Mechanism Analysis Simplified Graphical And Analytical Techniqu.pdf

MECHANICAL ENGINEERING/32

MCCHANISM ANALYSIS SIMPLIFIED GRAPHICAL AND ANALYTICAL TECHNIQUES

LXNDON O. BARTON

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MECHANISM ANALYSIS

MECHANICAL ENGINEERING A Series of Textbooks and Reference Books EDITORS

L. L. FAULKNER

S. B. MENKES

Department of Mechanical Engineering The Ohio State University Columbus, Ohio

Department of Mechanical Engineering The City College of the City University of New York New York, New York

1.

Spring Designer’s Handbook, by Harold Carlson

2.

Computer-Aided Graphics and Design, by Daniel L. Ryan

3.

Lubrication Fundamentals, by J. George Wills

4.

Solar Engineering for Domestic Buildings, by William

A. Himmelman 5.

Applied Engineering Mechanics: Statics and Dynamics,

6.

by G. Boothroyd and C. Poli Centrifugal Pump Clinic, by Igor J. Karassik

1.

Computer-Aided Kinetics for Machine Design, by

Daniel L. Ryan 8.

Plastics Products Design Handbook, Part A: Materials and Components; Part B: Processes and Design for Processes, edited by Edward Miller

9.

Turbomachinery: Basic Theory and Applications, by

Earl Logan, Jr. 10. 11.

Vibrations of Shells and Plates, by Werner Soedel Flat and Corrugated Diaphragm Design Handbook, by Mario

Di Giovanni 12.

Practical Stress Analysis in Engineering Design, by Alexander Blake

13.

An Introduction to the Design and Behavior of Bolted Joints, by

John H. Bickford 14.

Optimal Engineering Design: Principles and Applications,

by James N. Siddall 15. 16.

Spring Manufacturing Handbook, by Harold Carlson Industrial Noise Control: Fundamentals and Applications,

edited by Lewis H. Bell

17.

Gears and Their Vibration: A Basic Approach to Understanding

18.

Chains for Power Transmission and Material Handling: Design

19.

Corrosion and Corrosion Protection Handbook, edited by

Gear Noise, by J. Derek Smith and Applications Handbook, by the American Chain Association

Philip A. Schweitzer 20.

Gear Drive Systems: Design and Application, by Peter Lynwander

21.

Controlling In-Plant Airborne Contaminants: Systems Design and

22.

Calculations, by John D. Constance CAD/CAM Systems Planning and Implementation, by Charles S. Knox

23.

Probabilistic Engineering Design: Principles and Applications,

by James N. Siddall 24.

Traction Drives: Selection and Application, by Frederick W. Heilich III

and Eugene E. Shube 25.

Finite Element Methods: An Introduction, by Ronald L. Huston

and Chris E. Passerello 26.

Mechanical Fastening of Plastics: An Engineering Handbook, by

29.

Bray ton Lincoln, Kenneth J. Gomes, and James F. Braden Lubrication in Practice, Second Edition, edited by W. S. Robertson Principles of Automated drafting, by Daniel L. Ryan Practical Seal Design, edited by Leonard J. Martini

30.

Engineering Documentation for CAD/CAM Applications,

27. 28.

by Charles S. Knox 31.

Design Dimensioning with Computer Graphics Applications, by Jerome C.

Lange 32.

Mechanism Analysis: Simplified Graphical and Analytical Techniques,

by Lyndon O. Barton

OTHER VOLUMES IN PREPARATION

MECHANISM ANALYSIS Simplified Graphical and Analytical Techniques

Lyndon O. Barton Project Engineer E.I. du Pont de Nemours & Co. Wilmington, Delaware

MARCEL DEKKER, INC.

New York and Basel

Library of Congress Cataloging in Publication Data

Barton, Lyndon O., [date] Mechanism analysis. (Mechanical engineering ; 32) Bibliography: p. Includes index. 1. Machinery, Kinematics of. II. Series. TJ175.B29 1984 ISBN 0-8247-7086-2

I. Title.

621.8'll

COPYRIGHT © 1984 by MARCEL DEKKER, INC.

ALL RIGHTS RESERVED

Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER, INC. 270 Madison Avenue, New York, New York 10016 Current printing (last digit): 10

98765432

PRINTED IN THE UNITED STATES OF AMERICA

To my wife Olive, my children Rhonda, Loren, Carol, and Leon, and my mother Clarice.

Preface

This book is written primarily for mechanical design engineers and mechan¬ ical design engineering students who are concerned with the design of ma¬ chines, in general, and in particular with mechanism analysis, a subject which forms a principal part of the study of kinematics of mechanisms. The principal aim of this volume is to place at the disposal of the reader a prac¬ tical book that will serve (1) as a handy reference for simplified approaches to problems typically encountered in the analysis and synthesis of a mech¬ anism, and (2) as a supplementary textbook for independent study or quick review of both principles and applications in mechanism analysis. The book presents a wide assortment of graphical and analytical techniques, as well as complete listings in FORTRAN of computer programs and programmable calculator programs for the Hewlett Packard HP-41C for analysis of basic classes of mechanisms. Special emphasis has been given to relatively simple kinematic chains such as slider-crank, four-bar, quick-return, and sliding coupler mechanisms. These mechanisms have been selected because they form the basic elements of most machines and because they are easily adaptable to the teaching of fundamental kinematic principles. Once these principles are fully understood, it is comparatively easy to apply this knowledge to the analysis of more complex mechanisms. Several novel approaches for simplification of the analytic process are presented. These include the rectilinear and angular motion diagrams presented in Chapter 2, the link extension concept for velocity analysis by instant centers in Chapter 5, the generalized procedure for constructing the acceleration polygon in Chapter 9, the Parallelogram Method for slidercrank analysis in Chapter 12, and the Simplified Vector Method and modified version of same in Chapters 14-18. One important feature is that the Simplified Vector Method, unlike conventional methods which rely on calculus and other forms of sophisticated mathematics, relies mainly on basic algebra and tiigonometiy to obtain an analytical solution. This simplified mathematical approach has made it

v

vi

Preface

possible to include several analytical problem solutions rarely found in kinematic textbooks. Hopefully, this approach will not only make this mate¬ rial accessible to a wide body of readers, but will also help to provide the quick insight often needed by designers in the analysis of a linkage. The book is written for easy readability and comprehension, without reliance on any other source. Needed background material on topics such as Uniformly Accelerated Motion (Chapter 2), Properties of Vectors (Chap¬ ter 3), Complex Algebra (Chapter 13), and Trigonometry (Appendix B) is provided for review. Concepts are presented as concisely as possible, employing numerous illustrative examples and graphical aids, as well as step-by-step procedures for most graphical constructions. In addition, the topics are arranged in a logical sequence corresponding to that ordinarily followed in teaching a course in kinematics. Some of the material in this book is based on several technical papers which the author has previously published (see References). Much of the material has been drawn from class notes which have been developed and used over several years of teaching kinematics of mechanisms as part of an Engineering Technology college curriculum. The author gratefully acknowledges his indebtedness to the E. I. DuPont de Nemours and Company Engineering Department and the Delaware Technical Community College Mechanical Engineering Department for pro¬ viding the engineering and teaching opportunities, respectively, that have enabled him to pursue and accumulate the knowledge and experience that form the basis of this book. Grateful acknowledgments and appreciation are also extended to •

Penton Publishing Company, publishers of Machine Design magazine, for permission to reprint portions of previously published articles (including illustrations);

•

American Society for Engineering Education, publishers of Engineering Design Graphics journal for permission to reprint portions of previously published articles (including illustrations);

•

Mr. Albert A. Stewart, for the valuable assistance he has rendered in the calculator program development;

•

Teachers, relatives, and friends who have been a source of inspiration and encouragement in the author's career; and

•

A devoted family, for their love, understanding, and support, always. Lyndon O. Barton

Contents

PREFACE Part I.

v

INTRODUCTORY CONCEPTS 1.

KINEMATIC TERMINOLOGY

1 3

Mechanical Concepts Motion Classification Motion Characteristics

3 9 13

UNIFORMLY ACCELERATED MOTION

16

Rectilinear Motion Angular Motion Conversion Between Angular and Rectilinear Motion Velocity-Time Graph Solutions Summary of Motion Formulas

16 21

1.1 1.2 1.3

2.1 2.2 2.3 2.4 2.5

41

VECTORS 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

27 33 40

Properties of Vectors Vector Addition Vector Subtraction The Vector Polygon Vector Resolution Orthogonal Components Translational and Rotational Components Effective Components

vii

41 41 45 47 48 51 52 53

viii

Contents

Part H.

GRAPHICAL TECHNIQUES II. A.

4.

GRAPHICAL TECHNIQUES: VELOCITY ANALYSIS EFFECTIVE COMPONENT OF VELOCITY METHOD 4.1 4.2 4.3 4.4 4.5

Introduction The Rigid Body Principle Velocities of End Points on a Link Velocities of Points on a Rotating Body Velocity of Any Point on a Link

4.6 4.7 4.8 4.9

Velocity Analysis of a Simple Mechanism Velocities of Sliding Contact Mechanisms Velocity Analysis of a Compound Mechanism Summary

INSTANT CENTER METHOD 5.1 5.2 5.3 5.4 5.5 5. 6 5.7 5.8 5.9 5.10 5.11

6.7 6.8 6.9 6.10 6.11 6.12 6.13

59 60 60 60 60 61 63 65 68 72 74 75

Introduction Pure Rotation of a Rigid Body Combined Motion of a Rigid Body

75 76 81

Velocity of a Body with Rolling Contact Types of Instant Centers Locating Instant Centers

83 84 85 92 92 95 97 99

Velocity Properties of the Instant Axis Velocity Analysis by Instant Centers Velocity Analysis of a Simple Mechanism Velocity Analysis of a Compound Mechanism Summary

RELATIVE VELOCITY METHOD 6.1 6.2 6.3 6.4 6.5 6.6

57

Introduction Relative Motion Concept The Velocity Polygon The Velocity Polygon Convention Velocity Polygon: Linkage Application Relative Velocity of Two Points on a Rigid Body

101 101 101 103 104 104

Velocities of End Points on a Floating Link Velocity of Any Point on a Link

105 106 108

Velocity of Any Point on an Expanded Link Velocity Analysis of a Simple Mechanism Velocities of Sliding Contact Mechanisms Velocities of a Body with Rolling Contact Velocity Analysis of a Compound Mechanism

109 111 113 117 119

Contents

H.B.

7.

ix

GRAPHICAL TECHNIQUES: ACCELERATION ANALYSIS LINEAR ACCELERATION ALONG CURVED PATHS 7.1 7.2 7.3 7.4 7.5 7. 6 7.7 7.8

8.

EFFECTIVE COMPONENT OF ACCELERATION METHOD 8.1 8.2 8.3 8.4

9.

11.

Introduction Acceleration of End Points on a Link Slider-Crank Analysis Four-Bar Linkage Analysis

RELATIVE ACCELERATION METHOD 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8

10.

Introduction Normal Acceleration Tangential Acceleration Resultant Acceleration Proportionality of Accelerations Relative Acceleration of Two Points on a Rigid Body Acceleration of Any Point in a Floating Link Coriolis Acceleration

Introduction The Acceleration Polygon Acceleration Polygon Convention Generalized Procedure Mechanism with Expanded Floating Link Compound Mechanism Cam-Follower Mechanism Summary

VELOCITY DIFFERENCE METHOD

123 124 124 124 126 127 133 134 138 14 6

151 151 151 154 157 161 161 162 162 163 172 175 179 181 183

10.1 10.2

Introduction Slider-Crank Mechanism Analysis

183 184

10.3 10.4

Quick-Return Mechanism Analysis Four-Bar Mechanism Analysis

187 189

GRAPHICAL CALCULUS METHOD 11.1 11.2

Graphical Differentiation Graphical Integration

194 194 204

Contents

X

12.

SPECIAL METHODS 12.1 12.2 12.3

Part m. 13.

14.

223 229

COMPLEX ALGEBRA

231

13.1 13.2 13.3 13.4 13.5

Introduction Complex Vector Operations Geometric Representation of a Complex Vector Complex Forms The Unit Vector

13.6

Linkage Application

235 237

FOUR--BAR MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD

242

Introduction Scope and Assumptions Geometric Relationships Velocity Analysis Acceleration Analysis

SLIDER-CRANK MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD 15.1 15.2 15.3 15.4 15.5

16.

212 218

ANALYTIC TECHNIQUES

14.1 14.2 14.3 14.4 14.5 15.

Complete Graphical Analysis Method Equivalent Linkage Method Slider-Crank Acceleration: Parallelogram Method

212

231 231 232 234

242 245 245 246 249

260

Introduction Scope and Assumptions Geometric Relationships Velocity Analysis Acceleration Analysis

260 263 263 265 267

QUICK-RETURN MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD

274

16.1 16.2 16.3 16.4 16.5

Introduction

274

Scope and Assumptions Geometric Relationships Velocity Analysis

276 277 277 279

Acceleration Analysis

Contents

17.

xi

SLIDING COUPLER MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD 17.1 17.2 17.3 17.4 17.5

18.

Introduction Scope and Assumptions Geometric Relationships Velocity Analysis Acceleration Analysis

289 290 295 295 297

SLIDER-CRANK MECHANISM ANALYSIS: MODIFIED VECTOR METHOD

306

18.1 18.2 18.3 18.4 18.5 19.

289

Introduction Scope and Assumptions Geometric Considerations Velocity Analysis Acceleration Analysis

SLIDER-CRANK MECHANISM ANALYSIS: CALCULUS METHOD 19.1 19.2 19.3

Introduction Scope and Assumptions Displacement, Velocity, and Acceleration Analysis

PROBLEMS

314 314 314 314 319

Kinematic Terminology Uniformly Accelerated Motion Vectors Graphical Techniques: Velocity Analysis Graphical Techniques: Acceleration Analysis Graphical Techniques: Miscellaneous Analytical Techniques: Velocity and Acceleration APPENDIX A A.l A. 2

306 306 306 307 309

319 320 321 327 339 342 353 355

Introduction Calculator Operating Procedure

FORTRAN Programs Four-Bar: Simplified Vector Method Slider-Crank: Simplified Vector Method Quick-Return: Simplified Vector Method Sliding Coupler: Simplified Vector Method Slider-Crank: Modified Vector Method

355 355 358 364 371 377 383

xii

Contents

Calculator Programs Four-Bar: Simplified Vector Method Slider-Crank: Simplified Vector Method Quick-Return: Simplified Vector Method Sliding Coupler: Simplified Vector Method Slider-Crank: Modified Vector Method APPENDIX B B. 1 B.2 B.3

388 396 404 413 421 428

Nomenclature Trigonometry Review Table of Trigonometric Functions

428 431 437

SELECTED REFERENCES

438

INDEX

443

MECHANISM ANALYSIS

I INTRODUCTORY CONCEPTS

Mechanism analysis (or kinematics of machines) is inherently a vital part in the design of a new machine or in studying the design of an existing machine. For this reason, the subject has always been of considerable importance to the mechanical engineer. Moreover, considering the tremen¬ dous advances that have been made within recent years in the design of high-speed machines, computers, complex instruments, automatic controls, and mechanical robots, it is not surprising that the study of mechanisms has continued to attract greater attention and emphasis than ever before. Mechanism analysis may be defined as a systematic analysis of a mechanism based on principles of kinematics, or the study of motion of machine components without regard to the forces that cause the motion. To better appreciate the role of mechanism analysis in the overall design proc¬ ess, consider the following. Typically, the design of a new machine begins when there is a need for a mechanical device to perform a specific function. To fulfill this need, a conceptual or inventive phase of the design process is required to establish the general form of the device. Having arrived at a concept, the designer usually prepares a preliminary geometric layout of the machine or mechanism for a complete kinematic analysis. Here the designer is concerned not only that all components of the machine are prop¬ erly proportioned so that the desired motions can be achieved (synthesis phase), but also with the analysis of the components themselves to determine such characteristics as displacements, velocities, and accelerations (analy¬ sis phase). At the completion of this analysis, the designer is ready to pro¬ ceed to the next logical step in the design process: kinetic analysis, where individual machine members are analyzed further to determine the forces resulting from the motion. Mechanism analysis therefore serves as a necessary prerequisite for the proper sizing of machine members, so that they can withstand the

1

2

Introductory Concepts i-IDEAL FLUIDS

rFLUIDS- -VISCOUS FLUIDS

COMPRESSIBLE FLUIDS

rSTRENGTH OF MATERIALS MECHANICSDEFORMABLE BODIES-

•THEORY OF ELASTICITY

lTHEORY OF PLASTICITY

lSOLIDS-

[-STATICS lRIGID BODIES-

rKINETICS LDYNAMICS—

_MECHANISM 'synthesis lKINEMATICSMECHANISM ANALYSIS

Figure 1.1

Mechanism analysis and other branches of mechanics

loads and stresses to which they will be subjected. Figure 1.1 shows the relationship of mechanism analysis to other branches of mechanics.

1

Kinematic Terminology

1.1

MECHANICAL CONCEPTS

A mechanism is a combination of rigid bodies so connected that the motion of one will produce a definite and predictable motion of others, according to a physical law. Alternatively, a mechanism is considered to be a kine¬ matic chain in which one of the rigid bodies is fixed. An example of a mech¬ anism is the slider crank shown in Figure 1.1. Instruments, watches, and governors provide other examples of mechanisms. A machine is a mechanism or group of mechanisms used to perform useful work. A machine transmits forces. An example of a machine is the internal combustion engine shown in Figure 1.1. The term "machine" should not be confused with "mechanism" even though in actuality, both may refer to the same device. The difference in terminology is related primarily to function. Whereas the function of a machine is to transmit energy, that of a mechanism is to transmit motion. Stated in other words, the term "mechanism" applies to the geometric arrangement that imparts definite motions to parts of a machine.

B

3

4

Introductory Concepts

Figure 1.2

Four-bar linkage mechanism.

A pair is a joint between two bodies that permits relative motion. 1.

A lower pair has surface contact, such as a hinge or pivot. Surface contact is a characteristic of sleeve bearings, piston rings, screwed joints, and ball-and-socket joints. As an example, joints A, B, C, and D of Figure 1.2 are lower pairs. These joints are called revolute or turning pairs.

2.

A higher pair has line or point contact between the surface ele¬ ments. Point contact is usually found in ball bearings. Line contact is characteristic of cams, roller bearings, and most gears.

A link is a rigid body that serves to transmit force from one body to another or to cause or control motion, such as the connecting rod or crank arm in Figure 1.1. Alternatively, a link is defined as a rigid body having two or more pairing elements. A kinematic chain is a group of links connected by means of pairs to transmit motion or force. There are three types of chains: locked chain constrained chain, and unconstrained chain.

Figure 1.3

Locked chain.

Kinematic Terminology

12.

3.

5

A locked chain has no relative motion between the links. An example of this is the three-link chain shown in Figure 1.3. A constrained chain is one in which there is definite relative motion between the links. For example, if one link is fixed and another link is put in motion, the points on all the other links will move in definite paths and will always move in the same paths regardless of the number of lines the motion is repeated. An example is the four-bar mechanism shown in Figure 1.4. An unconstrained chain is one in which, with one link fixed and another link put in motion, the points on the remaining links will not follow definite paths. An example of an unconstrained chain is the five-link mechanism shown in Figure 1.5.

A linkage is a mechanism in which all connections are lower pairs. The four-bar and slider-crank linkages are typical examples. A simple mechanism is one that consists of three or four links, whereas a compound mechanism consists of a combination of simple mech¬ anisms, and is usually made up of more than four links.

Figure 1.5

Unconstrained chain.

6

Introductory Concepts

A structure is a combination of rigid bodies capable of transmitting forces or carrying loads, but having no relative motion in them. Alterna¬ tively, a structure may be thought of as a locked kinematic chain. A frame is a stationary structure that supports a machine or mechanism. Normally, it is the fixed link of a mechanism (e.g., link 1 in Figure 1.2). A driver is that part of a mechanism which causes motion, such as the crank in Figure 1.2. A follower is that part of a mechanism whose motion is affected by the motion of the driver, such as the slider in Figure 1.2.

Figure 1.7

Cam-follower mechanism.

Kinematic Terminology

7

8

Introductory Concepts

Modes of Transmission Motion can be transmitted from driver to follower by: 1.

Direct contact a. Sliding b. Rolling

2.

Intermediate connectors a. Rigid: links b. Flexible: belts, fluids

3.

Nonmaterial: magnetic forces

A gear is a machine member, generally circular, whose active sur¬ face is provided with teeth to engage a similar member to impart rotation from one shaft to another. An example is the spur gear shown in Figure 1.6. A cam is a rotating or sliding machine member whose function is to impart a predetermined motion to another part that rolls or slides along the surface of the member. An example is shown in Figure 1.7.

Inversion Inversion is the process of fixing different links in a chain to create different mechanisms. Many useful mechanisms may be obtained by the inversion of various kinematic chains. An example of such inversion can be seen in the slider crank chain shown in Figure 1.8: 1.

By making link 1 of the chain fixed, we obtain a steam engine mechanism (Figure 1.8a).

2.

By fixing the crank, link 2, we obtain the Whitworth quickreturn mechanism, used in various types of metal shapers (Figure 1.8b).

3.

By fixing the connecting rod, link 3, we obtain the oscillating cylinder engine, once used as a type of marine engine (Figure

4.

Finally, by fixing the slider, link 4, we obtain the mechanism shown in Figure 1.8d. This mechanism has found very little practical application. However, by rotating the figure 90° clockwise the mechanism can be recognized as part of a garden pump.

R is important to keep in mind that the inversion of a mechanism does not change the relative motion between the links, but does change their abso¬ lute motions.

Kinematic Terminology

1.2

9

MOTION CLASSIFICATION

Definitions Motion is the act of changing position. The change of position can be with respect to some other body which is either at rest or moving. Rest is a state in which the body has no motion. Absolute motion is the change of position of a body with respect to another body at absolute rest. Relative motion is the change of position of a body with respect to another body that is also moving with respect to a fixed frame of reference.

Types of Motion Plane Motion In plane motion all points on a body in motion move in the same plane or parallel planes. All points of the body or system of bodies remain at a con¬ stant distance from a reference plane throughout the motion. Typical ex¬ amples are the connecting rod on a slider-crank mechanism (Figure 1.1) and the side rod on a locomotive (Figure 1.9). There are three classes of plane motion: (1) rotation, (2) translation, and (3) combined translation and rotation. 1.

2.

Figure 1.9

Rotation: When one point in a body remains stationary while the body turns about that point, the body is said to be in rotation. That is, all points in the body describe circular paths about a stationary axis that is perpendicular to the plane of rotation. Crank AB is Figure 1.1 has rotary motion. Translation: When a body moves without turning, that body is said to be in translation: or, the distances between particles

Locomotive side rod drive. Link BC undergoes curvilinear

motion; link AD undergoes rectilinear translation.

CD

0)

*CJ

£ be

d

o PS


u 3

be

10

Kinematic Terminology

11

of the body remain unchanged. There are two types of trans¬ lation: a. Rectilinear translation, where all points in the body move along parallel straight paths, such as the slider C in Figure 1.1. b. Curvilinear translation, where all points in the body move along similar (or parallel) curved paths. Curvilinear trans¬ lation is not to be confused with rotation, where all paths on the body are in the form of concentric circles. A good ex¬

3.

ample is the locomotive side rod in Figure 1.9. Note that the paths of any two points on the rod, say B and C, have the same curvature. Combined motion: When a body undergoes simultaneous trans¬ lation and rotation, that body is said to be in combined motion. That is, all points in the body change position and all lines turn as the body moves. A common example of a body that undergoes combined motion is the connecting rod BC in Figure 1.1, where one end, B, rotates about the crank axle A, while the other end, C, translates along a straight path as defined by the slider motion. Hence every other point on the member experiences part rotatation and part translation. Another common example is that of the rolling wheel (Figure 1.10), where it can be seen that the resultant motion of point B is the summation of its translation and rotation motions.

Three-Dimensional Motion There are two types of three-dimensional motion: 1.

2.

Helical motion: When a body has rotation combined with trans¬ lation along the same axis of rotation, that body is said to be in helical motion. The most common example is the turning of a nut on a screw (see Figure 1.11). The nut rotates and at the same time, translates along the axis of rotation. Spherical motion: When a body moves in three-dimensional space such that each part of that body remains at a constant distance from a fixed point, the body is said to be in spherical motion. A common example is the ball-and-socket joint, shown in Figure 1.12. A point on the ball or on the handle, which is rigidly attached to the ball, moves in space without changing its distance from the center of the sphere.

12

Introductory Concepts

■■■■

"S

1

Figure 1.11

Screw and nut.

Other Types of Motion Additional types of motion include: 1•

2>

Figure 1.12

Continuous motion: When a point continues a move indefinitely along a given path in the same sense, such a motion is said to be continuous. An example of this is a rotating wheel, where the path of a point away from the axis returns on itself. Reciprocating motion: When a point traverses the same path and reverses its motion at the end of such path, the motion is

Ball joint.

Kinematic Terminology

3.

1.3

13

said to be reciprocating. An example is the slider in a typical slider-crank mechanism (Figure 1.1). Oscillation is recipro¬ cating circular motion, as in a pendulum. Intermittent motion: When a motion of a point is interrupted by periods of rest, such motion is said to be intermittent, as in a ratchet.

MOTION CHARACTERISTICS

The path is the locus of a point as it changes from one position to another, distance is a measure of the path through which a point moves, and displace¬ ment is a measure of the net change in position of a point. There are two types of displacement: 1.

2.

Linear displacement (As) is the change position of a point as it moves along a straight line. Linear displacement is typically expressed in terms of inches, feet, or miles in a specified direction. Hence it is a vector quantity. Angular displacement (Ad) is the angle between two positions of a rotating line or body. It has both magnitude and sense, either clockwise or counterclockwise. Angular displacement is typically expressed in terms of degrees, radians, or resolu¬ tions in a specified rotational sense.

Velocity is the rate of change of position of a point with respect to time, or displacement per unit time. 1.

Linear velocity (v) is displacement per unit of time of a point moving along a straight line. The average linear velocity is given by the expression As V “ At

2.

where As is the change in linear displacement and At is the time interval. Linear velocity is typically expressed in terms of inches per second, feet per second, or miles per hour in a specified direction. Hence it is a vector quantity. Angular velocity (w) is the angular displacement per unit time of a line or body in rotation and has both magnitude and sense, either clockwise or counterclockwise. The average angular velocity is given by the expression = ^ At

Introductory Concepts

14

where A6 is the average angular velocity and At is the time interval. Angular velocity is typically expressed in terms of radians per second or revolutions per minute in a specified rotational sense. Acceleration is the rate of change of velocity with respect to time, or the rate of speedup. 1.

Linear acceleration (a) is the change in linear velocity per unit of time. Average linear acceleration is given by the expression Av a = — At

2.

where Av is the change in linear velocity and At is the time interval. Linear acceleration is typically expressed in terms of inches per second per second or feet per second per second in a specified direction. Hence it is a vector quantity. Angular acceleration (a) is the change in angular velocity per unit of time. Average angular acceleration is given by the expression Aw “ “ At where Aw is the change in angular velocity and At is the time interval. Angular acceleration is typically expressed in terms of radians per second per second or revolutions per minute per minute.

Deceleration or retardation is negative acceleration or rate of slowing down. Speed is the rate of motion in any direction, or the total distance covered in one unit of time. Speed is not to be confused with velocity, which is a vector quantity. Speed is typically expressed in magnitude terms such as feet per second, inches per second, or miles per hour, without regard to direction. Hence it is a scalar quantity. Angular speed is the rate at which a body turns about an axis. Typical units are revolutions per minute and radians per second, without any refer¬ ence to rotational sense. Phase describes the relative positions of links in a mechanism or a machine at any instant. This is usually defined by the angle of one of the links of the mechanism, for example, the angle of crank AB in Figure 1.1. When a mechanism moves through all its possible phases and returns to its starting position, it has completed a cycle, or a motion cycle. Vir¬ tually all mechanisms have a cyclic pattern where the cycle repeats itself over and over again. The time required for a motion to complete one motion cycle is called a period.

Kinematic Terminology

15

1 radian Figure 1.13

Geometric representation of a radian.

A body that rotates in a counterclockwise direction is assumed to have a positive sense. This is because angular displacement is normally measured counterclockwise. Thus in Figure 1.1, the crank AB has a posi¬ tive sense, and its angular velocity is also positive. Conversely, a body that rotates in a clockwise direction is assumed to have a negative sense, in which case the angular velocity of crank AB in Figure 1.1 is negative. Finally, radian is the angle subtended by an arc that is equal in length to the radius of the circle (see Figure 1.13). There are 2ir radians in a circle.

2 Uniformly Accelerated Motion

2.1

RE CTILINEAR MOTION

Displacement, Velocity, Acceleration, and Time Relationships The analysis of rectilinear or straight-line motion with constant or uniform acceleration is relatively common in the study of kinematics. To establish some basic relationships, let us consider a body moving with uniform accel¬ eration where the velocity changes from an initial value to a final value after some time, while completing some distance. Let s Vj v2 a t

= = = = =

distance completed, ft initial velocity, ft/sec final velocity, ft/sec acceleration, ft/sec2 time, sec

By definition. Distance = average velocity x time 'l

After After After After

1 2 3 t

sec, sec, sec, sec,

+ 2

V2

initial initial initial initial

1 velocity velocity velocity velocity

(2.1) is is is is

increased increased increased increased

This means that

16

to to to to

vx vx vx vx

+ + + +

a. 2a. 3a. at.

Uniformly Accelerated Motion

v2 = Vj + at

17

(2.2)

Now, substitute for v2 in Equation (2.1) and obtain Vi + (Vj + at) s =

2 2vj + at

s = Vjt + ^ at2

(2.3)

From Equation (2.2), v2 - Vj = at

(2.4)

and from Equation (2.1),

v, + v. or V2 + vx =

2 s t

Now, multiply Equations (2.4) and (2.5) and obtain 2 (v2 -vx)(v2 + Vj) = atx-s which gives v2 - v2 = 2as or v2 = vf + 2as In summary, for uniform linear acceleration:

s = Vjt + ^-at2 v2 = Vj + at v 2 = v2 + 2as

(2.5)

18

Introductory Concepts

For uniform linear velocity (a = 0): S

= Vjt

Also, if the body is accelerating, a is positive; if the body is decelerating, a is negative. EXAMPLE 2.1 A car passes a certain point A with a velocity of 30 ft/sec and another point B 1 mile away with a velocity of 60 ft/sec. If the acceleration is uniform, determine: a. b. c.

The average velocity of the car The time taken to travel from A to B The acceleration of the car

Given Vj = 30 ft/sec v2 = 60 ft/sec s = 5280 ft Required: v = ?

t = ?

a = ?

SOLUTION Average velocity:

2 30 + 60 2 = 45 ft/sec Time taken: distance average velocity 5280 45 = 117.3 sec or 1 min 57.3 sec

Uniformly Accelerated Motion

19

Acceleration: v2 = vf + 2as 602 = 302 + 2a(5280)

a

_ 602 - 302 “ 2(5280) = 0.256 ft/sec2

EXAMPLE 2.2 In coming to a stop, a train passes one signal with a speed of 60 mph and a second signal 30 sec later. During this period the brakes are applied to give a uniform acceleration. If the signals are 2400 ft apart, find: a. b.

The velocity of the train passing the second sign The magnitude of the deceleration

Given: v1 = 60 mph = 88 ft/sec

t s

=30 sec = 2400 ft

Required: v2 = ?

a = ?

SOLUTION Final velocity v2 is obtained from Vj + v2 S = —t

88 + v2 2400 = ---30 Lj

= 2

_ 88 30

= 72 ft/sec Deceleration is obtained from 1

.2

s = Vjt + - at2 2400 = 88(30) + |a(302)

20

Introductory Concepts

a

(2400 - 2640)(2) 900 = -0.53 ft/sec2

Rectilinear Motion Relationship Diagram The importance to the study of kinematics of the rectilinear relationships just derived cannot be overemphasized. Yet for many students, and even designers, quick recall of these expressions can be difficult. Furthermore, if the appropriate references are not readily available, precious time can be lost in trying to derive the desired expressions. To aid in such situations, the simple diagram presented in Figure 2.1 can be useful. Here: Vj v2 at t

(side of the smaller square) represents the initial linear velocity (side of the larger square) represents the final linear velocity (a times t) represents the difference between vx and v2, where is time

as

(a times s) represents the area of each trapezoidal section, where is the displacement

s

Considering the larger square, we can write the expression for any one of its sides as follows: v2 = vj + at An expression for the midpoint of the side (or average linear velocity v) is given by _ Vj + (vj + at) v =--2 or -

2

V! + V

v = 2 Also, considering the area of the larger square, we can write an expression for this area in terms of its constituent parts: v2 = vf + 2as Finally, considering one of the trapezoids, we can determine its area from the relationship

Area =

2~(suln

of Parallel sides) x width

Uniformly Accelerated Motion

Figure 2.1

21

Rectilinear motion.

Therefore, as = |(Vj + v2)at or s = \ (vi + v2)t Alternatively, the area of a trapezoid may also be expressed as the sum of its rectangular and triangular portions, as follows: 1 , as = vxat + — (atr Li

s = Vjt + ^-at2

2.2

ANGULAR MOTION

Displacement, Velocity, Acceleration, and Time Relationships As with rectilinear motion, to establish the basic angular relationships, let us consider a rotating body a uniform angular acceleration where the angular

22

Introductory Concepts

velocity changes from an initial value to a final value after some time t, while turning through some angle. Let 9 coj co2 a t

= = = = =

angle turned through, rad initial angular velocity, rad/sec final angular velocity, rad/sec, after t sec angular acceleration, rad/sec2 time, sec

By definition, Displacement = average velocity x time CO, + co7

(2.6)

9 = ---t

After After After After

1 2 3 t

sec, sec, sec, sec,

initial initial initial initial

velocity velocity velocity velocity

is is is is

increased increased increased increased

to to to to

coj coj coj coj

+ + + +

a. 2a. 3a. to?.

This means that co2 = coj + at

(2.7)

Now substitute for co2 In Equation (2.6) and obtain (coj + (to, + ot) 9 = ---t 2co, + at 9 = ---t

9 = coj + — at2 z

(2.8)

Again from Equation (2.7), co2 - coj = at and from Equation (2.6), 29 t

to, + or

w2

, +

=

20 J-

(2.10)

Uniformly Accelerated Motion

23

Now, multiply Equations (2.9) and (2.10) and obtain

(w2 -

) (co2 + cox)

which gives cjj

=

2aQ

or

(2.11)

= cox + 2 ad In summary, for uniform angular acceleration:

co2 — goj + at

co2 = coj + 2 aQ For uniform angular velocity {a = 0): 0

-

coxt

Also, if the body is accelerating, a is positive; if the body is decelerating, a is negative. EXAMPLE 2.3 A motor starting from rest develops a speed of 3000 rpm in 15 sec. If the acceleration is uniform, determine: a. b.

The rate of acceleration The number of revolutions made in coming up to speed

Given: d>j = 0

(rest)

w2 = 3000 rpm t

=

15 sec

Required: 9=7

a

?

24

Introductory Concepts

SOLUTION Acceleration: oo2 = cjj + at 314.2 = 0 + a (15) 314.2 15

a

=20.9 rad/sec2 Number of revolutions:

e =

<°1 + ^2

2

t

0 + 314.2 (15) 2 = 2356.5 rad 2356.5 2 TT = 375 rev EXAMPLE 2.4 A crank shaft rotating at 50 rpm has an angular deceleration of 1 rad/min/ sec. Calculate its angular velocity after 20 sec and the number of revolu¬ tions it makes a.

In 40 sec

b.

In coming to rest

CO,

50 rpm - 50= 5.24 rad/sec

Given:

a = -1 rad/min/sec = — rad/sec/sec 60 t = 20 sec Required: oo2 = ?

e =

?

Uniformly Accelerated Motion

SOLUTION Angular velocity after 20 sec: w2

101

~

+

= 5.24 + ^(40) = 5.24 - 0.33 60 =4.9 rad/sec Number of revolutions in 40 sec: 9 = Wjt + ~crt2 z = 5.24(40) +|(^) (40)2 = 196.26 rad 196.26 27T

- 31.23 rev Number of revolutions in coming to rest: Data: coj = 5.24 rad/sec co2 = 0 a -

(rest)

60

rad/sec2

Required: = ?

(angular displacement)

Equations: co2 = co2 + 2 ol9

s = 5-24Z + 2(^)e ■ -»■«*(?) = 823.7 rad _ 823.7 27T

= 131.1 rev

25

26

Introductory Concepts

Angular Motion Relationship Diagram As in the rectilinear case, a similar diagram (Figure 2.2) can be used for the angular relationships. Here, coj

(side of the smaller square) represents the initial angular velocity

w2 at a t

(side of the larger square) represents the final angular velocity (a times t) represents the difference between cox and oo2, where is the uniform angular acceleration, and is time

aQ

(a times 9) represents the area of each trapezoidal section, where 9 is the angular displacement

Considering the larger square, we can write an expression for any one of its sides as follows: w2 = coj + at and an expression for the midpoint of the side (or average angular velocity co) is given by ojj + (coj + at) co = -

2

Also, considering the area of the larger square, we can write an expression for this area in terms of its constituent parts: oof = oof + 2 aO

Finally, considering the trapezoidal area, we can write

ae = |(wi + w2)o;t from which 0 =

+ «2)t

Alternatively, the area of one trapezoid can be expressed as the sum of its rectangular and triangular sections. Thus

Uniformly Accelerated Motion

Figure 2.2

27

Angular motion

016 = wjCrt + “-(crt)2 or 9 = “it + \ at2 O

2.3

CONVERSION BETWEEN ANGULAR AND RECTILINEAR MOTION

There are many situations where it is necessary to convert from angular to rectilinear motion, and vice versa. To establish the basic relationships, consider the pulley-and-belt arrangement in Figure 2.3. If the pulley turns through an angle 9, assuming that the belt moves without slipping, the cor¬ responding displacement of the belt is equivalent to the length of arc r 9, or s = r9

(2* 12)

If the pulley is rotating at uniform velocity w where the angular displacement 9

-

cot,

the total belt displacement is s = r(cot)

(2.13)

28

Introductory Concepts

PULLEY

Figure 2.3

Conversion from angular to rectilinear motion.

Also, the belt moves at uniform speed; its velocity v is related to the dis¬ placement by s = vt Therefore, Equation (2.13) can be written as Vt = Rot

or v = rto Ihis expression is probably the most useful in velocity analysis. In effect, it states that if a point has uniform motion in a circular path, the linear speed at any instant is equal to the distance of that point from the center of rotation multiplied by the angular speed. Also, since velocity is a directed speed, it is clear from the illustration that the only direction the velocity of the point can have at that instant is tangential to the circular path in the same sense as the angular motion. Another way of looking at the velocity direction is to consider any object, such as a ball, being rotated about one end of a string while the other end is held fixed. Suppose that the string were to be suddenly released, the ball will then "take off in a direction tangential to the circular path that it maintained before the release.

Uniformly Accelerated Motion

29

Similarly, it can be shown that for uniform acceleration of the pulley where the angular acceleration is a, the linear acceleration, a, of the belt is given by a - ra

This acceleration is known as the tangential acceleration since the linear velocity acts in a direction tangential to the path of rotation. Hence, to con¬ vert from angular motion to linear motion, we multiply the respective values by radius r. EXAMPLE 2.5 In Figure 2.4 the pulley D is belt-driven by pulley B, which is fastened to pulley C. Starting from rest, the body A falls 60 ft in 4 sec. For each pulley, determine: a. b. c.

The number of revolutions The angular velocity The angular acceleration

SOLUTION Calculate linear values for the pulleys, then convert to angular values.

Figure 2.4

Example problem.

30

Introductory Concepts

Given r. B

8 in

rc

6 “■

rQ = 10 in. S

= 60 ft

Vj

=

0 ft/sec

Required:

"b

' 7

aB = 7

"c “ ? “c =

7

Revolutions: Pulley B:

= 90 rad =

27r

= 14.3 rev Pulley C: Op = 14.3 rev

(same as

Pulley D: Q D _ radius of pulley C #c radius of pulley D

ft

D 14.3

6 10

"d '

7

“D ' 7

Uniformly Accelerated Motion

31

eD = 14'3(^) = 8.58 rev Angular velocities: Pulley B: The angular velocity of pulley B is obtained from the fol¬ lowing relationship: v

= r co B B

B

where v_ = linear velocity after 4 sec (unknown) B to

= angular velocity after 4 sec (required

B

Therefore, to find Vg, we first need to determine the linear acceleration ag. Since the motion is not uniform or free-falling, we use the relationship 1

7

s = Vjt + — am Ld

where s

= 60 ft

Vi

=

t

= 4 sec

0

a

Therefore, 60 = ° + |aB(4)2 and

aB

= 60(2) 16 = 7.5 ft/sec2

Linear velocity vg is given by the relationship v2 = Vj + at where

32

Introductory Concepts

vz = vB Vj = 0 a

-

= 7.5 ft/sec2

t

=4 sec

Therefore vB = 0 + 7.5(4) (2.14)

= 30 ft/sec

Required angular velocity wB can now be determined by substituting the value found for vB.

=45 rad/sec Pulley C: WC = 45 rad/sec

(same as coB> since pulleys C and B are attached)

Pulley D: CO.

f> _ radius of pulley C coc radius of pulley D

45 oo

10 = 27 rad/sec

Angular acceleration: Pulley B:

where aB “ 7*5ft/sec2

(found)

33

Uniformly Accelerated Motion

_ 8_ rB ~ 12 Therefore,

5 = Ii“ B and

= 11.25rad/sec2 Pulley C: - 11.25 rad/sec2

(same as a^ since pulleys B and C are attached)

Pulley D: 01

oi

D _ radius of pulley C radius of pulley D

11.25

10

®D = 11-25(To) =

2.4

6.75rad/sec2

VELOCITY-TIME GRAPH SOLUTIONS

In many cases it has been found more convenient and simpler to solve some motion problems graphically, using the velocity-time diagram concept. The velocity diagram is a graph in which the velocity of a point is plotted against a time base. Figure 2.5 shows the three possible conditions that can exist: 1. 2. 3.

Uniform velocity (Figure 2.5a) Uniform acceleration (Figure 2.5b) Variable velocity (Figure 2.5c)

Note that in case 1 the velocity-time curve has no slope, and therefore the point has no acceleration. This motion is normally referred to as uniform motion.

34

Introductory Concepts

Figure 2.5 Velocity-time curves: (a) uniform velocity; (b) uniform accel¬ eration; (c) variable velocity.

If the motion is not uniform, but the acceleration is constant, the point is said to have uniformly accelerated motion, as in case 2. Otherwise, the motion is variable, as in case 3, where the acceleration changes from one instant to another. Also, note that acceleration may be depicted as positive or negative depending on whether the slope of the velocity curve is positive (upward to the right or downward to the left) or negative (upward to the left or down¬ ward to the right). In each case, the distance covered by the point is given by the area under the curve corresponding to the time period during which the motion took place. EXAMPLE 2.6 The maximum acceleration of a Ferris wheel at a park is 1 rad/min/sec and the maximum deceleration is 2 rad/min/sec. Determine the minimum time it will take the wheel to complete 15 revolutions going from rest to rest. SOLUTION Let T = total time t = time to accelerate T - t = time to decelerate Given: “» = 1 rad/min/sec = ~ rad/sec2 A 60 “D = ~2 rad/min/sec =

rad/sec2

£» = 15 rev = 15(27r) = 94.24 rad

Uniformly Accelerated Motion

35

Required: T = ? Let triangle imf in Figure 2.6 represent the starting velocity vj through maximum velocity vm to final velocity Vf. This triangle then represents the total angular displacement 9 of the wheel in time T. That is, 9 = area of triangle imf Now for the acceleration portion of the curve, consider triangular segment ima and find gj^ in terms of t' using the relationship co2 = coj + at where

w2

“

“m

coj = coi = 0 a = a t

= t'

A

= ^-rad/sec2 60 (acceleration time)

wA

_Time f

Figure 2.6

Velocity-time graph.

(sec

36

Introductory Concepts

Hence O)

m = — rad/sec oU

(2.15)

Similarly, for the deceleration portion of the curve, consider triangular segment amf and find t' in terms of T, using the relationship to2 = coj + at

where co2 = cjf = 0 rad/sec t' wi = “m = go rad/sec O!

= a^ = ^-rad/sec2

t

= (T - t') sec

Hence 0

t')

60 " 30

t'^

from which T - t' = 2 t' T = - + t' 2

(2.16) t'

(2.17)

Using the area relationship for a triangle to define angular displacement we note that

Uniformly Accelerated Motion

After substituting for

94'24

37

and T from Equations (2.15) and (2.17), we obtain

= ICio) (|T)T

from which

T‘- 94 • 24(f) (?)(!) = 16,963.2 sec2 T = 130.2 sec or 2 min 10.2 sec EXAMPLE 2.7 A car, traveling between two stoplights 4 miles apart, does the distance in 10 min. During the first minute, the car moves at a constant acceleration, and during the last 40 sec, it comes to rest with uniform deceleration. For the remainder of the journey the car moves at a uniform speed. Find: a. b. c. d.

The The The The

uniform speed acceleration deceleration distances covered during uniform velocity, acceleration,

and deceleration SOLUTION Construct a velocity-time curve (Figure 2.7) to describe the motion of the car. Given: v

= 0

(at rest)

cl

T

= 600 sec

tj

= 60 sec

t2

= 600 - 60 - 40 = 500 sec

S

= 4(5280) ft (total area under curve)

Let the uniform velocity be v. Then the total distance covered, Srj., may be computed from ST = area of trapezium abed = ^ (be + ad)v 4(5280) = ~(500 + 600)v z

38

Introductory Concepts

= 4(5280)(2) 1100 = 38.4 ft/sec Acceleration: v2 = Vj + at where Vi = va = 0 v2 =38.4 ft/sec t

= tj = 60 sec

38.4 = 0 + a(60) and 38.4

-0.64 ft/sec2

Uniformly Accelerated Motion

Deceleration: v2 = Vj+ at where v2 = 0 Vj = vf = 38.4 ft/sec t

= t2 = 40 sec

0

= 38.4 + a(40)

a

“ "

and 38.4 40

= -0.96 ft/sec2 Note: Negative sign denotes deceleration. Distance covered during acceleration: Sj = area of triangle abe

= | (60) (38.4) = 1152 ft Distance covered during uniform velocity: 52 = area of rectangle ebcf = vt2 = 38.4(500) = 19,200 ft Distance covered during deceleration: 53 = area of triangle fed

= | (38.4) (40) = 768 ft

39

40

2.5

Introductory Concepts

SUMMARY OF MOTION FORMULAS

Linear and Angular Relationships

Linear Symbol Units

Angular Symbol Units

Displacement

s

ft

9

rad

Initial velocity

vi

ft/sec

COj

rad/sec

Final velocity

V2

ft/sec

w2

rad/sec

Average velocity

V

ft/sec

CO

rad/sec

Acceleration

a

ft/sec2

a

rad/sec2

Time

t

sec

t

sec

v2 = Vl + at

co2 = coj + at

v2 = vf + 2as

co2 = cof + 2 ad

Vj + v2 V

-

COj + “2

CO

-

s

V1 + V2 . = 2 4

9

U), + CO 9 —t 2

s

= Vjt + —at2

9

= cojt + ioV

2

Li

Conversion from Angular to Linear

e

Displacement:

s = r x

or (r<9)

Velocity:

v = r x co or (rco)

Acceleration:

a = r x a or (ra)

2

3 Vectors

3.1

PROPERTIES OF VECTORS

In mechanics, quantities are classified as either vectors or scalars. A vector has been defined as a quantity that has magnitude and direction. Ex¬ amples of vector quantities are displacement, velocity, acceleration, and force. A quantity that has magnitude but no direction is called a scalar. Examples of scalar quantities are time, volume, area, speed, and distance. Graphically, a vector is represented by an arrow, as in Figure 3.1, where the length, usually drawn to scale, represents the magnitude and the arrowhead indicates the direction. The arrowhead is commonly called the head or terminus of the vector, while the opposite end is called the tail or origin. The direction is usually specified as the angle in degrees which the arrow makes usually with some known reference line. For example, in Figure 3. la the vector V represents a velocity having a magnitude of 5 units in the direction of 9 and in Figure 3. lb the vector -V represents a velocity of the same magnitude but in opposite direction. Thus, to convert a vector from positive to negative, we reverse its direction. In this text, vector quantities are normally denoted by capital letters with bars above (e.g., R, V, and A) to distinguish them from their scalar counterparts, denoted by capital letters without bars. Using this notation, the normal addition and subtraction signs, + and -, can be used without risk of confusion between vector and scalar operations. Lowercase letters (e.g., v and a) are also used in some instances to denote scalar quantities, partic¬ ularly where vectors are not involved.

3.2

VECTOR ADDITION

There are two methods for adding two vectors.

41

42

Introductory Concepts (a)

Figure 3.1

Figure 3.2

Properties of a vector: (a) vector V; (b) vector -V.

Triangular method.

Vectors

43

1.

The triangular method, where the vectors are connected head to tail and the resultant is determined by a third vector, which extends from the tail to the head of the connected vectors.

For example, consider the vectors A and B in Figure 3.2a. To determine the sum or resultant of these vectors, start at some point O, called the origin, and connect vector A to vector B as shown in Figure 3.2b. Then, from the same origin or the tail of vector A, draw a_third vector, C, to close the triangle at the arrowhead of vector B. Vector C is therefore the required sum or resultant of vectors A and B . Note that the resultant vector always tends to oppose the general sense of the summed vectors. We could think of it as a "counterbalancing vector," since it appears to have a counterbalaneing effect on the loop, which in this case is a triangle. Note also that the resultant is the same for B + A as for A + B. This means that vector addition is commutative. 2.

The parallelogram method, where the vectors are connected tail to tail so that they form two adjacent sides of a parallelo¬ gram. The resultant is found by drawing a third vector extend¬ ing from the connected point to form a diagonal of the parallel¬ ogram.

For example, consider the same vectors A and B in Figure 3.2a. To determine the sum or resultant of these vectors, connect the vectors A and B tail to tail as shown in Figure 3.3. Define this point of connection as O (origin). Using these vectors as adjacent sides, complete the parallelo¬ gram as shown. Then from the point of origin O, draw a third vector, C, also originating at O, to form a diagonal of the parallelogram. This vector, C, is the required resultant of the summed vectors. Both the parallelogram and triangular methods are also applicable to vector addition involving more than two vector quantities. For example, consider the summation and vectors A, B, C, and D shown in Figure 3.4. From the triangular method, we note that

Figure 3.3

Parallelogram method.

44

Introductory Concepts

R2 = A + B R3 = (A + B) + C R4 = (A + B + C) + D =A+B+C+D Also, in Figure 3.5, we note, from the parallelogram method, that 52 = A + B 53 = (A + B) + C 54 = (A + B + C) + D = A+B+C + D

F igure 3.5

Paralle logram method.

Vectors

3.3

45

VECTOR SUBTRACTION

Triangular Method To subtract one vector from another we simply reverse the direction of that vector and sum both vectors normally. For example, consider vectors A and B in Figure 3.6a. To determine the resultant of A - B, we reverse the direction of vector B, which in effect changes the vector from +B and -B. With this change made, we now add both vectors by placing the tail of -B at the head of A and connecting the tail of A and head of B to obtain vector A - B, the resultant (see Figure 3.6b). In equation form, vector subtraction can be expressed as A - B = A + (-B) Note that vector subtraction is just a specialized case of vector addition, where the subtracted vector is reversed in direction and treated as a posi¬ tive vector.

Parallelogram Method An alternative method of vector subtraction is to join both vectors tail to tail, then draw a third vector to connect the termini of the given vectors. This third vector, when properly directed, represents the difference of the two vectors. The direction of the third vector is obtained by directing the arrow toward the vector from which the subtraction^ made. As an example, consider again the vectors A and B in Figure 3.7a. To find A - B, we join the tail of A to that of B, and_the magnitude of vector A - B is given by a line connecting the terminus of A to that of B (see Fig¬ ure 3.7b). The direction is given by directing the vector A - B from the

A

46

Introductory Concepts

A

Figure 3.8

A

Determining resultants of vector systems.

Vectors

47

terminus of B to the_terminus of A. In other words, the vector A - B is directed from B to A. Note that this method, in effect, is a variation of the parallelogram method used in vector addition. In that case, the difference is obtained by completing the "other" diagonal that joins the termini of the given vectors. EXAMPLE 3.1 Given vectors A, B, and C in Figure 3.8, where magnitudes and directions are as shown, determine the following: (a)

A + B - C

(b)

C - B + A

SOLUTION See construction given in Figures 3.8a and b.

3.4

THE VECTOR POLYGON

The vector polygon is the configuration that results from addition or sub¬ traction of more than two vectors graphically. The polygon can be considered to be a closed loop consisting of the vectors that are added or subtracted and the resultant vectors. Each vector polygon can therefore be represented by an algebraic expression in terms of the vector components and their re¬ sultant. For example, consider the vector polygon shown in Figure 3.9. Let it be required to write an algebraic expression for vector E. For convenience, it may be assumed that all vectors having one sense (clockwise or counterclockwise) with respect to the closing of the loop are positive, and vectors having the opposite sense are negative. Then it is easy to see that the sense of vector E opposes that of vectors D, B, and A, whereas it is the same as that of vector C. Therefore, from the rules of vector addition and subtraction discussed, we can immediately write the equation E - D + C - (-B) - A = 0

(3.1)

from which E-D+C + B- A = 0 or E=D-C-B+A

(3.2)

48

Introductory Concepts

E = D - C + Figure 3.9

(-B)

+ A

Vector polygon.

Similarly, the expressions for vectors A, B, C, and D can be derived from first principles, as follows: -A - (-B) + C + (-D) + E = 0

(3.3)

A = B+C-D + E

(3.4)

-(-B) +C-D+E-A = 0

(3.5)

B=D-E+A-C

(3.6)

C-D + E- A- (-B) = 0

(3.7)

C^D-E+A-B

(3.8)

-D + C - (-B) - A + E = 0

(3.9)

D = C + B- A + E

(3.10)

Alternatively, the expressions for A, B, C, and D can be obtained directly from Equation (3.2).

3.5

VECTOR RESOLUTION

Recalling the rule on summation of vector quantities, we saw that a number of vectors added together was equivalent to a single vector called a resultant.

Vectors

49

Or, stated another way, we can say that the resultant vector is a summation of a number of component vectors. When a vector is represented as a sum¬ mation of other vectors, that vector is said to be resolved, and the vectors being summed are the components of the resolved vector. The vectors A1; A2, and A3 in Figure 3.10, for example, are components of resolved vector A. Although the components of a vector can be limitless, it is normally more useful to resolve a vector into just two components along specific axes. In such a case, it is useful to recall the two methods of vector addi¬ tion—the triangular method and the parallelogram method—and note that the process of vector resolution is simply a reversal of the addition process. Suppose that we wish to resolve vector A in Figure 3.11a so that its two components, C and B, have the orientation shown by the dashed lines b-b and c-c. Either of the following methods can be employed. 1.

2.

Triangular method (Figure 3.11b): Through the origin and terminus of vector A, draw lines parallel to b-b and c-c to intersect at some point. This point, according to the triangular method, defines the terminus of vector C, which extends from the tail of vector A to the tail of vector B, which in turn extends to the terminus of vector A. Parallelogram method (Figure 3.11c): An alternative resolu¬ tion approach is to draw through the tail of vector A two axes, each parallel to lines b-b and c-c, then through the terminus or head of A draw a line parallel to axis c-c to intersect b-b and similarly, another line parallel to axis b-b through the terminus to intersect axis c-c. These points of intersection on axes b-b and c-c define, respectively, the termini of vector components B and C.

50

Figure 3.11a and b

Introductory Concepts

Vector resolution: triangular method.

Vectors

51

c Figure 3.11c

3.6

Vector resolution: parallelogram method.

ORTHOGONAL COMPONENTS

A specialized, but commonly encountered case is the resolution of a vector into orthogonal or rectangular components along mutually perpendicular axes. For example, given orthogonal axes x-x and y-y and vector V_in Figure 3.12, suppose that we wish to determine the components of V along x-x and y-y. Using the parallelogram method discussed above, we can readily determine these components to be Vx along axis x-x, and Vy along axis y-y,

Introductory Concepts

52

y

y

Figure 3.12

Orthogonal components of a vector.

by dropping perpendicular lines from the terminus of vector V to axes x-x and y-y, respectively. Here it is useful to note, from trigonometry, that the magnitudes of component vectors Vx and Vy are V

x

= V cos 9

and V

y

= V sin 9

where 9 is the angle that vector V makes with the x axis.

3.7

TRANSLATIONAL AND ROTATIONAL COMPONENTS

Orthogonal components, when applied to link motion, may be described in terms of translational and rotational components, where The translational component is defined as that component of the vector which tends to cause translation of the link along its own axis. The rotational component is defined as that component of the vector which tends to cause rotation of the link about some center located on the link axis. Consider the vector V depicted in Figure 3.13a and b. Observe that the translational component (V*) in Figure 3.13b is equivalent to the x component

Vectors

53

Figure 3.13

Translational and rotational components of a vector.

(Vx) in Figure 3.13a, or V

= V1

translational component

and the rotational component (Vr) (Figure 3.13b) is equivalent to the y com¬ ponent (Vy) in Figure 3.13a, or -

3.8

-r = V

rotational component

EFFECTIVE COMPONENTS

The effective component of a vector may be defined as the projection of that vector along the axis where the effect is to be measured. To determine the effective component of any vector, we simply drop a perpendicular line from the terminus of the vector to a line drawn through the origin of the vector along which the effect is to be measured. The point at which these two lines meet defines the terminus of the required effective component. For example, if we consider again the vector V shown in Figure 3.13, Vx is the effective component of this vector along the x axis. In other words, the effective component of vector V along axis x-x is the value Vx. Similarly, if we consider the same vector with respect to another axis, y-y, Vy is the effective component of V along that axis. In general, if we consider vector Vp oriented with respect to axes a-a and b-b as shown in Figure 3.14, then by dropping perpendiculars from the terminus of this vector to these axes, we obtain, respectively,

Introductory Concepts

54

a

Figure 3.14

Effective component of a vector.

Vp

effective component along a-a

-bb Vp

„ effective component along b-b

where the superscripts aa and bb are used to indicate the axes along which the effective components are considered. From this example, the following observations can be made: 1•

A perpendicular drawn from the terminus of an effective com¬ ponent of a vector to intersect the line of action of that vector will define the magnitude of that vector. Therefore, if one effective component and direction of a vector are known, the vector can readily be determined.

2.

The line drawn from the terminus of the absolute vector must be at a right angle to the effective component, not to the abso¬ lute vector. This is because the latter construction will yield an effective component that is "greater" in magnitude than the

Vectors

55 absolute vector, which is impossible. No component can be greater than its whole. A vector can have an effective component in any conceivable direction, except that which is perpendicular to the vector it¬ self. In this context, it might be well to note that the effective component in the direction of an absolute vector is the vector itself. The orthogonal components of a vector are merely a special¬ ized set of effective components of that vector along mutually perpendicular axes. Compare Figures 3.12 and 3..13. If two effective components of a vector are known, that vector can be completely determined by constructing perpendiculars from the terminus of each effective component vector such that they intersect each other. This point of intersection de¬ fines the terminus or magnitude of the required vector. For example, in Figure_3.14, the perpendiculars drawn from the termini of and Vbb define the terminus Vp. Note that this is precisely the reverse of the procedure that is followed in finding the two effective components Vbb and when the absolute vector Vp is known. P

II GRAPHICAL TECHNIQUES

Graphical techniques offer a convenient way to solve velocity and acceleration problems relating to mechanism. Compared to analytic techniques, they are simpler, faster, and more easily understood. Accuracy, however, depends of precision of line work, measurements, and a wise choice of scales for the vectors. In this section a variety of graphical methods are presented for ana¬ lyzing mechanisms to determine the velocities and accelerations of their members. These include such methods as instant centers, effective compo¬ nent of velocity and acceleration, relative velocity and acceleration, graph¬ ical differentiation and integration, and some special constructions. As one would expect, each method has advantages and disadvantages. No single method is suitable for solving all problems, even though some are more versatile than others. Generally, the most suitable method is one that yields the required information in the shortest time with the least amount of effort. The objective here is to illustrate the variety of graphical methods used to analyze a mechanism which are available to mechanical design engi¬ neers and students—hopefully, enabling them to make an appropriate selec¬ tion when confronted with a practical problem.

57

II.A GRAPHICAL TECHNIQUES: VELOCITY ANALYSIS Velocity is inherently an important factor in dynamic analysis. Since force is proportional to acceleration, which is the rate of change of velocity, velocity analysis becomes a necessary prerequisite to the acceleration and force analysis of a machine member. In high-speed machines, forces gen¬ erated during impact and during sudden changes of velocity can limit the operating speed of the machine. Also, as the operating speed increases, it requires greater and greater forces to make various links move through their intended cycles. Drive torques must be increased correspondingly as speeds are increased. As with impact, this can result in increased defor¬ mation and vibration within the machine. Also, as speeds increase, lubrication and wear become more critical. For example, in the crankshaft bearing of an automobile, wear depends on the speed of the crankshaft and the pressure between the crank pins and bearings. Similarly, the cutting speeds of machine tools and the flow rates of fluids in engines and pumps are all functions of the velocities of the output members. For these reasons, methods of determining relative and absolute velocities are of great importance in making a complete analysis of the motions of parts of a machine.

