tennis ball physics

MODELING OF IMPACT DYNAMICS OF A TENNIS BALL WITH A FLAT SURFACE A Thesis by SYED MUHAMMAD MOHSIN JAFRI Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE May 2004 Major Subject: Mechanical Engineering MODELING OF IMPACT DYNAMICS OF A TENNIS BALL WITH A FLAT SURFACE A Thesis by SYED MUHAMMAD MOHSIN JAFRI Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Approved as to style and content by: John M. Vance (Chair of Committee) Alan B.Palazzolo (Member) Guy Battle (Member) Dennis L.O’Neal (Head of Department) May 2004 Major Subject: Mechanical Engineering iii ABSTRACT Modeling of Impact Dynamics of a Tenni s Ball with a Flat Surface. (May 2004) Syed Muhammad Mohsin Jafri, B.E., NED University, Pakistan Chair of Advisory Committee: Dr. John M. Vance A two-mass model with a spring and a damper in the vertical direction, accounting for vertical tran slational motion and a tors ional spring and a damper connecting the rotational motion of two masses is used to simulate the dynamics of a tennis ball as it comes into contact with a flat surface. The model is supposed to behave as a rigid body in the horizontal direction. The model is used to predict contact of the ball with the ground and applies fr om start of contact to end of contact. The springs and dampers for both the vertical and the rota tional direction are linear. Differential equations of motion for the two-mass system ar e formulated in a plane. Two scenarios of contact are considered: Slip and no-slip. In the slip case, Coulomb’s law relates the tangential contact force acting on the outer mass with the normal contact force, whereas in the no-slip case, a kinematic constraint re lates the horizontal coordinate of the center of mass of the system with the rotational c oordinate of the outer mass. Incorporating these constraints in the differential equations of motion and applyi ng initial conditions, the equations are solved for kinematics and ki netics of these two different scenarios by application of the methods for the solutions of second-order linear differential equations. Experimental data for incidence and rebound kinematics of the tennis ball with incidence zero spin, topspin and backspin is available. The incidence angles in the data range from 17 degrees up to 70 degrees. Si mulations using the developed equations are performed and for some specific ratios of inner and outer mass and mass moments of inertia, along with the spring-damper coe fficients, theoretical predictions for the iv kinematics of rebound agree well with the experimental data. In many cases of incidence, the simulations predict transition from sliding to rolling during the contact, which is in accordance with the result s obtained from available experimental measurements conducted on tennis balls. Thus the two-mass model provides a satisfactory approximation of the tennis ball dynamics during contact. v DEDICATION To my Family and Teachers vi ACKNOWLEDGMENTS I would like to express my earnest gratitude to Dr. John M.Vance, my advisor, who provided me with an opportunity to work on this interesting thes is topic. His broad knowledge and patience, with gr eat insight and wisdom inspired me to work for the comprehensive formulation of this topic within the limitations of assumptions in the analysis. He has been an invaluable help to me and his great communication has been of enormous encouragement. I will like to thank Dr. Alan Palazzolo and Dr. Guy Battle for serving on my thesis committee. I have benefited immens ely from their teaching while attending classes under them. The concepts and analytical methods th at I gathered from those classes have been extremely valuable and useful for the completion of my thesis. Finally, I will like to thank all of my friends at the Turbomachinery lab and at home for their encouragement and help in ma ny aspects. I am thankful to all of you. vii NOMENCLATURE X Axis - Horizontal coordinate direction Y Axis - Vertical coordinate direction V 1 - Incident velocity of mass center of tennis ball [L/T] V 2 - Rebound velocity of mass center of tennis ball [L/T] θ i - Incident angle [-] θ r - Rebound angle [-] ω1 - Incident spin [1/T] ω 2 - Rebound spin [1/T] Vy1 - Vertical component of incident velocity [L/T] V x1 - Horizontal component of incident velocity [L/T] Vy2 - Vertical component of rebound velocity [L/T] Vx2 - Vertical component of rebound velocity [L/T] t - Time [T] tc - Time of contact [T] n - Dimensionless contact time [-] y(t) - Vertical motion coordinate of mass M 1 [L] )(ty• - Vertical velocity of mass M 1 [L/T] )(ty•• - Vertical acceleration of mass M 1 [L/T2] x(t) - Horizontal motion coordinate of system [L] viii )(tx• - Horizontal velocity of system [L/T] )(tx•• - Horizontal acceleration of system [L/T2] R - Outer radius of tennis ball [L] M - Mass of tennis ball [FT2/L] M 1 - Mass of inner core [FT2/L] M 2 - Mass of outer shell [FT2/L] K y - Stiffness of the spring in vertical direction [F/L] C y - Damping coefficient of the vertical damper [FT/L] ζy - Damping ratio of vibration in vertical direction [-] ωy - Natural frequency of vibration in vertical direction [1/T] ωdy - Damped natural frequency in vertical direction [1/T] I - Mass moment of inertia of tennis ball [FLT 2] I 1 - Mass moment of inertia of inner core [FLT2] I2 - Mass moment of inertia of outer shell [FLT2] Kθ - Torsional stiffness [FL] C θ - Torsional damping coefficient [FLT] ζ θ - Torsional damping ratio [-] ω θ - Torsional natural frequency [1/T] ω dθ - Damped torsional natural frequency [1/T] µ - Sliding coefficient of friction [-] µ - Time-averaged coefficient of friction [-] ix COR - Vertical coefficient of restitution [-] HCOR - Horizontal coefficient of restitution [-] F X(t) - Tangential or frictional contact force [F] F Y(t) - Normal contact force [F] θ 1(t) - Rotational motion coordinate of inner core [-] θ 2(t) - Rotational motion coordi nate of outer shell [-] )(1t• θ - Rotational velocity of inner core [1/T] )(2t• θ - Rotational velocity of outer shell [1/T] )(1t•• θ - Angular acceleration of inner core [1/T2] )(2t•• θ - Angular accelerati on of outer shell [1/T2] )(tθ - Relative rotational coordinate [-] )(t• θ - Relative rotational velocity [1/T] )(t•• θ - Relative rotational acceleration [1/T2] x TABLE OF CONTENTS P a g e ABSTRACT ....................................................................................................................iii DEDICATION ................................................................................................................. v ACKNOWLEDGEMENTS ............................................................................................ vi NOMENCLATURE........................................................................................................ vii TABLE OF CONTENTS ................................................................................................ x LIST OF FIGURES........................................................................................................xiii LIST OF TABLES .........................................................................................................xvi CHAPTER I INTRODUCTION..................................................................................1 BACKGROUND OF IM PACT DYNAMICS ................................ 1 LITERATURE REVIEW................................................................ 3 RESEARCH OBJECTIVE...............................................................9 RESEARCH METHOD................................................................. 9 II MODELING AND ANALYSIS OF A TENNIS BALL IMPACT............................................................................................... 10 IMPACT MODEL FOR TENNIS BALL ...................................... 10 MOTION IN Y-DIRECTION........................................................ 12 MOTION IN THE HORIZ ONTAL DIRECTION......................... 16 Kinematical constraint-rolling.................................................. 17 Kinetic constraint-sliding ......................................................... 17 ROTATIONAL EQUATI ONS OF MOTION ............................... 19 Inner core.................................................................................. 20 Outer core................................................................................. 21 Rolling motion.......................................................................... 21 Sliding motion .......................................................................... 24 VELOCITY OF CONTACT PO INT AND SIGN OF CONTACT FORCE .............................................................................................. 25 xi CHAPTER Page Topspin..................................................................................... 26 Backspin ................................................................................... 27 Zero spin................................................................................... 27 MOTION IN X-DIRECTION........................................................... 29 Rolling...................................................................................... 29 Sliding ...................................................................................... 30 EFFECT OF HIGH INCIDENT VERTICAL VELOCITY COMPONENT ON ROLLING MOTION........................................ 31 OFFSET DISTANCE AS A FUNCTION OF VERTICAL IMPACT VELOCITY....................................................................... 35 TRANSITION BETWEEN SLIDING AND ROLLING: TIME- AVERAGED COEFFICIENT OF FRICTION................................. 37 INNER AND OUTER CORE DYNAMIC PARAMETERS ........... 41 Inner core.................................................................................. 41 Outer core................................................................................. 42 III GRAPHICAL RESULTS OF SOLUTIONS OF EQUATIONS OF MOTION.............................................................................................. 45 SLIDING THROUGHOUT THE CONTACT ................................. 45 Vertical displacement as a function of time ............................. 47 Vertical velocity as a function of time ..................................... 48 Horizontal velocity as a function of time ................................. 49 Angular velocity as a function of time ..................................... 52 Normal contact force as a function of time .............................. 54 Tangential (frictional) contact fo rce as a function of time....... 55 NO-SLIP THROUGHOUT THE CONTACT .................................. 56 Angular velocity as a function of time ..................................... 58 Horizontal velocity as a function of time ................................. 63 Tangential contact force as a function of time ......................... 65 TRANSITION BETWEEN SLIDING AND ROLLING.................. 66 EXAMPLES FOR ILLUSTRATING THE APPLICATION OF EQUATIONS .................................................................................... 68 IV EXPERIMENTAL DATA ................................................................... 77 EXPERIMENTAL PROCEDURE ................................................... 77 xii CHAPTER Page MEASUREMENT OF THE MASS MOMENT OF INERTIA OF A TENNIS BALL ....................................................................... 91 Theoretical background of measurement for mass moment of inertia ................................................................................... 91 Experimental setup................................................................... 96 Results of the experiment......................................................... 98 V BEST RESULTS COMPARISONS WITH THE MEASUREMENTS ............................................................................. 100 VI CONCLUSIONS.................................................................................. 112 REFERENCES............................................................................................................. 114 APPENDIX A .............................................................................................................. 116 APPENDIX B .............................................................................................................. 119 APPENDIX C .............................................................................................................. 129 APPENDIX D .............................................................................................................. 156 VITA ........................................................................................................................... . 166 xiii LIST OF FIGURES Page Fig.1 Kinematic parameters of the tennis ball striking the non-smooth surface.......... 10 Fig.2 Linear spring-damper-mass model for vertical impact....................................... 12 Fig.3 Inner core and outer shell connect ed by linear spring-damper elements............ 13 Fig.4 The model, motion and force in X direction....................................................... 17 Fig.5 Model and coordinates of the rotational motion ................................................. 19 Fig.6 Free body diagrams of the i nner core and the outer core.................................... 20 Fig.7 Effect of high incident velo city on the ball during rolling.................................. 32 Fig.8 Vertical displacement during contact as a function of time................................ 47 Fig.9 Vertical velocity during co ntact as a function of time........................................ 48 Fig.10 Horizontal velocity during contact as a function of time.................................. 49 Fig.11 Horizontal velocity during contact as a function of time (effect of initial velocity)...................................................................................................................... .. 50 Fig.12 Horizontal velocity during contact as a function of time (high topspin) .......... 51 Fig.13 Angular spin as a function of time.................................................................... 52 Fig.14 Angular spin as a function of time.................................................................... 53 Fig.15 Normal contact force as a function of time....................................................... 54 Fig.16 Frictional force as a function of time................................................................ 55 Fig.17 Rolling angular velocity as a function of time.................................................. 58 Fig.18 Angular spin of outer shell as a function of time (S pecial case of topspin) ..... 60 Fig.19 Angular spin of outer sh ell as a function of time.............................................. 60 Fig.20 Angular spin velocity during contact as a functi on of time (corresponding to three offset distances)............................................................................................... 62 Fig.21 Horizontal velocity as a function of time (effect of initial conditions)............. 63 Fig.22 Horizontal velocity as a f unction of time (effect of spin)................................. 63 Fig.23 Tangential friction for ce as a function of time.................................................. 65 Fig.24 Transition from sliding to rolling motion ......................................................... 66 xiv Page Fig.25 Surface and center of ma ss velocities during contact ....................................... 