial T2 in terms α and β for the range α ≤ t ≤ β: T2 (t) = t 2 + p1 t + p0 p1 = −(α + β) p0 =

1 1 αβ + (α + β)2 2 8

There is no compatibility condition for T2 . Coefficients of T4 will be: T4 (t) = t 4 + p3 t 3 + p2 t 2 + p1 t + p0 p3 = −2(α + β) p2 = p1 = p0 =

3 1 (α + β)2 − (α − β)2 2 4

1 1 (α + β)(α − β)2 − (α + β)3 4 2

1 1 1 (α + β)4 − (α + β)2 (α − β)2 + (α − β)4 16 16 128

In this case we have two compatibility conditions as:

9.3 Application of Chebyshev Theorem to Straight-Line Motion Generation …

407

4 p2 p3 − 8 p1 = p33 1 1 and p0 = 18 p22 − 32 p2 p32 − 512 p34 Up to now the functions φj (t) in P(t) were considered  as φj (t) = tj . Instead, function 1 1   1 1 2 series such as: t 2 , t , 1, t, t . . . qr 1, 1+t , {1+t}2 , . . . can also be used to form a linear function P(t) as defined by Chebyshev Theorem. In Table 9.1 coefficients for the 3 Chebyshev polynomials which have been used in mechanisms synthesis are shown. In the following Sects. 9.4, 9.5 and 9.6 these functions will be directly used in three different mechanism synthesis problem. As an indirect use of Chebyshev polynomials, we utilize the precission points of Chebyshev polynomials as the points at which the error of the function to be generated is zero. During the precission point synthesis of mechanisms, usually the error function is not a Chebyshev polynomial. But when we assume that the error curve is close to a Chebyshev polynomial, then the precission points within the interval will be the zero points of the Chebyshev polynomial. Now if we look at the cylinder from the top plane the projection of zero points will be as Shown in Fig. 9.2 and will be given by:

xj =

   1  1 (2 j − 1) x f in + xin − x f in − xin cos π j = 1, 2, . . . n 2 2 2n

The points where the function is at extreme value will be:    1  j 1 j = 0, 1, 2 . . . n x j = x f in + xin − x f in − xin cos π 2 2 n

9.3 Application of Chebyshev Theorem to Straight-Line Motion Generation: Straight-Line Motion Generation by a Swinging Block Mechanism [3] Consider the swinging block mechanism shown in Fig. 9.3. Let A0 A = r, A0 B0 = a. Point M is a point along AB0 (AM = b). Using the triangles MQB0 and A0 AB0 we can write: (b − t)2 = y 2 + (a − x)2 r 2 = a 2 + t 2 − 2at cos β a−x cos β = t −b Eliminating b and rearranging terms yields:

sinθ2 sinθ3

p 1 = t1 t3

√ t1 t3

α ≤ t ≤ β t1 = α, t3 = β,

Compatibility condition: 4 p2 = t1 ( p1 − t1 )2

L = 21 (t1 + t3 ) −

α ≤ t ≤ β t1 = α, t4 = β,  √  t2 = 21 t1 + t1 t4  √  t3 = 21 t4 + t1 t4 √ √ 2 L = 18 (t1 − t4 ) t1 − t4

f (t) = t 2 + p1 t + p0 + p2 1t  √  p0 = 18 t12 + t42 + 10t1 t4 + 6(t1 + t4 ) t1 t4   √ p2 = − 41 t1 t42 + t4 t12 + 2t1 t4 t1 t4 √ p1 = −t1 − t4 − t1 t4

f (t) == t + p1 1t + p0 √ p0 = − t1 t3 − 21 (t1 + t3 )

θ2 = θ3 + δ

δ = 21 (θ1 − θ4 )   θ3 = cot−1 k−cosδ sinδ

θi = cot−1 (t) / 1 k = sinθ sinθ4 =

p3 = − 21 (t1 + t4 ) − (t2 + t3 ) L = 21 (t4 − t1 ) + t2 − t3

p2 = t1 + 2t3 − t1 t32 = t4 + 2t2 − t4 t22

p1 = 2t1 t3 + t32 − 1 = 2t2 t4 + t42 − 1

t 1 f (t) = t + p1 1+t 2 + p2 1+t 2 + p3

Table. 9.1 Three Chebyshev polynomials other than φj (t) = tj

408 9 Chebyshev Methods in Kinematic Synthesis: Russian School

9.3 Application of Chebyshev Theorem to Straight-Line Motion Generation …

n=4

409

Zeros of the cosine curve on the cylinder

Zeros of Chebyshev Polynomial

45° ° 45

x1

x2

x3

22.5°

x In

x

x4 xFin

Fig. 9.2 Zeros of cosine curve drawn on a cylinder projected to the diagonal plane for n = 4

Fig. 9.3 Swinging block mechanism

    b a2 − r 2  2 1 2 2 x= −t + bt + a + r + 2a t If the path of point M is to describe a vertical line x = x0, Then the deviation of the path of point M from this straight line is given by:    2 2   − r b a 1 ϕ(t) = x0 − x = x0 − x = −t 2 + bt + a 2 + r 2 + 2a t or:    2 2   2 − r b a 1 2 ϕ(t) = t − bt + 2ax0 − a + r 2 − 2a t

410

9 Chebyshev Methods in Kinematic Synthesis: Russian School

or    b a2 − r 2  f (t) = 2aϕ(t) = t 2 − bt + 2ax0 − a 2 + r 2 − t which is in the form: f (t) = t 2 + p1 t + p0 +

p2 t

For f(t) to deviate least from zero, it must be a Chebyshev polynomial and the coefficients must be related by: t 1 ≤ t ≤ t4 √  1 p0 = − t12 + t42 + 10t1 t4 + 6(t1 + t4 ) t1 t4 = 2ax0 − a 2 − r 2 8   √  1 p2 = −t4 t22 = −t1 t32 = − t1 t42 + t4 t12 + 2t1 t4 t1 t4 = −b a 2 − r 2 4 √ p1 = −t1 − t4 − t1 t4 = −t1 − 2t3 = −t2 − 2t4 = − b √ √ 1 L = (t1 − t4 )( t1 − t4 )2 8 and 4 p 2 = t 1 ( p 1 − t 1 )2 Due to the factor 2a between f(t) and ϕ(t), The actual maximum deviation, L1 is related to L as: L1 = L/2a In order to use the advantage of symmetry, let us select the first position of the mechanism as the symmetry position (A0 AB collinear). In such a case, the length of the straight line will be doubled and the number of precission points will be 6 instead of 3. Assuming b > a + r, a case as shown in Fig. 9.4, at the first position: t1 = a + r At the symmetry position b−r = x0 −L. At the other extreme position t = t4 = AB0 and point M is displaced most from the symmetry position (s/2). s is the length of the approximate straight line drawn (Fig. 9.4). The angular rotation of the crank during the straight-line motion, φ, is given by (using the cosine theorem for the triangle A0 AB0 ):

9.3 Application of Chebyshev Theorem to Straight-Line Motion Generation …

411

Fig. 9.4 Swinging block mechanism whose coupler curve describes straight line at its extreme positions

φ = 2 π − cos

−1



r 2 + a 2 − t42 2ra

(9.1)

When t = t4 , the limit position for the straight line is reached. The coupler point coordinates will be (x4 , ± s/2). Where x4 is given by:     2  b a2 − r 2 1 2 2 x4 = −t4 + bt4 + a + r + 2a t4

(9.2)

and the total length of the straight-line s is given by: √ s = 2 (b − t4 )2 − (a − x4 )2 Using the compatibility condition: 4 p 2 = t 1 ( p 1 − t 1 )2   2 4b a − r 2 = (a + r )(b − a − r )2 4b(a − r ) = (b − a − r )2 or b2 − 2b(3a − r ) + (a + r )2

(9.3)

412

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Solving for b yields (neglecting a solution where b < 0): b = 3a − r +

√

8a(a − r )

(9.4)

From p2 expression one can obtain the value of t3 as:   t1 t32 = b a 2 − r 2   (a + r )t32 = b a 2 − r 2 √ t3 = b(a − r ) Also, from p1 expression: t3 =

1 1 (b − t1 ) = (b − a − r ) 2 2

(9.5)

From p1 expression we also have: √ t1 t4 √ (b − a − r ) = t4 + (a + r )t4 2t3 = t4 +

Solving for t4 : t4 =

 √ 1 2b − a − r − (a + r )(4b − 3a − 3r ) 2

(9.6)

√ √ 2 1 (t1 − t4 ) t1 − t4 16a

(9.7)

and L1 =

x0 = b − r + L cos(α/2) =

t 2 + a2 − r 2 2(x0 + L − a) = 4 s 2t4 a

(9.8)

(9.9)

One can also express total stroke s (Fig. 9.5) as: s = 2(b − t4 )sin

α 2

(9.10)

If we are given the length a = A0 A, we can first assume r = 1 and calculate length b (Eq. 9.4), t4 (Eq. (9.6)), L1 (Eq. (9.7)), x0 (Eq. (9.8)), angle α (Eq. (9.9)), s (Eq. (9.10)) and angle φ (Eq. (9.1)). We can determine s/r ratio, and scale the mechanism dimensions such that the length of the straight line is satisfied. If the

9.3 Application of Chebyshev Theorem to Straight-Line Motion Generation …

413

Fig. 9.5 Swinging block mechanism describing straight line and the relations between the parameters

corresponding angular rotation φ is specified, we must change a (r = 1) till we achieve the required angle φ. We now have an indirect method for the synthesis. If we want to realize a straight-line motion of length s, when the crank is rotating by an angle φ degrees, we first assume a value of a and let r = 1 unit (scaling). Using these values we can determine b, t4 , L1, x0 , α, s and φ. We must change the value of a till the angle φ is the required amount. We then scale the whole mechanism to obtain the required dimensions. As seen in the following three figures, you can construct design charts [4]2 (Figs. 9.6a–c). We can use these charts as a rough initial guess for the solution. Example 9.2 Design a swinging block mechanism that generates 150 mm straight line within 1500 crank rotation. From Charts for 1500 crank rotation a/r ≈ 1.7 Let a = 1.7 and r = 1 unit and calculate b, t4 , L1, x0 , α, s and φ in an Excel sheet either in a row or in a column as seen in Fig. 9.7a. The resulting angular rotation is φ = 162.010 . Now, using “goal seek” we want to set Cell C11 to value 1500 by changing Cell C2. The result is as shown in Fig. 9.7b. We next determine the ratio of the value of s required to the value of s obtained when r = 1 unit (C10). The final resulting mechanism link lengths and maximum error L is shown in row F. During the rotation of the crank from 800 to 2300 , the deviation ε of point M from a straight line for the resulting mechanism is as shown in Fig. 9.8.

2

These Charts were first prepared by Volmer and published as VDI Guideline 21373.

414

9 Chebyshev Methods in Kinematic Synthesis: Russian School

(a) 360 330 300 270 240 210 180 150 120

α

90 60 30 0

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

a/r

(b)

10 9 8

b/r

s/r,b/r,x0/r

7

x0/r

s/r

6 5 4 3 2 1 0

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

a/r

(c)

6 5

%L/s

4 3 2 1 0 1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

a/r Fig. 9.6 Variation of a α and φ, b s/r, b/r, x0 /r, c %L/s as a function of a/r ratio for swinging bock mechanisms generating a straight line

9.4 Application of Chebyshev Theorem to Straight-Line Motion …

415

Fig. 9.7 a assuming a/r = 1.7 unit (r = 1), we calculate φ. b using “solver” or “goal seek” command we determine the value of a that makes φ = 150°. Scaling the mechanism results with 150 mm straight line motion (s) for a given 150° crank rotation

Fig. 9.8 Deviation from a straight line (ε in mm) of the swinging block coupler curve within 150 mm vertical displacement of the coupler point

9.4 Application of Chebyshev Theorem to Straight-Line Motion: Straight-Line Motion Generation by a Slider-Crank Mechanism [3] Consider an in-line slider crank mechanism as shown in Fig. 9.9. (A0 A = r, AB = l). x and y coordinates of point C can be written as: (l − b)2 = y 2 + (t − x)2 r 2 = t 2 + l 2 − 2lt cos β

416

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Fig. 9.9 Slider Crank mechanism with a coupler point along AB

cos β =

t−x l −b

Eliminating β and solving for x yields:    (l − b) l 2 − r 2 1 t(b + l) + x= 2l t The y coordinate is given by: b−l y=± 2l

/ 2l 2 + 2r 2 − t 2 −

(r 2 − l 2 )2 t2

If the path of point C is to describe a vertical line x = x0, Then the deviation of the path of point C from this straight line is given by: f(t) = x − x0 or:    (l − b) l 2 − r 2 1 t(b + l) − 2lx0 + φ(t) = 2l t   2 (l − b) l − r 2 2lx0 (b + l) φ(t) = t− + 2l b+l t(b + l) or   (l − b) l 2 − r 2 2l 2lx0 + f (t) = φ(t) = φ(t) = t − b+l t(b + l) (b + l) which is in the form:

9.4 Application of Chebyshev Theorem to Straight-Line Motion …

f (t) = t +

417

p1 + p0 t

for f(t) to deviate least from zero, the polynomial must be of Chebyshev type and the coefficients must be: √ 1 −2lx0 p0 = − t1 t3 − (t1 + t3 ) = 2 (l + b)   (l − b) l 2 − r 2 p 1 = t1 t3 = (l + b) √ 1 L = (t1 + t3 ) − t1 t3 2 Note that due to the factor 2 l/(b + l) between f(t) and f, (t), the actual maximum deviation L1 is related to L as: L1 =

b+l L 2l

To make use of symmetry, select t1 when the mechanism is at a dead center position (A0 AB collinear). At such position: t1 = l ± r Substituting into p1 yields t3 as: t3 =

(l − b) (l ∓ r ) (l + b)

(note that in t1 if the sign is positive, then the sign in t3 is negative). Substituting t1 and t3 into the expressions for L and p0 yields:

/   1 2 2 2 2 l(b − r ) − b − r l − r L1 = 2l

/   1 l(b − r ) + b2 − r 2 l 2 − r 2 = b ± r + L 1 x0 = 2l

(9.1) (9.2)

When t = t3 , point C has maximum displacement(s/2) in y direction, given by: s=

4 b+l

/

  b(b − l) br − l 2

(9.3)

One can solve for b from Eq. (9.1) and for l from Eq. (9.3) to yield: b=

√  2 12 2 l L (r +L ) l − r + 2L ± 2l (r ) 1 1 1 r2

(9.4)

418

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Fig. 9.10 4 different configurations for the slider-crank mechanism and the coupler point describing an approximate straight line [5]

√ 2b(b − r ) 4br − s 2 − bs 2 l= 4b(b − r ) + s 2

(9.5)

Equations (9.1), (9.2), (9.3) or Eqs. (9.2), (9.4), (9.5) gives us three relations between s, L1 , x0 , b, r and l. Actually, there are 3 free parameters. Volmer [5], instead of selecting 3 of the above parameters as free, defined certain other parameters and distinguished 4 configurations as shown in Fig. 9.10. The equations can be written in the following form (note that some of the parameters have changed their sign) (Fig. 9.11): f =l +b

(9.1)

r = b − x0 + L 1

(9.2)

m = r + b = f − x0 + L 1

(9.3)

/ k= L 1 = ( f + m)

 f2

−

1 s 2

2

√

1 m(k − f + m) − |2m − f + k| 2

(9.4)

(9.5)

9.4 Application of Chebyshev Theorem to Straight-Line Motion …

419

s

Fig. 9.11 Slider-crank mechanism at two limiting positions. Point C describes an approximate straight-line of length s while the input crank rotates by an angle φ0

c

o

+L1 -L1

x0 b

A0

r φ0

f

A l

B



1 f −m f 1− 2 k+m /   4 b(b − r ) br − l 2 s= b−l l=

(9.6) (9.7)

Example 9.3 We would like to design a slider crank mechanism to generate 200 mm straight line. We also select x0 = 150 mm and f = BC = 250 mm. Initially we can assume L1 ≈0 and select m ≈ f-x0 ≈ 100 mm. Then Determine k = 229.1288 (Eq. 9.4), L1 = 1.192 mm (Eq. 9.5), l = 68.03 (Eq. 9.6), b = 181.97 (Eq. 9.1), r = 33.16 (Eq. 9.2). When we evaluate s = 216.6 mm (Eq. 9.7) /= 200. we re-evaluate m = f-x0 + L1 = 101.192, l = 68.688, b = 181.312, r = 32.477. When we evaluate s = 199.597 mm /= 200 mm. We repeat the process till s has converged to 200 mm as required. This is shown in Fig. 9.12 (You can perform this iterative computation in Excel using circular reference, i.e., a formula in a cell refers to a cell whose value depends on the value of the formula typed. You must check “Iterative Calculations” in File- OptionsFormulas Menu). One can select different x0 and f values and obtain different results (for this case decreasing x0 and increasing f yields a lower error L1 ). The final result must be checked as usual. In Fig. 9.13, the resulting mechanism and the deviation, ε, of the coupler path from a vertical line is shown as a function of deviation of the coupler path from straight lie within 200 mm straight line path is shown. Note that Chebyshev theorem gives us a relation between the design variables. It does not yield a unique solution and it does not tell you which parameters to select. It is up to the designer.