59

4 Effective Component of Velocity Method

4.1 INTRODUCTION The effective component method of velocity is based on two principles. The first is the use of the effective components of a vector, and the second is the rigid body principle. As discussed earlier, the effective component of a vector in any direction is the projection of that vector along a line drawn through the vector origin in the direction of interest. This direction is usually defined for the convenience of applying the rigid body principle.

4.2 THE RIGID BODY PRINCIPLE The rigid body principle may be stated as follows: In a rigid body, the dis¬ tance between two points remains constant and the velocity components along a line joining these points must be the same at both points. This principle is easily explained, in that if the velocity components were different at the two points, the link would change in length, and would therefore not remain rigid. Thus, if we know the velocity of one point of a rigid body, we can find the velocity of any other point on that body by resolving the known veloc¬ ity into components along and perpendicular to the line joining the two points and making the velocity component of the unknown velocity equal to that of the known component along the joining line.

4.3 VELOCITIES OF END POINTS ON A LINK Consider link BC in Figure 4.1, where the velocity of point B is completely known (in magnitude as well as direction). The line of action for the velocity of point C is also known. We want to find Vq. To obtain the velocity at point C, we need the effective component of VC along BC, which has the same magnitude as the effective component of 60

Effective Component of Velocity Method

Figure 4.1

61

Velocities of end points on a link.

VB along BC. We must first obtain V§c, then V®c, and finally VA,. B C C PROCEDURE 1.

2.

We determine the effective component of Vg along BC, that is, V®^. This is determined by dropping a perpendicular from the terminus of Vg to intersect a line joining B and C (line BC). Since the link is a rigid body and therefore all points along BC must experience the same velocity as v5^, we can immediB ately lay out the effective component of Vr along BC, that is, VBC.

vBC = vBC B

3.

With the effective component V^ completely defined, and the direction of V^ also known, we simply construct a perpendicu¬ lar from the terminus of V®^ to intersect the line of action Vq.

4.4

C

This point of intersection defines the magnitude of Vq.

VELOCITIES OF POINTS ON A ROTATING BODY

Now consider body D, which rotates about point O (Figure 4.2). The velocity of point B or V-g is given as shown, and the velocity of point C (Vc) is required. A close look at this problem will suggest that the procedure should be identical to that used in the preceding problem, except that now the direc¬ tion of V(j is not given explicitly. However, as we noted earlier, since the body is rotating, the linear velocity of point C must be tangential to a circu¬ lar path described by C. Therefore, the direction of Vc is perpendicular to the radial line AC.

62

Figure 4.2

Graphical Techniques: Velocity Analysis

Velocities of points on a body with pure rotation.

Effective Component of Velocity Method

63

PROCEDURE 1.

2.

Determine the effective component of Vg along BC, that is, v?c. vB Here it should be noted that the effective component is obtained by dropping the perpendicular from the terminus of Vg to exten sion of a line joining B and C (line BC). Lay out the effective component of Vq along BC, that is, VqC

vBC C

3.

= VBC B

(From the rigid body principle, all points along C must have the same velocity.) Determine the required velocity, V_,. From the terminus of -BC C Vq , project a perpendicular line to intersect the line of action of V

.

This point of intersection defines the mag'nitude of V

v_/

4.5

VELOCITY OF ANY POINT ON A LINK

Sometimes it is necessary to determine the velocity of a point on a link other than the two end points. This can easily be obtained as follows. Consider link BC in Figure 4.3, where the velocities of points B and C are known. Find the velocity of point D located on the link. The velocity of point D (Vg) can be obtained from the summation of two components: (1) a translation component V^, which must be the same for all points along BC (according to the rigid body principle); and (2) a rota¬ tional component Vg, which must be proportional to Vg or Vc based on its distance from a center of rotation (according to the rotation principle). In summary,

V

where

D

= v1D + v1D

. L/

64

Graphical Techniques: Velocity Analysis

C

Figure 4.3

Proportionality of velocities.

PROCEDURE 1.

Resolve the given velocity vectors VB and Vc into their orthog¬ onal (rotational and translational) components with respect to link BC. The rotational components are obtained by dropping

Effective Component of Velocity Method

65

perpendicular lines from the termini of Vg and Vg to lines drawn normal to BC through points B and C. These components are des¬ ignated Vg and V£ (see Figure 4.3b). The translational compo¬ nents are obtained by dropping perpendiculars from the termini of Vg and Vg to line BC or BC extended. Note that these com¬ ponents, designated V* and V* , are the same as the effective ■D

C

components of Vg and Vg. That is, t B

V®° ±5

and

2. Lay out the translation component of Vg, that is, Vg or Vg , along BC. From the rigid body principle, all points experience the same velocity along a straight line. Therefore,

or

or

or —

3.

4.

component Vg is obtained by constructing a perpendicular from point D to meet the proportionality line (see Figure 4.3c). Determine the velocity Vg. Having determined both the rotational and translational components of VD, we can now obtain the re¬ sultant vector by graphically summing both the rotational and translational components of Vg. V

4.6

- x*

Determine the rotational component of Vg (Vg). Because the point D lies on the same straight line as B and C, its rotational component must be proportional to that of point B as well as point C. Therefore, draw a straight line to connect the terminus of the rotational component of Vg to the terminus of Vg. This line is the line of proportionality for rotational velocity compo¬ nents of all points on BC. Therefore, the required rotational

D

-1 -r = v + V D D

VELOCITY ANALYSIS OF A SIMPLE MECHANISM

Crank AB of the slider-crank mechanism shown in Figure 4.4 rotates clock¬ wise at 0.5 rad/sec. We want to determine the velocity of the slider. In the preceding sections, the linear velocity of point B in link BC was given, and the direction of the velocity of point C was either known or determinable from the constraints of the link motion. Here, although Vg is not given directly, it is determinable since we know that its magnitude is given by

66

Graphical Techniques: Velocity Analysis

V

B

ABoj

AB

and that its direction must be perpendicular to AB (in the same sense as co^g). Further, the direction of Vq, although not given explicitly, is obviously horizontal, considering that the line of action of the slider of which point C is a part must be along the slot, which is horizontal. Therefore, in principle, the procedure for determining the velocity of point C is basically the same as before. PROCEDURE 1.

Lay out the velocity of point B, that is, Vg (direction and mag¬ nitude), using a convenient scale. The magnitude of this vector is given as V

AB

B

X AB

= 0.5(5) =2.5 in./sec

.

2

Determine the effective component of Vg along BC. Drop a per¬ pendicular line from the terminus of Vg to link BC.

3.

Lay out the effective component of V

4.

C

along BC.

Determine V^. a. Construct a perpendicular line from the terminus of V intersect the known line of action of Vq. b. Scale the magnitude of Vq.

to

LINE OF ACTION

Figure 4.4

Slider-crank mechanism.

Effective Component of Velocity Method

V

-2.3 in./sec

V

= 2.3 in./sec

67

Hence (directed as shown)

EXAMPLE 4.1 Consider the four-bar linkage in Figure 4.5, where crank AB rotates counterclockwise with an angular velocity of 1 rad/sec, as shown. Let it be required to find the angular velocity of the follower CD. SOLUTION To find the angular velocity of the follower, we must first determine the linear velocity of point C. The linear velocity of point C is obtained basically in the same manner as for the slider crank in Figure 4.4. There Vc was obtained using the effective component of Vq along link BC and the line of action of Vc, which was known. Here the only difference to be noted is in the line of action of the velocity of point C, which must be perpendicular to follower arm CD. This is because C can rotate only about pivot D. Using this line of action of Vc and the effective component of Vc along BC, Vc is easily found as before.

68

Graphical Techniques: Velocity Analysis PROCEDURE !•

Lay out the velocity of point B, that is, Vg (magnitude and direc¬ tion), using a convenient scale. The magnitude of this vector is given as V

B

co

AB

X AB

- 1(1) = 1 in./sec 2.

3.

4.

Determine the effective component of VB along link BC (V^C). This is obtained by dropping a perpendicular line from the ter¬ minus of Vg to meet BC extended (at right angles). Locate the effective component of Vq along BC. Since from the rigid body principle, the velocities of all points along BC must be the same, along BC lay out VBC equal to VBC. Determine the velocity of point C (Vc). Since point C is con¬ strained to move in a circular path, the line of action of Vc is known to be perpendicular to link CD. Therefore, Vc is found by dropping a perpendicular line from the terminus of VBC to

c

5.

intersect the line of action of Vq. This point of intersection defines the magnitude of Vq. Scale the magnitude of V^. Vc - 0.45 in./sec

6. Determine the angular velocity of CD (wCD). This is found from

GJ

CD = 0.45 1.75 =0.26 rad/sec

4.7

VELOCITIES OF SLIDING CONTACT MECHANISMS

An important rule in the analysis of velocities in sliding contact states as follows: ILfo0 bodies are in sliding contact, their velocities perpendicular to the sliding path are equal. As an example, consider the Scotch yoke mechanism shown in Figure 4. 6. Here slider S and yoke Y are members in sliding contact, and the sliding path is T-T. Also, P is a point on slider S as well as on crank arm OP, hence no relative motion exists between these

Effective Component of Velocity Method

69

N I

SLIDER YOKE TANGENT LINE NORMAL LINE CRANK

T

VT

Figure 4.6

(C)

Scotch yoke mechanism.

two members at this point. This means that V

P(C)

= V

P(S)

Now, since Y can have only vertical motion, all points in contact with this member, including point P, must have identical velocities to that of Y in the same vertical direction. This means that Vy must be the same as the vertical component of Vp on S or Vp on C, or

v

Y

= VNN

P(C)

= vNN

PCS)

Thus, in accordance with the sliding contact rule, both bodies—slider S and yoke Y—have equal velocities in direction N-N perpendicular to sliding path T-T. EXAMPLE 4.2 In the quick-return mechanism shown in Figure 4.7, crank OP rotates clockwise at 7 rad/sec, while slider S, to which it is attached, slides on follower F. Determine the angular velocity of the follower.

70

Graphical Techniques: Velocity Analysis

\

\ Figure 4.7

Velocity analysis of a quick-return mechanism.

SOLUTION Like the Scotch yoke just discussed, point P on the slider is the same as point P on the crank arm. However, note that points P on S and P on F, although coincidentally located, do not have identical velocities. In fact, it is for precisely this reason that sliding occurs between the members. Nevertheless, according to the rule on sliding, these velocities do share a common component in the direction normal to the sliding path. Ac¬ cordingly, if we were to determine the direction normal to the sliding path and the effective component of the known velocity in this direction, we could use this component to determine the unknown velocity. PROCEDURE 1.

Determine and lay out the velocity of P on S (Vp,oD. The magni¬ tude of this vector is given as ^

Effective Component of Velocity Method

71

OP co - 7(1) = 7 in./sec 2.

Determine the direction of velocity of P on F (Vp^). Connect P to Q with a straight line. This defines the radius arm with which point P(F) on the follower is rotating at the given instant. Therefore, through point P, draw a line perpendicularly to QP to represent the direction of Vp/p\.

3.

Determine the effective component of the velocity Vp^g) in the direction normal to sliding. The rule states that the velocities or velocity components of Vp^ and Vp^ in the direction perpen¬ dicular to sliding must be equal. a. Accordingly, through point P, construct the coordinate axis TT to indicate the path of sliding and another coordinate NN to indicate the path perpendicular to the path of sliding. b. Then drop a perpendicular line from Vp^ to NN to define the effective component V NN Note that P(S)' -NN _ yNN P(S) P(F)

4.

Find the velocity magnitude of P on F (or Vp^pp. From the terminus of V^N construct a perpendicular to intersect the P(b) line of action of Vp^, determined in Step 2. This point of inter¬ section defines the magnitude of the velocity Vp^. V

4.4 in./sec

P(F) 5.

(scaled)

Find the angular velocity of the follower, or co

= WQP w

QP

QF

:

V

P(F) = 4.4 QP 2.1

=2.1 rad/sec

Note that the velocity of sliding is given by the vectorial difference between Vp(S) and Vp^p^, or the scaled distance between the termini of these two vectors.

72

4.8

Graphical Techniques: Velocity Analysis

VELOCITY ANALYSIS OF A COMPOUND MECHANISM

Consider the mechanism in Figure 4.8, where wheel W turns clockwise at w rad/sec. We want to determine the velocity of pin C. Earlier, we saw that if the direction of a velocity and its effective component were known, the magnitude of that velocity could be readily deter¬ mined. Alternatively, it has been shown (see Chapter 3) that if two effective components of a vector are known, that vector can be determined completely. In this case, since the direction of Vc is not known nor is the path of C readily determined, it is necessary to determine V^ using two effective components of this vector: (1) the effective component along link BC (VBC), —CD and (2) the effective component along link CD (Vc ). Hence we use the velocity VB to obtain VBC and the velocity V£ to obtain V^D via point D.

Figure 4.8

Velocities of points on a compound mechanism.

Effective Component of Velocity Method

73

PROCEDURE 1. Starting at point B, lay out the velocity Vg, whose magnitude is obtained as

V

B

) AB

2.

Find the effective component Vg*'’ along BC.

3.

Locate the effective component VgE on BC.

vBC = vBC C

B

Since the motion of C is not known, Vg cannot be determined directly from VEE. However, by returning to the wheel and

4. 5. 6.

stepping off from point E in the opposite direction, we can ob¬ tain the additional information needed to define Vg. Therefore: Lay out the velocity Vg. Find the effective component VEE along DE. E-DE Locate the effective component Vg along DE. yDE = yDE D E

7.

.

8

9.

10.

Find the velocity of slider Vg. Since the motion of slider D is known, Vg is determined directly from VEE. Using Vg, find the effective component of this vector along link CD, that is, V^D. U -CD Locate the effective component Vg along CD.

Now that we have determined two effective components for the velocity at pin C, VEC and V^D, the absolute velocity V is determined by projecting perpendicular lines from the terminus of these vectors until they intersect. This point of intersection defines the magnitude of Vg.

74

Graphical Techniques: Velocity Analysis

4.9 SUMMARY The effective component method is particularly suitable for analyzing veloc¬ ities of sliding members, and is very useful when the instant centers of a mechanism are outside the limits of the drawing paper. However, a disad¬ vantage of this method is the need for the analysis of velocities of points from link to link. For complex mechanisms, this could result in consider¬ able drawing time and some loss of accuracy. Another disadvantage of the method is that it does not provide relative velocities of points on the mech¬ anism, which are essential to the acceleration analysis.

5

Instant Center Method

5.1

INTRODUCTION

One of the most effective techniques for analyzing velocities of members or links in a mechanism is the method of instant centers. In simple terms, the instant center (or instantaneous center, as it is sometimes called) has been defined as that point about which a body may be considered to be rotating relative to another body at a given instant. Applying this concept to a moving link of a mechanism makes it convenient to describe its motion, at any given instant, in terms of pure rotation about an instant center. In this context it may be noted that even a link which undergoes translation may be considered as rotating about an instant center located at infinity. In other words, any straight line may be considered to be an arc of a circle of infinite radius. The ability to describe any motion in terms of pure rotation greatly simplifies the analysis of a complex mechanism by making it more convenient to determine the velocity of any point on such a mechanism. Consequently, it is clear that the key to successful application of the method of instant centers must depend on one's ability first, to locate all possible instant centers of a mechanism, and second, to use these centers effectively to determine the required velocities. Generally, for a simple mechanism that consists of four links, this analysis presents little or no problem. However, for a complex mechanism with more than four links, experience has shown that locating all of the instant centers from first principles can be a painstaking exercise. More¬ over, once the instant centers have been found and documented, the resulting diagram is often so complex that it does not allow straightforward analysis of velocities. In this chapter we show how the velocity analysis by instant centers can be simplified by employing graphical aids such as circle diagrams and link extensions. Circle diagrams are used to help locate instant centers

75

76

Graphical Techniques: Velocity Analysis

that cannot be found easily by inspection, while link extensions aid in the visualization of relationships between the links of a mechanism.

5.2

PURE ROTATION OF A RIGID BODY

To comprehend the concept of instant centers more fully, it is necessary first to consider some basic principles regarding pure rotation of a rigid body. We have already established the fact that if a body has rotary motion, that motion can be converted to rectilinear motion using the radius as a multiplying factor. Also, the linear velocity of any point on that body acts in a direction tangential to the path of rotation and has the same sense as the rotation. For example, consider point B on the rotating body in Figure 5. la. If A is fixed, the linear velocity of B is given by the relationship Vg = ABco

(directed perpendicular to AB)

(5.1)

where VB = linear velocity of B AB = radius of B w = angular velocity Therefore, the angular velocity for the same body is given by

CO

AB

(5.2)

Similarly, Vc = ACu

(5.3)

and

(5.4) From Equations (5.2) and (5.4),

AB or

AC

(5.5)

Instant Center Method

VB _ AB _ rB

77

(5.6)

VC = AC = rc which means that Vg is proportional to Vq, as rg is to r^. The velocities Vg and V£ are represented vectorially in Figure 5.1b.

78

Graphical Techniques: Velocity Analysis

The proportionality of the linear velocities Vg and to their respective distances from the center of rotation is easily verified graphi¬ cally (see Figure 5. lb) by rotating the velocity vector Vg about center A from point B to a point B' on the radial line AC and constructing a straight line through the center A to touch the termini of Vq and Vgt (Vg relocated). This straight line is normally called the line of proportionality between tri¬ angles AB'b and ACc, where B'b = ab; Cc AC or V

b1 = ab;

Vc

AC

or V

B AB V~ " AC

, , „ (aS before)

EXAMPLE 5.1 Consider the 24-in.-diameter rotating disk shown in Figure 5.2, where A is the axis of rotation. B and C are points located on the radial lines AB and AC, as shown. If the linear velocity Vg is 2 ft/sec, as shown, deter¬ mine the angular velocity of the disk and the linear velocity of point C. Also, determine the distance of any point E on the disk where the linear velocity of V£ is 1.3 ft/sec. Locate velocity vectors Vg, Vq, and Vg. SOLUTION The angular velocity is obtained from V^ = ABco

B AB

_

2 12/12

= 2 rad/sec The linear velocity of point C is obtained from

Instant Center Method

V

79

= ACco v_/

=

~(2)

= 0.66 ft/sec

J. Li

The radius of E (or AE) is obtained from Vr = AEco hj

co

= — = 0.66 ft = 8 in. 2 Figure 5.2 shows the location of point E and linear velocities VB, V^, and VE. Note that all other points (e.g., E' and E") on the same path described by point E have the same linear velocity.

80

Graphical Techniques: Velocity Analysis

(c) Figure 5.3

Locating the center of rotation.

Instant Center Method

81

In summary, then, it is useful to note the following principles: 1.

The linear velocity is always tangential to the path of rotation of the point or perpendicular to the radial line joining the point to the center of rotation.

2.

The direction of the velocity is always in the same general sense as that of rotation.

3.

The magnitude of the velocity is always proportional to the dis¬ tance of the point from the center of rotation.

Consequently, if the velocity of one point of a rotating body is known, the velocity of any other point on that body can be determined, provided that the center of rotation is known. To illustrate these principles further, if the directions of velocities and the two points are given, the center of rotation lies at the intersection of lines drawn from the two points perpendicular to the velocity directions, as shown in Figure 5.3a. However, if these directions are the same, the location of the center of rotation also depends on the relative magnitudes of the velocities. For example: If the velocities are unequal, the center of rotation lies at the inter¬ section of the common perpendicular drawn from the tails of the two vectors with a line joining the termini of the same vectors. See the construction in Figure 5.3b. If the velocities are equal, the center of rotation lies at infinity, which means that the body has no rotation and is therefore translating. See the construction in Figure 5.3c.

5.3

COMBINED MOTION OF A RIGID BODY

Although it is relatively simple to determine velocities on a body in pure rotation about a fixed axis, it is certainly not as straightforward to find the velocities on a body in combined motion. This is because a body in combined motion is simultaneously rotating and translating and therefore has no fixed axis of rotation. However, this problem can be simplified by considering the motion to conform, just for an instant, to that of pure rotation about some center of rotation, just for that instant, and thus the velocities in the body can be found using the same principles as those applicable to pure rotation. The center of rotation in this case is aptly called the instant center of rotation, since its position changes continuously from one instant to another. The validity of this approach can be demonstrated by considering link AB in Figure 5.4a, which moves with combined motion. Imagine that the link moves from position 1 to position 2 in a very small time interval.

82

Graphical Techniques: Velocity Analysis

Then, using the construction shown in Figure 5.4b, the same motion can be seen to conform to a circular path where point I is the center. This point is the instant center of link AB. Further, if we connect points A and B (or points A' and B') to the center I by radial lines as shown in Figure 5.5, we see that the velocity directions of these points are indeed perpendicular to the four radial lines drawn, which is consistent with the principles of pure rotation.

Figure 5.4 Instant center of link in plane motion: (a) link in plane motion; (b) construction.

Instant Center Method

Figure 5. 5

83

Velocity of a link in plane motion.

Thus, if we know the velocity directions of two points on a floating link, the instant center of rotation must lie at the intersection of lines drawn fi’om those points perpendicular to the velocity directions. Therefore, using the instant center, the combined motion of a link or any rigid body can be conveniently reduced to pure rotation, thereby simplifying the velocity analysis.

5.4

VELOCITY OF A BODY WITH ROLLING CONTACT

A common example of combined motion occurs with a rolling wheel. Con¬ sider the wheel rolling to the right in Figure 5. 6. Here, as the wheel moves from one position to another, there is both rotational and translational motion.

-P Q

Figure 5. 6

Velocities of points on a rolling wheel.

84

Graphical Techniques: Velocity Analysis

To visualize this combined motion, it is convenient to think of each motion as though it occurred independently, then superpose the two effects to obtain the final result. For example, the combined motion can be ex¬ pressed graphically as shown in Figure 5. 6, where Vr and are the rotational and translational velocities for the respective points on the rim. From the summation of the rotational and translational components, it is noted that: 1.

Point A is a point of zero velocity.

2.

The velocity of each point on the rim is perpendicular to a line joining that point to point A.

3.

The magnitude of the velocity of each point is proportional to its distance from point A.

Thus point A is considered the instant center of rotation of the rolling wheel. In summary, the instantaneous center of a wheel rolling without slip¬ ping lies at the point of contact. All points of the wheel have velocities per¬ pendicular to their radii from the instant center and these velocities are proportional in magnitude to their respective distances from that center.

5. 5

TYPES OF INSTANT CENTERS

Instant centers are of three types: 1.

Fixed (type 1), that is, a stationary point in one body about which another body actually turns. This is normally a fixed axis of rotation on a mechanism.

2.

Permanent (type 2), that is, a point common to two bodies having the same velocity in each body, such as a hinged joint connecting two moving links of mechanism. The term "permanent" implies that the relative position between the connected links is always the same, regardless of the change in position of the mechanism. Imaginary (type 3), that is, a point within or outside the mech¬ anism which can be visualized as having the same characteris¬ tics as either a fixed center (type 1) or a permanent center (type 2) at any given instant. When this center behaves like a fixed center about which the body tends to turn, it is considered an instant axis of rotation.

3.

The slider-crank mechanism shown in Figure 5.7 depicts these three types of instant centers. Usually, fixed and permanent instant centers can be readily identified by inspection. Typically, imaginary instant centers must be located by more detailed analysis. Generally, the circle diagram method (described in Section 5.6) is used to determine the imaginary centers.

Instant Center Method

85

Figure 5.7 Instant centers for a slider-crank mechanism. The location of type 1 and type 2 instant centers can be determined by inspection, whereas the location of type 3 instant centers requires additional analysis.

Note that the instant center 14 (read "one-four") for the path of slider 4 on frame 1 is indeterminate because it lies at infinity. This is be¬ cause the slider path is a straight line, and therefore the slider can be con¬ sidered as a body that actually turns about a point located at infinity. An instant center that lies at infinity can be located along an infinite number of lines perpendicular to a straight path.

5. 6

LOCATING INSTANT CENTERS

Obvious Instant Centers Obvious instant centers are those that can be readily located (by inspection) on the mechanism. These may be of either fixed or permanent type. There are four types of obvious instant centers: 1. 2. 3. 4.

Instant Instant Instant Instant

center for pin-connected links (see Figure 5.8). center for sliding body (see Figure 5.9). center for rolling body (see Figure 5.10). centers for direct contact mechanisms.

a. For sliding contact between 2 and 4, instant center 24 lies at the intersection of the common normal through the con¬ tact point and the line of centers (see Figure 5.11a). b. For rolling contact between bodies 2 and 4, both instant center 24 and the contact point are coincident and lie on the line of centers (see Figure 5.11b).

Graphical Techniques: Velocity Analysis

86

Figure 5.8

Instant center for pin-connected links.

12 Figure 5.9

Figure 5.10

°°

Instant center for a sliding body.

Instant center for a rolling body.

Instant Center Method

87

Circle Diagram Method The circle diagram method is based on Kennedy's theorem, which states that any three bodies having plane motion relative to one another have three instant centers, and they lie on a straight line. PROOF Consider any three bodies 1, 2, and 3 having plane relative motion as shown in Figure 5.12. Assuming that bodies 2 and 3 are pinned to body 1, and therefore that instant centers 12 and 13 are known, the problem is to show that the third instant center of 2 and 3 must be on the straight line connecting 12 and 13. First, suppose that the instant center of 2 and 3 were at P'. Then, as a point in body 2, P' must move at right angles to line 12-P' (V^)), and as a point in body 3, P' must move at right angles to line 13-P' (V^)). This means that point P' moves in two different directions at the same time,

Graphical Techniques: Velocity Analysis

88

which is impossible. Therefore, in order for the instant center of 2 and 3 to have the same velocity direction in both 2 and 3, it must lie on the straight line joining 12 and 13. This proof also applies if all three bodies are moving. Consider the four-bar linkage shown in Figure 5.13. Let it be re¬ quired to find all instant centers of this mechanism. PROCEDURE 1.

Locate all fixed and permanent instant centers on the mechanism in Figure 5.13. These centers are usually found by inspection. For example, instant center 12 is located where link 1 joins link 2, and instant center 23 is located where link 2 joins link 3. Note that the order of the digits used to designate the instant centers is not important. That is, either 23 or 32 may be used to designate the same instant center.

34

Figure 5.13

Mechanism.

Instant Center Method

89

Point represents link number.

Figure 5.14

Step 2.

2.

Lay out points 1, 2, 3, and 4, approximately equally spaced and in sequence on a circle (see Figure 5.14), each point repre¬ senting a link on the mechanism (Figure 5.13). Any straight line drawn to connect any two of these points represents an in¬ stant center on the mechanism.

3.

Using solid lines, indicate on the diagram (Figure 5.15) the instant centers that have been located so far on the mechanism (i.e., the fixed and permanent centers) by connecting the points that represent those centers. Lines 12, 23, 34, and 14 should therefore be drawn in as solid. Connect all other points in the diagram (Figure 5.15) using dashed lines to indicate the instant centers that remain to be found. For example, lines 13 and 24 indicate those outstanding centers.

4.

5.

To locate these centers, examine the diagram in step 4 (Fig¬ ure 5.15) to find one dashed line that, if it were solid, will com¬ plete two solid triangles. For example, the dashed line 13 com¬ pletes triangles 123 and 341 (Figure 5.15). Using these triangles, we can now locate instant center 13, noting that the sides of tri¬ angle 123 represent three instant centers—12, 23, and 13 — which by Kennedy's theorem, must lie on a straight line within

Dashed line represents instant center to be found. Solid line represents instant center already found. Figure 5.15

Steps 3 and 4.

Graphical Techniques: Velocity Analysis

90

Figure 5.16

Step 5.

or outside the mechanism. Hence the instant center 13 lies somewhere on a line joining points 12 and 23, or its extension in the mechanism (see Figure 5.16). Similarly, the sides of triangle 341 represent three instant centers—34, 14, and 13—which again by Kennedy's theorem must lie on a straight line within or outside the mech¬ anism. Hence instant center 13 must lie on a line joining points

Figure 5.17

Step 5.

Instant Center Method

91

\

Figure 5.18

6.

Step 6.

34 and 14, or its extension in the mechanism (see Figure 5.17). Since by both triangles, instant center 13 is given to be some¬ where along lines 12-23 and 34-14 (Figures 5.16 and 5.17), is must be located at the intersection of these lines. Accordingly, on the mechanism (Figure 5.18) draw two straight lines, 12-23-13 and 14-34-13, and at their intersection locate instant center 13. Then, on the diagram, change line 13 to a solid line before proceeding to find the other instant center 24 (Figure 5.19).

It is important that after an instant center has been located on the mecha¬ nism, it be immediately drawn in as a solid line on the diagram. Otherwise, when locating the remaining centers, it may not be possible to find addi¬ tional pairs of triangles which are solid except for a common dashed line.

Common line that completes two triangles

Fig. 19

Step 6.

92

5.7

Graphical Techniques: Velocity Analysis

VELOCITY PROPERTIES OF THE INSTANT AXIS

A link is considered to be rotating when all points in that link remain at fixed distances from an axis of rotation. In a machine, each member is rotating about a fixed axis or about a moving axis whose location varies from one instant to another. For purposes of analysis, this moving axis may be thought of as a stationary axis having properties similar to those of a fixed axis, at any given instant. Also, it should be noted that: 1.

There is one instant axis of velocity for each floating link of a machine.

2.

There is no single instant axis of velocity for all links of a machine. Instant centers are also axes of rotation when they have no abso¬ lute motion. That is, their velocity is zero with respect to the frame. Such instant centers can be readily identified because their numerical designation includes the number representing the frame. For instance, in the slider-crank example, link 1 represents the frame and instant centers 12 and 13 are axes of rotation for links 2 and 3, respectively. Instant 14 is also an axis of rotation but is indeterminate and lies at infinity. Instant centers 24, 23, and 34 are not axes of rotation and may or may not have absolute motion with respect to the frame. Note that 12 is a fixed axis of rotation, whereas 13 is a moving axis of rotation.

3.

4.

5.8

The instant center of velocity is not an instant center of accel¬ eration, although it moves as the link moves and may have an actual zero acceleration. Neither does it necessarily have zero acceleration as does the fixed center.

VELOCITY ANALYSIS BY INSTANT CENTERS

The determination of velocities in mechanisms by instant centers is based on three principles: 1. 2. 3.

The velocity magnitude at a point in a rotating body is directly proportional to the radius of rotation of that point. The velocity direction at a point in a rotating body is perpendic¬ ular to its radius of rotation. An instant center is a point common to two bodies and has the same linear velocity (both magnitude and direction) in both bodies.

Consequently, if the linear velocity of any point in a body relative to an instant center is known, the velocity of any other point in that body, relative

Instant Center Method

93

Figure 5.20 Velocity is proportional to radius of rotation. The velocity of any point in a link is proportion to its radius of rotation about an instant center. As a result, if the velocity of one point (such as B) is known, the velocity of any other point (such as A) in that link can be determined graph¬ ically by proportions.

to the same instant center, can be determined graphically by proportions, as illustrated by the radius of rotation method in Figure 5.20. Note that instant centers must be located first before velocities can be determined. Also, in mechanism analysis, velocities are usually deter¬ mined about fixed instant centers (either normally fixed or momentarily fixed) having no velocity with respect to the frame. To illustrate how these principles are applied, let us now consider two important concepts as they apply to a four-bar linkage (Figure 5.21): (1) fixed axis (or center), and (2) link extensions. The term fixed axis is used to refer to an instant center that is normally fixed or one that is momentarily fixed (an instant axis). To identify

LINKAGE Figure 5.21

Four-bar linkage.

94

Graphical Techniques: Velocity Analysis

the fixed axis of a mechanism, we look for those centers whose numerical designations include the fixed link of the mechanism, that is, the link whose velocity is zero. Since in this example the fixed link is 1, it immediately becomes evident that the fixed axes are (12), (13), and (14). Note also that a fixed axis typically has a numerical designation which includes both the frame and the link that rotates about the axis. For example, the fixed axes (12), (13), and (14) have, respectively, links 2, 3, and 4 rotating about them. The other centers—(24), (34), and (23)—can be considered as joints on the extended mechanism. These centers may or may not have a velocity. Link extensions are imaginary bodies defined by three instant centers, one of which typically is a fixed axis, and by a common link that rotates

Figure 5.22 Link extensions for a four-bar linkage. Link extensions 2(a) and 4(c) have fixed axes of rotation. Link extension 3(b) has a moving axis rotation, denoted by the imaginary instant center 13, which for analytical purposes is assumed to be momentarily fixed.

Instant Center Method

95

about this fixed axis. To identify the link extension for a given link, we simply locate the fixed axis (or center) about which that link rotates and the other centers whose numerical designations include that link. These centers, together, comprise the link extension for the given link. The four-bar mechanism has three rotating members: links 2, 3, and 4. Link extensions for these rotating links are defined by the following sets of centers: (12) (23) (24)

(13) (23) (34)

(14) (24) (34)

where link 2 is the common link and center (12) is the fixed axis of rotation. This link extension is depicted as triangular plate (2) in Figure 5.22a. where link 3 is the common link and center (13) is the fixed axis of rotation. This link extension is depicted as triangular plate ® in Figure 5.22b. where link 4 is the common link and center (14) is the fixed axis of rotation. This link extension is depicted as triangular plate © in Figure 5.22c.

For purposes of analysis, link extensions essentially reduce complex mech¬ anisms to several simpler mechanisms, thereby reducing the computation of velocities to simple graphics. Also, link extensions help to illustrate such concepts as axes of rotation, common links, and transfer points—all of which are essential elements of velocity analysis.

5.9

VELOCITY ANALYSIS OF A SIMPLE MECHANISM

Consider again the four-bar linkage shown in Figure 5.21. For the position shown, the velocity V23 is given and the velocity V34 is required. PROCEDURE First, we determine all instant centers for the mechanism. The total number (N) of instant centers of a mechanism is given by the equation

where n is the number of links in the mechanism. To locate these centers, the circle diagram method has been found to be most useful, particularly for the more complex mechanisms. Instant centers for the four-bar linkage are shown in Figure 5.23. Next, we examine the figure to identify the fixed centers and the link extensions that tend to have rotation about these centers. Again, these link extensions are shown in Figure 5.22.

96

Graphical Techniques: Velocity Analysis

(b)

Point represents

Line represents instant center 14

Figure 5.23 Instant centers for a four-bar linkage. Instant centers 24 and 13 are both imaginary. However, center 13 is also an axis of rotation for floating link 3. (a) Instant center diagram; (b) circle diagram.

Finally, having defined the link extensions and their respective axes of rotation, we are ready to apply principles 1 and 2, using the rotation of radius method (shown in Figure 5.20) to determine the required velocities of points in the linkage. In the present example, the velocity of link 4 (V4) can be determined by proportions using the link extension (13) (23) (34) and the known velocity V23. For instance, velocity V34 at 34 is simply propor¬ tional to the radius of rotation of 34 about 13. Therefore,

V

34

= V

23 13-23

where 13-34 and 13-23 denote the radii of rotation of 34 and 23, respectively. Note that V34 is a velocity common to both links 3 and 4. This observation leads to the concept of a transfer point. A transfer P°*rct is an instant center that has the same velocity in two different links or link extensions. In the four-bar linkage, 34 is a transfer point for links 3 and 4 and for link extensions 3 and 4. Also, 23 is a transfer point for links 2 and 3, and 24 is a transfer point for links 2 and 4.

Instant Center Method

97

The concept of a transfer point provides a powerful tool for velocity analysis whereby the velocity of any point in a link can be determined without knowing the velocity of another point in the same link. For example, suppose that only V23 is known; then the velocity at any point in link 4, or its exten¬ sion, can be determined without knowing the velocity at 34. This is accom¬ plished by first finding the velocity at 24, using link extension 2, which contains the given velocity V23, then transferring the velocity found at 24 (V24) to link extension 4 to find the velocity of any other point on that exten¬ sion. This means that the velocity at point 34, which is a point on link ex¬ tension 4, can be determined as follows: 1.

The velocity at 24, using link extension 2, is

V

2.

12-24 —— 23 12-23

The velocity at 34, using link extension 4, is

V

5.10

= V

24

34

=

v

14-34

24 14-24

VELOCITY ANALYSIS OF A COMPOUND MECHANISM

Consider a steam locomotive driven by a compound mechanism having six links and 15 instant centers (Figure 5.24a). As indicated by the circle dia¬ gram, nine of the instant centers were identified by inspection, while the remaining six imaginary instant centers had to be located by detailed analy¬ sis. Given a velocity at instant center 36 in link 6, find the velocity at point B in link 2, using extended links to simplify the analysis (Figure 5.24b). First, the velocity at instant center 26 must be determined. Instant center 26 is the transfer point between links 2 and 6. Therefore, 26 has the same velocity in both link extensions 2 and 6. For link extension 6 the axis of rotation is 16 and the common link is 6. By inspection it can be determined that link extension 6 consists of instant centers 16, 26, and 36 (Figure 5.25a). Thus the velocity at the transfer point 26 is determined by proportions as

V

26

V

16-26 36 16-36

For link extension 2 the axis of rotation is 12, the common link is 2, and the constituent instant centers are 12, 23, and 26 (Figure 5.25b). Note

98

Graphical Techniques: Velocity Analysis

that transfer point 26 is common to both link extensions. Given velocity ^g, the velocity 23 is

V

23

= V

— 26 12-26

Finally, the velocity at point B is determined graphically by propor¬ tions as

V

B

= V

23 12-23

It should be noted that V-g could have been determined more directly by recognizing that the velocity at 23 is the same as that of 36. However, transfer points were used deliberately in the analysis, for instructive purposes.

Figure 5.24 Determining velocities by link extensions: (a) mechanism; (b) circle diagram.

Instant Center Method

99

Figure 5.25 Velocity analysis by link extension. Given a velocity in link 6, velocities in link 2 can be determined once the transfer point (instant center 26) is identified and the transfer point velocity (V26) is determined, (a) Link extension 6; (b) link extension 2.

5.11

SUMMARY

To find the linear velocity of a point in a link, we first identify the link (or its extension) in which the point occurs and its axis of rotation. If the veloc¬ ity of any point in this link (or its extension) is known, the velocity of that point can readily be determined from the principles of rotation. However, if there is no known velocity in this link (or its extension), we (1) select a point (i.e., an instant center) common to this link and a second link (or its extension), where the velocity of such a point can be determined from

100

Graphical Techniques: Velocity Analysis

rotation principles; (2) determine this velocity; and (3) transfer the velocity found to the first link (or its extension) to find the required velocity, again using the principles of rotation.

6 Relative Velocity Method

6.1

INTRODUCTION

The relative velocity method is probably the most common among the graphical methods used in velocity analysis. Compared to the other graph¬ ical methods, it readily provides solutions not only for absolute velocities, but also for relative velocities of points in a mechanism without requiring the location of instant centers. This singular feature makes it most desir¬ able to use when determining the relative velocities needed for acceleration analysis. The relative velocity method is based on two important concepts: (1) relative motion, and (2) the rigid body principle, considered earlier.

6.2

RELATIVE MOTION CONCEPT

Relative motion has been defined as the motion of a body with respect to another body that is itself moving. If the motion of the body is with respect to a stationary frame of reference or fixed point such as the earth, the motion is defined as absolute motion. If we think of two bodies, A and B, having independent or absolute motion, the velocity of A relative to B is the velocity that A appears to have to an observer traveling on B. To illustrate the concept of relative motion, let us consider three ways in which we can observe a car A traveling at 50 mph while seated in a second car, B. 1.

2.

Our car (B) is parked on the shoulder of the road. Then, in the adjacent lane, car A comes speeding past at 50 mph and imme¬ diately we have the experience of seeing the vehicle moving at 50 mph. Our car (B) is moving at 50 mph in one lane, and in another lane, car A is moving in the opposite direction with a speed of

101

102

Graphical Techniques: Velocity Analysis

3.

50 mph. As it speeds past our car, it appears to be moving at a speed of 100 mph. Our car (B) is now moving at 40 mph, and along comes car A, still in another lane and moving at 50 mph, but in the same direction as our car. We now have the experience of seeing car A moving at 10 mph.

Note that the velocity of car A, although the same in each case, does in fact vary, from our point of view, and that we realize the true velocity only when we make our observation from a stationary position, as in case 1. This proves that the velocity which a moving body appears to have is always dependent on the observation point or frame of reference. Consider two ships, A and B, whose velocities, and Vg, in still water are in the direction shown in Figure 6. la. It is desired to find the velocity that B appears to have to an observer on A. Since the required velocity is to be relative to A, we must consider a method of bringing A to rest. To do this, we imagine the water to be a stream moving with a velocity equal and opposite to that of A (i.e., with a velocity in -V^). This means that as fast as A moves forward, the stream moves backward with the same velocity, and consequently A makes no progress as far as the earth is concerned. The same effect would be pro¬ duced if one were to run forward on an endless belt moving with a velocity equal and opposite to that of the runner. As far as the surroundings are concerned, the runner would be stationary.

Figure 6.1

Vectorial representation of relative velocity.

Relative Velocity Method

103

Now, consider what is happening to B. Not only is it moving with its original velocity Vg, but it has added to it the velocity of the stream, which is -VA. Therefore, Velocity of B relative to A = = = =

velocity of B velocity of B velocity of B difference of

+ velocity of stream + (-velocity of A) - velocity of A velocities

Hence we can conclude that: The velocity of B relative to A is the vectorial difference between the velocities of A and B. Symbolically, V

B/A

V

B

- V

A

(6.1)

Note the order of subscripts on both sides of the equation. An alternative form of Equation (6.1) is + V

B/A

It should be noted that whereas the relative velocity is the vectorial differ¬ ence between two velocities, the resultant of two velocities is the vectorial sum of those velocities. Figure 6. lb illustrates the vector polygon that represents Equation (6.1).

6.3

THE VELOCITY POLYGON

The velocity polygon is an alternative form of the vector polygon presented earlier. To illustrate the procedure, let us consider ships A and B dis¬ cussed in Section 6.2. PROCEDURE (refer to Figure 6.2) 1.

Define a point o, called the pole, as the origin for the construc¬ tion. This is a point of zero velocity and it represents all the fixed points on the mechanism. All absolute velocities originate from this point. By "absolute velocity" we mean the real and true velocity of a body, as observed from a stationary frame of

2.

reference such as the earth._ Since the velocities VA and Vg are absolute velocities, draw VA and Vg from point o and define their respective termini,

3.

a and b. From point a, the terminus of VA, draw a third vector to ter¬ minate at point b, the terminus of Vg, thereby_closing the poly¬ gon. This vector defines the relative velocity V^^.

104

Graphical Techniques: Velocity Analysis

Figure 6.2 Vectorial representation of relative velocity: (a) vector poly¬ gon; (b) velocity polygon.

Note the agreement in the results between the vector polygon and the velocity polygon given in Figure 6.2a and b, particularly the fact that Vg/A is the same in both cases. This agreement is also validated by the vector equations shown.

6.4

VELOCITY POLYGON CONVENTION

The convention just employed in developing the velocity polygon (Figure 6.2b) can be summarized as follows: oa represents the velocity of A or V^. ob represents the velocity of B or VB. ab represents the velocity of B relative to A (or VB/A). Note that the letter to which the arrow_points indicates the velocity under consideration. In the vectors oa and ob the arrows point toward a and b, respectively. Hence the vectors _represent the velocities of A and B, respec¬ tively. Similarly, in the vector ab, the arrow points toward b and away from a; hence the vector represents the velocity of B relative to A. If the arrow were reversed, pointing toward a and away from b, the vector would represent the velocity of A relative to B.

6.5

VELOCITY POLYGON: LINKAGE APPLICATION

Application of the velocity polygon to linkage analysis is based on the follow¬ ing two important principles covered earlier.

Relative Velocity Method

1. 2.

6. 6

105

If two points lie on the same rigid body, their relative velocity is the vectorial difference between their absolute velocities. If two points lie on the same rigid body, their relative velocity is perpendicular to the line connecting the two points.

RELATIVE VELOCITY OF TWO POINTS ON A RIGID BODY

Consider floating link AB in Figure 6.3. Imagine that we were seated at end A looking toward the other end, B. Since the distance between A and B does not change (i-e., it is a rigid body) and the link has motion, B would appear to move about us in a circular path. This is the only motion that B could have. Therefore, the linear velocity of B relative to A (Vg/^) must act perpendicular to link AB, and proportional to the distance from A to B, or Va = AB X oj

(in one direction)

Similarly, if we were seated at end B looking toward A, A would also appear to have the same circular motion about us. Therefore, the linear velocity of A relative to B (V^/g) must also act perpendicular to link AB, and pro¬ portional to the distance from A to B, or V^

= AB x oj

(in the opposite direction)

As a result, it is easy to see that the linear velocity of B relative to A is equal and opposite to the linear velocity of A relative to B, or

Figure 6.3

Relative velocity of points on a link.

106

Graphical Techniques: Velocity Analysis

V

B/A

- -V

A/B

In summary, if two points, A and B, lie on the same rigid body, and the velocity of B relative to A is required, we must assume that A is fixed and that B rotates about it. Similarly, if the velocity of A relative to B is required, we must assume that B is fixed and that A rotates about it.

6.7

VELOCITIES OF END POINTS ON A FLOATING LINK

If the velocity of one end of a floating link is completely known (in direction and magnitude) and that of the other end is known only in direction, the magnitude of the latter velocity can be determined from a velocity polygon. For example, consider link AB of Figure (6.4a). Given the velocity of point A (both magnitude and direction) and the velocity of point B (direction only), determine the complete velocity, VB. Since point A has motion, the velocity of point B is obtained from the vector equation

VB = VA + V B/A where the vector VByA is the vectorial difference between the completely known velocity VA and the partially known velocity VB. Also, since points A and B are on the same link, the velocity of B relative to A (VB/A) must act perpendicularly to the line joining A to B. Hence, by laying out the velocity of A, which is known, and the velocity directions of B and VB/A, the magnitudes of both VB and VB/A can be determined. PROCEDURE 1.

Define a point o as a pole for the velocity polygon.

2.

From point o, lay out the vector VA, terminating at a point a.

3.

Through point o, draw a line "b"-"b" parallel to the given direc¬ tion of vector VB. The magnitude and orientation of this vector are defined by a point b on this line, which is as yet unknown. To define VB, we must relate the velocity of point A, which is completely known, to that of B, which is only partially known. This means that we must seek to determine the velocity of B relative to A (or_VB/A). Referring to the link, we know that the velocity vector VB/A must be perpendicular to link AB; and from the velocity polygon convention, we also know that this velocity must be directed from point a to point b on the polygon. Therefore, draw a vector in the direction perpendicular to link AB, originating from point a and terminating at a point b in

4.

Relative Velocity Method

107

’b" (a)

(b)

Figure 6.4

Velocities of end points on a link: (a) link; (b) velocity polygon.

line "b"-"b". This line, ab, represents the vector VB/A, and the line ob represents the required velocity vector, VBA quick check of the completed velocity polygon (Figure 6.4b) reveals the balanced vector equation + V

B/A

Note that it would have made no difference in the results obtained for VB if we had considered the velocity of A relative to B (VA/B) instead of the velocity of B relative to A (VB/A)._ However, as far as the velocity polygon is concerned, since VA/B is directed

108

Graphical Techniques: Velocity Analysis

opposite to Vb/^, the direction of vector Vg/A would have to be reversed, in order to be consistent with the vector equation

VB = VA - Vb

6.8

VELOCITY OF ANY POINT ON A LINK

If the velocities of two points on a link are completely known, the velocity of any other point on that link can easily be determined from the velocity triangle using the method of proportionality. Let us consider again link AB in Figure 6.5a, where the velocities V^ and Vg are both known from the velocity triangle oab (Figure 6.5b). It is required to find the velocity of a point C in the link. The velocity of point C is given by the vector equation

v

= V c

Figure 6. 5

A

+ V C/A

Velocity of any point on a link: (a) link; (b) velocity polygon.

Relative Velocity Method

109

where Vc/a> the velocity of C relative to A, must be (1) perpendicular to AC and (2) proportional to AC. For example, if point C were located at the midpoint of the link between A and B, vector Vq/A would be defined by a distance ac on the velocity polygon, where point c is the midpoint between a and b. Hence, for any point C on link AB, the velocity Vc/A obtained by selecting a point c between a and b on the velocity polygon such that ac _ AC ab ~ AB The distance ac will then satisfy both conditions of direction and magnitude of vector 'Vq/A' Therefore, a line drawn from o to c, that is, oc, will represent V^, the vector sum of Va and Vq/^. PROCEDURE 1.

Determine a point c on line ab of the triangle such that cLC

— (on the polygon) =

2. 3.

^Q

(on the link)

Join oc. Now since all vectors originating from the pole o represent absolute velocities, line oc represents magnitude and direction of velocity of C (Vc). A quick check of the velocity polygon (Figure 6.5b) will verify the vector equation

6.9

= Va + ^C/A*

VELOCITY OF ANY POINT ON AN EXPANDED LINK

In the preceding section, the point of unknown velocity considered was located on the centerline connecting two points of known velocity. Let us now consider the expanded link ABD in Figure 6.6a, where the velocities of points A and B are both given as shown, and let it be required to find the velocity of point D, located away from centerline AB. The velocity of D (VD) is given by the vector equation

v

= V D

+ V A

D/A

or V

D

V

B

+ V

, D/B

where Va and Vb

can be obtained from a velocity polygon (Figure 6. 6b).

110

Graphical Techniques: Velocity Analysis

D

Figure 6.6

(a) Link; (b) velocity polygon.

PROCEDURE 1. 2.

Develop the velocity polygon, assuming a simple link AB. (Note that this polygon is the same as that of the preceding example.) Through point a on the polygon, draw a line perpendicular to side AD on the link to indicate the direction of V-q/A' Since A is i igidly connected to D, the only direction that the velocity of D relative to A could have is one perpendicular to a line ioinine A to D. 6

3.

Similarly, to indicate the direction of VD/B, draw a line through point b perpendicular to DB.

4.

Define the point where the two perpendiculars in steps 2 and 3

5.

intersect as point d. This point defines the terminus of vector Vr-). Draw a line to connect point d to the polar origin o. This line defines the magnitude of the required vector Vj-j.

Relative Velocity Method

111

The Velocity Image: Alternative Approach In the example of Figure 6.6, we note that the relative velocity VD/A is given by the vector extending from point a to point d, and the relative veloc¬ ity VD/B is given by the vector extending from b to d. We also note that triangle abd is similar to triangle ABD, since it was produced by lines drawn perpendicular to corresponding sides of the link. Because of this similarity (or proportionality), triangle abd is often referred to as the velocity image of link ABD. The velocity image is a useful concept in velocity analysis. If the velocities of any two points on a link are known in the velocity polygon, the velocity of a third point on that link can readily be determined by construct¬ ing the velocity image, making sure that the lines which define the image are perpendicular to the corresponding lines which form the link. Also, the letters used to designate both the link and the image must run in the same cyclic order.

6.10

VELOCITY ANALYSIS OF A SIMPLE MECHANISM

Consider the four-link mechanism shown in Figure 6.7a, in which the angular velocity of link 2 or w2 is 10 rad/sec. Determine the angular veloc¬ ities of links 4 and 3. Since point C is on link 4, the angular velocity of link 4 is the same as that of point C and is given by

AB = 1.0 in. BC = 2.0 in.

Figure 6.7

Velocity analysis of a four-bar mechanism: (a) four-bar mech¬

anism; (b) velocity polygon.

Graphical Techniques: Velocity Analysis

112

GO,

= CO,

where Vq is the magnitude of the linear velocity of point C. Hence we must first determine Vq, which is given by the vector equation

V

C

= y

B

+y

C/B

Similarly, the angular velocity of link 4 is found from V co

3

C/B CB

= CO_= ——r—

CB

PROCEDURE (see Figure 6.7b) 1. 2.

Define a point o as the pole or origin for a velocity polygon. Starting from point o and using a convenient scale, lay out vec¬ tor ob perpendicular to link 2 to represent velocity V , whose magnitude is computed as = co? X

AB = 10 in./sec

3.

Through point o, draw a line "c"-"c", perpendicular to link 4, to indicate the direction of velocity Vq. Note that Vq must be perpendicular to CD since link 4 could have circular motion only about the fixed axis D. However, the orientation and mag¬ nitude are as yet unknown.

4.

To define the magnitude of Vq, we must find the velocity Vq/b, which relates point b (known) to point c (unknown) on the polygon. By definition, this velocity must be perpendicular to link BC, and by convention, must be directed to point c from point b on the polygon. Therefore, draw a line perpendicular to BC extend¬ ing from point b and terminating at line "c"-"c". This point of termination defines point c or the magnitude of Vc; and it also defines the magnitude of Vq/b.

5.

Scale velocity vectors Vq and Vc/B to obtain the magnitudes. Vc = 12.0 in./sec VQ/-g = 8.5 in./sec

6.

Finally, as required angular velocities of links 4 and 3 are given as

Relative Velocity Method

Ic

12.0

CD

1.5

113

=8.0 rad/sec

and V oor

6.11

C/B CB

8.5 =4.25 rad/sec 2.0

VELOCITIES OF SLIDING CONTACT MECHANISMS

When one body slides on another, the difference in their absolute velocities (or their relative velocity) is defined as the velocity of sliding. The velocity of sliding is always directed along the common tangent drawn through the contact point. To illustrate this concept, consider the cam-and-follower mechanism shown in Figure 6.8a, where P is the point of contact between the two bodies. The angular velocity of the cam is known and Vp(Q/p(F)> the velocity of sliding, is required. By definition, the velocity of sliding is the velocity of P on C relative to the velocity P on F, or V

, = V - V P(C)/P(F) P(C) P(F)

The velocity of sliding must, by definition, be along the common tangent T-T. Therefore,

Vp(Q/P(F)

can be determined as follows.

PROCEDURE 1. 2.

Define polar origin o to start the velocity polygon (Figure 6.8b). From origin o, lay out known velocity Vp^Q (magnitude and direction) perpendicular to OP. OP x w

(directed perpendicular to OP)

3. 4.

Define the terminus of Vp^Q as p(C). Through the origin draw line "p(F)"-"p(F)n perpendicular to link OP to indicate the direction of Vp^.

5.

To define Vp(Q/P(F)» draw a line starting from p(Q extending toward line "p(F)"-"p(F)u in the known direction of sliding (i.e., parallel to tangent line T-T). The point at which the two lines intersect defines the terminus of Vp/pp and line p(F)-p(C) de¬ fines the magnitude of Vp^Q/p^pp Vector

Vp(C)/P(F)

is pointed

Graphical Techniques: Velocity Analysis

114

(a]

(b)

"p(F)"

Figure 6.8 Velocities of points on a cam mechanism: (a) mechanism; (b) velocity polygon.

Relative Velocity Method

115

toward p(C), in accordance with convention. A quick check of the completed polygon reveals the vector equation V

. = V - V P(C)/P(F) P(C) P(F)

EXAMPLE 6.1 Consider the quick-return mechanism shown in Figure 6.9a. Crank OP rotates at 2 rad/sec and slider S slides on follower F. Determine the angu¬ lar velocity of the follower and the velocity of sliding. SOLUTION To find the angular velocity of the follower, we must first find the linear

Figure 6.9

Velocity analysis of a quick-return mechanism: (a) mechanism

(b) velocity polygon.

116

Graphical Techniques: Velocity Analysis

velocity of point P on F (VP(Pp. This unknown velocity is related to the known velocity, Vp^, as follows: V

= V

P(F)

P(S)

+ V

P(F)/P(S)

where Vp/pwp/gx is the velocity of sliding (or the velocity of slip), which is directed along the instantaneous path of the slider. After determining the velocity of point P on F, we obtain the required angular velocity from the relationship V

w

JPiD F

QP

PROCEDURE 1. 2.

Define polar origin o to start the velocity polygon (Figure 6.9b). Lay out the velocity of point P on S (Vp/gA perpendicular to OP. The magnitude of this velocity is obtained from = co(OP) = 2(1.5) = 3 in. /sec

3.

Define the terminus of the vector VP(S) as p(S).

4.

Through point o draw a line "p(F)"-Mp(F)M to indicate the known direction of Vp^p^.

5.

Lay out the relative velocity (or velocity of slip) starting from and moving in the known direction of slip toward line "p(F)"-"p(F)M. This relative velocity vector intersects line "p(F)"-"p(F)" and defines the magnitude of VP/F\ as well as that of the slip velocity Vpjgyp^p^ at that point. Label the point of intersection p(F). Note that the orientation of the velocity of the sliding vector depends on whether we consider the velocity of P(S)

the slider relative to the follower, where vector Vp(S)/P(F) is is pointed toward p(S), or the velocity of the follower relative to the slider, where vector Vp,FwP(S) would be pointed in the opposite direction, toward p(F). 6.

Scale vectors VP(F) and Vp^gyp^p^ to obtain their magnitudes as follows:

VP(F) = 1,7 in-/sec and VP(S)/P(F)

2.0 m./sec

Relative Velocity Method

7.

117

Determine the angular velocity of F from the relationship

VP(F) = WQP X QP Therefore, V CO

QP

Pill

QP 1.7 2.35

6.12

0.72 rad/sec

VELOCITIES OF A BODY WITH ROLLING CONTACT

Consider wheel W in Figure 6.10a, which rolls, without slipping, on a track. As the wheel moves to the right, the center A is considered the moving frame of reference and the velocity of every other point on the wheel is relative to the velocity of this point. Therefore, we apply the rela¬ tive motion concept to determine the velocities of points B and C as follows. PROCEDURE 1. 2.

Define polar origin o (Figure 6.10b). Locate the velocity vector VA- The direction of this vector is perpendicular to a line joining A to the instant center I (or AI), and the magnitude is given by V

3. 4.

5.

6.

A

= AI X co

Define the terminus of VA as a. Lay out the direction of the vector Vg perpendicular to a line joining B to I (or BI). Therefore, draw a line "b"-"b", of undefined length, through pole o. Determine the vector Vg. a. Determine the vector Vg/^. By definition, Vg/A assumes that point B rotates about point A and is perpendicular to a line connecting B to A (i.e., AB) on the wheel. Also, this velocity is represented by a vector headed from a to b on the polygon. b. Therefore, from a draw a line in a direction perpendicular to AB to intersect line "b"-"b". This point of intersection defines the termini of vectors Vg/A and Vg. Label the point of intersection b. A quick check of the polygon should immediately reveal the vector equation:

Graphical Techniques: Velocity Analysis

118

Figure 6.10

Velocities of a body with rolling contact.

V

B

= V

A

+ V

, B/A

Following the procedure described above, the velocities of point C (Vc) can be determined as shown in Figure 6.10c.

Relative Velocity Method

6.13

119

VELOCITY ANALYSIS OF A COMPOUND MECHANISM

Consider the toggle mechanism in Figure 6.11a. Wheel W rotates at 10 rpm Find the velocities of slider S and point P.

Figure 6.11 Velocities of points on a toggle mechanism: (a) mechanism; (b) velocity polygon.

Graphical Techniques: Velocity Analysis

120

We first note the following: 1.

The velocity of the slider is the same as that of point E, that is, VE-

2.

3.

Linkage may be analyzed in two parts: a. The four-bar section ABCD, to determine V^. b. The slider-crank section DCE, using the value found for in part (a) to find VE. VE is found using the velocity image of link CE.

PROCEDURE 1. 2.

Define polar origin o to start the velocity polygon (Figure 6. lib). Layout, using a convenient scale, velocity vector Vg- The magnitude of this vector is given as

VB

AB x

AB

Therefore, w. ^ = lo(§~) = 1.05 rad/sec AB ' 60 '

3. 4. 5.

6.

Define the terminus of vector Vg as point b. Through the origin draw a line "c"-"c" perpendicular to CD to indicate the orientation of velocity vector V^. Lay out relative velocity Vc/g. Starting from point b, draw a line heading toward line "c"-"c" in a direction perpendicular to link BC. This line should meet "c"-"c" at point c, which defines the termini of both Vq and Vg/£. Scale the magnitude of Vq, that is, the line oc. Vc = 1.06 in. /sec

7. 8.

9.

Again, through the origin, draw the line "e"-"e" to indicate the orientation of the absolute velocity vector VE (horizontal). Lay out relative velocity vector VE/C- Starting from point c, draw a line heading toward line "e"-"e" in a direction perpen¬ dicular to link CE. This line should meet line "e"-"e" at point c to define the magnitudes of both VE and VE/£). Scale the magnitude of VE, that is, the line oe.

10. To find the velocity of point P (Vp), locate a point p in line ec on the polygon such that the ratio of cp to pe is the same as ratio of CP to PE on the link, or cp _ CP pe PE

Relative Velocity Method

11. 12.

Connect point p with a vector originating from point o. This vector, op, then defines the velocity Vp. Scale the magnitude of Vp. Vp = 0. 58 in./sec

121

II.B GRAPHICAL TECHNIQUES: ACCELERATION ANALYSIS Acceleration is a very important property of motion to a machine designer. By Newton's law, F = ma, the force imposed on machine members is directly proportional to the acceleration. Although fundamentally kinematics is not concerned with forces, the study of acceleration is vitally important to the dynamic analysis, because of its influence on stresses, bearing loads, vibration, and noise. In high-speed machines, the accelerations and resulting inertial forces can be very large compared to static forces which do useful work. For example, in modern automobile and aircraft engines, the stresses im¬ posed on the connecting rods as a result of these accelerations are consider¬ ably greater than those produced on the piston by gas pressure. Also, a small imbalance of the rotating members can produce forces that are many times larger than the weights of the members. Therefore, a complete accel¬ eration analysis is a prerequisite to stress analysis and proper design of machine members.