70 Fig.26 Normal and tangential contact forces ............................................................... 71 Fig.27 Horizontal velocities of the two-mass mode..................................................... 73 Fig.28 Horizontal and spin velocities during contact................................................... 75 Fig.29 Schematic drawing of experimental arrangement............................................. 80 Fig.30 Incident vs rebound kinematics for zero spin ................................................... 86 Fig.31 Incident vs rebound kinematics for topspin ...................................................... 87 Fig.32 Incident vs rebound kinematics for backspin.................................................... 88 Fig.33 Equivalent linear vibrating systems .................................................................. 92 Fig.34 Equivalent torsional vibrating systems ............................................................. 94 Fig.35 Experimental setup to measure the mass moment of inertia............................. 97 Fig.36 Incident vs rebound parameters for the zero spin (average COR = 0.765), case 1. (c y = 0.0225 lb-s/in).......................................................................................... 103 Fig.37 Incident vs rebound parameters for the zero spin (average COR = 0.765), case 2. (c y = 0.0239 lb-s/in).......................................................................................... 104 Fig.38 Incident vs rebound parameters for the zero spin (average COR = 0.765), case 3. (c y = 0.0185 lb-s/in).......................................................................................... 105 Fig.39 Incident vs rebound parameters fo r the top spin (average COR = 0.778), case 1. (c y = 0.0211 lb-s/in).......................................................................................... 106 Fig.40 Incident vs rebound parameters fo r the top spin (average COR = 0.778), case 2. (c y = 0.0224 lb-s/in).......................................................................................... 107 Fig.41 Incident vs rebound parameters fo r the top spin (average COR = 0.778), case 3. (c y = 0.0173 lb-s/in).......................................................................................... 108 Fig.42 Incident vs rebound parameters for the back spin (average COR = 0.732), case 1. (c y = 0.0262 lb-s/in).......................................................................................... 109 Fig.43 Incident vs rebound parameters for the back spin (average COR = 0.732), case 2. (c y = 0.0277 lb-s/in).......................................................................................... 110 xv Fig.44 Incident vs rebound parameters for the back spin (average COR = 0.732), case 3. (c y = 0.0215 lb-s/in).......................................................................................... 111 xvi LIST OF TABLES P a g e Table1. Incidence and rebound kinematics for incident zero spin............................... 83 Table 2. Incidence and rebound kinematic s for zero spin (w ith restitution coefficients and kinetic energies)................................................................... 83 Table 3. Incidence and rebound kine matics for incident topspin................................. 84 Table 4. Incidence and rebound kinematics fo r topspin impact (with restitution coefficients and kinetic energies)................................................................... 84 Table5. Incidence and rebound kinematics for tennis ball with incident backspin....................................................................................................................... . 85 Table 6. Incidence and rebound kinematics fo r backspin impact (with restitution coefficients and ki netic energies)................................................................................. 85 Table 7. Time-averaged coefficient of fr iction values for experimental data.............. 89 Table 8. Experimental results of th e twisting test on the tennis ball............................ 98 Table 9. Dynamic ratios for all cases of in cident angles and spins giving best results ........................................................................................................................ ... 101 1 CHAPTER I INTRODUCTION BACKGROUND OF IMPACT DYNAMICS Historically, the topic of impact dynami cs has been of both experimental and theoretical interest from the time of Newton to the present time. Impact dynamics has its importance in mechanical systems whenever two or more bodies, one of which is in motion with respect to the others, come into contact with each other for a short duration of time. This brief contact creates contact forces of significan t magnitudes that can change dramatically the kinematics of the bodi es involved in the contact. The impact can occur in mechanical systems, for instan ce, when a rotor supported on the magnetic bearings falls on its retainer bearings due to the failure of the magnetic bearings. In this situation, the rotor impacts the inside surf ace of the retainer bear ings and hence upon its rebound from the bearing surface, its velocity and spin changes significantly from what it was when it came into contact with the bearing. Another important problem where impact dynamics plays an important part is th e collision of a rotor with its stator, when the clearance between the rotor and the stator is very small and rotor contacts the stator due to the vibration induced by its imbalan ce. Contact forces of high magnitudes are usually developed when the rotating speeds are high, so that the contact forces have the destructive potential for both the rotor and th e stator. Analysis of this and all such problems with the application of impact dyna mics can help design the proper speeds and material selection for the movi ng parts, so as to devise some means of avoiding the contact and in case of cont act, avoiding the failure of the parts involved. Also measurements made on the rotating machin eries which show different vibration signatures can be compared with the theoretical predictions from an This thesis follows the style and format of ASME Journal of Applied Mechanics. 2 impact dynamics model of such machineries and hence used to diagnose the root cause of the problem. Impact can also be seen in a beneficial context, for the case of rotordynamic bumpers which are used to suppress vibrati on amplitudes in rotating machines operating near their critical speeds. In this case, the bumpers, which are mounted aro und the rotor, are impacted by the rotor near the critic al speed and their frictional and damping properties allow the suppression of vibration amplitudes. The impact with the bumpers also provides damping to the system wh en properly designed. Improperly designed, bumpers can create violent dynamic instab ilities, such as dry friction whip. From a theoretical point of view, impact dynamics can be analyzed in two ways: Rigid-body collisions and deformable body coll isions. The rigid-body collision approach is based on the classical Newtonian viewpoi nt of considering the impact as an instantaneous phenomena and de scribing the loss of kinetic energy of the colliding bodies in terms of a parameter called as co efficient of restitution. Newton’s laws of motion are applied to the body before and af ter collision, and then the kinematics is solved for the rebound in terms of the incide nce kinematics. There is no description of whatever happens during the collision, nor a ny description of the duration of contact, because these parameters are simply eliminated from the equations of motions while solving for the rebound kinematics in terms of a coefficient of restitution. From the flexible-body point of view, whic h is credited to Hertz for its development, the impact is not considered as an instantaneous phe nomenon but rather a phenomenon involving a finite duration of time, no matter however small. The co lliding bodies involved are considered to be deformable in a small region around the point of contact. During this contact time, the contact forces are develope d and the bodies change their kinematic and kinetic properties gradually, usually in fo rm of continuous mathematical functions. Newton’s laws are applied during the contac t based on the assumed form of contact force, which in turn is usually dependent on the geometric and elas tic properties of the 3 bodies involved in the collision. The equatio ns of motion are solved and the kinematics during the contact and at rebound is determined. The application of this concept of fin ite contact duration and the assumption of the contact force can then be applied to a problem where the experimental measurements are available regarding the rebound kinematics given the incident kinematics such as incident translational velocities and angular spins, so that the developed impact model can be implemented and the results devel oped thereof can be compared against the measurements to ascertain the accuracy of th e model. If such a problem can be modeled with a linear model of contact force, then it will definitely provide an efficient method of computation in terms of time and effort. One such problem of impact found in the literature is the incidence and rebound of tennis ball with flat surfaces of varying properties, the modeling and simulation of which is the objective of this thesis. The next section describes the relevant theoretical and experimental research appropriate to this objective. LITERATURE REVIEW Wang [1] conducted experiments on the impacts of tennis balls with acrylic surface, with varying incident conditions. In his experiments, he varied the angles of incidence from 17 degrees up to 70 degrees, and for all incident angles, the tennis ball was thrown on the surface with zero spin, topspin and backspin. The average incident translational speeds were around 17 m/s. From the measurements, he concludes that the tennis ball incident with backspin rebounds at a higher rebound angle than the ball with zero spin, which in turn rebounds at a highe r rebound angle than th e ball thrown with topspin. He also proposes a damped contac t force model to pred ict the velocities at rebound, for only one case of incidence angle. 4 Smith [2] conducted experiments on the tennis balls in order to study the effect of the angle of incidence, the velocity of incidence and the spin of incidence on the angle of rebound. He projected the tennis balls on a Laykold court surface and the angles of projections were 15, 30, 45, 60 and 75 degr ees. The tennis ball was incident with the zero spin, the topspin and the b ackspin. He concludes that in general, a ball thrown with the backspin will rebound at a high angle than with either the zero spin or the topspin. Further more, he concludes that the angle of rebound for the backspin and the zero spin will be greater than the corresponding angles of incidences for these incident spins. For the topspin, the angle of rebound, in general, will be either eq ual to or less than the angle of incidence. Considering the effects of in cident velocity on the angle of rebound, he concludes that the rebound angles for the balls incident with the topspin will be higher for higher incident translati onal velocities. For backspin , the opposite is concluded. From zero spin results, the conclusion for the backspin holds that the higher the incident velocity, the higher is the rebound angle. Cross [3] conducted measurements on the vertical bounce of various sports balls with no spin, including tennis balls, with the help of piezo disks. From his experiments on different balls, he obtained the time vary ing form of normal contact forces during impact, which generally showed asymmetry a bout the time axis. He estimated the time of contact for various balls and curve fitted the force wave-forms into mathematical functions to obtain the displacements and velo cities as functions of time during contact. He concludes that the impact of tennis balls can be approximated as the one in which the vertical coefficient of restitution can be trea ted as independent of incident velocity and also that the contact force for tennis ball co ntact can be modeled as linear force model. Cross [4] conducted further measurements on the tennis ball with oblique incidences with various flat surfaces but zero incident spin in all cases. He employs Brody’s [5] model of tennis ball impact and shows from his measurements that the model reasonably well predicts the rebound kinematics, including rebound spin. He concludes from the measurements of tangential to normal impulse s during the contact that for higher angles 5 of incidences and rougher surfaces, the tenni s ball during its contact with the surface does not continue to slide, but that there is a transition in the motion from sliding to rolling mode during the contact. Due to this occurrence of rolli ng mode in the motion during contact, the ball leaves at a higher rebound horizontal velocity and lower spin, then will be expected if it were slidin g throughout the contact. He verifies this by modifying Brody’s equations to account for this transition and shows that the theoretical results agree well with the observations. He defines another parameter analogous to the vertical coefficient of restitution, called the horizontal coefficient of restitution. Horizontal coefficient of restitution is defi ned as the ratio of rebound horizontal velocity of contact point to incident horizontal velo city of contact point. For the rolling mode, it will be expected that the horizontal coefficient of restitution will be equal to zero(because the contact point has no rela tive motion with respect to the ground in rolling mode), whereas his measurements reveal it does not when it should, as per Brody’s model. This forms some devi ation between Brody’s model and his measurements. He concludes that the deviati on can be explained if the tennis ball is considered as flexible in horizontal direction of motion as well. Brody [5] presented a model for the bounce of a tennis ball from a flat surface, in which the incident vertical ve locity component is not too hi gh. He models the tennis ball as a hollow sphere, with certain thickness. He considers the separate scenarios of sliding and rolling throughout the motion. His model for both cases is based upon the application of Newton’s impulse and moment um laws applied at incidence and rebound and incorporates the vertical coefficient of restitution and sliding coefficient of friction in the formulations. He concludes that for the sliding mode, the rebound horizontal velocity is dependent upon incident vertical velocity as well as coefficients of friction and restitution, whereas fo r the rolling mode, the hor izontal rebound velocity is dependent only on the incident horizontal velo city. He develops a relation for transition between sliding to rolling and concludes that the transition will o ccur depending strictly upon the coefficient of sliding friction and th e angle of incidence of the tennis ball. 6 Hubbard and Stronge [6] consider the analysis and experiments of the bounce of hollow balls on flat surfaces. They consider various dynamic properties of the table tennis balls such as the elastic deformation under the action of the in teraction force, the contact force itself and the velo city of rebound. They consider in detail the geometry of the ball and conduct their analysis using a finite element method. Hubbard [7] considers the impact phenomena of the ball with both the classic point of view considering both no duration of c ontact as well as finite duration of contact by using spring-damper models to model the impact of solids with each other. From these later simplified models, he develops the idea of the coefficient of restitution for the central as well as oblique impacts and expl ains the velocities of rebound. He also considers a number of contact models to model the normal contact force. His methods represent the nonlinear models of the contact; he accordingly solves most of the contact problems with the help of numerical techniques. Stevenson, Bacon and Baines [8] describe the measurements of the normal contact forces using the piezo di sks. They show the inelastic ity of the vertical incident collisions for the cases of collisions of steel ball bearings and plumber’s putty balls with pads made of different surfaces. They derived the relationship between the voltage output obtained from the measurements and th e normal contact force. They vary the height of drop as well the types of surface to determine the normal contact force. Also they estimate the duration of contact for va rious vertical incident collisions. Their experiments indicate that the normal contac t force can not be described as perfectly symmetrical about the time axis, rather it ha s some asymmetry. They concluded that the impulse imparted to the collid ing ball with a surface is de pendent on the type of surface as well as the height from which it is dropped. 