Advanced DynamicsAdvanced DynamicsRigid Body, Multibody,and Aerospace ApplicationsReza N. JazarSchool of Aerospace, Mechanical, and Manufacturing EngineeringRMIT UniversityMelbourne, AustraliaJOHN WILEY & SONS, INC.This book is printed on acid-free paper.Copyright c 2011 by John Wiley & Sons, Inc. All rights reserved.Published by John Wiley & Sons, Inc., Hoboken, New JerseyPublished simultaneously in CanadaNo part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form orby any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permittedunder Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permissionof the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright ClearanceCenter, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600, or on the webat www.copyright.com. Requests to the Publisher for permission should be addressed to the PermissionsDepartment, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201)748-6008, or online at www.wiley.com/go/permissions.Limit of Liability/Disclaimer of Warranty: While the publisher and the author have used their best efforts inpreparing this book, they make no representations or warranties with respect to the accuracy or completenessof the contents of this book and specically disclaim any implied warranties of merchantability or tnessfor a particular purpose. No warranty may be created or extended by sales representatives or written salesmaterials. The advice and strategies contained herein may not be suitable for your situation. You shouldconsult with a professional where appropriate. Neither the publisher nor the author shall be liable for any lossof prot or any other commercial damages, including but not limited to special, incidental, consequential,or other damages.For general information about our other products and services, please contact our Customer Care Departmentwithin the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317)572-4002.Wiley also publishes its books in a variety of electronic formats. Some content that appears in print maynot be available in electronic books. For more information about Wiley products, visit our web site atwww.wiley.com.Library of Congress Cataloging-in-Publication Data:Jazar, Reza N.Advanced dynamics : rigid body, multibody, and aerospace applications / Reza N. Jazar.p. cm.Includes index.ISBN 978-0-470-39835-7 (hardback); ISBN 978-0-470-89211-4 (ebk); ISBN 978-0-470-89212-1 (ebk);ISBN 978-0-470-89213-8 (ebk); ISBN 978-0-470-95002-9 (ebk); ISBN 978-0-470-95159-0 (ebk);ISBN 978-0-470-95176-7 (ebk)1. Dynamics. I. Title.TA352.J39 2011620.1