123

7 Linear Acceleration Along Curved Paths

7.1

INTRODUCTION

A body moving along a curved path is always subjected to an acceleration since its velocity changes from one position to the next. If the body moves with a constant angular speed, the velocity changes is one direction only, and the resultant acceleration is a normal acceleration. If the body moves with a variable angular speed, the velocity change is one of direction and magnitude, and the resultant acceleration is a summation of the normal acceleration (due to the direction change) and a tangential acceleration (due to the magnitude change).

7.2

NORMAL ACCELERATION

Consider a point A on a body which moves in a circular path of radius R with constant angular speed through an angle Ad (Figure 7.1a). The initial velocity at point A^ is V1 and the final velocity at point Af is Vf . A A The velocity polygon representing this motion is given in Figure 7.1b. Here the resultant change in velocity of point A (AVA = Vf - V1 ) is in direc—j _f A A tion only, since and are equal in magnitude. Also, for a small angu¬ lar displacement Ad, Ais oriented normal or perpendicular to the instantaneous linear velocity. The normal acceleration magnitude of point A or A a is obtained from

v[A - v1A AT

AT

where

124

Linear Acceleration Along Curved Paths

Figure 7.1

125

Disk with uniform rotation.

AV N = _A aa at AVa = VA Ad A A

(for small Ad)

N = VA A° at

aa But Ae AT

w

A^ = V go A A = Rco

X

co = Rco2

Also, V R

an = v ^ A A R

Ia R

(7.1)

This relationship states that the magnitude of the acceleration of any point on a body rotating at a constant speed is equal to the square of the angular velocity of the body multiplied by the distance of the point from the center

Graphical Techniques: Acceleration Analysis

126

of rotation, or the square of the linear velocity divided by the distance of the point from the axis of rotation. This acceleration called the normal acceleration, is always directed radially toward the center of rotation per¬ pendicular or normal to the instantaneous linear velocity. A familiar example of radial or normal acceleration is that of a car traveling around a circular track at a constant speed. Although the speedom¬ eter reading remains the same, there is acceleration because the direction of velocity is continually changing. Whenever there is a change of direction of velocity, there is acceleration. Furthermore, it is the normal accelera¬ tion that keeps the car on its circular path. If it ceases, the car will at once move in the direction tangent to the circular track. It should be noted from Equation 7.1 that if the point has rectilinear motion, the normal acceleration is always zero. That is.

=

CO

7.3

0

TANGENTIAL ACCELERATION

Consider a point A on a body that moves in a circular path of radius R with increasing angular speed u> through an angle Ad (Figure 7.2). As before, V1^ is the initial velocity at point A^ and is the final velocity at point Af. A

Here, for convenience, we consider the velocity change due to mag¬ nitude only. This gives the vector equation

AV

vf - v1

A

A

A

(in tangential direction only)

Therefore, the magnitude of the tangential acceleration of point A(A

AV A. =

A AT

£ * Rrn - Roj AT R Aco AT But

AjJ AT Therefore,

=

a

A

) is

Linear Acceleration Along Curved Paths

Figure 7.2

127

Disk with nonuniform rotation.

(7.2)

This relationship states that the magnitude of the tangential acceleration of any point on a rotating body whose angular velocity is changing is equal to the angular acceleration multiplied by the distance of the point from the center of rotation. (Note: This is the same relationship as that derived by an alternative method in Section 2.3.)

7.4

RESULTANT ACCELERATION

The total or resultant acceleration of a point A (or A

) (Figure 7.3)

is the vectorial sum of the normal or radial acceleration (A^) and the tan-T A gential acceleration (A^). Symbolically,

where the magnitude is given by n"2

(Aa)

t”1

+(Aa)

= \l( Roo2)2 + (Ra)Z

The direction of this acceleration or the angle cp which the vector makes with the radius R is obtained from

Graphical Techniques: Acceleration Analysis

128

Figure 7. 3 Resultant acceleration of a point with nonuniform rotation.

at

4 = tan"1 >

aa EXAMPLE 7.1 A, B, and C are points on a rigid body that rotates about center O as shown in Figure 7.4. Given that the angular velocity u is 2 rad/sec and the angular acceleration a is 3 rad/sec2, calculate and check graphically the magnitudes and phase angles of the resultant acceleration of A, B , and C . SOLUTION The normal acceleration magnitude A^ is given by aN 2 A = ror Therefore, A^ = 4(2)2 = 16 in./sec2

Ag = 1(2)2 = 4 in./sec2

A^ = 5(2)2 = 20 in./sec2

The tangential acceleration magnitude A

T

is given by

Linear Acceleration Along Curved Paths

129

aT A = ra

Therefore, T A^ =4(3) = 12 in./sec2

A^ = 1(3) =

3 in./sec2

-D

T A^ = 5(3) = 15 in./sec2

The resultant acceleration magnitude A is given by / N1 T^ A - N (A ) + (A ) Therefore, A^ = \f(16)2 + (12)2 = 20 in./sec A

15

A

= \T(4)2 + (3)z =

= 5 in./sec

20)2 + (15)2 = 25 in./sec

v_/

A

Figure 7.4

Example problem.

OA

=

4

OB

=

1 in

OC

=

5

in

in

130

Graphical Techniques: Acceleration Analysis

The phase angle 0 is given by

0 = tan-1 — A 0A = tan-1 || = 36.8° 3 0Q = tan-1 - = 36.8° -D

4

0c =

tan_1i=

36-8°

Figure 7.4 also shows the required graphical solution. Note that this problem could have been solved using the fact that both normal and tangential components are proportional to their distances from the center of rotation. That is, having found the values of components for point A, we could have determined the values of the corresponding compo¬ nents for points A and B by simple ratio. Also, note that the phase angle 0 is the same for all points on the link. This is because this angle depends only on the angular velocity and angular acceleration of the link, as can be seen from the following deri¬ vation:

0

tan-1

tan -l

ra rw2

tan-1 % GO

EXAMPLE 7.2 In Figure 7.5, pulleys P2 (2 in. in diameter) and P3 (6 in. in diameter) are driven by pulley Px (4 in. in diameter) via a continuous belt. At this instant Px has an angular velocity of 1 rad/sec counterclockwise, and an angular ’ acceleration of 5 rad/sec2. Determine the acceleration of points ABC and D on the belt. ’ ’ ’ SOLUTION For point A on pulley Px:

Linear Acceleration Along Curved Paths

131

C

Normal acceleration: A^ = rjco2

(directed toward center of pulley)

= 2(1)2 = 2 in./sec2

(directed toward center of pulley)

Tangential acceleration: -T A = rjd! A

(directed along the belt)

= 2(5) = 10 in./sec2

(directed along the belt)

Resultant acceleration: A

_N _T = A + A A A A

(vectorial sum)

= 10.2 in./sec2

(directed as shown)

132

Graphical Techniques: Acceleration Analysis

For point B on the belt:

Ag = A^

(same as tangential acceleration of point A)

= 10 in./sec2

(directed along the belt)

For point C on pulley P3: Normal acceleration: -N . A^ - r^

(directed toward center of pulley)

-

= 4/3 in./sec2 (directed toward center of pulley)

Tangential acceleration: -T Ac - Ag

(same as belt acceleration)

- 10 in./sec2

(directed along the belt)

Resultant acceleration: -N -T Ac = Ac + Ac

(vectorial sum)

Aq = 10.1 in./sec2

(directed as shown)

For point D on pulley P2: Normal acceleration: r-N _

,

ad ~ r2w _ Hi/

(directed toward center of pulley) -4 in./sec2

(directed toward center of pulley)

Tangential acceleration: - Ag

(same as belt acceleration)

= 10 in./sec2

(directed along the belt)

Resultant acceleration: - _ -N -T AD = AD + ad

(vectorial sum)

= 10.8 in./sec2

(directed as shown)

Figure 7.5 also shows the graphical solution.

Linear Acceleration Along Curved Paths

7.5

133

PROPORTIONALITY OF ACCELERATIONS

It was shown earlier that if a body rotates about a fixed point, the normal and tangential accelerations of a point located on that body at a distance r from the center of rotation are given by

Figure 7. 6

Proportionality of accelerations: link OBA.

134

Graphical Techniques: Acceleration Analysis

A

= rw2

-T A = ra

(directed along r toward the center) (directed perpendicular to r)

Therefore, if we consider points A and B on link OAB in Figure 7.6, it is easy to see that

"B A and N B N A

B ‘A

which indicates that both normal and tangential components of any two points along the radial line OA are proportional. Similarly, since the acceleration components of the points are proportional, the sum of these components, or the total acceleration, must be proportional. A graphical representation of the proportionality of accelerations Ab and Aa and their components can be seen in Figure 7.6, where the ter¬ mini of these vectors lie on respective straight lines (termed lines of pro¬ portionality) that pass through the center of rotation O. Note that the propor¬ tionality line for the normal components Ab and Aa is a straight line drawn through the center O to touch the termini of these vectors when rotated 90° from the radial line.

7.6

RELATIVE ACCELERATION OF WO POINTS ON A RIGID BODY

Just as in the velocity case, if the absolute accelerations of two points on a body are known, the relative acceleration between these points is the vec¬ torial difference of their absolute accelerations. For example, if in link AC shown in Figure 7.7a the accelerations of both points A and C are known, then (Figure 7.7b)

AA/C = AA~AC

(7-3)

ac/a

(7.4)

or

Linear Acceleration Along Curved Paths

135

C

Figure 7.7

c

Relative acceleration of points on a rigid body.

Normal and Tangential Accelerations Using the vector A^/c* we can resolve this vector into its normal and tan¬ gential components simply by dropping two perpendiculars from the terminus of this vector: one to the line AC and the other to a line normal to AC passing through point A. This construction yields along AC and A^ along the line normal to AC (see Figure 7.8). For the vector Aq/a, a similar construction can be made at point C. Consequently, the components for A^ and A^ will have opposite direc-N -T /A / tions to those of A^/c an<^ Aa/c* Normal and tangential acceleration relationships for a point relative to another point on a link are similar to those for a point that undergoes pure rotation. For example, in Figure 7.8, where the acceleration of point A relative to point C (which is assumed to have a nonzero acceleration) is known, the following relationships apply: 1.

Normal acceleration of A relative to C:

an aa/c

ACXuac

(directed from A to C)

136

Graphical Techniques: Acceleration Analysis

2.

Tangential acceleration of A relative to C: -T

~ AC x ap^Q 3.

(directed perpendicular to AC

Resultant acceleration of A relative to C:

AA/C = AA/C + AA/C 4.

(vectorially)

Magnitude of acceleration of A relative to C:

r

A/C1

. n ~~2 t 2 = "J(AA/C) +(A,^) a/c'

5. Direction of resultant acceleration A A/C' A (p = tan 1 A

T A/C N A/C

It should be evident that if the accelei-ation of point C were zero, Equation (7.4) would become

Linear Acceleration Along Curved Paths

137

and all relative relationships would be reduced to absolute relationships, as in pure motion. EXAMPLE 7.3 In link BC shown in Figure 7.9a, the angular velocity of point C relative to point B is 1 rad/min (counterclockwise) and the relative angular accelera¬ tion is 0.5 rad/min2 (clockwise). If the absolute linear acceleration of point B is 5 in./min, as shown, determine the absolute linear acceleration of point C. SOLUTION The absolute linear acceleration of point C can be represented in terms of the acceleration of point B and the relative motion of point C to point B, by the vectorial relationship C

Figure 7.9

Example problem: (a) floating link; (b) acceleration diagram.

138

Graphical Techniques: Acceleration Analysis

(7.5) where A

-N -T = A + A C/B AC/B AC/B

(7.6)

But

= 3(1.0)2 = 3 in./min2 and T ac/b = BC x “ = 3(0.5) = 1.5 in./min2 Therefore, application of Equations (7.5) and (7.6) yields

7.7

AC/B = 3'4 in,//min2

(directed as shown in Figure 7.9b)

AC

(directed as shown in Figure 7.9b)

= 4,2 in*/min2

ACCELERATION OF ANY POINT IN A FLOATING LINK

Method 1: Normal and Tangential Component Method Consider floating link AB shown in Figure 7.10a. Suppose that the absolute accelerations of two points, A and B, are known and the absolute accelera¬ tion of a third point, C, is required. The acceleration of point C can be found by applying the relationship

where Ac/A is unknown but can be determined by proportionality with A which is obtainable from

Linear Acceleration Along Curved Paths A

Figure 7.10

B

Method 1.

139 C

140

Graphical Techniques: Acceleration Analysis

The proportionality between AC/B and AB/A is based on the fact that A, B, and C are points on the same link and therefore have the same angular veloc¬ ity and acceleration. Hence

C/A an B/A

CA BA

at C/A _ CA T BA B/A PROCEDURE 1.

Determine the acceleration of point B relative to point A (Fig¬ ure 7.10b), using the vector equation

AB/A = \ " AA Resolve vector AB//A into its normal and tangential components, AB/A along the link and AB/A perpendicular to the link (Figure 7.10c). Determine A^/a (Figure 7.10c). Since A^/a is proportional to aB/A> complete the following steps: a. From point A, draw a line proportionality for the tangential accelerations, touching the terminus of . b.

From point C, construct a vector perpendicular to AB, ter¬ minating^ the proportionality line in step (a). This vector defines A^ , .. C/A

Again from point A, draw a second line of proportionality for the relative accelerations, touching the terminus of AB/A. d. From point C, construct a vector parallel to AB/A, termi¬ nating at the second line of proportionality in step (c). This vector defines Ac/A. Determine Aq (Figure 7. lOd) using the vector equation

Ac '

aa + ac/a

Method 2: Orthogonal Component Method Consider again floating link AB in Figure 7.11a, where the accelerations of points A and B are known and the acceleration of point C is required. In this

Linear Acceleration Along Curved Paths

141

method we make use of the fact that since A, B, and C are on the rigid body, their orthogonal (rotational and translational) components of acceleration must be proportional to each other. If we consider the rotational components for all points on the link centerline, the termini of these vectors must lie on a straight line. This line may be termed the line of proportionality for the rotational components of acceleration.

Graphical Techniques: Acceleration Analysis

142

Similarly, if we consider the translational components, the termini of these vectors, when rotated through a common angle away from the link centerline, must lie on a straight line. This line may be termed the line of proportionality for the translational components of acceleration. Therefore, to determine the acceleration of point C, we must first determine the rotational and translational components of acceleration at point C, using their respective lines of proportionality. Then, by adding these components, we obtain the required absolute acceleration of point C. Following is the procedure for construction. PROCEDURE

.

1

2.

3. 4.

Resolve vectors A^ and Ag into their translational and rotational components (along the link and perpendicular to the link). Establish the line of proportionality for rotational components of acceleration. This is a straight line drawn to touch vectors Ar , rr A and Ab. _r Determine Aq. This is a vector drawn from point C perpendicu¬ lar to the link and terminating at the proportionality line. Establish the line of proportionality for translational components of acceleration. This is obtained by first rotating vectors A1 -t A and Ag in a direction perpendicular to the link, then joining the termini with a straight line. Note that the line of proportional¬ ity is used to obtain the magnitudes only (not directions) of the translational components of accelerations of all points along the link.

5.

Determine the magnitude of Ac- This is obtained by drawing an equivalent vector A^, which extends from point C perpendicular to the link and terminates at the proportionality line. To obtain the true direction of A* the equivalent vector A^ must be ^ G t rotated 90° to act in the direction consistent with A^ and Ag along the link.

6.

Determine Ac- This is obtained by summing the tangential and translational components A?!, and A* obtained above.

Note also that if the directions of the given translational components of the two points oppose each other, this will indicate the the vectors are propor¬ tional with respect to a point that lies between the two points. Hence these vectors must be rotated to opposite sides of the link centerline to establish the line of proportionality. This line of proportionality must therefore inter¬ sect the line joining the two points.

Linear Acceleration Along Curved Paths

143

Method 3: Instant Center of Acceleration Method It has been established that the components of acceleration of a point on a rotating body are proportional to the distance of that point from the center of rotation. Therefore, if we consider any floating link, such as AB in Fig¬ ure 7.12a, having the same conditions as in methods 1 and 2, by locating the center of acceleration of the two given points A and B, we can readily determine the acceleration of any other point, such as point C, by proportion.

A

Figure 7.12

B

Method 3.

C

Graphical Techniques: Acceleration Analysis

144

Here it should be noted that the center of acceleration or point of zero acceleration is the center of rotation. This should not be confused with the instant center used for velocity analysis. Although the instant center as defined for velocity analysis does have zero velocity, it may or may not have zero acceleration. To determine the center of acceleration we employ the following construction, called the four-circle method. The Four-Circle Method (Figure 7.12b) 1. 2. 3.

4.

Extend the lines of action of vectors A and B until they intersect at a point K. Construct a circle to circumscribe the triangle formed by points A, B, and K. Construct a second circle to pass through point K and the ter¬ mini of vectors A^ and Ag. This circle will intersect the first circle at another point, J. This point locates the center of accel¬ eration of points A and B. Connect points A and B to point J, using straight lines. Lines AJ and BJ are, therefore, the radii of rotation for points A and B.

Note the proportionality between the absolute acceleration vectors Aj± and Ag and their respective radii of rotation. Also note the equality between phase angles JAK and JBK. PROCEDURE (for locating point C) Having determined the center of acceleration, complete the following steps: 1. 2. 3. 4.

Draw a straight line to connect point C and center J. Rotate vector Ag to a point B' on radius CJ. Using the rotated vector Ag> as a gauge, construct a propor¬ tionality line from point J to pass through the terminus of Agi. From point C, construct a vector parallel to Agt and terminating at the line of proportionality. This vector defines the required acceleration of point C (Ac) •

Note that although called the four-circle method, the method requires only two circles to locate the center of acceleration. The method takes its name from the four-circle theorem in geometry, which states that the four circles that circumscribe each side of a quadrilateral, and the intersection formed by the extension of two adjacent sides, intersect at a point.

Linear Acceleration Along Curved Paths

145

Method 4: Relative Acceleration (or Polygon) Method Consider again floating link AB where, as before, the accelerations of points A and B are known and the acceleration of point C is required (Figure 7.13a). This method employs the acceleration image concept, which is similar to the velocity image concept considered earlier. Based on the geometric similarities between points on the link and their relative accelerations on the polygon, we can use proportions to determine the rela¬ tive acceleration of any other point on the link, and hence the absolute acceleration of that point. PROCEDURE (Figure 7.13b) 1.

Define a point o', called the pole, from which all absolute accel¬ eration vectors originate.

2.

Lay out given acceleration vectors A^ and Ag originating from the pole and label the respective termini a' and b'.

3.

Draw a straight line to connect terminus a' to terminus b'. This line represents the magnitude of the relative acceleration be¬ tween points B and A (or A and B) and, as in the velocity case, is called the acceleration image of link AB. Locate point c on line a'b' in step 3 such that

4.

Figure 7.13

Method 4.

Graphical Techniques: Acceleration Analysis

146

a'c' _ AC a'b' " AB

5.

7.8

This point defines the terminus of vector A(j, the required accel¬ eration vector. Construct the vector Aq extending from the pole and terminating at point c' to determine its magnitude and direction.

CORIOLIS ACCELERATION

If one link slides radially on another link which is rotating, the sliding link (or slider) experiences an acceleration perpendicular to the radial line join¬ ing it to the center of rotation. Part of this acceleration is the effect of the changing distance of the slider from the center, and part is the effect of rotation of the radial sliding velocity vector. The tangential acceleration of the slider relative to the other link is called the Coriolis acceleration. Consider rotating link AD in Figure 7.14, which turns at a constant velocity w about a fixed axis A as slider S freely slides (radially outward) on it. B and C are two coincident points at a distance r from A. Point C is on link AD directly beneath point B, which is on the slider. FIRST CHANGE (Figure 7.15) Consider the relative velocity of point B to point C radially along the link as the slider rotates from B, C to B', C' (a change in the direction of VB/C due

Figure 7.14

Rotating link with slider.

Linear Acceleration Along Curved Paths

147

to rotation). Assuming that Ad is small, the magnitude of AV can be ex¬ pressed as AV = VB/C Ad and AV = A0_ AT VB/C AT Therefore,

ab 1' = Vc “ or A(1> B

B/C

co

(in the direction of AV)

SECOND CHANGE (Figure 7.16) Consider the tangential velocity of slider S as it moves outward from the center (a change in Vg due to the change in slider distance from A). Here the magnitude of AV is given by AV = VB, - VB = Arco and

148

Graphical Techniques: Acceleration Analysis

AY _ Ar AT AT Therefore,

ab2) = Vc “ or r( 2) Ag =

(in the direction of AV)

SUMMARY Combining the results of the two changes, we obtain the Coriolis acceleration:

AC°r = a£> + A® or rC/Or

A

-

(in the direction of AV)

The Coriolis acceleration vector is always directed perpendicular to the rotating link and pointed as if it had been rotated about its tail through an angle 90° with vector VB/c in the direction of rotating link on which the sliding occurs.

go,

the angular velocity of the

Linear Acceleration Along Curved Paths

PQ U

(a)

(b)

(c)

(d)

Figure 7.17 Coriolis acceleration for the four cases of Example 7.4: (a) case 1; (b) case 2; (c) case 3; (d) case 4.

150

Graphical Techniques: Acceleration Analysis

EXAMPLE 7.4 (Figure 7.17) Consider a block B which slides on a rotation rod AD and has a velocity of 5 ft/sec relative to a point C on the rod (Vg/Q = 5 ft/sec). The angular velocity of the rod is 4 rad/sec. Determine the Coriolis acceleration in terms of both magnitude and direction for the following cases. 1. 2. 3. 4.

Block B moves outward and toward D as AD rotates counter¬ clockwise (Figure 7.17a). Block B moves inward and toward A as AD rotates counter¬ clockwise (Figure 7.17b). Block B moves inward and toward A as AD rotates clockwise (Figure 7.17c). Block B moves outward and toward D as A rotates clockwise (Figure 7.17d).

SOLUTION The magnitude of the Coriolis acceleration in each case is given by Cor 2Vb/c“ad = 2(5)(4) = 40 ft/sec2 Figure 7.17 shows the direction of the Coriolis acceleration in each case. Note that the direction of ACor is obtained from rotating the vector VB/C through an angle 90° in the same direction (clockwise or counterclockwise) as that of rotating link AD.

8 Effective Component of Acceleration Method

8.1

INTRODUCTION

The effective component of acceleration method is similar to that of the effective component of velocity method in that it is also based on the rigid body principle. However, it is somewhat more complex for bodies with combined motion since it involves the concepts of relative acceleration and relative velocity. It will be seen that the effective component of acceleration along a line connecting two points, A and B, on a rigid body is composed of two parts:

8.2

1.

The acceleration component of point A relative to point B due to rotation of A about B

2.

The acceleration component of the reference point B due to translation

ACCELERATION OF END POINTS ON A LINK

Consider link AB shown in Figure 8.1. Both the velocity and acceleration of point A are known in magnitude and direction, whereas only the line of action of the acceleration of point B (or Ag) is known. This line of action is the line "b"-"b". Determine the complete acceleration of point B (Ag). The acceleration of point B is related to that of point A by the vector equation

B

aa + ab/a

and similarly, the effective components of A^ and A related by

151

along link AB are

152

Graphical Techniques: Acceleration Analysis

-AB -AB , -AB A„ = A . + A^ , 4 B A B/A To determine Ab/a> we must consider the motion of point B relative to point A. That is, we assume that point A is fixed while B rotates about it. Therefore, the only effective component of acceleration that B can have along link AB is the normal or radial acceleration (A^ ). In other words,

ab/a

ab/a

where N = B/A

V2 B/A AB

Therefore, AB = AB/A

V2 B/A AB _ AD

This means that we must find Vb/a bi order to find Ab/a- vB/A can determined from a velocity polygon.

I

Figure 8.1

Linkage.

Effective Component of Acceleration Method

153

"b"

Figure 8.2 polygon.

Acceleration of end points of a link: (a) linkage; (b) velocity

Graphical Techniques: Acceleration Analysis

154

PROCEDURE (Figure 8.2a) 1.

At point A, lay out the effective component of the given accel¬ eration Aa, that is, AAB. This is obtained, as in the velocity case, by dropping a perpendicular line from the terminus AA

2.

to the extended line AB. Find A^B on link AB. This is obtained from the relationship B -AB -AB aa + ab/a

3.

where Ag/A is the radial or normal component of acceleration of point B relative to point A. _ Determine the effective component of Ag on link AB (Ag ). a. Determine Vg/A. This may be obtained from a velocity poly¬ gon, as in Figure 8.2b. b. Calculate the magnitude of A^B .

AB AB/A

N AB/A

V2 B/A AB

-AB c. From point B, lay out vector Ag (the summation of vectors AB/A anc* Af> alonS Bnk AB. 4.

8.3

Knowing the path of acceleration Ag to be along line "b"-"b", this vector can now be defined by constructing a perpendicular line from the terminus of A^B to meet line "b"-"b", similar to B the procedure used in the velocity case.

SLIDER-CRANK ANALYSIS

Crank AB of the slider-crank mechanism shown in Figure 8.3 rotates with an angular velocity of 2 rad/sec (counterclockwise) and an angular accelera¬ tion of 2 rad/sec2 (counterclockwise). Determine the acceleration of the slider. The acceleration of slider C is related to that of point B by the vector equation

C

ab + a C/B

and similarly, the effective components of these accelerations along BC are related by

Effective Component of Acceleration Method

155

B

-BC

Ac

a-BC

=ab

-BC

+ac/b

where abc = an ac/b c/b and N = X/B

V2 C/B BC

PROCEDURE (Figure 8.4b) 1.

From point B, lay out acceleration vector Ag to a convenient scale using the vector equation

where A

N = AB B

,

X co2

= 1. 5(2)2 = 6.0 in./sec2 A

T = AB 13

X

a

= 1.5(2) = 3 in. /sec2 2-

3.

-BC Construct the effective component of Ag on BC (Ag ). Drop a perpendicular from the terminus of Ag to extended link BC. -BC Determine the effective component of A^ on link BC (A^, ). a- Determine the magnitude Vg/g from a velocity polygon (Figure 8.4a).

Graphical Techniques: Acceleration Analysis

156

fa)

b

fb)

Figure 8.4 Acceleration of points on a slider-crank mechanism: (a) acceleration diagram; (b) velocity polygon.

V= 2.3 in./sec

b. Lay out A^C^ from point C along link BC. The magnitude of this vector is obtained A^g - ■'

;

= 1.76 in./sec2

c. Lay out vector A®c added to vector A®^ along BC. 4.

5.

Determine the absolute acceleration of point C (Ac) • Since the path of the slider is a straight line, the direction of its accel¬ eration must be along the same path. Therefore, to find the magnitude of this acceleration, drop a perpendicular from the effective component A®c to intersect the line of action of A . The point of intersection defines the acceleration Scale the vector Ac to determine its magnitude. A^ = 7.5 in./sec2

Aq.

Effective Component of Acceleration Method

8.4

157

FOUR-BAR LINKAGE ANALYSIS

Crank AB of the four-bar linkage in Figure 8.5 has an angular velocity of 2 rad/sec (counterclockwise) and is accelerating at the rate of 1 rad/sec2. Determine the acceleration of point C. In the example of Section 8.4, the acceleration of point C was related to point B by the vector equation

ab + a C/B and the effective component of the acceleration of point C along link BC was obtained from

abc

Ac

abc ac/b

In that example, since the direction of acceleration of point C was known, only one effective component of that acceleration was required to define the acceleration completely. In this example, however, the direction of accel¬ eration of point C is not known. Therefore, another effective component is needed to define Ac- This leads us to consider the effective component of point C along link CD, which is given by the equation

C

Figure 8.5

Four-bar mechanism.

Graphical Techniques: Acceleration Analysis

158

-CD AC

-CD _ AD

-CD + ac/d

where -CD AD = °

(s 11106 P0^ D is fixed)

t-CD ac/d

t-N C/D

an ac/d

——

where

vc

(directed from C to D)

Therefore, -CD A =

V2 C

(directed from C to D)

—BC —CD With the two effective components Aq and Ac determined, the required acceleration of point C can readily be found. PROCEDURE 1. From point B (Figure 8.6b), lay out acceleration vector Ag to a convenient scale, applying the relationship

A

•D

= A^ + A 13

13

(vectorial sum)

where A

N 13

= AB

X to2

= 1.5(4) = 6 in./sec2 A

T

= AB x a

ID

=

2. 3.

1.5(1) = 1.5 in. /sec2

Construct the effective component of Ag on BC (Ag ). Drop a perpendicular from the terminus of Ag to extended link BC. Find the effective component of Ac on link BC (A^).

c

^BC = aBC + ABC

c

where

ab

ac/b

159

Effective Component of Acceleration Method

BC AC/B

N = AC/B

V2 . C/B BC

a. Therefore, construct a velocity polygon (Figure 8.6a) to find VC/B. V

C/B

4.15 in./sec

b. Determine BC = 4.152 AC/B ~ 2.5

6.8 in./sec2

Therefore, 6.8 in./sec2

(directed along BC toward B)

b

Figure 8.6

Accelerations of points on a four-bar mechanism: (a) velocity

polygon; (b) acceleration diagram.

Graphical Techniques: Acceleration Analysis

160

— ■pC1

c. From point C, add vector A , to vector Ar, along link BC -BC C/B B to obtain Ag • _ 4. Determine the effective component of Ac along link CD (Ac )•

A^° = A^ L

v_/

(radial acceleration of point C)

where V2 N C Ac = CD 2.152

(VC fr°m velocity polygon) =1.54 in./sec2

Therefore, -CD , , A^ = 1.54 in./sec^

(directed from C to D)

5.

From point C, lay out vector A^ the fixed axis D.

on link CD directed toward

6.

With the two effective components of point C now known, the

7.

required acceleration Ac Is obtained by projecting perpendicu¬ lars from the termini of these two components until they inter¬ sect. This point of intersection defines the magnitude and direc¬ tion of required vector Ac. Scale the magnitude of vector Ac • A^ = 14.5 in./sec2 Therefore, Ac = 14.5 in./sec2

(directed as shown)

9 Relative Acceleration Method

9.1 INTRODUCTION The relative acceleration method is probably the fastest and most common among the graphical methods used for acceleration analysis. This is merely an extension of the relative velocity method used for velocity analysis and utilizes an acceleration polygon that is very similar to the velocity polygon. The method is based on the following principles: 1. 2.

All motions are considered instantaneous. The instantaneous acceleration of a point A relative to another point B on a rigid link is obtained from the vectorial relationship + A

A/B

3. The instantaneous motion of a point may be considered one of pure rotation in which the acceleration can be resolved into two rectangular components: one normal and one tangential to the path of rotation. Thus the expression in step 2 can be repre¬ sented graphically as -N -T -N ,-T -N , -T aa + aa = ab + ab + aa/b + aa/b 4. The absolute as well as relative velocities of various points in the mechanism are known. This requirement makes it most desirable to use the relative velocity or instant center method to determine the velocities involved.

161

Graphical Techniques: Acceleration Analysis

162

9.2

THE ACCELERATION POLYGON

Let us consider the two points A and B on link AB shown in Figure 9.1. As we saw earlier, the acceleration of point B relative to point A is given by the vector expression

B

= A

+ A. B/A

Just as in the velocity case, an acceleration polygon can be developed to represent the vector expression. Furthermore, the construction procedure and convention employed for this development are the same in both cases. PROCEDURE 1.

2.

3.

Define a point o', called the pole, as the origin for the construc¬ tion of the polygon. This is the point of zero acceleration. All absolute acceleration vectors originate from this point. By "absolute acceleration" we mean the real or true acceleration of a body as observed from a stationary frame of reference such as the earth. Since the accelerations A^ and Ag are absolute accelerations, draw vectors A^ and Ag from point o' and define their respective termini as a' and b'. From point a', the terminus of A^i, draw a third vector to ter¬ minate at point b', the terminus of Agt, thereby closing the polygon. This vector defines the relative acceleration Ag/^.

9.3

ACCELERATION POLYGON CONVENTION

The convention used to develop the acceleration polygon of Figure 9.1 can be summarized as follows:

Figure 9.1 Acceleration polygon.

Relative Acceleration Method

163

o'a' represents the acceleration vector AA. o'b' represents the acceleration vector A3, a'b' represents the acceleration vector AB//A. As in the velocity polygon case (Section 6.4), note that the letter to which the vector is directed indicates the acceleration under consideration. For example, in the vectors o'a' and o'b', the arrows point toward a' and b\ respectively. Hence the vectors represent the accelerations of points A and B, respectively. Similarly, in the vector a'b', the arrow points away from a' toward b'; hence the vector represents the acceleration of point B rela¬ tive to point A (Ab/^). If the arrow were reversed, pointing toward a' and away from b', the vector b'a' would represent the acceleration of A relative to B (Aa/b)-

9.4

GENERALIZED PROCEDURE

The construction procedure for developing the acceleration polygon of a mechanism will now be generalized using the four-bar mechanism as a model. Consider the four-bar mechanism ABCD shown in Figure 9.2. In this mechanism, AB, the driven member, has a clockwise angular accelera¬ tion coAb and a counterclockwise angular acceleration «Ab- It is required to find the acceleration of point C on the follower CD (Aq). PROCEDURE 1.

2.

3.

Determine the velocities of all points on the mechanism, includ¬ ing those with combined motion. This may be done using the instant center method, the effective component method, or as in the present case, the relative velocity method (Figure 9.3). Define a starting point o', called the polar origin. All absolute acceleration vectors originate from the polar origin. By "abso¬ lute acceleration" we mean the real or true acceleration of the point as observed from a fixed frame of reference such as the earth. Lay out the known components of absolute acceleration of the drive member, starting with the normal acceleration, which has the magnitude A^ = AB x co^B and is directed parallel to AB. Then add the tangential acceleration, which has the magnitude A^ = AB x £(,„ and is directed perpendicular to the normal B AB _>p acceleration, or Ab 1 Ab* The summation of these^two compo¬ nents defines the absolute acceleration of point B (Ab = ab + _T Ab) (Figure 9.4).If an angular acceleration 0^3 is not given, _rp _rp _n Ab does not exist (AB = 0), and Ab become_s the absolute accel¬ eration of point B. Define the terminus of Ab as b'.

164

Figure 9.2

Figure 9.4

Graphical Techniques: Acceleration Analysis

Four-bar mechanism.

Acceleration polygon.

Relative Acceleration Method

165

4. Select another point on the mechanism that has absolute motion and is rigidly connected to point B (call it point C) and lay out its normal acceleration component A^, noting that this vector is defined by the magnitude

N = C

5.

V2 , C/D CD

and is directed parallel to link CD. The normal components of acceleration are readily determined from the velocity data and the link orientation. -N Through the terminus of vector A^, construct a perpendicular to represent the tangential acceleration (A^), whose direction v

6.

is known but those magnitude is as yet undefined. This line contains point c', the terminus of vector Ac• Consider next the relationship between point B, whose accelera¬ tion is completely known, and point C, whose acceleration is only partially known. This means that we must consider the ac¬ celeration of point B relative to point C.

an +at ab/c ab/c

^B/C rN where A ,

can be completely determined from the velocity ^

rp

analysis and Ag/^ is known in direction only. Therefore, lay out vector A^c knowing that: a. Its magnitude is given by

B/C

VB/C BC

Its direction is obtained by assuming that point C on link is fixed while point B rotates about it. A^^, therefore lies on the link and is directed toward the center of rotation assumed. Also, in accordance with the polygon convention, relative acceleration vectors normally do not originate at the pole o' but extend between the termini of the absolute acceleration vectors to which they relate. For example, the notation Ag/c means that polygon vector is directed from c' to b', and A^/g means that polygon vector is directed from b' to c', where b' and c' are the termini of absolute vectors on the accelera¬ tion polygon, and B and C are the corresponding points in the linkage.

166

Graphical Techniques: Acceleration Analysis

Figure 9.5

Slider-crank mechanism.

Figure 9.6

Velocity polygon—step 1.

Figure 9.7

Acceleration polygon.

Relative Acceleration Method

7.

.

8

167

Through the tail* of A^,^, construct a perpendicular line of undefined length to represent the acceleration Ab/c, whose direction is known but whose magnitude is yet to be determined. This line also contains point c1, the terminus of Ap Now since c' lies on both AT (step 7) and AT (step 5), it fol¬ lows that the intersection of these two lines will define that point. Therefore, extend Aand A£, until they intersect and

9.

.

10

label the point of intersection c'. Finally, lay out a vector from origin o' to terminus c' to define the required absolute acceleration Aq and a vector from ter¬ minus c' to point b' to define the relative acceleration Ab/c* A quick check of the completed polygon should reveal the bal¬ anced vector equation

ab

ac + ab/c

or -N

-T

-N

-T

-N

-T

ab + ab = Ac + Ac + ab/c + ab/c Special Cases Slider-Crank Mechanism The slider-crank mechanism shown in Figures 9.5 to 9.7 may be considered a special case of the four-bar linkage where the follower link—for example, CD in Figure 9.2—is infinitely long. This means that the velocity of point C must be a straight line. Therefore, in step 4 (Figure 9.7) the normal accel¬ eration of the slider A^ is zero, and

-T In effect, the tangential component of acceleration Ac becomes the absolute acceleration of C and originates at o'.

*Note that had the normal relative acceleration A^g been chosen instead of Ag/c, the perpendicular drawn to represent A^g would have been con¬ structed through the terminus of Ac/g. This is because, by polygon con¬ vention, the normal acceleration vector A^^ must head toward c' and away from b'.

168

Graphical Techniques: Acceleration Analysis

ORIGIN

2V_

Figure 9.10

Acceleration polygon.

Relative Acceleration Method

169

Quick-Return Mechanism Another special case is the quick-return mechanism shown in Figures 9.8 to 9.10. Here point B slides on link CD, which is itself rotating. This means that in addition to the normal acceleration of point B relative to a coincident point C on CD, we must consider an additional component of acceleration: the Coriolis acceleration. Also, since the path of B relative to C is a straight line, the accel¬ eration of B relative to C can only be tangential. That is, B can have no normal component of acceleration relative to C; or

Had the path of B on CD been a curve, there would have existed a normal acceleration component of B relative to C (A^5 , ) as will be seen later (see Section 9. 7). To determine the Coriolis acceleration, we note from earlier dis¬ cussion that the magnitude of this vector is given by

A

Cor

2V

B/C^CD

where Vg/g and wgg are obtained from the velocity analysis. Also, we note that in defining the direction of this vector, we always consider (1) the linear velocity of the sliding body relative to that of the rotating body, and (2) the angular velocity of the same body on which the sliding occurs. In this example, since body B slides on link CD, we consider the velocity of B relative to C or Vg/C (not Vg/g) and the angular velocity of link CD or cogg (not oj^g). Accordingly, in step 6 (Figure 9.10), connect the vector A^01" to terminus b' on the polygon, observing the following rules: 1.

2.

This vector has the orientation of vector Vg/q when rotated 90° about its tail in the direction of cogg (counterclockwise in this case). Since point B relative to point C on the link is being considered, the polygon vector must go from c' to b' (note the reversed letter sequence).

EXAMPLE 9.1 (see Figure 9.11) Let AB = 1.5 in., BC = 1.5 in., CD = 1 in., coAg = 1 rad/sec (clockwise), and «ab = 0*5 rad/sec1 2 (counterclockwise). Determine Ac-

Graphical Techniques: Acceleration Analysis

170

a

Figure 9.11

Four-bar mechanism.

SOLUTION 1.

Determine the velocities. Construct a velocity polygon (Figure 9.12) and obtain V

= AB x co

= 1.5(1) = 1.5 in./sec

Aid

-D

V

= 1.45 in./sec

VB/C = 0.8 in./sec

2. 3.

Define polar origin o' for the acceleration polygon (Figure 9.13). Lay out the acceleration of the drive point, Ag.

T Ag - AB x

o

V

~ 1.5(0. 5) = 0.75 in./sec2

c V. B/C SCALE: 1 in. = 1 in. sec.

Figure 9.12

Velocity polygon.

Relative Acceleration Method

A

171

B

The vector originates at o' and terminates at b'. 4. Lay out the normal acceleration of the driven point, A(j. The magnitude of this vector is given as V2 , N C 1.452 AC “ CD “ 1.0

2.13 in./sec2

The vector is directed parallel to link CD, originating from o'. 5. Add the tangential acceleration A^ (undefined length). -T C 6. Lay out the normal acceleration of the drive point relative to the driven point, A^ . .

-N = AB/C

V2 B/C = 0.82 BC 1.5

0.426 in./sec2

The vector is directed parallel to link BC, pointing to b' (or away from c'). ^ 7. Add the tangential acceleration Ag/C (undefined length).

IT

rN

ab/c 1 ab/c

SCALE: 1 in. = 1 in,, sec.

Figure 9.13

Acceleration polygon.

172

Graphical Techniques: Acceleration Analysis

8.

9.

Determine the magnitude of tangential accelerations Aq and A^ , Extend undefined lines for these acceleration components in steps 6 and 7, until the polygon is closed. Aq intersects A-q/q at c'. Determine the required acceleration Aq. Connect point c' to point o' and measure o'c', or magnitude Aq. A

=3.2 in./sec2

(directed as shown in Figure 9.13)

L-

9.5

MECHANISM WITH EXPANDED FLOATING LINK

It was shown earlier that if the accelerations of two points on a link are known, the acceleration of a third point on that link can be found by propor¬ tion. In the following example it will be seen that the polygon construction procedure can also be used to determine the acceleration of the third point. EXAMPLE 9.2 Consider the mechanism ABCE shown in Figure 9.14a, where the crank AB rotates with an angular velocity of 2 rad/sec (clockwise). It is required to find the acceleration of point E on the expanded floating link BEC. SOLUTION 1.

Develop the velocity polygon for the mechanism as shown in Figure 9.14b. From this polygon the absolute and relative veloc¬ ity magnitudes are obtained as follows: V

=2.5 in./sec ID

Vc =

2.8

in./sec

=

1.0

in./sec

V£/c = 1.5 in./sec

2.

Develop the acceleration polygon for the basic mechanism ABC (Figure 9.14c) (ignoring point E for the time being), following the construction procedure discussed earlier. From this polygon the accelerations of points B and C are obtained. A-g = 5 in. /sec2 = 3.9 in./sec2

(directed as shown) (directed as shown)

Relative Acceleration Method

173

Figure 9.14 Mechanism with expanded floating link: (a) mechanism; (b) velocity polygon; (c) basic acceleration polygon (for ABC); (d) complete acceleration polygon.

174

Graphical Techniques: Acceleration Analysis

To determine the acceleration of point E (Ag), complete the following steps (Figure 9.14d). a. Lay out the vector for the normal acceleration of E relative to B • The magnitude of this vector is obtained from

N = AE/B

V2 E/B = (1^ EB 1

= 1 in./sec2 This vector is directed from point E toward point B on the mechanism, but on the polygon, heads from point b' (which is defined) to point e' (which is undefined), still maintaining its basic orientation. _JS[ b. Through the terminus of vector Ag /g, draw a line of unde¬ fined length perpendicular to this vector to indicate the direc¬ tion of AT^. This line contains point e', the terminus of vector Ag. c. From point c' on the polygon, lay out the vector for the nor¬ mal acceleration of E relative to C. The magnitude of this acceleration is obtain from

N

=

E/C

V2 E/C = (1.5)2 EC

1.5

=1.5 in./sec2 This vector is directed from point E to point C on the mech¬ anism but on the polygon heads from point c' (which is de¬ fined) toward point e' (which is undefined), still maintaining its basic orientation. d. Through the terminus of vector Ag/C, draw a line of unde¬ fined length perpendicular to this vector to indicate the direc¬ tion of AT This line contains point e', the terminus of A . rp E e. Since point e' is to be found on both lines Ag/g [step (c)] and AE/C [s*"eP (d)]» it must be located at the intersection of these lines. This point e' defines the terminus of the vector Ag drawn from the pole o'. Therefore, A

=5.6 in./sec2

(directed as shown)

Relative Acceleration Method

175

Acceleration Image As in the relative velocity case, it should be observed from the polygon that the triangle formed by points b', e', c' is similar to link BEC, and for this reason it is often referred to as the acceleration image of link BEC. Conse¬ quently, the acceleration of point E could have been determined more directly by constructing the acceleration image on link b'c' of the polygon for the basic mechanism ABC. (Note that the letters used to designate both the links and the image must run in the same order and direction.)

9.6

COMPOUND MECHANISM

Let ABCDEF represent a compound mechanism in the form of a shaper, where crank AB rotates with a constant angular velocity of 1 rad/sec (counterclockwise) (Figure 9.15a). Determine the acceleration of point E on the slider. PROCEDURE 1

Develop the velocity polygon (Figure 9.15b) and determine the velocities as follows:

VB ~ VC =

AB x co = 1.5(1) = 1.5 in. /sec 1.2 in./sec

1.0 in./sec V. B/C =

V

2. 3.

VD

OCX VC = 4(1'2) = l*8 in-/sec

V

2.1 in./sec

D/E

0.75 in./sec

Define polar origin o' for the acceleration polygon_(Figure9.15c). Lay out the acceleration of the driver at point B (AB). A^ = AB x oj2 = 1.5(1)2 = 1.5 in./sec2 B A

T = AB x a = 0 B

_ A

_at _rp = A + A = 1.5 in./sec2 B B B

(directed as shown)

176

Graphical Techniques: Acceleration Analysis

z u Figure 9.15 Acceleration analysis of a shaper mechanism: (a) shaper mechanism; (b) velocity polygon; (c) acceleration polygon.

Relative Acceleration Method

177

-N 4. Lay out the normal acceleration of point-driven point C (Aq)

J Ac

V2

c oc

N = <1.2if AC 4 -N A^ = 0.36 in./sec2

(directed as shown)

5. Add the tangential acceleration of point C (undefined length) to AN

V

-T -N AC 1 AC . Determine the relative acceleration between drive point B and

6

related point C (Ab/c)*

ab/c = a.N B/C N Ag^c =

0

-T AB/C

-Cor AB/C

(since the path of the slider is a straight line)

and -Cor b/C

VB/CWOC

2

(Pr°Perly directed)

or

AB/C = 2(1)(^) =

0,6

hl'/se°2

a. Lay out the vector Ain accordance with convention. That is, the vector must have the same orientation as vector Vb/C when rotated 90° in the direction of wqq. _ b. Add the tangential relative acceleration A^^ (undefined -Cor length) to Ag/£.

AT 1 AC°r ab/c 1 ab/c Extend the tangential accelerations in steps 4 and 5 until the polygon is closed. The acceleration of point C on the polygon is defined at the intersection of A^ and AT that is, at point c'.

Graphical Techniques: Acceleration Analysis

178

8.

Draw vector Ac extending from the pole o' to c'. Note:

A

C

and A

ab/c A

9.

= A

ab/c

+ A

ab/c

=0.5 in./sec2

Determine magnitude of the acceleration of point D (Aq). This is obtained from the proportion

AP _ OP Ac-°c

A^ = -(0.5) = 0.75 in./sec2

10. 11.

Define the terminus of vector Ap> as point d' on polygon. Determine the acceleration of point E relative to point D. r AD/E

rN rT ad/e + AD/E

where V2 . D/E DE

rN

D/E

(properly directed)

(0.75)2 0.1875 in./sec2

(properly directed)

rN

a. Lay out vector Aj-j/g. This vector must be pointed toward d' on the polygon and be parallel to DE on the mechanism. b. Add vector A^ to vector AN D/E D/E’ rT

,

7n

ad/e 1 ad/e 12. Determine the acceleration of point E (AE). Note that since the sLyer Pa^ a straight line, there is no normal acceleration (ae = 0). so A£ is equal to the tangential acceleration AT,

E

which exists along the same path. Therefore:

179

Relative Acceleration Method

Lay out the direction of vector Ap by drawing a line "e"-"e" through pole o' parallel to the slider path, b. Extend vector A^,^ in step 11(b) until it intersects line "e"-"e"

13. 14.

(direction line). This point of intersection defines the mag¬ nitude of ApDefine the point of intersection in step 12(b) as e'. Measure line o'-e', the vector Ap. A

= 0.625 in./sec2

(directed as shown)

-L

9.7

CAM-FOLLOWER MECHANISM

Consider the cam-follower mechanism shown in Figure 9.16a. The cam (2) rotates counterclockwise at a constant angular velocity of 2 rad/sec. Find the acceleration of the follower (4). At first glance it would appear that to determine the acceleration of 4 it would be necessary first to determine the motion of the roller (3). However, since the path that 3 traces on 2 is generally not easily recogniz¬ able, because of the two curved surfaces in contact, a more direct approach is to do the following: 1. 2. 3. 4.

Assume that the roller does not turn. Then the acceleration of contact point C is the same as that of any point in 4. Construct a pitch line to represent the locus of the center of the roller as it rolls on the cam. Designate the new point of contact between 2 and 4 as P2 or P4. Proceed with the analysis based on the assumption that the fol¬ lower 4 actually slides on the expanded cam defined by the pitch line.

Figure 9.16b shows the velocity polygon for the mechanism. Note that since 4 rides on 2, the relative velocity VP4/p2 is considered for Coriolis accel¬ eration. This velocity has a direction tangent to the curvature of the cam at the contact point. Figure 9.16c shows the acceleration polygon for the mechanism. Here it should be noted that the acceleration of P4 relative to P2 (or AP4/P2) consists of three components: -N ,T + yCor AP4/P2 “ AP4/P2 + AP4/P2 P4/P2 This equation, incidentally, is unlike those of earlier examples where A^ not exist, because the sliding paths in those cases were stiaight

180

Graphical Techniques: Acceleration Analysis

Figure 9.16 Acceleration analysis of a cam-follower mechanism: (a) camfollower mechanism; (b) velocity polygon; (c) acceleration polygon.

Relative Acceleration Method

181

lines. In the present case, since the path of P4 on P2 is a curve (or 4 is forced to move_in a curved path as it rides on 2), there must exist a normal acceleration, -^p4/p2> whose magnitude is given by

N _ VP4/P2 P4/P2 r + 0.5 where r + 0.5 or R is the radius of curvature of the path. For the Coriolis acceleration A(-'or, we consider, as before, the motion of P4 relative to P2, computing its magnitude from Cor _ AP4/P2 “ 2VP4/P2W and defining its direction by rotating the vector Vp4/p2 through an angle of 90° in the same angular direction (counterclockwise) as co2. Following is a summary of calculations and results:

V V V

P2 P4

P4/P2 an AP2

Cor VP4/P2

1.6 in./sec

(from the velocity polygon)

3.35 in./sec

(from the velocity polygon)

1. 5(2)2 = 6 in./sec2

2(3.35)(2) = 13.4 in./sec2

N

= 5.6 in./sec2

P4/P2 T P4/P2 AP4

9.8

1. 5(2) = 3 in./sec

4.0 in./sec2

(from the acceleration polygon)

2.8 in./sec2

(from the acceleration polygon)

SUMMARY

The generalized procedure for constructing the acceleration polygon may be summarized as follows: 1.

Proceed from the "known" to the "unknown." That is: a. Lay out those absolute vectors whose magnitude and direction are known.

Graphical Techniques: Acceleration Analysis

182

2.

3.

4.

b. Lay out those components of absolute and relative vectors that are known (magnitude and direction) or can be deter¬ mined. These include normal accelerations and Coriolis acceleration. c. Add to the components in step (b) their corresponding tangen¬ tial accelerations (directions only), and extend these to close the polygon. All absolute vectors on the acceleration polygon originate from pole o', while relative vectors extend between the termini of the absolute vectors. a. A vector that originates from b' and terminates at c' repre¬ sents the relative acceleration of point C to that of point B on the link. b. The choice between Ag/g and A^/g makes no difference in the polygon configuration or the results, except that these vectors have opposite senses. Two undefined vectors such as A^^ and A^ will contain a common point c' on the polygon, and will define that point where they intersect.

10 Velocity-Difference Method

10.1

INTRODUCTION

The velocity-difference method of determining accelerations of points in a mechanism is probably the most straightforward of all the graphical methods used in kinematic analysis. Applicable to any type of mechanism—pin con¬ nected, rolling contact, or sliding contact—the method does not rely on the use of sophisticated formulas, but instead employs the simple relationship

based on the fundamental definition, which states: The acceleration (A) of a point is the rate of change of velocity (AV) of that point over the time interval (AT) in which the change occurs. In terms of mechanism analysis, if the linear acceleration is required for a point in a mechanism at any given position of the mechanism, this acceleration may be found by first consid¬ ering a small change in position (A0) of the point, then determining the change in velocity (AV) resulting from the change in position, and finally dividing the change of velocity by the time interval (AT) during which the change has taken place. In applying the velocity-difference method, experience has shown that, generally, for reasonably accurate results, a position change of the mechanism based on a crank angular displacement of 1/10 rad is accept¬ able. This displacement is normally measured such that the given crank position is centrally located between its initial and final positions. That is, the initial and final crank positions are each 1/20 rad on either side of the given position. The associated time interval AT depends on the angular velocity of the crank and its angular displacement. If, for example, the crank of a mechanism turns at a constant speed of 5 rad/sec, AT is the time it takes

183

184

Graphical Techniques: Acceleration Analysis

for the crank to move through 1/10 rad. In our earlier discussion it was noted that for constant angular speed, AT is obtained from

co

Ad AT

or AT = -

Ad co

1 /rad \ 5 rad/sec V 10 '

= 1/50 sec If, on the other hand, the crank turns with a uniform acceleration, which means that the angular velocity is changing, then again from our earlier discussion, AT is obtained from

a

Aco AT

or Af,'

AT = —

a

where Ago is the change in angular velocity between the initial and final positions of the mechanism. This uniform or constant acceleration case will be discussed in more detail in Section 10.4. The following two sections illustrate the procedure for applying the velocity-difference technique to mechanisms where the crank arm rotates with constant angular velocity.

10.2

SLIDER-CRANK MECHANISM ANALYSIS

Consider slider-crank mechanism ABC shown in Figure 10.1, where crank arm AB is rotating at a constant angular velocity of 1 rad/sec (counter¬ clockwise). Let it be required to find the acceleration of point C for the position shown. PROCEDURE 1.

Lay out the given mechanism in the position ABjCj (see Figure 10.2), where ABj indicates the initial position of the crank, dis¬ placed 1/20 rad (or 1/2 of 1/10 rad) clockwise from the given

Velocity-Difference Method

Figure 10.1

185

Slider-crank mechanism.

position. Note that the angular displacement 1/10 rad can be accurately laid out by applying the well-known relationship S = R A9 where S is the length of an arc that subtends angle A0 and R is the radius of the arc. Hence the angle 1/20 rad is obtained by laying out a segment of a circle having some convenient radius, say 6 in., in which case the arc length is

S = 6^2o) = 0,3 ln‘

A

Figure 10.2

Slider-crank mechanism.

Graphical Techniques: Acceleration Analysis

186

2.

Determine the linear velocity of point Cj using any of the meth¬ ods studied earlier. The relative velocity method, used in this example, yields V

C.

=1.0 in. /sec

(directed as shown in Figure 10.2)

l

3.

4.

Lay out the given mechanism in position ABfCf (see Figure 10.2), where ABf indicates the final position of the crank, dis¬ placed 1/20 rad counterclockwise from the given position. Determine the linear velocity from point Cf as in step 2. This velocity has been determined to be V

=1.17 in./sec

(directed as shown in Figure 10.2)

°f 5.

Determine the velocity difference AV^, which is given by AV

= V

- V I

(vectorial difference) 1

This equation, which is represented graphically in Figure 10.2, yields AV

= 0.17 in./sec

(directed to the left)

L/

6.

Note that the vectorial difference in this example yields the same results as the algebraic difference since Vq and Vq. both have the same line of action. Determine the acceleration of point C. The magnitude of this vector is given by AV, A„ =

AT

where A0 AT = — co

0.1 rad/1.0 rad/sec = 0.1 sec Therefore, 0.17 . A^ = in./sec2 C 0.1 and the vector A

L/

= 1.70 in./sec2 (directed as shown in

Velocity-Difference Method

187

Figure 10.2). Note that the direction of acceleration is always the same as that for AVq.

10.3

QUICK-RETURN MECHANISM ANALYSIS

Now consider the quick-return mechanism ABCD in Figure 10.3. Crank AB turns with a constant angular speed of 30 rad/sec (counterclockwise) and the linear acceleration of point C on the rod is required. PROCEDURE We follow the same procedure as that given in Section 10.2. 1.

2.

Lay out the mechanism AB^C^D (Figure 10.4), showing the initial position of the mechanism rotated back 1/20 rad from the given position. Determine the linear velocity of point C (Vq.) for this position (see Figure 10.4). VP

= 72 in. /sec

(directed as shown)

'■'i

3.

Lay out the mechanism ABfCfD (Figure 10.4), showing the final position of the mechanism rotated forward l/20 rad from the given position.

Figure 10.3

Quick-return mechanism.

(ON

BLOCK)

(ON

ROD)

188

Graphical Techniques: Acceleration Analysis

4. Determine the linear velocity of point C (Vc ) for this position (see Figure 10.4), Vq

5.

= 69.5 in./sec

Determine the velocity difference AVC from the relationship AV^, = Vc^ = 15.0 in./sec

6.

(directed as shown)

(vectorial difference) (directed as shown in Figure 10.4)

Determine the linear acceleration Ac. The magnitude of this vector is given by

Avc

A

= _—

C

AT AV

— C AO

15.0

'0.1/

4500 in./sec2 Therefore,

Velocity-Difference Method

189

= 4500 in. /sec2

7.

(directed like AV^ as shown)

To obtain the angular acceleration of CD (acE)), aPPty the relationship Aco

a

CD

CD AT

where CD " wCDf " ^CDi VCf

VCi

" CDr “ CD. f l = 69.5 _ 72 1.7 1.52 = 40.9 - 47.4 = -6.5 rad/sec AT = — GO

= -(-) 10^30^

1/300 sec

Therefore, = -6.5 1/300 = -1950 rad/sec2 Note that in this analysis there was no need to determine the Coriolis acceleration, which is ordinarily the case in the relative acceleration method. In this respect, the velocity-difference method offers additional simplifica¬ tion to the solution of a problem that can otherwise be more complicated.

10.4

FOUR-BAR ME CHANISM ANALYSIS

Consider the four-bar mechanism ABCD, where crank AB turns with a constant angular acceleration of 1 rad/sec2 and for the position shown has an angular speed of 2 rad/sec (Figure 10.5). It is required to find the linear acceleration of point C on the follower. The procedure for solving this problem is basically the same as that used in Sections 10.2 and 10.3, with the exception that in this case there is

Graphical Techniques: Acceleration Analysis

190

C

Figure 10.5

Four-bar mechanism.

an angular acceleration, and therefore the magnitudes of the linear velocities of Vj} and Vg^ are not the same. These velocity magnitudes are calculated from V td — ABco. Bi i

(10.1)

VBf = ABco{

(10.2)

and

where

is the initial angular velocity of ABj obtained from

2

co

=

2 + 2a AO

CO

(10.3)

and coj is the final angular velocity of ABf, obtained from

2 =

CJ

2 - 2a AO

GO

(10.4)

Also, since the angular acceleration is constant, the angular speed is changing. AT must be expressed in terms of the angular acceleration a and the changing speed w. That is,

Velocity-Difference Method

191

(10.5) where cof = coj. PROCEDURE 1. 2.

Figure 10.6 shows the mechanisms ABjC^D in the initial position and ABfCfD in the final position. Determine coj. This is found by applying Equation (10.3). co2 = co? + 2a A9 l

22 = “U 2(1>(ij;) co? = 4 - 0.1 = 3.9 l

co. =1.97 rad/sec l

3.

Determine VB . From Equation (10.1), the magnitude of this vector is computed as VB. = 1.5(1.9) = 2.95 in./sec Therefore, VBi = 2 .45 in./sec

4.

(directed as shown)

Determine toj . Since a is constant, cof is easily determined by recognizing that its increment from co at the given crank position is equal to that of the same co from co^, due to the same angular displacement of 1/20 rad in both cases. Therefore, a>

= 2 + (2 - 1.97) =2.03 rad/sec

5.

Determine VBf- The magnitude of this vector is computed as

VBf = 1.5(2.03) =3.04 in./sec

192

Graphical Techniques: Acceleration Analysis

Figure 10.6

Four-bar mechanism.

Therefore, = 3.04 in./sec 6.

(directed as shown)

Determine VCf using the mechanism ABfCfD, as before. vCf ~

7 in./sec

(directed as shown)

Velocity-Difference Method

7.

193

Determine Vc. using the mechanism ABjCjD, as before. Vc. = 1.25 in./sec

8.

Determine AVc from AV^ = Vc

- Vc.

= 0.46 in./sec 9.

(directed as shown)

(vectorial difference) (directed as shown)

Determine Ac- The magnitude of this vector is computed as

where AT, from Equation (10.5), is given by

AT

ACO

a 2.03 - 1.97 1 0.06 sec

Therefore, A_

0.46 0.06 7.7 in./sec2

Hence A„ = 7.7 in./sec2

(directed like AV^ as shown)

Despite the simplicity of the velocity-difference method, it should be understood that because of the need to measure small changes in dis¬ placements as well as velocities, a high degree of drawing precision is required if sufficiently accurate results are to be achieved. In some cases, this requirement can impose practical limitations such that the results obtained may not be reliable.

11 Graphical Calculus Method

11.1

GRAPHICAL DIFFERENTIATION

A common method for obtaining velocity and acceleration curves for a mechanism consists of constructing a time-displacement curve and then by graphical differentiation developing the velocity and acceleration curves. Graphical differentiation consists of obtaining the slopes of various tangents along one curve and plotting these values as ordinates to establish a second curve. The method, as applied to the displacement and velocity curves, is based on the following principles. 1.