420

s x0 f k m L1 l b r s

9 Chebyshev Methods in Kinematic Synthesis: Russian School

200 150 250 229.1288 100 1.19194 68.03142 181.9686 33.16052 216.6123

101.19194 1.16468596 68.6880397 181.31196 32.4766463 199.597392

101.164686 101.1653 1.16529767 1.1652839 68.6730789 68.673415 181.326921 181.32659 32.4922187 32.491869 200.009024 199.9998

101.1652839 1.165284246 68.67340721 181.3265928 32.49187704 200.0000046

Fig. 9.12 Iterative method of solution. Use an approximate equation for m and refine its value till s = 200 is reached

Fig. 9.13 a Slider crank mechanism generating an approximate straight line for 200 mm. b deviation (ε) from straight line as a function of vertical displacement of the coupler point

9.5 Application of Chebyshev Theorem to Obtain Constant Velocity …

421

Another important point is to make use of symmetry in design. By selecting a symmetric mechanism, the number of precission points and the range is increased by two-fold.

9.5 Application of Chebyshev Theorem to Obtain Constant Velocity for a Four-Bar Mechanism [3] Consider a slider-crank mechanism with A0 A = r, AB = b, B0 B = c, A0 B0 = d. If the crank is running at a constant angular velocity ω, the velocity polygon will be as shown in Fig. 9.14 and the following vector equation is satisfied: − → − → −−→ VB = V A + VB/A − → where the magnitude of V A is ωr. We can rotate the velocity polygon 90° counter clockwise and scale the drawing it so that VA = r and move the origin to B, as shown. On this scaled polygon let VB = v and V B/A = t (in order to determine the actual velocities, must multiply v and t by ω). Using the velocity polygon and the triangle A0 B0 q, we have the following relations between the link lengths and the variables. r 2 = t 2 + v 2 − 2tv cos μ d 2 = (b − t)2 + (c − v)2 − 2(b − t)(c − v) cos μ Solving for cosμ from the second equation and rearranging we have: Fig. 9.14 Four-bar mechanism and the velocity polygon. Velocity polygon is rotated 90° and scaled such that Va = r

vB B

vB/A vA =r ω

vA =r ω

b A

μ

vB

c

r

vB/A

μ A0

d

B0

q

422

9 Chebyshev Methods in Kinematic Synthesis: Russian School

      b(c − v) r 2 − v 2 b(c + v) 2 v b2 − d 2 + (c − v) cv − r 2 f (t, v) = t − t + t+ c c c =0 3

Which is in the form: f (t) = t 3 + p2 t 2 + p1 t + p0 If we require the output velocity to be equal to a constant, k, within t1 ≤ t ≤ t4, then the coefficients must satisfy Chebyshev theorem: 3 b(c + k) = − (t1 + t4 ) p2 = − 2    c  k b2 − d 2 + (c − k) ck − r 2 3 = p1 = (t1 + 3t4 )((3t1 + t4 ) c 16   2   b(c − k) r − k 2 1 = − (t1 + t4 ) t12 + t42 + 14t1 t4 p0 = c 32 1 L= (t4 − t1 )3 32 The first three equations give us a relationship between the parameters r, b, c, d, t1 , t4 . Since it is the output rotation which is required to be constant, we can select one of the link lengths, say c, arbitrarily. (VB = ωout *c and VA = ω*r. Since kv = velocity scale is 1/ω, k = ωOUT ∗ χ /ω). Given t1 , t4 . the coefficients p2 , p1 and p0 can be computed. Solving the above equations for r, b and d: p2 c b=− k+c /

p0 (k + c) p2 (k − c) /   (k − c) r 2 − ck − cp1 2 d= b + k r=

k2 +

Note that the error involved will increase by the cube of the difference between t1 and t4 meaning, as the interval within which constant output angular velocity is increased, the maximum error will increase. Example 9.4 Let c = 1 and t1 = 0.5 and t4 = 1.5. We would like to have a velocity ratio between the input and output as 2. Using the above values we obtain p2 = −3, p1 = 2.8125, p0 = −0.8125. Solving for the link lengths: b = 1, d = 1 and r = 2.194 units and L = 0.03125. The final mechanism is shown in Fig. 9.15. The displacement and angular velocity of the

9.6 Extension of Chebyshev Theorem to Function Generation

423

1

A

0.8 0.6 0.4 0.2

A0

B0

0

0

0.2

0.4

0.6

0.8

1

1.2

B 1.4

1.6

1.8

2

2.2

2.4

-0.2

Fig. 9.15 Four-bar mechanism with an approximate velocity ratio of 2 within a certain range

Fig. 9.16 Change in the output displacement and angular velocity as a function input crank rotation for the four-bar mechanism with an approximate velocity ratio of 2

output in terms of input crank angle is shown in Fig. 9.16 (ω12 = 1 unit) within 100 < θ < 450 range of the input (Actual range is 12.700 < θ < 41.510 ).

9.6 Extension of Chebyshev Theorem to Function Generation In general, function generation of a four bar is an approximation. A function φ = F(θ) is to be approximated by a function generated by the four-bar mechanism in the form φ, = G(θ) which contains three link lengths (a2 , a3 , a4 , Fixed link is assumed as 1 unit) and initial crank angles θs and φs (Fig. 9.17). The difference between the required and generated functions at any crank angle is (let a1 = 1):

424

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Fig. 9.17 A two degree of freedom mechanism. x and y components of the loop closure equation can be written as

ε = φ − φ , = F(θ ) − G(a2 , a3 , a4 , θs , φs , θ )

(9.1)

For a “good” approximation, the difference, ε must differ from zero as little as possible within a species range of the crank rotation. For mathematical treatment of the problem we need an optimization criterion. Using optimization techniques one can numerically either minimize the maximum ε within the range (minimax method, min εmax ) or consider a number of values within the interval range ∑ and minimize the sum of squares of the errors (least squares minimization, min εi2 ). The resulting equations can either be solved directly or if an iterative solution is required, an appropriate initial guess for the parameters a2 , a3 , a4 , θs , φs is required for convergence. Rather than using ε as the criterion, consider a two degree of freedom mechanism Shown in Fig. 9.17, [6–10] The coupler length AB (a3θ ) depends on the variables θ and φ. a3θ is the required variable length of the coupler to realize a given φ = F(θ) function. a3θ cosγ = 1 + a4 cos(φ + φs ) − a2 cos(θ + θs ) a3θ sinγ = a4 sin(φ + φs ) − a2 sin(θ + θs )

(9.2)

Squaring, adding and simplifying the two equations we can eliminate γ and obtain: 2 a3θ = a22 + a42 + 1 − 2a2 a4 cos(φ − θ + φs − θs ) − 2a2 cos(θ + θs )

+ 2a4 cos(φ + φs )

(9.3)

The aim is to minimize the change in the coupler length within the specified range of crank rotation. Now a new error function can be defined as: Δa3 = a3 − a3θ =

2 Δ a32 − a3θ ≈ a3 + a3θ 2a3

(9.4)

9.6 Extension of Chebyshev Theorem to Function Generation

425

where Δ is the square of the difference of the required and actual coupler lengths. Since the change in a3θ is expected to be small, a3 + a3θ 2a3 . Δ can be written as: Δ = 2a2 a4 cos(φ − θ +s φ − θs ) + 2a2 cos(θ + θs ) − 2a4 cos(φ + φs ) + a32 − a22 − a42 − 1

(9.5)

Assume the crank is fixed, thus reducing the system to a single degree of freedom. Consider the velocity polygon. Referring to the velocity polygon in Fig. 9.17, for a small time increment Δt, Δ2 = Δa3 is the change in the length of coupler link (parallel to AB). Δ1 = a4 Δφ. is the change of position of point B on the output link (perpendicular to B0 B). Δs is due to the angular rotation of the coupler link (perpendicular to AB). Velocity polygon forms a right angled triangle. Δ2 = sinμ Δ1 Δ2 = a4 Δφsinμ

(9.6)

From Eqs. (9.6) and (9.4): Δφ =

Δ 2a4 a3 sinμ

(9.7)

Referring to Fig. 9.18, when we take the projection of links A0 B0 , A0 A and AB perpendicular to B0 B we obtain: a3 sinμ = sin(φ + φs ) − a2 sin(φ − θ +s φ − θs )

(9.8)

Equation (9.6) can be written as: Δφ =

Δ 2a4 [sin(φ + φs ) − a2 sin(φ − θ + φs − θs )]

(9.9)

Equations (9.7) and (9.9) indicate that Δ, square of the differences of the required and actual coupler lengths, is closely related to Δφ when the transmission angle μ is close to 900 . when the transmission angle is close to 00 (or 1800 ), use of Δ as a minimization criterion is not permissible. If the transmission angle is within usable limits, minimization of Δ is parallel to the minimization Δφ Assuming the deviation of the transmission angle μ is within a reasonable magnitude, letting φ = F(θ), Eq. (9.5) can be written as a polynomial in the form: Δ = 2[P0 f 0 (θ ) + P1 f 1 (θ ) + P2 f 2 (φ) + P3 f 3 (φ) +P4 f 4 (θ ) + P5 f 5 (θ, φ) + P6 f 6 (θ, φ)] where:

(9.10)

426

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Fig. 9.18 Projections of A0 B0 , A0 A and AB perpendicular to B0 B

f 0 (θ) = cos θ, P0 = a2 cos θs f 1 (θ) = sin θ, P1 = −a2 sin θ5 f 2 (φ) = cos φ, P2 = −a4 cos φ5 f 3 (φ) = sin φ, P3 = a4 sin φ5   f 4 (θ) = 1, P4 = a23 − a22 − a24 − 1 /2 f 5 (θ, φ) = cos(φ − θ) P5 = a2 a4 (cos θs cos φs + sin θs sin φ5 ) = a2 a4 cos(φ5 − θs ) f 6 (θ, φ) = sin(φ − θ) P6 = a2 a4 (sin θ5 cos φ5 − cos θ5 sin φ5 ) = a2 a4 sin(φ5 − θ5 ) Coefficients Pj are functions of design parameters a2 , a3 , a4 , θs , φs . P5 and P6 are not independent parameters. They can be expressed in terms of other coefficients as: P5 = −(P0 P2 + P1 P2 ) P6 = P1 P2 − P0 P3 In its most general form, Eq. (9.10) contains 5 coefficients P0 , P1 , P2 , P3 , P4 as functions of 5 parameters of the mechanism a2 , a3 , a4 , θs and φs . If we let θs = φs = 0, i.e. when the angles are measured from the fixed link, in the above equations P5 = −P0 P2 , P1 = P3 = P6 = 0. Equation (9.10) simplifies to: Δ = 2[P0 f 0 (θ) + P2 f 2 (θ) + P4 f 4 (θ) + P5 f 5 (θ)] f 0 (θ) = cos θ, P0 = a2

9.6 Extension of Chebyshev Theorem to Function Generation

f 2 (θ) = cos φ, P2 = −a4   f 4 (θ) = 1, P4 = a32 − a22 − a42 − 1 /2 f 5 (θ) = cos(φ − θ) P5 = a2 a4

427

(9.11)

and P5 = −P0 P2 . There are 3 independent design parameters (a2 , a3 and a4 ). If we divide every term by a2 a4 :  1 1 1  2 Δ a3 − a22 − a42 − 1 + cos(φ − θ ) = cos θ − cos φ + a4 a2 2a4 a2 2a4 a2

(9.12)

The aim in approximation is to minimize the change in Δ as much as possible. One way is to assume the above equation is close to a Chebyshev polynomial. In such a case, the points where Δ is zero will be Chebyshev precission points and we will end up with Freudenstein’s equation in the form:   K 1 cos φ j − K 2 cos θ j + K 3 = cos φ j − θi ( j = 1, 2, 3)

(9.13)

a 2 −a 2 +a 2 +1

where K 1 = a12 , K 2 = a14 , K 3 = 2 2a3 2 a44 (fixed link = 1 unit) Let us next assume the output angle φ is measured from A0 B0 reference (φs = 0). In such a case, we have 4 design variables (a2 , a3 , a4 , and θs ). Equation (9.5) takes the form: Δ = 2a2 a4 cos(φ − θ − θs ) + 2a2 cos(θ + θs ) − 2a4 cos(φ) + a32 − a22 − a42 − 1

(9.14)

expanding: Δ = 2{a2 a4 [cos(φ − θ )cosθs + sin(φ − θ )sinθs ]} + 2a2 [cos(θ )cosθs − sin(θ )sinθs ] − 2a4 cos(φ) + a32 − a22 − a42 − 1 or: Δ = a2 cosθ cosθs − a2 sinθ sinθs − a4 cos 2   + a32 − a22 − a42 − 1 /2 + a2 a4 sin(φ − θ )sinθs + a2 a4 cos(φ − θ )cosθs Let us divide terms by −a2 a4 cosθs :

428

9 Chebyshev Methods in Kinematic Synthesis: Russian School

−Δ 1 1 1 cosφ = − cosθ + tanθs sinθ + a2 a4 cosθs a4 a4 a2 cosα a 2 − a32 + a42 + 1 + 2 − tanθs sin(φ − θ ) − cos(φ − θ ) a2 a4 cosθs define new functions and coefficients such that: Δ = P0 f 0 (θ ) + P1 f 1 (θ ) + P2 f 2 (θ ) + P3 f 3 (φ) + P4 f 4 (θ, φ) − cos(φ − θ ) =

4 ∑

P j f j (θ ) − cos(φ − θ )

(9.15)

j=0

where: 1 1 f1 = sin θ P1 = tan θs a4 a4 1 f2 = cos φ P2 = a2 cos θs a 2 − a32 + a42 + 1 f3 = 1 P3 = 2 2a2 a4 cos θs f4 = − sin(φ − θ ) P4 = tan θs

f0 = − cos θ P0 =

Note that P4 = PP01 . We have four design parameters and out of five coefficients only four are linearly independent. In order to solve for Pi , we use what is known as Lagrange multipliers [11]. Let P4 = λ and express the other coefficients Pj as Pj = mj + nj λ. If we are to design for four precission points, at the precission points the error must be zero: 3 ∑ 

 m j + n j λ f j (θk ) − λ sin(φk − θk ) − cos(φk − θk ) = 0 (k = 0, 1, 2, 3)

j=0

Equating the coefficients of 1 and λseparately, we have two sets of equations in the form:

9.6 Extension of Chebyshev Theorem to Function Generation

3 ∑

m j f j (θk ) − cos(φk − θk ) = 0 (k = 0, 1, 2, 3)

429

(9.16)

j=0

and 3 ∑

n j f j (θk ) − sin(φk − θk ) = 0 (k = 0, 1, 2, 3)

(9.17)

j=0

For four design parameters these two sets of equations can be solved for mi and ni. . Now: P4 = λ =

P1 m 1 + λn 1 = P0 m 0 + λn 0

A quadratic equation in terms of λ is obtained as: n 0 λ2 + (m 0 − n 1 )λ − m 1 = 0 When solved, Pj = mj + nj λ can be calculated. The link lengths and angle θs can be calculated from: 1 1 , θs = tan−1 [P4 ] = tan−1 [a4 P1 ], a2 = , P0 P2 cosθs /  a3 = a22 + a42 + 1 − 2P3 a2 a4 cosθs

a4 =

A similar result will be obtained if we let θs = 0 and use φs as a design variable. Example 9.5 Consider Example 8.4 in Chap. 8. A four-bar mechanism is to be designed to correlate the following four crank angles.