04dc222010039778Printed in the United States of America10 9 8 7 6 5 4 3 2 1The answer is waiting for the right question.To my daughterVazan,my sonKavosh,and my wife,MojganContentsPreface xiiiPart I Fundamentals 11 Fundamentals of Kinematics 31.1 Coordinate Frame and Position Vector 31.1.1 Triad 31.1.2 Coordinate Frame and Position Vector 41.1.3 Vector Denition 101.2 Vector Algebra 121.2.1 Vector Addition 121.2.2 Vector Multiplication 171.2.3 Index Notation 261.3 Orthogonal Coordinate Frames 311.3.1 Orthogonality Condition 311.3.2 Unit Vector 341.3.3 Direction of Unit Vectors 361.4 Differential Geometry 371.4.1 Space Curve 381.4.2 Surface and Plane 431.5 Motion Path Kinematics 461.5.1 Vector Function and Derivative 461.5.2 Velocity and Acceleration 511.5.3 Natural Coordinate Frame 541.6 Fields 771.6.1 Surface and Orthogonal Mesh 781.6.2 Scalar Field and Derivative 851.6.3 Vector Field and Derivative 92Key Symbols 100Exercises 1032 Fundamentals of Dynamics 1142.1 Laws of Motion 1142.2 Equation of Motion 1192.2.1 Force and Moment 1202.2.2 Motion Equation 125Note: A star () indicates a more advanced subject or example that is not designed for undergraduateteaching and can be dropped in the rst reading.viiviii Contents2.3 Special Solutions 1312.3.1 Force Is a Function of Time, F = F(t ) 1322.3.2 Force Is a Function of Position, F = F(x) 1412.3.3 Elliptic Functions 1482.3.4 Force Is a Function of Velocity, F = F(v) 1562.4 Spatial and Temporal Integrals 1652.4.1 Spatial Integral: Work and Energy 1652.4.2 Temporal Integral: Impulse and Momentum 1762.5 Application of Dynamics 1882.5.1 Modeling 1892.5.2 Equations of Motion 1972.5.3 Dynamic Behavior and Methods of Solution 2002.5.4 Parameter Adjustment 220Key Symbols 223Exercises 226Part II Geometric Kinematics 2413 Coordinate Systems 2433.1 Cartesian Coordinate System 2433.2 Cylindrical Coordinate System 2503.3 Spherical Coordinate System 2633.4 Nonorthogonal Coordinate Frames 2693.4.1 Reciprocal Base Vectors 2693.4.2 Reciprocal Coordinate Frame 2783.4.3 Inner and Outer Vector Product 2853.4.4 Kinematics in Oblique Coordinate Frames 2983.5 Curvilinear Coordinate System 3003.5.1 Principal and Reciprocal Base Vectors 3013.5.2 Principal Reciprocal Transformation 3113.5.3 Curvilinear Geometry 3203.5.4 Curvilinear Kinematics 3253.5.5 Kinematics in Curvilinear Coordinates 335Key Symbols 346Exercises 3474 Rotation Kinematics 3574.1 Rotation About Global Cartesian Axes 3574.2 Successive Rotations About Global Axes 3634.3 Global Roll PitchYaw Angles 3704.4 Rotation About Local Cartesian Axes 3734.5 Successive Rotations About Local Axes 3764.6 Euler Angles 379Contents ix4.7 Local Roll PitchYaw Angles 3914.8 Local versus Global Rotation 3954.9 General Rotation 3974.10 Active and Passive Rotations 4094.11 Rotation of Rotated Body 411Key Symbols 415Exercises 4165 Orientation Kinematics 4225.1 AxisAngle Rotation 4225.2 Euler Parameters 4385.3 Quaternion 4495.4 Spinors and Rotators 4575.5 Problems in Representing Rotations 4595.5.1 Rotation Matrix 4605.5.2 AxisAngle 4615.5.3 Euler Angles 4625.5.4 Quaternion and Euler Parameters 4635.6 Composition and Decomposition of Rotations 4655.6.1 Composition of Rotations 4665.6.2 Decomposition of Rotations 468Key Symbols 470Exercises 4716 Motion Kinematics 4776.1 Rigid-Body Motion 4776.2 Homogeneous Transformation 4816.3 Inverse and Reverse Homogeneous Transformation 4946.4 Compound Homogeneous Transformation 5006.5 Screw Motion 5176.6 Inverse Screw 5296.7 Compound Screw Transformation 5316.8 Pl ucker Line Coordinate 5346.9 Geometry of Plane and Line 5406.9.1 Moment 5406.9.2 Angle and Distance 5416.9.3 Plane and Line 5416.10 Screw and Pl ucker Coordinate 545Key Symbols 547Exercises 548x Contents7 Multibody Kinematics 5557.1 Multibody Connection 5557.2 Denavit Hartenberg Rule 5637.3 Forward Kinematics 5847.4 Assembling Kinematics 6157.5 Order-Free Rotation 6287.6 Order-Free Transformation 6357.7 Forward Kinematics by Screw 6437.8 Caster Theory in Vehicles 6497.9 Inverse Kinematics 662Key Symbols 684Exercises 686Part III Derivative Kinematics 6938 Velocity Kinematics 6958.1 Angular Velocity 6958.2 Time Derivative and Coordinate Frames 7188.3 Multibody Velocity 7278.4 Velocity Transformation Matrix 7398.5 Derivative of a Homogeneous Transformation Matrix 7488.6 Multibody Velocity 7548.7 Forward-Velocity Kinematics 7578.8 Jacobian-Generating Vector 7658.9 Inverse-Velocity Kinematics 778Key Symbols 782Exercises 7839 Acceleration Kinematics 7889.1 Angular Acceleration 7889.2 Second Derivative and Coordinate Frames 8109.3 Multibody Acceleration 8239.4 Particle Acceleration 8309.5 Mixed Double Derivative 8589.6 Acceleration Transformation Matrix 8649.7 Forward-Acceleration Kinematics 8729.8 Inverse-Acceleration Kinematics 874Key Symbols 877Exercises 878Contents xi10 Constraints 88710.1 Homogeneity and Isotropy 88710.2 Describing Space 89010.2.1 Conguration Space 89010.2.2 Event Space 89610.2.3 State Space 90010.2.4 StateTime Space 90810.2.5 Kinematic Spaces 91010.3 Holonomic Constraint 91310.4 Generalized Coordinate 92310.5 Constraint Force 93210.6 Virtual and Actual Works 93510.7 Nonholonomic Constraint 95210.7.1 Nonintegrable Constraint 95210.7.2 Inequality Constraint 96210.8 Differential Constraint 96610.9 Generalized Mechanics 97010.10 Integral of Motion 97610.11 Methods of Dynamics 99610.11.1 Lagrange Method 99610.11.2 Gauss Method 99910.11.3 Hamilton Method 100210.11.4 GibbsAppell Method 100910.11.5 Kane Method 101310.11.6 Nielsen Method 1017Key Symbols 1021Exercises 1024Part IV Dynamics 103111 Rigid Body and Mass Moment 103311.1 Rigid Body 103311.2 Elements of the Mass Moment Matrix 103511.3 Transformation of Mass Moment Matrix 104411.4 Principal Mass Moments 1058Key Symbols 1065Exercises 106612 Rigid-Body Dynamics 107212.1 Rigid-Body Rotational Cartesian Dynamics 107212.2 Rigid-Body Rotational Eulerian Dynamics 109612.3 Rigid-Body Translational Dynamics 1101xii Contents12.4 Classical Problems of Rigid Bodies 111212.4.1 Torque-Free Motion 111212.4.2 Spherical Torque-Free Rigid Body 111512.4.3 Axisymmetric Torque-Free Rigid Body 111612.4.4 Asymmetric Torque-Free Rigid Body 112812.4.5 General Motion 114112.5 Multibody Dynamics 115712.6 Recursive Multibody Dynamics 1170Key Symbols 1177Exercises 117913 Lagrange Dynamics 118913.1 Lagrange Form of Newton Equations 118913.2 Lagrange Equation and Potential Force 120313.3 Variational Dynamics 121513.4 Hamilton Principle 122813.5 Lagrange Equation and Constraints 123213.6 Conservation Laws 124013.6.1 Conservation of Energy 124113.6.2 Conservation of Momentum 124313.7 Generalized Coordinate System 124413.8 Multibody Lagrangian Dynamics 1251Key Symbols 1262Exercises 1264References 1280A Global Frame Triple Rotation 1287B Local Frame Triple Rotation 1289C Principal Central Screw Triple Combination 1291D Industrial Link DH Matrices 1293E Trigonometric Formula 1300Index 1305PrefaceThis book is arranged in such a way, and covers those materials, that I would haveliked to have had available as a student: straightforward, right to the point, analyzinga subject from different viewpoints, showing practical aspects and application of everysubject, considering physical meaning and sense, with interesting and clear examples.This book was written for graduate students who want to learn every aspect of dynam-ics and its application. It is based on two decades of research and teaching courses inadvanced dynamics, attitude dynamics, vehicle dynamics, classical mechanics, multi-body dynamics, and robotics.I know that the best way to learn dynamics is repeat and practice, repeat andpractice. So, you are going to see some repeating and much practicing in this book.I begin with fundamental subjects in dynamics and end with advanced materials. Iintroduce the fundamental knowledge used in particle and rigid-body dynamics. Thisknowledge can be used to develop computer programs for analyzing the kinematics,dynamics, and control of dynamic systems.The subject of rigid body has been at the heart of dynamics since the 1600sand remains alive with modern developments of applications. Classical kinematics anddynamics have their roots in the work of great scientists of the past four centuries whoestablished the methodology and understanding of the behavior of dynamic systems.The development of dynamic science, since the beginning of the twentieth century, hasmoved toward analysis of controllable man-made autonomous systems.LEVEL OF THE BOOKMore than half of the material is in common with courses in advanced dynamics,classical mechanics, multibody dynamics, and spacecraft dynamics. Graduate studentsin mechanical and aerospace engineering have the potential to work on projects thatare related to either of these engineering disciplines. However, students have not seenenough applications in all areas. Although their textbooks introduce rigid-body dynam-ics, mechanical engineering students only work on engineering applications whileaerospace engineering students only see spacecraft applications and attitude dynam-ics. The reader of this text will have no problem in analyzing a dynamic system in anyof these areas. This book bridges the gap between rigid-body, classical, multibody, andspacecraft dynamics for graduate students and specialists in mechanical and aerospaceengineering. Engineers and graduate students who read this book will be able to applytheir knowledge to a wide range of engineering disciplines.This book is aimed primarily at graduate students in engineering, physics, andmathematics. It is especially useful for courses in the dynamics of rigid bodies suchas advanced dynamics, classical mechanics, attitude dynamics, spacecraft dynamics,and multibody dynamics. It provides both fundamental and advanced topics on thexiiixiv Prefacekinematics and dynamics of rigid bodies. The whole book can be covered in twosuccessive courses; however, it is possible to jump over some sections and cover thebook in one course.The contents of the book have been kept at a fairly theoretical practical level.Many concepts are deeply explained and their use emphasized, and most of the relatedtheory and formal proofs have been explained. Throughout the book, a strong emphasisis put on the physical meaning of the concepts introduced. Topics that have beenselected are of high interest in the eld. An attempt has been made to expose thestudents to a broad range of topics and approaches.ORGANIZATION OF THE BOOKThe book begins with a review of coordinate systems and particle dynamics. Thisintroduction will teach students the importance of coordinate frames. Transformationand rotation theory along with differentiation theory in different coordinate frames willprovide the required background to learn rigid-body dynamics based on NewtonEulerprinciples. The method will show its applications in rigid-body and multibody dynam-ics. The Newton equations of motion will be transformed to Lagrangian equations asa bridge to analytical dynamics. The methods of Lagrange will be applied on particlesand rigid bodies.Through its examination of specialist applications highlighting the many differentaspects of dynamics, this text provides an excellent insight into advanced systemswithout restricting itself to a particular discipline. The result is essential reading forall those requiring a general understanding of the more advanced aspects of rigid-bodydynamics.The text is organized such that it can be used for teaching or for self-study. PartI Fundamentals, contains general preliminaries and provides a deep review of thekinematics and dynamics. A new classication of vectors is the highlight of Part I.Part II, Geometric Kinematics, presents the mathematics of the displacement ofrigid bodies using the matrix method. The order-free transformation theory, classi-cation of industrial links, kinematics of spherical wrists, and mechanical surgery ofmultibodies are the highlights of Part II.Part III, Derivative Kinematics, presents the mathematics of velocity and accel-eration of rigid bodies. The time derivatives of vectors in different coordinate frames,Raz acceleration, integrals of motion, and methods of dynamics are the highlights ofPart III.Part IV, Dynamics, presents a detailed discussion of rigid-body and Lagrangiandynamics. Rigid-body dynamics is studied from different viewpoints to provide differ-ent classes of solutions. Lagrangian mechanics is reviewed in detail from an appliedviewpoint. Multibody dynamics and Lagrangian mechanics in generalized coordinatesare the highlights of Part IV.METHOD OF PRESENTATIONThe structure of the presentation is in a factreasonapplication fashion. The fact isthe main subject we introduce in each section. Then the reason is given as a proof.Preface xvFinally the application of the fact is examined in some examples. The examples area very important part of the book because they show how to implement the knowledgeintroduced in the facts. They also cover some other material needed to expand thesubject.PREREQUISITESThe book is written for graduate students, so the assumption is that users are familiarwith the fundamentals of kinematics and dynamics as well as basic knowledge of linearalgebra, differential equations, and the numerical method.UNIT SYSTEMThe system of units adopted in this book is, unless otherwise stated, the InternationalSystem of Units (SI). The units of degree (deg) and radian (rad) are utilized for variablesrepresenting angular quantities.SYMBOLS Lowercase bold letters indicate a vector. Vectors may be expressed in an n-dimensional Euclidean space:r, s, d, a, b, cp, q, v, w, y, z, , , , , Uppercase bold letters indicate a dynamic vector or a dynamic matrix:F, M, I, L Lowercase letters with a hat indicate a unit vector. Unit vectors are not bolded: , , k, e, u, nI, J, K, e, e, e Lowercase letters with a tilde indicate a 3 3 skew symmetric matrix associatedto a vector: a =__0 a3 a2a3 0 a1a2 a1 0__ a =__a1a2a3__ An arrow above two uppercase letters indicates the start and end points of aposition vector:ON = a position vector from point O to point Nxvi Preface A double arrow above a lowercase letter indicates a 4 4 matrix associated to aquaternion:q =_____q0 q1 q2 q3q1 q0 q3 q2q2 q3 q0 q1q3 q2 q1 q0__q = q0+q1i +q2j +q3k The length of a vector is indicated by a nonbold lowercase letter:r = |r| a = |a| b = |b| s = |s| Capital letters A, Q, R, and T indicate rotation or transformation matrices:QZ, =__cos sin 0sin cos 00 0 1__ GTB =____c 0 s 10 1 0 0.5s 0 c 0.20 0 0 1__ Capital letter B is utilized to denote a body coordinate frame:B(oxyz), B(Oxyz), B1(o1x1y1z1) Capital letter G is utilized to denote a global, inertial, or xed coordinate frame:G, G(XYZ), G(OXYZ) Right subscript on a transformation matrix indicates the departure frames:TB = transformation matrix from frame B(oxyz) Left superscript on a transformation matrix indicates the destination frame:GTB = transformation matrix from frame B(oxyz)to frame G(OXYZ) Whenever there is no subscript or superscript, the matrices are shown in brackets:[T ] =____c 0 s 10 1 0 0.5s 0 c 0.20 0 0 1__ Left superscript on a vector denotes the frame in which the vector is expressed.That superscript indicates the frame that the vector belongs to, so the vector isexpressed using the unit vectors of that frame:Gr = position vector expressed in frame G(OXYZ)Preface xvii Right subscript on a vector denotes the tip point to which the vector is referred:GrP = position vector of point Pexpressed in coordinate frame G(OXYZ) Right subscript on an angular velocity vector indicates the frame to which theangular vector is referred:B = angular velocity of the body coordinate frame B(oxyz) Left subscript on an angular velocity vector indicates the frame with respect towhich the angular vector is measured:GB = angular velocity of the body coordinateframe B(oxyz)with respect to the global coordinate frame G(OXYZ) Left superscript on an angular velocity vector denotes the frame in which theangular velocity is expressed:B2G B1 = angular velocity of the bodycoordinate frame B1with respect to the global coordinate frame Gand expressed in body coordinate frame B2Whenever the subscript and superscript of an angular velocity are the same, weusually drop the left superscript:GB GGBAlso for position, velocity, and acceleration vectors, we drop the left subscriptsif it is the same as the left superscript:BBvP BvP If the right subscript on a force vector is a number, it indicates the number ofcoordinate frames in a serial robot. Coordinate frame Bi is set up at joint i +1:Fi = force vector at joint i +1 measured at the originof Bi(oxyz)At joint i there is always an action force Fi that link (i) applies on link (i +1)and a reaction force Fi that link (i +1) applies on link (i). On link (i) there isalways an action force Fi1 coming from link (i 1) and a reaction force Ficoming from link (i +1). The action force is called the driving force , and thereaction force is called the driven force. If the right subscript on a moment vector is a number, it indicates the number ofcoordinate frames in a serial robot. Coordinate frame Bi is set up at joint i +1:Mi = moment vector at joint i +1 measured at theorigin of Bi(oxyz)At joint i there is always an action moment Mi that link (i) applies on link(i +1), and a reaction moment Mi that link (i +1) applies on link (i). Onxviii Prefacelink (i) there is always an action moment Mi1 coming from link (i 1) and areaction moment Mi coming from link (i +1). The action moment is called thedriving moment , and the reaction moment is called the driven moment . Left superscript on derivative operators indicates the frame in which the derivativeof a variable is taken:Gddt x, GddtBrP, BddtGBrPIf the variable is a vector function and the frame in which the vector is dened isthe same as the frame in which a time derivative is taken, we may use the shortnotationGddtGrP = G rP,BddtBo rP = Bo rPand write equations simpler. For example,Gv =GddtGr(t ) = G r If followed by angles, lowercase c and s denote cos and sin functions in mathe-matical equations:c = cos s = sin Capital bold letter I indicates a unit matrix, which, depending on the dimensionof the matrix equation, could be a 3 3 or a 4 4 unit matrix. I3 or I4 are alsobeing used to clarify the dimension of I. For example,I = I3 =__1 0 00 1 00 0 1__ Two parallel joint axes are indicated by a parallel sign (). Two orthogonal joint axes are indicated by an orthogonal sign (). Two orthogonaljoint axes are intersecting at a right angle. Two perpendicular joint axes are indicated by a perpendicular sign (). Twoperpendicular joint axes are at a right angle with respect to their common normal.Part IFundamentalsThe required fundamentals of kinematics and dynamics are reviewed in this part.It should prepare us for the more advanced parts.1Fundamentals of KinematicsVectors and coordinate frames are human-made tools to study the motion of particlesand rigid bodies. We introduce them in this chapter to review the fundamentals ofkinematics.1.1 COORDINATE FRAME AND POSITION VECTORTo indicate the position of a point P relative to another point O in a three-dimensional(3D) space, we need to establish a coordinate frame and provide three relative coordi-nates. The three coordinates are scalar functions and can be used to dene a positionvector and derive other kinematic characteristics.1.1.1 TriadTake four non-coplanar points O, A, B, C and make three lines OA, OB, OC. Thetriad OABC is dened by taking the lines OA, OB, OC as a rigid body. The positionof A is arbitrary provided it stays on the same side of O. The positions of B and C aresimilarly selected. Now rotate OB about O in the plane OAB so that the angle AOBbecomes 90 deg. Next, rotate OC about the line in AOB to which it is perpendicularuntil it becomes perpendicular to the plane AOB. The new triad OABC is called anorthogonal triad.Having an orthogonal triad OABC, another triad OA