The instantaneous velocity of a moving point can be represented graphically as the slope of the displacement curve at that instant, or As v = —

2.

where

At is very small

The instantaneous acceleration can be represented as the slope of the velocity curve at that instant, or Av a = —

where

At is very small

Graphical differentiation can be an effective tool for the designer, particularly when used during the preliminary stages of a mechanism design since it affords an overall picture of the velocity and acceleration throughout the motion cycle. For example, from motion curves such as the s-t, v-t, and a-t curves, one can readily determine: The maximum absolute values of displacement, velocity, and accel¬ eration

194

Graphical Calculus Method

195

Where and when the maximum values of angular displacement occur Whether there is any abrupt change in displacement, velocity, and acceleration during the cycle Also, there are many instances where it is necessary to differentiate a curve for which an equation is difficult to obtain. In such instances, graph¬ ical differentiation is the most convenient approach.

Tangent Method Consider the displacement-time curve shown in Figure 11.1. Suppose that it is required to develop the velocity-time curve from this curve. Since the instantaneous velocity at any point is represented by the slope of the displacement curve at that point, a tangent drawn through any point of the curve defines the velocity at that point. Therefore, at point A, the velocity v^ is obtained from

s

Figure 11.1

Graphical differentiation.

196

Graphical Techniques: Acceleration Analysis

= slope of tangent at A

where kg = displacement scale, in./in. = time scale, sec/in.

Similarly,

v

= slope of tangent at B =

vc = slope of tangent at C =

B (k ) s s

w w

Here the negative sign indicates a negative slope of the curve and hence a negative velocity. Also, when the tangent to the curve is a horizontal line, this simply means that the velocity at the point of tangency is zero. In addition, it should be clear that by making the time intervals A^, B^, Ct, and so on, equal, the velocities v^, vg, v^, and so on, will be pro¬ portional to As, Bg, Cs, and so on. Hence, once the velocity v^ has been found, the velocities Vg, v^, and so on, are readily obtained as follows: B

s

s In this manner, the velocities of other points can be obtained to enable a smooth continuous curve to be drawn. EXAMPLE 11.1 The curve shown in Figure 11.2a represents a typical displacement-time curve for a cam mechanism. It is desired to obtain the velocity-time curve from this displacement curve, and subsequently, the acceleration-time curve from the velocity curve.

Graphical Calculus Method

Figure 11.2

197

Graphical differentiation—tangent method: (a) displacement

curve; (b) velocity curve.

Graphical Techniques: Acceleration Analysis

198

SOLUTION 1. 2.

Draw tangents to various points on the given displacement-time curve: points A, B, C, D, and E. Using the tangents as hypotenuses, construct right triangles at each point, making all the bases equal, that is, A

=

= C

etc.

3. Calculate the velocities at points A, B, C, D, and E as follows: k

s

=2 in. /in.

k^

= 3 sec/in.

k

=1 (in./sec)/in.

v

A k s s Atkt B A C

v.

A D =

A E

v„ =

/'0.5 \ 2 '1.0'3 ^

s s

VA

VO.5'

VA

(2.20\ VO.5 '

s s s s

'sec

/O.SA VA ' Vo. 5'

s

VA

( 0 ), Vo. 5'

Note that the remaining points on the velocity curve can be located by inspection based on the symmetrical shape of the displacement curve. 4. 5. 6. 7.

Draw the velocity-time axis using the same time scale as that used for the displacement-time axis (Figure 11.2b). Plot the velocities calculated in step 3 on the velocity-time axis. Draw a smooth curve connecting the plotted points. To obtain the acceleration curve, repeat steps 1 through 6, re¬ placing the displacement curve with the velocity curve just found (Figure 11.3a). The required acceleration curve is shown in Figure 11.3b.

Graphical Calculus Method

Figure 11.3

199

Graphical differentiation—tangent method: (a) velocity curve;

(b) acceleration curve.

Polar Method An alternative graphical differentiation method commonly used is the polar method. This method has been found to produce greater accuracy than the

200

Graphical Techniques: Acceleration Analysis

I

I

Figure 11.4 Graphical differentiation—polar method: (a) displacement curve; (b) velocity curve.

tangent method mainly because it requires less graphical construction. The method is best described by considering the displacement-time curve given in Figure 11.4a. Let it be required to develop the velocity and acceleration diagrams from the given curve.

Graphical Calculus Method

201

PROCEDURE 1.

2.

Lay out the axes for the velocity-time graph directly below those for the displacement curve, maintaining the same time scale in both cases (see Figure 11.4b). Through the point of steepest slope on the displacement curve (point M), draw a tangent line to define the point of maximum velocity (t^j). Now, since by definition, velocity equals slope, the maximum velocity at point M is given by As vmax = T7 At where As is the change of displacement over any interval At, or

v

max

= (—) ^ me4 5 * 7

where k

s

= displacement scale, in./in.

k^. = time scale, sec/in.

3.

Define a point P (called the pole) on the time axis of the velocity¬ time curve, at any convenient distance left of the origin, and from it draw line t]yp parallel to tangent line tM to intersect the velocity axis. This line locates a point M' on the velocity axis which repre¬ sents the point of maximum velocity, or V-,., = V

M

4.

5.

max

Plot the velocity of point M at the intersection of the horizontal line drawn from M' on the velocity curve and a vertical line drawn from M on the displacement curve. Label this point M". Use the value obtained for vmax to establish the scale for the velocity axis of the velocity-time graph. To do this, compare the actual measurement on the diagram with the computed maximum velocity and then by ratio determine the velocity value that is represented by 1 in. on the diagram. In this example we find that

Graphical Techniques: Acceleration Analysis

202

where k

s

= 5 in./in. = 2 sec/in.

Therefore,

v

max

5 2

4 in./sec

Figure 11.5 Graphical differentiation—polar method: (a) velocity curve; (b) acceleration curve.

Graphical Calculus Method

203

Hence 2 in. on the velocity curve represents 4 in./sec, or 1 in. represents 2 in./sec, or = 2 (in./sec)/in. 6.

To establish the remaining velocity points at ordinates A, B, C,

(B) PAPER

Figure 11.6

Use of a mirror in constructing tangents: (a) incorrect mirror

position; (b) correct mirror position.

204

Graphical Techniques: Acceleration Analysis

7.

D, and so on, follow the procedure outlined in steps 3 and 4. Figure 11.5a shows the completed velocity diagram. After the velocity diagram has been completed, the acceleration diagram can be developed by following the procedure outlined above. Figure 11.5b shows the required curve.

Note on Accuracy It should be emphasized that accuracy of the graphical differentiation meth¬ ods depends primarily on: 1. 2. 3.

How accurately the tangents can be drawn The number of intervals chosen The ability to fit a smooth curve to a given set of points

For greater accuracy in constructing tangents, any straight edge having a reflective surface, such as the rectangular mirror shown in Figure 11.6, may be used. The straight edge, held vertical to the paper, is placed across the curve and rotated such that the visible curve and its reflection form a continuous curve. In this position, the straight edge is then normal to the curve. Therefore, a line drawn perpendicular to the normal and touching the curve is the required tangent. Accuracy is increased with the number of intervals into which the curve is divided. The greater the number of intervals chosen, the greater the accuracy that is achieved. Finally, the ability to fit a smooth curve to a given set of points, particularly if double differentiation is required, greatly minimizes the risk of compounding errors from one curve to the other.

11.2

GRAPHICAL INTEGRATION

Just as it is possible to go from the displacement curve to the velocity curve and then to the acceleration curve using graphical differentiation, it is also possible to reverse the process, going from acceleration to velocity and then to displacement. The process by which this is achieved is called graphical integration. The process of integration can be thought of as the procedure for obtaining the area under a given curve. As applied to the motion of a mech¬ anism, the integration of the acceleration curve gives the velocity curve, and the integration of the acceleration curve gives the displacement curve. This is based on the motion laws, which state that Av = a At and

or

V]? - vQ = a(tp - tQ)

Graphical Calculus Method

As = v At

or

205

sp - sQ = v(tF - tQ)

where the subscripts F and O denote final and original conditions, and a and v denote average acceleration and velocity. Hence the following rules apply. 1.

To derive the velocity-time curve from the acceleration-time curve, the change in velocity between any two times equals the the area under the acceleration-time curve between the same two times.

2.

To derive the displacement-time curve from the velocity-time curve, the change in displacement between any two times equals the area under the velocity-time curve between the same two times.

Mid-ordinate Method There are several graphical methods available to determine the area under the curve. One of the simplest is the mid-ordinate method. The following example will illustrate the procedure. EXAMPLE 11.2 The curve shown in Figure 11.7a represents a typical acceleration curve for a cam mechanism. It is desired to obtain the velocity-time curve. SOLUTION 1.

2.

3.

Divide the given curve into an equal number of sections, S1; S2, S3, and so on, as shown in Figure 11.7a, and construct mean ordinates alt a2, a3, and so on, for each section to intersect the curve at 1, 2, 3, and so on. Through points 1, 2, 3, and so on, draw horizontal lines extending between the boundaries of alternate sides of the curve. For ex¬ ample, in section S1, triangles A and B are on alternate sides of the curve. From this it is easily seen that, provided that the slope of the curve within the section remains fairly constant, the alter¬ nate triangles are approximately equal. This means that in sec¬ tion Sj, the area of triangle A can be considered approximately equal to that of triangle B; similarly, in section S2, the area of triangle C is equal to that of triangle D; and so on. Therefore, the area under the curve in section Sj can be approximated by the rectangle aj Atx; the area under section S2 can be approximated by the rectangle a2 At2; and so on. Since the velocity equals the area under the curve, the velocities vx, v2, and v3 may be found as follows:

206

Graphical Techniques: Acceleration Analysis

Figure 11.7 Graphical integration—mid-ordinate method: (a) acceleration curve; (b) velocity curve.

Graphical Calculus Method

Vj

= aj

207

Atj

v2 = Vj + a2 At2 v3 = v2 + a3 At3

etc.

For example, in Figure 11.7a, Vj = 1.4(1) = 1.4 in./sec v2 = 1.4 + 2(1) = 3.4 inr/sec v3 = 3.4 + 2.4(1) = 5.8 in./sec

4.

The velocity curve is shown in Figure 11.7b. After the required velocity curve has been completed, repeat steps 1 to 3 to develop the required displacement curve.

Polar Method Another graphical integration method that is commonly used is the polar method, which is relatively fast and easy to apply. To illustrate the method, let us consider the acceleration curve shown in Figure 11.8a. From this curve we will develop the velocity curve and, in turn, use the velocity curve to develop the displacement curve. PROCEDURE 1.

Erect coordinates BB, CC, DD, and so on, to divide the curve into an equal number of line intervals, as shown in Figure 11.8a. In this case, five intervals have been chosen with ordinates a1 to a5 defined at the midpoints of the intervals. Also, since the horizontal axis is the time axis, each interval is defined at Atx, At2, At3, and so on. Note that all time scales for acceleration, velocity, and displacement curves must be the same, as with

2. 3.

4.

graphical differentiation. From points 1, 2, 3, and so on, project horizontals to meet a line parallel to the a axis at points 1', 2', 3', and so on. From a point P', located on the time axis, at any convenient distance left of the origin, draw connecting straight lines to points 1', 2', 3', and so on. From the origin of the velocity axis (Figure 11.8b) draw line ob' parallel to P'l' to meet ordinate BB at b', line b'c' parallel to P'2' to meet ordinate CC at c', line c'd' parallel to P'3' to meet ordinate DD at d, and so on, until the complete velocity curve is obtained. From this velocity curve, the velocities at points 1, 2,3, and so on, are given by

208

Graphical Techniques: Acceleration Analysis

v, = P'Q' (bb')k k 1 cl t v

= P'Q’(cc')k k ^

v

cL X

= P'Q'(dd')k k a t

etc.

where b, c, d, and so on, are the intercepts of the first, second, third, and so on, ordinates with the time axis, and ka [(in./sec2)/ in.] and k^ (sec/in.) are the acceleration and time scales, re¬ spectively. The relationships above can be verified by applying the properties of similar triangles to triangles P'Q'l' and obb',

Graphical Calculus Method

209

Figure 11. 8 Graphical integration—polar method: (a, facing page) acceleration curve; (b, facing page) velocity curve; (c) displacement curve.

where bb' is the vertical leg of the velocity curve segment ob. From these triangles, we obtain Q'l' _ P'Q' bb' - ob

(11.1)

But Q'l’ = ^ and ob = Atj Therefore, Equation (11.1) becomes _*i_ = P'Q' bb' Atj from which follows a.1 Atj = P'Q'(bb') But

(all parameters in inches)

Graphical Techniques: Acceleration Analysis

210

ai Atj(kakt) = Vj - v0 where v0 = 0 Therefore, vx = P'Q'(bb')kakt Similarly, from triangles P'Q'2' and b'c"c', where c"c' is the vertical leg of the velocity curve segment b'c", we obtain Q'2' = P'Q' c"c' b'c" from which follow a2 At2 = P'Q'(c"c') v2 - vx = P'Q'(c"c')kakt v2 = Vj + P'Q'(c"c')kakt v 2 = P'Q' (bb')kak|. + P'Q'(c"c')kakt = P'Q'(bb' + c"c')kak^ = P'Q'(cc')kakt and from triangles P'Q'3' and c'd"d', Q'3' = P'Q' d"d' c'd" from which follow a3 At3 = P'Q'(d"d') v3 - v2 = P'Q'(d"d')kakt v3 = v2 + P'Q'(d"d')kakt v3 = P'Q'(cc')kakt+ P'Q'(d"d')kakt = P'Q'(cc' + d"d')kakt - P'Q' (dd')kakt

Graphical Calculus Method

211

To obtain the displacement diagram, repeat steps 1 to 4, replacing the acceleration curve with the velocity curve. Figure 11.8c shows the required displacement diagram.

Note on Accuracy As in the case of the graphical differentiation methods, accuracy of the graphical integration method increases with the number or size of the inter¬ vals chosen. The smaller the interval chosen, the more likely the slope of the curve within that interval (or section) will remain constant, and hence the closer the approximation of equality between the alternate triangles. It is therefore desirable to make the intervals or sections as small as possible for greater accuracy. Also, if double differentiation is required, accuracy will depend on the ability to fit a smooth curve to a set of given points.

12 Special Methods

12.1

COMPLETE GRAPHICAL ANALYSIS METHOD

In the graphical methods for acceleration analysis presented so far, one of the required calculations to be performed was that used to determine the normal acceleration components for points in the mechanism. In this section we demonstrate a complete graphical method where, by proper choice of link, velocity, and acceleration scales, both the magnitude and direction of all normal acceleration components can be determined without the need for calculations. To illustrate this method, let us consider link AB, shown in Figure 12. la, which rotates at an angular velocity of to rad/sec. First, we obtain the magnitude of the velocity of point B relative to A (VB/A) by multiplying the length of the link by the angular velocity,

vb/a = abx"ab

(12-D

We then lay out the vector perpendicular to the link as shown. Here it is useful to recall that the magnitude of Vg/A is related to that of AN, by the expression r>/A

.N _ B/A

V2 B/A BA

(12.2)

Rearranging this equation, we obtain the relationship

A V

N B/A B/A

Va (12.3)

BA

212

Special Methods

213

c

(c) Figure 12.1

which can be represented as BD _ BC BC BA

(12.4)

Graphical Techniques: Acceleration Analysis

214

using two similar triangles, ACB and CDB, as shown in Figure 12. lb, where

BD and BC = V

B/A

Here is can be seen that if the scales of link AB, velocity Vg/^, and accel¬ eration A^, in Equation (12.3) are properly chosen, in accordance with the relationship expressed in Equation (12.4), the length of line BD will accurately represent the required normal component. Let the scales be defined as follows: kg = space scale, ft/in. k^ = velocity scale, (ft/sec)/in. k^ = acceleration scale, (ft/sec2)/in.

where the inch units in the denominators represent actual measurements of the drawing in inches. For example, if the space scale is given as ks = 5 in./in., a link length of 5 in. will be represented by a line 1 in. long on the drawing. Similarly, if the velocity scale is given as ky = 10 (ft/sec)/in., a velocity of 20 ft/sec will be represented by a line 2 in. long on the drawing; and for an acceleration scale of 50 (ft/sec2)/in. an acceleration of 150 ft/sec2 will be represented by a line 3 in. long. The required relationship of the scales is obtained by considering tiiangle abc in Figure 12.1c, where sides x, y, and z are proportional, respectively, to sides AB, BC, and BD of triangle ABC in Figure 12.1b. In other words, if x represents the link length, y the velocity, and z the acceleration, we can write xk

s

= BA

or

- M

x - k yk or

v

s

= BC

(12-5)

215

Special Methods

BC k v

(12.6)

BD

zk or

BD

(12.7)

and from similarity, z = JL y x

(12.8)

Substitution of Equations (12.5) to (12.7) into Equation (12.8) yields BD/k BC/k _a _ _v BC/k " BA/k

(12.9)

or

A* Vk B/A

VB/A/kv BA/k

a

VB/A/kv

(12.10)

from which

,N B/A

V2 k k B/A a s BA k2 v

(12.11)

Thus the scales must be related by the equation k k a s k2 v

(12.12)

or k k a s

k2 v

(12.13)

This means that any two scales may be chosen arbitrarily, but the third must be chosen from Equation (12.12).

216

Graphical Techniques: Acceleration Analysis

Figure 12.2

Four-bar mechanism.

EXAMPLE 12.1 For the four-bar mechanism ABCD shown in Figure 12.2, make a complete graphical acceleration analysis given that crank AB is rotating with an angular velocity of 2 rad/sec (counterclockwise) and an angular acceleration of 1 rad/sec2 (clockwise). SOLUTION 1.

Select a space scale ks = 2 in./in. or 1 in. (space scale) = 2 in. and velocity scale kv = 2 (in./sec)/in. or 1 in. (velocity scale) = 2 in./sec, and from the relationship

v obtain

b

VEL. 1"

Figure 12.3

Velocity polygon.

SCALE

= 2

in /sec

217

Special Methods

k

.

2

3.

4.

5.

a

(2/l)z 2/1

2 (in./sec2)/in.

or 1 in. (acceleration scale) = 2 in./sec2. Construct the velocity polygon (Figure 12.3) using the ky scale. This requires the calculation of the velocity magnitude Vg/^ as a first step, then the layout of the vectors in accordance with the velocity polygon procedure already discussed. Transfer vectors Vg, Vq, and Vg/C from the velocity polygon to points B and C on the linkage, maintaining the same orientation. Vg is drawn from B perpendicular to AB; Vq is drawn from C perpendicular to CD; and Vg/C is drawn from B perpendicular to BC. using the construction outlined Determine A^, A^J, and A^ B/C 13 C -N above. For example, in Figure 12.4, Ag is obtained by constructa right-angle triangle Abx in which z. b is 90°, Ax is the extended link AB, and the line Bb, perpendicular to AB, is the velocity magnitude of point B. The line segment Bx then represents the required value of Ag. Construct the acceleration polygon using the k~ scale (1 in. = 2 in./sec2), starting with the calculation of A^, then following the acceleration polygon procedure already discussed. Figure

Figure 12.4

Normal acceleration construction

Graphical Techniques: Acceleration Analysis

218

SCALE

n 2

ACC.

12.5 shows the acceleration polygon, from which the required acceleration Aq is determined to be A

= 12.2 in./sec2

(as directed)

In the example above, it is to be noted that only two calculations were neces¬ sary to complete the analysis after the scales were determined. These were to determine and A^, the velocity and acceleration magnitudes of the first link. If this link were to rotate with a constant angular velocity, calcu¬ lation of A^ would not have been necessary, since the value of this accelera¬ tion would be zei-o. Thus, for a complete graphical analysis, the maximum number of calculations necessary is two.

12.2

EQUIVALENT LINKAGE METHOD

Determining the acceleration of points on many higher-paired mechanisms, such as those with rolling and sliding contacts, can become rather involved if a point-to-point analysis is attempted. This is because of the need to know the curvature of the path traced by a point on one link relative to the

Special Methods

Figure 12.6

Kinematically equivalent four-bar linkages.

220

Figure 12.7

Graphical Techniques: Acceleration Analysis

Kinematically equivalent slider-crank linkages.

Special Methods

221

other and to apply the Coriolis law. If no easily recognized path is found, the analysis can be difficult. To simplify this problem, the use of equivalent linkages has been found most effective. In application, an equivalent linkage replaces a higherpaired contact with appropriate lower pairs that will produce the correct values of velocities and accelerations for the instantaneous phase under consideration. An equivalent linkage may then be defined as one that pro¬ duces identical motion as the part being analyzed for a given position or phase. Figures 12.6 and 12.7 show several mechanisms with their equivalent linkages depicted by dashed lines. Note that the rolling and sliding surfaces have been replaced by pin joints as part of a more simplified four-bar link¬ age. Note also that in each case, the floating link of the equivalent linkage is drawn along the common normal of the two contacting surfaces and con¬ nects the centers of curvature of the surfaces. Although an equivalent linkage is generally valid only for a given instant or phase and does not ordinarily apply to a complete cycle, there are some instances where the equivalent linkages of some higher-paired mechanisms will duplicate the input/output motion of those mechanisms throughout their motion cycle. Some examples are shown in Figures 12.6 and 12.7. EXAMPLE 12.2 Consider the cam mechanism shown in Figure 12.8a. The cam (2) rotates counterclockwise at a constant angular velocity of 2 rad/sec. Find the acceleration of the follower (4) using the equivalent linkage method. SOLUTION The equivalent mechanism for the cam mechanism given is the simple slider-crank ABC shown in Figure 12.8b, for which the velocity and accel¬ eration diagrams are readily obtained, as shown in Figure 12.8c and d. Applying the velocity polygon construction procedure, we obtain

B

V. V

C/B

0.9(2) = 1.8 in./sec 1.45 in./sec 0. 6 in./sec

(from the velocity polygon) (from the velocity polygon)

Applying the acceleration construction procedure, we obtain

A

13

= 0.9(2)2 =3.6 in./sec2

222

Graphical Techniques: Acceleration Analysis

Fig. 12.8 Acceleration analysis of a cam-follower mechanism: (a) camfollower mechanism; (b) equivalent linkage; (c) velocity polygon; (d) accel¬ eration polygon.

223

Special Methods

*Tb =

0

an ac/b

^

= 0.18 in./sec2

= 3.8 in./sec2 at C/B AC =

2.8 in./sec2

(from the acceleration polygon) (directed as shown)

Note that the resulting acceleration of point C (A^) is exactly the same as that obtained for point P4 (the same point) in Section 9.7, using an alter¬ native method.

12.3

SLIDER-CRANK ACCELERATION: PARALLELOGRAM METHOD

Despite the wide use of the relative acceleration method in solving linkage problems, it is not uncommon for one to experience some confusion in cor¬ rectly applying this method to the slider-crank mechanism. The confusion most often encountered arises from uncertainties or oversight as to the proper location of the relative radial acceleration vector in the acceleration polygon. As a result, many errors are made. To avoid such confusion or oversight, the graphical method presented here makes use of simple parallelogram constructions which serve as guides in laying out the vectors. The method is not only simple to apply, but also saves time.

Scope In a typical slider-crank mechanism, as shown in Figure 12.9a, crank AB has angular rotation to and angular deceleration o about point A. Angle 0q is the instantaneous angular position of the crank with respect to the line of action AC, and dc is the angle between the sli_der arm BC and line AC. It is requii-ed to determine slider acceleration Ac at any angle Qq. Construction Development 1.

Define o' as the point of zero acceleration coincident with pivot B on the mechanism, and draw vector o'b' (see the acceleration diagram in Figure 12.9c) to represent acceleration, AB, as determined from (12.14)

224

Graphical Techniques: Acceleration Analysis

(b)

KEY POINT: vN

,

Ag/C (°r

-rN

. _

_.

if used)

is

always located on the 'reflected' connecting rod.

(c)

Figure 12.9 Graphical construction procedure: (a) slider-crank mecha¬ nism; (b) velocity polygon; (c) acceleration polygon.

225

Special Methods

where the magnitudes of Ag and Ag are given by

OBw2

AT- = OBa Jd

2.

3.

With o'b' as the diagonal and 9q as_the subtended angle, construct a parallelogram o'Cb'c such that o'C is equal and parallel to cb', and b'C is equal and parallel to o'c. (Note the geometric identity between triangular sections o'b'c and o'b'C. Vector o'b' may be considered the axis of asymmetry to the parallelogram.) Determine Vg/c from the vectorial relationship (12.15) where V

B

OB co

(directed perpendicular to OB)

(12.16)

and Vc is known in direction only (along the slider path). Vg/c can therefore be obtained from a velocity diagram, as follows. From a point o (see the velocity diagram in Figure 12.9b), draw line ob scaled to represent Vg perpendicular to the instantaneous position of crank OB and pointing in the direction of the motion. Then draw a line parallel to the direction of the slider motion to represent velocity V^.. Finally, draw a line_from point b perpen¬ dicular to arm BC so that it intersects the Vc line at point c. The velocity magnitude Vg/C is determined by measuring line be and converting to the appropriate velocity according to the chosen scale. 4. Compute the acceleration magnitude A^c from

N = AB/C

V2 , B/C BC

(12.17)

Next, mark a segment of line b'c on the parallelogram to repre¬ sent vector Ag ^ scaled the same as o'b' and heading toward b . (This requirement is in keeping with the polygon convention, in which the acceleration vector relative to a point on a link must be be directed toward the corresponding point on the acceleration polygon. Alternatively, if aN/b is considered, this vector must point from b' to c' on the polygon.)

226

Graphical Techniques: Acceleration Analysis -N From the tail of vector Ag/^, draw a perpendicular line to inter¬ sect line o'c^at point c'. Point c' defines the acceleration, Ac. along line o'c, and the perpendicular drawn from a£J represents rji

B/ C

the tangential acceleration, Ag/C. The value of Ac may be checked with the vectorial relationship

aN4.

ab

TT ab

a-N rT ab/c ~ ab/c

(12.18)

(Note that for a uniform velocity of crank OB, triangle o'b'c is identical to the configuration of the given mechanism OBC, in which case C and C are coincident points. Note also that A^ alB/ C ways lies on the "reflected" slider arm.) As an example, consider a slider-crank mechanism where OB = 1.5 in., BC = 3 in., co = 1 rad/sec, and a - 0. Determine the slider acceleration, Ac, for eQ = 45° (Figure 12.10), 135° (Figure 12.11), 225° (Figure 12.12), and 315°(Figure 12.13). PROCEDURE 1*

From Equation (12.14), acceleration Ag = 1.5 in./sec and line o b is drawn 1.5 in. in length to represent this vector.

Figure 12.10 polygon.

Crank angle at 45°: (a) velocity polygon; (b) acceleration

Special Methods

Figure 12.11 polygon.

227

Crank angle at 135°: (a) velocity polygon; (b) acceleration

225°

b

Figure 12.12 polygon.

(b) Figure 12.13 polygon.

Crank angle at 315°: (a) velocity polygon; (b) acceleration

228

Graphical Techniques: Acceleration Analysis

.

2 3.

4.

Construct the parallelogram o'Cb'c, using o'b' as the diagonal. From Equation (12.16), Vg =1.5 in./sec. With line ob drawn 1.5 in. long to represent vector Vg, a velocity polygon is con¬ structed from which Vg/c = 1.12 in./sec. From Equation (12.17), A^^ = o.42 in. and a 0.42-in. segment

5.

of line b'c is denoted on the parallelogram to represent Ag/^. A perpendicular line is drawn from the tail of vector A^^ to intercept line o'c at point c'. The length of line o'c' is measured to be 1.05 in., so that Aq = 1.05 in./sec2 (to the left). Use of Equation (12.18) confirms this value as being accurate. Similarly, Ac for the crank position <90 = 135°, 225°, and 315° are found to be 1.05 in./sec2 (to the right), 1.05 in./sec2 (to the right), and 1.05 in./sec2 (to the left), respectively.

Ill ANALYTICAL TECHNIQUES

Analytical techniques or mathematical methods for velocity and accelerations determination can be a powerful tool in the analysis and design of a mecha¬ nism. Compared to graphical techniques, analytical techniques offer two major advantages. They are faster and more accurate, provided that a valid expression has been derived for the mechanism. Graphical methods typically involve numerous diagrams that must be constructed for each shift in the position of the mechanism. The use of mathematics, on the other hand, allows general expressions for velocity and acceleration to be developed in terms of link geometry for all positions. Once a general expression is obtained for a mechanism, the velocity and acceleration values for points on that mechanism can readily be determined. This expression also reveals how various parameters such as lengths and angular positions of the links affect motion characteristics. This information is most important in the synthesis of a mechanism where a specific output motion is desired. The main disadvantage of analytical techniques is that the mathematical analysis required to obtain general velocity and acceleration expressions for a mechanism are typically complex, lengthy, and error-prone. Conven¬ tional mathematical methods require both complex numbers and calculus, and generally, these methods are too theoretical to provide the quick insight the designer needs to develop a practical linkage. Fortunately, the mathematical analysis can be simplified using the simplified vector method presented in Chapters 14 through 17. Unlike the conventional methods, this method does not rely on calculus, but combines trigonometry, complex numbers, and principles of relative motion to obtain required motion relationships. Plane trigonometry and complex numbers are used to express motion relationships in concise vectorial forms, and relative motion principles help to simplify equation development.

229

230

Analytical Techniques

Besides the simplified vector method, two alternative mathematical methods are demonstrated in this part. These are the modified vector method, which is basically a variation of the simplified vector method, and the calculus method, which is conventional. In both cases it is clear that the application of calculus is essential.

13 Complex Algebra

13.1

INTRODUCTION

In mechanism analysis, a convenient way to describe the position of velocity or acceleration of a link is by the use of a complex number. A complex number consists of two parts, normally written in the form a + ib where a is the real part and b is the imaginary part, denoted by the letter i, which has a value of . Two complex numbers that differ only in the sign of their imaginary parts, such as a + ib and a - ib, are termed conjugates. Thus a - ib is the conjugate of a + ib; and conversely, a + ib is the conjugate of a - ib.

13.2

COMPLEX VECTOR OPERATIONS

Computations involving complex numbers follow the rules for algebra, with the additional requirement that all powers of i be reduced to the lowest terms by applying the following properties:

14 = +1 15 = +i

etc.

For example, if rj - a + ib and r2 = c + id, then

231

232

Analytical Techniques ri +

r2

~ (a + ib) + (c + id) = (a + c) + i(b + d)

1*1

- r2 = (a + ib) - (c + id) = (a + c) - i(b + d)

ri x r2 = (a + ib)(c + id) = (ac - bd) + i(bc + ad)

_ (a + ib) (c - id) (c + id)(c - id) = ac + bd [ i(bc - ad) _ (ac + bd) + i(bc - ad) c2 + d2 c2 + d2 c2 + d2 Note that in the division case, the quotient is conveniently found by multiply¬ ing both numerator and denominator by the conjugate of the denominator.

13.3

GEOMETRIC REPRESENTATION OF A COMPLEX VECTOR

A complex number is represented geometrically by a vector in the complex plane defined by two mutually perpendicular axes: the real and imaginary axes. For example, in Figure 13.1 the complex number (a + ib) is shown to be represented by the vector Rl5 which extends from the origin o of the complex axes to a point P in the plane such that the horizontal component along the positive real axis represents the real part of the number, a, and the vertical component along the positive imaginary axis represents the imaginary part, b. Thus the magnitude of Rj is given by Rj = I Rj | = si a2 + b2 and the direction is given by


(0°< 0j < 90°)

Note that the angle 9X is conventionally defined as the angular displacement of the vector from the positive real axis and is positive when measured counterclockwise.

233

Complex Algebra +VE IMAG AXIS

Figure 13.1

Graphical representation of complex numbers.

Similarly, the complex number (-a + ib) is represented by a vector R2 as shown in Figure 13.1. Here the horizontal distance -a along the nega¬ tive real axis represents the real part of the number, and the vertical dis¬ tance b along the positive imaginary axis represents the imaginary part. Therefore, the magnitude and direction of vector R are obtained as follows: R2 = I R21 = ^ (-a)2 + br

Q7 = tan"1 — L -a

(90° < 9Z <180°)

Note that care should be taken in interpreting the value of 9Z since in the range 0 to 360°, there are two values of 9? for each value of tan 9Z. To avoid ambiguity, it is usually advisable first to determine the quadrant in which the vector lies, using a simple sketch, before deciding which of the two angles applies. In general, to represent a complex quantity, say R — a + ib, as a vector: 1.

Find the absolute value of the components 1

| R I = (a2 + b2)2

234

Analytical Techniques

2.

Find the direction (9) from

0 = tan- to»&***y real

or 9 = tan

-ib

noting that: If If If If

13.4

a a a a

is is is is

positive and b positive, negative and b positive, negative and b negative, positive and b negative,

9 is in first quadrant. 9 is in second quadrant. 9 is in third quadrant. 9 is in fourth quadrant.

COMPLEX FORMS

The vector R may be expressed in one of four complex forms: (1) rectangu¬ lar, (2) trigonometric, (3) exponential, and (4) polar. In the rectangular form, R = a + ib where, as before, a and b are, respectively, the real and imaginary com¬ ponents of R. In the trigonometric form, R = R cos 9 + iR sin 9 where R cos 9 = a and R sin 9 = b, as shown in Figure 13.2. Alternatively, R = R (cos 9 + i sin 9) where (cos 9 + i sin 9) i_s the trigonometric form of the unit vector that defines the direction of R. In the exponential form,

where ei0 is the exponential equivalent of the unit vector (cos 9 + i sin 9) given above. In the polar form.

Complex Algebra

235

IMAG AXIS

IMAG AXIS

Figure 13.2

Graphical representation of the complex vector.

R = R jL9 where

13.5

4.

6 defines the angular position of R.

THE UNIT VECTOR

A unit vector is a vector that has a magnitude of unity. If V is a vector with magnitude V * 0, then V/V is a unit vector having the same direction of V. Defining this unit vector as V, we can write

236

Analytical Techniques

Figure 13.3

V V

Unit vector.

v

or V = Vv which means that any vector V can be represented by the magnitude of V multiplied by the unit vector v in the direction of V. The unit vector may be expressed in exponential form, such as ei9 and e1(0+90 ), where 9 and (9+ 90°) are the position angles of the vectors, measured from the real axis (see Figure B. 1 in Appendix B) and i is equal to \P-l. However, for the purpose of computation, it is more convenient to convert these unit vectors to their equivalent trigonometric forms, as follows:

e

id

= cos 9 + i sin 9

and e

i(0+ 90°)

= cos (9 + 90°) + i sin (9 + 90°)

Thus, from Figure 13.3,

vx = viei0 Vj = V1(cos 9+ i sin 9) and

237

Complex Algebra

V2=V2e‘<9+9°-> V2 = V2[cos(0 + 90°) + 1 sin(0 + 90°)]

13.6

LINKAGE APPLICATION

In linkage analysis it is convenient to express vector quantities such as displacement velocity and acceleration as complex numbers which can be written in any of the complex forms discussed previously. Consider the case of a link AB that rotates counterclockwise with angular velocity to and angular acceleration oi. The displacement or change in position of the point B can be described by a position factor R directed from the origin of the complex axes to that point (Figure 13.4a). That is, the displacement of link AB is given by

where R is the link length and e* ® is the exponential form of the unit reactor which defines the instantaneous angular position of the link. From previous velocity studies we know that the magnitude of velocity of point B can be obtained from the relationship VB

=

rco

and that the vector describing this velocity acts perpendicularly to the link in the same sense as 0. This velocity vector VB, shown in Figure 13.4b, can therefore be expressed in the complex form

Rcoe

where

i(0 +90°)

) is the unit vector used to indicate the direction of the

velocity. Similarly, the tangential, acceleration, shown in Figure 13.4c may be written as

aZB , i(0+9O°) . where Ra is the magnitude of the acceleration and e is the unit vector defining the direction. The normal acceleration, shown in Figure 13.4d, may be written as

238

Analytical Techniques

Figure 13.4 Graphical representations: (a) displacement; (b) velocity; (c) tangential acceleration; (d) normal acceleration.

239

Complex Algebra

Ruze

i(0+18O°)

-Roo2

e

id

whei'e Ru2 is the magnitude of the acceleration and -e1® is the unit vector defining the direction. EXAMPLE 13.1 Link AB of length R in Figure 13.5a rotates counterclockwise at an angular velocity

co.

a. b.

Develop a complex expression for the velocity Vg. Evaluate the expression found in part (a) given that R = 1.5 in., co —

1 rad/sec, and 9 = 120°.

SOLUTION Proceed as follows: 1.

Using point B as the intersection of the real and imaginary axes of the complex plane (Figure 13.5b), draw line Bb, perpendicular to AB, to represent Vg. Drop a perpendicular from point b or terminus of Vg to intersect

2.

the imaginary axis at point o (Figure 13.5c). From this construction we obtain the vector equation V

ob + Bo

B

(vectorial summation)

where ob = -V^ sin 9 B

(real)

Bo = iV

(imaginary)

13

cos 9

Therefore, V

B

= -V

B

sin 9 + iV

B

cos 9

= -V (sin 9 - i cos 9) B = -Rco(sin 9 - i cos 9)

240

Figure 13.5

Analytical Techniques

Velocity representation for Example 13.1.

Complex Algebra

V

= -Roj(sin 9 - i cos 9) ID

= -(1.5)(1)(sin 120° - i cos 120°) = -1. 5 [0.866 - i(-0. 5)] = -1.5(0.866 + i 0. 5) = -1.5 ^(0.866)2 + 0.57 = -1.5(1) at tan-' = -1.5 in./sec at 30° = 1.5 in./sec at 210°

241

14 Four-Bar Mechanism Analysis: Simplified Vector Method

14.1

INTRODUCTION

The four-bar mechanism is often considered the most basic of all kinematic mechanisms. Consisting of four rigid links (AB, BC, CD, and AD) and four turning pairs (A, B, C, and D) as shown in Figure 14.1, this mechanism can be arranged in three basic configurations and, accordingly, is classified as follows: The crank-rocker mechanism, where drive crank AB is capable of complete rotation while the follower CD oscillates. Here the drive crank must be the shortest link. The drag link mechanism, where both drive crank AB and follower CD are capable of complete rotation. Here the frame AD must be the shortest link.

C

Figure 14.1

Four-bar mechanism.

242

Four-Bar Mechanism Analysis: Simplified Vector Method

Figure 14.2

243

Beam pump mechanism.

The double rocker mechanism, where both drive crank AB and follower CD oscillate and neither is capable of complete rotation. Here the coupler BC must be the shortest link. The four-bar mechanism is important in many ways: 1. 2.

3.

It is the simplest possible plane linkage that can provide virtually any type of output motion. Variations and combinations of this linkage make possible an al¬ most limitless variety of mechanisms. Typical examples can be seen in some common applications shown in Figures 14.2 to 14.4. Analysis of many complex direct contact mechanisms can be simplified by replacing the mechanism with an equivalent four-bar linkage.

244

Figure 14.4

Analytical Techniques

Forklift truck mechanism. (From A. S. Hall, 1961.)

Four-Bar Mechanism Analysis: Simplified Vector Method

245

For these reasons, the motion characteristics of the four-bar linkage are studied more than those of other kinematic mechanism. The simplified mathematical method presented here quickly deter¬ mines linear velocity and acceleration relationships of key points in the four-bar linkage for any given position of the drive crank during its motion cycle.

14.2

SCOPE AND ASSUMPTIONS

The mechanism ABCD in Figure 14.1 represents a typical four-bar linkage mechanism in which links AB, BC, CD, and AD have known lengths, and #A> %> and 0C are their respective angular positions measured from base link AD. Crank AB rotates with an angular velocity co and an angular accel¬ eration a about pivot point A. In this analysis, equations for computing linear velocities and accel¬ erations of the mechanism for a given instantaneous crank angle 9^ will be determined. Note that all angular displacements, velocities, and accelera¬ tions are considered positive for counterclockwise rotation and negative for clockwise rotation.

14.3

GEOMETRIC RELATIONSHIPS

First, we determine the angular relationships of and 9q by constructing a diagonal BD, as in Figure 14.5, to form two triangles, ABD and BCD, where ABD - (j>B ADB = 4>d CBD - yB CDB = yD

Using basic trigonometric relationships, we can show that

4> D

(14.1)

and (14.2) *C = 18°° - ^D - YD where

Analytical Techniques

246

C

Figure 14.5

y

—

Geometric relationships.

COg

BC2 + BD2 - CD2 -

*B

2(BD)(CD)

,

.

I'd

/ AB

\

= sm (.55 sm V

•y_ = cos'D

, BD2 + CD2 - BC2 2(BD) (CD)

BD = [AB2 + AD2 - 2(AB)(AD) cos

14.4

VELOCITY ANALYSIS

To determine the velocities Vg, Vc, and Vc/b> we construct the velocity polygon Bcb for the mechanism, as in Figure 14.6, letting:

Figure 14.6

Velocity polygon.

Four-Bar Mechanism Analysis: Simplified Vector Method

247

Bb represent magnitude of the linear velocity of point B (VB) Be represent magnitude of the linear velocity of point C (Vc) be represent magnitude of the relative velocity of point C with respect to point B (V^/g) Then from this polygon, we determine the angular relationships of the veloc¬ ities as follows: = 0^ + 90° - 0 C

CD

A

= 180° + 0

B

- 0

- (3)

= 180° +

B

- 0. - (0_ + 90° - 0 )

= 180° + 0B - 0A - 0C - 9°° + 0A =

®

= 90‘

9O° + 0b-0c

= 90° - (90° + 9b - 0n) - (0r + 90° - 0A)

-

A'

= 90° - 90° “ 0B + 9C ~ dc ~ 90° + °A = e, -

A

^c = 90° - CD

- 90°

B

= 90° - (90° + eB - 0 ) = 90° - 90° - 0_ + 0_ JD

9C

z.b = 90° -

L/

°B

= 90° - (0A - 0_ - 90°) ' A B = 90° -0+0 + 90° A B = 180° -0+0^ A B

/ bBc =

CD +

(90° + 0B - 0n) + (0A - 0R - 90°) B = 9O°+0b-0c+0a-0b-9O<

0A-0C VB, VC, and Vc/B can now be found by apP^ing the rule of sines from trigonometry. V. B sin(0„ - 0„) B

V

V. sin(180° - 0

+ 0 )

B

sin(0

C/B - 0 )

(14.3)

248

Analytical Techniques

Noting that sin(180° - 0A + eB) = sin[180° -

- e^]

= sin(eA-oB) Equation 14.3 can be written

vc

VB sin(0c + dB)

sin(6»A -

VC/B

V

sln«,A-‘,C)

VB

vc

VC/B

sin(0B-6»c)

sin(6»B-0A)

sta<0c - eA)

or

Also, since V-3 = co AB 13

(14.4)

then

sm(6> V

C

e

)

= co AB —--A

(14.5)

sin(0B - eQ)

and

sin(0r ~ 6

)

VC/B = “AB sin (dg - eQ)

(14.6)

Since VB, Vc, and Vc/B are oriented, respectively, at angles (0A + 90°), (0C + 90°), and (% + 90°) from the real or reference axis, we can write the required vectorial expressions as follows: i(0A+9O°)

V

B

V

(14.7) i(0c+9O°)

V,

(14.8)

"V i(dB+90°)

C/B = VC/B6

(14.9)

Four-Bar Mechanism Analysis: Simplified Vector Method

3i(0A+9»-)j ei«fcM>
249

and ei(%+90-) are mit vectorSi usedto

define the direction of V-g, V^, and V^/g, respectively. (Note that e^ = cos 9 + i sin 9, where 9 is the position angle.)

14.5

ACCELERATION ANALYSIS

To find the linear acceleration of point C (Aq) , we apply the relative motion theory, which states that + A

(14.10)

C/B

Expanding this equation into its normal and tangential component form, we have rN ^

Ac

xT - yN u. *'T m x-N

Ac

ab

ab

ac/b

.

rT

(14.11)

ac/b

where

-N VC lSC AC = ~ CD 6

Ti(V90")

Ac = Ace AT aJJ = -co2 ABe ID

K A

= aABe

rN ac/b -T

Ac/B

V2 .„ ifL C/B B BC 6 T

i(V9°0)

AC/Be

or 10, V€ iSC T “V90”* CD6 +AC6

■co2 ABe

i(»A+90’) + a ABe

VC/B ‘"B T i(V9°‘) -e +AC/Be BC

(14.12)

Analytical Techniques

250

Examination of this complex equation indicates that the only unknown quanti¬ ties are the magnitudes A^ and A^^. All other quantities (magnitudes and directions) are either known or can readily be determined from the problem data. Note that the directions of A^ and A^^, although not precisely known, are assumed positive (or counterclockwise) for convenience only. If, in reality, any of the directions is reversed, the numerical solution of the equation will automatically produce a negative sign for the unknown quantity. To solve Equation (14.12), we equate the real and imaginary parts as follows: Real: V2 C T — cos ec + Ac cos (6>c + 90°) = -co2AB cos 9A + aAB cos (@A + 90°) V2 C/B

-^cos 9b + Ac/b cos(0b+90-)

(14.13) Imaginary: V2 C “

. sm

T + aQ sin (0p + 90°) = - co2 AB sin 9 + aAB sin (9 + 90°) A A V2 , C/B

T

-lTsm 0b + ac/b sin<0B + 9O°> (14.14) The solution of these simultaneous equations yields AT ... CjB2 - C2Bt

c

A:B2 - A2Bj

T

Aj C2 - A2Cx

lC/B

Aj B2 - A2Bj

(14.15)

(14.16)

where ■1 = cos (9C + 90°)

■2 = sin(0c + 90°) i = -cos (9

+ 90°)

2 = -sin (0

+ 90°)

(14.17) (14.18) (14.19) (14.20)

Four-Bar Mechanism Analysis: Simplified Vector Method

C ci

v cos

=

w2AB cos 9 + aAB cos (0 +90°) ^ A

-

251

/ BC

cos 9 B (14.21)

cz = CD Sin °C ~ w2AB Sin 9A + “AB sin(6lA + 90°) " • ~^B sin 9

Given that the values V^,,

A^, and A^

(14.22)

have been determined,

general equations for computing the linear accelerations are: i0 A

= -co2ABe

1(0+90°) + aABe

(14.23)

-D

VJ, AC

i0c

CD 6

T K0C+9O°) (14.24)

+AC6

— VC/B i0B T AC/B = ~ ~BC 6 +AC/B6

(14.25)

EXAMPLE 14.1: Crank Rocker Considering the four-bar linkage in Figure 14.1, let AB = 1.5 in., BC = 3 in., CD = 3 in., AD = 4 in., 0A = 30°, co = 2 rad/sec (counterclockwise), and a = 1 rad/sec2. It is required to find Vq, Vq/jj, A^, and A^/g. SOLUTION 1.

Determine 0g and 0C using Equations (14.1) and (14.2). 9

B

9

C 2.

= 62.14° - 15.51° = 46.63°

= 180° - 15.51° - 62.14° = 102.35°

Determine Vg, Vq, and Vc/B usinS Equations (14.4), (14.5), and (14.6). V

B

= 2(1.5) = 3.00 in./sec

v - 2(1 5) s*(46-6°-30_°1_ VC )sin(46.6°-102.3c sin(102.3° - 30.0°) sin(46.6° - 102.3°) VC/B = 2(1,5)~

-1.04 in./sec

-3.46 in./sec

Analytical Techniques

252

3.

Determine VB, Vc, and VC//B using Equations (14.7), (14.8), and (14.9).

= 3.0el(12° ) = 3.0(cos 120°+ i sin 120°)

13

= 3.0 in./sec / 120° i(T92 3°1 Vc = -1.04e v ; = -1.04(cos 192.3°+ i sin 192.3°) =

1.04 in./sec /12.3°

Vc/B = -3.46el(136’4 ) = -3.46(cos 136. 6°+i sin 136.6°) = 4.

3.46 in./sec /-43.4°

Determine the constants A1( A2, Blf B2, Cj, and C2 using Equations (14.17) through (14.22). Aj = cos (102.35 + 90°) = -0.98 A2 = sin (102.35 + 90°) =

Vq (in / sec)

0.21

-

Four-Bar Mechanism Analysis: Simplified Vector Method

^C/B

253

^'n / sec)

Figure 14.8

Velocity of C relative to B versus crank angle.

Bx = - cos (46.6°+ 90°) = 0.73 Bz = - sin (46.6°+ 90°) = -0.69 2

C1 -- (~--'-n4)

cos 102.35° - 2Z(1. 5) cos 30°

o. U

+ 1.0(1.5) cos (30°+ 90°) J-jM6.) cos 46.6° = -8.76 o• U

2

C2 =

o. u

sin 102.35° - 22(1.5) sin 30° /

o

+ 1.0(1.5) sin (30°+ 90°) - yy—sin 46.6° = -4.25

5.

Determine

and A^^ using Equations (14.15) and (14.16).

T _ (-8.76) (0.69) - (-4.25)(0.73) AC “ (-0.98)(-0.69) - (-0.21)(0.73) = 11.02 in./sec2

Analytical Techniques

254

T _ (-0.98)(-4.25) - (-0.21)(-8.76) AC/B (-0.98)(-0.69) - (-0.21)(0.73) =2.76 in./sec2 6.

T Substitute the values found for A^ and and solve for Aq and Aq/jj.

T

in Equations (14.24)

(-1.04)2 ei(102.3°) + n<02ei(192.3°) 3.0 0.36 /-77.6° + 11.02 /-167.6° 11.03 in./sec2 /-165.70 (~3.46)2 3.0

i(46.63°)

i(136.63°) + 2.76 e

3.99 /-133.4° + 2.76 /136.6° 4.85 in./sec2 /-168.04°

Four-Bar Mechanism Analysis: Simplified Vector Method

255

ac/b (in / sec^)

Figure 14.10

Acceleration of C relative to B versus crank angle.

See Figures 14.7 to 14.10 and Table 14.1 for velocity and acceleration profiles of the complete crank cycle (9A = 15 to 360°). EXAMPLE 14.2: Drag Link Considering the four-bar linkage in Figure 14.1, let AB ==2.0 in., BC = 1.5 in., CD = 2.5 in. , AD = 1.0 in., dA = 240°, go = 1 rad/sec (counter¬ clockwise), and a = 0.5 rad/sec2. It is required to find Vc, Vc/B, Ac, and Ac/b. SOLUTION 1.

Determine 0g and 0(j using Equations (14.1) and (14.2). 0

- 67.8° - (-40.9°) = 108.7°

e

= 180° - (-40.9°) - 33.7° = 187.14°

B

Analytical Techniques

256 O

O

o

—


ID

CD

ID

O

O

CD

cn

—

O

od

-

O

—

cn

c£>

in

ALPHA (A )

O CD

o o

o

o

o

O

O

CD

—

CD

—

CD

—

COGOCOCOCOOOODGOCDaDGOCOGOaoCDGOOOGOaOCO

2.00 * OMEGA ( A)

CO

O

O -

--

in

CD

co

4.00

O CD

■*-

rj

CD

—

CO

CD

»-

r-

O

3.00

AD

»

in

--

•

O

3.00

CD

O cd

o

o

CD

0888888888888888888 n

cn

cn

co

cn

*

>tnocnr>cntncntncncn

1.50 CD

IT)

AB

*

CD

O CD

■*“

CN

O 6

o

CD

O

TABLE 14.1

o

CDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDlDCDCDCDCDCDCD

BC

Velocity and Acceleration Profiles for Example Problem

o in

-

0

O

O

O

in

8 8 8 8 8 8 8

Four-Bar Mechanism Analysis: Simplified Vector Method

257

2. Determine Vg, Vq, and Vg/g using Equations (14.4), (14.5), and (14.6). Vg - 1(2.0) = 2.0 in./sec

V

V

C/B

=

U9 m sin (108.7° - 240°) _ . , ( '0) sin (108.7° - 187.14°) !• 53 m./sec sin (187.14° - 240°) ( '0) sin (108.7° - 187.14°)

, . . 1,63 m./see

3. Determine Vg, Vq, and Vq/b using Equations (14.7), (14.8), and (14.9).

VB = 2.0el(33° ) = 2.0 in./sec /-30°

V

= 1.53el(267-1 ) = 1.53 in./sec /-82.9

Vc/B = 1.63el(198'7 ) = 1.63 in. /sec /-161.3

4. Determine the constants Ax, Az, Bj, B2, Cj, and C2 using Equations (14.17) through (14.22). Aj = cos (187.1°+ 90°) =

0.12

A2 = sin (187.1°+ 90°) = -0.99 Bj = - cos (108. 7 + 90°) = 0.95 Bz = - sin (108.7 + 90°) = 0.32

1

_ U53^ cQg 187>1°+ (1)2(2.0) cos 240° 2.5 1 CQ 2 + 0.5(2.0) cos (240°+ 90°) -

cos 108.7°= 1-49

258

Analytical Techniques

1

S32

C2 = -1-— sin 187.1°+ (1)2(2.0) sin 240° Z• O

+ 0.5(2.0) sin (240°+ 90°) - ^-|3- sin 108.7° = -0.56 1.5

5.

Determine A^ and A^^ using Equations (14.15) and (14.16).

at = (l-49)(0.32) - (-0. 56)(0.95) AC (0.12) (0.32) - (-0.99) (0.95) = 1.03 in./sec2

A

T = (0.12)(-0. 56) - (-0.99)(1.49) C/B (0.12)(0.32) - (-0.99)(0.95) = 1.45 in./sec2

6.

Substitute the values found for A^ and A^

in Equations (14.24)

and (14.25) and solve for A^ and Aq/b.

Ac = -+fie1<187-14") + 1.03e1(277-l4“) = 0.94 /7.15° + 1.03 /-82.9° = 1.39 in./sec2 /-40.4° _ 1632 i(108.7°) i(198.7°) A-= - —— e*' ' + 1.45 e "C/B "1.5 = 1.76 /-71.3°+ 1.45 /-161.3° = 2.28 in./sec2 /-110.7° Note that the four-bar mechanism may be presented in a crossedphase configuration as shown in Figure 14.11, where, with crank AB in the first quadrant and rotating counterclockwise, link BC crosses the fixed link AD. In this configuration, the oscillation of the follower is below the fixed link and the equations for % and Qq do not apply directly. However, it is useful to note that this mechanism is the mirror image of the one in Figure 14.1 when the crank AB is in the fourth quadrant and its rotation is in the clockwise direction. Hence the equations derived above are applicable to the velocity and acceleration analysis, provided that the reflected crank angle and levei’sed rotation sense are considered. Alternatively, if we elect to analyze the mechanism in the crossed phase as presented, Equations (14.1) and (14.2) must be modified as follows:

Four-Bar Mechanism Analysis: Simplified Vector Method

9B = -yB “ 0D

ec

= i8«”+yD-^D

All other relationships remain unchanged.

259

15

Slider-Crank Mechanism Analysis: Simplified Vector Method

15.1

INTRODUCTION

The slider-crank mechanism is probably the most common kinematic element to be found in most machines. A basic variation of the four-bar linkage, in which the follower crank is replaced by a sliding block, this mechanism is capable of converting rotary motion to linear motion, and vice versa. The mechanism is generally found in three basic arrangements 2 1.

The central or in-line type, as in ABC of Figure 15.1, where the slider path of travel essentially passes through the crank pin or point A.

2-

The positive-offset type, as in ABCD of Figure 15.2, where the slider path is offset a distance CD (or eccentricity e) above a refeience line AD drawn parallel to the path and passing through point A.

3-

The negative-offset type, as in ABCD of Figure 15.3, where the slider path is offset a distance CD below the reference line AD, drawn parallel to the path and passing through point A.

Among the countless applications of this mechanism are piston engines, pumps, compressors, saws, and other forms of reciprocating machinery, ome examples of these applications can be seen in Figures 15.4 to 15.7. *?’ llke the fonr-bar linkage, this mechanism is generally used in combia ion with other basic mechanisms to produce a wide variety of output motions as well as a model to simplify the analysis of more complex ma-

nn11JeS' _°r ^ese reasons> the slider crank is one of the most frequently reiLuonshTos w"lIV3; ^ ^ ^ Vel°City and acceleration lelationships will be developed for any angular position of the drive crank.

260

Slider-Crank Mechanism Analysis: Simplified Vector Method

Figure 15= 1

Slider-crank mechanism. (Courtesy of Ingersoll Rand,

Woodcliff Lake, N.J.)

Figure 15.2

Slider crank with positive offset.

261

Analytical Techniques

262 B

Figure 15.3

Slider crank with negative offset.

Punch press

Figure 15.4

Typical slider-crank applications: machinery.

Figure 15.5

Toggle mechanism.

Slider-Crank Mechanism Analysis: Simplified Vector Method

Figure 15.6

15.2

263

Drag-link quick-return mechanism.

SCOPE AND ASSUMPTIONS

The mechanism ABC in Figure 15.1 represents a typical slider-crank mechanism in which AB, the length of the drive crank 2, and BC, the length of the connecting rod 3, are known. AC is a variable distance depending on the angular position d2 of the crank during the motion cycle. Given that the crank rotates with an angular velocity co2 and an angular acceleration oi2 (both in the counterclockwise direction), we will now derive general expres¬ sions to compute linear velocities and accelerations of points A, B, C, and point B relative to point C for any angular position of the crank. In this analysis, all angular displacements, velocities, and accelera¬ tions are considered positive for counterclockwise rotation and negative for clockwise rotation. All velocity vectors are positive with respect to the positions of their respective links.

15.3

GEOMETRIC RELATIONSHIPS

To begin the analysis, we first seek to determine the angle of the connecting rod 0, in terms of the crank angle 02. Considering the typical slider-crank mechanism represented by the triangle ABC in Figure 15.8, we note that

Analytical Techniques

264

Figure 15.8

Geometric relationships.

since the link lengths AB and BC are known, we can apply geometric and trigonometric relationships to obtain 63 = 180° + 03

(15. x)

where i

.

-1iAB

.

Cp ~ = sin sin-1 -si — sin

= - sin -l

(360° - 9Z)

AB sin 9, BC

or 03 - 180° - 03

(15.2)

where

03 = sin_1 (bc sin 90 For the positive-offset slider crank, represented by diagram ABCD in Figure 15.2, we proceed by dropping a perpendicular from point B to meet AD at L, then extending the slider path to intersect the perpendicular BL. From this, we find the angle 03 as follows: 6»3 - 180° - (p3

[Equation (15.2)]

sin 03 + e = AB sin 9Z

(15.3)

'AB . e \ 03 = sin-1 ( z— sm 07 - — BC 2 BC'' . —if AB 0 \ 03 = 180° - sm ‘(bc sm e2 - —)

(15.4)

(e > 0)

(15.5)

Similarly, for the negative-offset slider crank, represented by diagram ABCD in Figure 15.3, we obtain

Slider-Crank Mechanism Analysis: Simplified Vector Method

03 = 180° - sin

sin 9Z +

(e < 0)

265

(15.6)

Hence the general equation may be written as

d3 = 180° - sin

(AB VBC

)

(15.7)

where e has a positive value for positive offset and a negative value for negative offset.

15.4

VELOCITY ANALYSIS

To determine the linear velocity relationships for Vg, V^, and Mg/c* let Bcb in Figure 15.9 represent the velocity polygon of the mechanism ABC, where: Bb represents the magnitude of linear velocity of point B (VB). Be represents the magnitude of linear velocity of point C (Vc) • be represents the magnitude of the relative velocity of point B with respect to point C (Vb/C)* Then the internal angles of the polygon can be obtained as follows:

B

b Figure 15.9

Slider-crank analysis.

Analytical Techniques

266

y = 360° - 02 - 90° = 270°
ez

-03

- y

= 90°- (0j - 180°) - (270° - 02)

e3+ 180° - 270°+ e2

= 90° -

- 02 - 03 /bcB = 180° - y -z.b = 180° - (270° -

02)

= 180° - 270°+ e»2 -

-

(02 - e3)

e2 + d3

= 03 - 90° By applying the rule of sines from trigonometry, VB, Vc, and VB/C can be expressed in the form V. B

V,

sin (03 - 90°)

sin (92 - 03)

V

B/C sin (270° - Q2)

which reduces to V

V

B

sin (90° - e3)

sin (93 - d2)

B/C sin (90° - 02)

Thus the scalar expressions for VB, Vc, and VB/c are obtained as follows: V. = AB X o>2 B

V, V. B/C

sin (d3 - V _y

(15.8) -

B sin (90° -

d2) 03)

(15.9)

sin (90° - 02) B sin (90° - e3)

(15.10)

To convert these equations to vectorial forms, we return to Figure 15.9 and note the following:

e directions of velocities VB and VB/C are assumed to be oriented at angles (d2 + 90°) and (e3 + 90°), respectively, he direction of velocity Vc is known to act along a straight line only.

Slider-Crank Mechanism Analysis: Simplified Vector Method

267

Therefore, we can write the vectorial expressions as follows:

^B

B/C

15.5

= ABco2ei(02+9°°>

(15.11)

= *B^ Sin{/- « e“‘ 2 sin (90 - 03)

(15.12)

= ABU2Sto(90;-^ e1^90’* 2 sin (90 - 63)

(15.13)

ACCELERATION ANALYSIS

To determine the linear acceleration relationships for point C (A^), we apply the relative motion equation, which states that A

13

= A

C

+ A

, J3/ C

(vectorial sum)

(15.14)

Expanding this equation into its normal and tangential form, we obtain -N -T -N AB + ab “ Ac

-T

+ Ac

-N -T + ab/c + ab/c

(15.15)

where -N , 4 „ i d2 A^ = -co?ABe 2 B 2 -T AT3 i(02+9O°) A_ = a.ABe 2 B 2 V2 A^ = - —— ei9° = 0 C °° rT aT i0° Ac = Ace

an B/C -T

(straight line)

. .. (real)

V2 / B/C id3 -e 2 BC aT

i(6>3+90°)

AB/C " AB/C® Note that point C has no normal acceleration since the slider path is a straight line of infinite radius. Therefore, the absolute acceleration of C is the tangential acceleration. Thus Equation (15.15) becomes

268

Analytical Techniques

ji, ABe1 ri'

^ABe1^90"' = A*

c

L

V2 B/C BC '

i0.