430

9 Chebyshev Methods in Kinematic Synthesis: Russian School

i

θj (o )

φj (o )

1

20

45

2

60

50

3

90

60

4

120

75

Four equations will be: Am

+ λA

n − b 1 − λb 2 = 0 where: ⎡

− cos θ0 sin θ0 cos φ0 ⎢ − cos θ1 sin θ1 cos φ1 A=⎢ ⎣ − cos θ2 sin θ2 cos φ2 − cos θ3 sin θ3 cos φ3 ⎤ ⎡ cos(φ0 − θ0 ) ⎢ cos(φ1 − θ1 ) ⎥ ⎥ b1 = ⎢ ⎣ cos(φ2 − θ2 ) ⎦

⎤ ⎤ ⎡ 1 sin(φ0 − θ0 ) ⎥ ⎢ 1⎥ ⎥ b2 = ⎢ sin(φ1 − θ1 ) ⎥ ⎣ sin(φ2 − θ2 ) ⎦ 1⎦ 1

sin(φ3 − θ3 )

cos(φ3 − θ3 )

m

= [m 0 , m 1 , m 2 , m 3 ]T , n = [n 0 , n 1 , n 2 , n 3 ]T − → → The above equations result in two sets of linear equations: A− m = b 1 and − → → A− n = b 2 , Solving the two sets of equations for the given values: ⎡

⎤ ⎡ ⎤ −0.8111 0.0278 ⎢ 0.6599 ⎥ ⎢ −0.9818 ⎥ ⎥ ⎢ ⎥ m=⎢ ⎣ −1.3890 ⎦n = ⎣ 1.4618 ⎦ 0.9007 −0.2491 Now the second order equation in λ is: 0.0278λ2 + 0.1708λ − 0.6599 = 0 Which results in λ1 = 2.68685 and λ2 = −8.82040. Since θs = tan−1 (λ), we have θs = 69.58560 and −83.53180 , which is the same result obtained in example 8.4. The two mechanisms shown in Fig. 8.18 is obtained. Example 9.6 Consider the same problem discussed in Examples 8.2 and 8.5 (to generate a function y = ln(x) 1 ≤ x ≤ 2). Using the same range (Δθ = 900 ,Δφ = 600 ) and θin = 00 , φin = 1100 , four precission points according to Chebyshev spacing on x are:

9.6 Extension of Chebyshev Theorem to Function Generation θj (o )

431 φj (o )

x

y

0

1.03806

0.0374

3.425421

113.2334

1

1.30866

0.269

27.779246

133.2853

2

1.69134

0.5255

62.220754

155.4901

3

1.96194

0.6739

86.574579

168.3368

Solving matrix equations, we obtain two values of λ as λ1 = −0.30055 and λ2 = 3.44712. Since θs = tan−1 (λ), we have θs1 = −16.72820 and θs2 = 73.82270 . The results are as shown in Figs. 8.19 and 8.20. If we want to approximate a function in terms minimization of the squares of the errors (least squares minimization), consider n points within the interval where you want to correlate θk with φk . If we substitute these values into Eq. (9.2), square the error and take the sum: ∑ ∑ S= Δ2k = [P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (φk ) 2 k

k

+P3 f 3 (θk ) + P4 f 4 (θk , φk ) − cos(φk − θk )]2

(9.18)

For S to be a minimum a necessary (but not sufficient) condition is that the derivative of S with respect to four design variables must be equal to zero. In this equation the design variables are P0 , P1 , P2 , and P3 (since P4 = P1 /P0 , it is not a free design variable). This results in four equations ∑ 1 δS = f 0 (θk ) 2 δ P0 k P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (φk ) + P3 f 3 (θk ) + P4 f 4 (θk , φk ) − cos(φk − θk ) =0 ∑ 1 δS = f 1 (θk ) 2 δ P1 k P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (φk ) + P3 f 3 (θk ) + P4 f 4 (θk , φk ) − cos(φk − θk ) =0 ∑ 1 δS = f 2 (θk ) 2 δ P2 k P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (φk ) + P3 f 3 (θk ) + P4 f 4 (θk , φk ) − cos(φk − θk ) =0 ∑ 1 δS = f 3 (θk ) 2 δ P3 k

432

9 Chebyshev Methods in Kinematic Synthesis: Russian School

P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (φk ) + P3 f 3 (θk ) + P4 f 4 (θk , φk ) − cos(φk − θk ) =0 Again using Lagrange multipliers, letting P4 = λ, expressing the other coefficients

+λA

n − b 1 − λb 2 = 0, which results in two sets of 4 Pj = mj + λ nj we obtain A m linear equations: ⎤⎡ ⎤ ⎡ ⎤ m1 εf0k cos(φK − θκ ) εf20k εf0k f1k εf0k f2k εf0k f3k ⎢ εf0k f1k εf2 εf1k f2k εf1k f3k ⎥⎢ m2 ⎥ ⎢ εf1k cos(φκ − θκ ) ⎥ 1k ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ εf0k f2k εf1k f2k εf2 εf2k f3k ⎦⎣ m3 ⎦ = ⎣ εf2k cos(φκ − θκ ) ⎦ 2k f3k cos(φκ − θκ ) εf0k f3k εf1k f3k εf2 k3k εf23k m4 ⎡ ⎤ ⎤ ⎡ ⎡ ⎤ εf20k εf0k f1k εf0k f2k εf0k f3k n1 εf0k sin(φκ − θκ ) ⎢ εf0k f1k εf2 εf1k f2k εf1k f3k ⎥⎢ n2 ⎥ ⎢ εf1k sin(φκ − θκ ) ⎥ 1k ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ εf0k f2k εf1k f2k εf2 εf2k f3k ⎦⎣ n3 ⎦ = ⎣ εf2k sin(φκ − θκ ) ⎦ 2k εf3k sin(φκ − θκ ) εf0k f3k εf1k f3k εf2 k3k εf23k n4 ⎡

After mi and ni values are determined, the procedure explained for four precission point synthesis follows. The roots of the equation n 0 λ2 + (m 0 − n 1 )λ − m 1 = 0 are determined and the coefficients Pk (k = 0, 1, 2, 3, 4) are calculated, after which the design variables a2 , a3 , a4 , and θs can be found. Example 9.7 Consider the same problem discussed in Example 8.2 and 8.4 (to generate a function y = ln(x) 1 ≤ x ≤ 2). Using the same range (Δθ = 900 , Δφ = 60◦ ) and θin = 0◦ , φin = 105◦ . We want to approximate in the least squares sense. Taking 0.1 increments on x, we determine y = lnx and the corresponding θ and φ values: x

Y

θ

φ

1

0

0

105

1.1

0.09531

9

113.2502

1.2

0.182322

18

120.7821

1.3

0.262364

27

127.7107

1.4

0.336472

36

134.1256

1.5

0.405465

45

140.0978

1.6

0.470004

54

145.6843

1.7

0.530628

63

150.9321 (continued)

9.6 Extension of Chebyshev Theorem to Function Generation

433

(continued) x

Y

θ

φ

1.8

0.587787

72

155.8798

1.9

0.641854

81

160.56

2

0.693147

90

165

We obtain: ⎡

5.5 −3.15688 ⎢ −3.15688 5.5 A=⎢ ⎣ 4.097992 −5.66159 −6.8531 6.853102 ⎡ ⎤ −6.73373 ⎢ 6.76725 ⎥ ⎥ b2 = ⎢ ⎣ −7.65300 ⎦ 10.82155

⎤ ⎡ ⎤ 4.097992 −6.8531 0.99474 ⎢ ⎥ −5.66159 6.853102 ⎥ ⎥ b1 = ⎢ 0.21040 ⎥ ⎣ ⎦ 6.0335 −7.76198 0.03219 ⎦ −7.76198 11 −0.58894

The solution is: ⎡

⎡ ⎤ ⎤ 0.474876 −0.0798 ⎢ −0.01584 ⎥ ⎢ 0.332276 ⎥ ⎢ ⎥ ⎥ m

=⎢ ⎣ −0.08287 ⎦ n = ⎣ 0.356907 ⎦ 0.19371 0.978895 The quadratic equation in λ is: −0.079801λ2 + 0.142600λ + 0.015845 = 0 The roots of which: λ1 = −0.104949 and λ2 = 1.891888. Solution for λ1 does not yield a feasible result. Using λ2 we obtain θs = 62.1403◦ and a2 = 3.6125, a3 = 1.5027 a4 = 3.0874. The resulting mechanism and the error curve is shown in Fig. 9.19. Use of precission points in function generation by applying Chebyshev spacing is an indirect use of Chebyshev theorem for the minization of the maximum error. Another form of applying Chebyshev theorem is by noting the local maximum and minimum values of the error function obtained must be equal in magnitude and if there are n design variables, there must be n + 1 extreme points. Two of these extreme points are at the initial final points of the interval. Remaining n-1 points will be within the interval. These points we shall call “design points”. At these design points we hope that the errors will all be equal to an unknown value L. Using the correlation of x with and y with φ and selecting Δφ, Δθ, θin , φin as before, we can determine the crank angles θi , φi to be correlated using a four-bar mechanism. Similar to the precission point synthesis, this method of approximation will be called “Equal

434

9 Chebyshev Methods in Kinematic Synthesis: Russian School

0.0003 0.0002 0.0001 0 -0.0001

1

1.2

1.4

1.6

1.8

2

-0.0002 -0.0003

x

Fig. 9.19 Least square approximation of function y = ln(x) 1 ≤ x ≤ 2 using four design variables (Example 9.6). The resulting mechanism and the structural error

extreme local error synthesis”. This procedure can be used for 4, 5 or 6 extremum positions by measuring the φ and θ from two arbitrary reference angles, say θs and φs . Note that similar to function generation using precission points, equal extreme point synthesis is also a minimax method. In case of three design variables, for the minimization of the maximum error in Chebyshev sense, Δj must deviate by equal amount at 4 design points with alternating sign and the amount of deviation at these design points must be equal, say L. Two of these points must be at the start and end of the interval. Since our error function is not of Chebyshev type, we do not know the location of the other two maximum error points (which shall be used as design points). The design points will be x1 = xinitial , x4 = xfinal and x2 and x3 will be two points within the range xinitial ≤ x ≤ xfinal . At these design points we hope that the errors will be equal to a value L. Using the correlation of x with θ and y with φ and selecting Δφ, Δθ, θin , φin as before, we can determine the crank angles θi , φi to be correlated using a four-bar mechanism. Hence, Eq. (9.14) can be written as: K 1 cos φ1 − K 2 cos θ1 + K3 − cos(φ1 − θ1 ) = L K 1 cos φ2 − K 2 cos θ2 + K3 − cos(φ2 − θ2 ) = −L K 1 cos φ3 − K 2 cos θ3 + K3 − cos(φ3 − θ3 ) = L K 1 cos φ4 − K 2 cos θ4 + K3 − cos(φ4 − θ4 ) = −L where L = 2aΔ4 a2 . This forms a set of four linear equations in four unknowns (K1 , K2 , K3 and L). The assumed values x2 and x3 need not correspond to points where the error has a local maximum or minimum. As a wise guess we can assume the error function is a Chebyshev polynomial and select x2 and x3 in accordance with Chebyshev spacing for the extreme values. If these points do not yield extreme local values, we can locate points x2 , and x3 , where these local extremum values occur and replace x2 and

9.6 Extension of Chebyshev Theorem to Function Generation

435

x3 with these points and resolve the set of four equations. After a few steps solution converges such that at the design points all the local extremum values are almost equal. This iterative process is known as “Remez Algorithm”10 , [12]. Equal extreme point synthesis can easily be extended for cases with four or five design parameters. Example 9.8 The problem given in Example 8.2 of Chap. 8, namely the approximation of function y = lnx within the interval 1 ≤ x ≤ 2 will be used for comparison. Let θin = 300 , φin = 300 (corresponding to x = 1 and y = 0 respectively) Also let Δθ = 900 and Δφ = 600 . Then the crank angles that we want to correlate can be determined. Design points are x1 = 1, x4 = 2 and we shall determine x2 and x3 in accordance to Chebyshev polynomials as: x2,3 =

1 1 (x4 + x1 ) ± (x4 + x1 ) cos(π/3) 2 2

Hence, we can form a table as:

x 0 1 2 3

y

1 0 1.25 0.223144 1.75 0.559616 2 0.693147

θ 30 52.5 97.5 120

φ 30 49.32 78.44 90

We have a system of 4 linear equations: ⎡

0.8660 ⎢ 0.6519 ⎢ ⎣ 0.2004 0.0000

−0.8660 −0.6088 0.1305 0.5000

⎤ ⎤⎡ ⎤ ⎡ 1.0000 1 −1 K1 ⎥ ⎢ ⎥ ⎢ 1 1 ⎥ ⎥⎢ K2 ⎥ = ⎢ 0.9985 ⎥ ⎦ ⎣ ⎦ ⎣ 0.9452 ⎦ K3 1 −1 L 0.8660 1 1

The solution is: K1 = −0.8053; K2 = −0.5929, K3 = 1.17321 and L = −0.01073 The link lengths are: a1 = 1, a2 = −1.24175 a3 = 0.68728 a4 = 1.68656. The resulting mechanism and the error curve is as shown in Fig. 9.20. The extreme local errors are: ε1 = 0.016742, ε2 = 0.013396(at x2 , = 1.2786), ε3 = 0.015633 (at x3 , = 1.7507) and ε4 = 0.024551 (points 1 and 4 are the end points of the interval). Consider solving the same problem with four design variables. There will be 5 extremums, two of which are the end points. According to Chebyshev spacing, the internal extreme values will be assumed to be located at:

436

9 Chebyshev Methods in Kinematic Synthesis: Russian School

0.02 0.01

ε

0 -0.01

1

1.2

1.4

1.6

1.8

2

-0.02 -0.03

x

Fig. 9.20 The mechanism and the structural error for y = lnx within the interval 1 ≤ x ≤ 2, obtained using 4 equal extreme local error (three design variables)

x2,3,4 =

1 1 (x4 + x1 ) ± (x4 + x1 ) cos(π/4) 2 2

θs is now a free design variable hence we select θin = 00 . If θin is selected as 300 there is no solution. Therefore, we select θin = 1000 . The extreme points and their correlated crank angles are:

x 0 1 2 3 4

1 1.146447 1.5 1.853553 2

y

θ

φ

0 0 100 0.136667 13.18019 111.8302 0.405465 45 135.0978 0.617105 76.81981 153.4176 0.693147 90 160

Using Lagrange multipliers, letting P4 = λ, expressing the other coefficients Pj = mj + λ nj we obtain A m

+λA

n − b 1 − λb 2 = 0 where: ⎡ ⎤ ⎤ −0.17365 −1 0 −0.17365 1 −1 ⎢ −0.15040 ⎥ ⎢ −0.97366 0.228014 −0.37186 1 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ A = ⎢ −0.70711 0.707107 −0.70831 1 −1 ⎥ b1 = ⎢ −0.00171 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0.23178 ⎦ ⎣ −0.22801 0.973658 −0.89429 1 1 ⎦ 0.34202 −6.1E − 17 1 −0.93969 1 −1 ⎡ ⎤ 0.984808 ⎢ 0.988626 ⎥ ⎢ ⎥ ⎢ ⎥ b2 = ⎢ 0.999999 ⎥ ⎢ ⎥ ⎣ 0.972767 ⎦ 0.939693 ⎡

We obtain m

and n as: m

T = [0.98481, 0.98863, 1.00000, 0.97277, 0.93969] and: n = [0.10752, −0.34175, 0.36466, 0.941116, 0.00050]. T

9.6 Extension of Chebyshev Theorem to Function Generation

437

0.0008 0.0006 0.0004 0.0002 0 -0.0002 1

1.2

1.4

1.6

1.8

2

-0.0004 -0.0006

x

Fig. 9.21 The mechanism and the structural error for y = lnx within the interval 1 ≤ x ≤ 2, obtained using 5 equal extreme local error (four design variables)