BC may be derived by movingA to the other side of O to make the opposite triad OA

BC. All orthogonal triads canbe superposed either on the triad OABC or on its opposite OA

BC.One of the two triads OABC and OA

BC can be dened as being a positive triadand used as a standard. The other is then dened as a negative triad. It is immaterialwhich one is chosen as positive; however, usually the right-handed convention is chosenas positive. The right-handed convention states that the direction of rotation from OAto OB propels a right-handed screw in the direction OC. A right-handed or positiveorthogonal triad cannot be superposed to a left-handed or negative triad. Therefore,there are only two essentially distinct types of triad. This is a property of 3D space.We use an orthogonal triad OABC with scaled lines OA, OB, OC to locate a pointin 3D space. When the three lines OA, OB, OC have scales, then such a triad is calleda coordinate frame.Every moving body is carrying a moving or body frame that is attached to the bodyand moves with the body. A body frame accepts every motion of the body and mayalso be called a local frame. The position and orientation of a body with respect toother frames is expressed by the position and orientation of its local coordinate frame.34 Fundamentals of KinematicsWhen there are several relatively moving coordinate frames, we choose one of themas a reference frame in which we express motions and measure kinematic information.The motion of a body may be observed and measured in different reference frames;however, we usually compare the motion of different bodies in the global referenceframe. A global reference frame is assumed to be motionless and attached to the ground.Example 1 Cyclic Interchange of Letters In any orthogonal triad OABC, cyclicinterchanging of the letters ABC produce another orthogonal triad superposable on theoriginal triad. Cyclic interchanging means relabeling A as B, B as C, and C as A orpicking any three consecutive letters from ABCABCABC . . . .Example 2 Independent Orthogonal Coordinate Frames Having only two typesof orthogonal triads in 3D space is associated with the fact that a plane has just twosides. In other words, there are two opposite normal directions to a plane. This mayalso be interpreted as: we may arrange the letters A, B, and C in just two orders whencyclic interchange is allowed:ABC, ACBIn a 4D space, there are six cyclic orders for four letters A, B, C, and D:ABCD, ABDC, ACBD, ACDB, ADBC, ADCBSo, there are six different tetrads in a 4D space.In an nD space there are (n 1)! cyclic orders for n letters, so there are (n 1)!different coordinate frames in an nD space.Example 3 Right-Hand Rule A right-handed triad can be identied by a right-handrule that states: When we indicate the OC axis of an orthogonal triad by the thumb ofthe right hand, the other ngers should turn from OA to OB to close our st.The right-hand rule also shows the rotation of Earth when the thumb of the righthand indicates the north pole.Push your right thumb to the center of a clock, then the other ngers simulate therotation of the clocks hands.Point your index nger of the right hand in the direction of an electric current.Then point your middle nger in the direction of the magnetic eld. Your thumb nowpoints in the direction of the magnetic force.If the thumb, index nger, and middle nger of the right hand are held so thatthey form three right angles, then the thumb indicates the Z-axis when the index ngerindicates the X-axis and the middle nger the Y -axis.1.1.2 Coordinate Frame and Position VectorConsider a positive orthogonal triad OABC as is shown in Figure 1.1. We select a unitlength and dene a directed line on OA with a unit length. A point P1 on OA is ata distance x from O such that the directed line OP1 from O to P1 is OP1 = x . The1.1 Coordinate Frame and Position Vector 5yxPOABCiz1 rjkP1P2P3D23Figure 1.1 A positive orthogonal triad OABC, unit vectors , , k, and a position vector r withcomponents x, y, z .directed line is called a unit vector on OA, the unit length is called the scale, pointO is called the origin, and the real number x is called the -coordinate of P1. Thedistance x may also be called the measure number of OP1. Similarly, we dene theunit vectors and k on OB and OC and use y and z as their coordinates, respectively.Although it is not necessary, we usually use the same scale for , , k and refer to OA,OB, OC by , , k and also by x, y, z .The scalar coordinates x, y, z are respectively the length of projections of P onOA, OB, and OC and may be called the components of r. The components x, y, z areindependent and we may vary any of them while keeping the others unchanged.A scaled positive orthogonal triad with unit vectors , , k is called an orthogonalcoordinate frame. The position of a point P with respect to O is dened by threecoordinates x, y, z and is shown by a position vector r = rP:r = rP = x +y +zk (1.1)To work with multiple coordinate frames, we indicate coordinate frames by a capitalletter, such as G and B, to clarify the coordinate frame in which the vector r isexpressed. We show the name of the frame as a left superscript to the vector:Br = x +y +zk (1.2)A vector r is expressed in a coordinate frame B only if its unit vectors , , k belongto the axes of B. If necessary, we use a left superscript B and show the unit vectorsas B , B , Bk to indicate that , , k belong to B:Br = x B +y B +z Bk (1.3)We may drop the superscript B as long as we have just one coordinate frame.The distance between O and P is a scalar number r that is called the length,magnitude, modulus, norm, or absolute value of the vector r:r = |r| =_x2+y2+z2(1.4)6 Fundamentals of KinematicsWe may dene a new unit vector ur on r and show r byr = r ur (1.5)The equation r = r ur is called the natural expression of r, while the equation r =x +y +zk is called the decomposition or decomposed expression of r over the axes , , k. Equating (1.1) and (1.5) shows that ur = x +y +zkr= x +y +zk_x2+y2+z2= x_x2+y2+z2 + y_x2+y2+z2 + z_x2+y2+z2k (1.6)Because the length of ur is unity, the components of ur are the cosines of the angles1, 2, 3 between ur and , , k, respectively:cos 1 = xr= x_x2+y2+z2 (1.7)cos 2 = yr= y_x2+y2+z2 (1.8)cos 3 = zr= z_x2+y2+z2 (1.9)The cosines of the angles 1, 2, 3 are called the directional cosines of ur, which, asis shown in Figure 1.1, are the same as the directional cosines of any other vector onthe same axis as ur, including r.Equations (1.7)(1.9) indicate that the three directional cosines are related by theequationcos21+cos22+cos33 = 1 (1.10)Example 4 Position Vector of a Point P Consider a point P with coordinates x = 3,y = 2, z = 4. The position vector of P isr = 3 +2 +4k (1.11)The distance between O and P isr = |r| =_32+22+42= 5.3852 (1.12)and the unit vector ur on r is ur = xr + yr + zrk = 35.3852 + 25.3852 + 45.3852k= 0.55708 +0.37139 +0.74278k (1.13)1.1 Coordinate Frame and Position Vector 7The directional cosines of ur arecos 1 = xr= 0.55708cos 2 = yr= 0.37139 (1.14)cos 3 = zr= 0.74278and therefore the angles between r and the x-, y-, z -axes are1 = cos1 xr= cos10.55708 = 0.97993 rad 56.146 deg2 = cos1 yr= cos10.37139 = 1.1903 rad 68.199 deg (1.15)3 = cos1 zr= cos10.74278 = 0.73358 rad 42.031 degExample 5 Determination of Position Figure 1.2 illustrates a point P in a scaledtriad OABC. We determine the position of the point P with respect to O by:1. Drawing a line PD parallel OC to meet the plane AOB at D2. Drawing DP1 parallel to OB to meet OA at P1yxPOABCizrjkP1DFigure 1.2 Determination of position.The lengths OP1, P1D, DP are the coordinates of P and determine its position intriad OABC. The line segment OP is a diagonal of a parallelepiped with OP1, P1D, DPas three edges. The position of P is therefore determined by means of a parallelepipedwhose edges are parallel to the legs of the triad and one of its diagonal is the linejoining the origin to the point.8 Fundamentals of KinematicsExample 6 Vectors in Different Coordinate Frames Figure 1.3 illustrates a globallyxed coordinate frame G at the center of a rotating disc O. Another smaller rotatingdisc with a coordinate frame B is attached to the rst disc at a position GdO. Point Pis on the periphery of the small disc.xGyYXBGrPGdoPOBrPFigure 1.3 A globally xed frame G at the center of a rotating disc O and a coordinate frameB at the center of a moving disc.If the coordinate frame G(OXYZ) is xed and B(oxyz ) is always parallel to G,the position vectors of P in different coordinate frames are expressed byGrP = X +Y +Zk = GrP_cos +sin _ (1.16)BrP = x +y +zk = BrP_cos +sin _ (1.17)The coordinate frame B in G may be indicated by a position vector Gdo:Gdo = do_cos +sin _ (1.18)Example 7 Variable Vectors There are two ways that a vector can vary: length anddirection. A variable-length vector is a vector in the natural expression where its mag-nitude is variable, such asr = r(t ) ur (1.19)The axis of a variable-length vector is xed.A variable-direction vector is a vector in its natural expression where the axis of itsunit vector varies. To show such a variable vector, we use the decomposed expressionof the unit vector and show that its directional cosines are variable:r = r ur(t ) = r_u1(t ) +u2(t ) +u3(t )k_ (1.20)_u21+u22+u23 = 1 (1.21)1.1 Coordinate Frame and Position Vector 9The axis and direction characteristics are not xed for a variable-direction vector, whileits magnitude remains constant. The end point of a variable-direction vector slides ona sphere with a center at the starting point.A variable vector may have both the length and direction variables. Such a vectoris shown in its decomposed expression with variable components:r = x(t ) +y(t ) +z(t )k (1.22)It can also be shown in its natural expression with variable length and direction:r = r(t ) ur(t ) (1.23)Example 8 Parallel and Perpendicular Decomposition of a Vector Consider a linel and a vector r intersecting at the origin of a coordinate frame such as shown is inFigure 1.4. The line l and vector r indicate a plane (l, r). We dene the unit vectors u

parallel to l and u perpendicular to l in the (l, r)-plane. If the angle between rand l is , then the component of r parallel to l isr