T i(63+90c B/C

(15.16)

This equation contains two unknown quantities: the magnitudes Aq and Ag/^ All other quantities (magnitudes and directions) are either known or can readily be determined from problem data. Note that the directions of A^ _rp C and Ag/C, although not known precisely, are assumed to be positive for convenience. If the actual direction of either is reversed, the numerical solution of the equation will automatically produce a negative sign for the unknown quantity. To solve the unknowns, therefore, we equate the real and imaginary parts of the equation and rearrange as follows: Real:

Ac + ab/c cos

((?3

+ 900)

Imaginary:

^"q/q sin(03+ 90°) = -to3AB sin 0Z + a2AB sin(02 + 90°) +

V2 B/C sin 03 BC (15.18)

The solution of these simultaneous equations yields aT

CjB2 - C2Bj

‘C

a1b2-a2b1

aT

AjC2 - A2Cj

“B/C

A:B2 - A2B,

(15.19)

(15.20)

where

(15.21)

II O

>

A! = 1

(15.22) Bj = cos (e»3 + 90°)

(15.23)

B2 = sin (6*3 + 90°)

(15.24) V2 ,

Cj = -AB cos e2 + a2AB cos (dz + 90°) +

c2

cos 03

(15.25)

V2 = -W2AB sin 9Z + a2AB sin(02 + 90 ) +

sin 03

(15.26)

Slider-Crank Mechanism Analysis: Simplified Vector Method

Given that the values V.^^, A^, and A^

269

have been found, the

general equations for computing the linear accelerations can be summarized as follows:

Ab = -w2ABe102 + o2ABel(02+9°O)

(15.27)

7 T i0° AC = AC6

(15.28)

A B/C

(real)

= _ VB/C BC

i03 +

T AB/C

1(03+90°)

(15.29)

EXAMPLE 15.1: Central Slider Crank Considering the slider-crank mechanism in Figure 15.1, let AB = 1. 5 in., BC = 3 in., 02 = 150°, go2 = 1.0 rad/sec (counterclockwise), and a2 = 0 rad/sec2. It is required to find Vq, Vb/£, Aq, and AB/£. SOLUTION* 1.

Determine 03 using Equations (15.2) and (15.1).

03 = sin"1

sin 150°)

= 14.5C 6»3 = 180° - 14.5° - 165.5° 2.

Determine VB, Vc, and VB/C using Equations (15.8), (15.9), and (15.10). V

B

= 1. 5(1.0) = 1.50 in./sec sin (165.5° - 150°)

„

.

,

VC = (1'5) slnU'-ieTsV- = -°'41 m-/3e°

VB/C

U

sin (90° - 150°_j = ' sin (90° - 165.5°)

in./sec

*See Figure 15.10 for velocity and acceleration profiles of complete crank cycle from 0Z = 0°to 0Z - 360°.

Analytical Techniques

270

3.

Determine VB, Vc, and VB/C using Equations (15.11), (15.12), and (15.13).

VB =

VC = V

4.

B/C

, _ i(240°) 1.5e = 1.5 in./sec /-120°

-0.41e

i0°

= 0.41 in./sec / 180°

1 i(255.5°) l*34e - 1.34 in./sec /-104.5°

Determine the constants A1( A2, Bv B2, Cj, and C2 using Equations (15.21) through (15.26). A] = 1

[from Equation (15.21)]

A2 = 0

[from Equation (15.22)]

Slider Crank Mechanism Analysis: Simplified Vector Method

271

Bj = cos (165.5°+ 90°) = -0.25 B2 = sin (165.5° + 90°) = -0.97 ci = -(l)2 (1. 5) cos 150°+ 0(1.5) cos (150°+ 90°) 1.342 + —~— cos 165. 5° O

= -(-1.3) + 0 + (-0.58) = 0.72 C2 = -(1)2(1. 5) sin 150°+ 0(1.5) sin (150°+ 90°) 1.342 + —-— sin 165.5° O

= -0.75 + 0 + 0.15 = -0.60

5.

Determine

T

T and A-q/q using Equations (15.19) and (15.20).

T _ (0.72)(-0.97) - (-0.60)(-0.25) AC (1.00)(-0.97) - (0.00)(-0.25) = 0.87 in./sec2 T = (1.00)(-0. 60) - (0.00)(0.72) AB/C (1.00)(-0. 97) - (0.00)(-0.25) = 0.62 in./sec2

6.

T T Substitute the values found_for Ac_and AB/C into Equations (15.28) and (15.29) and solve for Ac and AB/C.

A

C = 0.87 in./sec2 /0° 1.342 i(165.5°) i(255.5°) = _-e + 0.62 e 3 = 0.60 /-14.5° +0.62 /-104.5° = 0.86 in./sec2 /-60.4°

Analytical Techniques

272

EXAMPLE 15.2: Offset Slider Crank Considering the offset slider-crank mechanism in Figure 15.2, let AB = 1.5 in., BC = 3 in., e = 0.5 in., 02 = 150°, u2 =1.0 rad/sec (counterclockwise), and az - 0 rad/sec2. It is required to find Vc, VB/c,

Ac, and AB/c« SOLUTION 1. Determine 03 using Equations (15.2) and (15.1).

= sin'‘(j^sini5o°-H) = 4.78° Q3 = 180° - 4.78° = 175.22° 2. Determine VB, Vc, and VB/C using Equations (15.8), (15.9), and (15.10). Vg = 1.5(1.0) - 1.50 in./sec

C

sin (175.22° - 150°) '3 4 5* sin(90° - 175.22°)

„

. . -0.64 m./sec

T7 _ sin(90° - 150°) . VB/C “ (1<5) sin (90° - 175.22°) = ^ WseC 3. Determine VB, Vc, and VB/c using Equations (15.11), (15.12), and (15.13).

Vg - 1.5e^240 ^ = 1.5 in./sec /-120°

Vc = -0.64e

i0°

^B/C ~ l*30e ^

=0.64 in./sec /180°

^ = 1.30 in./sec /-94.78°

4. Determine the constants Alf A2, Blf Bz, Clt and C2 using Equations (15.21) through (15.26). Aj = 1

[from Equation (15.21)]

A2 = 0

[from Equation (15.22)]

Slider-Crank Mechanism Analysis: Simplified Vector Method

= cos (175.2°+ 90°) = -0.08 B2 = sin (175.2° + 90°) = -0.99 Cj = -(12)(1. 5) cos 150° + 0(1.5) cos (150° + 90°) 1.302 + —-—cos (175.2°)

= -(-1.3) + 0 + (-0.56) = 0.73 C2 - -(12)(1.5) sin 150°+ 0(1.5) sin(150°+ 90°) 1302 + —— sin (175.2°) O

= -(0.75) + 0 + 0.084 = -0.70 5.

Determine A T and A T using Equations (15.19) and (15.20). C B/C T _ (0.73)(-0. 99) - (-0.70)(-0.08) AC (1.00)(-0.99) - (0.00)(-0.08) = 0.79 in./sec2 T = (1.00)(-0. 70) - (0.00)(0.73) AB/C (1.00)(-0. 99) - (0.00) (-0. 08) = 0.70 in./sec2

6.

T T Substitute the values found for A^_and A-q/q into Equations (15.28) and (15.29) and solve for Ac and AB//C' i0° Ac = 0.79 e1 = 0.79 in./sec2 / 0° (1.30)2 i(175.2°) AB/C

7Qei(265.2°)

3.0 = 0.57 /-4.78° + 0.70 /-94.8 = 0.90 in./sec2 /-56.0°

273

16 Quick-Return Mechanism Analysis: Simplified Vector Method

16.1

INTRODUCTION

Quick-return mechanisms, typified by ABCD in Figure 16.1, are a special class of sliding contact linkages that provide uniform velocity motion fol¬ lowed by a fast return stroke to an initial position. These mechanisms, often used in equipment such as machine tools and production machinery to pro¬ duce long, slow movements for cutting or feeding and fast return strokes in which no work is done, are generally found in two basic arrangements: !•

The crank-shaper (or oscillating-beam) type (Figure 16.2), where the crank arm AB is shorter than the base AD, and as a result, the follower CD oscillates as the crank arm makes a complete revolution.

2-

Tlie Whitworth type (Figure 16.3), where the crank arm AB is longer than the base AD, and as a result, both crank arm and follower make complete revolutions.

274

Quick-Return Mechanism Analysis: Simplified Vector Method

275

Figure 16.2 (a) Oscillating-beam shaper; (b) oscillating-beam linkage. (From Applied Kinematics by J. Harland Billings, © 1953 by D. Van Nos¬ trand Company, Inc. Reprinted by permission of Wadsworth Publishing Company, Belmont, CA 94002.) Quick-return mechanisms, like other basic mechanisms, can be used as models to simplify the analysis of more complex machines or they can be arranged in combination with other mechanisms to produce specific output motions.

Figure 16.3

Whitworth linkage.

276

Analytical Techniques

The Geneva mechanism shown in Figure 16.4 is a classical application of the quick-return mechanism. Motion characteristics of quick-return mechanisms are usually diffi¬ cult to analyze because of complexity of the combined linkage and slider movements which give rise to the Coriolis acceleration. The simplified mathematical method presented here quickly determines linear velocity and acceleration relationships for a quick-return mechanism for any given drive crank position in its motion cycle.

16.2

SCOPE AND ASSUMPTIONS

The mechanism ABCD in Figui’e 16.1 represents a typical quick-return mechanism in which the drive crank AB (or link 2) and the follower guide CD (or link 3) have known lengths. B and C are coincident points on the slider (link 4) and follower guide, respectively. AD (or link 1) is the dis¬ tance between crank and follower pivot points. 9Z is the angular position of link 2 measured from base link 1 or AD. The crank AB rotates with angular velocity w2 and angular acceleration a2 (both counterclockwise). In this analysis, equations for computing linear velocities and accel¬ erations in terms of the variable angular position are to be determined. All angular displacements, velocities, and accelerations are considered positive for counterclockwise rotation and negative for clockwise rotation.

Quick-Return Mechanism Analysis: Simplified Vector Method B,C

Figure 16.5

16.3

Geometric relationships.

GEOMETRIC RELATIONSHIPS

To determine 03, consider triangle ACD in Figure 16.5. Since AB, AD, and 9Z are known, basic geometric and trigonometric relationships yield 6>3

= 180° - 03

(16.1)

where _! CDZ + AD2 - AB2 2 (CD) (AD)

03 -

COS

03 -

-COS

_x CD2 + AD2 - AB2 2 (CD) (AD)

(92

180°)

(16.2)

(02

180°)

(16.3)

i CD = [AB2 + AD2 - 2(AB)(AD) cos

16.4

]2

(16.4)

VELOCITY ANALYSIS

To determine the relationships for velocities Vg, V^, and Vg/g, let Bcb in Figure 16. 6 represent the velocity polygon of the mechanism ABCD, where Bb represents the magnitude of the linear velocity of point B (Vg). Be represents the magnitude of the linear velocity of point C (Vg). be represents the magnitude of the relative velocity of point B with respect to point C (Vg/g) • Then Vg, Vg, and Vg/g can be determined in terms of the angular positions of the links as follows: ^Bcb = 90° /LbBc = 03 - 02 /Lb = 90° - (03 - 0Z)

278

Analytical Techniques

Figure 16.6

Velocity analysis.

By applying the rule of sines from trigonometry, the velocities can be expressed in the form V

B sin 90°

V,

V

B/C sin(03 - 02)

sin (90° - 03 + 02)

Since V

B

= co?AB £

(16. 5)

then Vc = w2AB sin (90

- 03 - 02) = io2AB cos (03 - 02)

(16.6)

and VB/C = “2AB

Sin(^3

-

(16.7)

02)

Also, since VB, Vc, and VB/C are assumed to be oriented at angles (02 + 90°), (03+ 90°), and (03), respectively, from the real or reference axis, the required vectorial expressions can be written as follows: at, i(02+9O°) V_, = co?ABe 2 ’ B 4

(16.8)

V c = u2AB cos (0 - 02) e i(03+9O°)

VB/C = w2AB sin (0 - 02)e

(16.9)

i03 (16.10)

Quick-Return Mechanism Analysis: Simplified Vector Method

279

ai(d2+90°) i( @3+ 90°) id. are unit vectors, where e , e 3 and e used to define the directions of Vg, Vq, and Vg/Q, respectively. Note that e*^ = cos d + i sin d, where d is the position angle of the unit vector.

16.5

ACCELERATION ANALYSIS

To determine the linear accelerations, we apply the relative motion theory, which states that

AB ~ AC + AB/C

(16.11)

Expanding this equation into its normal and tangential component form, we obtain

A„

B

+ A

B

Ac

Ac

ab/c

L rT , rCor ab/c ab/c

(16.12)

where ■rN 2 . ^ id, A — —co,ABe B -T AT3 i(d2+90°) A„ = ff.ABe 2 B eV2 C id, CD e

an

Ac

T i(d3+90°)

at

Ac rN

Vc -T ^/C

Ace B/C i = --e oO

=

T idAB/Ce

V. rCor = 2V _C i(d3+90°) B/C 2VB/C CD Here it should be observed that an important difference between Equation (16.12) and the corresponding equation for the four-bar linkage is that since points B and C are not rigidly connected, but slide relative to each other, it is necessary to determine the Coriolis acceleration Ar°£, an acceleration component which results from the sliding motion.

280

Analytical Techniques

Note that for the Coriolis acceleration, we consider the linear velocity of the slider relative to the rotating guide and the angular velocity of the guide, namely, to3 or Vc/CD. Also, the Coriolis acceleration has the orien¬ tation of V-g/Q when rotated 90° about its tail in the direction of w3. Also, in Equation (16.12), note that since the sliding path of B on C is a straight line, its radius of curvature is infinite. That is, R = °o. Hence A^ = 0. Thus Equation (16.12) becomes

■wlABe16*2 + a,ABel(02+9° ) =

^ei03+ATei(03+9O°) CD6 AC®

+ 2V

, B/C CD

- aT e‘^ B/C (16.13)

This equation contains only two unknown quantities, the magnitudes A^ and rp C Ab/C* All other quantities (magnitudes and directions) are either known or can_readily_be determined from the problem data. (Note that the directions of Ac and A^,, although not precisely known, are assumed to be positive for convenience. If any of these directions is reversed, the numerical solu¬ tion of the equation will automatically produce a negative sign for the unknown quantity.) To solve for the unknowns, therefore, we equate the real and imaginary parts of the equation and rearrange as follows: Real: T T Ac cos (ft, + 90°) + Ab/c

cos

d3 = -w2AB cos d2 + a2AB cos (02 + 90°) VC

V2

' 2VB/C CD COS (03 + 90°) + CT cos 03 (16.14) Imaginary: T t Ac sin(03 + 90°) + Ab/,c sin 03 = -oj2AB sin Qz + q;2AB sin(02 + 90°)

vc

v2

" 2VB/C CD sin(03+ 90°) + — Sin d3 (16.15) The solution of these simultaneous equations yields

Quick-Return Mechanism Analysis: Simplified Vector Method

at = CiB2 ~ C?B, C axb2 - a2b3 AiC2

B/C

281

(16.16)

A2Cj (16.17)

AjB2 - A2Bj

where Aj = cos (03 + 90°)

(16.18)

A2 = sin (03 + 90°)

(16.19)

Bj = cos 03

(16.20)

B2 = sin 03

(16.21)

Cj = -co|AB

cos

02 + a2AB cos (02 + 90°)

V,

vc

- 2V. B/CCDOOS<93 + 90”) + ciCOS

(16.22)

C2 = -w2AB sin 02 + azAB sin(02 + 90°)

V.

vc

2VB/CCDSto(<,’+90 ) + 55sta *

(16.23)

Given that the values V^, V^^,, A^, and A^^ are found, the gen¬ eral equations for computing the linear accelerations can be summarized as follows:

ab

= -ufABe18’ + a2ABei(^9°">

(16.25)

A^ = -l1^ + Alei<^+90‘> CD C C

ab/c

(16.24)

2(VB/C)(VC) i(03+9O°) , aT „i03 CD 6 AB/CG

(16.26)

EXAMPLE 16.1: Shaper Mechanism Considering the quick-return mechanism in Figure 16.1, let AB — 2 in., AD = 4 in., 02 = 30°, oo2 = 62.83 rad/sec (counterclockwise), and a2 = 0. It is required to find V^, V-q/q>

^B/C*

282

Analytical Techniques

SOLUTION 1. Determine d3 using Equations (16.4), (16.2), and (16.1). i

CD = [0.1672 + 0.3332 - 2(0.167)(0.333) cos 30°]2 = 0.21 ft

3

tfi

cos-1

0.212 + 0.332 - 0.1672 2(0.21) (0.33)

= 23.8° d3 = 180° - 23.8° = 156.2° 2. Determine Vg, Vq, and Vg/£ using Equations (16.5), (16.6), and (16. 7). V„ = (62.83)(0.167) = 10.47 ft/sec Jb> Vc - (62.83)(0.167) cos (156.2° - 30°) = -6.18 ft/sec

VB/C - (62.83)(0.167) sin (156. 2° - 30°) -

3.

8.45 ft/sec

Determine Vg, Vc, and VB/C using Equations (16.8), (16.9), and (16.10).

VB = 10.47 e

i('12001

= 10.47 ft/sec /120°

Vc = -6.18e^246'2 * = 6.18 ft/sec /66.2°

Vg/c =

4.

8.45e^156*2 * = 8.45 ft/sec / 156.2°

Determine constants Alt Az, Blf Bz, Cj, and C2 using Equa¬ tions (16.18) through (16.23). Aj = cos (156.2°+ 90°) = -0.4033 A2 = sin (156.2°+ 90°) = -0.9151

Quick Return Mechanism Analysis: Simplified Vector Method

283

Bj = cos (156.2°) = -0.9151 B2 = sin (156.2°) = 0.4033 Cj = -62.832(0.1666) cos 30°+ 0 - 2(8.45) (,(-6.18) 0.21

cos (156.2°+ 90°)

sin 156.2C

= -942.95 C2 = -62.832(0.1666) sin 30°+ 0

(6 18 6.18 - 7r-J7j \

0.21

(’6'18) 0.21

sin (156.2° + 90°)

sin 156.2°

= -716.92 Determine

C

and Aj using Equations (16.16) and (16.17). B/ C

T = (-942.95)(0.40) - (-716.92)(-0.92) AC “ (-0.40)(0.40) - (-0.92) (-0.92) = 1036.3 ft/sec2 T _ (-Q.40H-716.92) - (-0■ 92)(-942■ 95) AB/C " (-0.40)(0.40) - (-0.92)(-0.92) = 573.8 ft/sec2 6.

Substitute the values found for A^ and A^^ into Equations (16.25) and (16.26) and solve for Ac and A-q/q(-6-18)2 i(156. 2°) + A

C

1036.3e

i(246.2°)

0.207 6 185.2 7-23.8° + 1036.3 7-113.8° = 1052.7 ft/sec2 /-103.7

Figure 16.7

Shaper mechanism characteristics. Example 16.1.

ACCELERATION

284 Analytical Techniques

Quick-Return Mechanism Analysis: Simplified Vector Method

B/C

285

= 2(8.45)(-6.18) i(246.2°) i(156.2°) + 573.8e 0.207 = 505.7 / 66.2° + 573.8 /156.20 = 764.8 ft/sec2 / 114.8°

See Figure 16.7 for velocity and acceleration profiles for complete crank cycle from 0°to 360°.

EXAMPLE 16.2: Whitworth Mechanism (Figure 16.8) Considering the quick-return mechanism in Figure 16.1, let AB = 3 in., AD = 2 in. , 9Z =60°, co2 = 30 rad/sec (counterclockwise), and a2 = 0. It is required to find Vc, VB/C, Ac, and AB/C. SOLUTION 1. Determine 03 using Equations (16.4), (16.2), and (16.1). 1

CD = [0.252 + 0. 1662 - 2(0.25)(0.166) cos 60°]2 = 0.22 ft

COS

03

0.222 + 0.1662 - 0.252 2(0.22)(0.166)

d3 = 180° - 79.1° = 100.9° 2. Determine VB, Vc, and VB/C using Equations (16.5), (16.6), and (16.7). V V

B

c

V

= 30(0.25) = 7.5 ft/sec = 30(0.25) cos (100.9° - 60°) = 5.67 ft/sec = 30(0.25) sin (100.9° - 60°) = 4.9 ft/sec

B/C 3. Determine VB, Vc, and VB/C using Equations (16.8), (16.9), and (16.10). V

=

7-5eil5°°

= 7.5 ft/sec /150°

B v

= 5.67ei190'10 = 5.67 ft/sec /-169.1° C

V

=4 9eil0°’9 B/C

=4.9 ft/sec / 100.9°

Analytical Techniques

286

4. Determine constants Aj, A2, Bx, B2, Cj, and C2 using Equations (16.18) through (16.23). Aj

= cos (100.9°+90°) = -0.98

A2 = sin (100.9° + 90°) = -0.19 Bj = cos 100.9° = -0.19 B2 = sin 100.9° = 0.98 C1 = -302(3) cos (60°) + 0(3) cos (60°+ 90°) " 2(4-9)(^||) cos (100.9°+ 90°) + (^~) cos (100.9°) = 107.9 C2 = -302(3) sin (60°) + 0(3) sin (60°+ 90°) - 2(4‘9)(f^||) sin (100.9°+ 90°) + (p|~ ) sin (100.9°) = -3.96 5.

Determine A? and A^/c using Equations (16.16) and (16.17).

aT = (107.94)(0.98) - (-3.96)(-0.19) C (-0.98)(0. 98) - (-0.19)(-0.19) = -105.3 ft/sec2 aT = (-0.98)(-3.96) - (-0.19)(107. 94) B/C (-0.98)(0.98) - (-0.19)(-0.19) = -24.25 ft/sec2 6.

Substitute the values found for A^ and Ajjand (16.26) and solve for A

^

and A_,,^. B/ C

into Equations (16.25)

287

Figure 16.8

(IPS

)

VEL

(IPS)

Whitworth mechanism characteristics. Example 16.2.

ACC

288

Analytical Techniques

|^iei(100.9»)+(1053)ei(190.9o) u • zz

145.9 /-79.10 + 105.3 / 10.9° 179.9 ft/sec2 /-43.3°

ab/c

2(4.9M5.67)ei(190.9»)+(-24.25)e1(1°0.9-) u • zz 252.5 /-169.1° + 24.25 /-79.1° 253.7 ft/sec2 /-163.6°

See Figure 16.8 for velocity and acceleration profiles for complete crank cycle from 0°to 360°.

17

Sliding Coupler Mechanism Analysis: Simplified Vector Method

17.1

INTRODUCTION

The sliding coupler mechanism, represented by ABC in Figure 17.1, is an important class of the slider crank chain where the connecting rod or coupler BC (link 3) slides through a cylinder or block which is free to rotate, via trunnions, about a fixed axis. The mechanism is generally found in two basic arrangements: an oscillating cylinder (or rocking block) type, where the crank arm AB is shorter than the base AC, and the rotating cylinder (or rotating block type), where the crank arm is longer than the base. In the first arrangement, the cylinder (or block) oscillates as the crank arm rotates, whereas in the second arrangement, the cylinder (or block) rotates with the crank. Typical applications for the sliding coupler include steam engines, some pumps and compressors, hydraulic actuators (such as front-end load¬ ers), variable-speed indexing drives, and various compound mechanisms (see Figures 17.2 to 17.9). Motion characteristics of the sliding coupler mechanism are usually difficult to analyze because of complexity of the coupler and slider motions

289

290

Analytical Techniques

(a)

Figure 17.2 (a) Compressor. Signal)

(Courtesy of Kinney Vacuum, Unit of General

which give rise to the Coriolis acceleration. The simplified method pre¬ sented here quickly determines linear velocity and acceleration relation¬ ships for the sliding coupler mechanism for any angular position of the crank cycle.

17.2

SCOPE AND ASSUMPTIONS

The sliding coupler mechanism ABC in Figure 17.1 consists of a crank AB of fixed length, a coupler BC of variable length, and a slider C which is

Sliding Coupler Mechanism: Simplified Vector Method

Figure 17.2

(Continued)

291

(b) mechanism.

constrained to move within a cylinder pivoted at a fixed point C. Given that the crank rotates with an angular velocity and an angular acceleration (both in a counterclockwise direction), we will now derive general expres¬ sions to compute linear velocities and accelerations of points A, B, and C relative to B for any angular position 0^ of the crank arm. All angular dis¬ placements, velocities, and accelerations are considered positive for counterclockwise rotation and negative for positive rotation.

Figure 17.3

Application: foot pump.

292

Figure 17.5

Analytical Techniques

Application: dump truck.

Sliding Coupler Mechanism: Simplified Vector Method

Figure 17.6

Rack and pinion mechanism.

Figure 17.7

Cam-follower mechanism.

293

Figure 17.9

Figure 17.10

Walking mechanism. (From A. S. Hall, 1961.)

Geometric relationships.

294

Sliding Coupler Mechanism: Simplified Vector Method

17.3

295

GEOMETRIC RELATIONSHIPS

First, we determine the angle 9(2 of the coupler BC in terms of crank angle dA. Considering triangle ABC in Figure 17.10, we note that

BC = [AB2 + AC2 - 2(AB)(AC) cos 9 .f A

9C = 180° + c^c

(17.1)

(17.2)

where it can be shown that

17.4


"

cp

= cos

-cos

BC2 + AC2 - AB2 2(BC)(AC)

_! BC2 + AC2 - AB2 2(BC) (AC)

(9 < 180c A (17.3) (dA> 180°)

VE LO CITY ANA LYSIS

To determine the relationships for velocities Vg, V^-., and V(-yg, let Bcb in Figure 17.11 represent the velocity polygon of the mechanism ABC, where Bb represents the magnitude of Vg (Vg) Be represents the magnitude of V^ (Vq) be represents the magnitude of V^/g (V^/g) Applying the rule of sines from trigonometry, the velocity magnitudes Vg, Vc, and Vc/g can be determined in terms of angular positions of the links as follows:

VB = VC = VC/B sin 90° sin(0A - 6>c) sin (90° - 9A + 9Q)

(17.4)

where sin(&A - 9Q) = sin(0A - c - 180°) = -sin (180° ~ 9A +
(17.5)

Analytical Techniques

296

sin (90° - 0

A

+ 0 ) = sin(90° - 0 L

A

+
L

+ 180°)

- sin (270°- 0A+ 0C) ■sin (90° - 9A + (pQ)

(17.6)

Applying Equations (17.5) and (17.6) to (17.4) yields V. B

V,

sin 90°

-sin(0

V

-0 )

C/B - sin (90° - 0

+ $ )

(17.7)

where

VB = ABt0

Vc =

“

(17.8)

VB sin(0A " ^790^

= -ABcosin(0A-^>c)

(17.9)

VB sin (90° - 0A +
sin 90°

ABw cos ^A "

(17,10^

An inspection of the velocity polygon shows that VB and Vc/B are oriented, respectively at angle (0A + 90°) and (
Figure 17.11

Velocity analysis.

Sliding Coupler Mechanism: Simplified Vector Method

i(0

297

+ 90°)

= ABcoe

(17.11)

-D

i(t> c Vc = -ABu sin(0A - 4>c)e

(17.12)

m c+9o°) VC/B = "ABt° COS (0A " Ve

17.5

(17.13)

ACCELERATION ANALYSIS

To determine the acceleration relationships, we apply the relative motion equation, which states that

AC = AB+AC/B

<17'14>

Expanding this equation into its normal and tangential form, we obtain -N

-T

-Cor

Ac + Ac + Ac

-N -T -N' -T = ab + ab + ac/b + ac/b

where 4-n vc ^c*90^ n A = --e = 0 C °° -T

T i(^C

Ac - Ace rCor

Ac

„

VC/B ‘‘V90’*

= 2Vc^e

A^ = -co2ABe" A _T A

A

£>

i(0A+9O°) = ffABe

N C/B

_T AC/B

VC/B ‘*C BC T AC/B6

i(0C+9°O)

(17-15>

298

Analytical Techniques

As in the case of the quick-return mechanism, since the slider C is constrained to move relative to a rotating guide, the cylinder, it is neces¬ sary to determine Coriolis acceleration A^or resulting from this motion. In this case, for Coriolis acceleration, we consider the linear velocity of the slider or Vc and the angular velocity of the guide (or connecting rod), namely cocg or V^/g/CB. Also, Coriolis acceleration has the orientation of the Vc/g vector when rotated 90° about its tail in the direction of u>cg. Expansion of Equation (17.15) yields

.T e ‘^C

A

2aE ‘V+ aABe ‘ - -VC/B e

= -oo2ABe

i + AC/B6

Vc/B 2VC

BC

6

(17.16)

Equation (17.15) contains two unknown quantities: the magnitudes A^ and rp

L/

Ac/b* All other magnitudes and unit vectors are either known or can be determined from the geometry and operating characteristics of the mech¬ anism. Although the directions for the vectors A^ and A^ are not pre¬ cisely known, they can be assumed for convenience to be positive. If any of of these directions are reversed, the numerical solution of the equation will automatically produce a negative sign, indicating the correct direction. To solve for the unknowns, therefore, we equate the real and imaginary parts of the equation and rearrange as follows: Real: T

Ac cos ^c

T " ac/b cos ^c + 9°0) = _t°2AB cos °A + "AB cos (9a + 90°) V2 , C/B cos BC

V

«C " 2Vc

C/B

BC

cos (d> + 90°) VVC

(17.17)

Imaginary:

Ac sin 'f’c

sin (0

^

+ 90°) = -co2 AB sin 9. + aAB sin(0 + 90°) A A

V2 , V C/B C/B — sin 0C- 2Vc-i^-sin(^c + 90°)

(17.18)

Solution of the simultaneous equations (17.17) and (17.18) is obtained by

Sliding Coupler Mechanism: Simplified Vector Method

299

aT = CtB2 - C2Bt C aT

(17.19)

AjB2 - A2Bj = AtC2 - A2Ct

C/B

(17.20)

AjB2 - A2Bj

where Ai = cos 4>c

(17.21)

A2 - sin 4>q

(17.22)

Bj = -cos (
+ 90°)

(17.23)

B2 = -sin (cpc + 90°)

(17.24)

Cj = -o2AB

6

cos

+ aAB cos (0

A

A

v2C/-R -

vv c/b
COS

C2 = -w2AB sin 0

A

-

v2C/B , BC

+ 90°)

cos (0C + 90°)

+ aAB sin(0

A

(17.25)

+ 90°)

V sin
C/B BC

sin (
+ 90°)

(17.26)

Given that the values Vc> Vc/B> Aj, and a£/b have been determined, the general equations for computing the linear accelerations can be summa¬ rized as follows:

A

B

= -w2ABe

i0A A

2VC

BC

6

VC/B ‘^C ^

Ac/B _ ~ BC

i(0A+9°O) A

1 ,

VC/E

Ac

+ oABe

T l*C

(17.27)

(17.28)

Ce T

V90'>

(17.29)

AC/Be

EXAMPLE 17.1: Oscillating Cylinder Considering the oscillating cylinder mechanism in Figure 17.1, let AB 8 in., AC = 10 in., 0A = 120°,yA = 18 rad/sec (counterclockwise), and a * = 0. It is required to find Vc, Vc/B, Ac, and Ac/B-

O

- CT>

O

Oscillating cylinder characteristics. Example 17.1.

O CO

Figure 17.12

VELOCITY

(V

300 Analytical Techniques

W U 2

<

S u

O rCM

CM

o

Sliding Coupler Mechanism: Simplified Vector Method

301

SOLUTION* 1.

Determine (pc using Equations (17.1) and (17.2).

BC = [0.672 - 0.832 - (0.67)(0.83) cos 120°]^ = 1.30 ft
= C

rn~-i 1»32 + 0.832 - 0.672 2(1.3)(0.83)

= -26.33° 2.

Determine Vg, V^, and V^/g using Equations (17.8), (17.9), and (17.10). VB = 0.67(18) = 12 ft/sec Vc = -0.67(18) sin[120° - (-26.33°)] - -6. 65 ft/sec vc/g = -(0.67)(18) cos [120° - (-26.33°)] = 9.98 ft/sec

3.

Determine Vg, V^, and V^/g using Equations (17.11), (17.12), and (17.13).

V

13

V

= 12.0el(12° +9° ) - 12.0 ft/sec /-150°

= -6.65e^-26'3 ) = 6. 65 ft/sec /153. 6°

Vc/b - 9.98el(“26*3 +9° * = 9.98 ft/sec /63.7°

4.

Determine constants Au Az, Bj, B2, Clf and C2 using Equations (17.21) through (17.26). A1 = cos (-26.3°) = 0.89 A2 = sin(-26.3°) = -0.44 Bj = -cos (-26.3°+90°) = -0.44 B2 = -sin (-26.3°+90°) - -0.89

*See Figure 17.12 for velocity and acceleration profiles for complete crank cycle from 0° to 360°.

Analytical Techniques

302

Cx = -(182)(0.67) cos 120°+ 0(0.67) cos (120°+ 90°) _ (9^9§)_

cos(_26>3o) _

2(-6.

65)(7^) cos (-26.3°+90°)

= 84.61 C2 - -(182)(0. 67) sin 120°+ 0(0.67) sin (120°+ 90°) -

-L • o

sin(-26.3°) - 2(-6.65)(t1|^) sin(-26.3°+ 90°) >*1.3 7

= -61.58 5.

Determine A^ and A^

using Equations (17.19) and (17.20).

.T = (84.61)(-0. 89) - (-61. 58)(0.44) C (0.89)(-0. 89) - (-0.44)(-0.44) = 103.14 ft/sec2 aT = (0.89)(-61.58) - (~0.44)(84.61) C/B (0. 89)(-0.89) - (-0.44)(-0.44) = 17.66 ft/sec2 6.

Substitute the values found for A^ and A^ .

into Equations (17.28)

and (17.29) and solve for Aq and A^/g. A C

= 2.(.-6-65)(9.98)ei(-26.3°+90°) + 103> 1.3

14

ei(-26.3°)

= 102.1 /-116.3° + 103.14 /-26.3° = 145.12 ft/sec2 /-71.04° A

= C/B

(9.98)2 1.3

i(-26.3°)

j(-26.3°+90°) + 17 •

66

e

= 76.6 /153.6° + 17.66 /63.7° = 78.64 ft/sec2 / 140.7° EXAMPLE 17.2: Rotating Cylinder Considering the rotating cylinder version of mechanism in Figure 17.1, let AB = 3 in., AC = 2 in., 0A - 210°, _wA = 30 rad/sec (counterclockwise), and aA = 0. It is required to find Vc> Vc/B> Ac> and Ac/B.

Sliding Coupler Mechanism: Simplified Vector Method

303

SOLUTION 1. Determine
BC = [3.02 - 2.02 - (3.0)(2.0) cos 210°]^ = 4.8 in.


.1 4.82 + 2.02 - 3.02 2(4.8) (2.0)

= 18.06° 2. Determine magnitudes Vg, Vq, and V^/g using Equations (17.8), (17.9), and (17.10). V

= 3.0(30) = 90.0 in./sec JD

Vc = -3.0(30) sin (210° - 18.06°) = 18.6 in./sec Vc/B = -3.0(30) cos (210° - 18.06°) = 88.06 in./sec

3. Determine complete vectors Vr, V/-’, and Vp/-R using Equations (17.11), (17.12), and (17.13). C/B VB = 90el(21° +9° ) = 90 in./sec /-60°

V„ = 18.6el(18'06 } = 18.6 in./sec /18.06° v

Vc/B = 88.06el(18‘°6 +9° ^ = 88.06 in./sec /108.6°

4. Determine constants Alt A2, Bj, B2, C:, and C2 using Equations (17.21) through (17.26). Aj = cos(18.06°) = 0.95 A2 = sin(18.06°) = 0.31 Bx = -cos (18.06°+ 90°) = 0.31 B2 = -sin (18.06°+ 90°) = -0.95

Figure 17.13

Rotating cylinder characteristics. Example 17.2.

VELOCITY

(V

)^

304 Analytical Techniques

Sliding Coupler Mechanism: Simplified Vector Method

305

ci-(30.0)2 (3.0) cos 210°+ 0(3.0) cos (210°+ 90°) (88.06)2

-

4 8

/rr nm

cos (18.06°) - 2(18. 6)

cos (18.06°+90°)

= 1024.3 C2 = -(30.0)2(3.0) sin210° + 0(3.0) sin (210°+ 90°) “

(88.06)2 „ /88.06\ 4>g sin (18.06°) - 2(18.6)(— ■—j sin (18.06°+ 90°)

= 208.64 5.

Determine

and

using Equations (17.19) and (17.20).

T = (1024.3)(-0.95) - (208.6)(0.31) AC (0.95)(-0.95) - (0.31)(0.31) = 1038.5 in. /sec2 T = (0.95)(208.6) - (0.31)(1024.3) AC/B (0.95)(-0. 95) - (0.31)(0.31) = 119.32 in./sec2

.

6

Substitute the values found for A^ and A^^ into Equations (17.28) and (17.29) and solve for Ac and Aq/^. -

_ 2(18.6)(88j_06^ i(l8.06°+90°) + 1038> 5ei(18-06°) C

4.8 = 677.57 /108.07°+ 1038.5 /18.06° = 1240 in./sec2 /51.18°

A AC/B

= .M^+ei<18-°«') + 119.32ei<18-06"+9°”) 4.8 = 1603.15 /-161.9° + 119.32 7108.06° = 1607.58 in./sec2 /166-18°

See Figure 17.3 for velocity and acceleration profiles for complete crank cycle from 0°to 360°.

18 Slider-Crank Mechanism Analysis: Modified Vector Method

18.1 INTRODUCTION In the slider-crank analysis by the simplified vector method (Chapter 15), the motion relationships were obtained completely by applying relative motion principles. There the velocity relationships were derived with the aid of the velocity polygon and the acceleration relationships were derived by expressing the relative acceleration equation in terms of the velocity relationships. In this analysis, the procedure is basically the same except that the acceleration equations are not obtained by relative motion but in¬ stead by differentiation of the velocity relationships.

18.2 SCOPE AND ASSUMPTIONS Let ABC in Figure 18.1 represent a typical slider-crank mechanism where the crank AB rotates counterclockwise at an angular velocity co about joint A, and Qp^ is its instantaneous angular position away from top dead center.
18.3 GEOMETRIC CONSIDERATIONS As in previous analysis, we first seek to determine the angle of the connect¬ ing rod Op^. There is was shown that

°C = 180°+V

306

Slider-Crank Mechanism Analysis: Modified Vector Method

Figure 18.1

307

Slider-crank analysis: mechanism.

where = -sin-1

AB • „ 5c sm "a

or

ec = iso - 0C where • iAB . „ *C ’ sm BC Sln 6A

18.4

VELOCITY ANALYSIS

Let Bcb be the velocity polygon for the mechanism ABC (Figure 18.2). In Section 15.4 it was shown that the scalar velocity equation for a typical slider slider crank may be expressed as V. B in (90° - 0 ) sm

V. sin (0

V -0.)

B/C

(18.1)

sin(9O°-0)

where V. B

magnitude of linear velocity of point B

V.

magnitude of linear velocity of point C

V. B/C

magnitude of linear velocity of point B relative to point C

Then we would obtain

308

Analytical Techniques

Figure 18.2

Slider-crank analysis: velocity polygon.

sin(90° - 0 ) = sin (90° - 180°+ A) C C = -sin (90° -
sin(6>c - eA) = sin (180°-

- 6>A)

= sin (<£c + dA)

Hence Equation (18.1) can be rewritten as V. B ■sin(90 - tf>c)

V, sin(0c+0A)

from which = ABco ID

V

= - V C

V

sin ( + 0 ) ---— B sin (90° -

sin(90° - e.) . = -V -2B/C B sin (90° - $ )

c

V. B/C sin (90 - 9 )

Slider-Crank Mechanism Analysis: Modified Vector Method

309

In vector form (magnitudes and directions considered), these equations can be expressed in polar form as follows: VB - -VB(si„ 0A - i cos 9 ) A Vc - Vc cos 0°

V

B/C

= v _ V B C

where - (sin 9& - i cos 0^) and cos 0° are unit vectors used to orient the vectors Vg and Yq. Finally, we obtain the following:

VB =

-ABco (sin 9

A

- i cos 9.) A

(18.2)

sin ((pQ + 9a)

vc =

-ABco

sin (90° -
sin (c + 9a) = -ABco sm 0, — . /nr\° V. \ - icos 9, A sm (90 -

18.5

(18.3)

(18.4)

ACCELERATION ANALYSIS

With the velocity equations for the mechanism determined, the corresponding acceleration expressions are obtained by differentiating these equations with respect to time, noting that 9j^ = « xt.

Acceleration of B (Ag)

A

B

3 dt

Using Equation (18.2), we have V

A

B

= -AB X co (sin 9. - i cos 9 ) A -fi = -AB x co2(cos 9

+ i sin 9.) - AB Xw(sin 6>A - i cos 9A)

(18.5)

where the first and second terms on the right-hand side represent the normal

310

Analytical Techniques

and tangential accelerations, respectively, of point B, and w (or a) is the angular acceleration of the same point.

Acceleration of C (Ac) dV, AC

dt

Using Equation (18.3), we have sin(0

+ 0 )

Vc = -AB x os sin(90o _ ,

)

and substituting for 0C> it follows that

V v

— -AB X co sin 9, + A

A

AB cos 9 A sin 9. A

(BC2 - AB2 sin2 9 )2

from which AB (cos2 9 A

= -AB

A

-sin2 9.) A'

cos 9. +

X co2

1/2

L/

(BC2 - AB2 sin2 9A)

AB x w sin 9 . + A

AB3 cos2 9. sin2 9,

,3/2 (BC2 - AB2 sin2 fl )

AB cos 9 . sin 9. A A (BC2 - AB2 sin2 9 )

L

A

(18.6)

1/2

Acceleration of B Relative to C (Ab/c)

ab/c

ab "

Ac

Therefore, using Equations (18.5) and (18.6), we have

B/C

= AB

AB(cos2 9 -sin2 9.) _A_A_

X co2

AB3cos2 9, sin2 9 _A_A

1 /9

(BC2-AB2sin2 9.) 3?2 "isin0A

(BC2 - AB2sin2 9 )

_

A

A

AB cos 9. sin 9. + AB

X w

A

A

(BC2 - AB2 sin2 9 )

l/2+iC0S SA

(18.7)

Slider-Crank Mechanism Analysis: Modified Vector Method

311

Note that when there is no angular acceleration (or S> = 0), the second terms on the right-hand side of both Equations (18.6) and (18.7) vanish. EXAMPLE 18.1 Let AB = 1.5 in. BC = 3.0 in. - 1 rad/sec (clockwise) 0A = 30° A Find Vc, VB/C, Ac, and AB/C. SOLUTION Using Equation (18.3), we have sin(6>A + tf>c) V

-AB x co

C

sin (90° -
where co = -co..,,

AB

=

-1

rad/sec

eA)

0c "

= sin-1(0.5 sin 30°) = 14.5°

V

sin (30°+ 14.5°) sin (90° - 14.5°)

c

= 1.09 in. /sec / 0° Using Equation (18.4), we have sin(6»A +
B/C

-AB

X co

= -AB X co

(sin

eA

(sin 30

„

- i cos

eA) - -in(9OO_0c)

. sin 44.5° - i cos 30 ) - gin 75^0

Analytical Techniques

312

= -1. 5(-l)(0.5 - 0.866i - 0.729) = 1. 5(-0.229 - 0.866i) - 1.34 tan-1

777^

in./sec /-104.50

Using Equation (18.6), we have „„0 . 1.5(cos2 30°-sin2 30°) , 1. 53 cos2 30° sin2 30° Ac = -1.5(1) cos 30 +-1^ (32 - 1. 52 sin2 30°) (32 - 1. 52 sin2 30°)

= -1.5(0.866 + 0.258 + 0.0258) = 1.72 in./sec2 /180° Using Equation (18.7), we have

ab/c = 1*5^

1. 5(cos2 30°-sin2 30°) 1.53 cos2 30° sin2 30° 1/2 +-3/2 ' (32 - 1.52 sin2 30°) (32 - 1.52 sin2 30°) 7

= 1.5(0.289 - 0.5i) =

„ 0.86

, 500 , , tan-1 - 77777 in./sec2 /-60.40 Zou

EXAMPLE 18.2 Let AB = 1. 5 in. BC = 3.0 in. =

1

rad/sec (counterclockwise)

e>A = 120° A Find Vc, VB//C, Ac, and Ag/C.

SOLUTION Using Equation (18.3), we have

-AB

where

X co

(^a +

'V

sin (90° -
1

Sin

30

Slider-Crank Mechanism Analysis: Modified Vector Method

313

w - +t0AB ~ 1 rad/sec •

= Sin
iMB . - o VBC sto °A

)

= sin-1 (0.5 sin 120°) = 25.6° V c

= -AB X „ 2111112^+25,61 sin (90° - 25.6°) = -1. 5(1) (t~~) v ;Vn.90 '0.90 / = 0.94 in./sec /180°

Using Equation (18.4), we have sin(0 +

- -AB x

(sin 120° - i cos 120° - —n 145'6 sin 64.4'

-!• 5(1) ( 0.866 + 0.51

0.565^ 0.90 /

= 1. 5(-0.239 - 0. 5i) = 0.830 tan-1

in./sec /-115.6' Zo
Using Equation (18.5), we have 0 1. 5(cos2 120° - sin2 120 °) 1. 53 cos2 120° sin2 120° Ac = -1.5(1) cos 120° + -i--7^ +-jp: (32 - 1.52 sin2 120°) 7 (32 - 1.52 sin2 120°) = -1.5(-0.5 - 0.277 + 0.032) -

1.12 in./sec2 /0°

Using Equation (18.7), we have

Vc = 1-5(1)

~1.5(cos2 120°-sin2 120°) + 1.52 cos2 120° sin2 120°_ . sin 120' i/9 ' 3/2 (32 - 1. 52 sin2 120°) 7 (32 - 1.52 sin2 120°)

= 1. 5(-0.245 - 0. 866i) = 1.35 tan-1 ~~ in./sec2 /-105.8' o

19 Slider-Crank Mechanism Analysis: Calculus Method

19.1 INTRODUCTION An analytical approach, commonly used in analyzing the motion characteris¬ tics of a mechanism consists of: 1.

Writing a mathematical expression to describe the displacement or position of the mechanism

2.

Differentiating the displacement expression with respect to time to obtain the velocity expression

3.

Differentiating the velocity expression with respect to time to obtain the acceleration expression

This method is illustrated with the familiar slider-crank mechanism.

19.2 SCOPE AND ASSUMPTIONS Let the mechanism ABC in Figure 19.1 represent a typical slider-crank where AB in turn represents the crank, BC the connecting rod, and C the slider. Given that the angular displacement of the crank AB at any instant is 9, we will now develop general expressions to compute the linear dis¬ placement, velocity, and acceleration of the slider C in terms of 0.

19.3 DISPLACEMENT, VELOCITY, AND ACCELERATION ANALYSIS Let AB = R

and

BC = L.

314

Slider-Crank Mechanism Analysis: Calculus Method

315

B

Figure 19.1

Slider-crank model.

Then x = R cos 9 + L cos cp

(19.1)

dx „ . . d <9 _ . , dd> — = -R sm 9 — - L sin

(19.2)

L sin

(19.3)

L cos cp ^ = R cos 9^r dt dt

(19.4)

dcp _ R cos 9 d 9 dt L cos (p dt

^g ^

where

and

or

Therefore, by substituting Equations (19.3) and (19.5) in (19.2), we obtain dx dt

R cos 9 dQ d9 -R sin 9 — - R sin 9 L cos

R sin 9 cos 9' L cos (p '

(19.6)

(19.7)

or R sin 29 v = -R ^yfsin 9 + 2L cos cp dt v

(19.8)

316

Analytical Techniques

Let

A = -R^ dt

and

B = sin 0 +

R sin 20 2L cos cp

(19.9)

Then V H d_ _ = b_(A) + a-(B) dt

, sin 0+

-R

R sin 29 \d / 2L cos cp'dt '

d2 9( .

d0\ dt '

R sin 2 9 \ 2L cos cp'

/ V

d0\ d_/ R sin 29 \ sin 9 + dt' dtv 2L cos cp'

d9 d f . „ , R sin 20 \ dt dt^Sm 2L cos cp'

(19.10)

Let u = R sin 2 0

and

v - 2L cos
(19.11)

Then d R sin 20 _ v du - u dv dt 2L cos cp v2 = 2L cos p(2R cos 20) d0/dt - R sin 20(-2L sin cp) dd/dt 4L2 cos2 p (4L cos

0

= (cos

+

d2x dt2

2L cos cp'

\

l^(sl

sin

-R

0

+ 5^12129 ) d0

+ L cos cp

+

4L3 cos3 cp' dt

(19.12)

(19.13)

R sin 2 0 2L cos cp-

„ , R cos 2 0 R 3sin2 26 \ COS 0 + --1 L cos cp 4L3 cos3 cp'

2 '

If L » R, we can set cp = 0. This yields

(19.14)

Slider-Crank Mechanism Analysis: Calculus Method

a

d2x dt2

D d2 e ( .

R

^dt5" VSU1 0+

~

2L

317

\ Sin 20)

^/dfA2 R R3 sin2 2 0 R\^J cos 0 + - cos 20 + 4L3

(19.15)

Note that the first term of this equation becomes zero for constant angular velocity of the crank (i.e., no acceleration). Also, by making use of the relationship

L cos (p = (L2 - R2 sin2 0)‘

(19.16)

we can rewrite the general expressions for velocity and acceleration Equa¬ tions (19.8) and (19.15) as follows: R sin 20

v = -Rw sin 0 +

-Ra

(19.17)

1

2(L2 - R2 sin2

6>)2

R sin 2 0

sin 0 +

i

2(L2 - R2 sin2 0)2 - Rco2

cos 0 +

R cos 2 0 7

(L2 -

■>

7 1/2 R2 sin2 0)

,

4(L2

R3 sin2 2 0 , , 3/2 - R2 sin2 0) (19.18)

Problems

KINEMATIC TERMINOLOGY 1.

Define (a) kinematics, and (b) kinetics.

2.

Distinguish between the terms "mechanism" and "machine."

3.

How does a mechanism differ from a structure?

4.

Name and give examples of the three types of plane motion.

5.

Define (a) displacement, (b) velocity, and (c) acceleration.

6.

Distinguish (a) between the terms "speed" and "velocity," and (b) between the terms "distance" and "displacement."

7-

Explain mechanism inversion. How does it affect (a) relative motion, and (b) absolute motion of the components ?

8.

How does rotation differ from curvilinear motion?

9.

Distinguish between absolute motion and relative motion.

10.

Define radian. Determine the number of radians in 30°, 45°, 150°, and 330°.

11.

What are the three basic modes of transmitting motion? Indicate using sketches, one example in each case.

12.

Distinguish between the terms "reciprocal motion" and "oscillatory motion."

319

Problems

320

13.

Define (a) higher pair, and (b) lower pair. Give an example of each, using sketches.

14.

What is a kinematic chain? Describe the three types.

UNIFORMLY ACCELERATED MOTION 1.

A train traveling at 50 mph speeds up to 70 mph in 1 min and 30 sec. Determine its acceleration and the distance traveled.

2.

A flywheel turning at 200 rpm attains a speed of 300 rpm in 1 min with constant acceleration. Determine the acceleration and number of revolutions taken to attain the higher speed.

3.

An engine crank pin has a linear velocity of 2400 ft/min while rotating at 150 rpm. What is the length of the crank?

4.

Starting from rest, a body A held by a string, wrapped around a 12-in. -diameter pulley, falls 60 ft in 4 sec. Determine for the pulley the following: a. Number of revolutions b. Angular velocity after 4 sec c. Angular acceleration after 4 sec

5.

An automobile accelerates from a speed of 20 mph to 55 mph in a distance of 300 ft. If the acceleration is constant, find the time taken.

6.

The speed of an automobile is 55 mph. If the outside diameter of the tires is 27 in., determine the rpm of the wheels and the angular speed in rad/sec.

7.

A train traveling between two stations 5 miles apart takes 10 min. It uniformly accelerates to a maximum speed at 2 ft/sec2 and uniformly decelerates at 6 ft/sec2. What is the maximum speed of the train and the distance traveled at this speed ? What are the distances covered during the first and last minutes of the train's motion?

8.

Determine the minimum time for a car to travel between two stop signs, 1 mile apart if its acceleration is 3 ft/sec2, its deceleration is 4 ft/sec2, and its maximum speed is 50 mph.

9.

Develop a velocity-time curve to depict the motion of a body between two points, A and D, as follows:

Vectors

321 (1)

passing point A, its velocity is 15 mph.

(2)

During the next 30 sec it accelerates uniformly to 50 mph to reach a point B.

(3) (4)

It then continues at 50 mph for 4 min to another point C. Finally, it comes to rest at D 6 min after passing point A.

Determine also the total distance traveled between points A and D. 10.

A rotating fan, 6 ft in diameter, comes to rest with uniform accelera¬ tion from a speed of 600 rpm. If it turns 15 revolutions while stopping, determine the time it takes to stop.

11.

Two points, A and B, lie on a radial line of a rotating disk 2 in. apart. Determine the radius of rotation of each of these points if the velocity at A is 700 ft/min and at B 800 ft/min.

12.

A 20-in. -diameter wheel turns at 200 rpm. Determine: a. b. c.

The angular velocity in rad/sec The linear velocity of a point on the rim The linear velocity of a point 12 in. from the center

If the wheel speeds up to 300 rpm with uniform acceleration in 2 min, determine: d. e.

The angular acceleration The linear acceleration of a point on the rim

VECTORS 1.

Using data in Figure P. 1, determine graphically the following:

1 B = 1.5

Figure P. 1

322

Problems

a.

A + B

b.

A - B

c.

A + B - C

d.

A + B + C

2. Using the vector polygons given in Figure P.2, complete the following vector equations: a.

V =

A =

b.

B =

D =

c.

E =

H =

Figure P.2 3.

A hiker desiring to go to a point northeast, because of various ob¬ stacles, goes I5 miles due east, then turns left 120°, and goes straight to the point. How far was he originally from the point, and how far did he travel to arrive at the point ?

4.

Graphically determine the resultant of vectors A + B + C in Figure P.3. Then find the effective component of this vector along axes c-c, x-x, and y-y.

Figure P.4 323

324

5.

Problems

Find the effective components of vector A in Figure P.4, along axes x-x, y-y, a-a, and b-b.

6.

7.

The effective component of a velocity vector V along the axis S-S (Figure P.5) is known. Locate this vector and also its effective com¬ ponents along axes T-T and R-R.

Determine: a.

8.

The sum of two vectors: one 10 units due north and the other 20 units northeast

b.

The resultant of 5 (at 90°) - 4 (at 180°)

c.

The value of a vector quantity which when added to 100 units due south gives 100 units northwest

In Figure P. 6, N and M are effective components of a vector R, along axes n-n and m-m. Determine this vector.

Vectors

325

Figure P. 6

9.

Resolve the vector A in Figure P.7 into its components along axes b-b and c-c.

c b

c Figure P.7

326

Problems

10.

Indicate the directions for all vectors shown in Figure P.8, based on the equations a.

R = T- S + V- U

b.

A=B-C+D-E

A

-E

Figure P.8

If-

The velocity of point B on the link BC in Figure P.9 is shown to act 60° with respect to the link centerline. Determine (a) the rotational and tianslational effects of this vector, and (b) the effective component of the same vector along link BD oriented 15° with respect to link BC.

V

Figure P.9

B

=

1.5

in /sec

Graphical Techniques: Velocity Analysis

327

12. From data given on vectors A, B, and C in Figure P. 10, determine the magnitude and sense of all vectors when a. b. c.

C is the resultant. A is the resultant. B is the resultant.

/

/

Figure P.10

13.

An airplane is flying at an airspeed of 300 mph heading N45°. There is a tail wind from the west at 40 mph. What is the ground speed of the plane, and what is its actual flight direction?

14.

In order to cross a stream flowing at 5 mph in a boat that travels at 12 mph, at what angle upstream should the boat be headed in order to reach a point directly opposite ? What is the resultant speed of the boat ?

GRAPHICAL TECHNIQUES: VELOCITY ANALYSIS Effective Components 1.

Using the effective component method, determine the linear and angu¬ lar velocities of point C in Figure P. 11 for any of the following conditions:

Problems

328

a. b. c. d.

9 0 9 0

= = = =

30°, 30°, 60°, 60°,

to to to to

1 1 1 1

rad/sec rad/sec rad/sec rad/sec

(clockwise) (counterclockwise) (clockwise) (counterclockwise)

B

Figure P.11 2.

Repeat Problem 1 but use Figure P. 12.

Figure P. 13

AB

1.5"

BC

3.0”

Graphical Techniques: Velocity Analysis

4.

Repeat Problem 1 but use Figure P.14.

Figure P.14

5.

Repeat Problem 1 but use Figure P. 15.

Figure P.15

330

6.

Problems

Repeat Problem 1 but use Figure P. 16.

AB

7.

=

1.5"

Repeat Problem 1 but use Figure P.17.

3.0" 2.0"

Figure P.17

Graphical Techniques: Velocity Analysis

8.

Repeat Problem 1 but use Figure P. 18.

9.

Repeat Problem 1 but use Figure P.19.

331

AB

Figure P.19

=

1.0"

332

Problems

10.

Repeat Problem 1 but use Figure P.20.

11.

Determine the linear velocities of points C and E in Figure P.21.

Figure P.21

AB

=

3.0"

AD

=

1.25

DF

=

3.0"

FE

=

2.5"

Graphical Techniques: Velocity Analysis

12.

333

Repeat Problem 11 but use Figure P.22.

1.0" 1.25 1.0" 2.0"

Figure P.22 13.

Repeat Problem 11 but use Figure P.23.

E

Figure P.23

AB

=

1.5"

BC

=

3.0"

CF

=

1.5"

FE

=

2.0"

DE

=

1.5"

334

14.

Problems

Repeat Problem 11 but use Figure P.24.

AB

15.

=

1.0"

Repeat Problem 11 but use Figure P.25.

AB

=

1.0

II

It

Figure P.25

Graphical Techniques: Velocity Analysis

16.

335

Repeat Problem 11 but use Figure P.26.

AB

1.5

BC

3.0 1.5 1.5

Figure P.26

17.

Repeat Problem 11 but use Figure P.27.

Figure P.27

AB

=

1.0"

BC

=

2.0"

CD

=

2.0"

AD

=

3.0"

BE

=

1.5"

CE

=

0.75"

Figure P.28

19.

Repeat Problem 11 but use Figure P.29.

1.75

BC

=

1.0"

CD

=

BE

=

2.0"

AD

=

2.0"

O

=

i—1

Figure P.29

AB

Graphical Techniques: Velocity Analysis

20.

337

Repeat Problem 11 but use Figure P.30.

AB

=

0.8"

AD

=

2.0"

DE

=

2.6"

Figure P.30

Instant Center 1.

First, locate all the instant centers; then, using the instant center method, determine the linear and angular velocities of point C in Figure P. 11 for any of the following conditions: a. b. c. d.

0 = 30°, co = 0 = 30°, co = 0 = 60°, co = 9 = 60°, co =

1 1 1 1

rad/sec rad/sec rad/sec rad/sec

(clockwise) (counterclockwise) (clockwise) (counterclockwise)

2.

Repeat Problem 1 but use Figure P.12.

3.

Repeat Problem 1 but use Figure P.13.

4.

Repeat Problem 1 but use Figure P.14.

5.

Repeat Problem 1 but use Figure P.15.

6.

Repeat Problem 1 but use Figure P. 16.

7.

Repeat Problem 1 but use Figure P.17.

.

Repeat Problem 1 but use Figure P.18.

8

Problems

338

9.

Repeat Problem 1 but use Figure P.19.

10.

Repeat Problem 1 but use Figure P.20.

11.

Locate all instant centers in Figure P.21.

12.

Locate all instant centers in Figure P.22.

13.

Locate all instant centers in Figure P.23.

14.

Locate all instant centers in Figure P.24.

15.

Locate all instant centers in Figure P.25.

Relative Velocity 1.

Using the relative velocity method, determine the linear and angular velocities of point C in Figure P. 11 for any of the following conditions: a. b. c. d.

9 = 30°, oo = 1 rad/sec (clockwise) 9

- 30°,

9 = 60°, 9

go

= 1 rad/sec (counterclockwise)

co = 1 rad/sec (clockwise)

- 60°, oj = 1 rad/sec (counterclockwise)

2.

Repeat Problem 1 but use Figure P.12.

3.

Repeat Problem 1 but use Figure P. 13.

4.

Repeat Problem 1 but use Figure P. 14.

5.

Repeat Problem 1 but use Figure P.15.

6.

Repeat Problem 1 but use Figure P.16.

7.

Repeat Problem 1 but use Figure P. 17.

8.

Repeat Problem 1 but use Figure P.18.

9.

Repeat Problem 1 but use Figure P.19.