The quadratic equation is: −0.10752λ2 + 0.12264λ − 0.01479 = 0 The roots are: λ1 = 0.13711 and λ2 = 1.00352. Solution for λ1 does not yield a feasible result. Using λ2 we obtain θs = 45.10072° and a2 = 4.45056, a3 = 2.65867 a4 = 2.80510. The resulting mechanism and the error curve is shown in Fig. 9.21. Maximum error is 6.19 × 10–4 . Comparing the solutions with 3 and 4 design variables will be unfair since the arbitrary initial crank angle φin has been changed from 30° to 100°. The solution obtained with 4 design variables can be obtained with 3 design variables if we select θin and φin accordingly. Selection of the correct solution usually will not depend on structural error ε only. One has to consider the transmission angle, link length ratios, the volume and the shape of the occupied space etc. Even the shape of the structural error curve may be important in a certain design. If five design variables are to be used, in Eq. (9.10), P0, P1, P2, P3 and P4 will be five parameters from which we can determine the link lengths a2 , a3 , a4 (a1 = 1) and the reference angles θs and φs . Referring to Eq. (9.10), dividing the terms by P5 = a2 a4 cos(φs− θs ) we have: Δ = [P0 f 0 (θ) + P1 f 1 (θ) + P2 f 2 (φ) + P3 f 3 (φ) + P4 f 4 (θ) 2a2 a4 cos(φs − θs ) (9.19) +P5 f 5 (θ, φ) − cos(φ − θ)] where:

438

9 Chebyshev Methods in Kinematic Synthesis: Russian School

cos θs a4 cos(φs − θs ) sin θs f 1(θ) = sin θ, P1 = a4 cos(φs − θs ) cos φs f 2 (φ) = cos φ, P2 = a2 cos(φs − θs ) sin φs f 3 (φ) = sin φ, P3 = − a2 cos(φs − θs )   1 + a22 − a32 + a42 f 4 (θ) = 1, P4 = 2a2 a4 cos(φs − θs ) f 5 (θ, φ) = sin(φ − θ) P5 = tan(φs − θs )

f 0 (θ) = cos θ, P0 = −

note that: P1 P3 = −tanθs and = −tanφs P0 P2 and P1 − PP23 tanφs − tanθs P0 P5 = tan(φs − θs ) = = 1 + tans tanθs 1 + PP01 PP23

Or P5 (P0 P2 + P1 P3 ) = P1 P2 − P3 P0

(9.20)

For five precission points, again letting P5 = tan(φs − θs ) = λ and using Pj = mj + nj λ in the compatibility Eq. (9.20): λ[(m0 + n0 λ)(m2 + n2 λ) + (m1 + n1 λ)(m3 + n3 λ)] = (m1 + n1 λ)(m2 + n2 λ) − (m0 + n0 λ)(m3 + n3 λ) Expanding: (no n2+ n1 n3 )λ3 + (m0 n2 + n0 m2 + m1 n3 + n1 m3 + n0 n3 − n1 n2 )λ2 + (m 0 m 2 + m 1 m 3 + m 0 n 3 + n 0 m 3 − m 1 n 2 − n 1 m 2 )λ + (m 0 m 3 − m 1 m 2 ) =0 (9.21) There are three roots of the cubic equation. There is at least one real root and the other two roots can be real or imaginary. Using Pj = mj + nj λ in Eq. (9.19) for five precission points, we end up two sets of five linear equations in the form:

9.6 Extension of Chebyshev Theorem to Function Generation 4 ∑

439

m j f j (θk ) − cos(φk − θk ) = 0 (k = 0, 1, 2, 3, 4)

(9.22)

j=0

and 4 ∑

n j f j (θk ) − sin(φk − θk ) = 0 (k = 0, 1, 2, 3, 4)

j=0

When we solve for λ from Eq. (9.20), we can now determine Pj = mj + nj λ(j = 0, 1, 2, 3, 4)and P5 = λ. The reference angles θs and φs and the link lengths a2 , a3 , a4 can be determined as:     P1 P1 or θs = tan−1 + π, θs = tan−1 P0 P0     P3 P3 or φs = tan−1 +π φs = tan−1 P2 P2 sin θs a4 = , P1 cos(φs − θs ) cos φs , a2 = − P2 cos(φs − θs ) / a22 + a42 + 1 − 2P4 a2 a4 cos(φs − θs ) a3 = The same method of solution is also applicable for least squares minimization with five design parameters. Example 9.9 Consider the same problem discussed in Example 9.7 (to generate a function y = ln(x) within 1 ≤ x ≤ 2). The same range (Δθ = 90°, Δφ = 60°) and θin = 0°, φin = 0° is used. We use precission point synthesis and apply Chebyshev spacing.

(2 j + 1) π x j = 1.5 − 0.5 cos 10

j = 0, 1, 2, 3, 4

The resulting five equations will be: We obtain the following precission points and their corresponding crank angles: θj (o )

φj (o )

x

y

0

1.0245

0.0242

2.2025

2.0928

1

1.2061

0.1874

18.5497

16.2215

2

1.5000

0.4055

45.0000

35.0978 (continued)

440

9 Chebyshev Methods in Kinematic Synthesis: Russian School

(continued) x

y

θj (o )

φj (o )

3

1.7939

0.5844

71.4503

50.5856

4

1.9755

0.6808

87.7975

58.9343

The resulting equations will be: Am

+ λA

n − b 1 − λb 2 = 0 where: ⎡

− cos θ0 sin θ0 cos φ0 ⎢ − cos θ sin θ cos φ ⎢ 1 1 1 ⎢ A = ⎢ − cos θ2 sin θ2 cos φ2 ⎢ ⎣ − cos θ3 sin θ3 cos φ3 − cos θ4 sin θ4 cos φ4 ⎤ ⎡ sin(φ0 − θ0 ) ⎢ sin(φ − θ ) ⎥ ⎢ 1 1 ⎥ ⎥ ⎢ b2 = ⎢ sin(φ2 − θ2 ) ⎥ ⎥ ⎢ ⎣ sin(φ3 − θ3 ) ⎦ sin(φ4 − θ4 )

sin φ0 sin φ1 sin φ2 sin φ3 sin φ4

⎤ ⎡ ⎤ cos(φ0 − θ0 ) 1 ⎢ cos(φ − θ ) ⎥ 1⎥ ⎥ ⎢ 1 1 ⎥ ⎥ ⎢ ⎥ 1 ⎥b1 = ⎢ cos(φ2 − θ2 ) ⎥ ⎥ ⎢ ⎥ ⎣ cos(φ3 − θ3 ) ⎦ 1⎦ cos(φ4 − θ4 ) 1

− → → The above equations result in two sets of linear equations: A− m = b 1 and − → → A− n = b 2 . When solved: Solving the two sets of equations fort he given values: ⎤ ⎡ ⎤ 0.03649 −0.25621 ⎢ 0.41624 ⎥ ⎢ −0.03109 ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ ⎥ n = ⎢ −0.42359 ⎥m = ⎢ 0.28330 ⎥ ⎥ ⎢ ⎢ ⎥ ⎣ −0.42946 ⎦ ⎣ 0.07291 ⎦ 0.68011 0.67977 ⎡

When we determine the coefficients of Eq. (9.21) we obtain the cubic equation: 0.106259λ3 −0.07619λ2 + 0.129391λ − 0.13359 = 0 The plot of this cubic compatibility equation is shown in Fig. 9.22. we see that it has one real root in between 0.5 < λ < 1. When we use “Goal Seek”, λ = 0.905525 is obtained. Now using Pj = mj + λnj we obtain: P0 = −0.195513 P1 = 0.388087 P2 = −0.100267 P3 = −0.3634 P4 = 1.295658 P5 = 0.905525

9.6 Extension of Chebyshev Theorem to Function Generation

441

0.2

Fig. 9.22 Plot of the cubic compatibility equation obtained for precission synthesis

0.1 0 -1.5

-1

-0.5

-0.1

0

0.5

1

1.5

2

-0.2 -0.3

θs = 63.261625 or 243.261625 φs = −74.576741 or 105.423258 and the link lengths: a2 = 3.5782472 a3 = 1.450514 a4 = −3.104482 These values are the same as the values obtained in Example 8.6. Let us now solve the same problem by considering the local extreme values of the function to be equal with alternating sign at the design points.       P0 f 0 θ j + P1 f 1 θ j + P2 f 2 (φi ) + P3 f 3 (φi ) + P4 f 4 θ j + P5 f 5 (θi , φi ) − cos(φ5 − θ5 ) = ±L If the above polynomial was of Chebyshev type, the design points where the function is at an extreme value would be given by:    1  j 1 j = 0, 1, 2 . . . n x j = x f in + xin − x f in − xin cos π 2 2 n In this case: 

j x j = 1.5 − 0.5 cos π 5

 j = 0, 1, 2, 3, 4, 5

The six design points and correlated crank angles will be ((Δθ = 90°, Δφ = 60°).

0

x

y

1.0000

0.0000

θj (o ) 0.0000

φj (o ) 0.0000

1

1.0955

0.0912

8.5942

7.8947

2

1.3455

0.2968

31.0942

25.6880

3

1.6545

0.5035

58.9058

43.5842

4

1.9045

0.6442

81.4058

55.7651

5

2

0.6931

90.0000

60.0000

442

9 Chebyshev Methods in Kinematic Synthesis: Russian School

Treating L as another coefficient we have: L = m6 + λn6 we again have a matrix equation in the form: Am

+ λA

n − b 1 − λb 2 = 0 where: ⎡

⎤ − cos θ0 sin θ0 cos φ0 sin φ0 1 1 ⎢ − cos θ sin θ cos φ sin φ 1 −1 ⎥ 1 1 1 1 ⎢ ⎥ ⎢ ⎥ ⎢ − cos θ2 sin θ2 cos φ2 sin φ2 1 1 ⎥ A=⎢ ⎥ ⎢ − cos θ3 sin θ3 cos φ3 sin φ3 1 −1 ⎥ ⎢ ⎥ ⎣ − cos θ4 sin θ4 cos φ4 sin φ4 1 1 ⎦ − cos θ5 sin θ3 cos φ5 sin φ5 1 −1 ⎤ ⎤ ⎡ ⎡ cos(φ0 − θ0 ) sin(φ0 − θ0 ) ⎢ cos(φ − θ ) ⎥ ⎢ sin(φ − θ ) ⎥ 1 1 ⎥ 1 1 ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ ⎢ cos(φ2 − θ2 ) ⎥ ⎢ 1(φ2 − θ2 ) ⎥ b1 = ⎢ ⎥ b2 = ⎢ ⎥ ⎢ cos(φ3 − θ3 ) ⎥ ⎢ sin(φ3 − θ3 ) ⎥ ⎥ ⎥ ⎢ ⎢ ⎣ cos(φ4 − θ4 ) ⎦ ⎣ sin(φ4 − θ4 ) ⎦ cos(φ5 − θ3 ) sin(φ5 − θ3 ) Solving the two sets of equations with given values: ⎡ ⎤ ⎤ −0.25548 0.041107 ⎢ −0.03146 ⎥ ⎢ 0.413795 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ −0.42529 ⎥ ⎢ 0.272596 ⎥ m=⎢ ⎥ ⎥ n=⎢ ⎢ 0.073191 ⎥ ⎢ −0.42749 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0.680741 ⎦ ⎣ 0.686224 ⎦ 2.1E − 05 7.41E − 05 ⎡

When we determine the coefficients of Eq. (9.21) we obtain the cubic equation: 0.06434λ3 − 0.20712λ2 −0.05482λ − 0.24653 = 0 The plot of this cubic equation is shown in Fig. 9.23. we see that it has one real root in between 0.5 < λ < 1. When we use “Goal Seek”, λ = 0.883978 is obtained as the root. Now using Pj = mj + λnj we obtain: P0 = −0.42876 P1 = 0.500117 P2 = 0.462245 P3 = −0.45107 P4 = 0.966966 P5 = 0.883978 L = −0.00045 θs = 64.424699 or 244.424699 φs = −74.099329 or 105.900671

9.6 Extension of Chebyshev Theorem to Function Generation

443

Fig. 9.23 Plot of the cubic compatibility equation for equal extreme local error

and the link lengths: a2 = 3.538210 a3 = 1.395768 a4 = −3.119101 The mechanism and the error curve is as shown in Fig. 9.24. Maximum error is 1.5 × 10–4 . Least square method of approximation can also be extended using five design parameters. Consider n points within the interval where you want to correlate θk with φk (k = 1, 2,n). If we substitute these values into Eq. (9.19), square the error and take the sum: ∑ Δ2k = εk [P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (φk ) S= k

+P3 f 3 (θk ) + P4 f 4 (θk , φk ) + P5 f 5 (θk , φk ) − cos(φk − θk )]2

(9.18)

0.0003 0.0002 0.0001 0 -0.0001

1

1.2

1.4

1.6

1.8

2

-0.0002 -0.0003 -0.0004

x

Fig. 9.24 Mechanism and its structural error curve for approximating function y = ln(x) within 1 ≤ x ≤ 2, for equal extreme local error condition with five design parameters

444

9 Chebyshev Methods in Kinematic Synthesis: Russian School

For S to be a minimum a necessary (but not sufficient) condition is that the derivative of S with respect to the five design parameters must be equal to zero. In this equation the design variables are P0 , P1 , P2 , and P3 and P5 . This results in five equations: ∑ 1 δS = f 0 (θk ) 2 δ P0 k

P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (k ) + P3 f 3 (θk ) + P4 f 4 (θk ,k ) + P5 f 5 (θk ,k ) − cos(k − θk ) =0 ∑ 1 δS = f 1 (θk ) 2 δ P1 k

P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (k ) + P3 f 3 (θk ) + P4 f 4 (θk ,k ) + P5 f 5 (θk ,k ) − cos(k − θk ) =0 ∑ 1 δS = f 2 (θk ) 2 δ P2 k

P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (k ) + P3 f 3 (θk ) + P4 f 4 (θk ,k ) + P5 f 5 (θk ,k ) − cos(k − θk ) =0 ∑ 1 δS = f 3 (θk ) 2 δ P3 k

P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (k ) + P3 f 3 (θk ) + P4 f 4 (θk ,k ) + P5 f 5 (θk ,k ) − cos(k − θk ) =0 ∑ 1 δS = f 4 (θk ) 2 δ P4 k

P0 f 0 (θk ) + P1 f 1 (θk ) + P2 f 2 (k ) + P3 f 3 (θk ) + P4 f 4 (θk ,k ) + P5 f 5 (θk ,k ) − cos(k − θk ) =0

Letting P5 = λ and expressing Pj = mj + λnj we again obtain a system of equations: Am

+ λA

n − b 1 − λb 2 = 0 where ⎡

εf20k ⎢ εf f ⎢ 0k 1kk ⎢ A = ⎢ εf0k f2k ⎢ ⎣ εf0k f3k εf0kk f4k

εf0k f1k εf21k εf1k f2k εf1k f3k εf1k f4k

εf0k f2k εf1k f2k εf22k εf2 k3k εf2k f4k

εf0k f3k εf1k f3k εf2k f3k εf23k εf3k ffk

⎤ ⎤ ⎡ εf0k cos(φk − θκ ) εf0k f4k ⎢ εf cos(φ − θ ) ⎥ εf1k f4k ⎥ ⎥ ⎢ 1k k κ ⎥ ⎥ ⎥ ⎢ εf2k f4k ⎥ b1 = ⎢ εf2k cos(φk − θκ ) ⎥ ⎥ ⎥ ⎢ ⎣ εf3k cos(φk − θκ ) ⎦ εf3k f4k ⎦ εf4k cos(φk − θκ ) εf24k

9.6 Extension of Chebyshev Theorem to Function Generation

445

⎤ εf0k sin(φx − θk ) ⎢ εf sin(φ − θ ) ⎥ ⎢ 1k x k ⎥ ⎥ ⎢ b2 = ⎢ εf2k sin(φx − θk ) ⎥ ⎥ ⎢ ⎣ εf3k sin(φx − θk ) ⎦ εf4k sin(φk − θK ) ⎡

Again, we solve to sets of equations Am

= b 1 and A

n = b 2 and obtain mj and nj . We determine the coefficients of the cubic equation (Eq. 9.21) and solve for the roots to obtain the λ values. We follow the same steps discussed previously to determine Pj , θs , φs and the link lengths a2 , a3 , a4 . Example 9.10 Consider the same problem discussed in Example 9.7 (to generate a function y = ln(x) within 1 ≤ x ≤ 2). The same range (Δθ = 90°, Δφ = 60°) and θin = 0°, φin = 0° is used. We are to use least squares approximation. Taking 0.1 increments on x, we determine y = lnx and the corresponding θ and φ values: x