7 Malcolm [9] describes the propertie s of piezoelectric films for measurements and shows an example of impact. Minnix and Carpenter [10] use the piezoelectric film to perform experiments on different balls and obtai n the impact force as a function of time. The International Tennis Federation (ITF) [11] provides the standards of some of the parameters for the tennis balls that need to be met for them to give the level of performance expected to be met out of th em. In their standards, they mention the procedures for the testing of the balls as well as the forward and return deformations values for the testing of the balls. They me ntion the range of the deformation values for the balls. Also they mention the coefficient of restitution range values to be expected out of the tennis balls. They also mention the construction of tennis balls and how the internal pressure is maintained inside some of the designs, and what must be the values of the internal pressure s under different conditions. Lankarani and Nikravesh [12] employ the Hertz theory of contact to model the two-body and multi-body impact, neglecting the e ffect of friction. They elaborate the idea that the kinetic energy loss during an imp act can be interpreted as the damping term in the contact force equation. Thus the contact force will consist not only of the elastic term, but also a damping term. The form of the normal contact force in their analysis is non-linear. They apply the Newton’s impulse and momentum laws, and express the deformation of the colliding bodies, which th ey call indentation, and the indentation velocities in terms of the initial approach ve locities, normal coefficient of restitution and the masses of the bodies. Thus they deri ve a contact force equation and thence investigate the kinematics occurring during and at the end of contac t. They apply this theory to multi-body impact. Stronge [13] performs a theoretical analysis of impact problem and calculates how friction does mechanical work to dissipate energy. Sonderbaarg [14] conducted 8 measurements on the rebound of solid sphere s made of steel and pyrex glass from various surfaces of varying dimensions. He c oncludes that the vertical coefficient of restitution is dependent on both the incident surface and the bounce surface. The vertical coefficient is more or less a constant, varyi ng initially with some geometric parameters of the system. The horizontal coefficient of re stitution, in his pape r defined as the ratio of rebound horizontal velocity of center of mass to incident horizontal velocity of center of mass, is not a constant value and varies significantly with the incident angles Keller [15] considers the problem of impact of two bodies with friction. He derives the equations of motion for the bodies and considers the impact as a phenomenon involving a finite duration of contac t. He shows in the analysis that if the slip velocity between two bodies changes di rection, the friction force reverses in direction, so that the energy at re bound is lesser than at incidence. Hudnut and Flansburg [16] have modeled the collision problem of gliders on air tracks as masses connected by linear springs. G liders are specially shaped metal objects that are used for experiments on air tracks and they levitate on the air tracks when air is passed through the track. They have performed the experiments on the collisions of gliders with various initial separations and they show that the elastic modeling of the collisions of gliders provides a more comp rehensive and realistic picture of the rebound phenomenon as compared to the classi cal approach, employing the impulse and momentum equations applied to the impact and rebound. Bayman [17] derives the wave equation for th e contact modeling of two bodies which collide elastical ly and which can be modeled as linear Hooke’s springs. He considers the wave propagati on through the bodies and estima tes the duration of contact during collision. He also conc ludes that the elastic wave reflections from both bodies play an important role in determining the time of contact. 9 Johnson [18] applies the theory of elasticity at the contact point of the super ball to explain the unusual bounce phenomenon it show s when it is thrown with backspin. He shows that these phenomenon can not at a ll be explained by the rigid-body impact theory. He concludes that it is the tangential flexibility at the interface of super ball and the flat surface which imparts the super ball its unusual properties of bounce. Hunt and Crossley [19] consider a non-linear force model base d on Hertz theory of contact to the case of solids impacting at low speeds. From energy considerations and the work done by the contact force, they conclude that the velocity -dependent coefficient of restitution can be regarded as damping in a second-order, non-linear differential equation describing the contact motion of two bodies. RESEARCH OBJECTIVE The objective is to develop a computati onally efficient model and method of analysis to predict impact dynamics and kinematics of a tennis ball. RESEARCH METHOD The method investigated is to apply a pi ecemeal theory of linear vibrations and impact dynamics to the phenomena of the impact, contact and rebound of a tennis ball striking ground or any other flat surface. The ma in predictions of interest are going to be the rebound spin and translational velocitie s, the contact forces during contact, the translational and spin velocity variations w ith time during contact and the identification of an important dynamic parameter called co efficient of restitution that relates the velocities before and after impact, given the mass, stiffness (translational and torsional) of the ball, amount of damping, incoming spin and translational veloci ties and angles of impact. 10 CHAPTER II MODELING AND ANALYSIS OF A TENNIS BALL IMPACT IMPACT MODEL FOR A TENNIS BALL Impact of a tennis ball with a surface is considered in detail with the dynamic parameters taken into account. Consider a te nnis ball striking a surface, as shown in the following figure: Fig. 1 Kinematic parameters of the tennis ball striking the non-smooth surface. Kinematic parameters of the tennis ball in cident on ground surface are shown in figure 1. The incident parameters are: 1. Translational speed of the center of mass, V 1 2. Angle of the velocity vector with respect to ground (incident angle of the ball), θ1 C iθrθ1V2V YX C ω1 +θω2 11 3. Spin of the ball about its centroidal axis, ω1 Similarly, rebound parameters of the ball are: 4. Translational speed of the center of mass, V 2 5. Angle of the velocity vector with resp ect to ground (rebound angle of the ball), θ2 6. Spin of the ball about its centroidal axis perpendicular to plane of the page, ω2 The kinematics on rebound is related to the kinematics at impact with the help of dynamics, or Newton-Euler equations of motion. Physical modeling of the impact phenome na of the ball with the ground and the derivation of the equations of motion consists in modeling the ball as a linear spring- damper system as soon as it contacts the gr ound. This linear spring-damper system is for modeling the vertical translation motion as well as rotational motion of the ball. The motions before and after impact are dete rmined by two physical possibilities during impact: (a) No Slip or Rolling condition (b) Sliding or slipping condition For modeling of spin of the ball, the ball is modeled as a two-mass system connected by torsional spring and damper. This torsiona l spring-damper is also considered as a linear pair. From these modeling elements, equations of motion for the ball during the impact are derived and then the results are compared with the available experimental data. In this thesis, by incident conditions is meant the kinematics of the ball before it contacts the ground, whereas initial conditi ons will be the values of the kinematic parameters as soon as it comes into contact with ground. 12 MOTION IN Y-DIRECTION Physical model for the impact of the ba ll with the ground is i llustrated in figure 2: Fig. 2 Linear spring-damper-mass model for vertical impact. For physical description of the problem, the ball is modeled as having an inner core and an outer core. As soon as the ball t ouches the ground, the outer core comes into contact with the ground and is stopped whereas the inner core continues to move down. The outer and the inner core are connected with a linear spring-damper as shown in following figure on next page: M K C 13 Fig. 3 Inner core (mass M 1) and outer shell (mass M 2) connected by linear spring- damper elements. Initial conditions of the kinematic s can be expressed as follows: 1 )0(0)0( yV yy == • i.e., measured from the position of the inner core as soon as outer core touches the ground, initial displacement of the inner core is zero whereas it carries the vertical velocity component indicated above. This ver tical component of velocity is vertical velocity of the ball seen as a whole just befo re impact. The above two relations are initial conditions for vertical motion during the contact. Equation of motion in the vertical direction is: M1 M2 Ky Cy y 14 0)( )( )(1 = + +• •• tyKtyCtyMy y (1) Equation (1) describes motion of the inner core of the ball and it helps to model contact duration time when the ball is going to be rebounded, as well as the velocity of the ball at rebound. Solution of equation (1) is given as follows: ) sin( )(1t eVtydyt dyyyyωωωζ−= (2) after applying the initial conditions of velo city and displacement. Definitions of the symbols are given as: yy yMC ωζ 12= 1MKy y=ω 21y y dy ζ ω ω − = It can be seen from equations (1) and (2) how the spring-damper approach is employed to model a collision, or impact problem to one of the vibration problem. The physical significance of the above model can be explored by observing that the time of rebound for the ball, i.e., the ti me in which downward motion is completely reversed, inner core mass M 1 reaches its original position and with negative velocity in the Y-direction is simply half of the peri od of this simple vertical spring-mass-damper system. The time of contact is given as follows: 15 dyCtωπ= ( 3 ) Differentiating equation (2) with respect to time, it is seen that: ) cos( ) sin( )(11t eVt eVtydyt y dyt dyyyy yω ωωζωζω ζω − −• +−= (4) It can be immediately seen that this a pproach used to describe the collision process is not only applicable in finding out the rebound vertical velocity component but also the variation of the ve rtical velocity with time during the contact. For evaluating the vertical velocity at rebound for inner mass, simply substituting the value of contact time in equation (4), it can be seen that: 112 )(y c V e tyyy ζπζ −−• −= ( 5 ) Now for finding out the vertical velocity at rebound of the tennis ball, or in other words, of the two-mass system, it is nece ssary to apply the conservation of linear momentum between the time the contact ends (at which the inner mass moves with the rebound velocity given by equation (5)) and some time after the ball has left the ground such that the inner and outer mass are moving with some common value of vertical velocity. Expressing it in mathematical form: 2 2 1 1 ) ()(y c VM M tyM + =• ( 6 ) 16 In equation (6), v y2 is the velocity of rebound and is related to the vertical coefficient of restitution as follows: 1 2 ) (y y V COR V −= ( 7 ) Combining equations (5), (6) and (7), the following equation is obtained which can be used to determine the damping ratio and damping coefficient of the two-mass model in vertical direction: ) () ( 12 112 CORMM Meyy +=−− ξπξ ( 8 ) It can be seen that the coefficient of restitution is dependent on the system dynamic parameters such as mass, stiffness and damping and these parameters determine as to what fraction the veloci ty of rebound is of the incomi ng velocity in Y-direction. MOTION IN THE HORIZONTAL DIRECTION Motion in the x-direction presents two po ssible scenarios of impact for the ball, that is, rolling and slipping. For either cas e, the basic equation of motion in the x- direction is given as: )() ( 2 1 tx M M FX•• + =∑ (9) 17 The model can be presented in Figure 4 below: Fig. 4 The model, motion and force in x-direction. Depending on whether there is rolling or s lipping condition, we accordingly have the following equations: Kinematical constraint- rolling )( )(2t Rtx θ= (10) Kinetic constraint- sliding ))0( )0( sgn()( )(2• • − = θ µ R x tF tFY X (11) X Y Fx(t) R Fy(t) θ2 M2, I2 M1, I1 18 Equation (8) represents the rolling condition-it relates the kinematic parameters, namely the translational displacement of the inner core with the rotational displacement or coordinate of the outer core, as soon as the ball contacts the ground. Equation (9), on the other hand, expresse s the relationship between the horizontal force acting on the outer shell and the norma l reaction on the outer core as soon as contact is made. It is the statement of C oulomb’s law of sliding friction, with ‘ µ’ as the coefficient of sliding friction. In equation (9 ), “sgn” represents sign function, which is defined as follows: 1 ) sgn( +=A , A>0 1 ) sgn( −=A , A<0 (12) Equations in (A) above indi cate that when this function acts on an argument, which in turn itself can be a function, then it either attaches +1 or –1 with the argument. Physically, its use in equation (9) means that the friction force during the sliding motion can be either in the positive coordinate direction or in the negative coordinate direction, depending upon the velocity of point of cont act with the ground. The velocity of the point of contact is not only determined by the translational velocity but also the incoming spin velocity and will be discussed later in this chapter. Once the rotational equation of motion for the ball is formulated, these conditions can be used in conjunction w ith the vertical motion to de scribe the dynamics of the ball during contact with ground in either of the tw o modes of motion i.e., sliding or rolling. 