= r cos (1.24)and the component of r perpendicular to l isr = r sin (1.25)These components indicate that we can decompose a vector r to its parallel and perpen-dicular components with respect to a line l by introducing the parallel and perpendicularunit vectors u

and u:r = r

u

+r u = r cos u

+r sin u (1.26)yxrPlzOuuFigure 1.4 Decomposition of a vector r with respect to a line l into parallel and perpendicularcomponents.10 Fundamentals of Kinematics1.1.3 Vector DenitionBy a vector we mean any physical quantity that can be represented by a directed sectionof a line with a start point, such as O, and an end point, such as P. We may show avector by an ordered pair of points with an arrow, such as OP. The sign PP indicatesa zero vector at point P.Length and direction are necessary to have a vector; however, a vector may haveve characteristics:1. Length. The length of section OP corresponds to the magnitude of the physicalquantity that the vector is representing.2. Axis. A straight line that indicates the line on which the vector is. The vectoraxis is also called the line of action.3. End point . A start or an end point indicates the point at which the vector isapplied. Such a point is called the affecting point .4. Direction. The direction indicates at what direction on the axis the vector ispointing.5. Physical quantity. Any vector represents a physical quantity. If a physical quan-tity can be represented by a vector, it is called a vectorial physical quantity.The value of the quantity is proportional to the length of the vector. Havinga vector that represents no physical quantity is meaningless, although a vectormay be dimensionless.Depending on the physical quantity and application, there are seven types ofvectors:1. Vecpoint . When all of the vector characteristicslength, axis, end point, direc-tion, and physical quantityare specied, the vector is called a bounded vector,point vector, or vecpoint . Such a vector is xed at a point with no movability.2. Vecline. If the start and end points of a vector are not xed on the vector axis,the vector is called a sliding vector, line vector, or vecline. A sliding vector isfree to slide on its axis.3. Vecface. When the affecting point of a vector can move on a surface whilethe vector displaces parallel to itself, the vector is called a surface vector orvecface. If the surface is a plane, then the vector is a plane vector or veclane.4. Vecfree. If the axis of a vector is not xed, the vector is called a free vector,direction vector, or vecfree. Such a vector can move to any point of a speciedspace while it remains parallel to itself and keeps its direction.5. Vecpoline. If the start point of a vector is xed while the end point can slideon a line, the vector is a point-line vector or vecpoline. Such a vector has aconstraint variable length and orientation. However, if the start and end pointsof a vecpoline are on the sliding line, its orientation is constant.6. Vecpoface. If the start point of a vector is xed while the end point can slideon a surface, the vector is a point-surface vector or vecpoface. Such a vectorhas a constraint variable length and orientation. The start and end points of avecpoface may both be on the sliding surface. If the surface is a plane, thevector is called a point-plane vector or vecpolane.1.1 Coordinate Frame and Position Vector 11yyxrAzOByxrzOxzOr(a) (b) (c)yxzOr(d)Figure 1.5 (a) A vecpoint, (b) a vecline, (c) a vecface, and (d) a vecfree.yxrzOyxrzOyxzOr(a) (b) (c)Figure 1.6 (a) a vecpoline, (b) vecpoface, (c) vecporee.7. Vecporee. When the start point of a vector is xed and the end point canmove anywhere in a specied space, the vector is called a point-free vector orvecporee. Such a vector has a variable length and orientation.Figure 1.5 illustrates a vecpoint, a vecline, vecface, and a vecfree and Figure 1.6illustrates a vecpoline, a vecpoface, and a vecporee.We may compare two vectors only if they represent the same physical quantity andare expressed in the same coordinate frame. Two vectors are equal if they are compara-ble and are the same type and have the same characteristics. Two vectors are equivalentif they are comparable and the same type and can be substituted with each other.In summary, any physical quantity that can be represented by a directed sectionof a line with a start and an end point is a vector quantity. A vector may have vecharacteristics: length, axis, end point, direction, and physical quantity. The length anddirection are necessary. There are seven types of vectors: vecpoint, vecline, vecface,vecfree, vecpoline, vecpoface, and vecporee. Vectors can be added when they arecoaxial. In case the vectors are not coaxial, the decomposed expression of vectorsmust be used to add the vectors.Example 9 Examples of Vector Types Displacement is a vecpoint. Moving from apoint A to a point B is called the displacement. Displacement is equal to the differenceof two position vectors. A position vector starts from the origin of a coordinate frame12 Fundamentals of Kinematicsand ends as a point in the frame. If point A is at rA and point B at rB, then displacementfrom A to B isrA/B = BrA = rArB (1.27)Force is a vecline. In Newtonian mechanics, a force can be applied on a body atany point of its axis and provides the same motion.Torque is an example of vecfree. In Newtonian mechanics, a moment can be appliedon a body at any point parallel to itself and provides the same motion.A space curve is expressed by a vecpoline, a surface is expressed by a vecpoface,and a eld is expressed by a vecporee.Example 10 Scalars Physical quantities which can be specied by only a numberare called scalars. If a physical quantity can be represented by a scalar, it is calleda scalaric physical quantity. We may compare two scalars only if they represent thesame physical quantity. Temperature, density, and work are some examples of scalaricphysical quantities.Two scalars are equal if they represent the same scalaric physical quantity and theyhave the same number in the same system of units. Two scalars are equivalent if wecan substitute one with the other. Scalars must be equal to be equivalent.1.2 VECTOR ALGEBRAMost of the physical quantities in dynamics can be represented by vectors. Vector addi-tion, multiplication, and differentiation are essential for the development of dynamics.We can combine vectors only if they are representing the same physical quantity, theyare the same type, and they are expressed in the same coordinate frame.1.2.1 Vector AdditionTwo vectors can be added when they are coaxial . The result is another vector on thesame axis with a component equal to the sum of the components of the two vectors.Consider two coaxial vectors r1 and r2 in natural expressions:r1 = r1 ur r2 = r2 ur (1.28)Their addition would be a new vector r3 = r3 ur that is equal tor3 = r1+r2 = (r1+r2) ur = r3 ur (1.29)Because r1 and r2 are scalars, we have r1+r2 = r1+r2, and therefore, coaxial vectoraddition is commutative,r1+r2 = r2+r1 (1.30)and also associative,r1+(r2+r3) = (r1+r2) +r3 (1.31)1.2 Vector Algebra 13When two vectors r1 and r2 are not coaxial, we use their decomposed expressionsr1 = x1 +y1 +z1k r2 = x2 +y2 +z2k (1.32)and add the coaxial vectors x1 by x2 , y1 by y2 , and z1k by z2k to write the resultas the decomposed expression of r3 = r1+r2:r3 = r1+r2=_x1 +y1 +z1k_+_x2 +y2 +z2k_=_x1 +x2 _+_y1 +y2 _+_z1k +z2k_= (x1+x2) +(y1+y2) +(z1+z2)k= x3 +y3 +z3k (1.33)So, the sum of two vectors r1 and r2 is dened as a vector r3 where its componentsare equal to the sum of the associated components of r1 and r2. Figure 1.7 illustratesvector addition r3 = r1+r2 of two vecpoints r1 and r2.Subtraction of two vectors consists of adding to the minuend the subtrahend withthe opposite sense:r1r2 = r1+(r2) (1.34)The vectors r2 and r2 have the same axis and length and differ only in having oppositedirection.If the coordinate frame is known, the decomposed expression of vectors may alsobe shown by column matrices to simplify calculations:r1 = x1 +y1 +z1k =__x1y1z1__ (1.35)X YZGr1r2r3y1y2y3z1z2z3x3Figure 1.7 Vector addition of two vecpoints r1 and r2.14 Fundamentals of Kinematicsr2 = x2 +y2 +z2k =__x2y2z2__ (1.36)r3 = r1+r2 =__x1y1z1__+__x2y2z2__=__x1+x2y1+y2z1+z2__ (1.37)Vectors can be added only when they are expressed in the same frame. Thus, avector equation such asr3 = r1+r2 (1.38)is meaningless without indicating that all of them are expressed in the same frame,such thatBr3 = Br1+ Br2 (1.39)The three vectors r1, r2, and r3 are coplanar, and r3 may be considered as thediagonal of a parallelogram that is made by r1, r2.Example 11 Displacement of a Point Point P moves from the origin of a globalcoordinate frame G to a point at (1, 2, 0) and then moves to (4, 3, 0). If we express therst displacement by a vector r1 and its nal position by r3, the second displacementis r2, wherer2 = r3r1 =__430____120__=__310__ (1.40)Example 12 Vector Interpolation Problem Having two digits n1 and n2 as the startand the nal interpolants, we may dene a controlled digit n with a variable q such thatn =_n1 q = 0n2 q = 1 0 q 1 (1.41)Dening or determining such a controlled digit is called the interpolation problem.There are many functions to be used for solving the interpolation problem. Linearinterpolation is the simplest and is widely used in engineering design, computergraphics, numerical analysis, and optimization:n = n1(1 q) +n2q (1.42)The control parameter q determines the weight of each interpolants n1 and n2 in theinterpolated n. In a linear interpolation, the weight factors are proportional to thedistance of q from 1 and 0.1.2 Vector Algebra 15XZOq(1 q)bar1r2rGFigure 1.8 Vector linear interpolation.Employing the linear interpolation technique, we may dene a vector r = r (q) tointerpolate between the interpolant vectors r1 and r2:r = (1 q)r1+qr2 =__x1(1 q) +qx2y1 (1 q) +qy2z1(1 q) +qz2__ (1.43)In this interpolation, we assumed that equal steps in q results in equal steps in r betweenr1 and r2. The tip point of r will move on a line connecting the tip points of r1 andr2, as is shown in Figure 1.8.We may interpolate the vectors r1 and r2 by interpolating the angular distance between r1 and r2:r = sin_(1 q)_sin r1+ sin (q)sin r2 (1.44)To derive Equation (1.44), we may start withr = ar1+br2 (1.45)and nd a and b from the following trigonometric equations:a sin (q) b sin_(1 q)_= 0 (1.46)a cos (q) +b cos_(1 q)_= 1 (1.47)Example 13 Vector Addition and Linear Space Vectors and adding operation makea linear space because for any vectors r1, r2 we have the following properties:1. Commutative:r1+r2 = r2+r1 (1.48)2. Associative:r1+(r2+r3) = (r1+r2) +r3 (1.49)16 Fundamentals of Kinematics3. Null element:0 +r = r (1.50)4. Inverse element:r +(r) = 0 (1.51)Example 14 Linear Dependence and Independence The n vectors r1, r2, r3, . . . , rnare linearly dependent if there exist n scalars c1, c2, c3, . . . , cn not all equal to zerosuch that a linear combination of the vectors equals zero:c1r1+c2r2+c3r3+ +cnrn = 0 (1.52)The vectors r1, r2, r3, . . . , rn are linearly independent if they are not linearly dependent,and it means the n scalars c1, c2, c3, . . . , cn must all be zero to have Equation (1.52):c1 = c2 = c3 = = cn = 0 (1.53)Example 15 Two Linearly Dependent Vectors Are Colinear Consider two linearlydependent vectors r1 and r2:c1r1+c2r2 = 0 (1.54)If c1 = 0, we haver1 = c2c1r2 (1.55)and if c2 = 0, we haver2 = c1c2r1 (1.56)which shows r1 and r2 are colinear.Example 16 Three Linearly Dependent Vectors Are Coplanar Consider three linearlydependent vectors r1, r2, and r3,c1r1+c2r2+c3r3 = 0 (1.57)where at least one of the scalars c1, c2, c3, say c3, is not zero; thenr3 = c1c3r1 c2c3r2 (1.58)which shows r3 is in the same plane as r1 and r2.1.2 Vector Algebra 171.2.2 Vector MultiplicationThere are three types of vector multiplications for two vectors r1 and r2:1. Dot, Inner, or Scalar Productr1 r2 =__x1y1z1____x2y2z2__= x1x2+y1y2+z1z2= r1r2 cos (1.59)The inner product of two vectors produces a scalar that is equal to the productof the length of individual vectors and the cosine of the angle between them.The vector inner product is commutative in orthogonal coordinate frames,r1 r2 = r2 r1 (1.60)The inner product is dimension free and can be calculated in n-dimensionalspaces. The inner product can also be performed in nonorthogonal coordinatesystems.2. Cross, Outer, or Vector Productr3 = r1r2 =__x1y1z1____x2y2z2__=__y1z2y2z1x2z1x1z2x1y2x2y1__= (r1r2 sin ) ur3 = r3 ur3 (1.61) ur3 = ur1 ur2 (1.62)The outer product of two vectors r1 and r2 produces another vector r3 thatis perpendicular to the plane of r1, r2 such that the cycle r1r2r3 makes aright-handed triad. The length of r3 is equal to the product of the length ofindividual vectors multiplied by the sine of the angle between them. Hence r3is numerically equal to the area of the parallelogram made up of r1 and r2.The vector inner product is skew commutative or anticommutative:r1r2 = r2r1 (1.63)The outer product is dened and applied only in 3D space. There is noouter product in lower or higher dimensions than 3. If any vector of r1and r2is in a lower dimension than 3D, we must make it a 3D vector by adding zerocomponents for missing dimensions to be able to perform their outer product.3. Quaternion Productr1r2 = r1r2r1 r2 (1.64)We will talk about the quaternion product in Section 5.3.18 Fundamentals of KinematicsIn summary, there are three types of vector multiplication: inner, outer, and quater-nion products, of which the inner product is the only one with commutative property.Example 17 Geometric Expression of Inner Products Consider a line l and a vectorr intersecting at the origin of a coordinate frame as is shown in Figure 1.9. If the anglebetween r and l is , the parallel component of r to l isr