Y

θ

1

0.0000

0

1.1

0.0953

9

8.2502

1.2

0.1823

18

15.7821

1.3

0.2624

27

22.7107

1.4

0.3365

36

29.1256

1.5

0.4055

45

35.0978

1.6

0.4700

54

40.6843

1.7

0.5306

63

45.9321

1.8

0.5878

72

50.8798

1.9

0.6419

81

55.5600

2

0.6931

90

60.0000

We obtain: ⎡

5.5 3.2 ⎢ 3.156876 5.5 ⎢ ⎢ A = ⎢ 6.044775 4.9 ⎢ ⎣ 2.622861 4.6 6.853102 6.9 ⎡ ⎤ 6.761736 ⎢ 6.482203 ⎥ ⎢ ⎥ ⎢ ⎥ b2 = ⎢ 8.479378 ⎥ ⎢ ⎥ ⎣ 5.389586 ⎦ 10.60524

6.044775 4.854813 7.215593 4.038495 8.716463

φ 0.0000

⎤ ⎤ ⎡ 2.622860647 6.853 0.7819758 ⎢ 1.9547268 ⎥ 4.560463477 6.853 ⎥ ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ 4.038495379 8.716 ⎥ b1 = ⎢ 1.4635165 ⎥ ⎥ ⎥ ⎢ ⎣ 1.6262761 ⎦ 3.784406912 5.7 ⎦ 5.700227401 11 2.2319528

446

9 Chebyshev Methods in Kinematic Synthesis: Russian School

When solved: ⎡

⎤ ⎡ ⎤ 0.0415416 −0.25509 ⎢ 0.4136804 ⎥ ⎢ −0.03164 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ m = ⎢ 0.2715578 ⎥n = ⎢ −0.42619 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −0.427517 ⎦ ⎣ 0.073305 ⎦ 0.6868612 0.681264 We obtain the quadratic equation: 0.106300λ2 −0.075308λ + 0.1301 = 0 There is one real root: λ = 0.881053. Using Pk = mk + λnk (k = 0, 1, 2, 3, 4, 5) we obtain P0 = −0.18320, P1 = 0.385806, P2 = −0.103936, P4 = 1.287090 and P5 = 0.881053. The initial crank angles will be: θin = 64.598850 and φin = −74.019390 . The link lengths are: a1 = 1, a2 = 3.53031, a3 = 1.38670 and a4 = −3.120534. The resulting mechanism and the error curve are as shown in Fig. 9.25. Maximum error is 1.4 × 10–4 . The same problem has been solved for 5 design variables using three different methods. The mechanisms and the structural error curves for the three results are comparable and very close. In Fig. 9.26 results for the approximation of the function function y = logx 1 ≤ x ≤ 2, with Δθ = Δφ = 60° are shown using 3 different approximation methods. In Fig. 9.27 results for approximation of the function y = sinx for 00 ≤ x ≤ 900 with Δθ = Δφ = 900 are shown. In Fig. 9.28 results for y = tanx, 00 ≤ x ≤ 450 with Δθ = Δφ = 900 are shown. For precission point synthesis and equal extreme point synthesis Chebyshev spacing has been used (no respacing is applied). For Least squares minimization 10 equally spaced points are selected.

0.00015 0.0001 0.00005 0 1

1.2

1.4

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9.6 Extension of Chebyshev Theorem to Function Generation

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Fig. 9.26 Structural error curves in approximating y = logx 1 ≤ x ≤ 2. Δθ =Δφ = 600 , using least squares, equal extreme local error, precission point synthesis (Five design variables. a1 = 1)

Problems 1. Design a swinging Block mechanism that generates 200 mm straight line within 140° crank rotation. Plot the error involved in describing the straight line within 200 mm. 2. Using a slider crank mechanism, generate a straight line of 200 mm length. The straight line must be located 50 mm with respect to A0 and it must be in between A0 and B (type 3). Let f = 250 mm. Plot the error involved in describing the straight line within 200 mm. Determine the angular rotation of the crank when the coupler is describing the straight line. 3. Using a slider crank mechanism, generate a straight line of 200 mm length. The straight line must be located 400 mm with respect to A0 and it must be of type 5. Let f = 250 mm. Plot the error involved in describing the straight line within 200 mm. Determine the angular rotation of the crank when the coupler is describing the straight line. Also compare the error curves involved in problems 2, 3 and the result obtained in Example 9.2.

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9 Chebyshev Methods in Kinematic Synthesis: Russian School

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4. Generate function (a) y = ex 0 ≤ x ≤ 1 (b) y = 1/ × 1 ≤ x ≤ 2 (c) y = x1.5 0 ≤ x ≤ 1 Use: A. Three, four or five design variables. B. Chebyshev Spacing for the precission points C. Least squares minimization, selecting 5, 8, 10, 20 equally spaced data or selecting 6 data points using Chebyshev spacing. D. Equal extreme point realization. As expected, there are several different ways for the formulation of the problem. Also, you can arbitrarily select the range for the crank angles. Try different cases and compare the results for each of the functions.

References

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Fig. 9.28 Structural error curves in approximating y = tanx, 00 ≤ x ≤ 450 , Δθ= Δφ = 900 , using least squares, equal extreme local error, Precission point synthesis (five design variables. a1 = 1)

References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

F. Freudenstein, Mathematics of Approximation, Lecture Notes (1971) Chebyshev polynomials—Wikipedia S. Sch Bloch, Angenaherte Synthese von Mechanismen. (Verlag Technik, Berlin, 1951) J. Volmer, Konstruktion eines Gelenkgetriebe für eine Geradführung VDI Berichte, Bd 12, 12 1956, and Published as VDI guideline 2137 in 1959 VDI Guideline 2136 (1959) H.И. ЛEBИTCКИЙ, CИHTEЗ MEXAHИЗMOB ПO ЧEБыШEBУ. (Moscow, 1946) ИИ Aptoбoлebcкий, H.И Лebиtckий, C.A. Чepkудиhob, CИHTEЗ ПЛOCКИX MEXAHИЗMOB (ФИЗMATTИЗ, Moscow, 1959) N.I. Levitskii, Y. Sarkisyan, On the special properties of lagrange’s multipliers in the leastsquare synthesis of mechanisms. J. Mech. 3, 3–10 (1968) N.I. Levitskii, Y.L. Sarkisyan, G.S. Gekchian, Optimum synthesis of function generating mechanisms, Mech. Mach. Theory. 7, 387–398 (1972) H.И. ЛEBИTCКИЙ, TEOPИЯ MEXAHИЗMOB И MAШИH. (Moscow, 1990) G. Kiper, Approximation Synthesis of Mechanisms. Lecture notes (2016) E. Remez, Sur la calcul effectif des polynomes d’approximation de Tchebyscheff. C. R. Acad. Sci. Paris 199, 337–340 (1934)

Chapter 10

Optimization Methods in the Synthesis of Mechanisms Using Excel®

Abstract Optimization is an important topic in every field of engineering. With the advance of computers, different algorithms have been developed in different platforms to solve engineering problems. In this chapter, some optimization problems in mechanism design are solved as examples. Rather than preparation of optimization programs, the two algorithms (engines) that are available in Excel® . Solver will be used. Both constraint and unconstrained mechanism optimization problems will be solved. Keywords Excel® solver in mechanism design · Numerical optimization · Optimization of mechanisms

10.1 Introduction “Optimization” (sometimes spelled as optimisation) “is the selection of a best element, with regard to some criterion from some set of available alternatives”.1 One can say Chebyshev method is an optimization method specifically developed for mechanism design in nineteenth century. Before the extensive use of computers Chebyshev methods and trial-and-error techniques were used to obtain the best possible mechanism proportions for a certain application. Starting from 1960s with the advance of computers, optimization has been an important topic in science, engineering, finance, economics and in all other fields. New algorithms have been developed in recent years. In machine design optimization has been an important issue even in old ages due to the conflicting nature of any design process. Kinematic synthesis of mechanisms using optimization methods was one of the first fields in engineering where optimization tools were used. The aim of this chapter is not to discuss optimization in detail or explain different optimization algorithms. The aim is to give some optimization examples in kinematic synthesis using an existing Excel® tool: “Solver”.

1

Wikipedia “Mathematical Optimization” Mathematical optimization—Wikipedia.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 E. Söylemez, Kinematic Synthesis of Mechanisms, Mechanisms and Machine Science 131, https://doi.org/10.1007/978-3-031-30955-7_10

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Many different algorithms using different optimization methods have been used in motion, function and path generation problems. Especially in recent years some of the methods have been very successful even for complex problems. Also, in recent years there is a big interest in mechanisms with elastic parts-known as compliant mechanisms. In such mechanisms the rigid body assumptions fail and one cannot use the theories developed for the synthesis methods discussed and optimization methods are used to obtain satisfactory solutions. Due to highly nonlinear characteristics of kinematic synthesis problems, all optimization algorithms depend on the initial estimate and usually what is found is the local optimum close to the initial estimate. To eliminate this defect in recent years heuristic methods such as genetic algorithms are being used to a good effect. The heuristic methods are mainly wise search algorithms where the design variables are changed within a certain design region according to certain rules. The solution obtained is the “best” result obtained within a certain amount of search. An optimization problem in general is defined as [1]: “Find x, which minimizes or maximizes f (x) subject to di (x) ≤ air i = 1, 2, ..m ei (x) = bi i = 1, 2, .., p where x is an n-dimensional design vector, f (x) is the objective function, d i (x) are inequality constraints, ei (x) are equality constraints, and ai and bi are constants.” In mechanism synthesis the objective function is highly nonlinear and the number of design variables are usually more than one. We may have both equality or inequality constraints depending on how we formulate the problem and these constraints can be linear or nonlinear. Such problems are known as multivariable nonlinear (constrained or unconstrained) optimization. Excel “Solver” tool possesses three optimization algorithms: GRG nonlinear, LP Simplex and Evolutionary. In Excel there is a simple note stating: “Select GRG nonlinear Engine for Solver Problems that are smooth nonlinear. Select LP Simplex for linear Solver Problems, and evolutionary engine for Solver Problems that are non-smooth”. It turns out that GRG nonlinear works quite efficiently in the synthesis problems considered. One can also use Evolutionary engine in the synthesis, provided that limits are placed on all the design parameters. GRG Nonlinear method uses “Generalized Reduced Gradient” method developed by L. S. Larson, et al in 1978 [2]. It is useful to solve small to medium size problems with equality and inequality constraints. The global extremum is not guaranteed. When the solution converges to a local optimum, algorithm stops. Therefore, depending on the initial condition, the solution may converge to a different result or it may diverge.

10.2 Lifting a Load

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Evolutionary method is a genetic programming method. It is able to find a good solution but this solution need not be local or global optimum. When using evolutionary method, one must specify the lower and upper bounds of all the design variables used. Evolutionary algorithm may end with different results even when the initial conditions and the constraints are the same. Therefore, it is wise to run evolutionary algorithm several times to come up with a good result. Experience shows that using GRG nonlinear and Evolutionary algorithms in kinematic synthesis yields good results. Usually, Evolutionary algorithm takes a little more computer time. In the following sections some applications using Excel Solver will be shown. In kinematic synthesis we usually use two different objectives depending on the problem. One method is the minimization of the maximum error, known as “minimax” in short. The other objective that we can use to a good effect is the minimization of the sum of squares of the errors, known as “least squares minimization”. The error can be the deviation of a coupler point curve from a desired path or the deviation of a function generated by the mechanism from the desired function. Although the two objectives are closely related, they may yield different results. The designer must decide on the objective and must critically evaluate the result obtained. In certain cases, if the constraints are not set correctly, the result may converge to a solution that cannot be useful in practice. In the following sub-sections optimization methods are illustrated mainly via examples.

10.2 Lifting a Load A piston-cylinder assembly used to lift a heavy weight by rotating an arm about a fixed axis is commonly used in different applications as shown in Fig. 10.1 (Figures are from Wikipedia). The structure used is an inverted slider crank mechanism. You can see some applications ranging as a loading device in a ship to lifting a patient in a hospital or as a machine shop lift to a dump truck. The piston-cylinder assembly can be hydraulically or pneumatically actuated. In recent years a screw and nut driven with an electric motor (i.e. linear actuator) is becoming very common. All the above problems involve an inverted slider crank mechanism driven by the prismatic joint in between link 3 and 4 in Fig. 10.2. The problem is to lift the load W with a beam A0 Q of length L from a certain height h1 to h2 . The beam will rotate about A0 within the range θ1 ≤ θ ≤ θ2 . The rotation of the beam will be imparted by a piston-cylinder attached between B0 and A. The coordinates of A (u, v) measured with respect to a moving frame attached to the beam and the coordinates of A0 (x, y) are measured from the fixed frame as shown. The coordinates of both A and B0 are limited depending on the application. i.e. x1 ≤ x ≤ x2 , y1 ≤ y ≤ y2 , u1 ≤ u ≤ u2 , v1 ≤ v ≤ v2 . Let us assume that we have selected the coordinates of B0 and A within the feasible region where the constraints are satisfied. Then we can perform kinematic analysis.

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

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(a)

(b)

https://hidrokon.com/

https://trad4u.eu/fr/grue-atelier-hydraulique-pliable-2-tonnes (c)

https://proletarsky.ru/product_catalog/15 Fig. 10.1 Piston-cylinder assembly to lift load for different applications a mobile crane, b engine lift, c jib crane, d patient lift, e manlift

10.2 Lifting a Load

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https://medmartonline.com/advance-e-340-portable

https://hidrokon.com/ Fig. 10.1 (continued)

The coordinates of A with respect to the fixed frame (coordinates (u, v) are measured positive in accordance with the coordinate shown): xa = u cos θ − v sin θ ya = u sin θ − v cos θ

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

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Fig. 10.2 Common structure for a simple piston and cylinder lift. An inverted slider-crank mechanism

We can now determine the distance s = |B0 A| and the angle of B0 A with respect to horizontal, φ as: /    s= (xa − x)2 + (ya − y)2 , φ = tan−1 (xa − x), (ya − y) Now consider the static force analysis of the mechanism by simply taking the free body diagram of the beam (link 2, Fig. 10.3). The clockwise moment due to the weight, W, about A0 is: Mw = L W cos(θ) This moment will be in equilibrium due to the counterclockwise moment about A0 by the piston force FP : MP = a2 FP sin(φ − (θ − α2 )) For equilibrium, these two moments must be equal. Solving for FP /W, we have: L FP cos(θ ) = W a2 sin(φ − θ + α2 ) here a2 =

√

u2 + v 2 and α2 = tan−1 (u; v).