19 ROTATIONAL EQUATIONS OF MOTION Fig. 5 Model and coordinates of the rotational motion. Figure 5 shows the model and the coordina tes used for the formulation of the rotational equations of motion. In this figure, it is seen that a torsional spring and damper connect the inner and outer core. As in the case of vertical motion, these elements are considered as linear. As can be guessed, the damping element will give rise to the concept of rotational coefficient of restitution. Applying Newton-Euler equations of motion, the rotational equations of motions are written as follows: G G Mt I =•• )(θ (13) FY(t) θ2 θ1 Sgn (F X(t)) X YKθ Cθ +θI1 I2 20 Applying equation (10) to the inner and th e outer core, considering the torsional spring, damper and finally, the horizontal friction force acting on the outer shell, the equations of motion are derived as follows in Figure 6: (a) Inner core ( b ) O u t e r s h e l l Fig. 6 Free body diagrams of the inner core and the outer core. Inner core 0))( )(( ))( )(( )(2 1 2 111 = − + − +• • •• t t Kt t Ct I θ θ θ θ θθ θ (14) X Y+θ))( )(( 2 1 t t C• • − − θ θθ ))( )((2 1 t t K θ θθ − − ))( )(( 2 1 t t C• • −θ θθ ))( )((2 1 t t K θ θθ − FX(t) R N = F Y(t)+M 2g FY(t) M1g 21 Outer core 0 )( ))( )(( ))( )(( )(1 2 1 222 = − − + − +• • •• RtF t t Kt t Ct IX θ θ θ θ θθ θ (15) In order to solve equations (11) and (12) completely, we need one more equation, that relates F x(t) to one of the other parameters, namely θ2(t) or the vertical motion. These are the conditions of rolli ng and sliding, respectively. First, consider the case of rolling. Rolling motion Using equation (7) and equation (8), it can be seen that: )( ) ( )(2 2 1 t RM M tFX•• + −= θ (16) The negative sign is due to the reason that in the rolling moti on, velocity of the contact point is zero at all instants, and he nce the friction force then becomes dependent only on the sign of the incident horizontal velocity. Since in th e chosen coordinate system, the incident horizontal velocity is always positive, accordingly the direction of the rolling friction force is always in the negati ve x-direction. This is not necessarily so, however, for the sliding friction force. Using equations (12) and (13), it can be seen that the resulting differential equation of rotational motion is: 0))( )(( ))( )(( )()) ( (1 2 1 222 2 1 2 = − + − + + +• • •• t t Kt t Ct RM M I θ θ θ θ θθ θ (17) 22 Now it is possible to solve equati ons (11) and (14) completely. Defining: )( )( )(2 1 t t t θ θ θ − = (18) Dividing equations (11) and (14) by thei r respective coefficients of the first terms, and then subtracting from each other, using equation (15), th e following rotational equation of motion in one singl e coordinate is obtained: 0)() ) (1 1( )() ) (1 1( )(2 2 1 2 12 2 1 2 1= + ++ + + ++ +• •• t RM M I IKt RM M I ICt θ θ θθ θ (19) It is obvious that the term M 1+M 2 is the total mass of the ball. This is a second-order linear differe ntial equation and it can be solved analytically. The initial cond itions are given as follows: 0)0(= θ i.e., no relative motion between the i nner and outer core at instant of contact In order to determine the initial condition regarding velocity, it is first necessary to determine the initial spin of the outer shell upon c ontacting the ground. It can be determined by applying the conservation of kinetic energy between the incident and initial impact of the tw o-mass model as follows: 2 222 112 2 2 12 1 0 ))0((21 21))0((21))0((21 21 21• • • + + + = + = θ ω ω I I yM xM I MV T (20) 23 In equation (17) above, T 0 is the total incident kinetic energy of the ball. This energy can be expressed in terms of the ball as a whole, as well as in terms of the two- mass model, as expressed by the terms on right hand side. During the no slip case, from equation (8 ), the initial conditio ns are related as: )0( )0( 2• • =θR x Hence the energy equation can be solved for the initial spin of the outer core as follows: 2 22 2 112 12 1 2))0(()0( MRIyM I I MV +− − +=• • ω ωθ (21) where ) 0(1ω ω= The initial condition for θ related to spin velocity will be: 2 22 12 2 1 ))0(( ))0(()0( )0( MRII I MVX +− +− =• ω ωω θ Thus the solution of equation (16) is given as follows: ) sin()0()( t e tdt dn θωζ θωωθθθ θ−• = (22) where the following definitions are given: 24 )1 1(2 2 1 MRI IKn++ =θ θω θθ θωζ nC 2= 21θ θ θ ζ ω ω − =n d As can be seen from equation (17), this equation has a similar form as the one for the case of vertical motion with a linear spring-mass-damper. Differentiating equation (17) with respect to time, the relative spin velocity is obtained i.e., ) cos( )0() sin()0()( t e t e tdt dt dn n n θωζ θωζ θθ θω θ ωωθωζθθ θ θ θ −•−• • +−= (23) So the spin velocity at rebound can be de termined by substituting for ‘t’ the value of t c from equation (3). This is an based on assumption that the spin velocities of both the inner and the outer masses attain the same values well before the contact is over and hence the spin velocities at rebound can be determined by using the contact time of vertical motion. Sliding motion In the sliding scenario, as indicated by equation (9), the tangential force (friction) and the normal contact force are related by the coefficient of dynamic friction. Using this equation in equation 12, the following equation for the spin of the outer core is obtained: 0 ))( sgn())( )(( ))( )(( )( 1 2 1 222 = + − + − +• • •• RtF t t Kt t Ct IYµ θ θ θ θ θθ θ (24) 25 During contact between the ball and the ground, the normal contact force can be expressed as follows: )( )( )( tKytyC tF Y − −=• (25) Substituting equation 20 into equation 19, the following differential equation results: )])( )([ sgn())( )(( ))( )(( )(1 2 1 222 tKytyCR t t Kt t Ct I + = − + − +• • • •• µ θ θ θ θ θθ θ (26) VELOCITY OF CONTACT POINT AND SIGN OF CONTACT FORCE In the two-mass model of the tennis ball, the ball moves as a rigid body in the x- direction. The direction of th e sliding friction force from th e ground to the outer core of the ball depends upon the direction of the velo city of the contact point on the ball at the instant the outer core comes into contact wi th the ground; i.e. the sliding friction force acts opposite to the initial contact point velocity. The contact point initial velocity can be determined as follows: CAXA ViV V /^ 1→ → + = (27) In equation (23), the first term on the right hand side is the velocity of the center of mass of the ball in x-dire ction, as soon as it contacts the ground, whereas the second term is the relative velocity of the contact point with respect to the center of mass, considering center of mass as fixed and th e contact point moving about point C with angular spin ω. 26 However, this computation of the cont act point velocity requires careful consideration with regards to the possibiliti es of incoming angular spin as topspin, backspin or zero spin. Consequently, each one is considered on the next pages: Topspin In topspin, the direction of spin is su ch that the surface velocity of the bottom point due to angular spin alone is in opposition to the center of mass velocity. Consequently, from equation (22), the cont act point velocity can be evaluated as follows: ^ 1 1 ) ( i R V VXA ω− =→ (28) Hence the direction determination of slidi ng friction force is now strictly a matter of whether the center of mass translational ve locity is greater than or less than the rotational surface velocity, which is the R ω1 term (In case they are equal at the very onset of impact, then V A is zero and it is then a case of rolling motion). Based on equation (23), then the sign f unction relevant to the sliding friction force as described in equation (9 ) can be evaluated as follows: 1 1 ωR VX> , sgn = +1 ,1 1 ωR VX< sgn = -1 27 Backspin From equation (23), the contact point veloc ity can be determined for the case of incident backspin as follows ( ω1 is positive for backspin in the chosen coordinate system): ^ 1 1 ) ( i R V VXA ω+ =→ (29) Physically, in backspin, the direction of incoming angular spin is such that the velocity of the bottom point of the ball due to spin alone is in the same direction as the center of mass velocity at the moment of impact. Inspecting equation (24), it can be immedi ately concluded that the sign of the sliding friction force acting on the ball for the case of incoming backspin can be determined as follows: Sgn = +1 Zero spin For incident zero spin, equa tion (22) indicates that: ^ 1iV VXA=→ (30) Consequently, sign of the sliding friction fo rce in this case can be determined as follows: S g n = + 1 28 Now, in equation 21, results of equation 2 and equation 4 can be substituted to yield an equation of motion for the relative spin, which is a non-homogenous differential equation and hence must be solved for transi ent and forced respons e (forced response is in the form of the normal force wh ich is also a function of time). With the same procedure as used for the rolling motion and using the notation developed in equation 15, the equation of motion for the relative spin becomes: )]) cos( ) )( sin( [ sgn()()1 1( )()1 1( )( 22 2 t eVCVK V Ct eIRtI IKtI ICt dyt yy dyyy dyy oyyy dytG G oyy oyyωω ωωξωµθ θ θ ωξ ωζθ θ − −• •• − −= + + + + (31) Equation 22 is of the following form: ) cos( ) sin( )( )( )(2 1 t e t e tctbtat tβ γ β γ θ θ θα α+ = + +• •• (32) where the following notation has been used: 1=a )1 1( 2I ICb G+ =θ )1 1( 2I IKc G+ =θ ]) [ sgn( 21 dyyy yy dyyy V C VK IR ωωξ ωµγ − = ) sgn( 22 yyVCIRµγ= y yωξ α−= dyωβ= The particular solution of equation 26 is as follows: 29 ) cos( ) sin( )( t Bet Aett t p β β θα α+ = (33) where: ]) 2() ([ ]) 2(} ) ([{) 2(2 2 1 2 2 2 2 2β αβα β αγγ β αβ α β αβ αβ b ac b a a b a c b ab aA++ + −+ + + + + −+= (34) ) 2() (12 2 β αβγ α β α b ac b a aAB+−+ + −= ( 3 5 ) The homogenous solution of equation 26 is as follows: 0)(=t hθ (36) since: 0 )0( )0( )0( = − =• ω ω θ for sliding motion. The complete solution to equation (22) is, therefore, )( )( )( )( t t t tp p h θ θ θ θ = + = (37) MOTION IN X-DIRECTION Rolling From equation 14, it is seen that: 30 )( )())( )(( ))( )(( )()) ( (1 2 1 222 2 1 2 t Kt Ct t Kt t C t RM M I θ θθ θ θ θ θ θ θθ θ − −= − − − −= + + •• • •• (38) From equation (18) and equation (19), th e solution for the relative spin has already been determined. This can be subs tituted in equation (34) above to obtain a differential equation for outer shell, which can then be solved for angular spin and displacement by successive integrations. Once th e angular spin velocity of the outer shell is obtained, it can be related to the translationa l velocity of the center of mass of the ball in X-direction using equation (8). Hence a complete solution for rebound is obtained in the no-slip case, with the solution for vertical velocity already determined earlier on. Sliding From equations 21, 22 and 28, it is seen that: )]) cos() )( sin( [ sgn()( )( )( 222 t eVCVK V Ct eIRt Kt C t I dyt yydyyy dyyy yy dyt oyyoyy ωω ωωξωµθ θ θ ωξωζ θ θ −−• •• −− + − −= (39) Hence from equations 21, 26 and 28, angul ar acceleration of the outer shell is determined and thus by successive integrations, its angular velocity as well as displacement as a function of time during contact can be determined. For the sliding motion, motion in the X direction is independent of the spin velocity of the outer shell. Using equati ons 7, 9 and 20, it can be seen that the acceleration in X direction can be simply expressed for the sliding case as follows: 31 • • • •• − + −= ))0( )0( sgn())( )(( )(2θµR x tkytyCMtx (40) Then by utilizing equations 2 and 4, and substituting the results in equation 31, the acceleration of the ball in X direction during sliding, as a function of time, is obtained. Then successive integrations lead to the velocity and displacement of the ball in the X direction. Equations developed thus for the tennis ba ll in both scenarios i.e., slip and no- slip, then represent the two dimensional m odel and the motion during the contact with the ground. With the knowledge of the motion, the contact forces can be formulated and the contact problem completely solved. EFECT OF HIGH INCIDENT VERTICAL VELOCITY COMPONENT ON ROLLING MOTION In rolling motion, when there is high inci dent vertical velocity, it is possible that the ball squashes asymmetrically and as a result, the normal force of contact does not pass through the center of the ball but has so me x-eccentricity about the center, as shown in the figure below: 32 Fig. 7 Effect of high incident velocity on the ball during rolling. As shown in figure 7, the way in which th is effect is taken into account in the present model is simply that since the ball is considered as a rigid body in the horizontal motion (x-direction), the normal contact force F Y(t) is offset from the common center of mass X-coordinate by a distance ‘ ε’. This results in a moment about the center of outer shell. Accordingly, the equations of rotationa l motion for the outer core need to be re- formulated and then the coupled rotational equa tions of motion need to be solved to get the solution considering this effect. 2 θ F X (t)) FY(t) θ1S gn( X Y KΘ Cθ 33 Equations (1), (2), (4) and (8 ) still hold. Equation of motion for the outer shell can be re- written as follows: 0 )( )( ))( )(( ))( )(( )(1 2 1 222 = + − − + − +• • •• ε θ θ θ θ θθ θ tFRtF t t Kt t Ct IY X (41) Equation (36) is same as equation (12) except for the moment term due to normal contact force F Y(t). Utilizing equations (13) and (20) , equation (36) can be written as: ))( )( ( ))( )(( ))( )(( )() (1 21 2 22 2 tyKtyC t t K t t Ct MRIy y + = − + − + +• • • •• ε θ θ θ θ θθ θ (42) Adopting the same procedure as in analysis of rolling without the eccentric force, the two rotational equations of motion can be written as: )( )( )( 1 11 tIKtICt θ θ θθ θ− −=• •• (43) ))( )( ( )( )( )(2 22 22 22 tyKtyC MRIt MRIKt MRICty y + ++ ++ +=• • •• εθ θ θθ θ (44) where )( )( )(2 1 t t t θ θ θ − = Subtracting equation (39) from (38), the following equation of relative rotational motion between the inner core and th e outer shell is obtained: 34 )) cos( ) sin( ) (()()1 1( )()1 1( )( 11 2 22 2 12 2 1 t eVCt eVC K MRIt MRI IKt MRI ICt dyt yy dyt dyy y yy yyy yyω ωωωζεθ θ θ ωζ ωζθ θ − −• •• + − +−= ++ + ++ + (45) Equation (40) is a second-order, non- homogenous, linear differential equation with constant coefficients. Solution to (40) consists of complementary function plus particular integral. Complementary solution to equation (40) is identical to that of equation (16) and is given by equation (17). In order to obtain the particular integral , equation (40) is written in the following notation to facilitate the description of solution: ) cos( ) sin( )( )( )( 2 1 t e t e tctbtat tβ γ β γ θ θ θα α+ = + +• •• (46) In equation (41), upon comparison with equation (40), the coefficients are defined as follows: 1=a )1 1(2 2 1 MRI ICb ++ =θ )1 1(2 2 1 MRI IKc ++ =θ ) (1 2 21 y yy y dyyC KV MRIωζωεγ − +−= 1 2 22 yyVC MRI+−=εγ y yωζ α−= dyωβ= Then the particular solution to e quation (41) is given as follows: ) cos( ) sin( )( t Bet Aett t p β β θα α+ = (47) 35 where: ]) 2() ([ ]) 2(} ) ([{) 2(2 2 1 2 2 2 2 2β αβα β αγγ β αβ α β αβ αβ b ac b a a b a c b ab aA++ + −+ + + + + −+= (48) ) 2() (12 2 β αβγ α β α b ac b a aAB+−+ + −= (49) Hence, complete solution to the rolling problem then becomes: ) cos( ) sin( ) sin()0()( t Bet Aet e tt t dt dnβ β ωωθθα α θωζ θθ θ+ + =−• (50) with A and B given by equations (44) and (45) above. OFFSET DISTANCE AS A FUNCTION OF VERTICAL IMPACT VELOCITY As the vertical component of impact velocity increases, the flatness becomes more pronounced in the tennis ball [1,5]. It can be assumed that the offset distance ‘ ε’ of the normal reaction from center of mass of the tennis ball dur ing rolling is a function of the vertical component of imp act velocity. If a linear func tional relationship is assumed, this can be expressed as: 2 1 )( C VC Vy y + = ε (51) While performing the simulations (Chapter V) for the tennis ball , it was observed that ‘ ε’ varies from 0.02 to 0.5 for vertical component velocities of 5 m/s and 36 16 m/s, respectively. Thes e values are selected based on good agreement with experimental observations. It is then possible to calculate the values of coefficients C 1 and C 2 using these extreme values. Thus, the following linear equations in C 1 and C 2 result: 2 12 1 )16( 5.0)5( 02.0 C CC C + =+ = (52) Solving these two equations: 189.00431.0 21 −== CC Thus the offset distance as a linear function of vertical component of impact velocity is expressed as: 189.0 0431.0)( − =y y V Vε (53) Using equation (49), the offset distance can now be obtained at any value of vertical componet of impact velocity between 5 m/s and 16 m/s or beyond 16 m/s. Equation (49) has been utilized in Chapter V in simulations whenever the ball’s motion changes from sliding to rolling. Then the ‘ ε’ value is needed and can be calculated using equation (49) to use in rolling motion simulation. 