10.

Repeat Problem 1 but use Figure P.20.

11.

Determine the linear velocities of points C and E in Figure P.21.

12.

Repeat Problem 11 but use Figure P.22.

Graphical Techniques: Acceleration Analysis

13.

Repeat Problem 11 but use Figure P.23.

14.

Repeat Problem 11 but use Figure P.24.

15.

Repeat Problem 11 but use Figure P.25.

16.

Repeat Problem 11 but use Figure P.26.

17.

Repeat Problem 11 but use Figure P.27.

18.

Repeat Problem 11 but use Figure P.28.

19.

Repeat Problem 11 but use Figure P.29.

20.

Repeat Problem 11 but use Figure P.30.

GRAPHICAL TECHNIQUES: ACCELERATION ANALYSIS Effective Components 1.

Determine the linear acceleration of point C in Figure P.ll for any of the following conditions: a. b. c. d.

9 =

45° rad, to = 0.5 rad/sec (clockwise), ol = 1 rad/sec2 (clockwise) 9 = 45° rad, to = 1 rad/sec (counterclockwise), a = 0.5rad/sec2 (clockwise) 9 - 135° rad, to = 0.5 rad/sec (clockwise), a = 1 rad/sec2 (counterclockwise) 9 = 135° rad, to = 1 rad/sec (counterclockwise), a = 0.5rad/sec2 (counterclockwise)

2.

Repeat Problem 1 but use Figure P.12.

3.

Repeat Problem 1 but use Figure P.13.

4.

Repeat Problem 1 but use Figure P.14.

5.

Repeat Problem 1 but use Figure P.15.

6.

Repeat Problem 1 but use Figure P.16.

7.

Repeat Problem 1 but use Figure P.17.

8.

Repeat Problem 1 but use Figure P. 18.

340

Problems

9. Repeat Problem 1 but use Figure P. 19. 10.

Repeat Problem 1 but use Figure P.20.

11.

Determine the linear acceleration of points C and E in Figure P.21.

12.

Repeat Problem 11 but use Figure P.22.

13.

Repeat Problem 11 but use Figure P.23.

14.

Repeat Problem 11 but use Figure P.24.

15.

Repeat Problem 11 but use Figure P.25.

16.

Repeat Problem 11 but use Figure P.26.

17.

Repeat Problem 11 but use Figure P.27.

18.

Repeat Problem 11 but use Figure P.28.

19.

Repeat Problem 11 but use Figure P.29.

20.

Repeat Problem 11 but use Figure P. 30.

Relative Acceleration 1.

Determine the linear acceleration of point C in Figure P. 11 for any of the following conditions: a.

0 = 45° rad, co = 0.5 rad/sec (clockwise),

b.

9 = 45° rad, co = 1 rad/sec (counterclockwise),

a = 1 rad/sec2 (clockwise) a = 0.5rad/sec2 (clockwise)

c.

9 - 135

rad, co = 0.5 rad/sec (clockwise),

a ~ 1 rad/sec2 (counterclockwise)

d.

9 = 135° rad, co = 1 rad/sec (counterclockwise), oi - 0.5rad/sec2 (counterclockwise)

2.

Repeat Problem 1 but use Figure P.12.

3.

Repeat Problem 1 but use Figure P. 13.

4.

Repeat Problem 1 but use Figure P.14.

5.

Repeat Problem 1 but use Figure P.15.

Graphical Techniques: Acceleration Analysis

341

6.

Repeat Problem 1 but use Figure P.16.

7.

Repeat Problem 1 but use Figure P.17.

8.

Repeat Problem 1 but use Figure P. 18.

9.

Repeat Problem 1 but use Figure P. 19.

10.

Repeat Problem 1 but use Figure P.20.

11.

Determine the linear acceleration of points C and E in Figure P.21.

12.

Repeat Problem 11 but use Figure P.22.

13.

Repeat Problem 11 but use Figure P.23.

14.

Repeat Problem 11 but use Figure P.24.

15.

Repeat Problem 11 but use Figure P.25.

16.

Repeat Problem 11 but use Figure P.26.

17.

Repeat Problem 11 but use Figure P.27.

18.

Repeat Problem 11 but use Figure P.28.

19.

Repeat Problem 11 but use Figure P.29.

20.

Repeat Problem 11 but use Figure P.30.

Velocity Difference 1.

Determine the linear acceleration of point C in Figure P. 11 for any of the following conditions: a.

9 = 45° rad, to = 0.5 rad/sec (clockwise),

b.

9 = 45° rad, to = 1 rad/sec (counterclockwise),

a ~ 0 rad/sec

(clockwise)

a = 0 rad/sec

c. d.

2.

(clockwise) 9 = 135° rad, w = 0.5 rad/sec (clockwise), a = 1 rad/sec (counterclockwise) 9 = 135° rad, to = 1 rad/sec (counterclockwise), a = 0.5 rad/sec (counterclockwise)

Repeat Problem 1 but use Figure P.12.

342

Problems

3.

Repeat Problem 1 but use Figure P. 13.

4.

Repeat Problem 1 but use Figure P. 14.

5.

Repeat Problem 1 but use Figure P.15.

6.

Repeat Problem 1 but use Figure P. 16.

7.

Repeat Problem 1 but use Figure P. 17.

8.

Repeat Problem 1 but use Figure P.18.

9.

Repeat Problem 1 but use Figure P.19.

10.

Repeat Problem 1 but use Figure P.20.

GRAPHICAL TECHNIQUES: MISCELLANEOUS 1.

Determine the linear velocity of point C (Figure P.31).

C

in /sec

Figure P.31

Graphical Techniques: Miscellaneous

2.

Determine the linear velocity of point B (Figure P.32).

Figure P.32

3.

Determine the linear velocities of points C and A (Figure P.33).

Figure P.33

343

344

4.

Problems

Determine the linear velocities of points E and F (Figure P.34).

Figure P.34

5.

Determine the linear acceleration of point B (Figure P.35).

Figure P.35

Graphical Techniques: Miscellaneous

6.

345

Determine the following (Figure P.36):

a.

The linear acceleration of point B

b.

The linear acceleration of point C

c.

The linear acceleration of point B relative to point C

i

2

rad./sec.

rad./sec.

Figure P.36

2

346

7.

Problems

Using instant center 24, determine the linear velocity of the pivot 34 (Figure P.37).

13

8.

Determine the linear velocities of points B and C and the linear veloc¬ ity C relative to B (Figure P.38).

Figure P.38

Graphical Techniques: Miscellaneous

9.

347

Determine the linear acceleration of point C (Figure P.39) if the wheel rolls without slipping. /

10.

Determine the linear acceleration of point P on the follower (Figure P.40), using a.

The relative acceleration method

b.

The equivalent linkage method

Figure P.40

348

11.

12.

Problems

Determine the following (Figure P.41): a.

The linear acceleration of point B

b.

The angular acceleration of point C relative to point A

c.

The angular velocity of point C relative to point A

Determine the following (Figure P.42): a.

The linear acceleration of point B

b.

The angular acceleration of point A relative to point C

c.

The angular velocity of point A relative to point C

Figure P.42

Graphical Techniques: Miscellaneous

Figure P.43

Figure P.44

350

13.

14.

Problems

Determine the following (Figure P.43): a.

The linear acceleration of point C

b. c.

The angular acceleration of point A relative to point C The angular velocity of point A relative to point C

Determine the linear acceleration of the cam follower, link 4 of Figure P.44, for the following positions: a. b.

9 = 45° 9 = 75°

15. Determine the linear acceleration of the cam follower, link 4 of Figure P.45, using: a. b.

The relative acceleration method The equivalent linkage method

1 rad /sec

Figure P.45

Graphical Techniques: Miscellaneous

16.

351

Develop the linear velocity versus angular displacement curve for he complete cycle of the Scotch yoke mechanism (Figure P.46) based on a crank angular velocity of 2 rad/sec and a crank dis-’ placement of 30°. From linear velocity versus angular displacement curve in part (a) eve op the linear acceleration-displacement curve, using graph¬ ical differentiation.

Figure P.46

17.

a. Develop the complete linear velocity-time curve for point E on the slider-crank mechanism (Figure P.47), using convenient time intervals, based on a crank angular velocity of 6.28 rad/sec. b. From the linear velocity-time curve in part (a), develop the linear displacement-time curve, using graphical integration.

352

Problems AB

18.

=

1.5"

Develop the linear acceleration versus angular displacement curve for a complete cycle of the drag-link mechanism (Figure P.48), based on a crank angular velocity of 1 rad/sec and a crank angular accel¬ eration of 0.5 rad/sec.

AB

=

2.0”

BC

=

1.5"

CD

=

2.5"

AD

=

1.0"

Figure P.48

19.

Develop the complete linear acceleration-time curve for the sliding coupler mechanism (Figure P.49), based on a crank angular velocity of 30 rad/sec.

Analytical Techniques: Velocity and Acceleration

353

AB

3.0"

AC

2 0"

20. A vehicle starting from rest is observed to have the following speeds at the times given: Time (sec)_1 Velocity (mph)

1.5

2_4_6_8_10 3.0

8.5

16

21.5

25

12 26.5

Draw the velocity diagram and obtain from it the acceleration and displace¬ ment curves.

ANALYTICAL TECHNIQUES: VELOCITY AND ACCELERATION 1.

Using the simplified vector method equations, calculate the accelera¬ tion of point C and the acceleration of point C relative to B in Figure P. 11 for any of the following conditions: a. b. c. d.

9 = 45° rad, to = 1 rad/sec (clockwise), a = 0 rad/sec2 (clockwise) e = 135° rad, to = 1 rad/sec (counterclockwise), a = 0.5 rad/sec2 (clockwise) 9 = 225° rad, to = 1 rad/sec (clockwise), a = 0 rad/sec2 (counterclockwise) 9 = 315° rad, to = 1 rad/sec (counterclockwise), a - 0.5 rad/sec2 (counterclockwise)

354

Problems

Check the results using alternative equations, where possible, or a graphical method. 2.

Repeat Problem 1 but use Figure P.12.

3.

Repeat Problem 1 but use Figure P.13.

4.

Repeat Problem 1 but use Figure P. 14.

5.

Repeat Problem 1 but use Figure P.15.

6.

Repeat Problem 1 but use Figure P. 16.

7.

Repeat Problem 1 but use Figure P.17.

8.

Repeat Problem 1 but use Figure P. 18.

Appendix A

A. 1

INTRODUCTION

The computer programs listed in this section were developed to analyze the linkages covered in Part III, based on the mathematical methods discussed in that section. The programs are written in two common languages: 1. 2.

Fortran language applicable to a Univac 1108 digital computer Calculator Keystroke language applicable to a Hewlett-Packard HP-41C calculator

To apply these programs, typically, the link lengths of the mechanism, the crank angle, the angular velocity, and the angular acceleration must be known. In the slider-crank case, additional information on slider offset (or eccentricity) must also be known. From these data, the required velocities and accelerations can be computed for any specified angular position of the crank in its motion cycle. As illustrations of typical outputs obtained from these programs, printouts of example problem results are given following each program listing. The Fortran programs begin on page 358.

A. 2

CALCULATOR OPERATING PROCEDURE

This procedure is based on use of the Hewlett-Packard HP-41C program¬ mable calculator in conjunction with the Math Pack Module for computation involving complex numbers and a HP peripheral printer for printing out the results. The calculator programs begin on page 388.

355

356

Appendix A

PROCEDURE 1. 2.

Install the Math Pack Module and connect the printer. Turn on the calculator.

3.

Allocate the number of storage registers required for program. Press: XEQ alpha SIZE alpha 080

4.

Set a flag to facilitate use of the Math Pack Module which com¬ putes complex numbers. Press: □ SF04

5.

Prepare to load the program. It may be necessary to remove a previously stored program from program memory to create room for the new program. Press:

6.

□ GTO..

Load the program. Press: PRGM Write the program steps into the program memory following the step-by-step instructions given in the printed program. Upon entering the last instruction, "END," exit the program mode. Press: PRGM

7.

Execute and run the program. Press: XEQ alpha "PROGRAM NAME" alpha If the program name is spelled incorrectly, the display panel will flash "NONEXISTENT." To correct this, repeat the procedure more carefully. Also, if the printer is not on-line, the signal "PRINTER OFF" will be flashed on the display panel. Be sure that the printer is on-line, switched on, and set in the manual

Appendix A 357 position or mode. Check to be sure that the correct program is be mg executed. 6 b 8.

Input the data for variables as requested in the display panel. Press: (numerical keys for the appropriate value of each variable displayed), then Press: R/S (after each variable input)

9.

Note that, depending on the nature of the problem, some variable values are given, whereas others are calculated by the program. Program execution and printing will begin after the data for the last requested variable are entered.

10. At the completion of program execution and printing, turn the calculator and printer to the "OFF" position.

358

Appendix A

FOUR-BAR: SIMPLIFIED VECTOR METHOD LINKAGE‘ANALYSIS ( 1).FOUR-BAR/ANALYSISI 2) C FOUR-BAR LINKAGE ANALYSIS SIMPLIFIED VECTOR METHOD C AB - CRANK C BC » COUPLER C CD * FOLLOWER C AD = FRAME C C * DEGREES/RADIAN C THETAA THETA(A), POSITION ANGLE OF LINK AB C THETAB THETAIB), POSITION ANGLE OF LINK BC C THETAC THE T A(C) , POSITION ANGLE OF LINK CD 10 C THETAX THET A(A) 90 DEGS. 11 C THETAY THETAIB) 90 DEGS. 12 c THETAZ THETAIC) 90 DEGS. 13 c PHID PHI(D),ANGLE BETWEEN CB AND AD 14 c GAMMAB GAMMA IB ) 15 c GAMMAD GAMMA!D) 16 c OMEGAA = OMEGA!A),ANGULAR VELOCITY OF CRANK OMEGAA 17 c ALPHAA = ALPHA!A),ANGULAR ACCELERATION OF CRANK ALPHAA 18 c SMB = THETAfB )-THETA(C) 19 c SMC * THETA(B)-THETA(A) 20 c SMCB * THETA(C)-THETA(A) 2 1 c VB ' VELOCITY OF B 22 c VC » VELOCITY OF C 23 c VCB * VELOCITY OF C RELATIVE TO B 24 c C VB ■ VEL OF B (COMPLEX) 25 c CVC = VEL OF C (COMPLEX) 26 c C VCB = VEL OF C WITH RESPECT TO B (COMPLEX) 27 c ABSVB = ABSOLUTE VELOCITY OF B 28 c AB SVC = ABSOLUTE VELOCITY OF C 29 c ABSVCB " absolute velocity of c with respect to B 30 c ACCNB * NORMAL ACC OF B (COMPLEX) 3 1 c ACCTB = TANGENTIAL ACC OF B (COMPLEX) 32 c ACCNC = NORMAL ACC OF C (COMPLEX) 33 c ACCTC = TANGENTIAL ACC OF C (COMPLEX) 34 c ACCNCB = NORMAL ACC OF C WITH RESPECT TO B(COMPLEX) 35 c ACCTCB - TANGENTIAL ACC OF C WITH RESPECT TO B(COMPLEX) 36 c ATC * TANGENTIAL ACC OF C 37 c ATCB ' TANGENTIAL ACC OF C WITH RESPECT TO B 38 c ACCB = ACC OF B (COMPLEX) 39 c ACCC = ACC OF C (COMPLEX) 40 c ACCCB = ACC OF C WITH RESPECT TO B (COMPLEX) 4 1 c ABSNB - ABSOLUTE NORMAL ACC OF B 42 c ABSTB * ABSOLUTE TANGENTIAL ACC OF B 43 c ABSNC - ABSOLUTE NORMAL ACC OF C 44 c ABSTB * ABSOLUTE TANGENTIAL ACC OF C 45 c ABSNCB - ABSOLUTE NORMAL ACC OF C WITH RESPECT TO B 46 c ABSTCE = ABSOLUTE TANGENTIAL ACC OF C WITH RESPECT TO B 47 c ABSAB - absolute acceleration of B 48 c ABSAC = ABSOLUTE ACCELERATION OF C 49 c ABSACB “ ABSOLUTE ACC OF C RELATIVE TO B 50 c PHZ VB = PHASE ANGLE OF VELOCITY OF B 5 1 c PHZVC = PHASE, ANGLE OF VELOCITY OF C 52 c PHZVCB PHZVCB = PHASE ANGLE OF VELOCITY OF C WITH RESPECT TO B 53 c PHZNB = PHASE ANGLE OF NORMAL ACC OF B 54 c PHZTB PHASE ANGLE OF TANGENTIAL ACC OF B 55 c PHZNC PHASE ANGLE OF NORMAL ACC OF C 56 c PHZ TC PHASE ANGLE OF TANGENTIAL ACC OF C 57 c PHZNCB = PHASE ANGLE OF NORMAL ACC OF C WITH RESPECT TO P 58 c PHZTCB = PHASE ANGLE OF TANGENTIAL ACC OF C WITH RESPECT TO R 59 c PHZAB PHASE ANGLE OF ABSOLUTE ACC OF B ASPECT TO B 60 c PHZAC = PHASE ANGLE OF ABSOLUTE ACC OF C 61 c PHZACB = PHASE ANGLE OF ABSOLUTE ACC OF C WITH RESPECT TO B 62

c

63 64 65

66 67

68

C = 57.: COMPLEX COMPLEX

c 1 100

69 70

71

104

RE AD(5, 1 FORMAT(1 IF(AB.EC WRITE(6. FORMAT( 1

Appendix A 359 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104

WRIT E(6, 105) 105

106

140 141 142

WRITE(6.106)AB,BC,CD,AD,THETAA.OMEGAA FORMAT)/2X,8F12.4)

c THETAA=

thetaa/c

BD = SQRT(AB**2+AD**2-2.0*AB» AD * COS(THE PHID= ASIN)AB/BD*SIN(THETAA)) PHID =PHID *C S2=(BC+BD+CD)/2.0

c

GAMMAB = ACOS((BC**2+BD**2-CD** 2)/(2.0*1 GAMMAD = ACOS((BD**2+CD**2-BC* *2)/(2 0*1

c

GAMMAB = GAMMAD =

GAMMAB*C GAMMAD *C

c THETAA

=

THETAA *C

THETAB THETAC

= =

GAMMAB 180. -

- PHID PHID - GAMMAD

c WRITE(6,901) 901

FORMAT)//12X'BD'6X 'PH I(D)'4X'GAMMA ( B 14X'THETA)A)'4X'THETA(B ) '4X'THETA)C)' ) WRITE(6. 107)BD.PH ID.GAMMAB,GAMMAD.THE'

c SMB = THETAB - THETAC SMC - THETAB - THETAA SMCB = THETAC - THETAA SRB=SMB/C SRC=SMC/C SRCB = SMCB/C SNB = SIN(SRB) SNC = SIN)SRC) SNCB = SIN(SRCB) VB = OMEGAA * AB VC = VB*SNC/SNB VCB = VB*SNCB/SNB

105 106 107 108 109 1 10 1 1 1 1 12 1 13 1 14 1 15 1 16 1 17 1 18 1 19 120 121 122 123 124 125 126 127 128 129 1 30 131 132 133 134 135 136 137 138 139

FORMAT(//12X, 'AB'10X'BC'10X'CD'10X'AD 10X'AD'4X'THETA(A)'4X'OMEGA(A)' 14X'ALPHA( A ) ' )

c 902

903 904

WRIT E(6,902) FORMAT)//1IX,'SMB'9X'SMC'8X'SMCB') WRITE(6,107)SMB,SMC,SMCB WRIT E(6,903) F0RMAT(/10X,'LINEAR VELOCITIES') WRIT E(6,904) FORMAT)/10X'V(B)'8X'V(C)'7X'V(CB)') WRITE(6,107)VB,VC,VCB

c THETAA THETAB THETAC

= = =

(THETAA/C) (THETAB/C) (THETAC/C)

THETAY THETAX THETAZ

= = =

(THETAB (THETAA (THETAC

c + + +

90.0/C) 90.0/C) 90.0/C)

c X = COS(THET AX) Y = SIN(THETAX) CVB = VB * CMPLX(X.Y) X= COS(THE T A Z) Y = SIN(THETAZ) CVC = VC* CMPLX(X.Y) X = COS(THETAY) Y = SIN)THE T AY) CVCB = VCB * CMPLX(X,Y)

c ABSVB = CABS(CVB) PHZVB=C*ATAN2(AIMAG)CVB).REAL(CVB))

c

ABSVC = CABS(CVC) TEST ABSVC

360 Appendix A 143 144 145 146 147 148 149

CN CN

150 151 152 153 154 155 156 157 158 159 160 16 1 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 18 1 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 10

11

2 12 2 13 2 14

IF(ABSVC.GT. PHZVC=0.0 GO TO 85

0.0001)G0

TO

80 PHZVC=C*ATAN2(AIMAG(CVC) 85 CONTINUE

80

REAL(CVCI)

ABSVCB = CABS(CVCB) TEST ABSVCB

C

IF(ABSVCB.GT. PHZVCB=0.0 GO TO 95

0.0001)GO

TO

90

90 PHZVCB=C*ATAN2(AIMAG(CVCB),REAL(CVCB)) 95 CONTINUE WRITE(6.905) 905 F0RMAT(/10X,'REAL'8X'IMAG'9X'ABS'7X'PHASE') WRITE(6,301)CVB,ABSVB,PHZVB WRITE(6,302)CVC,ABSVC,PHZVC WRITE(6,303)CVCB,ABSVCB.PHZVCB D1 = VC**2*C0StTHETAC)/CD D2 03 04 05 06 D7 08 Cl C2 A1 A2 B 1

B2

= -OMEGAA**2*AB*C0S(THETAA) = ALPHAA*AB*COS(THETAX) =-VCB**2*C0S(THETAB)/BC = VC**2*SIN(THETAC)/CD - - OMEGAA * *2 *AB *SIN( TF1ETA A ) - ALPHAA*AB*SIN(THETAX) =-VCB**2+SIN(THETAB)/BC = D1+D2+D3+D4 = 05+D6+D7+D8 = COS(THETAZ) = SIN(THETAZ) = - COS(THET AY) =-SIN(THETAY)

C

C

ATC ATCB X Y

= =

(C1*B2 - C2-B1)/(A1*B2 (A 1+C2 - A 2 * C 1 ) / ( A 1 * B 2

A2*B1) A2*B1)

COS(THE T AA) SIN(THET AA)

ACCNB = -OMEGAA**2*AB*CMPLX(X Y) X=COS(THETAX) ’ ' Y=SIN(THETAX) ACCTB = ALPHAA *AB *CMPLX(X ACCB = ACCNB + ACCT8 X = COS(THET AC) Y - SIN(THETAC) ACCNC = -VC+*2*CMPLX(X X = COS(THET AZ) Y = SIN(THETAZ) ACCTC = ATC*CMPLX(X,Y) ACCC = ACCNC + ACCTC X = COS(THETAB) Y = SIN(THETAB)

C C

Y)

Y)/CD

ACCNCB =-VCB**2*CMPI_X(X X = COS(THETAY) Y = SIN(THE T A Y)

Y )/BC

ACCTCB = ATCB * CMPLXtX ACCCB = ACCNCB + ACCTCB

Y)

0tSSNB = CABS( ACCNB )

C

PHZNB = C*ATAN2(AIMAG(ACC ABSTB=CABS(ACCTB) TEST ABSTB IF(ABSTB.GT.0.OOOI)G0 PHZTB =0.0 GO TO 15

.REALt ACCNB ) )

TO

10 15 contT?nueC‘ATAN2(MMAG(ACC' C

.REAL(ACCTB ) )

ABSNC=CABS(ACCNC) TEST ABSNC IFtABSNC.GT. PHZNC=0.0 GO TO 25

0.0001)G0

TO

20

Appendix A 361 2 15 2 16 217 2 18 2 19

25

ABSTC=CABS(ACCTC) TEST ABSTC

C

IF(ABSTC.GT. PHZTC = 0.O GO TO 35

220 22 1

222 223 224 225 226 227 228 229 2 30 23 1 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 26 1 262 263 264 265 266 267 268 269 270 27 1 272 273 274 275 276 277 278 279 280 28 1 282 283 284 285 286

c0NTINUEC*ATAN2
0.0001)GO

TO

30

30 PHZTC = C*ATAN2(AIMAG(ACCTC),REAL(ACCTC)) 35 CONTINUE AB$NCB=CABS(ACCNCB) TEST ABSNCB

C

IF(ABSNCB. GT. PHZNCB = 0.O GO TO 45 40 PHZNCB = 45 CONTINUE

O.OOOIJGO

TO 40

C*ATAN2(AIMAG)ACCNCB),REAL)ACCNCB))

ABSTCB=CABS(ACCTCB) TEST ABSTCB

C

IF(ABSTCB.GT. O.OOOIJGO TO 50 PHZT CB =0.0 GO TO 55 50 PHZT CB = C*ATAN2(AIMAG(ACCTCB).REAL(ACCTCB)) 55 CONTINUE WRITE(6,906) 906 9 15

F0RMAT(/10X,'NORMAL WRITE(6,915)

ACCELERATIONS'38XTANGENTIAL

ACCELERATIONS')

FORMAT(/10X, 'REAL'8X'I MAG'9X'ABS'7X'PHASE'18X'REAL ' 8X ' I MAG' 19X'ABS'7X'PHASE') WRITE(6.301)ACCNB.ABSNB.PHZNB,ACCTB,ABSTB.PHZTB WRITE(6,302)ACCNC,ABSNC,PHZNC,ACCTC,ABSTC,PHZTC WRI TE(6,303(ACCNCB.ABSNCB,PHZNCB.ACCTCB.ABSTCB,PHZTCB

C WRITE(6,907) F0RMAT(/1OX'ABSOLUTE ACCELERATIONS') ABSAB = CABS(ACCB) PHZAB = C*ATAN2(AIMAG)ACCBI,REAL)ACCB)) ABSAC = CABS(ACCC) TEST ABSAC IF(ABSAC.GT. 0.0001) GO TO 60 PHZAC=0.0 GO TO 65 60 PHZAC = C*ATAN2(AIMAG(ACCC),REAL)ACCC)) 65 CONTINUE ABSACB = CABS(ACCCB) TEST ABSACB I F ( ABSACB . GT . 0.000OG0 TO 70 PHZACB=0.0 GO TO 75 70 PHZACB = C*ATAN2(AIMAG)ACCCB).REAL(ACCCB))

907

C

C

75

CONTINUE

C WRIT E(6,301 JACCB.ABSAB.PHZAB WRITE(6.302)ACCC.ABSAC,PHZAC WRITE(6,303)ACCCB,ABSACB.PHZACB THETAX=THETAX*C THETAY=THETAY*C THETAZ=THETAZ*C 908 107 909

WRIT E(6,908) FORMAT(/6X, 'THETA(X)'4X'THETA(Y)'4X'THETA(Z) ' ) WRITE(6,107)THETAX,THETAY,THETAZ F0RMAT(/2X,8F12.4) WRIT E(6,909) F0RMAT(/12X, 'A 1' 10X'A2'lOX'BI'10X'B2'lOX'CI'10X'C2'9X'ATC

18X'ATCB' ) WRITE(6,107)A 1,A2,B1 ,B2,C1 ,C2,ATC,ATC8 301 FORMAT)/2X,4F12.4,8X'B '.4F12.4) 302 303

FORMAT)/2X,4F12.4.8X'C '.4F12.4) FORMAT)/2X,4F12.4,8X'CB',4F12.4)

999

GO TO STOP END

C 1

Appendix A

362 PROBLEM DATA

AB

BC

CD

AD

THETA(A)

OMEGA(A)

ALPHA(A)

1.5000

3.0000

3.0000

4.0000

30.0000

2.0000

1.0000

BD

PHI(D)

GAMMA(B)

GAMMA(D)

THETA(A )

THETA(B)

THET A(C)

2.8032

15.5188

62 . 1478

62. 1478

30.0000

46.6289

102.3334

SMB

SMC

SMCB

-55.7045

16.6239

72.3334

LINEAR

VELOCITIES

V (B )

V(C)

V (CB )

3.0000

- 1.0392

-3.4601

REAL

IMAG

ABS

PHASE

- 1 .5000

2.598 1

3.0000

120.0000

B

1.0152

. 2220

1.0392

12.3334

C

-2.3761

3.4601

-43.3711

2.5152 NORMAL

CB

ACCELERATIONS

TANGENTIAL

ACCELERATIONS

REAL

IMAG

ABS

PHASE

REAL

IMAG

ABS

PHASE

-5.1962

- 3.0000

6.0000

-150.0000

B

-.7500

1.2990

1.5000

120.0000

.0769

-.3517

. 3600

-77.6666

C

-10.7699

2.3548

11.0244

- 167.6666

-2.7405

-2.9009

3.9907

-133.3711

CB

-2.0064

1 . 8954

2.7601

136.6289

ABSOLUTE

ACCELERAT][ONS

-5.9462

-1.7010

6. 1847

- 164.0362

B

- 10.6930

-2.7065

11.0302

- 165.7964

C

-4.7469

- 1.0055

4.8522

- 168.0402

CB

.THETAfX )

THETA(Y )

THET A(Z )

120.0000

136.6289

192.3334

A1

A2

B1

B2

Cl

C2

ATC

ATCB

- . 9769

-.2136

.7269

-.6867

-8.7635

-4.2502

1 1.0244

2.7601

Appendix A 363 PROBLEM DATA

AB

BC

CD

AD

THETA(A)

OMEGA(A)

ALPHA(A)

2.OOOO

1.5000

2.5000

1.0000

240.0000

1.0000

. 5000

BD

PHI(D)

GAMMA(B)

GAMMA(D)

THETA(A)

THET A(B)

THET A(C)

2.6458

-40.8934

67.7923

33.7446

240.0000

108.6857

187. 1488

SMB -78.4630 LINEAR

SMC

SMCB

-131.3143

-52.8512

VELOCITIES

V( B )

v(c)

V (CB )

2.0000

1.5332

1.6270

REAL

IMAG

ABS

PHASE

1.732 1

- 1.0000

2.0000

- 30.0000

B

. 1908

-1.5213

1.5332

-82.8512

C

-1.5413

-.5213

1.6270

- 161.3143

NORMAL

CB

ACCELERATIONS

TANGENTIAL

ACCELERATIONS

REAL

IMAG

ABS

PHASE

REAL

IMAG

ABS

PHASE

1.0000

1.732 1

2.0000

60.0000

B

. 8660

-.5000

1. 0000

- 30.0000

C

. 1280

1.0202

1. 0282

-82.8512

- 1.3705

-.4635

1. 4468

- 161.3143

. 9329

.1170

. 9403

7.1488

. 5654

-1.6718

1.7648

-71.3143

2.2361

33.4349

ABSOLUTE 1.8660

CB

ACCELERATIONS 1 .232 1

B

1.0609

-.9032

1.3933

-40.4102

- . 8051

-2.1353

2.2820

- 1 10.6594

THET A(X)

THETA(Y )

THET A(Z)

330.0000

198.6857

277.1488

A1

A2

B1

B2

Cl

C2

ATC

ATCB

. 9473

. 3204

1 .4985

- . 5567

1.0282

1 . 4468

. 1244

-.9922

C CB

364

Appendix A

SLIDER-CRANK: SIMPLIFIED VECTOR METHOD LINKAGE' ANAL VS I S < 1 ).SL-CRANK/ANALYSIS-1(2) 1 C SLIOER-CRANK MECHANISM ANALYSIS-SIMPLIFI ED VECTOR METHOD 2 C C DEGREES/RADI AN(CONSTANT) 3 C AB * CRANK 4 C BC ' CONNECTING ROD 5 C PHI = PHI(3),ANGLE BETWEEN CONNECTING ROD AND DEAD CENTER 6 c THETA2 * POSITION ANGLE OF LINK AB 7 c THE T A3 = POSITION ANGLE OF LINK BC 8 c THETAX = THETA(2)+90DEGS. 9 c THETAY * THET A(3) + 90DEGS. 10 c 0MEGA2 = ANGULAR VELOCITY OF LINK AB 1 1 c ALPHA2 = ANGULAR ACC OF LINK AB 12 c SNB : 90 - THETA(DEGS) 13 c SNC ; THETA(2)-THETA( 3) 14 c SNBC ' 90 - THET A(2) 15 c VB : VELOCITY OF B 16 c VC VELOCITY OF C 17 c VBC VELOCITY OF B RELATIVE TO C 18 c C VB VEL OF B (COMPLEX) 19 c CVC VEL OF C (COMPLEX) 20 c C VBC VEL OF B RELATIVE TO C (COMPLEX) 2 1 c ABSVB ABSOLUTE VELOCITY OF B 22 c ABSVC ABSOLUTE VELOCITY OF C 23 c ABSVBC ABSOLUTE VELOCITY OF B RELATIVE TO C 24 c ACCNB NORMAL ACC OF B (COMPLEX) 25 c ACCTB TANGENTIAL ACC OF B (COMPLEX) 26 c ACCNC NORMAL ACC OF C (COMPLEX) 27 c ACCTC TANGENTIAL ACC OF C (COMPLEX) 28 c ACCNBC NORMAL ACC OF B REL TO C (COMPLEX) 29 c ACCTBC TANGENTIAL ACC OF B RELATIVE TO C(COMPLEX) 30 c ATC TANGENTIAL ACC OF C 3 1 c ATBC TANGENTIAL ACC OF B RELATIVE TO C 32 c ACCB ACC OF B (COMPLEX) 33 c ACCC ACC OF C (COMPLEX) 34 c ACCBC rel="nofollow"> ACC OF B RELATIVE TO C (COMPLEX) 35 c ABSNB ABSOLUTE NORMAL ACC OF B 36 c ABSTB = ABSOLUTE TANGENTIAL ACC OF B 37 c ABSNC = ABSOLUTE NORMAL ACC OF C 38 c ABSTC = ABSOLUTE TANGENTIAL ACC OF C 39 c ABSNBC ■ ABSOLUTE NORMAL ACC OF B REL TO C 40 c ABSTBC = ABSOLUTE TANGENTIAL ACC OF B RELATIVE TO C 4 1 c ABSAB = ABSOLUTE ACC OF B 42 c ABSAC = ABSOLUTE ACC OF C 43 c ABSABC = ABSOLUTE ACC OF B RELATIVE TO C 44 c PHZV8 = PHASE ANGLE OF VELOCITY OF B 45 c PHZVC = phase ANGLE OF VELOCITY OF C 46 c PHZVBC = PHASE ANGLE OF VELOCITY OF B RELATIVE T C 47 c PHZNB = PHASE ANGLE OF NORMAL ACC OF B 48 c PHZTB = PHASE ANGLE OF TANGENTIAL ACC OF B 49 c PHZNC = PHASE ANGLE OF NORMAL ACC OF C 50 c PHZTC = PHASE ANGLE OF TANGENTIAL ACC OF C 51 c PHZNBC = PHASE ANGLE OF NORMAL ACC OF B REL TO C 52 c PHZTBC « PHASE ANGLE OF TANGENTIAL ACC OF B RELATIVE TO C 53 c PHZAB = PHASE ANGLE OF ABSOLUTE ACC OF B 54 c PHZAC = PHASE ANGLE OF ABSOLUTE ACC OF C 55 c PHZABC = PHASE ANGLE OF ABSOLUTE ACC OF B RELATIVE TO C 56 COMPLEX CVB.CVC,CVBC.ACCC.ACCNC.ACCTC 57 COMPLEX ACCNB,ACCTB,ACCB,ACCNBC.ACCTBC ACCBC 58 C=57.29578

Appendix A 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 1 10 1 1 1 1 12 1 13 1 14 1 15 1 16 1 17

365 1 99

READ(5,99)AB,BC,THETA2.0MEGA2,ALPHA2 FORMAT(6F12.4) IF(AS.EO.0.0) WRIT E(6,899)

899

GO

TO

E

999

FORMAT(1H1,9XPROBLEM DATA') WRITE(6,900)

900 ^FORMAT)//12X. 'AB'10X'BC'4X'THETA(2)'4X'OMEGA(2)'4X'ALPHA ( 2)'6X'ECC

100

901

902

903 904

905

WRITE(6,100)AB,BC,THETA2,0MEGA2 ALPHA2 FORMAT(/2X,6F12.4) THETA2=THETA2/C PHI=ASIN(AB/BC*SIN(THETA2)-E/BC) PHI=PHI+C

THETA3=180.0-PHI WRITE(6,901) FORMAT)//8X,'PHI(3)'4X'THETA(3)') WRITE!6.101)PHI.THETA3 THETA2=THETA2*C SNB=90.0-THETA3 SNOTHETA3-THETA2 SNBC=90.0-THETA2 WRIT E(6,902) FORMAT)/I IX, ' SNB'9X'SNC'8X'SNBC') WRIT E(6, 101)SNB.SNC,SNBC SMB = SNB/C SMC=SNC/C SMBC = SNBC/C VB = AB *0MEGA2 VC= VB+SIN)SMC)/SIN(SMB) VBC=VB*SIN(SMBC)/SIN(SMB) WRIT E(6,903) F0RMAT(/10X,'LINEAR VELOCITIES') WRITE(6.904) FORMAT)/10X,'V(B)'8X'V(C)'7X'V(BC)') WRITE(6,101)VB,VC,VBC THETAX=(THETA2+90.0)/C THETAY=(THETA3+90.0)/C THETA2=THETA2/C THETA3=THETA3/C X=COS(THETAX) Y=SIN(THETAX) CVB=VB*CMPLX(X,Y) ABSVB=CABS(CVB) PHZVB=C*ATAN2(AIMAG(CVB),REAL(CVB)) WRIT E(6,905) FORMAT)/10X, 'REAL'8X'IMAG'9X'ABS'7X'PHASE' ) WRITE(6,301)CVB,ABSVB,PHZVB. CVC=VC ABSVC=CABS(CVC) TEST FOR ABS VEL OF C IF(ABSVC.GT.O.0001)GO

C

20 25

E

TO

20

PHZVC=0.0 GO TO 25 PHZVC=C*ATAN2(AIMAG(CVC),REAL(CVC)) CONTINUE WRITE(6,302)CVC,ABSVC,PHZVC X = COS(THET AY) Y=SIN(THETAY) CVBC=VBC*CMPLX(X,Y)

Appendix A 1 18 1 19 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 14 1 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 16 1 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176

ABSVBC=CABS(CVBC) TEST VEL OF B REL TO C IF(ABSVBC.GT.O.0001) GO PHZVBC=0.0 GO TO 35

C

TO

30

30 PHZVBC=C*ATAN2(AIMAG(CVBC),REAL(CVBC)) 35 CONTINUE WRITE (6,303 )CVBC , ABS-VBC , PHZVBC A 1 = 1 A2 = 0 B1=COS(THE T A Y) B2=SIN(THETAY) C1=-(OMEGA2**2)*AB*COS(THETA2)+(ALPHA2+AB)*COS(THETAX)+(VBC*»2/BC) 1*COS(THE T A3) C2=-(0MEGA2**2)»AB*SIN(THETA2)+(ALPHA2*AB)*SIN(THETAX)+(VBC**2/BC) 1 * SIN(THE TA3 ) ATC=(C1»B2-C2*B1)/(A1»B2-A2*B1) ATBC=(A1*C2-A2*C1)/(A1*B2-A2*B1) X = COS(THET A2) Y=SIN(THETA2) ACCNB = -(VB**2/AB)*CMPLX(X,Y) ABSNB=CABS(ACCNB) PHZNB=C+ATAN2(AIMAG(ACCNB),REAL(ACCNB)) X = COS(THET AX) Y=SIN(THETAX) ACCTB=AB*ALPHA2*CMPLX(X,Y) ABSTB=CABS(ACCTB) TEST ABS TAN ACC OF B

C

I F(ABSTB.GT.0.OOOI ) PHZTB=0.O GO TO 45

GO

TO

40

40 PHZTB = C*ATAN2(A I MAG(ACCTB),REAL(ACCTB ) ) 45 CONTINUE WRITE(6,906) 906 9 15

C

F0RMAT(/10X.'NORMAL WRITE(6,915)

ACCELERAT IONS'40X. 'TANGENT IAL

FORMAT(/10X, 'REAL'8X'I MAG'9X'ABS'7X'PHASE'20X, 'REAL'8X'I MAG' 19X'ABS'7X'PHASE') WRITE(6,301)ACCNB,ABSNB,PHZNB,ACCTB,ABSTB ACCNC = 0.0 ABSNC=0.0 PHZNC=0.0 ACCTC=ATC ABSTC=CABS(ACCTC) TEST FOR TAN ACC OF C IF(ABSTC.GT.0.0001)G0 PHZTC= 0.0 GO TO 95

TO

PHZTB

90

90 PHZTC = C*ATAN2(AIMAG(ACCTC) ,RE A L(ACCT C ) ) 95 CONTINUE WRITE(6,302)ACCNC,ABSNC.PHZNC,ACCTC X=C0S(THETA3) Y = SIN(THET A3)

C

ACCELERATIONS')

ACCNBC=-(VBC**2/BC)*CMPLX(X,Y) ABSNBC = CAB S(ACCNBC) TEST FOR ABS NOR ACC OF B REL TO IF(ABSNBC.GT.0.0001) PHZNBC =0.0 GO TO 55

GO

TO

50

C

ABSTC

PHZTC

Appendix A 177 178 179 180 18 1 182 183 184 185 186 187 188 189 1 90 19 1 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 2 10 2 1 1 2 12 2 13 2 14 215 2 16 217 2 18 2 19 220 22 1 222 223 224 225 226 227 228 229 230 231 232

367 50 PHZNBC=C*ATAN2(A I MAG(ACCNBC),REAl(ACCNBC)) 55 CONTINUE X = COS(THE T AY) Y=SIN(THETAY) ACCTBC=ATBC*CMPLX(X,Y) ABSTBC=CABS(ACCTBC) TEST TAN ACC B REL TO C

C

IF(ABSTBC.GT.0.0001) PHZTBC=0.0 GO TO 65

GO

TO 60

60 PHZTBC = C*ATAN2(AIMAG(ACCTBC).REAL(ACCTBC)) 65 CONTINUE WRITE(6,303)ACCNBC,ABSNBC,PHZNBC,ACCTBC,ABSTBC,PHZTBC ACCB=ACCNB+ACCTB ABSAB=CABS(ACCB) PHZAB = C*ATAN2(A I MAG(ACCB).REAL!ACCB)) WRITE(6,907) 907

C

C

FORMAT(/10X,'ABSOLUTE ACCELERATIONS') WRITE(6,301)ACCB.ABSAB.PHZAB ACCC=ACCNC+ACCTC ABSAC=CABS(ACCC) TEST FOR ABS ACC OF C IFIABSAC.GT.O.0001)GO TO 10 PHZAC=0.0 GO TO 15 10 PHZAC = C*ATAN2(A I MAG(ACCC),REAL(ACCC)) 15 CONTINUE WRITE(6,302)ACCC,ABSAC,PHZAC ACCBC=ACCNBC+ACCTBC ABSABC=CABS(ACCBC) TEST ABS ACC OF B REL TO C IFIABSABC.GT.0.0001) GO TO 70 PHZABC = 0.0 GO TO 75 PHZABC=C*ATAN2(A I MAG(ACCBC).REAL(ACCBC)) 70 75 CONTINUE WRITE(6,303)ACCBC,ABSABC.PHZABC THETA2=THETA2*C THETA3=THETA3*C THETAX = THET AX *C THETAY=THETAY*C 908

WRITE(6.908) FORMAT( /6X,'THETA(X)'4X'THETA(Y)')

201

WRIT E(6.201 ) THETAX.THETAY FORMAT(/2X,4F12.4,12X,4F12.4 )

909

WRITE(6.909) F0RMAT(/12X,'A1'10X'A2'lOX'BI'10X'B2'lOX'CI'10X'C2'9X'ATC'8X 1 'ATBC' ) WRITE(6. 101 )

A 1 ,A2,B 1 ,B2,C1,C2.ATC,ATBC

101 301 302 303

FORMAT(/2X,8F12.4) F0RMAT(/2X,4F12.4,8X'(B) ' .4F12.4 ) F0RMAT(/2X.4F12.4.8X'(C) '.4F12.4) FORMAT(/2X,4F12.4,8X'(BC)',4F12.4)

999

GO TO STOP END

1

368

Appendix A

PROBLEM

DATA

AB

BC

THET A(2)

OMEGA(2)

ALPHA(2)

ECC(E )

1.5000

3.0000

150.0000

1.0000

.0000

.0000

PHI(3)

THETA(3)

14.4775

165.5225

SMB

SNC

SNBC

15.5225

-60.0000

-75.5225 LINEAR

VELOCITIES

V(B)

V(c)

V (BC )

1.5000

-.4146

1.3416

REAL

IMAG

ABS

PHASE

-.7500

- 1.2990

1.5000

- 120.0000

-.4146

.0000

.4 146

180.0000

-.3354

- 1.2990

1.3416

- 104.4775

NORMAL

(B) (C) (BC)

ACCELERATIONS

TANGENTIAL

ACCELERATIONS

REAL

IMAG

ABS

PHASE

1 . 2990

-.7500

1.5000

- 30.0000

(B)

.0000

.0000

.0000

.0000

.0000

.0000

.0000

.0000

(C)

. 8730

.0000

.8730

-OOOO

. 6000

- 14.4775

(BC)

-.1549

.6000

.6 197

- 104.4775

. 5809

-.1500

REAL

IMAG

ABS

ABSOLUTE ACCELERATIONS 1.2990

-.7500

1.5000

- 30.0000

(B)

. 8730

.0000

. 8730

.0000

(C)

. 4260

-.7500

. 8626

-60.4018

THETA(X)

THETA(V)

240.0000

255.5225

A1

A2

B1

B2

Cl

C2

ATC

ATBC

1.0000

.0000

-.2500

-.9682

.7181

-.6000

. 8730

.6197

(BC)

PHASE

Appendix A

PROBLEM

369

DATA

AB

BC

THET A(2)

OMEGA(2)

ALPHA(2)

ECC(E)

1.5000

3.0000

150.0000

1.0000

.0000

. 5000

PHI(3)

THETA(3)

4.7802

175.2198

SNB

SNC

SNBC

-85.2198

25.2198

-60.0000

LINEAR

VELOCITIES

V (B )

V(C)

V (BC )

1.5000

-.6414

1.3036

REAL

IMAG

ABS

PHASE

-.7500

- 1.2990

1.5000

-120.0000

(B)

-.6414

.0000

.64 14

180.0000

(C)

-.1086

- 1.2990

1.3036

-94.7802

(BC)

NORMAL

ACCELERATIONS

TANGENTIAL

REAL

IMAG

ABS

PHASE

1.2990

-.7500

1.5000

-30.0000

(B)

.0000

.0000

.0000

.0000

(C)

. 5645

-.0472

. 5664

-4.7802

ABSOLUTE

(BC)

ACCELERATIONS

REAL

IMAG

ABS

PHASE

.0000

. 0000

0000

.0000

. 7933

. 0000

7933

.0000

-.0588

-.7028

7053

-94.7802

ACCELERATIONS

1.2990

-.7500

1.5000

-30.0000

(B)

. 7933

.OOOO

. 7933

.0000

(C)

. 9046

-56.0099

(BC)

. 5057

-.7500

THETA(X)

THETA(V)

240.0000

265.2198

A1

A2

B1

B2

Cl

C2

ATC

ATBC

1.0000

.0000

-.0833

- .9965

. 7346

-.7028

.7933

7053

370

Appendix A

PROBLEM

DATA

AB

BC

THET A(2)

OMEGA(2)

ALPHA(2)

ECC(E)

1.5000

3.0000

150.0000

1.0000

.0000

-.5000

' PHI(3 )

THETA(3)

24.6243

155.3757

SNB

SNC

SNBC

-65.3757

5.3757

-60.0000

LINEAR

VELOCITIES

V( B )

V(C)

V (BC )

1.5000

- . 1546

1.4290

REAL

IMAG

ABS

PHASE

-.7500

- 1.2990

1.5000

- 120.0000

(B )

-.1546

.0000

. 1546

180.0000

(c)

- 1.2990

1.4290

-114.6243

-.5954 NORMAL

(BC)

ACCELERATIONS

TANGENTIAL

REAL

IMAG

ABS

PHASE

1 . 2990

-.7500

1.5000

-30.0000

.0000

.0000

.0000

.0000

- . 2836

. 6807

-24.6243

(BC)

.6 188 ABSOLUTE

ACCELERATIONS

REAL

IMAG

ABS

(B)

.0000

. 0000

0000

.0000

(C)

. 8940

. 0000

8940

.0000

-.2138

.4664

5130

-114.6243

ACCELERATIONS

1 . 2990

- . 7500

1.5000

- 30.0000

(B)

. 8940

.0000

. 8940

.0000

(c)

. 4050

-.7500

. 8524

-61.6308

THE TA(X)

THETA(Y )

240.0000

245.3757

(BC)

A 1

A2

B1

B2

Cl

C2

ATC

ATBC

1.0000

.0000

-.4167

-.9091

. 6803

-.4664

.8940

5130

PHASE

Appendix A

371

QUICK-RETURN: SIMPLIFIED VECTOR METHOD LINKAGE *ANALYSI S(

6 7 8 9 10 1 1 12 13 14 15 16 17 18 19 20 2 1 22 23 24 25 26 27 28 29 30 3 1 32 33 34 35 36 37 38 39 40 4 1 42 43 44 45 46 47 48 49 50 51 52 53 54

PHASE ANGLE OF CORIOLIS ACCELERATION PHZCOR = PHASE ANGLE OF ABSOLUTE ACC OF B PHZAB = PHASE ANGLE OF ABSOLUTE ACC OF C PHZAC PHZABC = PHASE ANGLE OF ABSOLUTE ACC OF B RELAT] COMPLEX CVB.CVC,CVBC.ACCNC.ACCTC,ACCC COMPLEX ACCNB,ACCTB.ACCB,ACCCOR.ACCTBC,ACCBC

c

1 99

COMPLEX ACCNBC C = 57.29578 RE AD(5, 99)AB,AD,THETAA,OMEGAA.ALPHAA

o

UJ

U.

FORMAT(5F12.4) 0.0)GO CO <

55 56 57 58 59 60 61 62

c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c

').OUICK-RETURN/ANALYSIS(6) OUICK- RE TURN LINKAGE ANALYSIS - SIMPLIFIED VECTOR METHOD AB.CO.AD * LINK LENGTHS C DEGREE S/RAD IAN(CONST ANT) THETAA THETAI2), POSITION ANGLE OF LINK AB PHID = PHI(3) THETAD = THETA(3), POSITION ANGLE OF LINK CD OMEGAA OME GA(2 ) , ANGULAR VELOCITY OF LINK AB ALPHAA = ALPHA(2 ) . ANGULAR VELOCITY OF AB SNB = 90(DEGS) SNC = 90.0+THETA(D)+THETA(A) = THETAfD)-THETA(A) SNBC VB = VELOCITY OF B VC = VELOCITY OF C VBC = VELOCITY OF B RELATIVE TO C = VEL OF B (COMPLEX) C VB CVC = VEL OF C (COMPLEX) = VEL OF B RELATIVE TO C (COMPLEX) C VBC ABSVB ABSOLUTE VELOCITY OF B = ABSOLUTE VELOCITY OF C ABSVC ABSVBC ABSOLUTE VELOCITY OF B RELATIVE TO C = NORMAL ACC OF B (COMPLEX) ACCNB = TANGENTIAL ACC OF B (COMPLEX) ACCTB = NORMAL ACC OF C (COMPLEX) ACCNC = TANGENTIAL ACC OF C (COMPLEX) ACCTC ACCNBC - NORMAL ACC OF B RELATIVE TO C (COMPLEX ACCTBC = TANGENTIAL ACC OF B RELATIVE TO C(COMP ACCCOR = CORIOLIS ACCELERATION (COMPLEX) = TANGENTIAL ACC OF C ATC = TANGENTIAL ACC OF B RELATIVE TO C A TBC = ACC OF B (COMPLEX) ACCB = ACC OF C (COMPLEX) ACCC - ACC OF B RELATIVE TO C (COMPLEX) ACCBC ABSNB ABSOLUTE NORMAL ACC OF B = ABSOLUTE TANGENTIAL ACC OF B ABSTB = ABSOLUTE NORMAL ACC OF C ABSNC = ABSOLUTE TANGENTIAL ACC OF C ABSTC absnbc = ABSOLUTE NORMAL ACC OF B RELATIVE TO C abstbc = ABSOLUTE TANGENTIAL ACC OF B RELATIVE TO C ABSCOR = ABSOLUTE CORIOLIS ACCELERATION = ABSOLUTE ACC OF B ABSAB = ABSOLUTE ACC OF C ABSAC ABSABC = ABSOLUTE ACC OF B RELATIVE TO C = PHASE ANGLE OF VELOCITY OF B PHZ VB PHASE ANGLE OF VELOCITY OF C PHZVC PHASE ANGLE OF VELOCITY OF B RELATIVE TO C PHZVBC = PHASE ANGLE OF NORMAL ACC OF B PHZNB PHASE ANGLE OF TANGENTIAL ACC OF B PHZTB = PHASE ANGLE OF NORMAL ACC OF C PHZNC = PHASE ANGLE OF TANGENTIAL ACC OF C PHZTC PHASE ANGLE OF NORMAL ACC OF B RE LATIVI TO C PHZNBC TO C phztbc = PHASE ANGLE OF TANGENTIAL ACC OF B REL/

TO

999

TO C

Appendix A 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 1 10 1 1 1 1 12 1 13 1 14 1 15 1 16 1 17 1 18 1 19 120 12 1 122 123 124 125

WRIT E(6,8 99) F0RMAT(1H1,9X,'PROBLEM DATA') WRITE(6.900) 900 FORMAT ( /12X,'AB'10X'AD'4X'THETA(2)'4X'OMEGA(2)'4X'ALPHA(2)') WRITE(6, 100)A B,A D,THETAA,OMEGAA.ALPHAA 100 F0RMAT(/2X,5F12.4) THETAA=THETAA/C CALCULATE CD USING COSINE RULE CD = SQRT()AB**2 + AD**2-2.0*AB*AD*C0S(THETAA ) ) CALCULATE PHID AND THETAD AA = (CD**2 + AD**2 - AB**2)/(2.0*CD*AD) IF(ABS(AA).LE.1.0) GO TO 11 IF(AA.LT.O.O) AA = - 1 .0 IF(AA.GT. 1.0) AA = 1.0 11 PHID = C *ACOS(AA) THE TAA = THETAA »C IF(THETAA.LT.18O.OIG0 TO 111 PHID = -PHID 899

C C

111

901

IF(ABS(PHID) .LT. 0.01)PHID=0.O THETAD = 180.0 - PHID THETAA = THETAA/C WRITE(6,901) FORMAT( /8X,'PHI(3)'4X'THETA(3)') WRITE(6.101)PHID,THETAD THETAD=THETAD/C SNB=90.0 SNC=90.0-THETAD*C+THETAA*C SNBC=THETAD*C-THETAA*C WRITE(6,902)

902

FORMAT(/12X,'CD'9X'SNB'9X'SNC'8X'SNBC'lOX'AA') WRITE(6,888)CD,SNB,SNC.SNBC,AA 888 F0RMAT(/2X,4F12.4.F12.9) SMB=90.0/C SMC=SNC/C SMBC = SNBC/C VB=AB*OMEGAA VC=VB*SIN(SMC) VBC=VB*SIN(SMBC) WRITE(6,903) 903

FORMAT(/10X,'LINEAR WRITE(6,904)

VELOCITIES')

904

FORMAT(/10X,'V(B)'8X'V(C)'7X'V(BC)') WRIT E(6, 101)VB,VC,VBC THETAX=THETAA+90.0/C THETAY = THET AD + 90.0/C X = COS(THE TAX) Y = SIN(THET AX) CVB=VB*CMPLX(X,Y) ABSVB=CABS(CVB) PHZVB=C*ATAN2(AIMAG(CVB),REALlCVB)) WRITE(6,905)

905

C

FORMAT(/I OX, 'REAL'8X'IMAG'9X'ABS'7X'PHASE') WRITE(6,301)CVB,ABSVB,PHZVB X = COS(THET AY) Y = SIN(THETAY) CVC=VC*CMPLX(X,Y) ABSVC=CABS(CVC) TEST VEL OF C IF (ABSVC.GT.O.0001)G0 PHZVC = 0.0 GO TO 25

TO

20

20 PHZVC = C*ATAN2(AI MAG(CVC),REAL(CVC)) 25 CONTINUE

Appendix A 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 16 1 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 18 1 182 183 184 185 186 187

373 WR I TE ( 6,302)CVC,ABSVC,PHZVC X = COS(THET AD) Y = S I N(THETAD ) CVBC=VBC*CMPLX(X,Y) ABSVBC=CABS(CVBC) TEST ABS VEL OF B WRT

c

IF(ABSVBC.GT.0.0001)G0 PHZVBC=0.0 GO TO 35

TO

30

30 PHZVBC=C*ATAN2(A I MAG(CVBC),REAL(CVBC)) 35 CONTINUE WRITE(6,303)CVBC,ABSVBC,PHZVBC CALCULATE ATC AND ATBC A 1=COS(THE T AY) A2 = SIN(THET AY) B1=C0S(THETA0) B2-SIN(THETAD) C1 = -VB **2/AB*C0S(THETAA )+AB*ALPHAA*COS(THETAX) 1TAY)+VC«*2/C0*C0S(THETAD) C2 = -VB**2/AB * SIN(THETAA)+AB*ALPHAA*SIN(THETAX) - : 1TAY)+VC**2/CD*SIN(THETAD) ATC=(C1*B2-C2*B1)/(A1*B2-A2*B1) ATBC=(A1*C2-A2«C1)/(A 1*B2-A2*B1) CALCULATE ACCELERATION COMPONENTS X=COS(THETAA)

c

c

c

Y = SIN( THE.TAA ) ACCNB=-VB**2/AB*CMPLX(X,Y) ABSNB=CABS(ACCNB) PHZNB=C*ATAN2(A I MAG(ACCNB),REAL!ACCNB)) X=COS(THETAX) Y = SIN(THE T AX) ACCTB=AB*ALPHAA *CMPLX(X,Y) ABSTB=CABS(ACCTB) TEST ABS TAN ACC OF B IF(ABSTB.GT.0.0001)G0 TO 40 PHZTB = 0.0 GO TO 45 40 PHZTB-C*ATAN2(AIMAG(ACCTB),RE AL(ACCTB)) 45 CONTINUE WRITE(6,906) 906 FORMAT(/lOX,'NORMAL AND CORIOLIS ACCELERATIONS' 2C E LERAT IONS' ) WRITE(6,915) 9 15 FORMAT(/10X, 'REAL'8X'I MAG'9X'ABS'7X'PHASE' 19X, ' 2'7X'PHASE') WRITE(6,301)ACCNB,ABSNB,PHZNB.ACCTB.ABSTB.PHZTB X=COS(THETAD) Y=SIN(THETAD) ACCNC=-VC**2/CD*

CMPLX(X.Y)

ABSNC=CABS(ACCNC) TEST ABS NOR ACC OF C IF(ABSNC.GT.0.0001)G0

c

TO

440

PHZNC = 0.0 GO TO 445 PHZNC=C*ATAN2(AIMAG(ACCNC).REAL(ACCNC))

440 445 CONTINUE X=COS(THETAY) Y=SIN(THETAY) ACCTC=ATC»CMPLX(X,Y) ABSTC=CABS(ACCTC)

c

TEST ABS TAN ACC OF C IF(ABSTC.GT.0.0001)G0

188

1 89 190 191 192 193 194 195

C

50 55

TO

50

PHZTC = 0.0 GO TO 55 i v PHZTC=C*ATAN2(AIMAG(ACCTC),REAL(ACCTC)) WRITE(6,302)ACCNC.ABSNC,PHZNC,ACCTC,ABSTC,PHZTC ACCNBC ABSNBC PHZNBC

= 0.0 = 0.0 = 0.0

'TANGENTIAL

AC

Appendix A 196 197 198 1 99 200 201 202 203 204 205 206 207 208 209 210 21 1 2 12 213 2 14 215 2 16 217 2 18 2 19 220 22 1 222 223 224 225 226 227 228 229 230 23 1 232 233 234 235 236 237 238 239 240 24 1 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 26 1 262

C

X=C0S(THETAD) Y=SIN(THETAD) ACCTBC=ATBC*CMPLX(X,Y) ABSTBC=CABS(ACCTBC) TEST TAN ACC B WRT C IF(ABSTBC.GT.0.0001)G0 TO 70 PHZTBC=0.0 GO TO 75 70 PHZTBC=C*ATAN2(AIMAG!ACCTBC),REAL(ACCTBC)) 75 CONTINUE WRITE(6,303)ACCNBC.ABSNBC,PHZNBC.ACCTBC.ABSTBC,PHZTBC X = COS(THET AY) Y=SIN(THETAY) ACCC0R=(2*VBC*VC/CD)*CMPLX(X,Y) ABSCOR-CABS(ACCCOR) TEST ABS CORIOLIS ACC IF(ABSCOR.GT.0.0001)G0 TO 60 PHZCOR =0.O GO TO 65

C

60 PHZC0R=C*ATAN2(AI MAG(ACCCOR).REAL(ACCCOR) ) 65 CONTINUE WRITE(6,304)ACCCOR,ABSCOR.PHZCOR ACCB=ACCNB+ACCTB ABSAB=CABS(ACCB) PHZAB=C*ATAN2(AIMAG(ACCB),REAL(ACCB)) WRITE(6,907) 907

F0RMAT(/10X,'ABSOLUTE WRITE(6.905)