= OA = r cos (1.65)This is the length of the projection of r on l . If we dene a unit vector ul on l by itsdirection cosines 1, 2, 3, ul = u1 +u2 +u3k =__u1u2u3__=__cos 1cos 2cos 3__ (1.66)then the inner product of r and ul isr ul = r

= r cos (1.67)We may show r by using its direction cosines 1, 2, 3,r = r ur = x +y +zk = r__x/ry/rz/r__= r__cos 1cos 2cos 3__ (1.68)Then, we may use the result of the inner product of r and ul,r ul = r__cos 1cos 2cos 3____cos 1cos 2cos 3__= r (cos 1 cos 1+cos 2 cos 2+cos 3 cos 3) (1.69)to calculate the angle between r and l based on their directional cosines:cos = cos 1 cos 1+cos 2 cos 2+cos 3 cos 3 (1.70)yxrPOAlBzulFigure 1.9 A line l and a vector r intersecting at the origin of a coordinate frame.1.2 Vector Algebra 19So, the inner product can be used to nd the projection of a vector on a given line. Itis also possible to use the inner product to determine the angle between two givenvectors r1 and r2 ascos = r1 r2r1r2= r1 r2r1 r1r2 r2(1.71)Example 18 Power 2 of a Vector By writing a vector r to a power 2, we mean theinner product of r to itself:r2= r r =__xyz____xyz__= x2+y2+z2= r2(1.72)Using this denition we can write(r1+r2)2= (r1+r2) (r1+r2) = r21+2r1 r2+r22 (1.73)(r1r2) (r1+r2) = r21r22 (1.74)There is no meaning for a vector with a negative or positive odd exponent.Example 19 Unit Vectors and Inner and Outer Products Using the set of unit vectors , , k of a positive orthogonal triad and the denition of inner product, we conclude that 2= 1 2= 1 k2= 1 (1.75)Furthermore, by denition of the vector product we have = _ _= k (1.76) k = _k _= (1.77)k = _ k_= (1.78)It might also be useful if we have these equalities: = 0 k = 0 k = 0 (1.79) = 0 = 0 k k = 0 (1.80)Example 20 Vanishing Dot Product If the inner product of two vectors a andb is zero,a b = 0 (1.81)then either a = 0 or b = 0, or a and b are perpendicular.20 Fundamentals of KinematicsExample 21 Vector Equations Assume x is an unknown vector, k is a scalar, and a,b, and c are three constant vectors in the following vector equation:kx +(b x) a = c (1.82)To solve the equation for x, we dot product both sides of (1.82) by b:kx b +(x b) (a b) = c b (1.83)This is a linear equation for x b with the solutionx b = c bk +a b (1.84)providedk +a b = 0 (1.85)Substituting (1.84) in (1.82) provides the solution x:x = 1kc c bk (k +a b)a (1.86)An alternative method is decomposition of the vector equation along the axes , , k of the coordinate frame and solving a set of three scalar equations to nd thecomponents of the unknown vector.Assume the decomposed expression of the vectors x, a, b, and c arex =_xyz_ a =_a1a2a3_ b =_b1b2b3_ c =_c1c2c3_ (1.87)Substituting these expressions in Equation (1.82),k__xyz__+____b1b2b3____xyz______a1a2a3__=__c1c2c3__ (1.88)provides a set of three scalar equations__k +a1b1 a1b2 a1b3a2b1 k +a2b2 a2b3a3b1 a3b2 k +a3b3____xyz__=__c1c2c3__ (1.89)that can be solved by matrix inversion:_xyz_=__k +a1b1 a1b2 a1b3a2b1 k +a2b2 a2b3a3b1 a3b2 k +a3b3__1__c1c2c3__=________kc1a1b2c2+a2b2c1a1b3c3+a3b3c1k (k +a1b1+a2b2+a3b3)kc2+a1b1c2a2b1c1a2b3c3+a3b3c2k (k +a1b1+a2b2+a3b3)kc3+a1b1c3a3b1c1+a2b2c3a3b2c2k (k +a1b1+a2b2+a3b3)__(1.90)Solution (1.90) is compatible with solution (1.86).1.2 Vector Algebra 21Example 22 Vector Addition, Scalar Multiplication, and Linear Space Vector addi-tion and scalar multiplication make a linear space, becausek1 (k2r) = (k1k2) r (1.91)(k1+k2) r = k1r +k2r (1.92)k (r1+r2) = kr1+kr2 (1.93)1 r = r (1.94)(1) r = r (1.95)0 r = 0 (1.96)k 0 = 0 (1.97)Example 23 Vanishing Condition of a Vector Inner Product Consider three non-coplanar constant vectors a, b, c and an arbitrary vector r. Ifa r = 0 b r = 0 c r = 0 (1.98)thenr = 0 (1.99)Example 24 Vector Product Expansion We may prove the result of the inner andouter products of two vectors by using decomposed expression and expansion:r1 r2 =_x1 +y1 +z1k__x2 +y2 +z2k_= x1x2 +x1y2 +x1z2 k+y1x2 +y1y2 +y1z2 k+z1x2k +z1y2k +z1z2k k= x1x2+y1y2+z1z2 (1.100)r1r2 =_x1 +y1 +z1k__x2 +y2 +z2k_= x1x2 +x1y2 +x1z2 k+y1x2 +y1y2 +y1z2 k+z1x2k +z1y2k +z1z2k k= (y1z2y2z1) +(x2z1x1z2) +(x1y2x2y1)k (1.101)We may also nd the outer product of two vectors by expanding a determinant andderive the same result as Equation (1.101):r1r2 = kx1 y1 z1x2 y2 z2(1.102)22 Fundamentals of KinematicsExample 25 baccab Rule If a, b, c are three vectors, we may expand their triplecross product and show thata (b c) = b(a c) c (a b) (1.103)because__a1a2a3______b1b2b3____c1c2c3____=__a2 (b1c2b2c1) +a3(b1c3b3c1)a3 (b2c3b3c2) a1(b1c2b2c1)a1(b1c3b3c1) a2 (b2c3b3c2)__=__b1(a1c1+a2c2+a3c3) c1(a1b1+a2b2+a3b3)b2(a1c1+a2c2+a3c3) c2(a1b1+a2b2+a3b3)b3(a1c1+a2c2+a3c3) c3(a1b1+a2b2+a3b3)__ (1.104)Equation (1.103) may be referred to as the baccab rule, which makes it easy toremember. The baccab rule is the most important in 3D vector algebra. It is the keyto prove a great number of other theorems.Example 26 Geometric Expression of Outer Products Consider the free vectors r1from A to B and r2 from A to C, as are shown in Figure 1.10:r1 =__130__=10__0.316230.948680__ (1.105)r2 =__102.5__= 2.6926__0.3713900.92847__ (1.106)xyr1zr2r3A BCDFigure 1.10 The cross product of the two free vectors r1 and r2 and the resultant r3.1.2 Vector Algebra 23The cross product of the two vectors is r3:r3 = r1r2 =__7.52.53__= 8.4558__0.886970.295660.35479__= r3 ur3 = (r1r2 sin ) ur3 (1.107) ur3 = ur1 ur2 =__0.886970.295660.35479__ (1.108)where r3 = 8.4558 is numerically equivalent to the area A of the parallelogram ABCDmade by the sides AB and AC:AABCD = |r1r2| = 8.4558 (1.109)The area of the triangle ABC is A/2. The vector r3 is perpendicular to this plane and,hence, its unit vector ur3 can be used to indicate the plane ABCD.Example 27 Scalar Triple Product The dot product of a vector r1 with the crossproduct of two vectors r2 and r3 is called the scalar triple product of r1, r2, and r3.The scalar triple product can be shown and calculated by a determinant:r1 (r2r3) = r1 r2r3 =x1 y1 z1x2 y2 z2x3 y3 z3(1.110)Interchanging two rows (or columns) of a matrix changes the sign of its determinant.So, we may conclude that the scalar triple product of three vectors r1, r2, r3 is alsoequal tor1 r2r3 = r2 r3r1 = r3 r1r2= r1r2 r3 = r2r3 r1 = r3r1 r2= r1 r3r2 = r2 r1r3 = r3 r2r1= r1r3 r2 = r2r1 r3 = r3r2 r1 (1.111)Because of Equation (1.111), the scalar triple product of the vectors r1, r2, r3 can beshown by the short notation [r1r2r3]:[r1r2r3] = r1 r2r3 (1.112)This notation gives us the freedom to set the position of the dot and cross product signsas required.If the three vectors r1, r2, r3 are position vectors, then their scalar triple productgeometrically represents the volume of the parallelepiped formed by the three vectors.Figure 1.11 illustrates such a parallelepiped for three vectors r1, r2, r3.24 Fundamentals of Kinematicsyxr1Ozr2r3Figure 1.11 The parallelepiped made by three vectors r1, r2, r3.Example 28 Vector Triple Product The cross product of a vector r1 with the crossproduct of two vectors r2 and r3 is called the vector triple product of r1, r2, and r3.The baccab rule is always used to simplify a vector triple product:r1(r2r3) = r2(r1 r3) r3(r1 r2) (1.113)Example 29 Norm and Vector Space Assume r, r1, r2, r3 are arbitrary vectorsand c, c1, c3 are scalars. The norm of a vector r is dened as a real-valued functionon a vector space v such that for all {r1, r2} V and all c R we have:1. Positive denition: r >0 if r = 0 and r = 0 if r = 0.2. Homogeneity: cr = c r.3. Triangle inequality: r1+r2 = r1 +r2.The denition of norm is up to the investigator and may vary depending on theapplication. The most common denition of the norm of a vector is the length:r = |r| =_r21 +r22 +r23 (1.114)The set v with vector elements is called a vector space if the following conditionsare fullled:1. Addition: If {r1, r2} V and r1+r2 = r, then r V.2. Commutativity: r1+r2 = r2+r1.3. Associativity: r1+(r2+r3) = (r1+r2) +r3 and c1(c2r) = (c1c2) r.4. Distributivity: c (r1+r2) = cr1+cr2 and (c1+c2) r = c1r +c2r.5. Identity element: r +0 = r, 1r = r, and r r = r +(1) r = 0.Example 30 Nonorthogonal Coordinate Frame It is possible to dene a coordi-nate frame in which the three scaled lines OA, OB, OC are nonorthogonal. Dening1.2 Vector Algebra 25three unit vectors b1, b2, and b3 along the nonorthogonal non-coplanar axes OA, OB,OC, respectively, we can express any vector r by a linear combination of the threenon-coplanar unit vectors b1, b2, and b3 asr = r1b1+r2b2+r3b3 (1.115)where, r1, r2, and r3 are constant.Expression of the unit vectors b1, b2, b3 and vector r in a Cartesian coordinateframe isr = x +y +zk (1.116)b1 = b11 +b12 +b13k (1.117)b2 = b21 +b22 +b23k (1.118)b3 = b31 +b32 +b33k (1.119)Substituting (1.117)(1.119) in (1.115) and comparing with (1.116) show that__xyz__=__b11 b12 b13b21 b22 b23b31 b32 b33____r1r2r3__ (1.120)The set of equations (1.120) may be solved for the components r1, r2, and r3:__r1r2r3__=__b11 b12 b13b21 b22 b23b31 b32 b33__1__xyz__ (1.121)We may also express them by vector scalar triple product:r1 = 1b11 b12 b13b21 b22 b23b31 b32 b33x y zb21 b22 b23b31 b32 b33= r b2 b3b1 b2 b3(1.122)r2 = 1b11 b12 b13b21 b22 b23b31 b32 b33b11 b12 b13x y zb31 b32 b33= r b3 b1b1 b2 b3(1.123)r3 = 1b11 b12 b13b21 b22 b23b31 b32 b33b11 b12 b13b21 b22 b23x y z= r b1 b2b1 b2 b3(1.124)The set of equations (1.120) is solvable provided b1 b2 b3 = 0, which means b1,b2, b3 are not coplanar.26 Fundamentals of Kinematics1.2.3 Index NotationWhenever the components of a vector or a vector equation are structurally similar, wemay employ the summation sign, , and show only one component with an index tobe changed from 1 to 2 and 3 to indicate the rst, second, and third components. Theaxes and their unit vectors of the coordinate frame may also be shown by x1, x2, x3 and u1, u2, u3 instead of x, y, z and , ,k. This is called index notation and may simplifyvector calculations.There are two symbols that may be used to make the equations even more concise:1. Kronecker delta ij:ij =_1 i = j0 i = j_= ji (1.125)It states that jk = 1 if j = k and jk = 0 if j = k.2. Levi-Civita symbol ijk:

ijk = 12(i j)(j k)(k i) i, j, k = 1, 2, 3 (1.126)It states that ijk = 1 if i , j , k is a cyclic permutation of 1, 2, 3, ijk = 1 if i ,j , k is a cyclic permutation of 3, 2, 1, and ijk = 0 if at least two of i , j , k areequal. The Levi-Civita symbol is also called the permutation symbol .The Levi-Civita symbol ijk can be expanded by the Kronecker delta ij:3

k=1

ijk

mnk = imjninjm (1.127)This relation between and is known as the edelt a or delt a identity.Using index notation, the vectors a and b can be shown asa = a1 +a2 +a3k =3

i=1ai ui (1.128)b = b1 +b2 +b3k =3

i=1bi ui (1.129)and the inner and outer products of the unit vectors of the coordinate system as uj uk = jk (1.130) uj uk = ijk ui (1.131)Example 31 Fundamental Vector Operations and Index Notation Index notationsimplies the vector equations. By index notation, we show the elements ri,i = 1, 2, 3 instead of indicating the vector r. The fundamental vector operations byindex notation are:1.2 Vector Algebra 271. Decomposition of a vector r:r =3

i=1ri ui (1.132)2. Orthogonality of unit vectors: ui uj = ij ui uj = ijk uk (1.133)3. Projection of a vector r on ui:r uj =3

i=1ri ui uj =3

i=1riij = rj (1.134)4. Scalar, dot, or inner product of vectors a and b:a b =3

i=1ai ui 3

j=1bj uj =3

j=13

i=1aibj_ ui uj_=3

j=13

i=1aibjij=3

i=1aibi (1.135)5. Vector, cross, or outer product of vectors a and b:a b =3

j=13

k=1

ijk uiajbk (1.136)6. Scalar triple product of vectors a, b, and c:a b c = [abc] =3

k=13

j=13

i=1

ijkajbjck (1.137)Example 32 Levi-Civita Density and Unit Vectors The Levi-Civita symbol ijk, alsocalled the e tensor, Levi-Civita density, and permutation tensor and may be denedby the clearer expression

ijk =___1 ijk = 123, 231, 3120 i = j or j = k or k = 11 ijk = 321, 213, 132(1.138)can be shown by the scalar triple product of the unit vectors of the coordinate system,

ijk =_ ui uj uk_= ui uj uk (1.139)and therefore,

ijk = jki = kij = kji = jik = ikj (1.140)28 Fundamentals of KinematicsThe product of two Levi-Civita densities is

ijk

lmn =il im injl jm jnkl km kni, j, k, l, m, n = 1, 2, 3 (1.141)If k = l, we have3

k=1

ijk

mnk =im injm jn= imjn injm (1.142)and if also j = n, then3

k=13

j=1

ijk

mjk = 2im (1.143)and nally, if also i = m, we have3

k=13

j=13

i=1

ijk

ijk = 6 (1.144)Employing the permutation symbol ijk, we can show the vector scalar tripleproduct asa b c =3

i=13

j=13

k=1

ijkaibjck =3

i,j,k=1

ijkaibjck (1.145)Example 33 Einstein Summation Convention The Einstein summation conventionimplies that we may not show the summation symbol if we agree that there is a hiddensummation symbol for every repeated index over all possible values for that index.In applied kinematics and dynamics, we usually work in a 3D space, so the range ofsummation symbols are from 1 to 3. Therefore, Equations (1.135) and (1.136) may beshown more simply asd = aibi (1.146)ci = ijkajbk (1.147)and the result of a b c asa b c =3

i=1ai3

j=13

k=1

ijkbjck =3

i=13

j=13

k=1

ijkaibjck= ijkaibjck (1.148)The repeated index in a term must appear only twice to dene a summation rule. Suchan index is called a dummy index because it is immaterial what character is used forit. As an example, we haveaibi = ambm = a1b1+a2b2+a3b3 (1.149)1.2 Vector Algebra 29Example 34 A Vector Identity We may use the index notation and verify vectoridentities such as(a b) (c d) = c (d a b) d(c a b) (1.150)Let us assume thata b = p = pi ui (1.151)c d = q = qi ui (1.152)The components of these vectors arepi = ijkajbk (1.153)qi = ijkcjdk (1.154)and therefore the components of p q arer = p q = ri ui (1.155)ri = ijkpjqk = ijk

jmn

krsambncrds= ijk

rsk

jmnambncrds=_irjs isjr_

jmnambncrds= jmn_(crir)_dsjs_ambn_crjr_(dsis) ambn_= jmn_ambncidj ambncjdi_= ci_

jmndjambn_di_

jmncjambn_ (1.156)so we haver = c (d a b) d(c a b) (1.157)Example 35 baccab Rule and Delta Identity Employing the delta identity(1.127), we can prove the baccab rule (1.103):a (b c) = ijkaibkcm

njm un = ijk

jmnaibkcm un= (imkn inkm) aibkcm un= ambncm unanbmcm un= amcmb bmcmc = b(a c) c (a b) (1.158)Example 36 Series Solution for Three-Body Problem Consider three pointmasses m1, m2, and m3 each subjected to Newtonian gravitational attraction from theother two particles. Let us indicate them by position vectors X1, X2, and X3 withrespect to their mass center C. If their position and velocity vectors are given at a timet0, how will the particles move? This is called the three-body problem.30 Fundamentals of KinematicsThis is one of the most celebrated unsolved problems in dynamics. The three-bodyproblem is interesting and challenging because it is the smallest n-body problem thatcannot be solved mathematically. Here we present a series solution and employ indexnotation to provide concise equations. We present the expanded form of the equationsin Example 177.The equations of motion of m1, m2, and m3 areXi = G3

j=1mjXi XjXji3 i = 1, 2, 3 (1.159)Xij = Xj Xi (1.160)Using the mass center as the origin implies3

i=1GiXi = 0 Gi = Gmi i = 1, 2, 3 (1.161)G = 6.67259 1011m3kg1s2(1.162)Following Belgium-American mathematician Roger Broucke (19322005), we usethe relative position vectors x1, x2, x3 to derive the most symmetric form of the three-body equations of motion:xi = ijk_Xk Xj_ i = 1, 2, 3 (1.163)Using xi, the kinematic constraint (1.161) reduces to3

i=1xi = 0 (1.164)The absolute position vectors in terms of the relative positions aremXi = ijk_mkxjj mjxk_ i = 1, 2, 3 (1.165)m = m1+m2+m3 (1.166)Substituting Equation (1.165) in (1.161), we have xi = Gmxi|xi|3 +Gi3

j=1xjxj3 i = 1, 2, 3 (1.167)We are looking for a series solution of Equations (1.167) in the following form:xi (t ) = xi0 + xi0 (t t0) + xi0(t t0)22! +...xi0(t t0)33! + (1.168)xi0 = xi (t0) xi0 = xi (t0) i = 1, 2, 3 (1.169)Let us dene = Gm along with an -set of parameters = Gm i = 1|xi|3 i = 1, 2, 3 (1.170)1.3 Orthogonal Coordinate Frames 31to rewrite Equations (1.167) as xi = ixi +Gi3