37 TRANSITION BETWEEN SLIDING AND ROLLING: TIME-AVERAGED COEFFICIENT OF FRICTION Recall from the moment equation writte n for the outer shell in pure rolling (equation (42) in Chapter II), that due to th e eccentricity of the normal force with respect to the center of mass of the two-mass syst em, there is a counter-clockwise moment, which is expressed as: () EYM Ftε=− ( 5 4 ) In tennis ball dynamics, the moment e xpressed in equation (54) corresponds to the opposing moment that tends to retard the ball while it is in rolling motion. The opposing moment given by equation (54) can be converted to an equivalent “rolling friction force F R”, as follows: 0 0()c ct Y R tFt d t F Rdtε− =∫ ∫ ( 5 5 ) Equation (55) represents the average opposing moment acting on the tennis ball during contact. The radius used to evaluate the average friction force is the radius of the outer shell,’R’. In a similar way, the averag e sliding friction force can be expressed as: 0 0()c ct Y SS tFt d t F dtµ=∫ ∫ ( 5 6 ) 38 Since the rolling friction de fined by equation (55) is less than sliding friction, RSFF< It follows from equations (55) and (56) that: SRεµ< (57) Or: RSµ µ< ( 5 8 ) In (57), µR can be defined as rolling friction co efficient and is given as the ratio of the eccentricity distance fr om the center of mass to the radius of the ball. The inequality (56) is obtained because ‘ ε’ is not a function of tim e (equation (55)) and hence the remaining terms in equations (54) and (55) simply cancel out. Recall from equation (53) that ‘ ε’ is a function of the vertical component of incident velocity. From inequality (57), it can be seen that for the values of coefficient of sliding friction 0.55 and radius of the tennis ba ll 1.3 in., the offset distance must satisfy the following constraint: 0.715 in ε< From equation (53), the maximum value of ‘ ε’ for the simulations in case of maximum value of vertical component of inci dent velocity of arounf 16 m/s turns out to be 0.500 in. Thus constraint (58) is satisfied in all the simulations whenever the rolling takes place during motion of the tennis ball with the ground. 39 In tennis ball dynamics, collision problems that involve rolling during contact always start from pure sliding contact and then during contact, the motion of the tennis ball changes from sliding to pure rolling. In such problems , an expression for a time-averaged coefficient of friction, can be defined as follows: ∫∫ = cc t Yt X dttFdttF 00 )()( µ (59) Thus, a problem that involves transition from sliding to the pure rolling during contact can be identified or its accuracy of solution can be checked by evaluating the time-averaged coefficient of friction in equa tion (59) and comparing its value against the coefficient if kinetic friction in pure slid ing. If there is a motion transition during contact, then the value evaluated from equation (59) will be appreciably less than the coefficient of kinetic friction of the surface involved in contact. Hence the time-averaged value calculated using e quation (59) serves as a check that shows if s µµ< ,then rolling probably occurs . Equation (59) can be simplified, si nce the integrals in numerator and denominator on the right hand side can be expr essed in terms of the mass and velocities of the ball in the two coordi nate directions, using Newt on’s second law of motion as follows: dtdvm tFX X −=)( (60) 40 dtdvm tFY Y −=)( (61) Integrating equations (60)and (61) from st art of collision (t = 0) to the end of collision (t = t c), following expressions are obtained: ) ( )(1 02 Xt X X V Vm dttFc ∫− −= (62) ) ( )(1 02 Yt Y Y V Vm dttFc ∫− −= (63) In equation (61), the rebound component of velocity in the vert ical direction can be related to the incident component of ve locity in the vertical direction by an experimentally determined coefficient of restitution as follows: 1 2 ) (Y Y V COR V −= (64) When equations (62) to (64) are in serted in equation (59), the following expression for the time-averaged coefficient of friction results: 12 1 ) 1(YX X V CORV V +−=µ (65) Thus, if a tennis ball impact problem is solved with transi tion of motion from pure sliding to pure rolling dur ing the contact (as determin ed by the condition when the contact point velocity goes to zero as moti on changes from pure sl iding to pure rolling), the solution of the velocities can be incorpor ated in equation (65) to calculate the time- averaged coefficient of friction during the c ontact. If transition occurs during the contact, then the value calculated from equation (65) wi ll be less than the coefficient of kinetic 41 friction for the pure sliding. If no rolling o ccurs during contact and there is only sliding motion, the value calculated from equation (65) will then be almost equal to the value of the coefficient of kinetic friction. Equation (65) can also be applied to experimentally determined data for the tennis ball dynamics to find out whether tran sition from sliding to pure rolling occurs during particular cases of impact. Equation (65) is used in Chapter IV to calculate the time-averaged coefficient of friction for an experimental data on tennis ball impact [1], for estimates of transition from slid ing to pure rolling during contact. INNER AND OUTER CORE DYNAMIC PARAMETERS In order to perform dynamic simulation of the tennis ball using the two-mass model, it is first necessary to ascertain physic ally feasible values of the inner and outer core dynamic parameters. These parameters include masses, mass moments of inertia, and radii of the two masses in the model. There are various possibilities as regard s to the feasible dynamic parameters. First of all, it is necessary to describe the equations for various dynamic parameters which are interrelated to each other. After that, it is possible to derive various combinations of dynamic parameters for the two masses. Inner core Mass of the inner core can be written in terms of its weight density and volume as follows: gR gVM)34(3 1 1 11 1π γγ= = ( 6 6 ) 42 Mass moment of inertia of the inner core can be described by th e following equations: 2 11 132RM I= (67) Outer core In a similar manner, mass of the outer shell can be described as: gR R gVMi )) (34(3 23 2 22 2− = =π γγ (68) And its mass moment of inertia can be expressed as: 2 22 252RM I= (69) In order that this model truly represents a tennis ball, there are simple constraints on the inner and outer cores’ masses and mass moments of inertia and these are that the sum of the inner and outer core masses a nd mass moments of inertia should equal, respectively, to the mass and mass moment of inertia of a real tennis ball. These constraints can be expressed as follows: M M M = +2 1 (70) I II = +2 1 (71) For a tennis ball, the aver age mass is about 0.000329 lb-s2/in whereas its mass moment of inertia about its ce ntroidal axes is 0.00028 lb-s2-in (Chapter IV). 43 Final constraints to be placed are dimensional constraints on the radii of inner and outer core. For a tennis ball, maxi mum radius of outer shell is 1.3 in. [11] and for inner core, the radius must be less than the outer shell. Mathematically, this can be expressed as: 3 .1 2=R in. (72) 2 1R R< Based on equations (66) to (72), the following physically possible sets of dynamic parameters can be obtained: (a) M 1 = M 2 = M/2 I1 = 0.0001008 lb-s2-in = 0.353I I2 = 0.0001853 lb-s2-in = 0.647I I = 0.00028 lb-s2-in R1 = 1.24 in. γ1 = 0.008 lb/in3 γ2 = 0.06 lb/in3 (b) M 1 = 2M/3 M 2 = M/3 I1 = 0.0001403 lb-s2-in = 0.532I I2 = 0.0001235 lb-s2-in = 0.468I I = 0.000264 lb-s2-in R1 = 1.26 in. γ1 = 0.01 lb/in3 γ2 = 0.07 lb/in3 44 ( c ) M 1 = 3M/4 M 2 = M/4 I1 = 0.000165 lb-s2-in = 0.641I I2 = 0.00009267 lb-s2-in = 0.359I I = 0.000257 lb-s2-in R1 = 1.29 in. γ1 = 0.0105 lb/in3 γ2 = 0.300 lb/in3 In calculating the weight density for the outer shell in case (a), the inside radius of the shell has been taken as 1.25 in. For cases (b) and (c), this value has been taken as 1.27 in. and 1.295 in., respectively. In Chapter III and Chapter V, simulations of the tennis ball have been performed using these dynamic parameters. It should be not iced that in cases (b) and (c), the sum total of mass moments of iner tia of inner and outer masse s are not equal to 0.00028 lb- s 2-in, but instead these values tu rn out to be 0.000264 and 0.000257 lb-s2-in, respectively. These values obtained for total mass moment of inertia are not very far off the experimentally determined value of 0.00028 lb-s 2-in (Chapter IV), with percentage difference as 6 % and 7%, respectively. This might as well be the uncertainty in experimental measurements. The reason these dynamic parameters have been used in Chapter III and Chapter V is that thes e parameters give good agreements with experimental rebound motion of the tennis ball, and the error in mass moment of inertia is not large. 45 CHAPTER III GRAPHICAL RESULTS OF SOLU TIONS OF EQUATIONS OF MOTION In the previous chapter, the equations of motion for the tennis ball impact with the ground have been derived and their solutions fo rmulated for the cases of the tennis ball’s motion as the slip and the no- slip scenarios during its contact with the ground. Kinematic and kinetic parameters of impor tance during the contact, as found from the analysis of the equations of motion for both cases of the slip and the no-slip are described below: (a) Vertical displacement of the ba ll as a function of time ( y(t)) (b) Vertical velocity of the ball as a function of time ( )) (ty• (c) Horizontal velocity of the ball as a function of time ( ) (tx• ) (d) Angular spin of the ball as a function of time ( )) (2t• θ (e) Normal contact force as a function of time ( ))(tFY (f) Tangential contact force (friction force) as a function of time ( )) (tFX Solutions for the equations of motion with re gards to the above me ntioned kinematic and kinetic parameters are described as follows: SLIDING THROUGHOUT THE CONTACT For reference, solutions of the equations of motion for the sliding case are repeated below: 46 ) sin( )(1t eVtydyt dyyyyωωωζ−= (2) ) cos( ) sin( )(11t eVt eVtydyt y dyt dyyy yyy yyω ωωωζωζ ωζ − −• +−= (4) Equations (2) and (4) are equally valid fo r the no-slip case, since the y-motion, according to the two-mass elastic model as well as per Newton’s second law of motion, is not affected by either scenario. )]) )() () cos( )() () sin( )() (() sin( [ sgn( )( 2 22 2 2 211 1 dy y ydydyt dy y ydy dyt dy y yy y dyy ydyt dyy y X t e t eVKt eVCMVtx yy yyyy ω ωξωω ω ωξωω ω ωξωξ ωωωµ ωξ ωξωξ +++− +−+ − = − −−• (73) dtt e t e dtt Bet AeIKBt Bet AeICt dyttt dyt dyt dytdyt dyt yy yy yy yyyy yy )] cos( () sin( [( )) cos( ) sin( ()) cos( ) sin( ( )0( )( 2 001 2222 ω γ ω γ ω ωω ω θ θ ωξ ωξ ωξ ωξ θωξ ωξ θ − − − −− −• • + + ++− + + = ∫∫(74) )( )( )( tyKtyC tFy y Y − −=• = ) cos( ) )( sin(11 1t eVCVK V Ct edyt yy dyyy dyyy yy dytyy yyωω ωωξωωξ ωζ − −− − (75) 2 ( ) ( )sgn( (0) (0))XYFt F t x R µθ•• =− ( 1 1 ) 47 These equations are plotted in MathCAD using the codes shown in Appendix B. For each physical variable, the curves are plotted parametrically in MathCAD. The results and their descriptions ar e described on the following pages. Vertical displacement as a function of time 0 0.2 0.4 0.6 0.8 100.20.40.60.8 Vy1=5.33 m/s Vy1= 10 m/s Vy1 = 15 m/sIndentation as a function of timeIndentation, inches Dimensionless contact time, n Fig. 8 Vertical displacement during contact as a function of time. Figure 8 shows the predicted vertical di splacement of the inner core mass as a function of time, during the contact of the outer core with the ground. The graphs are plotted for certain values of the inner core mass, a damping coefficient of the vertical +θ -θ X Y+ω 48 damper, and a stiffness coefficient of the ve rtical damper. The thr ee curves correspond to three different incident vertical velocity components. It can be seen that maximum displacement of the inner core mass, co rresponding to maximum compression of the tennis ball during contact, increases as the in cident vertical veloci ty increases. Also it can be observed from figure 8 that due to presence of the damper in the model, the curves are not symmetrical but are rather asymmetrical. This agrees well with the physical measurements of the tennis ball displacements [3]. Vertical velocity as a function of time 0 0.2 0.4 0.6 0.8 16420246 Cy = 0.01 (e=0.888) Cy = 0.02 (e=0.788) Cy = 0.03 (e=0.699)Vertical Velocity During ContactVertical velocity , m/s Dimensionless contact time, n Fig. 9 Vertical velocity during contact as a function of time. Figure 9 shows the curves of the vertical velocity during the c ontact as a function of time. The three curves correspond to different coefficients of restitution, which in turn is determined by a combination of the inner co re mass, the stiffness of the spring and the +θ -θ X Y+ω 49 vertical damping coefficient, as can be s een from equation (6). The mass and stiffness are the same in the curves in figure 9, only the damping coefficients are different. Figure 9 indicates that the higher the coefficien t of restitution in the vertical direction, the higher is the value of the rebound velocity in the same di rection and vice versa. If the coefficient of restitution approaches unity, the rebound velocity in th e vertical direction will be almost the same as the incide nt velocity in vertical direction. Horizontal velocity as a function of time 0 0.2 0.4 0.6 0.8 110121416 Mu = 0.35 Mu = 0.55 Mu = 0.75X-Velocity in Sliding vs Contact Time nHorizontal velocity, m/sIncident horizontal velocity = 16 m/s (Vx1 > R ω1) Dimensionless contact time,n Fig. 10 Horizontal velocity during c ontact as a function of time. +θ -θ X Y+ω 50 0 0.2 0.4 0.6 0.8 105101520 Vx1 = 16 m/s Vx1 = 10 m/s Vx1 = 5 m/sX-Velocity in Sliding vs Contact Time nHorizontal velocity, m/sFriction coefficient = 0.55 (Vx1 > R ω1) Dimensionless contact time,n Fig. 11 Horizontal velocity during contact as a f unction of time(effect of initial velocity). Figures 10 and 11 show the horizontal velo city of the center of mass of the ball during the sliding scenario as a function of time. There are two graphs for the horizontal velocity during the contact. In figure 10, the incident horizontal velo city is a constant whereas the coefficient of sliding friction is different for the three curves. In figure 11, the coefficient of sliding friction is same, whereas the incident horizontal velocities are different. In figure 10, it can be seen that the hori zontal velocity is decreasing during the contact. This is due to th e direction of friction force which is acting in an opposite direction to the incident horizontal velocity for this particular case considered. This is always the case with the zero spin or backspin incidences, but not always for the topspin. The higher the coefficient of sliding friction, the greater is the decl ine in the horizontal velocity during the contact and subseque ntly the rebound horizontal velocity. From figure 11, it can be seen that the fric tion slows down the ball as expected. 