10.2 Lifting a Load

457

Fig. 10.3 Free body diagram of the main beam

Now we can perform both the kinematic and static force analyses within θ1 ≤ θ ≤ θ2 , using a certain increment value, Fp /W ratio for the whole range is evaluated and its maximum value is determined. We can also determine maximum and minimum lengths of the piston-cylinder assembly (B0 A), (smax = s (θ2 ) and smin = s (θ1 )), from which we calculate the stroke s0 = smax − smin . If we are not going to use a telescoping piston cylinder, then s0 ≤ smin + a must be satisfied. The length a (a ≥ 0) is determined in accordance with the load and type of cylinder used. In addition, one can limit the stroke size. We have an optimization problem: Minimize the maximum force applied to the piston (FP /W) subject to the constraints, since you are going to select an actuator according to the maximum load. Depending on the application, the constraints will be different. As an example, let us design a hydraulic lift as shown in Fig. 10.4. You want the boom to be rotated in between −15° and + 70° with respect to a horizontal line. Assume a weight is to be lifted at 1500 mm away from the beam hinge. The closed length of the piston and cylinder must be 250 mm larger than the stroke. The fixed pivot for the cylinder must be within 100 × 500 mm area on the vertical post and the moving pivot of the piston must be within (400 × 70 mm area on the rotating column. Assuming coordinates of A and B0 within the feasible region, we select the link lengths and perform the analysis of the mechanism and calculate FP /W for every degree of the beam within −15° ≤ θ ≤ 70°. The excel sheet is as shown in Fig. 10.5 (calculation for each position is made in rows 15 to 100 (similar to the calculation shown in row 9) are not shown. The maximum value of FP /W is at cell F11. Now we use Excel® “solver”. We are to minimize the value in cell F11 by changing u, v, x, y the coordinates of points A and B0 (values in cells B2-B5). Subject to constraints umin

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10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

1500 300 350

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Fig. 10.4 Beam to rotate −15° ≤ θ ≤ 70°. the fixed and moving pivots of the piston cylinder must be within the rectangular regions shown

≤ u ≤ umax , vmin ≤ v ≤ vmax , xmin ≤ x ≤ xmax , ymin ≤ y ≤ ymax and the extra length value (a) in cell K12 ≥ 200. With the initial conditions for u, v, x, y the maximum Fp /W value is 4.3454. The condition for the extra length is not satisfied (value in cell K12 must be greater than 200!). We select Solver tool (Fig. 10.6). We select cell F11 to be minimized. The variable cells are the cells in which the initial value parameters are placed. These are cells B12, B13, B14 and B15. Next, we add all the constraints. You can either use “GRG Nonlinear” or “Evolutionary” algorithm for this case (remember evolutionary

Fig. 10.5 Excel sheet for the evaluation of the Fp /W ratio for given design parameters u, v, x, y and beam position (θ)

10.2 Lifting a Load

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Fig. 10.6 Solver menu

algorithm requires upper and lower bounds for each parameter and for this case, the bounds are specified). (Keep in mind that the Evolutionary algorithm requires more computer time). When we apply Solver and accept the solution the Excel sheet is as shown in Fig. 10.7. (GRG nonlinear algorithm is used). For this problem when we apply Evolutionary algorithm, the same result is obtained. Consider another similar problem as shown in Fig. 10.8, where the constraints are different. The motion range is 0° ≤ θ ≤ 70° Assuming an initial condition u = 400, v = 100, x = 800 y = 350 and L = 2000 we have (Fp/W)max = 15.7233 but the extra length is −43 mm which is less than 200 mm required. Constraints are as shown in the figure. Using GRG Nonlinear we obtain u = 651.478, v = 50, x = 1450 and y = 400 and we have (Fp/W)max = 9.2706 where the extra length requirement of 200 mm is satisfied. However, when we use Evolutionary algorithm and start with the same initial conditions we have u = 1250, v = 50, x = 443.781, y = 400 and (Fp/W)max = 3.679. A very close result is reached with GRG algorithm when the initial conditions are: u = 900, v = 100, x = 400, y = 350. It is obvious that GRG nonlinear algorithm converges to the closest local minimum and stops, whereas evolutionary algorithm searches the whole region and returns the best result found within the time limit set. This is why it usually takes more time. Another type of lifting device can be seen in ships and docks as shown in Fig. 10.9. These cranes are also known as “Level Luffing Jib Crane” which, during the luffing

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Fig. 10.7 Case when maximum (Fp /W) is minimized satisfying the constraints

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motion (raising or lowering of the boom of the crane) the weight is level with the ground (i.e., Point P shown in Fig. 10.9 moves horizontally while the boom is rotated). Length of the boom depends on the application. We have two design parameters a/L and α. During luffing motion, the hoisting drum which raises the cable is not operated. Depending on the application, there can be several wounds of cable between B0 A and a single cable length in between AP. The length B0 A can be determined from cosine theorem as: √ s1 = B0 A = [a2 + L2 − 2aL cos(α − θ )] and the length AP is: s2 = AP = L sin θ Hence, the total length of the cable is: S = ns1 + s2 where n is the number of wounds in between B0 A (n = 1, 3 …). For point P to remain horizontal as θ is changing, length S must be constant or must not change too much from a constant

10.2 Lifting a Load

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Fig. 10.9 Simple level luffing jib crane

value. Let us arbitrarily select a/L and determine S in between θin ≤ θ ≤ θfin where initial and final values of θ depend on the application, Let us determine the maximum and minimum values of S and calculate the mean value as: S=

1 (Smin + Smax ) 2

And define the error as the deviation of point P from a horizontal line as: ε =S −S ε is the amount the hook will deviate from a horizontal path for any luffing angle θ. Now, our aim is to minimize the work done while doing work against the gravitational effect when moving the load. During luffing motion, when the load moves upwards, we must do work against gravity and when it moves downward, we must apply brakes. This amount of work done is proportional to the area under the error curve ε Rather than minimization of the maximum error, least squares minimization is a better choice in this problem. As a constraint, a/L ratio must be less than a certain amount and the angle α must be greater than a certain value. We shall select a/L ≤ 0.7 and α ≥ 60°. Consider the number of wounds between B0 A as n = 2 and let us assume a/L = 0.2, α = 100°. Let us also assume the limits for the luffing angle as: 10° ≤ θ ≤ 80°.

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

Fig. 10.10 Deviation of point P from a horizontal line (ε) as a function of boom rotation angle (θ) when n = 2, 10° ≤ θ ≤ 80°, a/L = 0.2 and α = 100°

0.3 0.2 0.1

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∑ In such a case (ε/L)max = 0.2007 and εj2 = 1.3807. ε/L as a function of θ will be as shown in Fig. 10.10. For a 10 m boom the deviation from a horizontal line will be ± 2 m! When Solver GRG nonlinear algorithm is used to minimize the maximum deviation of point P from a horizontal line, with constraints ∑ a/L ≤ 0.7 and α ≥ 60°, maximum deviation reduces to (ε/L)max = 0.0140 and εj2 = 0.0025. The design parameter values are: a/L = 0.4407, α = 80.7860°. The error curve is as shown in Fig. 10.11. (For a 10-m boom, now the load will only move vertically ± 15 cm and the wasted energy will be far less). Similar dimensions and error curve are obtained when the maximum error is minimized or when evolutionary algorithm is used. If n is increased, the error will decrease (for example, when n = 8 for a 10 m boom the maximum vertical movement of the load during luffing is only ± 27 mm (using the previous initial conditions the optimum solution result is a/L = 0.1205 and α = 87.7112°, (ε/L)max = 0.0027). The optimum will also change depending on initial and final angular positions of the boom (i.e. for n = 1 and 25°∑ ≤ θ ≤ 85°, optimum εj2 = 0.0077 for a is when a/L = 0.7 and α = 79.803°, (ε/L)max = 0.0283 and 10-m boom the maximum vertical displacement will be ± 28 cm. Error curve is as shown in Fig. 10.12). 0.02 0.015 0.01 0.005 0 -0.005 0 -0.01 -0.015 -0.02

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Fig. 10.11 Optimized error curve: n = 2, 10° ≤ θ ≤ 80°, a/L = 0.4407, α = 80.7860°

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Fig. 10.12 Optimized error curve when n = 1, 25° ≤ θ ≤ 85°, a/L = 0.7 and α = 79.803°

10.3 Function Generation Problems In Chaps. 8 and 9 we have seen function generation for a four-bar mechanism using Freudenstein’s equation. We have used Chebyshev spacing for three positions and found solutions that approximate the required function. If we are free to select the initial crank angles, usually we end up with a good result by simply using brute force approach to change the two crank angles. Instead, one can use “Solver” tool in Excel. When generating function y = ln(x) within 1 ≤ x ≤ 2 using Δθ = 90° and Δφ = 60° and selecting initial crank angles as θin = 30° and φin = 30°; using three positions and Chebyshev spacing on x we had: a1 = 1, a2 = −1.383, a3 = 0.672, a4 = −1.844 with a maximum structural error εmax = 0.040 as the solution. Starting with this solution, we can optimize by considering the design variables as the initial angles θin and φin . Also, we can apply constraints on link dimensions by requiring 0.1 < | a2 | < 10 and 0.1 < |a4 | < 10. As an optimization criterion we can try to minimize the maximum error εmax , or we can perform analysis for every 0.01 units of x within 1 ≤ x ≤ 2 and determine the error for 100 points within the interval. Square the errors and determine the square root of the sum of these errors. The objective will be to minimize the square root of the sum of squares of the errors. When we use “GRG nonlinear” engine in “Solver” tool for the minimization of the maximum error, the method converges to a solution: θin = 35.576°, φin = 17.497°, a1 = 1, a2 = −0.1, a3 = 0.906, a4 = −0.183 with a maximum structural error εmax = 0.004 which is an order of magnitude better than what we had started. This result is shown in Fig. 10.13. If we minimize in least square sense, we end up with: θin = 33.380°, φin = 15.506°, a1 = 1, a2 = −0.1, a3 = 0.902, a4 = −0.188 with a maximum structural error εmax = 0.007, The two results are close but not the same. If we require the link lengths to be positive for the minimization of maximum error, we have: θin = 69.11°, φin = 45.08°, a1 = 1, a2 = 0.315, a3 = 1.224, a4 = 0.478 with a maximum structural error εmax = 0.005. This result is shown in Fig. 10.14. Although we have performed three precision points synthesis, the maximum error is comparable to four precision points. Another optimization that we can perform is to consider the link lengths a2 , a3 , and a4 as design variables while keeping the initial crank angles at: θin = 30° and

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10 Optimization Methods in the Synthesis of Mechanisms Using Excel® ε

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Fig. 10.13 Four-bar mechanism and the error curve in generating function y = ln(x) within 1 ≤ x ≤ 2 using Δθ = 90° and Δφ = 60°. θin = 35.576°, φin = 17.497°, a1 = 1, a2 = −0.1, a3 = 0.906, a4 = −0.183. (Minimization of the maximum error) ε

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Fig. 10.14 Four-bar mechanism and the error curve in generating function y = ln(x) within 1 ≤ x ≤ 2 using Δθ = 90° and Δφ = 60°. θin = 69.11°, φin = 45.08°, a1 = 1, a2 = 0.315, a3 = 1.224, a4 = 0.478. (Link lengths are made positive, Least squares minimization)

φin = 30°. Now we have 3 design parameters and we are no longer using Chebyshev spacing. As a result, we end up with a solution a1 = 1, a2 = −1.163, a3 = 0.696, a4 = −1.587 with εmax = 0.015. The mechanism and its structural error are shown in Fig. 10.15. Note that in the error curve the maximum and minimum values of the error are almost equal. The resulting maximum structural errors obtained in all the solutions are comparable with the manufacturing errors and tolerances that will occur during the construction of such systems. For example; normally if high precision is used, the link lengths

ε

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10.4 Path Generation Problems 0.004

5P

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Fig. 10.16 Comparision of the error curves obtained using five precision point synthesis (SP) and minimization of the maximum error (OP) for generating function y = sin x, 0° ≤ x ≤ 90° using a four-bar mechanism

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can be manufactured to at most ± 0.01 mm tolerance. Any structural error below this value will not be important. As another example consider y = sin x function generation within 0° ≤ x ≤ 90°. In Chap. 9 (Fig. 9.32) 5 design variable synthesis was performed using three different objectives. Using the results of 5 precision point and extreme point synthesis as an initial guess, Excel Solver was used to improve these solutions. Using Evolutionary algorithm in solver, when extreme point synthesis result was used as an initial guess no improvement was possible. When precision point synthesis result was used as an initial guess, a small improvement on the maximum error was possible as shown in Fig. 10.16. θs = 116.9904°, φs = 75.1051°, a1 = 1, a2 = 2.2465 a3 = 2.5444 and a4 = 0.7950 was obtained as a solution. The error for this optimum solution is εmax = yreq - ygen = 3.0399 × 10–03 , which is a slight improvement when compared with five precision point synthesis using Chebyshev spacing (εmax = 3.2027 × 10–03 ). Especially for cases where we have constraints on the link lengths, rather than obtaining a limited number of solutions using four or five design variables, use of an optimization routine starting from a three-design variable seems a better choice.

10.4 Path Generation Problems When the aim is coupler point path generation, one can either find some dimensions of a mechanism from the existing machine or from the literature or one can solve a three- or four-point path generation problem with prescribed timing to come up with a feasible solution. An initial estimate for the mechanism will be required. A swinging block mechanism to describe a straight line is given by Artobolevsky [3] is shown (Vol 2. Part1, P.377 Fig. 1356) in Fig. 10.17. For point E to generate an approximate straight line, link length ratios are given as: |AC| = 1.81 |AB| and |BD = 1.64 |AB|. The deviation of E from a vertical line ε = x − x (x = (xmax + xmin )/2) is shown (εmax = 0.0472 |AB|) within −45° ≤ θ ≤ +45°. There are two design variables |AC| and |BD|. The length of straight line is l = 3.1845|AB| (take |AB| = 1 unit).

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466 D B A

θ

C

The length of straight line is l = 3.1845|AB| (take |AB|=1 unit) (a)

(b)

Fig. 10.17 a Swinging block mechanism generating an approximate straight line8 . |AC| = 1.8, |BD| = 1.64. |AB| = 1 unit. b error curve as a function of vertical displacement y

When Excel Solver GRG nonlinear is used (with no constraints) and minimization of the maximum error is set as the objective, we obtain an optimum solution as |AC| = 2.1182 |AB| and |AD| = 2.7639 |AB|. εmax = 0.0317|AB| and the length of the straight line is l = 3.5858 |AB|. If we set limit on the lengths of AC and AD to be less than 3|AB|, the result is |AC| =2.1877|AB|, |AD| =3. εmax = 0.0301 |AB| and the length of the straight line is l = 4.0000 |AB|. This means GRG nonlinear algorithm without any constraint converged to a local minimum and stopped. When constraint on variables were imposed, solution converged to another best solution found within the set time limit. Using Evolutionary search algorithm, the problem was resolved and the result |AC| = 2.1877|AB|, |AD| =3|AB|. εmax = 0.0301 was found. In Fig. 10.18 The final error curve (solid line) and the error curve with original link lengths (dashed line) are shown. Consider four-bar mechanism to generate a straight line within a certain range. As an initial guess mechanism shown in Fig. 10.19a is assumed with link lengths A0 A = 100, AB = 50, |B0 B| = 40, AC| = 40, |A0 B0 | = 70 α1 = 40° and point A0 is located at A0 (xa ,ya ) where xa = 20 and ya = 10 units. As a limit for the link lengths let 0 ≤ |A0 A| ≤ 150, 0 ≤ |AB| ≤ 100, 0 ≤ |B0 B| ≤ 100, 0 ≤ |A0 B0 | ≤ 120, 0 ≤ |AC| ≤ 100, 0 ≤ xa ≤ 50, 0 ≤ ya ≤ 50, 0° ≤ α1 ≤ 90°. The deviation of point C from a straight-line y = 120 is as shown in Fig. 10.19 for a range 3° ≤ θ ≤ 51° There is only one precision point. The range of x is −19.7 ≤ x ≤ 84.1 (Δx = 103.79 units) units and maximum deviation from the line y = 120 is ε = 2.836 units (ε/Δx = 2.73%). Starting with above initial conditions, we determine the error ε = yc − 120 for every 1° within 3° ≤ θ ≤ 51°, square the errors and sum up (with the initial conditions given this sum is ∑εi 2 = 62.5115). Now, using GRG nonlinear algorithm (specifying the limits) we try to find an optimum by minimizing ∑εi 2 subject to all the constraints. The result is ∑εi 2 = 0.0616 and the design parameters are: xa = 20, ya = 20.7972, |A0 A| = 92.0851, |AB| = 23.1075, |B0 B| = 61.5255, |AC| = 44.3828, |A0 B0 | = 35.0782 and α1 = 49.9217°. The error curve is as shown in Fig. 10.20. Maximum

10.4 Path Generation Problems

467 0.06

ε

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Fig. 10.18 Comparison of the error curve for the original configuration (Fig. 10.17) and the swinging block mechanism with lengths: |AC| = 2.1877, |AD| = 3|, |AB| unit obtained using Evolutionary algorithm for the minimization of the maximum error

y 4

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Fig. 10.19 a Four-bar mechanism: A0 A = 100, AB = 50, |B0 B| = 40, AC| = 40, |A0 B0 | = 70 α1 = 40°, A0 (20, 10) and b deviation of the coupler curve C from y = 120. Δx = 103.79 units

error is εmax = 0.0606 units. However, within 3° ≤ θ ≤ 51°, Δx = 80.29 units. There is decrease in the range of the straight line. In order to keep Δx ≥ 100 units, keeping the resulting design parameter values the same and increasing the range of crank rotation as −1° ≤ θ ≤ 60°, The error curve is as shown in Fig. 10.21 (εmax = 1.1712 and ∑εi 2 = 3.0242). When solving the problem initially, if we had specified a constraint on the range of Δx such as 100 ≤ Δx ≤ 120, we could have satisfied the required range. One can also change the total angular rotation during the straight-line motion. Instead of using GRG algorithm, one can use Evolutionary algorithm as well. The problem shown in Fig. 10.19 was solved using GRG and Evolutionary algorithms with and without limits on Δx. Range 3° ≤ θ ≤ 51° was used and both least square optimization and minimax optimization were used. The same initial conditions and constraints were used for all cases (|A0 A| = 100, |AB| = 50, |B0 B| = 40, |AC| = 40, |A0 B0 | = 70 α1