ACCELERATIONS')

WRITE(6,301)ACCB,ABSAB,PHZAB ACCC=ACCNC+ACCTC ABSAC=CABS(ACCC) TEST ABS ACC OF C I F ( ABSAC.GT.O.0001)GO TO 770 PHZAC = 0.0 GO TO 775

C

770 PHZAC=C*ATAN2(AIMAG(ACCC).REAL)ACCC)) 775 CONTINUE WRITE!6,302)ACCC,ABSAC,PHZAC ACCBC=ACCCOR+ACCTBC+ACCNBC ABSABC=CABS(ACCBC) TEST ABS ACC B WRT C

C

IF(ABSABC.GT.O.0001)G0 PHZABC = 0.0 GO TO 85

TO

80

80 PHZABC = C*ATAN2(AI MAG(ACCBC),REAL!ACCBC) ) 85 CONTINUE

908 201

WRITE(6,303)ACCBC,ABSABC,PHZABC THETAA=THETAA*C THETAD=THETAD*C THETAX=THETAX*C THETAY=THETAY*C WRITE(6,908) FORMAT! /6X,'THETA(X)'4X'THETA(Y)') WRITE(6,201)THETAX,THETAY FORMAT (/2X.4F12.4,12X.4F12 4) WRIT E(6,909)

909^FORMAT!/12X, 'A1' 10X'A2'10X'B1' 10X'B 2'10X'Cl'10X'C2'9X'ATC'8X'ATBC'

101

WRITE(6,101)A 1,A2,B1,B2,C1,C2,ATC,ATBC F0RMAT(/2X,8F12.4)

301 302 303

FORMAT(/2X,4F12.4,8X'(B) '.4F12.4) F0RMAT(/2X.4F12.4,8X'(C) '.4F12.4) FORMAT(/2X,4F12.4,8X'(BC)'.4F12.4)

304

FORMAT!/2X,4F1-2.4.8X'(C0R)',4F12 GO TO 1 STOP END

999

4)

Appendix A

PROBLEM

DATA

AB

AD

THET A(2)

OMEGA(2)

ALPHA(2)

. 1666

. 3333

30.0000

62.8300

.0000

PHI(3)

THET A(3)

23.7828

156.2172

CD

SNB

SNC

SNBC

AA

. 2066

90.0000

-36.2172

126.2172

. 9 15080868

LINEAR

VELOCITIES

V(B)

V(C)

V (BC )

10.4675

-6.1847

8.4450

REAL

IMAG

ABS

PHASE

-5.2337

9.0651

10.4675

120.0000

(B)

2.494 1

5.6595

6. 1847

66.2172

(c)

-7.7278

3.4056

8.4450

156.2172

NORMAL

AND CORIOLIS

(BC)

ACCELERATIONS

TANGENTIAL

ACCELERATIONS

REAL

IMAG

ABS

PHASE

-569.5603

-328.8358

657.6716

-150.0000

(B)

.0000

.0000

.0000

.0000

169.4519

-74.6764

185. 1770

-23.7828

(C)

-417.9108

-948.3019

1036.3040

- 1 13.7828

.0000

.0000

.0000

.0000

(BC)

-525.0377

231.3808

573.7610

156.2172

203.9362

462.7617

505.7058

66.2172

ABSOLUTE

REAL

ABS

(COR )

ACCELERATIONS

REAL

IMAG

ABS

PHASE

-569.5603

-328.8358

657.6716

-150.0000

(B)

-248.4589

1022.9783

1052.7186

- 103.6515

(C)

-321.1015

694. 1425

764.8137

1 14.8247

THE T A(X )

THET A(Y)

120.0000

246.2172

A1

A2

-.4033

IMAG

- .9151

(BC)

B1

B2

Cl

C2

ATC

ATBC

-.9151

. 4033

-942.9485

-716.9210

1036.3040

573.7610

PHASE

376

Appendix A

PROBLEM

DATA

AB

AD

THETA(2 )

OMEGA(2 )

ALPHA(2)

. 2500

. 1666

60.0000

30.0000

.0000

PHI(3)

THETA(3)

79.1236

100.8764

CD

SNB

SNC

SNBC

AA

. 2205

90.0000

49. 1236

40.8764

.188690612

LINEAR

VELOCITIES

V (B )

V(C)

V (BC)

7.5000

5.6709

4.9082

REAL

IMAG

ABS

PHASE

-6.4952

3.7500

7.5000

150.OOOO

(B)

-5.5691

- 1.0700

5.6709

- 169. 1236

(C)

- .9261

4.8201

4.9082

100.8764

NORMAL

AND CORIOLIS

(BC)

ACCELERATIONS

TANGENTIAL ACCELERATIONS

REAL

IMAG

ABS

PHASE

-112.5000

- 194.8557

225.0000

-120.0000

(B)

.0000

.0000

.0000

.0000

27.5242

-143.2493

145.8696

-79.1236

(C)

103.3647

19.8607

105.2554

10.8764

4.5773

-23.8224

24.2581

-79.1236

.0000

.0000

.0000

.0000

-247.9662

-47.6447

252.5020

- 169.1236

REAL

(BC)

IMAG

ABS

(COR)

ABSOLUTE ACCELERATIONS REAL

IMAG

ABS

PHASE

-112.5000

-194.8557

225.0000

- 120.0000

130.8889

- 123.3886

179.8795

-43.3105

-243.3889

-7 1.467 1

253.6646

- 163.6360

THETA(X)

THETA(Y)

150.0000

190.8764

A1

A2

B1

B2

- .9820

-.1887

- . 1887

. 9820

(B) (C) (BC)

107

Cl

C2

ATC

ATBC

9420

-3.9617

- 105.2554

-24.2581

PHASE

Appendix A

377

SLIDING COUPLER: SIMPLIFIED VECTOR METHOD LINKAGE * ANA IVSIS( 1 2

3 4 5 6

7 8 9 10 1 1 12

13 14 15 16 17 18 19 20 2 1 22

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 4 1 42 43 44 45 46 47 48 49 50 5 1 52 53 54 55 56

C C C C

) .SLG-CPLR/ANALYSIS(3) SLIDING-COUPLER MECHANISM - SIMPLIFIED VECTOR METHOD AB.BC.AC = LINK LENGTHS c = DEGREES/RADIAN!CONSTANT) PHIC PH 1(C) THETAA = THETA(A), POSITION ANGLE OF LINK AB ALPHAA - ALPHA!A).ANGULAR ACCELERATION OF CRANK OMEGAA = OMEGA!A).ANGULAR VELOCITY OF CRANK VB VELOCITY OF B = VELOCITY OF C VC VCB VELOCITY OF C RELATIVE TO B C VB = VEL OF B (COMPLEX) CVC VEL OF C (COMPLEX) = VEL OF C RELATIVE TO BE (COMPLEX) C VCB = ABSOLUTE VELOCITY OF B ABSVB = ABSOLUTE VELOCITY OF C ABXVC ABSVCB - ABSOLUTE VELOCITY OF C RELATIVE TO B = NORMAL ACC OF B (COMPLEX) ACCNB = TANGENTIAL ACC OF B (COMPLEX) ACCTB = NORMAL ACC OF C (COMPLEX) ACCNC = TANGENTIAL ACC OF C (COMPLEX) ACCTC ACCNCB NORMAL ACC OF C RELATIVE TO B(COMPLEX) ACCTCB = TANGENTIAL ACC OF C RELATIVE TO B(COMPLEX) = TANGENTIAL ACC OF C ATC = TAGENT IAL ACC OF C RELATIVE TO B ATCB = ACC OF B (COMPLEX) ACCB = ACC OF C (COMPLEX) ACCC = ACC OF C RELATIVE TO B (COMPLEX) ACCCB = ABSOLUTE NORMAL ACC OF B ABSNB = ABSOLUTE TANGENTIAL ACC OF B ABSTB = ABSOLUTE NORMAL ACC OF C ABSNC = ABSOLUTE TANGENTIAL ACC OF C ABSTB ABSOLUTE NORMAL ACC OF C RELATIVE TO B ABSNCB ABSTCB = ABSOLUTE TANGENTIAL ACC OF C RELATIVE TO B * ABSOLUTE ACCELERATION OF B ABSAB = ABSOLUTE ACCELERATION OF C ABSAC ABSACB = ABSOLUTE ACC OF C RELATIVE TO B = PHASE ANGLE OF VELOCITY OF B PHZVB = PHASE ANGLE OF VELOCITY OF C PHZVC PHASE ANGLE OF VELOCITY OF C RELATIVE TO B PHZVCB = PHASE ANGLE OF NORMAL ACC OF B PHZNB PHASE ANGLE OF TANGENTIAL ACC OF B PHZTB = PHASE ANGLE OF NORMAL ACC OF C PHZNC s PHASE ANGLE OF TANGENTIAL ACC OF C PHZTC PHZNCB = PHASE ANGLE OF NORMAL ACC OF C RELATIVE TO B PHZTCB = PHASE ANGLE OF TANGENTIAL ACC OF C RELATIVE PHASE ANGLE OF ABSOLUTE ACC OF B PHZAB = PHASE ANGLE OF ABSOLUTE ACC OF C PHZAC PHZACB = PHASE ANGLE OF ABSOLUTE ACC OF C RELATIVE TO

c c c c c c c c c c c c *c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c

C * 57. 29578 COMPLEX CVB.CVC,CVCB COMPLEX ACCNB,ACCTB,ACCB,ACCNC.ACCTC.ACCCOR.ACCC COMPLEX ACCNCB.ACCTCB,ACCCB 1 RE AD(5, 100)AB,AC,THETAA.OMEGAA.ALPHAA 100 FORMAT!7F10.4) IF(AB.EO .0.0) GO TO 999

57

58

1

104

WRITEI6.104) FORMAT(1H1,9XPROBLEM DATA')

Appendix A WRIT E(6, 105) 105 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 8 1 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 1 10 1 1 1 1 12 1 13 1 14 1 15 1 16 1 17

106

11

FORMAT(//12X, 'AB'10X'AC'4X'THETA(A)'4X'OMEGA(A)'4X'ALPHA (A)' ) WRITE(6, 106 JAB,AC.THETAA,OMEGAA,ALPHAA F0RMAT(/2X,8F12.4) THETAA=THETAA/C BC=SQRT(AB**2+AC**2-2*AB*AOCOS(THETAA)) AA = (BC**2 + AC**2 - AB»*2)/(2.0*BC*AC) IF(ABS(AA).LE.1.0) GO TO 11 IF(AA.LT.O.O) AA = - 1.0 I F(AA.GT. 1.0) AA= 1.0 PHI = C*ACOS(AA) PHIC = -PHI TyETAA=THETAA*C IF(THETAA.LT.180.O)G0 PHIC = PHI

111 901

TO

111

IF(ABS(PHIC) .LT. O.O1)PHIC=0.0 WRIT E(6,901) F0RMAT(/1IX,'PHI'6X'PHI(C)') WRITEI6, 101)PH I .PHIC SNB = 90.0 SNC =(THETAA-PHIC) SNCB=(90.0-THETAA + PHIC ) WRITE(6.902)

902 888

FORMAT(/12X,'BC'9X'SNB'9X'SNC'8X'SNCB'lOX'AA') WRITE(6.888)BC,SNB,SNC.SNCB,AA F0RMAT(/2X,4F12.4,F12.9) VB = AB *OMEGAA VC = -VB * SIN(SNC/C)/SIN(SNB/C) VCB = -VB*SIN(SNCB/C)/SIN(SNB/C) X=C0S((THETAA+90.0)/C) V=SIN((THETAA+90.0)/C) CVB=VB*CMPLX(X,Y) ABSVB = CABS(CVB) PHZVB=C*ATAN2(AIMAGfCVB),RE A L(CVB)) X=C0S(PHIC/C) Y = SIN(PH I C/C) CVC=VC*CMPLX(X,Y) ABSVC = CABS(CVC) TEST ABSVC

C

IF(ABSVC.GT. PHZVC =0.0 GO TO 85

0.OOO1)GO

TO

80

80 PHZVC=C*ATAN2(AIMAG(CVC),REAL!CVC)) 85 CONTINUE X=COS((PHIC+90.0)/C) Y = SIN((PHIC + 90.0)/C) CVCB=VCB*CMPLX(X,Y ) ABSVCB = CABS(CVCB) TEST ABSVCB

C

IF(ABSVCB.GT. PHZVCB = 0.O GO TO 95

0.0001)G0

TO

90

90 PHZVCB=C*ATAN2(AIMAG(CVCB).REAL(CVCB)) 95 CONTINUE WRITE(6,903) 903

F0RMAT(/10X,'LINEAR WRIT E(6,904)

VELOCITIES')

904

FORMAT(/10X'V(B)'8X'V(C)'7X'V(CB)') WRITE(6,107)VB.VC.VCB

Appendix A 1 18 1 19 120 121 122 123 124 125 126 127 128 129 130 131 1 32 133 1 34 135 136 1 37 138 139 1 40 14 1 142 143 144 145 146 147 148 149 150 151 152 153 1 54 155 156 157 158 159 160 16 1 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176

379 WRIT E(6,905)

905

C

C

FORMAT(/10X, 'REAL'8X'I MAG'9X'ABS'7X'PHASE' ) WRITE(6,301)CVB,ABSVB,PHZVB WRITE(6,302)CVC,ABSVC,PHZVC WRITE(6,303)CVCB,ABSVCB,PHZVCB A1=C0S(PHIC/C) A2=SIN(PHIC/C) B1 = - COS((PH IC + 90.0)/C) B2 = -SIN((PHIC+90.0)/C)

C1=-0MEGAA**2*AB*C0S(THETAA/C)+ALPHAA*AB+C0S((THETAA+90.0)/C) 1-VCB**2/BC*C0S(PHIC/C)-2.0»VC*VCB/BC*COS((PHIC + 90- 0)/C) C2=-0MEGAA**2*AB*SIN(THETAA/C) + ALPHAA *AB*SIN((THETAA + 90.0)/C) 1 -VCB**2/BC*SIN(PHIC/C)-2.0*VC *VCB/BC»SIN((PHIC + 90.0)/C) ATC=(C1*B2-C2*B1)/(A1*B2-A2+B1) ATCB=(A1*C2-A2*C1)/(A1*B2-A2*B1) X=COS(THETAA/C) Y=SIN(THETAA/C) ACCNB=-0MEGAA**2*AB*CMPLX(X,Y) ABSNB=CABS(ACCNB) PHZNB = C*ATAN2(AIMAG(ACCNB),REAL(ACCNB)) X=COS((THETAA+90.0)/C) Y=SIN((THETAA+90.0)/C) ACCTB=ALPHAA*AB*CMPLX(X,Y) ABSTB=CABS(ACCTB) TEST ABSTB IF(ABSTB.GT.0.0001)G0 TO 10 PHZTB = 0.0 GO TO 15 10 PHZTB = C*ATAN2(AIMAG(ACCTB ).REAL(ACCTB)) 15 CONTINUE ACCNC = 0.0 PHZNC=0.0 ABSNC=0.0 X = COS(PH I C/C) Y = SIN(PHIC/C) ACCTC=ATC*CMPLX(X,Y) ABSTC=CABS(ACCTC) TEST ABSTC IF(ABSTC.GT. 0.0001)G0 TO 30 PHZTC=0.0 GO TO 35 30 PHZTC = C*ATAN2(AIMAGlACCTC).REAL!ACCTC)) 35

CONTINUE X=COS((PHIC+90.01/C) Y=SIN((PHIC+90.0)/C) ACCC0R=2.0*VC*VCB/BC*CMPLX(X,Y) ABSCOR=CABS(ACCCOR) TEST ABS CORIOLIS ACC IF(ABSCOR.GT.0.0001)G0

C

TO

110

PHZC0R=0.O GO TO 1 15 110 PHZCOR=C*ATAN2(AIMAG(ACCCOR),REAL(ACCCOR)) 115

CONTINUE X = COS(PH I C/C) Y=SIN(PHIC/C) ACCNCB=-VCB*»2/BC*CMPLX(X,Y) ABSNCB=CABSiACCNCB)

C

TEST ABSNCB IFIABSNCB.GT.0.0001)G0

TO

40

Appendix A 177 178 179 180 18 1 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199

C

PHZNCB =0.0 GO TO 45 40 PHZNCB = C*ATAN2(AIMAG(ACCNCB),REAL(ACCNCB)) 45 CONTINUE X=COS((PHIC+90.0)/C) Y = SIN((PHIC + 90.0)/C) ACCTCB=ATCB*CMPLX(X,Y) ABSTCB=CABS(ACCTCB) TEST ABSTCB IF(ABSTCB.GT.0.0001)G0 PHZT CB = 0.0 GO TO 55

906

F0RMAT(/1OX,'NORMAL 1 ACCELERATIONS') WRITE(6,915)

915

FORMAT(/10X, 'REAL'8X'IMAG'9X'ABS'7X'PHASE'18X'REAL'8X'I MAG' 19X'ABS'7X'PHASE')

22 1 222 223 224 225 226 227 228 229 2 30 23 1 232 233 234 235 236 237 238

AND CORIOLIS

ACCELERATIONS'26X'TANGENT IAL

WRITE (6,301 ) ACCNB , ABSNB , PHZNB , ACCTB . ABSTB , PHZTB WRITE(6.302)ACCNC,ABSNC,PHZNC,ACCTC.ABSTC,PHZTC WRITE(6,303)ACCNCB.ABSNCB,PHZNCB,ACCTCB,ABSTCB,PHZTCB WRITE(6,304)ACCCOR,ABSCOR,PHZCOR ACCB = ACCNB + ACCTB ABSAB = CABS(ACCB) PHZAB - C*ATAN2(AIMAG(ACCB).REAL(ACCB) ) ACCC= ACCNC + ACCTC + ACCCOR ABSAC = CABS(ACCC) TEST ABSAC

C

IF(ABSAC.GT.0.0001)G0 PHZAC =0.0 GO TO 65

2 10 21 1 2 12 213 2 14 2 15 2 16 217 2 18 219 220

50

50 PHZTCB = C*ATAN2(AIMAG(ACCTCB),REAL)ACCTCB ) ) 55 CONTINUE WRITE(6,906)

200 201 202 203 204 205 206 207 208 209

TO

TO

60

60 PHZAC = C*ATAN2(AIMAG(ACCC),REAL)ACCC ) ) 65 CONTINUE ACCCB = ACCNCB + ACCTCB ABSACB = CABS(ACCCB) TEST ABSACB

C

IFfABSACB.GT.0.0001)G0 PHZACB=0.0 GO TO 75

TO

70

70 PHZACB = C*ATAN2(AIMAG(ACCCB),REAL)ACCCB)) 75 CONTINUE WRITE(6,907) 907

FORMAT(/lOX'ABSOLUTE WRITE(6,905)

ACCELERATIONS')

WRITE(6,301)ACCB,ABSAB,PHZAB WRITE(6,302)ACCC.ABSAC.PHZAC WRITE(6,303)ACCCB.ABSACB,PHZACB WRIT E(6,909) 909

FORMAT(//12X,'A1'10X'A2',lOX'BI'10X'B2'lOX'CI'10X'C2'9X'ATC' 18X'ATCB ' )

101 107

WRITE(6, 107)A 1 ,A2.B1 ,B2,C1,C2,ATC F0RMAT(/2X,8F12.4) F0RMAT(/2X,8F12.4)

301 302

F0RMAT(/2X,4F12.4,8X'B FORMAT(/2X,4F12.4,8X'C

303 304

F0RMAT(/2X,4F12.4,8X'CB'.4F12.4) FORMAT(/2X,4F12.4,8X'COR' 4F12 4) GO TO 1 STOP END

999

'.4F12.4) '.4F12.4)

ATCB

Appendix A PROBLEM DATA

AB

AC

THETA(A)

OMEGA(A)

ALPHA(A)

. 6667

. 8333

120.0000

18.0000

.0000

PHI

PHI(C)

26.3310

-26.3310

BC

SNB

SNC

SNCB

AA

1.3017

90.0000

146.3310

-56.3310

.896246806

LINEAR

VELOCITIES

V(B)

V(C)

V ( CB)

12.0006

-6.6531

9.9875

REAL

IMAG

ABS

PHASE

- 10.3928

-6.0003

12.0006

-150.0000

B

-5.9628

2.9510

6.6531

153.6690

C

9.9875

63.6690

4.4300 NORMAL

8.9513 AND CORIOLIS

TANGENTIAL

REAL

IMAG

ABS

PHASE

108.0054

- 187.0708

216.0108

-60.0000

.0000

.OOOO

.0000

.0000

-68.6803

33.9901

76.6310

153.6690

-45.284 1

-91.5009

102.0934

-1 16.3310

ABSOLUTE

CB

ACCELERATIONS

ACCELERATIONS

REAL

IMAG

ABS

B

.0000

.0000

.0000

.0000

C

92.4432

-45.7505

103.1448

-26.3310

7.8340

15.8293

17.6618

63.6690

CB COR

ACCELERATIONS

REAL

IMAG

ABS

PHASE

108.0054

- 187.0708

216.0108

-60.0000

B

47.1591

-137.2514

145. 1273

-71.0375

C

-60.8463

49.8195

78.6400

140.6902

CB

A1

A2

B1

B2

Cl

C2

ATC

ATCB

. 8962

-.4436

-.4436

-.8962

84.6092

-6 1.5798

103.1448

17.6618

PHASE

382

Appendix A

PROBLEM DATA

AB

AC

THETA(A )

OMEGA(A)

ALPHA(A)

3.0000

2.0000

210.0000

30.0000

.0000

PHI

PHI(C)

18.0675

18.0675

BC

SNB

SNC

SNCB

AA

4.8366

90.0000

191.9325

- 101.9325

.950691596

LINEAR

VELOCITIES

Vf B )

V(C)

V (CB)

90.0000

18.6083

88.0553

REAL

IMAG

ABS

PHASE

45.0000

-77.9423

90.0000

-60.0000

B

17.6907

5.7711

18.6083

18.0675

C

-27.3093

83.7134

88.0553

108.0675

.NORMAL

CB

AND CORIOLIS ACCELERATIONS

TANGENTIAL

ACCELERATIONS

REAL

IMAG

ABS

PHASE

REAL

IMAG

ABS

2338.2687

1349.9998

2700.0000

30.0000

B

.0000

.0000

.0000

.0000

.0000

.0000

.0000

C

987.3007

322.0806

1038.5079

18.0675

CB

-37.0066

113.4394

1 19.3230

108.0675

.0000 ■1524 . 1017

-497.1977

1603.1505

- 161.9325

-210.1404

644.1609

677.5709

108.0675

ABSOLUTE

COR

ACCELERATI ONS

REAL

IMAG

ABS

PHASE

2338.2687

1349.9998

2700.0000

30.0000

B

777.1604

966.2415

1240.0004

51 . 1898

C

1561.1083

-383.7583

1607.5850

- 166. 1892

CB

A1

A2

B1

B2

C1

C2

ATC

ATCB

. 9507

. 3101

.3101

-.9507

1024.3073

208.6412

1038.5079

1 19.3230

PHASE

Appendix A 383

SLIDER-CRANK: MODIFIED VECTOR METHOD LINK AGE *ANALYSIS( 1 ).SL- CRANK/ANA LYSIS- 2 ( 2) 1 C SLIDER-CRANK MECHANISM ANALYSIS-MOD IF I ED VECTOR METHOD 2 C C - DEGREES/RADIANICONSTANT) 3 C AB = CRANK 4 C BC = CONNECTING ROD 5 c PHI = PHI(C).ANGLE BETWEEN CONNECTING ROD AND DEAD CENTER 6 c THETAA = POSITION ANGLE OF LINK AB 7 c THETAC = POSITION ANGLE OF LINK BC 8 c OMEGAA = ANGULAR VELOCITY OF LINK AB 9 c ALPHAA = ANGULAR ACC OF LINK AB 10 c VB = VELOCITY OF B 1 1 c VC = VELOCITY OF C 12 c VBC = VELOCITY OF B RELATIVE TO C 13 c CVB = VEL OF B (COMPLEX) 14 c CVC = VEL OF C (COMPLEX) 15 c CVBC = VEL OF B RELATIVE TO C (COMPLEX) 16 c ABSVB = ABSOLUTE VELOCITY OF B 17 c ABSVC = ABSOLUTE VELOCITY OF C 18 c ABSVBC = ABSOLUTE VELOCITY OF B RELATIVE TO C 19 c ACCNB = NORMAL ACC OF B (COMPLEX) 20 c ACCTB = TANGENTIAL ACC OF B (COMPLEX) 2 1 c ACCNC = NORMAL ACC OF C (COMPLEX) 22 c ACCTC = TANGENTIAL ACC OF C (COMPLEX) 23 c ACCB = ACC OF B (COMPLEX) 24 c ACCC = ACC OF C (COMPLEX) 25 c ACCBC » ACC OF B RELATIVE TO C (COMPLEX) 26 c ABSNB = ABSOLUTE NORMAL ACC OF B 27 c ABSTB » ABSOLUTE TANGENTIAL ACC OF B 28 c ABSNC = ABSOLUTE NORMAL ACC OF C 29 c ABSTC = ABSOLUTE TANGENTIAL ACC OF C 30 c ABSAB = ABSOLUTE ACC OF B 31 c A8SAC = ABSOLUTE ACC OF C C ABSABC = ABSOLUTE ACC OF B RELATIVE TO C 32 = PHASE ANGLE OF VELOCITY OF B PHZVB 33 C = PHASE ANGLE OF VELOCITY OF C 34 PHZVC C PHZVBC = PHASE ANGLE OF VELOCITY OF B RELATIVE T C 35 C = PHASE ANGLE OF NORMAL ACC OF B PHZNB 36 C = PHASE ANGLE OF TANGENTIAL ACC OF B 37 C PHZTB = PHASE ANGLE OF NORMAL ACC OF C PHZNC 38 C = PHASE ANGLE OF TANGENTIAL ACC OF C PHZTC 39 C = PHASE ANGLE OF ABSOLUTE ACC OF B PHZAB C 40 = PHASE ANGLE OF ABSOLUTE ACC OF C 4 1 PHZAC C PHZABC = PHASE ANGLE OF ABSOLUTE ACC OF B RELATIVE C 42 REAL NUMR1,NUMR2,NUMR3 COMPLEX CVB.CVC,CVBC,ACCC,ACCNC,ACCTC,CXY COMPLEX ACCNB.ACCTB.ACCB.ACC1BC,ACC2BC.ACCBC

43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73

1 99 19

899 900 100

901

C = 57.29578 READ(5,99)AB.BC,THETAA.OMEGAA,ALPHAA FORMAT(5F12.4) IF(AB.EQ.O.O) GO TO 999 WRITE(6,899) FORMAT(1H1,9X,'PROBLEM DATA') WRITE(6,900) , , FORMAT(//12X, 'AB' 10X'BC ' 4X ' THETA(A ) '4X'OMEGA(A)'4X'ALPHA(A) WRITE(6,100)AB,BC,THETAA,OMEGAA.ALPHAA FORMAT(/2X,5F12.4) THETAA^THETAA/C PHI=ASIN(AB/BC*SIN(THETAA)) PHI=PHI*C WRITE(6,901) FORMAT(//8X,'PHI(C)') WRITE(6, 101)PH I THETAA=THETAA*C SNB=90.O-PHI SNC=THET AA + PHI SNBC=90.O-THETAA

902

WRITE (6,902.) FORMAT(/11X,'SNB'9X'SNC'8X

SNBC

WRITE{6,101)SNB.SNC,SNBC VB=AB*OMEGAA VC=-VB*SIN(SNC/C)/SIN(SNB/C) VBC=-VB*SIN(SNBC/C)/SIN(SNB/C) THETAA=THETAA/C PH I=PHI/C

)

, )

Appendix A 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

WRITE(6,903) 903

F0RMAT(/10X,'LINEAR VELOCITIES') WRITE!6,904) 904 FORMAT(/10X,'V(B)'8X'V(C)'7X'V(BC)') WRITE(6,101)VB,VC.VBC X=SIN(THETAA) Y = -COS(THET AA) CXY=CMPLX(X,Y) CVB=-AB*OMEGAA*(CXY) ABSVB=CABS(CVB) PHZVB=C*ATAN2(AIMAG(CVB),REALfCVB)) WRITE(6,905) 905

ABSVC=CABS(CVC) TEST FOR ABS VEL

c

125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 14 1 142 143 144 145 146 147

OF

C

IF(ABSVC.GT.0.0001)G0 PHZVC=0.0 GO TO 25

TO

20

20 PHZVC=C*ATAN2(AIMAG(CVC),REAL(CVC)) 25 CONTINUE WRITE(6,302)CVC,ABSVC,PHZVC X=SIN(THETAA)-SIN(SNC/C)/SIN(SNB/C) Y = -COS(THE T AA) CXY = CMPLX(X,Y ) CVBC=-AB*OMEGAA*(CXY) ABSVBC=CABS(CVBC) TEST VEL OF B REL TO C IF(ABSVBC.GT.0.0001) GO TO 30 PHZVBC =0.0 GO TO 35

100 101 102 103 104 105 106 107 108 109 1 10 1 1 1 1 12 1 13 1 14 1 15 1 16 1 17 1 18 119 120 12 1 122 123 124

FORMAT(/10X, 'REAL'8X'I MAG'9X'ABS'7X'PHASE' ) WRITE(6,301)CVB.ABSVB.PHZVB. CVC=VC

c

30 PHZVBC=C*ATAN2(AIMAG(CVBC).REAL(CVBC)) 35 CONTINUE WRITE(6,303)CVBC,ABSVBC,PHZVBC WRITE(6.906) 906 915

F0RMAT(/10X, 'NORMAL WRITE(6,915)

ACCELERAT IONS'40XTANGENT IAL

ACCELERATIONS')

FORMAT(/10X, 'REAL'8X'I MAG'9X'ABS'7X'PHASE'20X. 'REAL'8X'I MAG' 19X'ABS'7X'PHASE') X = COS(THET AA) Y = SIN(THET AA) CXY=CMPLX(X,Y) ACCNB = - AB *OMEGA A * »2* ( CX Y ) ABSNB = CAB S(ACCNB) PHZNB=C*ATAN2(AIMAG(ACCNB).REAL(ACCNB)) X=SIN(THETAA) Y = -COS(THE TAA) CXY = CMPLX(X,Y )

c

ACCTB=-AB*ALPHAA*(CXY) ABSTB=CABS(ACCTB) TEST ABS TAN ACC OF B IF(ABSTB.GT.O.0001) PHZTB =0.0 GO TO 45

GO

TO

40

40 PHZTB=C*ATAN2(AIMAG(ACCTB),REAL(ACCTB)) 45 CONTINUE WRITE(6,301)ACCNB,ABSNB.PHZNB,ACCTB NUMR1 = AB*COS(THETAA)*SIN(THETAA)

ABSTB

PHZTB

NUMR2=AB*(C0S(THETAA)*+2-SIN(THETAA)**2) NUMR3 = AB**3* COS(THETAA)* * 2 *SIN(THETAA)**2 DNUMR=SQRT(BC**2-AB«*2*(SIN(THETAA))**2) ,^pu:;A?:?^5AA*t2t(COS(THETAA)+NUMR2/D^MR + NUMR3/DNUMR**3)-AB* 1 ALPHAA*(SIN(THETAA)+NUMR1/DNUMR) ACCNC =0.0 ABSNC = 0.0 PHZNC = 0.0 ACCTC=ACCC

c

ABSTC=CABS(ACCTC) TEST FOR TAN ACC OF

C

IF(ABSTC.GT.O.0001)G0 PHZT C = 0.0 GO TO 95

TO

90

90 PHZTC = C*ATAN2(AIMAG(ACCTC),REAL(ACCTC ) )

Appendix A 148 149 150 151 152 153 154 155 156 157 158 159 160 16 1 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206

95

CONTINUE WRITE(6,302)ACCNC,ABSNC,PHZNC.ACCTC,ABSTC X=NUMR2/DNUMR+NUMR3/DNUMR**3 Y=-SIN(THETAA) CXY=CMPLX(X,Y) ACC1BC=AB*0MEGAA**2*(CXY) ABS1BC=CABS(ACC1BC) TEST FOR TERM 1 OF ABS ACC OF B IF(ABS1BC.GT.0.0001) GO TO 50 PHZ1BC=0.0 GO TO 55

c

REL

TO

C

50 55

c

c

c

PHZ1BC=C*ATAN2(A I MAG(ACC1BC),REAL ( ACC 1 BC ) ) CONTINUE X=NUMR1/DNUMR Y = COS(THET AA) CXY=CMPLX(X,Y) ACC2BC=AB*ALPHAA*(CXY) ABS2BC=CABS(ACC2BC) TEST TERM 2 OF ACC OF B REL TO C IF(ABS2BC.GT.O.0001) GO TO 60 PHZ2BC=0.0 GO TO 65 60 PHZ2BC=C*ATAN2(A I MAG)ACC2BC),REAL(ACC2BC)) 65 CONTINUE WRIT E(6,907) 907 FORMAT(/10X, 'ABSOLUTE ACCELERATIONS' ) WRIT E(6,905) ACCB=ACCNB+ACCTB ABSAB=CABS(ACCB) PHZAB=C*ATAN2(AIMAG(ACCB),REAL(ACCB)) WRITE(6,301)ACCB,ABSAB,PHZAB ACCC=ACCNC+ACCTC ABSAC=CABS(ACCC) TEST FOR ABS ACC OF C IFIABSAC.GT.0.0001)G0 TO 10 PHZAC=0.0 GO TO 15 10 PHZAC=C*ATAN2(AIMAG(ACCC).REAL(ACCC)) 15 CONTINUE WRITE(6,302)ACCC,ABSAC,PHZAC ACCBC=ACC1BC+ACC2BC ABSABC=CABS(ACCBC) TEST ABS ACC OF B REL TO C IF(ABSABC.GT.O.0001) GO TO 70

70 75

201 101 301 302 303 999

PHZABC =0.0 GO TO 75 PHZABC=C*ATAN2(AIMAG(ACCBC),REAL(ACCBC)) CONTINUE WRITE)6,303)ACCBC,ABSABC,PHZABC THETAA=THETAA*C PHI= PHI*C FORMAT(/2X,4F12.4,12X,4F12.4) FORMA T(/2X.8F12.4) FORMAT(/2X,4F12.4,8X'(B) ' ,4F 1 2.4) FORMAT)/2X,4F12.4,8X'(C) '.4F12.4) FORMAT)/2X,4F12.4,8X'(BC)',4F12.4) GO TO STOP END

1

PHZTC

Appendix A

386

PROBLEM DATA

AB

BC

THETA(A)

OMEGA(A )

ALPHA(A)

1.5000

3.0000

120.0000

1.0000

.0000

SNC

SNBC

145.6589

- 30.0000

PHI(C) 25.6589 SNB 64.3411 LINEAR

VELOCITIES

V(B)

V(C)

V (BC )

1.5000

- . 9387

. 8321

REAL

IMAG

ABS

PHASE

- 1 .2990

-.7500

1.5000

-150.0000

(B)

- .9387

.0000

. 9387

180.0000

(C)

-.3603

-.7500

. 832 1

-115.6589

NORMAL

(BC)

ACCELERATIONS

TANGENTIAL

REAL

IMAG

ABS

PHASE

. 7500

- 1 .2990

1.5000

-60.0000

.0000

.0000

.0000

.0000

ABSOLUTE

IMAG

ABS

PHASE

(B)

.0000

.0000

0000

.0000

(C)

1.1180

.0000

1. 1180

.0000

ACCELERATIONS

REAL

IMAG

ABS

PHASE

. 7500

- 1 .2990

1.5000

-60.0000

(B)

1.1180

. 0000

1.1180

.0000

(C)

-.3680

- 1 .2990

1.3502

- 105.8 176

N

ACCELERATIONS

REAL

(BC)

Appendix A PROBLEM DATA

AB

BC

THETA(A)

OMEGA(A)

ALPHA(A)

1.5000

3.0000

30.0000

- 1.0000

.0000

SNC

SNBC

44.4775

60.0000

PHI(C) 14.4775 SNB 75.5225 LINEAR

VELOCITIES

V( B )

V(C)

V (BC )

- 1.5000

1.0854

1.3416

REAL

IMAG

ABS

PHASE

. 7500

- 1.2990

1.5000

-60.0000

(B)

1.0854

.0000

1.0854

.0000

(C)

- . 3354

- 1.2990

1.3416

- 104.4775

NORMAL

(BC)

ACCELERATIONS

TANGENTIAL

REAL

IMAG

ABS

PHASE

- 1.2990

-.7500

1.5000

-150.0000

.0000

.0000

.0000

.0000

ABSOLUTE

ACCELERATIONS

REAL

IMAG

(B)

.0000

. 0000

0000

.0000

(C)

- 1.7251

.0000

1. 7251

-180.0000

ACCELERATIONS

REAL

IMAG

ABS

PHASE

- 1 .2990

-.7500

1.5000

-150.0000

(B)

- 1 .7251

.0000

1.7251

1 80.0000

(C)

. 4260

-.7500

. 8626

-60.4018

(BC)

ABS

PHASE

388

Appendix A

FOUR-BAR: SIMPLIFIED VECTOR METHOD 0DLBL -4-BftR L' 02 • 4-BAR LINKAGE’ 03 XEQ 09 04 ‘-05 XEQ 04 06 FIX 4 07 SF 04 08 -AB?* 09 PROMPT 10 STO 10 11 -BC?12 PROMPT 13 STO 11 14 -CP15 PROMPT 16 STO 12 17 -AD’’ 18 PROMPT 19 STO 13 20 -THETA A’21 PROMPT 22 STO 14 23 -OMEGA A?" 24 PROMPT 25 STO 15 26 -ALPHA A?27 PROMPT 28 STO 16 29 RCL 14 30 COS 31 STO 17 32 RCL 10 33 * 34 RCL 13 35 * 36 2 37 * 38 CHS 39 RCL 18 40 Xt2 41 ♦ 42 RCL 13 43 Xt2 44 ♦ 45 SORT 46 STO 18 47 Xt2 48 RCL 11 49 Xt2 50 +

51 RCL 12 52 Xt2 53 54 RCL 11 55 / 56 2 57 / 58 RCL 18 59 / 60 ACOS 61 STO 19 62 RCL 18 63 Xt2 64 RCL 12 65 Xt2 66 + 67 RCL 11 68 Xt2 69 70 RCL 18 71 / 72 2 73 / 74 RCL 12 75 / 76 ACOS 77 STO 23 78 RCL 14 79 SIN 80 STO 28 81 RCL 10 82 * 83 RCL 18 84 / 85 ASIH 86 STO 21 87 CHS 88 RCL 23 89 90 180 91 ♦ 92 STO 24 93 RCL 19 94 RCL 21 95 % STO 25 97 RCL 15 98 RCL 10 99 * 180 STO 26

Appendix A

181 RCL 25 102 RCL 14 103 104 STO 69 105 SIN 106 STO 27 107 RCL 25 108 RCL 24 109 110 STO 70 111 SIN 112 ST/ 27 113 STO 28 114 RCL 24

389 151 •-* 152 XEO 04 153 RCL 24 154 90 155 + 156 STO 34 157 XEO 05 158 RCL 30 159 XEO 07 160 XEO 08 161 -:C=162 RSTO 33 163 XEO 03 164 RCL 14

115 RCL 14

165 90

116 -

166 +

117 STO 68

167 STO 32

118 SIN

168 XEO 05 51

119 RCL 28

169 RCL 26

120

170 XEO 07

/

121 STO 28

171 XEO 08

122 RCL 26

172 -•B=-

123 *

173 RSTO 33

124 STO 29

174 XEO 03

125 RCL 27

175 RCL 25

126 RCL 26

176 90

127 *

177 +

128 STO 38

178 STO 35

129 * IN/SEC-

179 XEO 05

130 RSTO 31

180 RCL 29

131 ‘LIHERR VELOCITY-

181 XEO 07

132 XEO 09

182 XEO 08

133 '--

183 * C/B=”

134 XEO 04

184 RSTO 33

135 *V:C=-

185 XEO 03

136 RRCL 30

186 RCL 34

137 RRCL 31

187 COS

138 XEO 04

188 STO 36

139 -V B=-

189 LRSTX

140 RRCL 26

190 SIN

141 RRCL 31

191 STO 37

142 XEO 04

192 RCL 35

143 -V:C/B=‘

193 COS

144 RRCL 29

194 STO 38

145 RRCL 31

195 CHS

146 XEO 04

1% STO 39

147 -IPS4-

197 LRSTX

148 RSTO 71

198 SIN

149 *VEC LIN VEL-

199 STO 40

150 XEO 09

200 CHS

390

281 STO 41 282 RCL 24 283 COS 284 STO 42 285 STO 43 286 RCL 38 287 Xt2 288 RCL 12 289 / 218 STO 44 211 ST* 43 212 RCL 15 213 Xt2 214 RCL 18 215 • 216 STO 45 217 RCL 17 218 * 219 ST- 43 228 RCL 16 221 RCL 18 222 * 223 STO 46 224 RCL 32 225 COS 226 STO 47 227 * 228 ST+ 43 229 RCL 29 238 Xt2 231 RCL 11 232 / 233 STO 48 234 RCL 25 235 COS 236 STO 49 237 * 238 ST- 43 239 RCL 24 248 SIH 241 STO 58 242 RCL 44 243 • 244 STO 51 245 RCL 28 246 RCL 45 247 * 248 ST- 51 249 RCL 14 258 98

Appendix A

251 ♦ 252 SIH 253 STO 52 254 RCl 46 255 * 256 ST+ 51 257 RCL 25 258 SIH 259 STO 63 268 RCL 48 261 * 262 ST- 51 263 RCL 43 264 RCL 41 265 * 266 STO 54 267 RCL 51 268 RCL 39 269 * 278 ST- 54 271 RCL 36 272 RCL 41 273 * 274 STO 55 275 RCL 37 276 RCL 39 277 * 278 ST- 55 279 RCL 36 288 RCL 51 281 * 282 STO 56 283 RCL 37 284 RCL 43 285 * 286 ST- 56 287 RCL 54 288 RCL 55 289 / 298 STO 57 291 RCL 56 292 RCL 55 293 / 294 STO 58 295 -HORN ACC2% XEC 89 297 -298 XEO 84 299 -IPSt2/388 ASTO 63

Appendix A

381 382 383 384 385 386 387 388 389 318 311 312 313 314 315 316 317 318 319 328 321 322 323 324 325 326 327 328 329

RCL 24 XEQ 85 RCL 44 CHS XEQ 87 XEQ 88 vOASTO 33 XEQ 16 RCL 14 XEQ 85 RCL 45 CHS XEQ 87 XEQ 88 vB/C=ASTO 33 XEQ 16 RCL 25 XEQ 85 RCL 48 CHS XEQ 87 XEQ 88 -:C/B=* ASTO 33 XEQ 16 -TAN ACC XEQ 89

77ft

•_ _'

331 332 333 334 335 336 337 338 339 348 341 342 343 344 345 346 347 348 349 358

XEQ RCL XEQ RCL XEQ XEQ

84 34 85 57 87 88

VO-

ASTO 33 XEQ 16 RCL 32 XEQ 85 RCL 46 XEQ 87 XEQ 88 vB=‘ ASTO 33 XEQ 16 RCL 35 XEQ 85 RCL 58

391

351 XEQ 87 352 XEQ 88 353 -:C/B=354 ASTO 33 355 XEQ 16 356 -ABS ACC' 357 XEQ 89 358 --359 XEQ 84 368 RCL 24 361 XEQ 85 362 RCL 59 363 STO 64 364 RCL 68 365 STO 65 366 RCL 27 367 RCL 26 368 * 369 Xt2 378 RCL 12 371 / 372 CHS 373 ST* 64 374 ST* 65 375 RCL 34 376 XEQ 85 377 RCL 59 378 STO 66 379 RCL 68 388 STO 67 381 RCL 57 382 ST* 66 383 ST* 67 384 XEQ 86 385 1 386 XEQ 87 387 XEQ 88 388 - O389 ASTO 33 398 XEQ 16 391 RCL 14 392 XEQ 85 393 RCL 59 394 STO 64 395 RCL 68 396 STO 65 397 RCL 26 398 RCL 15 399 * 488 CHS

392

481 ST* 64 482 ST* 65 483 RCL 32 484 XEQ 85 485 RCL 59 486 STO 66 487 RCL 68 488 STO 67 489 RCL 46 418 ST* 66 411 ST* 67 412 XEQ 86 413 1 414 XEQ 87 415 XEQ 88 416 ":B= 417 fiSTO 33 418 XEQ 16 419 RCL 25 428 XEO 85 421 RCL 59 422 STO 64 423 RCL 68 424 STO 65 425 RCL 28 426 RCL 26 427 * 428 Xt2 429 RCL 11 438 / 431 CHS 432 ST* 64 433 ST* 65 434 RCL 35 435 XEQ 85 436 RCL 59 437 STO 66 438 RCL 68 439 STO 67 448 RCL '58 441 ST* i66 442 ST* i67 443 XEQ i86 444 1 445 XEQ i87 446 XEQ l88 447 -:C/B=448 ASTO 33 449 XEQ 16 458 -PROBLEM IOTA

Appendix A 451 XEQ 89 452 *453 XEQ 84 454 -f»e=455 ARCL 16 456 XEQ 64 457 "BC=" 458 ARCL 11 459 XEQ 64 468 -CB=461 fiRCL 12 462 XEQ 84 463 *AD=' 464 fiRCL 13 465 XEQ 84 466 ‘THETA ft467 fiRCL 14 468 XEQ 84 469 ‘OMEGA fl=' 478 fiRCL 15 471 XEQ 84 472 ‘ALPHA ft=' 473 fiRCL 16 474 XEQ 84 475 *BD=‘ 476 ARCL 18 477 XEQ 84 478 ‘PHI B=‘ 479 fiRCL 21 488 XEQ 84 481 ‘GAMMA B=‘ 482 fiRCL 19 483 XEQ 84 484 ‘GAMMA »=‘ 485 fiRCL 23 486 XEQ 84 487 ‘THETA B=488 fiRCL 25 489 XEQ 84 498 "THETA C=* 491 fiRCL 24 492 XEQ 84 493 *SMB=" 494 ARCL 78 495 XEQ 84 496 *SMC=497 ARCL 69 498 XEQ 84 499 -SMCB=588 ARCL 68

Appendix A 581 XEQ 84 582 ■P 1=* 583 PRCL 36 584 XEQ 84 585 -P 2=* 586 PRCL 37 587 XEO 84 588 ■B 1=’ 589 PRCL 39 518 XEQ 04 511 •B 2=’ 512 PRCL 41 513 XEQ 84 514 •c 1=* 515 PRCL 43 516 XEQ 84 517 *C 2=* 518 PRCL 51 519 XEQ 04 528 •PTC=521 PRCL 57 522 XEQ 84 523 -PTCB=524 PRCL 58 525 XEQ 04 526 XEQ 13 527*LBL 85 528 57.296 529 / 538 8 531 XROfl -etZ 532 STO 59 533 XOY 534 STO 68 535 RTN 536*LBL 88 537 RCL 59 538 x 539 GTO 14 540 RTN 541*LBL 14 542 RCL 68 543 x 544 GTO 15 545 180 546 ST* 62 547 RTH 548*LBL 15 549 188 550 ST- 62

393 551 RTN 552*LBL 86 553 RCL 65 554 RCL 64 555 RCL 67 556 RCL 66 557 XROtl *C+* 558 STO 59 559 XOY 568 STO 68 561 RTH 562*LBL 87 563 ST* 59 564 ST* 68 565 RCL 68 566 RCL 59 567 XR08 -HflGZ 568 STO 61 569 RCL 59 578 X=8? 571 QTO 12 572 RCL 68 573 RCL 59 574 / 575 PTPN 576 STO 62 577 RTN 578*LBL 12 579 8 580 PTPN 581 STO 62 582 RTN 583*LBL 04 584 RVIEH 585 PDV 586 CLP 587 RTN 588*LBL 89 589 PV1EH 598 RTN 591*LBL 03 592 FIX 2 593 26 594 PCCHR 595 CLP 5% PRCL 33 597 PRCL 61 598 PRCL 71 599 PRCL 62 680 PCP

394

Ml PRBltF M2 MV M3 FIX 4 M4 RTN 6B5»LBL 16 M6 FIX 2 M7 22 6§8 flCCHR 60S afl 610 ORCl 33 611 ARCL 61 612 WRCL 63 613 ARCL 62 614 ACO 615 PRBUF 616 MV 617 FIX 4 618 RTN 61W.BL 13 620 AOFF 621 STOP 622 .END.

Appendix A

Appendix A PROBLEM MTfl

*=1.5880

4-BOR LINKAGE

LINEAR VELOCITY

BC=3.0000 CJN3.0000 AIM. 8000

V:C=-1.0392IN/SEC V:B=3.0008 IN/SEC V:C/B=-3.4601 IN/SEC

THETfl 0=30.0000 VEC LIN VEL OMEGA 0=2.0000 0LPH0 0=1.0000

uC=1.04IPS4l2.33

BB=2.8B32

0 B=3.00IPS4l20.00

PHI 1=15.5188

u:C/B=3.46IPS^-43.37

G0MH0 B=62.1478

NORM ACC

COMMA 8=62.1478 o C=0.36IPSt2^-77.67 THET0 B=46.6289 d:B/C=6.80IP$t22-150.00 THET0 C=102.3334 o:C/B=3.99IPSt2^-133.37 SMB=-55.7045 TON ACC 910=16.6289

--

SMCB=72.3334

«C=11.02IPSt24-167.67

0 l=-0.9769

iB=1.50IPSt24l20.00

0 2=-0.2136

i:C/8=2.76IPSt2^136.63

B 1=0.7269

0BS 0CC

6 2=-0.6867 C 1=-8.7635 C 2=-4.2502 0TC=11.0244 0TCB=2.7601

*C=11.03IPSt24-165.80 «B=6.181PSt2^-164.04 d:C/B=4.85IPSt24-168.04

396

Appendix A

SLIDER-CRANK: SIMPLIFIED VECTOR METHOD 81♦LBL "SL-CR V 02 • SLIDER CR ANL* 03 XEQ 89 §4 ’-• 05 XEQ 04 06 FIX 4 07 SF 04 08 -AF5* 09 PROMPT 10 STO 16 11 *BC?" 12 PROMPT 13 STO 11 14 'THETA 2’' 15 PROMPT 16 STO 12 17 'OMEGA 2?' 18 PROMPT 19 STO 13 20 'ALPHA 2*>21 PROMPT 22 STO 09 23 -ECC’24 PROMPT 25 STO 14 26 RCL 12 27 SIN 28 STO 15 29 RCL 10 30 * 31 RCL 14 32 33 RCL 11 34 / 35 ASIH 36 STO 16 37 180 38 RCL 16 39 40 STO 17 41 RCL 10 42 RCL 13 43 * 44 STO 18 45 RCL 17 46 RCL 12 47 48 SIN 49 STO 19 50 90

51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 88 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

RCL SIN STO RCL RCL / STO RCL • STO 98 RCL SIN STO RCL

17

20 19 28 21 18 22 12

23 20

/

STO 24 RCL 18 * STO 25 -LINEAR VELOCITYXEQ 09 '-‘ XEQ 04 •IN/SECASTO 26 -V;B=ARCL 18 ARCL 26 XEQ 04 -V:C=ARCL 22 ARCL 26 XEQ 04 -VB/C=ARCL 25 ARCL 26 XEQ 84 RCL 12 90 + STO 27 RCL 17 90 ♦ STO 28 -VEC LIN VELXEQ 09

Appendix A 397

iei -192 XEQ 94 193 RCL 27 194 XEQ 95 195 RCL 18 198 XEQ 97 197 XEQ 98 198 -1PS4’ 199 BSTO 29 119 *:B=111 BSTO 39 112 XEQ 93 113 9 114 XEQ 95 115 RCL 22 118 XEQ 07 117 XEQ 08 118 VC*’ 119 BSTO 38 128 XEQ 93 121 RCL 28 122 XEQ 85 123 RCL 25 124 XEQ 87 125 XEQ 08 126 vB/C=127 BSTO 38 128 XEQ 83 129 1 138 STO 31 131 8 132 STO 32 133 RCL 28 134 COS 135 STO 33 136 LBSTX 137 SIH 138 STO 34 139 RCL 12 148 COS 141 STO 35 142 RCL 18 143 RCL 13 144 * 145 STO 48 148 RCL 35 147 • 148 CHS 149 STO 38 158 RCL 09

151 152 153 154 155 156 157 158 159 188 161 162 163 164 165 166 167 168 169 178 171 172 173 174 175 176 177 178 179 188 181 182 183 184 185 186 187 188 189 198 191 192 193 194 195 196 197 198 199 288

RCL * STO RCL COS STO RCL * ST+ RCL COS STO RCL Xt2 RCL / STO RCL * ST+ RCL RCL * CHS STO RCL SIH STO RCL * ST+ RCL SIH STO RCL * ST+ RCL RCL * STO RCL RCL * STRCL RCL * STO RCL

18 37 27 38 37 36 17 39 25 11 43 39 36 15 48

41 27 42 37 41 17 44 43 41 36 34 45 41 33 45 31 34 46 32

398

281 RCL 33 282 * 283 ST- 48 284 RCL 31 285 RCL 41 286 * 287 STO 47 288 RCL 32 288 RCL 36 218 * 211 ST- 47 212 RCL 45 213 RCL 46 214 / 215 STO 48 216 RCL 47 217 RCL 46 218 / 219 STO 49 228 *HORH PCC* 221 XEQ 89 222 *-* 223 XEQ 84 224 RCL 12 225 XEQ 85 226 RCL 48 227 CHS 228 XEQ 87 229 XEQ 88 238 *IPSt24" 231 PSTO 58 232 *:B=233 PSTO 38 234 XEQ 16 235 8 236 STO 59 237 STO 68 238 1 239 XEQ 87 248 XEQ 88 241 -:C=242 PSTO 38 243 XEQ 16 244 RCL 17 245 XEQ 85 246 RCL 43 247 CHS 248 XEQ 07 249 XEQ 88 258 * B/C=*

Appendix A 251 PSTO 38 252 XEQ 16 253 -TQH PCC 254 XEQ 09 255 *-' 256 XEQ 04 257 RCL 27 258 XEQ 85 259 RCL 37 268 XEQ 07 261 XEQ 08 262 *:B=' 263 PSTO 38 264 XEQ 16 265 8 266 XEQ 85 267 RCL 48 268 XEQ 07 269 XEQ 88 278 ■:C=* 271 PSTO 38 272 XEQ 16 273 RCL 28 274 XEQ 85 275 RCL 49 276 XEQ 07 277 XEQ 88 278 -:B/C=‘ 279 PSTO 38 280 XEQ 16 281 *PBS PCC282 XEQ 09 283 *-• 284 XEQ 04 285 RCL 12 286 XEQ 85 287 RCL 59 288 STO 51 289 RCL 60 290 STO 52 291 RCL 48 292 CHS 293 ST* 51 294 ST* 52 295 RCL 27 296 XEQ 05 297 RCL 59 298 STO 53 299 RCL 60 388 STO 54

Appendix A

381 382 383 384 385 386 387 388 389 318 311 312 313 314 315 316 317 318 319 328 321 322 323 324 325 326 327 328 329 338 331 332 333 334 335 336 337 338 339 348 341 342 343 344 345 346 347 348 349 358

RCL 37 ST* 53 ST* 54 XEQ 86 1 XEQ 87 XEQ 88 *:B= RSTC! 38 XEQ 16 8 XEQ 85 RCL 48 XEQ 87 XEQ 88 •:C=:• RSTC) 38 XEQ 16 RCL 17 XEQ 85 RCL 59 STO 51 RCL 68 STO 52 RCL 25 Xt2 RCL 11 / CHS ST* 51 ST* 52 RCL 28 XEQ 85 RCL 59 STO 53 RCL 68 STO 54 RCL 49 ST* 53 ST* 54 XEQ 86 1 XEQ 87 XEQ 88 •:B/C=ftSTO 38 XEQ 16 • PROBLEM MTftXEQ 89

399

351 352 353 354 355 356 357 358 359 368 361 362 363 364 365 366 367 368 369 378 371 372 373 374 375 376 377 378 379 388 381 382 383 384 385 386 387 388 389 398 391 392 393 394 395 396 397 398 399 488

XEQ 84 ‘AB=* ARCL 18 XEQ 84 ■BC=* ARCL 11 XEQ 84 ‘THETA ; ARCL 12 XEQ 84 •OMEGA ! ARCL 13 XEQ 84 •ALPHA ; ARCL 89 XEQ 84 *ECC=* ARCL 14 XEQ 84 •PHI 3= ARCL 16 XEQ 84 •THETA ; ARCL 17 XEQ 84 98 RCL 17 -

STO 55 RCL 17 RCL 12 -

STO 56 98 RCL 12 -

STO 57 -SHB=ARCL 55 XEQ 84 •SNC=ARCL 56 XEQ 84 •SH8C=* ARCL 57 XEQ 84 •A 1=* ARCL 31 XEQ 84 •A 2='

400

401 MCI 32 402 XEQ 04 403 -B 1=* 404 PRCL 33 405 XEQ 04 406 -B 2=* 407 PRCL 34 40B XEQ 04 409 -C 1=’ 410 PRCL 36 411 XEQ 04 412 *C 2=* 413 PRCL 41 414 XEQ 04 415 *PTC=* 416 PRCL 48 417 XEQ 04 418 *QTBC=419 PRCL 49 428 XEQ 04 421 XEQ 13 422*LBL 05 423 57.2% 424 / 425 8 426 XROfl *etZ 427 STO 59 428 XOY 429 STO 60 430 RTN 431*LBL 08 432 RCL 59 433 X<8^ 434 GTO 14 435 RTN 436*LBL 14 437 RCL 60 438 X<0’ 439 GTO 15 440 188 441 ST+ 62 442 RTN 443*LBL 15 444 180 445 ST- 62 446 RTN 447*LBL 06 448 RCL 52 449 RCL 51 450 RCL 54

Appendix A

451 RCL 53 452 XROM *C+453 STO 59 454 XOY 455 STO 60 456 RTN 457*LBL 07 458 ST* 59 459 ST* 60 460 RCL 60 461 RCL 59 462 XRON -HOCZ 463 STO 61 464 RCL 59 465 X=0? 466 GTO 12 467 RCL 68 468 RCL 59 469 / 470 PTPN 471 STO 62 472 RTN 473*LBL 12 474 0 475 RTQN 476 STO 62 477 RTN 478*LBL 04 479 QVIEN 480 RBV 481 CLP 482 RTN 483*LBL 89 484 RVIEU 485 RTN 486*LBL 03 487 FIX 2 488 26 489 ACCHR 490 CLP 491 W?CL 30 492 PRCL 61 493 PRCL 29 494 PRCL 62 495 PCP 4% PRBUF 497 PBV 498 FIX 4 499 RTN 500*LBL 16

Appendix A CD OP (D A si

FIX 2 22 RCCHR

an

Si

RRCL 36 RRCL 61 RRCL 56 RRCL 62 RCR PRBUF 511 ABV 512 FIX 4 513 RTN 514*LBL 13 515 ROFF 516 STOP 517 .END.