51 0 0.2 0.4 0.6 0.8 116182022 Mu = 0.35 Mu = 0.55 Mu = 0.75X-Velocity in Sliding vs Contact Time nHorizontal velocity, m/sIncident horizontal velocity = 16 m/s ( Rω1>Vx1) Dimensionless contact time,n Fig. 12 Horizontal velocity during contact as a function of time(high topspin). The horizontal velocity during the slidi ng scenario can also increase during the contact, as shown by figure 12. This can ha ppen only when the ball is incident with a high topspin so that the initial surface velocity due to the spin is greater than the incident horizontal velocity. In this case, the physic al situation is revers e of the other cases presented in the other graphs. From figure 12, the higher the fric tion (corresponding to a higher sliding coefficient of friction), the greater the rebound horizontal velocity. +θ -θ X Y+ω 52 Angular velocity as a function of time 0 0.2 0.4 0.6 0.8 12000200400600800 Topspin Zero Spin Back SpinAngular spin as a function of timeAngular spin, rad/s Dimensionless contact time, n Fig. 13 Angular spin during contact as a function of time. The angular spin of the outer shell, interpreted as the a ngular velocity of the ball, is shown in figure 13 for three different incide nt cases, namely the zero spin, the top spin and the back spin. It can be s een that the angular spin duri ng the contact is increasing as a function of time for all the three cases. Th e angular spin decreases to zero and then reverses its direction for the back spin, in this particular case considered. For the top spin, the increase in the angular spin occurs when the surface velocity due to the incident spin is less than the incident horizontal velocity, but the angular velocity decelerates for the top spin if the rotational surface velocity at th e incidence is greater than the incident horizontal ve locity of the center of mass. The reason for the increment of the angular velocity during the contact for all th e cases considered here is that the +θ -θ X Y+ω 53 torque exerted on the outer shell of the mode l due to the friction force is acting in an anti-clockwise direction (the positive direction as per the c oordinate system selected for the analysis of the model) and due to this torque, the outer shell moves in the positive rotational direction and this is the reason th at the angular spin increases with time. 0 0.2 0.4 0.6 0.8 1100200300400500600700 Mu = 0.35 Mu = 0.55 Mu = 0.75Angular spin as a function of timeHorizontal velocities, m/sIncident spin = 140 rad/s (Vx1 > R ω1) Dimensionless contact time, n Fig. 14 Angular spin as a function of time. Figure 14 shows the angular spin as a func tion of time with an incident top spin. The three curves in figure 14 co rrespond to three different s liding friction coefficients and indicate that the higher the friction, th e more spin the ball will acquire during the contact. This argument always holds true fo r the zero spin and the back spin cases, but not always necessarily for the op spin incidence. In the top spin case, if the incident +θ -θ X Y+ω 54 surface velocity due to the spin is greater than the incident horizontal velocity, the friction force acts in a direction opposite to the surface velocity and due to the frictional torque, the angular spin decrea ses as a function of time, wher eas the horizontal velocity increases as a function of time. In that case, then the higher the friction, the lower the rebound angular spin and vice versa. Normal contact force as a function of time 0 0.2 0.4 0.6 0.8 16040200 Vy1 = 5 m/s Vy1 = 10 m/s Vy1 = 15 m/sNormal contact force as function of timeNormal contact force, lb Dimensionless contact time, n Fig. 15 Normal contact force as a function of time. Figure 15 shows the normal contact force as a function of time. The three curves correspond to different incident ve rtical velocities. It can be s een from the curves that the higher the incident vertical velocity, the larg er the magnitude of the compressive normal force developed in the spring-damper combin ation. Physically, it will correspond to a +θ -θ X Y+ω 55 greater magnitude of normal force exerted on the ball’s surface as it comes in contact with the ground. The three curves all corres pond to fixed inner core mass, stiffness and damping coefficients. Also, it should be obs erved from the curves that due to the presence of a finite amount of the damping in the vertical direction in the system, the normal force curves are not symmetrical and th eir maxima do not occur at n = 0.5, but rather they occur before n = 0.5. This prediction is consistent with the physical measurements of the normal contact forces conducted on the tennis balls [3,8,10]. Tangential (frictional) contact force as a function of time 0 0.2 0.4 0.6 0.8 13020100 Mu = 0.35 Mu = 0.55 Mu = 0.75Frictional force as a function of timeFrictional force, lbIncident vertical velocity = 10 m/s Dimensionless contact time, n Fig. 16 Frictional force as a function of time. +θ -θ X Y+ω 56 Figure 16 shows the tangentia l contact force as a function of time for three different values of the sliding friction coeffi cients. Due to the direct relationship between the tangential force and the normal force as implied by Coulomb’s law for sliding friction, it can be seen that the shape of th e tangential force curve during the sliding is the same as the normal contact force curve, only that it is scaled by the friction coefficient factor. The higher the value of the sliding friction coefficient, the greater the magnitude of tangential force during the contac t and vice versa. It s hould be noted that the sliding friction force can also be in the positive coordinate dire ction, as opposed to the curves shown in figure 16. This can happen when the relative tangential velocity at the point of contact is directed in the negative x-direction. This itself is possible only for the topspin incident, when the surface velocity due to the spin is higher than the incident horizontal component of the translational ve locity of the center of mass of the ball. NO-SLIP THROUGHOUT THE CONTACT In the no-slip case, the equations for the horizontal velocity, a ngular spin of outer shell (and the inner core) and the tangential contact force are different from the ones presented for the sliding case. These are presented as follows: ) cos( ) sin( ) sin())0( )0(()(2t Bet Aet e tt t dt dnβ β ωωθ ωθα α θωζ θθ θ+ +−=−• 57 ) sin( ) cos( ) cos() sin( ) cos( ))0( )0(() sin())0( )0(()( 22 t Be t Be t Aet Ae t e t e t t t tt dt dt dn n n β β β α β ββ α ω θ ω ωωθ ω ωζθ α α αα θωζ θωζ θθ θ θ θ θ θ − + ++ − +− −=−•−• • ))( )( ( )( )( )(2 22 22 22 tyKtyCMRItMRIKtMRICty y ++++++=• • •• εθ θ θθ θ ∫∫++++ =• • •tt dttMRIKdttMRICt 002 22 222 )( )( )0( )( θ θ θ θθ θ )( )( 2t Rtx• • =θ )( )( 2t MRtFX•• = θ The expressions for the vertical displacement, vertical velocity, and the normal contact force remain the same as in the sliding case. The above solutions are implemented in MathCAD software using the code given in Appendix B. These equations are plotted on the following pages: 58 Angular velocity as a function of time 0 0.2 0.4 0.6 0.8 1200300400500600 Vx1 = 20 m/s Vx1 = 16 m/s Vx1 = 10 m/sAngular Spin VelocityAngular Spin, rad/sZero incident spin Dimensionless contact time, n Fig. 17 Rolling angular velocity as a function of time. Figure 17 shows the rolling angular velo city curves as a function of time, corresponding to three different incident hor izontal velocity com ponents and neglecting the effect of the eccentricity of the normal force through the center of the ball. All the curves correspond to the case of zero spin. The curves indicate that the higher the incident horizontal component of the translati onal velocity, the larger will be the rolling angular velocity. Initial conditions of the spins as shown on n = 0 correspond to the initial condition obtained for the outer shell as indicated on page 20, Chapter II. Thus the curves simply do not start from the initial c ondition of Vx1/R, but rather from the initial condition as defined in equation (18) for the out er shell. From the curves in figure 17, it is also observed that the spin decreases and well before the contact is over (which occurs at n =1), the angular spin increases and then attains a steady state value. An explanation of this behavior in terms of the model is that since the inner and the outer cores, as far as +θ -θ X Y+ω 59 the rotational motion is concerned, are connect ed with a linear torsional spring –damper system, then as soon as the half-period of vibr ation of the outer core is completed, it is acted upon by torques from the torsional spring and the damper and this causes acceleration of its angular velocity. After this , the angular spin attains the steady-state value. This is due to the high damping of th e torsional damper as well as the reason that the frictional force keeps on decreasing in magnitude so that frictional torque is balanced by the internal torques, namely from the spring and the damper. As a result, any further vibrations die out and the outer core rotates as a rigid body with almost a constant value of the angular spin, towards the end of the c ontact. This initial deceleration of the spin and attainment of a steady-state value is always the case with the zero spin and backspin impacts, but not always necessarily for the topspin impacts. It should be noticed that the curves as shown in figure 17are not the only possibility of the motion of the outer shell. If the initial angular spin of inner core is higher than the initial spin velocity of the outer shell compatible with the rolling, which is determined in Chapter II, only for the case of the topspin, then in that case, the outer shell will be accelerated instead of being re tarded, because the initial torque acting on the outer shell from the torsional damper will be acting in the positive rotational direction. The graphs for such a case of topspin are presented in figure 18 on the next page. 60 0 0.2 0.4 0.6 0.8 1450500550600650700750 Vx1 = 20 m/s Vx1 = 15 m/s Vx1 = 10 m/sAngular Spin VelocityAngular Spin, rad/sIncident top spin Dimensionless contact time, n Fig. 18 Angular spin of outer shell as a func tion of time (Special case of topspin). 0 0.2 0.4 0.6 0.8 1250300350400450 Incident top spin (200 rad/s) Incident zero spin Incident back spin (-200 rad/s)Angular Spin VelocityAngular Spin, rad/sInitial horizontal velocity = 15 m/s Initial vertical velocity = 6 m/s Dimensionless contact time, n Fig. 19 Angular spin of outer shell as a function of time. +θ -θ X Y +θ -θ X Y+ω +ω 61 In Figure 19 are shown the angular spin curves of the outer shell corresponding to three different incident spins of the ba ll. The curves correspond to the zero spin, the topspin and the backspin impact, when the angles of incidence and the incidence translational velocities are the same for all the three cases. The curves indicate that the outer shell will rotate at a higher spin for the topspin impact as compared to the zero spin and the backspin. The backspin yields the lowest value of the rolling angular spin. This is due to the reason that in the model consid ered, the initial torque opposing the spin of the outer shell will be highest for the backspin, due to its spin direction as compared to either of the zero spin or the topspin. For the top spin, the initial torque will be lower than the zero spin, because this torque is pr oportional to the differe nce in initial angular spins of the inner mass and the outer mass. Sin ce the difference in initi al angular spins is less for the topspin case and hence the value of the initial torque oppos ing the spin of the outer shell is less as compared to the zero spin, the outer shell will spin faster as compared to the zero spin. As the outer shell rotates, the torsional spring connecting the inner and the outer masses starts winding. When it is unwinded, it starts exerting a torque on the two masses causing an acceleration for the outer shell and deceleration for the inner core. In figure 19, this acceleration of the outer shell is indicated as a hump in the curves of the angular spin velocities. Finally, the effect of offset distance fr om the horizontal coordinate of the center of mass of the system to the normal reaction force, neglecting the weight of the outer shell, can be observed on the angular spin of the outer shell as shown in figure 20 on the next page. From figure 20 on the next page, it can be seen that the higher the offset distance, the lower the rolling spin, keeping the rest of the dynamic parameters as same. The effect of the offset is often referred to as “rolling friction”. 62 0 0.2 0.4 0.6 0.8 1500550600650 Epsilon = 0 Epsilon = 0.1 Epsilon = 0.5Rolling Spin Velocity Rolling Spin Velocity, rad/s Dimensionless contact time, n Fig. 20 Angular spin velocity as a function of time (effect of offset distances). Figure 20 shows curves corresponding to th ree different offset distances. The graphs indicate clearly that the torque developed by the normal force due to the offset (which happens pronouncedly when the incident vertical component of the velocity gets higher and higher) slows down the angular spin during the rolling motion. It develops a torque which acts on the outer shell and opposes its angular spin. This is true for the cases of zero spin, topspin and backspin. For the zero offset distance, it can be seen that the rolling angular velocity a ttains almost a steady-state va lue well before the contact ends. 63 Horizontal velocity as a function of time 0 0.2 0.4 0.6 0.8 18101214161820 Vx1 = 20 m/s Vx1 = 15 m/s Vx1 = 10 m/sX Velocity during Rolling ContactHorizontal velocity,m/sIncident spin = 200 rad/s Incident vertical velocity = 6 m/s Dimensionless contact time, n Fig. 21 Horizontal velocity as a function of time (effect of in itial conditions). 