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Fig. 10.20 Least square minimization of the deviation of the coupler curve from y = 120. xa = 20, ya = 20.7972, |A0 A| = 92.0851, |AB| = 23.1075, |B0 B| = 61.5255, |AC| = 44.3828, |A0 B0 | = 35.0782 and α1 = 49.9217°. 3° ≤ θ ≤ 51°, Δx = 80.29 units (Using GRG engine)

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Fig. 10.21 Error curve for the four-bar mechanism with link lengths as in Fig. 10.20. -1° ≤ θ ≤ 60°

y

= 40°, xa = 20 and ya = 10). The results are shown in Figs. 10.22, 10.23, 10.24, 10.25, 10.26, 10.27, 10.28 and 10.29. In this particular problem GRG algorithm seemed to converge to a better result (this conclusion is problem specific. In some other cases evolutionary algorithm may be a better choice.). When running, evolutionary algorithm takes more computer

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Fig. 10.22 (MiniMax method using GRG): xa =20, ya = 11.4515, |A0 A| = 104.4929, |AB| = 51.0625, |B0 B| = 52.0331, |AC| = 38.4907, |A0 B0 | = 68.0278 and α1 = 43.5306°. εmax =0.3024, Δx=95.8153 (3° ≤θ≤ 51°)

y

10.4 Path Generation Problems

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Fig. 10.23 (MiniMax method using GRG): xa = 20, ya = 10.0255, |A0 A| = 100.3564, |AB| = 49.2135, |B0 B| = 42.0346, |AC| = 40.0103, |A0 B0 | = 68.2488 and α1 = 42.4721°. εmax = 0.7109, Δx = 100 (3° ≤ θ ≤ 51°)

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Fig. 10.24 (Least squares using GRG): xa = 20, ya = 17.5496, |A0 A| = 92.5340, |AB| = 20.4216, |B0 B| = 65.7359, |AC| = 50.0053, |A0 B0 | = 92.5340 and α1 = 48.8047°. εmax = 0.0526, ∑εi 2 = 0.0184 Δx = 79.9091 (3° ≤ θ ≤ 51°)

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Fig. 10.25 (Least squares using GRG): xa = 20, ya = 0, |A0 A| = 114.5065, |AB| = 43.6050, |B0 B| = 55.2940, |AC| = 37.5006, |A0 B0 | = 67.8883 and α1 = 51.0166°. εmax = 0.2040, ∑εi 2 = 0.2207 Δx = 100 (3° ≤ θ ≤ 51°)

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Fig. 10.26 (MiniMax method using evolutionary): xa = 18.1769, ya = 11.8056, |A0 A| = 100.7705, |AB| = 48.6424, |B0 B| = 43.3351, |AC| = 37.0173, |A0 B0 |= 68.1695 and α1 = 44.1143°. εmax = 0.5311, Δx = 96.9022 (3° ≤ θ ≤ 51°)

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Fig. 10.27 (MiniMax method using evolutionary): xa = 25.1436, ya = 12.4778, |A0 A| = 101.6262, |AB| = 53.8000, |B0 B| = 45.9530, |AC| = 38.5600, |A0 B0 | = 71.3901and α1 = 40.9122°. εmax = 0.6822, Δx = 100.0012 (3° ≤ θ ≤ 51°)

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Fig. 10.28 (Least squares using evolutionary): xa = 31.5493, ya = 115.9500, |A0 A| = 99.5844, |AB| = 48.8762, |B0 B| = 47.1213, |AC| = 34.7921, |A0 B0 | = 65.9328 and α1 = 44.3919°. εmax = 0.5618, ∑εi 2 = 1.4223 Δx = 92.2426 (3° ≤ θ ≤ 51°)

y

10.4 Path Generation Problems

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1 0.8 0.6 0.4 0.2 0 -0.2 0 -0.4 -0.6

471

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Fig. 10.29 (Least squares using Evolutionary): xa = 28.7076, ya = 6.9629, |A0 A| = 103.5054, |AB| = 49.7678, |B0 B| = 49.6939, |AC| = 44.1009, |A0 B0 | = 64.7756 and α1 = 41.5675°. εmax = 0.8985, ∑εi 2 = 2.7710 Δx = 100 (3° ≤ θ ≤ 51°)

time than GRG algorithm. Also, the evolutionary algorithm may stop searching for a solution due to the maximum number of iterations, convergence, maximum time without improvement. Hence you may obtain different results for each run. In certain cases, if you think the result may be improved, reapply the solver with the previous result obtained as your initial guess. You can adjust certain parameters using “options” key in the solver. Since Excel Solver gives designer a very simple and useful tool, one can try different algorithms and solve the problem using different constraints and change optimization criteria in a very short time. One can then compare the results to come up with the best solution possible for a particular problem. Now, starting with the initial conditions (|A0 A| = 100, |AB| = 50, |B0 B| = 40, |AC| = 40, |A0 B0 | = 70 α1 = 40°, xa = 20 and ya = 10) besides the minimization of deviation of point C from a straight line (minimization of ε1i = yCi − 120), if we also want the displacement of C in x direction to be linearly related to the angular rotation of the crank angle (i.e. xC = aθ + b), then we can define another error function ε2j = xCj -(aθj + b). the coefficients of a and b determined applying linear regression. With the initial values a = −2.0733 b = 81.7947 and the error curve for ε2 is as shown in Fig. 10.30 (with the initial values, error curve ε1 is as shown in Fig. 10.17). In ∑ terms of least squares ∑ 2 error minimization, we have two terms to be minimized: 2 ε1 = ε1i and ε2 = ε2i . In order to minimize both, we can define a new objective function ε = k1 ε1 + k2 e2 , where k1 and k2 are the two weighing factors that we may arbitrarily select. In this particular case, let k1 = 10 and k2 = 1. Initially ε1 = 62.5115 and ε2 = 446.3448 and ε = 1071.46. When we use GRG nonlinear to minimize the objective function we obtain: |A0 A| =120.2969, |AB| =28.5626, |B0 B| = 76.1541, |AC| = 42.7422, |A0 B0 | = 52.6075, α1 = 57.2416°, xa = 20.0187 and ya = 0. The mechanism and the two error curves are shown in Fig. 10.31. As another example “Chebyshev Four-Bar Approximate Straight-Line Mechanism” given by I. I. Artobolevsky [3] (Vol 1, p. 436 Fig. 657) will be considered (Fig. 10.32). The link dimensions are given as: |A0 B0 | = 2.16 |A0 A|, |AB| = |B0 B| = |BC| = 4.34 |A0 A|, < ABC = β = 100° and < xB0 A0 = α = 40° (CW).

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

472

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Fig. 10.30 Error curve ε2 = x − (2.0733θ + 81.7947) as a function of coupler curve horizontal displacement for the four-bar mechanism given in Fig. 10.17 140

Fig. 10.31 a Four-bar mechanism: |A0 A| = 120.2969, |AB| =28.5626, |B0 B| = 76.1541, |AC| = 42.7422, |A0 B0 | = 52.6075, α1 = 57.2416°, xa = 20.0187 and ya = 0 obtained using Solver GRG engine for the minimization∑ of the 2+ function: ε = 10 ε1i ∑ 2 ε2i . b deviation of the coupler curve (ε1 ) from the horizontal line y = 120. c deviation of the x displacement (ε2 ) from a linear relation with the crank rotation θ

120

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10.4 Path Generation Problems Fig. 10.32 Chebyshev four-bar approximate straight line mechanism8 .; |A0 B0 | = 2.16 |A0 A|, |AB| = |B0 B| = |BC| = 4.34 |A0 A|, < ABC = β = 100° and < xB0 A0 = α = 40° (CW)

473 y C B β

x B0

α

A

θ

A0

ε

Mechanism generates a symmetric coupler curve and within crank rotation − 100° ≤ θ ≤ 20° The variation of yc from a straight-line y = 21 (yCmax + yCmin ) = 4.2015 is ε = yC − y is shown in Fig. 10.33. Maximum error is 0.0568 and S = ∑εi 2 = 0.181228 (|A0 A| = 1). The length of the approximate straight line traced in this interval is l = 2.78|A0 A|. Coupler curve is symmetric and the symmetry point is when the crank and the fixed link are collinear. The straight-line portion is traced when the crank rotates ± 60° from the symmetry position. As objective we can minimize the maximum error ε or minimize the sum of the squares of the errors within the range −90° ≤ θ ≤ +20°. Using the evolutionary method in Excel solver, the deviation of the coupler path from a straight line when the crank rotates −90° ≤ θ ≤ +20° was tried to be optimized in least squares and minimax sense. In Figs. 10.34 and 10.35 disregarding symmetry, we select |AB|, |BC|, |B0 B|, |A0 B0 |, β and α as the six design parameters (|A0 A| = 1). In Fig. 10.36 and 10.37, we select |AB| = |BC| = |BB0 | thus reducing the design parameters to four. As constraint: 0 ≤ |A0 B0 | ≤ 3, 0 ≤ |AB| ≤ 6, 0 ≤ |BB0 | ≤ 6, 0 ≤ |BC| ≤ 6, 0 ≤

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Fig. 10.33 Error curve for the original four-bar mechanism

1

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10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

474

ε

β ≤ 130° and 0 ≤ α ≤ 60° is used. For comparison, εmax and ∑εi 2 values are shown regardless whether the problem is optimized in minimax or least squares sense. When we use GRG algorithm with the initial conditions |A0 A| = 1, |A0 B0 | = 2.16, |AB| = |B0 B| = |BC| = 4.34, β =100° and α =40°, the program stops with the initial values. You have a message “All constraints and the optimality conditions are satisfied”, since the initial condition is a local minimum. To overcome this critical position, the initial conditions are changed to: |A0 A| = 1, |A0 B0 | =2.1, |AB| =4.3, |B0 B| = 4.2, |BC| = 4.35, β =103° and α =40°. Initially, ∑εi 2 = 0.7400, εmax =0.1032. Again, the problem is solved for sum of the squares of the errors (least squares) and minimization of the maximum (minimax) using 6 or 4 design variables. The results are as shown in Figs. 10.38, 10.39, 10.40 and 10.41. The results are all comparable. All of the solutions show considerable improvement over the initial mechanism in terms of straight-line generation. Of course, one can impose additional constraints such as the location of the straight-line with respect

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Fig. 10.34 (Least squares using evolutionary algorithm.) |A0 A| = 1, |A0 B0 | = 1.9012, |AB| = 4.4702, |B0 B| = 4.4524, |BC| = 4.4935, β = 103.3543°, α = 39.0762°, ∑εi 2 = 1.35 × 10–5 , εmax = 0.0006, l = 3.486, = y 4.8339

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Fig. 10.35 (MiniMax using evolutionary algorithm) |A0 A| = 1, |A0 B0 | = 1.6150, |AB| = 43.6737, |B0 B| = 3.9638, |BC| = 2.6179, β = 92.7702°, α = 15.8237°, ∑εi 2 = 0.0005, εmax = 0.0030, l = 2.514, = y 3.1708

10.4 Path Generation Problems

475

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Fig. 10.36 (Least squares using evolutionary algorithm) |A0 A| = 1, |A0 B0 | = 1.7605, |AB| = |B0 B| = |BC| = 4.4330, β = 94.7541°, α = 42.6420°, ∑εi 2 = 6.11 × 10–5 , εmax = 0.0013, l = 3.3175, = y 4.3311

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Fig. 10.37 (MiniMax using evolutionary algorithm. |A0 A| = 1, |A0 B0 | = 1.7773, |AB| = |B0 B| = |BC| = 4.4024, β = 96.4176°, α = 41.67854°, ∑εi 2 = 6.13 × 10–5 , εmax = 0.0012, l = 3.3302, = y 4.3311

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Fig. 10.38 (Least squares using GRG algorithm.) |A0 A| = 1, |A0 B0 | = 1.9109, |AB| = 4.3423, |B0 B| = 4.2618, |BC| = 4.5101, β = 107.4012°, α = 40.2130°, ∑εi 2 = 1.395 × 10–5 , εmax = 0.0007, l = 3.5211, y=4.9016

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

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Fig. 10.39 (MiniMax using GRG algorithm) |A0 A| = 1, |A0 B0 | = 1.69552, |AB| = 4.5895, |B0 B| = 4.5123, |BC| = 4.8118, β = 104.651°, α = 41.8433°, ∑εi 2 = 1.15E−05, εmax = 0.0004, l = 3.6259, y=5.1395

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Fig. 10.40 (Least squares using GRG algorithm) |A0 A| = 1, |A0 B0 | = 1.8670, |AB| = |B0 B| = |BC| = 4.6067, β = 98.1475°, α = 40.9281°, ∑εi 2 = 1.7 × 10–5 , εmax = 0.0006, l = 3.4666, y=4.7380

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Fig. 10.41 (MiniMax using GRG algorithm). |A0 A| = 1, |A0 B0 | = 1.9875, |AB| = |B0 B| = |BC| = 4.45951, β = 106.1041°, α = 36.9479°, ∑εi 2 = 2.28E−06, εmax = 0.0002, l = 3.5853, y=5.1502

10.4 Path Generation Problems

477

to reference frame (y), limit on the transmission angle (μ = ∠ ABB0) and the length of straight line generated. For example, example let us select α = 0° and determine a mechanism such that the length of the straight line generated is greater than 4 (in order to give an upper bound, let 4 ≤ l ≤ 4.2) and the straight line to be approximated is y = 4. In this case, the range for the crank rotation is selected as: −90° ≤ θ ≤ + 90°. Error curve and the resulting mechanism is shown in Fig. 10.42. As another example, let us consider a six-link dwell mechanism given in reference [3] (volume 1, p. 504, Fig. 748) and shown in Fig. 10.43. Link lengths are given as |BC| = 4.22 |AB|, |DC| = |GF| = |EM| = 3|AB|, |EF| = 2.33 |AB|, |AD| = 3 |AB|, |GD| = 5.44 |AB|, |BM| = |MC| and |AG| = 2.4 |AB| (The labeling of the reference is used). 5

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Fig. 10.42 (Least squares, GRG algorithm l ≥ 4) |A0 A| =1, |A0 B0 | = 2.0209, |AB| = 4.0963, |B0 B| = 4.8669, |BC| = 2.6356, β = 93.4164°, α = 0.0000°, ∑εi 2 = 0.0008, εmax = 0.0050, l = 4.000, y=4. Mechanism and the error curve Fig. 10.43 6-link mechanism dwell from reference [3]

478

10 Optimization Methods in the Synthesis of Mechanisms Using Excel® 6 4

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Fig. 10.44 Motion curve and the angular motion of the output during dwell period for the six link mechanism given in reference [3] (4 )

Noting the three points ADG on the fixed link forms a triangle and |AG|+|AD| > |GD|. However, with the given values |AG|+|AD| = 5.4 |AB| < 5.44 |AB|. To eliminate this conflict, |AG| = 2.5 |AB| is selected (In such a case the coordinates of G is (−2.349, 0.731) with respect to a coordinate frame with A as the origin). The angular rotation of the output link GF with respect to input crank rotation and the angular motion of the output link during 50° < θ < 250° is shown in Fig. 10.44, During the dwell period the output link rotates εmax = ± 5.26°. Total angular oscillation of the output link is approximately 31.8°. The aim is to minimize the amount of rotation of the output link when the input rotates within 50° ≤ θ ≤ 250°, while keeping the total output oscillation 32°. Again, we can apply minimization of the maximum error or least squares minimization. 8 design Variables [Lengths of |AD|, |BC|, |DC|, |EM|, |GF|, |EF| and coordinates of point G (Gx, Gy)] are to be used. As constraints we can require all the link lengths to be positive and letting |AB| = 1 unit, we apply upper limits |AD| ≤ 5, |BC| ≤6, |DC| ≤5, |ME| ≤ 5, |GF| ≤5, |EF| ≤ 5. Gx ≥ −5 (negative value of Gx is taken as the parameter to change this constraint to an upper bound) and Gy ≤ 3. Besides the constraints on the eight design variables, the total angular oscillation of the output to be greater than 30° is set as another constraint. Using Solver Evolutionary algorithm, we obtain a result shown in Figs. 10.45. The link lengths are: |AB| = 1, |BC| = 4.11, |DC| = 3.05, |GF| = 2.95, |EM| =3.00, |EF| = 2.31, |AD| =2.82, G(−2.32, 0.74). (|AG| = 2.44, |GD| = 5.19). During the dwell period the output link oscillates εmax = ± 1.231° within 200° crank rotation! (Output link dwells when the angular position of the output is at 34.1739°). If we want the dwell period for half the crank rotation i.e., dwell between 60° ≤ θ ≤ 240°, and 50° oscillation of the output link within the remaining crank rotation, Using Evolutionary algorithm in Excel Solver and minimizing ε in least squares, we obtain a result whose output and error curves are given in Fig. 10.46. Another solution for 90° dwell (60° ≤ θ ≤ 240°) period and 50° oscillation of the output is shown in Fig. 10.47. Minimax method and Evolutionary algorithm was used.