j=1jxj i = 1, 2, 3 (1.171)We also dene three new sets of parametersaijk = xi xj|xk|2 bijk = xi xj|xk|2 cijk = xi xj|xk|2 (1.172)whereaiii = 1 aijk = ajik cijk = cjik (1.173)The time derivatives of the -set, a-set, b-set, and c-set are i = 3biiii (1.174) aijk = 2bkkkaijk +bijk +bjik aiii = 0 (1.175)bijk = 2bkkkbijk +cijk iaijk +Gi3

r=1rarjk (1.176) cijk = 2bkkkcijk _ibjik +jbijk_+Gi3

r=1rbjrk +Gi3

s=1saisk (1.177)The -set, a-set, b-set, and c-set make 84 fundamental parameters that are indepen-dent of coordinate systems. Their time derivatives are expressed only by themselves.Therefore, we are able to nd the coefcients of series (1.168) to develop the seriessolution of the three-body problem.1.3 ORTHOGONAL COORDINATE FRAMESOrthogonal coordinate frames are the most important type of coordinates. It is compati-ble to our everyday life and our sense of dimensions. There is an orthogonality conditionthat is the principal equation to express any vector in an orthogonal coordinate frame.1.3.1 Orthogonality ConditionConsider a coordinate system (Ouvw) with unit vectors uu, uv, uw. The conditionfor the coordinate system (Ouvw) to be orthogonal is that uu, uv, uw are mutuallyperpendicular and hence uu uv = 0 uv uw = 0 (1.178) uw uu = 032 Fundamentals of KinematicsIn an orthogonal coordinate system, every vector r can be shown in its decomposeddescription asr = (r uu) uu+(r uv) uv +(r uw) uw (1.179)We call Equation (1.179) the orthogonality condition of the coordinate system (Ouvw).The orthogonality condition for a Cartesian coordinate system reduces tor = (r ) +(r ) +(r k)k (1.180)Proof : Assume that the coordinate system (Ouvw) is an orthogonal frame. Using theunit vectors uu, uv, uw and the components u, v, and w, we can show any vector r inthe coordinate system (Ouvw) asr = u uu+v uv +w uw (1.181)Because of orthogonality, we have uu uv = 0 uv uw = 0 uw uu = 0 (1.182)Therefore, the inner product of r by uu, uv, uw would be equal tor uu =_u uu+v uv +w uw__1 uu+0 uv +0 uw_= ur uv =_u uu+v uv +w uw__0 uu+1 uv +0 uw_= v (1.183)r uv =_u uu+v uv +w uw__0 uu+0 uv +1 uw_= wSubstituting for the components u, v, and w in Equation (1.181), we may show thevector r asr = (r uu) uu+(r uv) uv +(r uw) uw (1.184)If vector r is expressed in a Cartesian coordinate system, then uu = , uv = , uw = k, and therefore,r = (r ) +(r ) +(r k)k (1.185)The orthogonality condition is the most important reason for dening a coordinatesystem (Ouvw) orthogonal. Example 37 Decomposition of a Vector in a Nonorthogonal Frame Let a, b,and c be any three non-coplanar, nonvanishing vectors; then any other vector r can beexpressed in terms of a, b, and c,r = ua +vb +wc (1.186)provided u, v, and w are properly chosen numbers. If the coordinate system (a, b, c)is a Cartesian system (I , J, K), thenr = (r I)I +(r J) J +(r K) K (1.187)1.3 Orthogonal Coordinate Frames 33To nd u, v, and w, we dot multiply Equation (1.186) by b c:r (b c) = ua (b c) +vb (b c) +wc (b c) (1.188)Knowing that b c is perpendicular to both b and c, we ndr (b c) = ua (b c) (1.189)and therefore,u = [rbc][abc] (1.190)where [abc] is a shorthand notation for the scalar triple product[abc] = a (b c) =a1 b1 c1a2 b2 c2a3 b3 c3(1.191)Similarly, v and w would bev = [rca][abc] w = [rab][abc] (1.192)Hence,r = [rbc][abc]a + [rca][abc]b + [rab][abc]c (1.193)which can also be written asr =_r b c[abc]_a +_r c a[abc]_b +_r a b[abc]_c (1.194)Multiplying (1.194) by [abc] gives the symmetric equation[abc] r [bcr] a +[cra] b [rab] c = 0 (1.195)If the coordinate system (a, b, c) is a Cartesian system (I, J, K), then_I J K_= 1 (1.196)I J = K J I = K K I = J (1.197)and Equation (1.194) becomesr =_rI_ I +_r J_ J +_r K_ K (1.198)This example may considered as a general case of Example 30.34 Fundamentals of Kinematics1.3.2 Unit VectorConsider an orthogonal coordinate system (Oq1q2q3). Using the orthogonality condi-tion (1.179), we can show the position vector of a point P in this frame byr = (r u1) u1+(r u2) u2+(r u3) u3 (1.199)where q1, q2, q3 are the coordinates of P and u1, u2, u3 are the unit vectors alongq1, q2, q3 axes, respectively. Because the unit vectors u1, u2, u3 are orthogonal andindependent, they respectively show the direction of change in r when q1, q2, q3 arepositively varied. Therefore, we may dene the unit vectors u1, u2, u3 by u1 = r/q1|r/q1| u2 = r/q2|r/q2| u3 = r/q3|r/q3| (1.200)Example 38 Unit Vector of Cartesian Coordinate Frames If a vector r given asr = q1 u1+q2 u2+q3 u3 (1.201)is expressed in a Cartesian coordinate frame, thenq1 = x q2 = y q3 = z (1.202)and the unit vectors would be u1 = ux = r/x|r/x| = 1 = u2 = uy = r/y|r/y| = 1 = (1.203) u3 = uz = r/z|r/z| =k1 = kSubstituting r and the unit vectors in (1.199) regenerates the orthogonality conditionin Cartesian frames:r = (r ) +(r ) +(r k)k (1.204)Example 39 Unit Vectors of a Spherical Coordinate System Figure 1.12 illustratesan option for spherical coordinate system. The angle may be measured from theequatorial plane or from the Z-axis. Measuring from the equator is used in geographyand positioning a point on Earth, while measuring from the Z-axis is an appliedmethod in geometry. Using the latter option, the spherical coordinates r, , arerelated to the Cartesian system byx = r cos sin y = r sin sin z = r cos (1.205)1.3 Orthogonal Coordinate Frames 35XYZrPGS u ur uFigure 1.12 An optional spherical coordinate system.To nd the unit vectors ur, u, u associated with the coordinates r, , , we substitutethe coordinate equations (1.205) in the Cartesian position vector,r = x +y +zk= (r cos sin ) +(r sin sin ) +(r cos )k (1.206)and apply the unit vector equation (1.203): ur = r/r|r/r| = (cos sin ) +(sin sin ) +(cos )k1= cos sin +sin sin +cos k (1.207) u = r/|r/| = (r sin sin ) +(r cos sin ) r sin = sin +cos (1.208) u = r/|r/| = (r cos cos ) +(r sin cos ) +(r sin )kr= cos cos +sin cos sin k (1.209)where ur, u, u are the unit vectors of the spherical system expressed in the Cartesiancoordinate system.Example 40 Cartesian Unit Vectors in Spherical System The unit vectors of anorthogonal coordinate system are always a linear combination of Cartesian unit vectorsand therefore can be expressed by a matrix transformation. Having unit vectors of anorthogonal coordinate system B1 in another orthogonal system B2 is enough to nd theunit vectors of B2 in B1.36 Fundamentals of KinematicsBased on Example 39, the unit vectors of the spherical system shown in Figure 1.12can be expressed as__ ur u u__=__cos sin sin sin cos sin cos 0cos cos sin cos sin ____ k__ (1.210)So, the Cartesian unit vectors in the spherical system are__ k__=__cos sin sin sin cos sin cos 0cos cos sin cos sin __1__ ur u u__=__cos sin sin cos cos sin sin cos cos sin cos 0 sin ____ ur u u__ (1.211)1.3.3 Direction of Unit VectorsConsider a moving point P with the position vector r in a coordinate system (Oq1q2q3).The unit vectors u1, u2, u3 associated with q1, q2, q3 are tangent to the curve tracedby r when the associated coordinate varies.Proof : Consider a coordinate system _Oq1q2q3_ that has the following relations withCartesian coordinates:x = f (q1, q2, q3)y = g (q1, q2, q3) (1.212)z = h(q1, q2, q3)The unit vector u1 given as u1 = r/q1|r/q1| (1.213)associated with q1 at a point P (x0, y0, z0) can be found by xing q2, q3 to q20, q30and varying q1. At the point, the equationsx = f_q1, q20, q30_y = g_q1, q20, q30_ (1.214)z = h_q1, q20, q30_provide the parametric equations of a space curve passing through (x0, y0, z0). From(1.228) and (1.358), the tangent line to the curve at point P isx x0dx/dq1= y y0dy/dq1= z z0dz /dq1(1.215)1.4 Differential Geometry 37and the unit vector on the tangent line is u1 = dxdq1 + dydq1 + dzdq1k (1.216)_dxdq_2+_dydq_2+_dzdq_2= 1 (1.217)This shows that the unit vector u1 (1.213) associated with q1 is tangent to the spacecurve generated by varying q1. When q1 is varied positively, the direction of u1 iscalled positive and vice versa.Similarly, the unit vectors u2 and u3 given as u2 = r/q2|r/q2| u3 = r/q3|r/q3| (1.218)associated with q2 and q3 are tangent to the space curve generated by varying q2 andq3, respectively. Example 41 Tangent Unit Vector to a Helix Consider a helixx = a cos y = a sin z = k (1.219)where a and k are constant and is an angular variable. The position vector of amoving point P on the helixr = a cos +a sin +k k (1.220)may be used to nd the unit vector u: u = r/q1|r/q1| = a sin +a cos +k k_(a sin )2+(a cos )2+(k)2= a sin a2+k2 + a cos a2+k2 + ka2+k2k (1.221)The unit vector u at = /4 given as u = 2a2a2+k2 +2a2a2+k2 + ka2+k2k (1.222)is on the tangent line (1.255).1.4 DIFFERENTIAL GEOMETRYGeometry is the world in which we express kinematics. The path of the motion ofa particle is a curve in space. The analytic equation of the space curve is used todetermine the vectorial expression of kinematics of the moving point.38 Fundamentals of Kinematics1.4.1 Space CurveIf the position vector GrP of a moving point P is such that each component is a functionof a variable q,Gr = Gr (q) = x (q) +y (q) +z (q) k (1.223)then the end point of the position vector indicates a curve C in G, as is shown inFigure 1.13. The curve Gr = Gr (q) reduces to a point on C if we x the parameter q.The functionsx = x (q) y = y (q) z = z (q) (1.224)are the parametric equations of the curve.When the parameter q is the arc length s, the innitesimal arc distance ds on thecurve isds2= dr dr (1.225)The arc length of a curve is dened as the limit of the diagonal of a rectangular boxas the length of the sides uniformly approach zero.When the space curve is a straight line that passes through point P(x0, y0, z0)where x0 = x(q0), y0 = y(q0), z0 = z(q0), its equation can be shown byx x0= y y0= z z0(1.226)2+2+2= 1 (1.227)where , , and are the directional cosines of the line.The equation of the tangent line to the space curve (1.224) at a point P(x0, y0, z0) isx x0dx/dq = y y0dy/dq = z z0dz /dq (1.228)_dxdq_2+_dydq_2+_dzdq_2= 1 (1.229)X YZG Cdrdydxdzdsr2r1Figure 1.13 A space curve and increment arc length ds1.4 Differential Geometry 39Proof : Consider a position vector Gr = Gr (s) that describes a space curve using thelength parameter s:Gr = Gr (s) = x (s) +y (s) +z (s) k (1.230)The arc length s is measured from a xed point on the curve. By a very small changeds, the position vector will move to a very close point such that the increment in theposition vector would bedr = dx (s) +dy (s) +dz (s) k (1.231)The length of dr and ds are equal for innitesimal displacement:ds =_dx2+dy2+dz2(1.232)The arc length has a better expression in the square form:ds2= dx2+dy2+dz2= dr dr (1.233)If the parameter of the space curve is q instead of s, the increment arc length would be_dsdq_2= drdq drdq (1.234)Therefore, the arc length between two points on the curve can be found by integration:s =_ q2q1_drdq drdq dq (1.235)=_ q2q1__dxdq_2+_dydq_2+_dzdq_2dq (1.236)Let us expand the parametric equations of the curve (1.224) at a point P(x0, y0, z0),x = x0+ dxdqq + 12d2xdq2q2+ y = y0+ dydqq + 12d2ydq2q2+ (1.237)z = z0+ dzdqq + 12d2zdq2q2+ and ignore the nonlinear terms to nd the tangent line to the curve at P:x x0dx/dq = y y0dy/dq = z z0dz /dq = q (1.238)