0 0.2 0.4 0.6 0.8 113141516171819 Omega 1 = 100 rad/s Omega 1 = 200 rad/s Omega 1 = 300 rad/sX Velocity during Rolling ContactHorizontal velocity,m/sIncident horizontal velocity = 20 m/s Incident vertical velocity = 6 m/s Dimensionless contact time, n Fig. 22 Horizontal velocity as a func tion of time (effect of spin). +θ -θ X Y+ω 64 Figures 21 and 22 on the previous page (f igures 21 and 22) show the horizontal velocity of the center of mass of the two-ma ss elastic system as a function of time during the no-slip case. Figure 21 corresponds to three different incident hori zontal velocities of the center of mass. Figure 22 corr esponds to three different incident spins. All cases are for the topspins. For the zero spin and backspin, trends of the curves will be the same. From figure 21, it can be seen that th e horizontal velocity during the contact decreases as a function of time. The higher the initial incident horizontal velocity, the larger will be the rebound value of the horiz ontal velocity. In this graph, only the incident horizontal speeds are different; the rest of the parameters used to generate the curves are identical. From figure 22, it can be seen that the ball incident at a higher spin but with the same value of incident ho rizontal velocity, will rebound at a higher horizontal velocity. Thus its angle of rebound will be lower for higher incident spin, if its starts rolling. 65 Tangential contact force as a function of time 0 0.2 0.4 0.6 0.8 13002001000100 Omega 1 = 100 rad/s Omega 1 = 200 rad/s Omega 1 = 300 rad/sTangential contact forceTangential force, lb Dimensionless contact time, n Fig. 23 Tangential friction force as a function of time. The tangential friction force(maintaining th e no-slip constraint) as a function of time for three different incident spins is shown in figure 23. From the graphs in figure 23 it can be seen that the tangential force magnitude increases with less spin and vice versa. It can be seen that the frictional force starts from negative values, t hus validating that it will retard the horizontal velocity during the contact, as can be seen from the previous graphs. The force graph eventually attains th e steady value very near to zero. Thus the friction force decreases with time to such an extent that its magnitude becomes almost equal to zero and then the outer shell star ts rotating as a rigid body subjected to no torques (either from the spring-damper or the friction). 66 TRANSITION BETWEEN SLIDING AND ROLLING DURING CONTACT Based on kinematic parameters presented for the cases of the sliding and the rolling, it is now possible to determine transition from the sliding to the rolling motion during the contact of the ball with the ground. This transitio n usually takes place when the ball is incident at large angles of incidence. The transition from the sliding to the rolling motion during the contact can occur for all of the zero spin, the topspin and the backspin impacts. Fig. 24 Transition from sliding to rolling motion. Figure 24 shows curves of the angular spin during the sliding scenario, multiplied by R corresponding to three different incident spins and the horizontal velocity of the 0 0.2 0.4 0.6 0.8 1 10 5 0 5 10 15 20 Omega 1 = 200 rad/s Omega 1 = -200 rad/s Omega 1 = 0 rad/s Center of mass horizontal velocityOnset of transition Dimensionless contact time, nHorizontal velocities, m/s +θ -θ X Y+ω 67 center of mass of the system. The curves correspond to a specific sliding friction coefficient and the other parameters are the same for each curve. It can be seen from the curves that the horizontal velocity curve intersects the topspin curve at a smaller value of ‘n’ as compared to the zero spin. For this particular case, it does not intersect the backspin curve at all. The intersection poi nt of the curves physically implies the introduction of the kinematic no-slip constr aint, due to which the sliding motion ceases and the ball starts rolling during the contact fo r remainder of the contact. It can be seen that the topspin actually helps in getting to the rolling mode earlier than either of the zero spin or the backspin. The rolling can occur for the backspin case, but then it is a strong function of the sliding friction co efficient of the surface, the horizontal component of incident velocity as well as th e vertical component (angle of incidence) and the magnitude of the incident backspin itself . If the ball is incident at a high value of backspin and the angle of incidence is low with a high value of the incident horizontal velocity, it will keep on sliding throughout the contact, unless the sliding friction coefficient is very high (around 0.7 to 0.9). This is all easily deducible from the velocity and the angular spin curves. In the sliding, it has been observed that the rebound horizontal velocity decreases as the coefficient of friction increases and the incident horizontal velocity is decreases. This physically means a rough surface with a high friction force and a lesser initial velocity (high incidence angle). Due to the high friction force, the velocity reduces at a higher rate (more deceleration) and a stage is reached when the horizontal velocity of the center of mass and the surface velocity due to spin become equal (the angular spin is accelerating at the same time due to the friction torque acting on it). When this stage is reached, as shown by the inte rsection points of the curves, the relative velocity of the contact point with respect to the ground becomes zero and it starts rolling. This can be taken into account in the solutions of the equations of motion for the spin of the outer shell and the horizontal velocity of the center of mass by using in the solutions for the rolling the limits from the instant ‘n rtc’ to t c, instead of from 0 to ‘t c’ where ‘n r’ corresponds to the fraction of time of the total contact time where the curves intersect. When the two solutions are combined in this piecewise 68 manner, this will represent a complete solution including the transition from the sliding to the rolling mode of motion. EXAMPLES FOR ILLUSTRATING THE APPLICATION OF EQUATIONS In order to illustrate the application of the equations of moti on and their solutions as outlined in Chapters II and the beginning of Chapter III, two examples of impact of the tennis ball with the ground are consid ered, which are simulated by the two-mass model. The examples will be the impact of the ball with the ground having friction at a low angle of incidence and a hi gh angle of incidence with the topspin in each case, such that the incident horizontal velo city component is greater than the surface velocity due to the spin. Thus in both cases, the initial contact point velocity will be in the positive direction. For the first case of impact simulation, consider the case of a ball incident with translational velocity of 17 m/s of the cente r of mass, with the topspin of 100 rad/s and an angle of incidence of around 18 degrees, measured with respect to surface of the ground. This is a really shallow angle of inci dence. The initial conditions for impact are calculated as follows: 13.16)18cos(17 cos0 1 = = =i XV V θ m/s 33.5)18sin(17 sin0 1 = = =i YV V θ m/s The values calculated are the initial conditions for the motion in X and Y directions, respectively. The given value of the topspin of 100 rad/s is the initial condition for the spin motion. Suppose further th at the surface with which the ball strikes is acrylic surface for which the coefficient of sliding friction is around 0.55. Also, the vertical coefficient of restitution on this surface can be taken as around 0.76. 69 In order to simulate the above impact problem with the help of the two mass model, first of all, a selection of the dynamic parameters for the two-mass model is required such that their combination makes the vertical coefficient of restitution as 0.76. Accordingly, the dynamic parameters are selected as follows: Inner core mass, 2M/3, outer shell mass M/3. Stiffness coefficient , 80 lb/in (14040 N/m). However, in order to determine the damping coefficient C y, equation (6) must be used. Using equation (8) in Chapter II, the value of C y that gives the vertical coefficient of restitution as 0.76 is found out as: 0113.0=yC lb-s/in (1.984 N-s/m) For selecting the rotational parameters, the inner and outer core mass moments of inertia are selected as 0.532I and 0.468I, resp ectively (Chapter II). The torsional stiffness coefficient of the torsional spring is se lected as 1000 lb-in/rad (113.26 N-m/rad). On the basis of assumption that the damped periods of the relative spin and the vertical motion are equal, the damping ratio in the rotational motion can be calculated as follows: dy dω ωθ= 2 21 1y ny n ξ ω ξ ωθ θ − = − 70 0 0.2 0.4 0.6 0.8 10200400600800 Surface velocity Center of mass velocityFinding out onset of rolling Horizontal velocities, m/s Dimensionless contact time, n Fig. 25 Surface and center of mass velocities during contact. Next, the horizontal velocities of the center of mass of the two-mass system and surface velocity of the outer shell are plotted on the same graph as shown in figure 25. This is done to find out if with the choice of the selected parameters, there is an onset of the rolling motion during the contact or that there is simply the sliding motion right till the end of contact. From figure 25, it can be seen that the graphs do not intersect during the contact time, indicating that the rolling do es not occur and the ball keeps on sliding during the contact. Hence there is no need fo r the piecewise solution during the contact in this scenario, since there is no rolling. Thus in this case, the horizontal com ponent of the velocity at rebound is 12.667 m/s, as read from the graph, at n=1 and found out from MathCAD software. 461.12 2=XV m/s The rebound value of the spin of the outer shell is: 71 586.3712=ω rad/s Positive sign of the spin velocity indicates that the rebound spin will be an increase of topspin. Comparing the value of the rebound with the incident spin, it can be seen that the spin value increases by more than twice, due to the friction torque acting on the outer shell which accelerates the spin. The value of rebound angle is: 0 22644.16 )461.12053.4tan( ) tan( −= − = = aVVa XY rθ The contact force graphs are plotted as follows: 0 0.2 0.4 0.6 0.8 13020100 Normal force Tangential forceNormal and tangential contact forcesContact forces, lb Dimensionless contact time, n Fig. 26 Normal and tangential contact forces. 72 Figure 26 predicts the variation of the normal and the tangential contact forces with the contact time. The value of the time-averaged coefficient of friction for this case can be calculated using equation (63) in Chapter II as follows: 400.0=µ This value is not greatly less than the value of the coefficient of sliding friction, which is 0.5. This shows that indeed in this case, the sliding occurs throughout the motion of the ball. This is also verified fr om figure 25, which shows that the ball does not start pure rolling during contact. Consider the next case of impact simula tion, in which the incident angle is increased from 18 degrees to 42 degrees, measured with respect to the ground surface. The incident topspin is 100 rad/s, whereas th e incident translational velocity is 17 m/s. Thus the initial conditions for impact are calculated as follows: = = = )42cos(17) cos(0 1 i XV V θ 12.63 m/s 37.11)42sin(17) sin(0 1 = = =i YV V θ m/s Values of the dynamic parameters for both the vertical as well as the rotational parameters are the same as for the first case. The vertical component of the velocity at the rebound can be evaluated from the value of vertical coefficient of restitution as: 646.82−= ∴YV m/s 73 Next, the center of mass velocity an d the surface velocity, both in the X direction, are plotted on the same graph to ascertain whether there is an onset of the rolling during the ball’s contact with the ground: 0 0.2 0.4 0.6 0.8 1051015202530 Surface velocity Center of mass velocityFinding out onset of rolling Horizontal velocities, m/s Dimensionless contact time, n Fig. 27 Horizontal velocities of the two-mass model. From figure 27, it can be seen that unlike the first case, the graphs intersect at around n = 0.35 (exact value is n = 0.342 as found from MathCAD code). Hence the rolling motion begins during the contact time as early as before half of the contact time is elapsed. This indicates that the ball will not keep on sliding during the contact: more than 50 % of its contact time is spent in the rolling or the no-slip mode, in this particular case. 74 It is important to accommodate this transi tion and then connect the piecewise solutions of the sliding and the rolling to form a complete solution. In order to complete the solution, the solution for the sliding velocities for both the horizontal velocity of the center of mass of the model as well as the angular spin of the outer shell should start from time n = 0 to n = 0.342. At n = 0.342, the rolling or the no-slip solution takes over from the slidi ng solution. The values of the horizontal velocity of mass center as well as the spin, at this particular intersecting point of the graphs, form the initial conditions for the rolling equations, as well as the value n = 0.342 which becomes the lower limit of integrat ion for the solutions of the angular spin in the rolling contact. Appl ying these conditions and using the code in MathCAD, the graphs of the horizontal veloci ty of the mass center and the a ngular spin of the outer core are depicted in figure 28 on next page. As can be seen from figure 28, the solutions are piecewise. The break in the graphs indicates the instant of time during th e contact at which the sliding motion ceased and the rolling took place. 75 0 0.2 0.4 0.6 0.8 1910111213 Sliding RollingX Velocity transition Sliding to RollingHorizontal velocity, m/s Dimensionless contact time, n 0 0.2 0.4 0.6 0.8 1100150200250300350 Sliding RollingAngular Vel.transition Sliding toRollingAngular spin, rad/s Dimensionless contact time, n Fig.28 Horizontal and spin velocities during contact. 76 The rebound values can be obtained from the MathCAD code (as well as less accurately from end points of the graphs) as: 813.9 2=XV m/s 181.2972=ω rad/s The angle of rebound with respect to the surface is thus: 0 2238.41 )813.9646.8tan( ) tan( −= − = = aVVa XY rθ The magnitude of rebound velocity is: = + =2 22 2 2 Y X V V V 15.99 m/s The value of average coefficient of fricti on, as calculated from equation (63), is: 127.0=µ The value is appreciably less than the coeffici ent of sliding friction, which is 0.55. This shows that the transition indeed takes place and hence the motion changes from sliding to rolling as indicated by figure 28. This completes the rebound motion predictions for both the cases. 77 CHAPTER IV EXPERIMENTAL DATA In this chapter, two major experimental measurements for the tennis ball are described: the impact and rebound kinematics for three spin options, namely the zero spin, the topspin and the backspin [1]. These results are taken from the doctoral thesis of J.C.Wang, 1989, who conducted the experiment s on the dynamics of the tennis balls and evaluated the rebound kinematics of the tennis balls given varying conditions of incident kinematics. The second experimental measurement is that of the mass moment of inertia of the tennis ball using a simple setup. This second measurement has been conducted by the author to determine an estimate of the num erical value of mass moment of inertia of the tennis ball, in order to use this value for the simulations of the tennis ball (Chapter V). EXPERIMENTAL PROCEDURE As mentioned in Wang [1], the experimental procedure used to conduct the study of the incident and the rebound kinematics cons ists of an experimental design, test equipment, and apparatus. Following this, the data obtained is digitized and analyzed to convert the experimental results into the numerical values. The experimental test equipment consists of a ball pitching machin e, an acrylic test surface, a camera in a single frame mode, two strobotacs for providi ng flashes of light at the impact point, operating at 20,000 flashes per minute, a function generator used to control the time period between the flashes of the strobotacs a nd a counter-timer, which is used to check the period of the flashes of the strobotac lights and check if there is any error in the period. 