401

Si tt S

402

PROBLEM BATP

ab=i.5800

Appendix A

SLIDER CR ANL

LINEAR VELOCITY

BC=3.0000 THETA 2=150.0000 OMEGA 2=1.0000 ALPHA 2=0.0000 ECC=0.0800

VB= 1.5080 IN/SEC V:C=-0.4146 IN/SEC V:B/C=1.3416IN/SEC VEC LIN VEL

PHI 3=14.4775

OB=1.50IPS4-120.00

THETfl 3=165.5225

U:C=0.41IPSil80.00

SHB=-75.5225

u:B/C=1.34IPSi-104.48

SNC=15.5225

NORM ACC

SHBC=-60.0000 fl 1=1.0000 A 2=0.0000 6 1=-0.2500 B 2=-0.%82 C 1=0.7181 C 2=-0.6000 ATC=0.8730 ATBC=0.6197

aB=1.50IPSt24-30.00 fl:C=8.00IPSt240.00 «B/C=0.60IPSt2^-14.48 TflN ACC

a:B=0.00IPSt2^0.00 a C=0.87IPSt240.00 e:B/C=0.62IPSt24-104.48 ABS ACC

oB=1.50IPSt24-30.00 aC=8.87IPSt240.00 a B/C=0.86IPSt2^-60.40

Appendix A

PPOBLEH DATA

AB=1.5B00

SLIDER CR ftHL

LINEAR VELOCITY

BC=3.0000 THETfl 2=150.0008 ONEGA 2=1.0000 ALPHA 2=0.0000

V:B=1.50001N/SEC V:C=-0.6414IN/SEC VB/C= 1.3036 IN/SEC VEC LIN VEL

ECC=0.5000 PHI 3=4.7802

0 B=1.50IPS^-120.00

THETfl 3=175.2198

u C=0.64IPS^180.00

SHB=-85.2198

O:B/C=1.30IPS^-94.78

SHC=25.2198

NORN ftCC

SHBC=-68.0000 i:B=1.501PSt24-30.00 fl 1=1.0000 dC=0.00IPSt2^0.00 A 2=8.0000 d:B/C=0.57IPSt2^-4.78 B 1=-0.0833 TflN ACC B 2=-0.9965

—

C 1=0.7346

d:B=0.00IPSt2*0.00

C 2=-0.7028

d C=0.79IPSt2^6.00

ATC=0.7933

i B/C=0.71IPSt2^-94.78

ATBC=0.7053

BBS ACC

d B=1.50IPSt2*-30.08 dC=8.79IPSt240.00 dB/C=0.90IPSt2^-56.01

404

Appendix A

PROBLEM BATA SLIBER CR ftHL (®=1.5M0 LIHEAR VELOCITY BC=3.0000

--

THETfl 2=150.0000

VB= 1.5000 IN/SEC

OMEGA 2=1.0000

V:C=-0.1546IN/SEC

ALPHA 2=0.0000

VB/C=1.4290 IN/SEC

ECC=-0.5000

VEC LIN VEL

PHI 3=24.6243 O:B=1.50IPS4-120.00 THETfl 3=155.3757 OC=0.15IPS^180.00 SH8=-65.3757 G:B/C=1.43IPS*-114.63 SHC=5.3757 NORM OCC SHBC=-60.0000

—

A 1=1.0006

aB=1.50IPSt2^-30.00

fl 2=0.0000

aC=0.00IPSt240.00

B 1=-0.4167

a B/C=0.68IPSt2^-24.62

B 2=-0.9091

TAN ACC

C 1=0.6803 c B=0.00IPSt2^0.00 C 2=-0.4664 aC=0.89IPSt240.00 ATC=0.8940 aB/C=8.51IPSt2<(-114.63 flTBC=0.5130 ABS ACC

a:B=1.50IPSt2*-30.00 aC=0.89IPSt240.00 a'B/C=0.85IPSt2^-61.63

Appendix A

405

QUICK-RETURN: SIMPLIFIED VECTOR METHOD

11

/

12 STO 10 13 'AD’’ 14 PROMPT 15 12 16 / 17 STO 11 18 -THETA V 19 PROHPT 28 STO 12 21 -OHEGA 2?22 PROHPT 23 STO 13 24 -ALPHA 2?25 PROHPT 26 STO 14 27 RCL 12 28 COS 29 ST0 08 30 RCL 11 31 * 32 RCL 10 33 * 34 2 35 * 36 CHS 37 RCL 11 38 Xt2 39 + 48 RCL 10 41 Xt2 42 + 43 SORT 44 STO 15 45 Xt2 46 RCL 11 47 Xt2 48 + 49 RCL 10 50 X+2

51 52 2 53 / 54 RCL 15 55 r 56 RCL 11 57 / 58 ACOS 59 STO 65 68 180 61 RCL 12 62 63 GTO 10 64 -1 65 ST* 65 66*LBL 10 67 180 68 RCL 65 69 70 STO 17 71 RCL 12 72 SIN 73 STO 16 74*LBL 02 75 RCL 10 76 RCL 13 77 * 78 STO 09 79 RCL 17 80 RCL 12 81 82 STO 18 83 COS 84 STO 19 85 RCL 09 86 * 87 STO 20 88 RCL 18 89 SIN 90 STO 21 91 RCL 09 92 * 93 STO 22 94 •LIN VELOCITY 95 XE0 09 96 •—--1 97 XEO 04 98 FIX 2 99 -FPS100 ASTO 23 X /N II -c o

0DLBL -0-R NEC* 02 • A-R HECHANISH' 03 XEO 09 04 *-’ 05 XE6 04 06 FIX 4 07 SF 04 08 ’AB?’ 09 PROHPT 10 12

406

181 -VB=182 ftRCL 09 183 ftRCL 23 184 XEC 04 185 -V:C=* 106 ftRCL 28 187 ftRCL 23 108 XEQ 04 189 -V:B/C=110 ftRCL 22 111 ftRCL 23 112 XEQ 84 113 FIX 4 114 -VEC LIN VEL115 XE8 09 116 *-' 117 XEG 04 118 RCL 12 119 98 128 + 121 STO 24 122 XEG 05 123 RCL 09 124 XEG 07 125 XEG 08 126 *:&=■ 127 ftSTO 25 128 -FPS4129 fiSTO 26 138 XEG 83 131 RCL 17 132 98 133 + 134 STO 27 135 XEG 85 136 RCL 20 137 XEG 07 138 XEG 88 139 *:C=* 140 ftSTO 25 141 XEG 03 142 RCL 17 143 XEQ 05 144 RCL 22 145 XEG 07 146 XEG 08 147 • 8/C=148 ftSTO 25 149 XEG 03 150 RCL 27

Appendix A

151 COS 152 STO 28 153 LfiSTX 154 SIN 155 STO 29 156 RCL 17 157 SIN 158 STO 30 159 LfiSTX 168 COS 161 STO 31 162 RCL 20 163 Xt2 164 * 165 RCL 15 166 / 167 STO 32 168 RCL 28 169 RCL 20 170 * 171 RCL 15 172 / 173 RCL 22 174 * 175 2 176 * 177 ST- 32 178 RCL 24 179 COS 180 STO 33 181 RCL 10 182 * 183 RCL 14 184 * 185 ST+ 32 186 RCL 08 187 RCL 10 188 * 189 RCL 13 190 Xt2 191 * 192 ST- 32 193 RCL 38 194 RCL 20 195 Xt2 196 * 197 RCL 15 198 / 199 STO 34 200 RCL 29

Appendix A 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 248 241 242 243 244 245 246 247 248 249 250

RCL * RCL / RCL * 2 * STRCL SIH STO RCL • RCL * ST+ RCL RCL * RCL Xt2 * STRCL RCL * STO RCL RCL * STRCL RCL * STO RCL RCL • STRCL RCL / STO RCL RCL * STO RCL RCL

407

20 15 22

34 24 35 10 14 34 16 10 13

34 32 30 36 34 31 36 28 30 37 29 31 37 36 37 38 28 34 39 29 32

251 * 252 ST- 39 253 RCL 39 254 RCL 37 255 / 256 STO 40 257 -NORN ACC258 XEQ 09 259 ’-• 260 XEO 04 261 RCL 12 262 XEQ 05 263 RCL 09 264 RCL 13 265 * 266 STO 41 267 CHS 268 XEO 07 269 XEO 08 278 -8=271 ASTO 42 272 -FPSt24' 273 ASTO 43 274 XEO 16 275 RCL 17 276 XEO 05 277 RCL 20 278 Xt2 279 RCL 15 288 / 281 STu 44 282 CHS 283 XEO 07 284 XEO 08 285 -:C=‘ 286 ASTO 42 287 XEQ 16 288 RCL 27 289 XEQ 05 290 0 291 XEQ 07 292 XEQ 08 293 -:B/C=* 294 ASTO 42 295 XEQ 16 296 -TANG ACC" 297 XEQ 09 298 ■-' 299 XEQ 04 300 RCL 24

408

•

XEQ RCL RCL * STO XEQ XEQ

85 10 14 45 07 08

9?•

£ CO

3«1 302 303 304 305 306 307

Appendix A

309 PSTO 42 310 XEQ 16 311 RCL 27 312 XEQ 05 313 RCL 38 314 XEO 07 315 XEO 08 316 vC=317 PSTO 42 318 XEO 16 319 RCL 17 320 XEO 05 321 RCL 48 322 XEO 07 323 XEQ 08 324 •:B/C=’ 325 PSTO 42 326 XEQ 16 327 ■COR PCC328 XEQ @9 329 "---" 330 XEO 04 331 RCL 27 332 XEQ 05 333 RCL 20 334 RCL 15 335 / 336 RCL 22 337 * 338 2 339 * 340 STO 46 341 XEQ 07 342 XEQ 08 343 XEQ 16 344 •PB$i PCC 345 XEQ 09 346 •---" 347 XEQ 04 348 RCL 24 349 XEQ 05 350 RCL 59

351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 388 38! 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400

STO 48 RCL 60 STO 49 RCL 45 ST* 48 ST* 49 RCL 12 XEQ 05 RCL 59 STO 50 RCL 60 STO 51 RCL 41 CHS ST* 50 ST* 51 XEQ 06 1 XEO 07 XEQ 08 vB=PSTO 42 XEO 16 RCL 27 XEQ 05 RCL 59 STO 48 RCL 60 STO 49 RCL 38 ST* 48 ST* 49 RCL 17 XEQ 85 RCL 59 STO 58 RCL 68 STO 51 RCL 19 RCL 10 * RCL 13 • Xt2 RCL 15 /

CHS ST* 50 ST* 51 XEQ 06

Appendix A 481 1 482 XEO 87 483 XEO 88 484 *:C=‘ 485 ASTO 42 486 XEO 16 487 RCL 17 488 XEO 85 488 RCL 59 418 STO 48 411 RCL 68 412 STO 49 413 RCL 48 414 ST* 48 415 ST* 49 416 RCL 27 417 XEC 85 418 RCL 59 419 STO 58 428 RCL 68 421 STO 51 422 RCL 19 423 RCL 21 424 * 425 RCL 41 426 * 427 RCL 18 428 * 429 2 438 * 431 RCL 15 432 / 433 ST* 58 434 ST* 51 435 XEO 86 436 1 437 XEO 87 438 XEO 88 439 -:B/C=* 448 PSTO 42 441 XEO 16 442 -PROBLEM BATA* 443 XEO 89 444 ■-* 445 XEO 84 446 ‘«B=447 ARCL 18 448 XEO 84 449 •«»=■ 458 ARCL 11

409 451 XEO 84 452 "THETA 2453 ARCL 12 454 XEO 84 455 "OHEQA 2= 456 ARCL 13 457 XEO 84 458 "ALPHA 2= 459 ARCL 14 468 XEO 84 461 "PHI 3=* 462 ARCL 65 463 XEO 84 464 •THETA 3465 ARCL 17 466 XEO 84 467 ■CD=' 468 ARCL 15 469 XEO 84 478 -A 1=' 471 ARCL 28 472 XEO 84 473 •A 2=’ 474 ARCL 29 475 XEO 84 476 ■8 1=* 477 ARCL 31 478 XEO 84 479 •B 2=* 488 ARCL 38 481 XEO 84 482 •c i=483 ARCL 32 484 XEO 84 485 •C 2=‘ 486 ARCL 34 487 XEO 84 488 •RTC=* 489 ARCL 38 498 XEO 84 491 -ATBC=492 ARCL 48 493 XEO 84 494 XEO 13 495*LBL 85 496 57.296 497 / 498 8 499 XROfl *etZ588 STO 59

410

sei xoy 582 STO 60 583 RTH 504*LBL 88 585 RCL 59

586 x 587 GTO 14 588 RTH 509*LBL 14 518 RCL 68 511 X<8? 512 GTO 15 513 188 514 ST+ 62 515 RTH 516*LBL 15 517 188 518 ST- 62 519 RTH 528*LBL 86 521 RCL 49 522 RCL 48 523 RCL 51 524 RCL 58 525 XROH -C+526 STO 59 527 XOY 528 STO 68 529 RTH 530*LBL 87 531 ST* 59 532 ST* 68 533 RCL 68 534 RCL 59 535 XROH *NBGZ 536 STO 61 537 RCL 59 538 X=87 539 GTO 12 548 RCL 68 541 RCL 59 542 / 543 BTBH 5<4 STO 62 545 RTH 546*LBL 12 547 8 548 BTBH 549 STO 62 558 RTH

Appendix A

551*LBL 04 552 BY I EM 553 BDV 554 CLB 555 RTH 556*LBL 09 557 BVIEH 558 RTH 559*LBL 83 568 FIX 1 561 26 562 BCCHR 563 CLP 564 BPCL 25 565 BRCL 61 566 BPCL 26 567 BRCL 62 568 BCB 569 PRBUF 578 BDV 571 FIX 4 572 RTH 573*LBL 16 574 FIX 1 575 22 576 BCCHR 577 CLP 578 BRCL 42 579 BRCL 61 580 BRCL 43 581 BRCL 62 582 BCB 583 PRBUF 584 BDV 585 FIX 4 586 PTH 587*LBL 13 588 BOFF 589 .END.

Appendix A PROBLEM DATA

06=0.1667

Q-P HECHAHISM

LIN VELOCITY

AIM). 3333 THETA 2=30.0000

V:B=1e.47FPS V:C=-6.19FPS

OMEGA 2=62.8389 PLPHA 2=0.0008

VB/C=8.45FPS VEC LIN VEL

PHI 3=23.7940 O:B=10.5FPS2120.0 THETA 3=156.2060 0C=6.2FPS*66.2 CB=0.2066 u:B/C=8.4FPS2l56.2 A 1=-0.4834 NORM ACC A 2=-0.9150 B 1=-8.9150

o B=657.9FP3t24-150.0

B 2=0.4034

o:C=185.2FPSt2^-23.8

C l=-943.4515

d:B/C=0.0FPSt2^6.0

C 2=-717.2896

TANG ACC

PTC=1.036.9562 o:B=8.0FPSt240.0 ATBC=573.8702 e C=1;037.0FPSt22-l13.8 «:B/C=573.9FPSt24l56.2 COP ACC

o:B/C=506.1FPSt2466.2 ABS ACC

o 8=657.9FPSt24-150.0 d:C=1.853.4FPSt2*-103.7 i:0/C=765.1FPSt24l14.8

412

PROBLEM DATA

08=0.2509

Appendix A 0-R MECHANISM

LIH VELOCITY

flB=0,1667 V:B=7.50FPS THETA 2=80.0000 VC=5.67FPS OMEGA 2=30.0080 VB/C=4.91FPS ALPHA 2=0.0000 VEC LIN VEL PHI 3=79.1066

—

THETA 3=100.8934

0-8=7.5FPS4150.0

C8=0.2285

0 C=5.7FPS4-169.1

0 l=-0.9820

u:B/C=4.9FPS^100.9

0 2=-0.1890

NORM ACC

8 l=-0.1890 d8=225.0FPSt24-120.0 B 2=0.9820 d:C=145.8FPSt24-79.1 C 1=107.9082 a:B/C=0.0FPSt240.0 C 2=-3.9766 TANG ACC OTC=-105.2122 0TBC=-24.2977

«B=0.0FPSt240.0 6C=105.2FPSt24l0.9 d B/C=24.3FPSt24-79.1 COR ACC

d B/C=252.5FPSt24-169.1 ABS ACC

d B=225.OFPSt 24-120.0 d:C=179.8FpSt24-43.3 d:BA>253.7FPSt24-163.6

Appendix A

SLIDING COUPLER: SIMPLIFIED VECTOR METHOD 0HLBL ‘SL-CPLR* 51 Xt2 02 -SLIDING COUPLER 52 RCL 11 03 XE0 09 53 Xt2 94 .54 + 05 XEQ 04 55 RCL 10 06 SF 04 56 Xt2 07 FIX 4 57 08 "AB9" 58 2 09 PROMPT 59 / 10 12 60 RCL 17 11 / 61 / 62 RCL 11 12 STO 10 63 / 13 -AC?* 64 ACOS 14 PROMPT 65 CHS 15 12 66 STO 18 16 / 67 188 17 STO 11 68 ENTERt 18 -THETA A?69 RCL 12 19 PROMPT 78 X<=Y? 20 STO 12 71 GTO 02 21 -OMEGA A?72 -1 22 PROMPT 73 ST* 18 23 STO 13 74 GTO 82 24 -ALPHA A?75*LBL 01 25 PROMPT 76 180 26 STO 14 77 ENTERt 27 RCL 12 78 RCL 12 28 SIN 79 X< =Y? 29 STO 15 GTO 18 80 30 LASTX 81 0 31 COS 82 ACOS 32 STO 16 83 CHS 33 RCL 11 84 STO 18 34 * 85*LBL 18 35 RCL 10 86 0 36 * 87 ACOS 37 2 88 STO 18 38 * 89*LBL @2 39 CHS 90 RCL 13 40 RCL 11 91 RCL 10 41 Xt2 92 * 42 + 93 STO 19 43 RCL 10 94 RCL 12 44 Xt2 95 RCL 18 45 + 96 46 SORT 97 STO 20 47 STO 17 98 SIN 48 X=0? 99 STO 21 49 GTO 01 100 RCL 19 50 RCL 17

413

414 101 » 102 CHS 103 STO 22 104 RCL 20 105 COS 106 STO 23 107 RCL 19 108 * 109 CHS 110 STO 24 111 -LIN VELOCITY' 112 XEQ 09 113 *-' 114 XEQ 84 115 -FPS118 fiSTO 25 117 *V'B=* 118 ORCL 19 119 ORCL 25 120 XEQ 04 121 -V:C=‘ 122 flRCL 22 123 fiRCL 25 124 XEQ 04 125 -VC/B=126 ORCL 24 127 fiRCL 25 128 XEQ 04 129 -VEC LIH VEL" 130 XEQ 09 131 . 132 XEQ 04 133 RCL 12 134 90 135 + 136 STO 26 137 XEQ 05 138 RCL 19 139 XEQ 07 140 XEQ 08 141 -FPS4142 fiSTO 27 143 ’:B=144 fiSTO 28 145 XEQ 03 146 RCL 18 147 XEQ 05 148 RCL 22 149 XEQ 07 150 XEQ 08

Appendix A 151 -:C=’ 152 fiSTO 28 153 XEQ 03 154 RCL 18 155 90 156 + 157 STO 29 158 XEQ 05 159 RCL 24 160 XEQ 07 161 XEQ 08 162 *:C/B=* 163 fiSTO 28 164 XEQ 03 165 RCL 18 166 COS 167 STO 30 168 LflSTX 169 SIH 176 STO 31 171 RCL 29 172 COS 173 CHS 174 STO 32 175 LfiSTX 176 SIH 177 CHS 178 STO 33 179 RCL 13 180 Xt2 181 RCL 10 182 * 183 STO 34 184 RCL 24 185 Xt2 186 RCL 17 187 / 188 STO 35 189 RCL 14 190 RCL 10 191 * 192 STO 36 193 RCL 24 194 RCL 17 195 / 196 STO 37 197 RCL 22 198 * 199 2 208 *

Appendix A

201 STO 38 202 RCL 16 203 RCL 34 204 * 205 CHS 206 STO 39 207 RCL 26 208 COS 209 STO 40 210 RCL 36 211 * 212 ST+ 39 213 RCL 38 214 RCL 35 215 * 216 ST- 39 217 RCL 32 218 CHS 219 RCL 38 220 * 221 ST- 39 222 RCL 15 223 RCL 34 224 * 225 CHS 226 STO 41 227 RCL 26 228 SIH 229 STO 42 230 RCL 36 231 * 232 ST+ 41 233 RCL 31 234 RCL 35 235 * 236 ST- 41 237 RCL 33 238 CHS 239 RCL 38 240 * 241 ST- 41 242 RCL 39 243 RCL 33 244 * 245 STO 43 246 RCL 41 247 RCL 32 248 * 249 ST- 43 250 RCL 30

415

251 RCL 33 252 * 253 STO 44 254 RCL 31 255 RCL 32 256 * 257 ST- 44 258 RCL 38 259 RCL 41 260 * 261 STO 45 262 RCL 31 263 RCL 39 264 * 265 ST- 45 266 RCL 43 267 RCL 44 268 / 269 STO 46 270 RCL 45 271 RCL 44 272 / 273 STO 47 274 -NORM ACC 275 XEO 89 276 ’277 XEO 04 278 RCL 12 279 XEO 05 280 RCL 34 281 CHS 282 XEO 07 283 XEO 08 284 -FPSt24285 OSTO 48 286 ":B=" 287 flSTO 49 288 XEO 16 289 RCL 29 290 XEO 85 291 0 292 XEO 87 293 XEO 08 294 -C=295 RSTO 49 296 XEO 16 297 RCL 18 298 XEO 05 299 RCL 35 300 CHS

Appendix A

416 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 328 321 322 323 324 325 326 327 328 329 338 331 332 333 334 335 336 337 338 339 348 341 342 343 344 345 346 347 348 349 350

XEQ 07 XEQ 88 vC/B=* PSTO 49 XEQ 16 -TPNG OCC” XEQ 09 ■-' XEQ 04 RCL 26 XEQ 05 RCL 36 XE6 87 XEQ 08 ‘ B=* PSTO 49 XEC 16 RCL 18 XEQ 05 RCL 46 XEQ 07 XEQ 08 vC=‘ PSTO 49 XEQ 16 RCL 29 XEQ 05 RCL 47 XEQ 87 XEQ 08 •: C/8PSTO 49 XEQ 16 -CORR PCC XEQ 09 ■XEQ 04 RCL 29 XEQ 05 RCL 38 XEQ 07 XEQ 08 -:C=' PSTO 49 XEQ 16 “PBS PCCXEQ 09 •-* XEQ 04 RCL 12

351 XECI 05 352 RCL. 59 353 STCl 50 354 RCL 68 355 STOi 51 356 RCL 34 357 CHS 358 ST* 50 359 ST* 51 360 RCL 26 361 XEQ 05 362 RCL 59 363 STO 52 364 RCL 60 365 STO 53 366 RCL 36 367 ST* 52 368 ST* 53 369 XEQ 06 370 1 371 XEQ 87 372 XEQ 88 373 ■ - B:=374 PSTO 49 375 XEQ 16 376 RCL 29 377 XEQ 05 378 RCL 59 379 STO 58 388 RCL 68 381 STO 51 382 RCL 21 383 RCL 23 384 * 385 RCL 34 386 * 387 RCL 10 388 * 389 2 390 * 391 RCL 17 392 / 393 ST* 50 394 ST* 51 395 RCL 18 396 XEQ 85 397 RCL 59 398 STO 52 399 RCL 60 400 STO 53

Appendix A

417 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450

RCL 46 ST* 52 ST* 53 XEQ 06 1 XEC 07 XEQ 08 vC=ASTO 49 XEQ 16 RCL 18 XEQ 05 RCL 59 STO 50 RCL 60 STO 51 RCL 23 Xt2 RCL 34 * RCL 10 * CHS RCL 17 / ST* 50 ST* 51 RCL 29 XEQ 05 RCL 59 STO 52 RCL 60 STO 53 RCL 47 ST* 52 ST* 53 XEQ 06 1 XEQ 07 XEQ 08 -:C/B=flSTO 49 XEQ 16 -PROBLEM DATA XEQ 09 ‘XEQ 04 *AB=" ARCL 10 XEQ 04

451 -AC=‘ 452 ARCL 11 453 XEQ 04 454 'THETA A= 455 ARCL 12 456 XEQ 84 457 -OHEGA A458 ARCL 13 459 XEQ 04 460 'ALPHA A=' 461 ARCL 14 462 XEQ 04 463 -PHI C=" 464 ARCL 18 465 XEQ 04 466 -A 1=467 ARCL 30 468 XEQ 64 469 -A 2=" 478 ARCL 31 471 XEQ 04 472 *B 1=473 ARCL 32 474 XEQ 04 475 -B 2=" 476 ARCL 33 477 XEQ 04 478 -C 1=’ 479 ARCL 39 488 XEQ 04 481 -C 2=" 482 ARCL 41 483 XEQ 04 484 *ATC=’ 485 ARCL 46 486 XEQ 04 487 -ATBC=* 488 ARCL 47 489 XEQ 04 490 XEQ 13 491*LBL 05 492 57.296 493 / 494 0 495 XROM -etZ496 STO 59 497 XOY 498 STO 60 499 RTN 500*LBL 08

418

Ml RCL 59 502 X<0-> 503 GTO 14 504 RTN 505*LBL 14 506 RCL 60 507 X<0? 508 GTO 15 509 180 510 ST+ 62 511 RTN 512*LBL 15 513 180 514 ST- 62 515 RTN 516*L8L 06 517 RCL 51 518 RCL 50 519 RCL 53 520 RCL 52 521 XROH -C+522 STO 59 523 XOY

524 STO 60 525 RTN 5264LBL 07 527 ST* 59 528 ST* 60 529 RCL 60 530 RCL 59 531 XROH -HfiGZ 532 STO 61 533 RCL 59 534 X=0T 535 GTO 12 536 RCL 60 537 RCL 59 538 / 539 ATAN 540 STO 62 541 RTN 542*LBL 12 543 0 544 ATAN 545 STO 62 546 RTN 547*LBL 04 548 AVI EH 549 ADV 550 CLA

Appendix A

551 RTN 552HBL 09 553 AVIEH 554 RTN 555*LBL 03 556 FIX 2 557 26 558 ACCHR 559 CLA 560 ARCL 28 561 ARCL 61 562 ARCL 27 563 ARCL 62 564 ACA 565 PRBUF 566 ADV 567 FIX 4 568 RTN 569*LBL 16 578 FIX 1 571 22 572 ACCHR 573 CLA 574 ARCL 49 575 ARCL 61 576 ARCL 48 577 ARCL 62 578 ACA 579 PRBUF 580 ADV 581 FIX 4 582 RTN 583*LBL 13 584 AOFF 585 .END.

Appendix A

419 SLIDING COUPLER

PROBLEM DPTP

LIN VELOCITY

RB=0.6667 PC=8.8333

V B=12.00B0FP$ THETP 0=120.0000 VC=-6.6530FPS ONEGO 0=18.0000 VC/B=9.9869FPS OLPHO 0=0.0000 VEC LIN VEL PHI C=-26.3295 0 B=12.00FPS^-150.00

0 1=0.8963

u C=6.65FPS<£153.67

0 2=-0.4435

0:C/B=9.99FPSrt3.67

B 1=-0.4435

NORN OCC

B 2=-0.8963 C 1=84.6063

dB=216.0FPSt2*-60.0 C 2=-61.5830 d C=0.0FPSt2^0.0 OTC=103.1432 d:C/B=76.6FPSt2^153.7 OTBC-17.6686 TONG fiCC

d B=0.0FPST2^0.0 dC=103. lFPSt2^-26.3 d:C/B=17.7FPSt2^63.7 CORR fiCC

d C=182.lFPSt2^-l16.3 OBS ACC

d B=216.0FPSt2<-60.0 d:C=145.1FPSt2^-71.0 d:C/B=78.6FPSt2^140.7

420

Appendix A

SLIDING COUPLER

PROBLEM DPTO

LIN VELOCITY

06=0.2500

VB=7.5000FPS V:C=1.5507FPS V-C/B=7.3379FPS VEC LIN VEL

PC=0.1867 THETO 0=210.0000 ONEGO 0=30.0000 OLPHO 0=0.0000 PHI C=18.0675

0 B=7.50FPS^-60.00

0 1=0.9507

G-C=1.55FP$<18.07

0 2=0.3101

u:C/B=7.34FPS4l03.@7

B 1=0.3101

NORN ACC

d:B=225.0FPSt2430.0 dC=0.0FPSt240.8 d:C/B=133.6FPSt24-161.9 TONG PCC

d B=0.0FPSt240.0 d C=86.5FP$t2^18.1 d:C/B=9.9FPSt24l08.1 CORR ACC

dC=56.5FPSt24l08.1 PBS ACC

d B=225.0FPSt2430.0 dC=103.3FPSt2*51.2 a C'B=134.0FPSt24-166.2

B 2=-0.9507 C 1=85.3589 C 2=17.3868 0TC=86.5423 0TBC=9.9436

Appendix A

421

SLIDER-CRANK: MODIFIED VECTOR METHOD •1*LBL -$L-CP C*

•2 * SI-CP mi Cl 93 XEO 99

94 *•5 XEO 84 96 FIX 4

97 SF 94 98 *98’* 99 PRWPT 19 STO 19

11 *BC>12 PPWPT 13 14 15 16 17 18 19 29 21 22 23 24 25 26 27 28 29 39 31 32 33

STO 11 *WECA7PPWPT STO 12 -THETA «“>* PRWPT STO 13 SIN STO 14 -ALPHA R7* PRWPT STO 99 PCL 14 PCL 18 • PCL 11 / ASIA STO 15 CHS 189

35 36 37 38 39 49 41 42 43 44 45 46 47 48 49 58

STO PCL PCL * STO PCL PCL 4 SIN STO 99 PCL SIN STO 1/X

34 ♦

16 19 12 17 15 13

18 15

19

51 52 53 54 55 56 57 58 59 68 61 62 63 64 65 66 67 68 69 78 71 72 73 74 75 76 77 78 79 89 81 82 83

RCL • PCL • CHS STO 98 PCL

18 17

28 13

-

SIN STO 21 98 RCL 15 -

SIN STO 22 RCL 21 PCL 22 /

PCL 17 * CHS STO 23 ■IN/SECASTt) 24 •LINEAP VELOCITY XEO 89 *XEP 84 *V B=’ APCL 17 APCL 24 XEe 84

84 -V:C=*

85 86 87 88 89 99 91 92 93 94 95 96 97 98 99 188

APCL 29 APCL 24 XEQ 84 -V:B/C=APCL 23 APCL 24 XEO 84 *VEC UN VEL* XEO 89 *-‘ XEO 84 RCL 13 XEO 85 PCL 68 STO 28 RCL 59

422

If] STO 27 lf2 RCl 17 183 CHS 184 XEO 87 185 XEO 88 188 -IPS/* 187 RSTO 25 188 vB=‘ 188 RSTO 28 118 XEO 83 111 8 112 XEO 18 113 RCL 59 114 STO 29 115 RCL 88 118 STO 38 117 RCL 28 118 XEO 87 119 XEO 88 128 VO121 RSTO 28 122 XEO 83 123 RCL 17 12* CHS 125 ST* 27 128 ST* 28 127 RCL 28 128 ST* 29 129 ST* 38 138 XEO 86 131 1 132 XEO 87 133 XEO 88 134 -:B/C=* 135 RSTO 26 138 XEO 83 137 -5 138 STO 49 139 -HORN RCC 148 XEO 89 141 *142 XEO 84 143*LBL 18 144 RCL 13 145 XEO 16 146 RCL 59 147 STO 27 148 RCL 68 149 STO 28 158 RCL 18

Appendix A

151 RCL 12 152 Xt2 153 • 154 STO 31 155 CHS 158 ST* 27 157 ST* 28 158 RCL 13 159 XEO 85 188 RCL 59 161 STO 29 162 RCL 68 163 STO 38 164 RCL 18 165 RCL 89 166 * 167 STO 32 168 CHS 169 ST* 38 178 ST* 29 171 XEO 86 172 1 173 XEO 87 174 XEO 88 175 -lPSt2/176 RSTO 33 177 * B=178 RSTO 34 179 XEO 17 188 RCL 49 181 X>0'> 182 CTO 28 183 8 184 STO 61 185 STO 62 186 *:C=* 187 RSTO 34 188 XEO 17 189 “TRW RCC 198 XEO 89 191 *-■ 192 XEO 84 193 8 194 STO 61 195 STO 62 196 *:B=* 197 RSTO 34 198 XEO 17 199*LBL 19 288 RCL 13

Appendix A 2*1 2*2 283 2W 285 286 287 288 289 218 211 212 213 214 215 216 217 218 219 228 221

CCS STO 35 LPSTX SIH STO 36 Xt2 CHS RCL 35 Xt2 ♦ STO 37 RCL 18 * STO 38 RCL 36 Xt2 RCL 18 XT2 * CHS RCL 11

222 XT2 223 + 224 STO 39 225 SOFT 226 STO 47 227 1/X 228 RCL 38 229 * 238 STO 48 231 RCL 35 232 ♦ 233 STO 41 234 3 235 ENTEPt 236 2 237 / 238 STO 42 239 RCL 39 248 RCL 42 241 YtX 242 STO 43 243 RCL 18 244 ENTEPT 245 3 246 YTX 247 RCL 35 248 XT2 249 • 258 RCL 36

251 Xt2 252 * 253 STO 44 254 RCL 43 255 / 256 STO 45 257 ST+ 41 258 RCL 41 259 RCL 31 268 CHS 261 * 262 STO 27 263 8 264 STO 28 265 RCL 18 266 RCL 35 267 * 268 RCL 36 269 * 278 STO 46 271 RCL 47 272 / 273 STO 48 274 RCL 36 275 ♦ 276 RCL 32 277 CHS 278 * 279 STO 29 288 8 281 STO 38 282 XEO 86 283 1 284 XEO 87 285 XEO 88 286 - C=’ 287 PSTO 34 288 XEO 17 289 RCL 49 298 X>87 291 CTO 21 292 *P6S PCC 293 XEO 89 294 *295 XEO 84 296 18 297 STO 49 298 XEO 18 299HBL 28 388 XEO 19

424

3® 1♦LBL 21 382 RCL 38 383 RCL 47 384 / 385 STO 58 388 RCL 44 387 RCL 43 388 / 388 ST+ 58 318 RCL 31 311 ST* 58 312 RCL 58 313 STO 27 314 RCL 31 315 RCL 36 316 * 317 CHS 318 STO 28 319 RCL 48 328 RCL 32 321 * 322 STO 29 323 RCL 35 324 RCL 32 325 • 326 STO 38 327 XEO 81 328 1 329 XEQ 87 338 XEQ 88 331 * 6/C=* 332 fiSTO 34 333 XEQ 17 334 ‘PROBLEM MTfi* 335 XEQ 89 336 ’-* 337 XEQ 84 338 *fiB=' 339 fiRCL 18 348 XEQ 84 341 -ec=342 fiRCL 11 343 XEQ 84 344 -THETP fi=* 345 fiRCL 13 346 XEQ 84 347 ‘OHEGfl fl=* 348 fiRCL 12 349 XEQ 84 358 ‘fiLPHfi fi=‘ •

Appendix A 351 fiRCL 89 352 XEQ 84 353 -PHI C=* 354 fiRCL 15 355 XEQ 84 356 98 357 RCL 15 358 355 STO 51 368 RCL 13 361 RCL 15 362 ♦ 363 STO 52 364 98 365 RCL 13 366 367 STO 53 368 *SHB=’ 369 fiRCL 51 378 XEQ 84 37| *SHC=" 372 fiRCL 52 373 XEC 84 374 -SHBC=” 375 fiRCL 53 376 XEQ 84 377 CTO 13 378*LBL 85 379 57.296 388 / 381 8 382 XROH VtZ 383 CHS 384 STO 68 385 XOY 386 STO 59 387 RTN 388»LBL 88 389 RCL 59 398 x<e"> 391 CTO 14 392 RTH 393*LBL 14 394 RCL 68 395 X<89 396 CTO 15 397 188 398 ST+ 62 399 RTN 488*LBL 15

Appendix A 425 461 186 482 ST- 62 483 RTN 484*LBL 66 485 PCI 28 486 RCL 27 487 PCI 38 488 PCI 28 488 XROH *C-* 418 STO 58 411 XOY 412 STO 68 413 PTH 414*LBL 87 415 ST* 58 416 ST* 68 417 RCL 68 418 PCI 58 418 XROH *HAG? 428 STO 61 421 RCL 58 422 X=8’ 423 GTO 12 424 RCL 68 425 RCL 59 426 / 427 ATAH 428 STO 62 429 RTH 438*LBL 12 431 8 432 ATAH 433 STO 62 434 RTH 435*LBL 84 436 AVIEH 437 ABV 438 CLA 439 PTH 448*LBL 89 441 AVIEU 442 RTH 443*LBL 16 444 57.286 445 / 446 6 447 XROH *etZ‘ 448 STO 58 449 XOY 458 STO 68

451 RTH 452*LBL 81 453 RCL 28 454 RCL 27 455 RCL 38 456 RCL 29 457 XROH *C+ 458 STO 59 459 XOY 468 STO 68 461 RTH 462*LBL 83 463 FIX 2 464 26 465 ACCHP 466 CLA 467 ARCL 26 468 ARCL 61 469 ARCL 25 478 ARCL 62 471 ACA 472 PRBI.IF 473 APV 474 FIX 4 475 PTH 476*LBL 17 477 FIX 2 478 22 479 ACCHP 488 CLA 481 ARCL 3< 482 ARCL 61 483 ARCL 33 484 ARCL 62 485 ACA 486 PRBUF 487 APV 488 FIX 4 489 RTN 498*LBL 13 491 AOFF 492 STOP 493 EHD

426

SL-CP ONAL CL

LIHEPP VELOCITY

Appendix A PPOBLEH PTp

OB=1.5W0 BC=3.O060

V:6=1.5088IN/$EC THETA 0=120.0000

V C=-«.938?IH/$EC OHEGA 0=1.0000

V:B/C=0.8321IH/SEC OLPHP 0=0.0000

VEC LIh VEL PHI C=25.6589 G:B=1.50IPS*-150.00

SNP=04.3411

o c=e.wiPS2iBB.e0

$VC=145.6589

O:B/C=8.83IPS2-1!5.00

SHBC=-30.O000

HOPH OCC

« b=i.501pst22-00.ee eC=€.WIPSt2/8.80 TON ACC

a B=8.88IPSt2rt.08 o-c=i.121pst220.ee OPS OCC

a 8=1.58lPSt22-60.ee c C=1.12IP$t220.00 e:B/C=1.35!PSt2^-!05.82

Appendix A

PPOBLEH MTfl

08=1.5000

427

SI-CP ONOL CL

LIHEOR VELOCITY

BC=3.8800 theto

0=30.0000

ONEGO ft=-1.0000 ALPHA 0=0.0000 PHI C=14.4775 SWB=75.5225 $HC=44.4775 SHBC=00.0000

VB=-1.50001N/SEC V:C=1.0854IH/SEC VBA=1.3416IN/SEC VEC LIH VEL

GB=1.50IPS2-00.00 u:C=l.WIPS<0.00 u B/C=1.34IPS^-104.48 HORN OCC

eB=!.50IP$t2^-150.00 *C=8.00IPST2*0.00 TOH OCC

iB=0.00IPSt240.06 6 C=1.73IPSt24l80.00 OBS OCC

i B=1.50IPSt24-150.00 iC=1.73IPSt24l86.00 i8/C=0.86IPSt2<-60.40

Appendix B

B.l

NOMENCLATURE

a

linear acceleration

a

average linear acceleration

aB

linear acceleration of point B

a, b, c, etc.

termini of velocity vectors VA, VB, Vc, etc. on velocity polygon

a', b', c', etc.

termini of acceleration vectors AA, AB, on acceleration polygon

A

linear acceleration (magnitude)

A

Aq,

etc.

linear acceleration vector (magnitude and direction)

A. B B/C CD

linear acceleration of point B linear acceleration of point B relative to point C effective component of acceleration along CD

^Cor Coriolis acceleration ,N normal acceleration rotational component of acceleration translational component of acceleration tangential acceleration A, B, C, etc. CCW

cw

pivot points on a linkage counterclockwise direction clockwise direction 428

Appendix B

e

429 eccentricity of a slider-crank mechanism

i I

instant center

k

k

a

acceleration scale (actual acceleration represented by unit length of acceleration vector or acceleration axis of motion curve)

s

space scale (actual length of machine member or dis¬ placement represented by unit length of vector or dis¬ placement axis of motion curve

kt

time scale (actual time represented) by unit length on time axis of motion curve

k

velocity scale (actual velocity represented by unit length of velocity vector or velocity axis of motion curve)

n

number of links of a mechanism

N

number of instant centers; number of revolutions per minute

o

pole or origin of velocity polygon

o’

pole or origin of acceleration polygon

p

point of contact between two sliding bodies

P(C)

contact point P on C

P(F)

contact point P on F

P(C)/P(F)

point P on C relative to point P on F

R, r

radius

s

linear displacement

t

time

V

linear velocity

V

average linear velocity, unit vector

Vi

initial linear velocity

V2

final linear velocity

V

VB

linear velocity of point B

V

linear velocity (magnitude)

V

linear velocity vector (magnitude and direction)

430

VB VB/C

vCD vr V*

vi vi

Appendix B

linear velocity of point B linear velocity of point B relative to point C effective component of velocity along CD rotational component of velocity translational component of velocity initial velocity of point A final velocity of point A initial velocity of point B

V1

initial velocity of point B

\

final velocity of point B

Oi

(alpha)

angular acceleration, or other angle angular acceleration of link 2

“ab (3 (beta)

angular acceleration of link AB

y (gamma)

angle

angle


(phi)

angle

9

(theta)

angular displacement

cj (omega)

COj

angular velocity initial angular velocity

0)2

angular velocity of link 2, final angular velocity

WAB

angular velocity of link AB

1, 2, 3, etc.

links 1, 2, 3, etc.

23

instant center of links 2 and 3

1

perpendicular to

Appendix B

B.2

431

TRIGONOMETRY REVIEW

Functions of a Right Triangle (Figure B. 1) opposite hypotenuse

sin a

cos

adjacent hypotenuse

tan a =

opposite adjacent

a c

b c a b

sin a cos a

1 c cosec a = —- = — sin a a 1 c sec a = - = r cos a b 1 b cos a cot a = -- = — = —:tan a a sin a

Functions of an Angle in the Interval 0° < 9 < 360° If a vector AB is rotated through the four quadrants as shown in Figure B.2, any function of the angle 9 is numerically equal to the same function of the acute angle a between the terminal side of the vector and the x axis. That is, Function of 9 = ± same function a where the positive or negative sign depends on the quadrant in which the angle a falls. Signs are determined as shown in Figure B.2. These signs of the functions of 9 in the four quadrants may be sum¬ marized as follows:

432

Appendix B

o° ^ © y qn° sin

©

= Z

= r

r

(a) cos

9

=

X

+— r

tan

9 =

r

+Z = Y X

90

X

180°

°

sin 9

= y

= r

cos

9

=

(b)

—

r _

r tan

9

X

r =_z

= —X

X

18O°^0^27O°

(c)

sin

9

=

^ = _Y r r

cos

9

=

—

X

r tan 9 =

—

-x

r =

z X

270^0^ 360 0

sin

© =

¥-

r cos

9

=

+— r

(d) tan

9

Vector in rotation.

r X

r

=_z

= X

Figure B. 2

=_y

X

Appendix B

433

Figure B. 3 Sign diagram.

a = 9

sin 9 is positive cos 9 is positive tan 9 is positive

90° < 9 < 180°:

a = 180° - 9

sin 9 is positive cos 9 is negative tan 9 is negative

180° < 9 < 270°:

a = 9 - 180°

sin 9 is negative cos 9 is negative tan 9 is positive

270° < 9 < 360°:

a = 360° -9

sin 9 is negative cos 9 is positive tan 9 is negative

0° < 9 < 90°:

Signs of the functions of 9 are most conveniently remembered using Fig¬ ure B. 3.

434

Appendix B B

Figure B.4

Oblique triangle.

Laws for Oblique Triangles Laws of Cosines In any triangle, the square of any side is equal to the sum of the squares of the other sides minus twice their product times the cosine of their included angle. For example, in triangle ABC in Figure B.4, a2 = b2 + c2 - 2bc cos a b2 = a2 + c2 - 2ac cos /3 c2 = a2 + b2 - 2ab cos y where a + (3 + y = 180° Law of Sines In any triangle, any two sides are proportional to the sides of the opposite angles. In triangle ABC in Figure B.4, a sin ot

_

b sin p

_

c sin y

Laws of Tangents In any triangle, the difference of the opposite angles divided by their sum equals the tangent of one-half the difference of the opposite angles divided by the tangent of one-half their sum. In triangle ABC in Figure B.4, a - b _ tan (1/2)(a - P) a+ b tan (1/2) (a + p) a - c _ tan (1/2) (a - y) a + c tan (1/2) (a + y)

Appendix B

435

b - c _ tan (1/2)(/j - y) b + c tan (1/2) ((3 + y)

Other Useful Relationships sin2 a + cos2 a - l sin

(a

+ /3) = sin

a

cos /3 + cos

a

sin (3

sin (a - (3) = sin a cos (3 - cos a sin (3 cos (a + (3) = cos a cos (3 - sin a sin /3 cos (a - i3) = cos a cos /3 + sin a sin tan (o' + j8)

tan a + tan (3 1 - tan a tan j3

tan (a - /3)

tan a - tan (3 1 + tan a tan (3

sin 2cv = 2 sin a cos a cos 2a = cos2 a - sin2 a tan 2a =

2 tan a 1 - tan2 a

ot

1 - cos oi

sin

a cos 2 ~ tan

oi

2

=

11 + cos a 2

J

/ sin ol —V 1 + cos a

Oi + (3 a - (3 sin a + sin (3-2 sin —cos —-— u

L*

a + (3 . a - [3 sin a - sin (3=2 cos —-— sin —-— a + (3 a - (3 cos a + cos (3=2 cos —-— cos —-

z

z

a + (3 . a - (3 cos Oi - cos (3 = -2 sin —-— sin 2 2

436

Appendix B

sin (90° - 9) = +cos 9 sin (90°+ 9) = +cos 9 sin (180° - 9) - +sin 9 sin (180°+ 9) = -sin 9 sin (270° - 9) = -cos 9 sin (270°+ 9) = -cos 9 sin (360° - 9) = -sin 9 sin (360°+ 9) = +sin 9 sin(-0) = -sin 9 cos(90° - 9) = +sin 9 cos(90° + 9) = -sin 9 cos(180° - 9) = -cos 9 cos(180°+ 9) = -cos 9 cos(270° - 9) = -sin 9 cos(270°+ 9) = +sin 9 cos(360° - 9) = +cos 9 cos(360°+ 9) = +cos 9 cos(—9) = +cos 9

N

Appendix B

B.3

437

TABLE OF TRIGONOMETRIC FUNCTIONS

deg

rad

sin

COS

0 1 2 3 4 5

.000 .017 .035 .052 .070 .087

.000 .017 .035 .052 .070 .087

6 7 8 9 10

. 105 .122 .140 .157 .175

11 12 13 14 15

tan

deg

rad

sin

cos

1.000 1.000 .999 .999 .998 .996

.000 .017 .035 .052 .070 .087

46 47 48 49 50

.803 .820 .838 .855 .873

.719 .731 .743 .755 .766

.695 .682 .669 .656 .643

1.036 1.072 1.111 1.150 1.192

.105 .122 .139 .156 .174

.995 .993 .990 .988 .985

.105 .123 .141 .158 .176

51 52 53 54 55

.890 .908 .925 .942 .960

.777 .788 .799 .809 .819

.629 .616 .602 .588 .574

1.235 1.280 1.327 1.376 1.428

.192 .209 .227 .244 .262

.191 .208 .225 .242 .259

.982 .978 .974 .970 .966

.194 .213 .231 .249 .268

56 57 58 59 60

.977 .995 1.012 1.030 1.047

.829 .839 .848 .857 .866

.559 .545 .530 .515 .500

1.483 1.540 1.600 1.664 1.732

16 17 18 19 20

.279 .297 .314 .332 .349

.276 .292 .309 .326 .342

.961 .956 .951 .946 .940

.287 .306 .325 .344 .364

61 62 63 64 65

1.065 1.082 1.100 1.117 1.134

.875 .883 .891 .899 .906

.485 .470 .454 .438 .423

1.804 1.881 1.963 2.050 2.145

21 22 23 24 25

.367 .384 .401 .419 .436

.358 .375 .391 .407 .423

.934 .927 .921 .914 .906

.384 .404 .424 .445 .466

66 67 68 69 70

1.152 1.169 1.187 1.204 1.222

.914 .921 .927 .934 .940

.407 .391 .375 .358 .342

2.246 2.356 2.475 2.605 2.747

26 27 28 29 30

.454 .471 .489 .506 .524

.438 .454 .470 .485 .500

.899 .891 .883 .875 .866

.488 .510 .532 .554 .577

71 72 73 74 75

1.239 1.257 1.274 1.292 1.309

.946 .951 .956 .961 .966

.326 .309 .292 .276 .259

2.904 3.078 3.271 3.487 3.732

31 32 33 34 35

.541 .559 .576 .593 .611

.515 .530 .545 .559 .574

.857 .848 .839 .829 .819

.601 .625 .649 .675 .700

76 77 78 79 80

1.326 1.344 1.361 1.379 1.396

.970 .974 .978 .982 .985

.242 .225 .208 . 191 .174

4.011 4.331 4,705 5. 145 5.671

36 37 38 39 40

.628 .646 .663 .681 .698

.588 .602 .616 .629 .643

.809 .799 .788 .777 .766

.727 .754 .781 .810 .839

81 82 83 84 85

1.414 1.431 1 .449 1.466 1.484

.988 .990 .993 .995 .996

.156 .139 . 122 .105 .087

6.314 7.115 8.144 9.514 11.430

41 42 43 44 45

.716 .733 .751 .768 .785

.656 .669 .682 .695 .707

.755 .743 .731 .719 .707

.869 .900 .933 .966 1.000

86 87 88 89 90

1.501 1.518 1.536 1.553 1.571

.998 .999 .999 1.000 1 .000

.070 .052 .035 .017 .000

14.301 19.081 28.636 57.290

tan

Selected References

Albert, C. D. , and F. S. Rogers, Kinematics of Machinery, Wiley, New York, 1931. Annand, W. J. D., Mechanics of Machines, Chemical Publishing, New York 1966. Barton, L. O., "Finding Slider-Crank Acceleration Graphically," Machine Design, Vol. 50, No. 28, December 7, 1978. Barton, L. O. , "Simplified Slider-Crank Equations," Machine Design, Vol. 51, No. 8, April 1979. Barton, L. O. , "The Acceleration Polygon—A Generalized Procedure," Engineering Design Graphics Journal, Vol. 43, No. 2, Spring 1979. Barton, L. O. , "Painless Analysis of Four-Bar Linkages," Machine Design. Vol. 51, No. 17, July 26, 1979. Barton, L. O., "Simplifying Velocity Analysis for Mechanisms," Machine Design, Vol. 53, No. 13, June 11, 1981. Barton, L. O., "Simplified Analysis of Quick Return Mechanisms," Machine Design, Vol. 52, No. 18, August 7, 1980. Barton, L. O., "A Diagrammatic Representation of the Basic Motion Equations," Engineering Design Graphics Journal, Vol. 44, No. 3 Fall 1980. Barton, L. O., "A New Way to Analyze Slider-Cranks," Machine Design Vol. 17, July 22, 1982. -—— Barton, L. O., "Simplifying the Analysis of Sliding Coupler Mechanisms, " Machine Design, Vol. 55, No. 19, August 25, 1983.

438

Selected References

439

Beggs, J. S., Mechanism, McGraw-Hill, New York, 1955. Bickford, J. H., Mechanisms for Product Design. Industrial Press, New York, 1972. Billings, J. H. , Applied Kinematics, 2nd ed., Van Nostrand Reinhold York, 1943.

New

Chironis, N. P., Mechanisms, Linkages and Mechanical Controls, McGrawHill, 1965. Chironis, N. P. , Machine Devices and Instrumentation, McGraw-Hill, New York, 1966. Dent, J. A., and A. C. Harper, Kinematics and Kinetics of Machinery, Wiley, New York, 1921. Durley, R. J., Kinematics of Machines, Wiley, New York, 1911. Esposito, A., Kinematics for Technology. Charles E. Merrill, Columbus, Ohio, 1973. Guillet, G. L. , Kinematics of Machines, Wiley, New York, 1934. Hain, K., Applied Kinematics, ed. by D. P. Adams and T. P. Goodman, McGraw-Hill, New York, 1967. Hall Jr., A. S., Kinematics and Linkage Design, Prentice-Hall, Englewood Cliffs, N.J., 1961. Ham, C. W., E. J. Crane, and W. L. Rogers, Mechanics of Machinery, 4th ed. , McGraw-Hill, New York, 1958. Heck, R. C. H. , Elementary Kinematics, D. Van Nostrand, New York, 1910. Hinkle, R. T., Kinematics of Machines, 2nded., Prentice-Hall, Englewood Cliffs, N.J., 1960. Hirschhorn, J. , Kinematics and Dynamics of Plane Mechanisms, McGrawHill, New York, 1962. Holowenko, A. R., Dynamics of Machinery, Wiley, New York, 1955. Horton, H. L. , Mathematics at Work, Industrial Press, New York, 1949. Hunt, K. H., Mechanisms and Motion, Wiley, New York, 1959. James, W. H., and M. C. Mackenzie, Principles of Mechanism, Wiley, New York, 1918. Keown, R. McA., and V. M. Faires, Mechanism, McGraw-Hill, New York, 1939. Kepler, H. B., Basic Graphical Kinematics, 2nded., McGraw-Hill, New York, 1973.

440

Selected References

Klein, A. W., Kinematics of Machinery, McGraw-Hill, New York, 1917. Kolstee, H. M., Motion and Power, Prentice-Hall, Englewood Cliffs, N.J., 1982. Lent. D-, Analysis and Design of Mechanisms, 2nd ed., Prentice-Hall, Englewood Cliffs, N.J., 1970. Mabie, H. H., and F. W. Ocvirk, Mechanisms and Dynamics of Machinery, Wiley, New York, 1975. Martin, G. H., Kinematics and Dynamics of Machines, 2nd ed., McGrawHill, New York, 1982. Maxwell, R. L., Kinematics and Dynamics of Machinery, Prentice-Hall, Englewood Cliffs, N.J., 1960. Nielsen, K. L., Modern Trigonometry, Barnes and Noble, New York, 1966. Patton, W. J., Kinematics, Reston, Reston, Va., 1979. Paul, B., Kinematics and Dynamics of Planar Machinery, Prentice-Hall, Englewood Cliffs, N.J., 1979. Pearce, C. E., Principles of Mechanism, Wiley, New York, 1934. Prageman, I. H., Mechanism, International Textbook Press, Scranton 1943. Ramous, A. J., Applied Kinematics, Prentice-Hall, Englewood Cliffs 1972.

Pa.

N.J

Reuleaux, F., Kinematics of Machinery (trans. and ed. by A. B. W. Kennedy), reprinted by Dover, New York, 1963. Rosenauer, N., and A. H. Willis, Kinematics of Mechanisms, Associated General Publications, Sydney, Australia, 1953. Sahag, L. M., Kinematics of Machines. Ronald Press, New York, 1952. Schwamb, P., A. L. Merrill, et al., Elements of Mechanism. 6th ed., Wiley, New York, 1947. Shigley, J. E., Kinematic Analysis of Mechanisms. 2nded., McGraw-Hill New York, 1969. Shigley, J. E., and J. J. Vicker, Jr., Theory of Machines and Mechanisms McGraw-Hill, New York, 1980. '-’ Smith, W. G., Engineering Kinematics. McGraw-Hill, New York, 1923. Soni, A. H., Mechanism Synthesis and Analysis. McGraw-Hill, New York 1974. Tao, D. C., Fundamentals of Applied Kinematics, Addison-Wesley, ReadingMass., 1967. 6’

Selected References

441

Tuttle, S. B., Mechanisms for Engineering Design. Wiley, New York, 1967. Walker, J. D., Applied Mechanics. English Universities Press, London 1959. Wilson, C. E., Mechanism Design-Oriented Kinematics. American Tech¬ nical Society, Chicago, 1969. Woods, A. T., and A. W. Stahl, Kinematics, revised and rewritten by Philip K. Slaymaker, D. Van Nostrand, New York, 1926.

Index

Acceleration angular, 14, 21-29 center of, 143-144 Coriolis, 146-150, 169, 280, 298 definition, 14 image, 145, 175 linear, 14, 16-20, 29, 124-150 normal (radial), 125, 135 polygon, 16, 163-169 proportionality of, 133-134 relative, 16, 134-146 resultant, 127-132, 162 tangential, 29, 124, 126, 135 -time curve, curve, 196-211 uniform, 33

[Complex forms] polar, 234 rectangular, 234 trigonometric, 234 Complex number, 231 Component (of a vector), 49 effective, 52-55, 60 orthogonal (rectangular), 51 rotational, 52, 63-65, 140-142 translational, 52, 63-65, 140-142 Computer programs, 355 Conjugate, 231 Convention acceleration polygon, 162 velocity polygon, 104

Calculator keystroke language, 355

Deceleration (retardation), 14 Direct contact mechanism rolling contact, 83, 85 sliding contact, 68, 85 Displacement angular, 13, 16-40 definition, 13 linear, 13, 16-40 -time curve, curve, 195-211

Cam, 8 Cam-follower mechanism, 6, 113, 179, 221 Chain constrained, 5 kinematic, 4, 8 locked, 5 slider-crank, 5 unconstrained, 5 Circle diagram, 75, 87-91 Complex forms exponential, 234

Distance, 13 Double-rocker mechanism, 243 Drag-link mechanism, 242, 255 Driver, 6

443

444

Effective component (see Component) Equivalent linkage, 218-223

Fixed axis (center), 93, 95 FORTRAN language, 355 Four-bar mechanism (linkage), 4, 67, 88, 93, 95, 111, 157, 163, 189, 216, 242 crank-rocker, 242, 251 crossed phase, 258 double rocker, 243 drag link, 242, 255 Four-circle method, 144 Frame, 6

Gear, 8 Geneva mechanism, 276 Graphical differentiation, 194-204 Graphical integration, 204-211

Higher paired mechanism, 218

Instant axis, 84 Instant center definition, 75, 81 fixed, 84, 95 permanent, 84 imaginary, 84 obvious, 85 of acceleration, 143-144 Inversion, 8

Kennedy's theorem, 87, 90 Kinematic chain (see Chain) Kinematics, 1 Kinetic analysis, 1

Line of action, 66

Index

Line of centers, 85 Line of proportionality, 65, 78, 134, 141-142 Link expanded, 109, 172 floating, 106, 138, 172 Linkage, 5 Link extension, 75, 94

Machine, 3 Mechanics branches of, 2 Mechanism analysis, 1, 2 compound (complex), 5, 72, 75, 97, 175 definition, 3 direct contact, 8, 85 rolling contact, 8, 85 simple, 5, 75, 111 sliding contact, 8, 68, 85 synthesis, 2 Motion absolute, 9, 101 angular, 21, 26, 27 combined, 11, 81 continuous, 12 helical, 11 intermittent, 13 oscillation, 13 reciprocating, 12 rectilinear, 16, 20, 27 relative, 9, 87, 101 spherical, 11 three-dimensional, 11 uniform, 33

Normal acceleration (see Accel¬ eration) Normal acceleration construction, 213 Normal, common, 85

Index

Offset slider-crank (see Slidercrank mechanism) Orthogonal component (see Com¬ ponent) Oscillating cylinder (rocking block) engine, 8 mechanism, 289, 299

Pair higher, 8 lower, 8 Path, 13 Period, 14 Phase, 14 Pole (origin), 103, 162 Polygon acceleration, 161-163, 169 vector, 47, 104 velocity, 103-104 Program listings (see Computer programs)

Quick-return mechanism, 8, 69, 115, 169, 187 crank-shaper, 274, 281 oscillating beam, 274 Whitworth linkage, 274, 285

Radian, 15 Radius of rotation method, 93 Rest, 9 Resultant, 43, 48, 103 Rigid body principle, 60 Rolling contact, 83, 117 Rotating cylinder (rotating block) mechanism, 289, 302 Rotation, 9, 76

Scalar, 41 Scotch yoke, 68

445 Sense negative, 15 positive, 15 Slider-crank mechanism (linkage) central (in-line), 3, 65, 154, 167, 184, 223, 226, 269, 306, 314 negative offset, 260 positive offset, 260 Sliding contact, 68, 71 rule, 69 Sliding coupler mechanism (linkage) oscillating cylinder (rocking block), 289, 299 rotating cylinder (rotating block), 289, 302 Speed angular, 14 definition, 14

Tangent, common, 113 Tangential acceleration (see Accel¬ eration) Toggle mechanism, 119 Transfer point, 96 Translation, 9 curvilinear, 11 rectilinear, 11 Transmission, modes of, 8 Trigonometry review functions of an angle, 437 law of cosines, 440 law of sines, 440 law of tangents, 440 useful relationships, 441

Unit vector, 234, 235, 237, 249, 279, 309

Vector components (see Component) polygon, 165 properties of, 41

446

Velocity absolute, 103 angular, 13, 21, 26, 29 definition, 13 image, 111 linear, 13, 16, 20, 28 of sliding (slip), 71, 113 relative, 103, 105

Index

[Velocity] -time curve, curve, 33, 195-211 uniform, 33 variable, 33

Whitworth mechanism (linkage), 8, 274, 285

DATE DUE

1R2V95

DEMCO 38-297

TJ

175

Barton,

.529

1984

Lyndon O.,

1934-

Mec hanism analysis

IJ

-US9

barton

1984

i '

Lyndor' 0. ,

Meeha"ls"> analysl.

7'/-^7 ^71 I H

HR 2 4'95

ISOfeif

193tl

about the book . . . Arranged as a step-by-step study of kinematics of mechanisms, this impressive volume offers novel, simplified means of solving typical problems that arise in mechanism syn¬ thesis and analysis. Unlike many conventional sources, this self-contained work mainly uses basic algebra and trigonometry—minimizing the need for complicated calculus meth¬ ods! Mechanism Analysis begins with an introduction of key concepts: kinematic terminology, uniformly accelerated motion, and properties of vectors. The book then delves into graphical techniques for both velocity and acceleration analysis, and analytic techniques. There is a complete listing of ready-to-use computer and calculator programs for analyz¬ ing basic classes of mechanisms. Generous examples and practice problems combined with over 250 graphic illustrations make the entire work easy to master. Establishing an important, practical foundation in basic skills, Mechanism Analysis serves as a major reference for mechanical and design engineers. Advanced undergraduate students of mechanical engineering, engineering technology, mechanical analysis, and kinematics will benefit enormously from this volume as a supplement to their studies. Additionally, the book is an ideal source of information for in-house training programs, professional seminars, and society or association courses.

about the author... Lyndon O. Barton is a project engineer at E. I. du Pont de Nemours & Company, Wilmington, Delaware, and an adjunct instructor in the Mechanical Engineering Depart¬ ment at Delaware Technical Community College, Stanton. He received the B.S. degree in mechanical engineering (1966) from Howard University, Washington, D.C., and M.S. degree in mechanical and aerospace engineering (1972) from the University of Delaware. Mr. Barton’s industrial experience encompasses mechanical and process engineering as¬ pects in the design of du Pont’s textile and chemical plants. He has published various articles on mechanism analysis, strength of materials, and heat transfer, and holds four U.S. patents. Mr. Barton is a member of the Society of American Inventors. ' » Printed in the United States of America

ISBN: 0—8247—7086—2

marcel dekker, inc./newyork • basel