40 Fundamentals of KinematicsExample 42 Arc Length of a Planar Curve A planar curve in the (x, y)-planey = f (x) (1.239)can be expressed vectorially byr = x +y (x) (1.240)The displacement element on the curvedrdx = + dydx (1.241)provides_dsdx_2= drdx drdx = 1 +_dydx_2(1.242)Therefore, the arc length of the curve between x = x1 and x = x2 iss =_ x2x1_1 +_dydx_2dx (1.243)In case the curve is given parametrically,x = x(q) y = y(q) (1.244)we have _dsdq_2= drdq drdq =_dxdq_2+_dydq_2(1.245)and hence,s =_ q2q1drdq=_ q2q1__dxdq_2+_dydq_2dq (1.246)As an example, we may show a circle with radius R by its polar expression usingthe angle as a parameter:x = Rcos y = Rsin (1.247)The circle is made when the parameter varies by 2. The arc length between = 0and = /2 would then be one-fourth the perimeter of the circle. The equation forcalculating the perimeter of a circle with radius R iss = 4_ /20__dxd_2+_dyd_2d = R_ /20_sin2 +cos2 d= 4R_ /20d = 2R (1.248)1.4 Differential Geometry 41Example 43 Alternative Space Curve Expressions We can represent a space curveby functionsy = y (x) z = z (x) (1.249)or vectorr (q) = x +y (x) +z (x) k (1.250)We may also show a space curve by two relationships between x, y, and z ,f (x, y, z) = 0 g(x, y, z) = 0 (1.251)where f (x, y, z) = 0 and g(x, y, z) = 0 represent two surfaces. The space curve wouldthen be indicated by intersecting the surfaces.Example 44 Tangent Line to a Helix Consider a point P that is moving on a helixwith equationx = a cos y = a sin z = k (1.252)where a and k are constant and is an angular variable. To nd the tangent line tothe helix at = /4,x0 =22 a y0 =22 a z0 = k4 (1.253)we calculate the required derivatives:dxd= a sin = 22 adyd= a cos =22 a (1.254)dzd= kSo, the equation of the tangent line is2a_x 122a_=2a_y 122a_= 1k_z 14k_ (1.255)Example 45 Parametric Form of a Line The equation of a line that connects twopoints P1(x1, y1, z1) and P2(x2, y2, z3) isx x1x2x1= y y1y2y1= z z1z2z1(1.256)42 Fundamentals of KinematicsThis line may also be expressed by the following parametric equations:x = x1+(x2x1) ty = y1+(y2y1) t (1.257)z = z1+(z2z1) tExample 46 Length of a Roller Coaster Consider the roller coaster illustrated laterin Figure 1.22 with the following parametric equations:x = (a +b sin ) cos y = (a +b sin ) sin (1.258)z = b +b cos fora = 200 m b = 150 m (1.259)The total length of the roller coaster can be found by the integral of ds for from 0to 2:s =_ 21_drd drdd =_ 21__x_2+_x_2+_x_2d=_ 2022_2a2+3b2b2cos 2 +4ab sin d= 1629.367 m (1.260)Example 47 Two Points Indicate a Line Consider two points A and B with positionvectors a and b in a coordinate frame. The condition for a point P with position vectorr to lie on the line AB is that r a and b a be parallel. So,r a = c (b a) (1.261)where c is a parameter. The outer product of Equation (1.261) by b a provides(r a) (b a) = 0 (1.262)which is the equation of the line AB.Example 48 Line through a Point and Parallel to a Given Line Consider a point Awith position vector a and a line l that is indicated by a unit vector ul. To determinethe equation of the parallel line to ul that goes over A, we employ the condition thatr a and ul must be parallel:r = a +c ul (1.263)1.4 Differential Geometry 43We can eliminate the parameter c by the outer product of both sides with ul:r ul = a ul (1.264)1.4.2 Surface and PlaneA plane is the locus of the tip point of a position vectorr = x +y +zk (1.265)such that the coordinates satisfy a linear equationAx +By +Cz +D = 0 (1.266)A space surface is the locus of the tip point of the position vector (1.265) such that itscoordinates satisfy a nonlinear equation:f (x, y, z) = 0 (1.267)Proof : The points P1, P2, and P3 at r1, r2, and r3,r1 =____DA00__ r2 =____0DB0__ r3 =____00DC__ (1.268)satisfy the equations of the plane (1.266). The position of P2 and P3 with respect toP1 are shown by 1r2 and 1r3 or r2/1 and r3/1:1r2 = r2r1 =______DADB0__1r3 = r3r1 =______DA0DC__(1.269)The cross product of 1r2 and 1r3 is a normal vector to the plane:1r2 1r3 =______DADB0________DA0DC__=________D2BCD2ACD2AB__(1.270)The equation of the plane is the locus of any point P,rP =__xyz__ (1.271)44 Fundamentals of Kinematicswhere its position with respect to P1,1rP = rP r1 =___x + DAyz__ (1.272)is perpendicular to the normal vector:1rP (1r2 1r3) = D +Ax +By +Cz = 0 (1.273)

Example 49 Plane through Three Points Every three points indicate a plane.Assume that (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) are the coordinates of three pointsP1, P2, and P3. The plane made by the points can be found byx y z 1x1 y1 z1 1x2 y2 z2 1x3 y3 z3 1= 0 (1.274)The points P1, P2, and P3 satisfy the equation of the planeAx1+By1+Cz1+D = 0Ax2+By2+Cz2+D = 0 (1.275)Ax3+By3+Cz3+D = 0and if P with coordinates (x, y, z) is a general point on the surface,Ax +By +Cz +D = 0 (1.276)then there are four equations to determine A, B, C, and D:____x y z 1x1 y1 z1 1x2 y2 z2 1x3 y3 z3 1______ABCD__=____0000__ (1.277)The determinant of the equations must be zero, which determines the equation ofthe plane.Example 50 Normal Vector to a Plane A plane may be expressed by the linearequationAx +By +Cz +D = 0 (1.278)1.4 Differential Geometry 45or by its intercept formxa+ yb+ zc= 1 (1.279)a = DAb = DBc = DC(1.280)In either case, the vectorn1 = A +B +Ck (1.281)orn2 = a +b +ck (1.282)is normal to the plane and may be used to represent the plane.Example 51 Quadratic Surfaces A quadratic relation between x, y, z is called thequadratic form and is an equation containing only terms of degree 0, 1, and 2 in thevariables x, y, z. Quadratic surfaces have special names:x2a2 + y2b2 + z2c2 = 1 Ellipsoid (1.283)x2a2 + y2b2 z2c2 = 1 Hyperboloid of one sheet (1.284)x2a2 y2b2 z2c2 = 1 Hyperboloid of two sheets (1.285)x2a2 + y2b2 + z2c2 = 1 Imaginary ellipsoid (1.286)x2a2 + y2b2 = 2nz Elliptic paraboloid (1.287)x2a2 y2b2 = 2nz Hyperbolic paraboloid (1.288)x2a2 + y2b2 z2c2 = 0 Real quadratic cone (1.289)x2a2 + y2b2 + z2c2 = 0 Real imaginary cone (1.290)x2a2 y2b2 = 1 y2= 2px Quadratic cylinders (1.291)46 Fundamentals of Kinematics1.5 MOTION PATH KINEMATICSThe derivative of vector functions is based on the derivative of scalar functions. To ndthe derivative of a vector, we take the derivative of its components in a decomposedCartesian expression.1.5.1 Vector Function and DerivativeThe derivative of a vector is possible only when the vector is expressed in a Cartesiancoordinate frame. Its derivative can be found by taking the derivative of its components.The Cartesian unit vectors are invariant and have zero derivative with respect to anyparameter.A vector r = r (t ) is called a vector function of the scalar variable t if there isa denite vector for every value of t from a certain set T = [1, 2]. In a Cartesiancoordinate frame G, the specication of the vector function r (t ) is equivalent to thespecication of three scalar functions x (t ), y (t ), z (t ):Gr (t ) = x (t ) +y (t ) +z (t ) k (1.292)If the vector r is expressed in Cartesian decomposition form, then the derivativedr/dt isGddtGr = dx (t )dt + dy (t )dt + dz (t )dtk (1.293)and if r is expressed in its natural formGr = r ur = r (t )_u1 (t ) +u2(t ) +u3 (t ) k_ (1.294)then, using the chain rule, the derivative dr/dt isGddtGr = drdt ur +rddt ur= drdt_u1 +u2 +u3k_+r_du1dt + du2dt + du3dtk_=_drdt u1+rdu1dt_ +_drdt u2+rdu2dt_ +_drdtu3+rdu3dt_k (1.295)When the independent variable t is time, an overdot r (t ) is used as a shorthand notationto indicate the time derivative.Consider a moving point P with a continuously varying position vector r = r (t ).When the starting point of r is xed at the origin of G, its end point traces a continuouscurve C as is shown in Figure 1.14. The curve C is called a conguration path thatdescribes the motion of P, and the vector function r (t ) is its vector representation.At each point of the continuously smooth curve C