10.4 Path Generation Problems

479

Fig. 10.45 Motion curve and the angular motion of the output during dwell period for the mechanism optimized in least squares sense using Solver, Evolutionary algorithm 20

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Fig. 10.46 Motion curve and the angular rotation of the output during dwell.|AB| = 1, |BC| = 3.9154, |DC| = 2.7641, |GF| = 2.3377, |EM| = 3.1045, |EF| = 2.6100, |AD| =2.6734, G (−1.8726, 0.1.4421). (|AG| = 2.3635, |GD| = 4.4793), εmax = 0.8812°, ∑εi 2 = 36.7998 (dwell period = 180°, 50° oscillation of the output. Output link dwells at 14.0547°) 0 0

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Fig. 10.47 Motion curve and the angular rotation of the output during dwell. |AB| = 1, |BC| = 3.9390, |DC| = 3.0972, |GF| = 3.2380, |EM| = 2.5595, |EF| = 2.6480, |AD| =2.3525, G (−2.8018, 1.5952). (|AG| = 3.2241, |GD| = 5.3946), εmax = 0.0705°, ∑εi 2 = 0.2166 (90° dwell, 50° oscillation of the output. Output link dwells at −0.9089°)

10.4.1 Ackermann Steering When horse drawn carts were used, people have realized if all four-wheel axes are fixed, the cart cannot take an easy turn. A remedy was to pivot the front axle at the midpoint as shown in Fig. 10.48 or have carts with two wheels. Even with the front

480

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

Fig. 10.48 Cart with front axle pivoted at the midpoint

axle midpoint pivoting, when a four wheeled cart takes a relatively fast turn, tipping of the cart is inevitable. Although the name implies Rudolph Ackermann as the inventor, it is the German carriage builder Georg Lankensperger who invented in 1817 the system known as “Ackermann Steering” today2 (Erasmus Darwin is also claimed to be the first inventor in 1758). Ackermann steering simply states that when a vehicle is taking a turn at any angle, in order to reduce the slippage, the axes of all the wheels must intersect at one point (Fig. 10.49). The angular relation between the rotation of the two wheels is given by: tan φ =

1 tan θ

1 or cot φ = R + cot θ +R

Of course, a simple mechanism cannot realize the above function exactly. In nineteenth century carriage manufacturers found that this requirement can be approximately satisfied by a four-bar mechanism which when the wheels are straight, extension of the cranks A0 A and B0 B intersect at the midpoint of the rear axle (Fig. 10.50). If we let a1 = 1 unit and R = a1 /L for this approximate design (|A0 A| = |B0 B| = a, |AB| = b): 2

https://en.wikipedia.org/wiki/Ackermann_steering_geometry.

10.4 Path Generation Problems

481 φ

L

θ

φ

θ

a1 1

Fig. 10.49 Ackerman steering rule Fig. 10.50 Approximate realization of Ackerman steering in early years

Α0

α

α

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Q

a1   2 and b = 1 − 2a cos α α = tan R −1

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

Fig. 10.51 Symmetric four-bar mechanism used in Ackerman steering

Α0 Α

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482

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Today, rather than the classical approximation, a four-bar mechanism design for function generation is used [4–7]. However, the four-bar mechanism to be used must be symmetric. Therefore, |A0 A| = |B0 B| = a and b = 1 − 2acosα Instead of three design parameters, one is left with a single link length, a, and the symmetry position must be preserved, i.e. one precision point is when θ = φ = 0 (Fig. 10.51). Let us assume a = 0.16 (note that a1 = 1 unit. If a1 = 1200 mm then a = 192 mm) and α = 60° for a ratio of R = a1 /L = 14/23 = 0.60869 (If α was determined in accordance to the classical approximation as shown in Fig. 10.51: α = 73.072°). b = 1 − 2acosα = 0.88354. Although it will depend on the type of vehicle to be designed, let 0° < θ < 40°. With the assumed values above, the mechanism can be analyzed, the required relation between φ and θ for Ackermann steering and the relation between φ and θ that is generated by the four-bar mechanism with the given dimensions can be determined (φreq , φgen ). Also, the error ε = φreq - φgen is evaluated for every input angle as shown in Fig. 10.52 (εmax = 4.8486°). We have two design parameters a and α. Let us assume 0.1 ≤ a ≤ 0.2 and 20° ≤ α ≤ 90° as constraints imposed on the design parameters. The amount of rotation of the steering angle depends on the vehicle for which it is designed. For the examples shown, 40° maximum steering angle will be used. As an objective, if we want to minimize the maximum error within the interval (mini-max method) we determine εmax using the tabular data and using Solver tool in Excel and selecting GRG nonlinear algorithm, we set the objective εmax to be minimized by changing a and α subject to the given constraints on a and α. The solution is a = 0.15154, α = 68.2805°, b = 0.88785 with εmax = 0.583610°. The result is as shown in Fig. 10.53. Note that the minimum and maximum deviations are equal. If we use the Evolutionary algorithm, we end up with the same result. However, the time it takes for a solution is comparatively longer than GRG algorithm.

10.4 Path Generation Problems

483 5

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Fig. 10.52 Required and generated wheel rotation, φ, in terms of the other wheel angular rotation, θ, and the difference between these values (error = ε)

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Fig. 10.53 Error curve ε as a function of angular rotation of a wheel obtained for the minimization of maximum error using Solver GRG nonlinear. a = 0.15154, α = 68.2805°, b = 0.88785 with εmax = 0.583610°, a1 = 1 unit

Instead of minimizing the maximum error within the interval, we can square the error for every degree of crank angle, sum these errors squared and take the square root. Now the objective function to be minimized is: F=

n ∑

(φreq − φgen )2

i=1

Again, using GRG algorithm in Solver and again starting with the previous initial conditions this time we try to minimize the error in the least square sense. The result is as shown in Fig. 10.54 with a = 0.1, α = 66.6316°, b = 0.9207 with εmax = 0.845°. (the value a is at its limit) Maximum and minimum values of the error are not equal. For vehicles such as fork-lift, the error throughout the range is important. However, for high-speed vehicles, large angular rotation of the front wheels is used only when the vehicle is moving slowly during parking or passing through a narrow lane. For this reason, error for small steering angles are considered to be more important than

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

error ( 0)

484

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Fig. 10.54 Error curve ε as a function of angular rotation of a wheel obtained for the minimization of maximum error using Solver GRG nonlinear. a = 0.1, α = 66.6316°, b = 0.9207 with εmax = 0.845°, a1 = 1 unit

the error for large steering angles. In such a case, we can multiply the error for each point with a linear function, known as the “weighing function”, w = (1 − i/n), where n is the number of data points. Note that for i = 0: w = 1 and for i = n: w = 0. The objective function to be minimized is: F=

n ∑ i=1

i (1 − )(φreq − φgen )2 n

The result is as shown in Figs. 10.55. The mechanism parameters are: a = 0.1, α = 65.2696°, b = 0.91633 with εmax = 1.4666°. Although the maximum error at 40° steering angle has increased, up to 31° the maximum error is less than 0.254°. If we want to decrease the error for lower steering angles more, one can use a second order weighing function w = (1 - i/n)2 . The error curve is as shown in Fig. 10.56 and a = 0.1, α = 64.427°, b = 0.91367 with εmax = 1.8509°. However, up to 27° the maximum error is less than 0.161°. One can also neglect the total angular rotation and determine the sum of the squares of the errors, say up to 30° steering angles, perform the optimization. As you can see, the solution depends on the objective function and the initial conditions. The selection of the objective function and initial condition may depend on the application and on the intuition, experience and judgement of the designer. Another mechanism structure that is used to approximate the Ackermann steering is to use two symmetric slider-crank linkages. This is known as “rack and pinion steering mechanism” where a rack is fixed to the slide and is driven by a pinion (Fig. 10.57). The mechanism can be used in two different configurations named as “leading” (Fig. 10.58a) or “trailing” (Fig. 10.58b).

error ( 0)

10.4 Path Generation Problems

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Fig. 10.55 Ackerman steering error optimized for the linear weighed sum of the squares of the errors using Solver GRG nonlinear

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Fig. 10.56 Ackerman steering error optimized for the second order weighed sum of the squares of the errors using Solver GRG nonlinear

Note that depending on the vehicle dimensions and its type, steering system must be designed for that particular vehicle. One can find quiet a diverse literature about this topic [8–10]. The limits for the variable dimensions are: 0.14 ≤ a ≤ 0.2, a ≤ b ≤ 0.4, −0.1 ≤ c ≤ 0.1 and α ≤ 90°. We next assume a set of initial guess values for the design parameters a, b, c and α, perform kinematic analysis of the mechanism for every degree of the input wheel rotation (If required, one can perform kinematic analysis for every tenth of a degree) and determine the amount of rotation of the outer wheel. This is the function generated by the mechanism (φgen ), the required function is determined from the

486

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

Fig. 10.57 Rack and pinion steering mechanism

Ackermann steering rule (φreq ). The difference is the error function ε. Again, we can (A) minimize the maximum error, (B) minimize the sum of the errors squared, (C) minimize the weighted sum of the errors squared. Since we have leading and trailing configurations, we can perform this analysis for both cases. R = a1 /L = 14/ 23 = 0.6087 is assumed for all the problems. When leading configuration is to be designed and depending on the initial conditions, we may have different solutions. The solution that has the best minimum error characteristics will be given here. Of course, one can obtain a better solution. For the minimization of the maximum error, we start with the initial condition: a = b = 0.2, c = 0.05, α = 90° (εmax = 7.1080°. ∑εi 2 = 512.1155, Fig. 10.59a). when we minimize the maximum error, we have a = 0.1816, b = 0.1816, c = 0.0909 and a = 90° (εmax = 0.09039°). The error curve is as shown in Fig. 10.58b. When the sum of the squares of the errors are minimized, starting with the same initial conditions: a = b = 0.2 c = 0.0999, α = 90° and εmax = 0.1270°, ∑εi 2 = 0.1299 (Fig. 10.58c). Note the error is less than 0.08 up to θ = 36°. And when weighed least square error is minimized, a = 0.1407, b = 0.1407, c = 0.0697, α = 90°, εmax = 0.2296° and error is less than 0.05° up to 32° steer angle (Fig. 10.58d).

10.4 Path Generation Problems Fig. 10.58 Rack and Pinion steering mechanism (leading configuration) a Initial error curve, optimized for b Minimization of maximum errors, c minimization of the sum of the squares of the errors, d minimization of the weighed sum of the squares of the errors

487 Initial Error curve

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For trailing configuration, starting with initial conditions: a = b = 0.2, c = −0.05 and α = 90° when minimization of maximum is used as the optimization criteria: a = 0.14, b = 0.14, c = −0.0648 and α = 61.2386° is obtained for εmax = 1.0159° (Fig. 10.59a). when we use the evolutionary algorithm a = 0.14, b = 0.4, c = − 0.0240 and α = 58.7763° is obtained with εmax = 0.70844° (Fig. 10.59b). When we start with the initial condition: a = b = 0.2, c = −0.05 and α = 90° and apply GRG algorithm to minimize the sum of the square of the errors, solution diverged. However, starting with the result obtained when minimization of the maximum error (Fig. 10.59b), using GRG algorithm for the minimization of the

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

Fig. 10.59 Rack and Pinion steering mechanism (trailing configuration) a minimization of maximum error using GRG nonlinear algorithm (εmax = 1.0159°). b Minimization of maximum error using evolutionary algorithm (εmax = 0.70844°.)

Minimization of the maximum error

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sum of the square of the errors, a solution shown in Fig. 10.60 is obtained. Repetitive application of optimization using different objective functions (and different algorithms) seems to yield good results. In this chapter several examples in mechanism optimization using Excel Solver are shown. From these examples it can be seen that with little effort a mechanism that is analyzed in Excel can be very easily optimized according to certain design criteria

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Fig. 10.60 Optimization using GRG algorithm for the minimization of the maximum error using the solution of minimization of maximum error using evolutionary algorithm as the initial condition. a = 0.2, b = 0.2, c = −0.0010 and α = 60.6843° with εmax = 0.1915°

10.4 Path Generation Problems

489

while satisfying the necessary design constraints. Depending on the problem, one can use different number of design variables and impose different constraints. As long as one makes an initial guess and performs the analysis of the mechanism, one can use Excel solver to a good effect to improve the initial guess made in accordance to the criteria with which you value the mechanism. Another important point is that the result to a problem will depend on your computer settings and on options settings on both Excel and Solver tool. Therefore, one should not expect to find exactly the same result given in this text for the same problem. If the problem is set correctly, an improvement from the initial conditions and results comparable to the ones given in this text can be expected. Please also note that in great number of problems proper selection of the constraints and objective function is the main task. Problems 1. In the literature there are lots of examples where a slider-crank coupler point is used to generate a straight line. “De Jonge Slider-Crank Approximate StraightLine Mechanism” [3]4 (vol 2, part 1, p 512, Fig. 1548) has link length proportions A0 A = 1 unit, AB = 1.52, BC = 3.7, BC = 2.26 and eccentricity a = 0.61 units as shown in Fig. 10.61. A slider-crank mechanism coupler point describes an exact straight line when a = 0, A0 A = AB and point C is a point on the circle with center A, radius A0 A. The straight lines pass through A0 . This is a well-known Cardan Mechanism (any other point on the coupler link describes an ellipse). (a) Within −45° ≤ θ ≤ 45°, De Jonge slider crank approximates a straight line x = −1.207AB with maximum error ± 0.0854AB. And the length of straight line is l = 3.7939AB. Considering AB, AC, BC and a as design variables, minimize maximum deviation of the coupler path from a vertical straight line, subject to constraints: 0 ≤ AB ≤3, 0 ≤AC ≤ 4, 0 ≤BC ≤ 6, 0 ≤a ≤ 1 and b = 1. using Excel Solver Evolutionary algorithm.

C

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Fig. 10.61 De Jonge slider-crank approximate straight-line mechanism

θ B

10 Optimization Methods in the Synthesis of Mechanisms Using Excel®

490

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Fig. 10.62 Evans-De Jonge four-bar approximate straight-line mechanism

(b) Let θin ≤ θ ≤ θin +90°. Starting with De Jonge Slider Crank, consider AB, AC, BC, a and θin as the design variables. Use the constraints given in (a) and take −90° ≤ θin ≤ 0°. (c) Repeat (a) and (b) with the condition that the straight linet o be generated is greater than 4AB. 2. Four-bar mechanism given by Artobolevsky [3]4 (volume 1, p.447, Fig. 678) “Evans-De Jonge four-Bar approximate straight-line mechanism” has link lengths AB = 1 unit, AD = 3.1, BC = 1.55, DC = BE = 1.9, CE = 3.45. The mechanism is shown in Fig. 10.62. Point E decribes a straight line when crank AB rotates in between −30° ≤ θ ≤50° (measured from x axis). Maximum error is ± 0.0231AB Using Excel Solver (keep AB = 1). (using GRG Nonlinear, maximum error was reduced to ± 0.0011AB) 3. The mechanism shown is given by Artobolevsky [3]8 (Vol. 1, page 509, Fig. 753) as “Multiple Bar Two-Dwell Mechanism”. When the coupler point E traces approximate circular arcs, the output link GF will have approximate dwell (Fig. 10.63) Link lengths are given as: AB = 1 unit, BC = 2.52, DC = 1.44, BE = 4.44, CE = 4.23, EF = 7.45, GF = 2.52, AD = 2.35, AG = 2.77 and DG = 2.48. When the crank angle