78 The stroboscopic photography is a relative ly modern technique for understanding and analyzing the motion of bodies. Since a ca mera usually can not take many pictures in a short duration of time, for example, of an impact (the impact duration is of the order of milliseconds), the strobotacts provide basi s of what is called as stroboscopic photography. As soon as the ball hits the impact point on the surface, the camera shutter is opened for around one second and at the same time, a picture is taken. However, the strobotact lights are also flashing at the object of interest, in this case, a tennis ball, for as short as duration of 1 microsecond and about 20,000 times per minute. This illuminates the object for a very short duration and hence in a sense, divides 1 second picture of the camera, which otherwise would be a still pict ure, in a sequence of images due to the strobotac flash images. This produces the perception of a continuous motion as a sequential motion in one single picture of the camera. This produces what is called as the stroboscopic effect. The schematic of the arrangement is shown in Figure 29. Based on this procedure, the tennis ball’s images before, during and after the contact were developed. The images were th en transferred into a digitizing board and a coordinate system was established there, that marked coordinates of the center of mass of the ball, point where the ball touches the test surface, and two-cross points across the diameter of the ball (that were marked before the experiment to ascertain the angular spin). Once the coordinates were located, the incident and the rebound velocities were calculated using the sequential images from the stroboscopes and dividing the difference of the respective X and Y coordinates by the strobe period, which was 0.003 seconds. 79 However, there is a possibility of an error in analysis of the data in a sense that for the impact and the rebound kinematics, the coordinates were located for the tennis balls that were slightly before impact and slightly after impact, not at the exact point where the incidence and rebound occurred [1]. In order to get to the impact points an d the rebound points, the linear curve fit based on the images before the impact and after the impact was utilized. This may introduce some error in the estimation of the horizontal and vertical velocities, because even if the force in X direction is neglected, the ball will follow a parabolic trajectory, instead of a linear path, as is assumed in the data analysis. This in turn affects the angle of in cidence and the rebound, and based on the equations mentioned in [1], this as well a ffects the incident and the rebound spin. This might be a source of error in the data analysis as shown in [1] by the values of standard deviation in the experimental values of the kinematics. 80 Fig. 29 Schematic drawing of experimental arrangement [1]. Tennis Ball Test Surface ELEVATION Timer Function Generator Thrower CameraStroboscopes PLAN 81 Wang’s data has been used by this author to construct Tables 1 to 6 and Figures 30 to 32. Experimental results for the impact and rebound kinematics are shown in tables 1 through 6. The experimental results are arranged such that the first five columns in each table indicate the incident kinematics, a rranged in the order of incidence angle, the horizontal component of velocity at inciden ce, the vertical compone nt of velocity at incidence, the resultant translational velocity of the center of mass of the ball at incidence, and the angular ve locity of the ball at the incidence. The remaining five columns indicate counter-part kinematics pa rameters at rebound from the ground, in exactly the same order. All the angles are measured in degrees, the translational velocities and the components are all measured in meters pe r second, and the angular velocities are measured in radians per second. For each of the cases of the zero spin, the topspin, and the backspin, there are two tables. The first table has already been descri bed, whereas in the second table, along with the incoming and the rebound kinematics are show n the coefficients of restitution in the vertical and the horizontal dir ections, and kinetic energies, immediately before and after the impact. The vertical coefficient of restitution for each angle of incidence is calculated by dividing the negative of the vertical rebound ve locity by the vertical incident velocity, whereas the horizontal coefficient is obt ained by the ratio of the rebound horizontal velocity to the incident horizontal velocity. 82 The kinetic energies immediately before and after the impact are determined from the resultant values of the translational velocities of the center of mass of the ball (V 1 and V 2) and the spin speeds ( ω1 and ω2). The equations used to evaluate the kinetic energies from the given data are: 2 12 1 121 21ωI MV T + = 2 22 2 221 21ωI MV T + = It will be seen from the experimental data that the calculated values of the kinetic energies indicate there is a loss of kinetic energy associated with each of the impacts. These losses are a result of the friction force acting on the ball as well as the damping associated with the bounce of the ba ll in the vertical direction. Figures 30 through 32 show the experime ntal results on the tennis ball in graphical form. 83Table 1 Incidence and rebound kinematics with incident zero spin Table 2 Incidence and rebound kinematics for zero spin impact (with restitution coefficients and kinetic energies) θi VX1 VY1 V1 ω1 θr VX2 VY2 V2 ω2 (Degrees) (m/s) (m/s) (m/s) (rad/s) (Degrees) (m/s) (m/s) (m/s) (rad/s) 17.13 15.62 4.81 16.36 -5.34 - 20.83 11.09 -4.22 11.73 239.1 22.85 15.62 6.71 17.00 2.61 - 32.39 8.84 -5.76 10.55 313.2 34.47 14.49 9.95 17.59 -1.1 - 45.04 7.66 -7.66 10.84 293.31 41.1 13.46 11.74 17.87 23.46 - 50.33 7.25 -8.73 11.36 239.99 48.55 11.7 13.25 17.70 10.63 - 55.37 6.47 -9.37 11.4 213.17 58.94 9.16 15.19 17.74 10.63 - 70.42 3.87 -10.87 11.54 144.69 67.84 6.57 16.14 17.44 1.15 - 76.12 2.74 -11.09 11.44 96.25 θi V1 ω1 θr V 2 ω2 εy εx T 1 T2 (Degrees) (m/s) (rad/s) (Degrees) (m/s) (rad/s) (lb-in) (lb-in) 17.13 16.36 -5.34 -20.83 11.73 239.1 0.87734 0.70999 68.181 44.823 22.85 17.00 2.61 -32.39 10.55 313. 2 0.85842 0.56594 73.62 45.125 34.47 17.59 -1.1 -45.04 10.84 293.31 0.76985 0.52864 78.813 44.642 41.1 17.87 23.46 -50.33 11.36 239. 99 0.74361 0.53863 81.436 42.721 48.55 17.70 10.63 -55.37 11.4 213. 17 0.70717 0.55299 79.821 40.874 58.94 17.74 10.63 -70.42 11.54 144.69 0.7156 0.42249 80.182 37.501 67.84 17.44 1.15 -76.12 11.44 96. 25 0.68711 0.41705 77.475 34.921 84Table 3 Incidence and Rebound kinematics for te nnis ball with incident topspin Table 4 Incidence and rebound kinematics for topspin impact (w ith restitution coefficien ts and kinetic energies) θi V1 ω1 θr V2 ω2 εy εx T1 T2 (Degrees) (m/s) (rad/s) (Degrees) (m/s) (rad/s) (lb-in) (lb-in) 18.3 16.99 138.2 -23.09 11.94 386. 1 0.8761726 0.680719 76.794 61.806 22.64 17.54 158.06 -30.54 12.03 398. 11 0.9051852 0.639901 82.638 63.965 33.94 17.89 152.53 -40.58 12.71 323. 74 0.8276553 0.650708 85.503 59.071 42.38 18.64 136.32 -43.33 13.3 285. 46 0.7266932 0.702762 91.681 58.992 47.7 17.81 147.58 -51.88 12.56 236. 54 0.7501898 0.646912 84.521 49.751 61.07 17.48 146.74 -62.41 11.89 163. 87 0.6886854 0.652071 81.513 40.603 70.91 17.36 145.88 -67.6 11.95 119. 09 0.6741916 0.802469 80.405 38.800 θi VX1 VY2 V1 ω1 θr VX2 VY2 V2 ω2 (Degrees) (m/s) (m/s) (m/s) (rad/s) (Degrees) (m/s) (m/s) (m/s) (rad/s) 18.3 16.13 5.33 16.99 138.2 - 23.09 10.98 -4.67 11.94 386.1 22.64 16.19 6.75 17.54 158.06 - 30.54 10.36 -6.11 12.03 398.11 33.94 14.83 9.98 17.89 152.53 - 40.58 9.65 -8.26 12.71 323.74 42.38 13.76 12.55 18.64 136.32 - 43.33 9.67 -9.12 13.3 285.46 47.7 11.98 13.17 17.81 147.58 - 51.88 7.75 -9.88 12.56 236.54 61.07 8.45 15.29 17.48 146.74 - 62.41 5.51 -10.53 11.89 163.87 70.91 5.67 16.39 17.36 145.88 - 67.6 4.55 -11.05 11.95 119.09 85Table 5 Incidence and Rebound kinematics for tennis ball with incident backspin Table 6 Incidence and rebound kinematics for backspin impact (with restitution coefficients and kinetic energies) θi VX1 VY2 V1 ω1 θr VX2 VY2 V2 ω2 (Degrees) (m/s) (m/s) (m/s) (rad/s) (Degrees) (m/s) (m/s) (m/s) (rad/s) 17.42 16.18 5.08 16.96 -168.32 -20.31 10.56 -3.91 11.26 105.5 22.44 15.95 6.59 17.26 -148.05 -28.25 9.82 -5.27 11.14 170.35 34.79 14.35 9.97 17.48 -157.58 -49.78 6 -7.09 9.29 261.54 40.48 13.8 11.78 18.15 -179.12 -58.19 5.36 -8.64 10.17 208.97 45.21 12.02 12.11 17.07 -164.57 -65.58 4.12 -9.08 9.99 175.12 58.76 9.12 15.04 17.61 -184.04 -74.78 2.81 -10.29 10.68 106.79 67.62 6.57 15.95 17.25 -165.53 -81.72 1.57 -10.84 10.98 71.43 θi V1 ω1 θr V2 ω2 εy εx T1 T2 (Degrees) (m/s) (rad/s) (Degrees) (m/s) (rad/s) (lb-in) (lb-in) 17.42 16.96 -168.32 -20.31 11.26 105.5 0.769685 0.652658 78.114 34.199 22.44 17.26 -148.05 -28.25 11.14 170.35 0.799697 0.615674 79.632 36.573 34.79 17.48 -157.58 -49.78 9.29 261.54 0.711133 0.418118 82.076 33.681 40.48 18.15 -179.12 -58.19 10.17 208.97 0.733447 0.388406 89.397 33.813 45.21 17.07 -164.57 -65.58 9.99 175.12 0.749794 0.342762 78.854 30.665 58.76 17.61 -184.04 -74.78 10.68 106.79 0.684176 0.308114 84.784 31.004 67.62 17.25 -165.53 -81.72 10.98 71.43 0.679624 0.238965 80.481 31.582 86 Fig. 30 Incident vs rebound kinematics for zero spin [1]. Rebound Angle vs Incidence Angle-80-70-60-50-40-30-20-100 0 1 02 03 04 05 06 07 0Incidence Angle, degreesRebound Angle, degrees Rebound vs Incident Horizontal Velocity Components 024681012 0 5 10 15 20 Incident Horizontal Velocity, m/sRebound Horizontal Velocity, m/s Rebound vs Incident Vertical Velocity Components -12-10-8-6-4-20 05 1 0 1 5 Incident Vertical Velocity, m/sRebound Vertical Velocity, m/s 87 Fig. 31 Incident versus rebound kinematics for topspin [1]. Rebound Angle vs Incidence Angle -80-70-60-50-40-30-20-100 0 1 02 03 04 05 06 07 0 Incidence Angle, degreesRebound Angle, degrees Rebound vs Incident Horizontal Velocity Components 024681012 05 1 0 1 5 2 0 Incident Horizontal Velocity, m/sRebound Horizontal Velocity, m/s Rebound vs Incident Vertical Velocity Components -12-10-8-6-4-20 05 1 0 1 5 Incident Vertical Velocity, m/sRebound Vertical Velocity, m/s Rebound vs Incident Angular Speed 050100150200250300350400450 135 140 145 150 155 160 Incident Spin Speed, rad/sRebound Spin Speed, rad/s 88 Fig. 32 Incident versus rebound kinematics for backspin [1]. Rebound Angle vs Incidence Angle -90-80-70-60-50-40-30-20-100 0 1 02 03 04 05 06 07 0 Incidence Angle, degreesRebound Angle, degrees Rebound vs Incident Horizontal Velocity Components 024681012 0 5 10 15 20 Incident Horizontal Velocity, m/sRebound Horizontal Velocity, m/s Rebound vs Incident Vertical Velocity Components -12-10-8-6-4-20 05 1 0 1 5 Incident Vertical Velocity, m/sRebound Vertical Velocity, m/s Rebound vs Incident Angular Speed 050100150200250300 -200 -150 -100 -50 0 Incident Spin Speed, rad/sRebound Spin Speed, rad/s 89 From the data presented in Tables 1 through 6, values of the time-averaged coefficient of friction are calculated using e quation (63) in Chapter II, and presented in the following table for three cases of zero spin, top spin and back spin. Table 7. Time-averaged coefficient of fric tion values for experimental data Incidence Angles (Degrees) Zero Spin Top Spin Back Spin 18 0.502 0.515 0.625 23 0.543 0.453 0.516 34 0.388 0.284 0.489 42 0.303 0.188 0.413 48 0.231 0.184 0.373 60 0.203 0.114 0.249 70 0.141 0.04 0.186 From the values presented in Table 7 for the time-averaged friction factors, it can be seen that for zero spin incidence, the fi rst three incidences, st arting roughly from 18 degrees to 35 degrees, the friction factor is near the value of 0.55. However, after this, the value of the friction factor decreases gradually with each incidence until it reaches the value of 0.141 at an angle approximately equal to 70 degrees, at which there is not only sliding during motion but also pure rolling. For top spin incidences at same angles as zero spin, the values of the friction factors are seen to be decreasing more with increasing incidence angle. This shows that for top spin, rolling during cont act occurs earlier as compared to zero spin. Note the last two values for top spin in Table 7, which are appreciably less than 0.55, thus proving that the ball changes its motion from sliding to rolling during these cases. 90 For back spin, the friction factor is higher as compared to both top spin and zero spin, thus showing that the ball incident w ith back spin undergoes rolling only at high angles of incidence. For zero and top spin in cidence, transition of motion from sliding to rolling will start as early when angle of incidence reaches around 35 to 40 degrees, but for the ball incident with strong back spin, like in the experimental data presented, the transition will start only at around 55 to 60 degrees angle of incidence. The results of the experimental measurem ents on the tennis ball as shown in the previous graphs in figures 31 to 33 show th at the angle of rebound is greater than the angle of incidence in case of the backspin im pact. In case of the topspin, its trend is the same as the backspin case. For the zero spin impact, the angles of rebound are very close to those for the topspin impacts. The vertical coefficient of restitution vari es for each case of the zero spin, the top spin and the backspin, but trend still follows n early a straight line, which indicates that the vertical coefficient of restitution for each spin case can be taken, on an average, as a constant. The average vertical coefficient of restitution for all cases comes in a range of 0.74 to 0.78, which is a normally accepted range for this parameter [1]. The horizontal coefficient of restitution which is calculated and presented in the tables and can be directly deduced from th e graphs of the rebound and the incident horizontal velocities, does show some scatter. It is not as constant as the vertical coefficient of restitution, but for zero spin a nd topspin impacts, values for this parameter are not as scattered as is the case with the backspin impact. 91 MEASUREMENT OF THE MASS MOMENT OF INERTIA OF A TENNIS BALL The mass moment of inertia of a tennis ba ll is determined experimentally as a part of this thesis to obtain a numerical value of this important rotational parameter of the tennis ball which is used in the impact simulations (Chapter V). Theoretical background of measur ement for mass moment of inertia The mass moment of inertia of an obj ect can be determined by performing a twisting test on the object. A twisting test is basically a torsional vibration of the object so that it is hung or suppor ted vertically downwards from a ceiling with two or more strings of any appropriate material that can withstand the weight of the object. The torsional vibration is induced by giving an a ngular twist of a small amplitude about the vertical axis of the object th at passes through its center of mass of it and then measuring the period of the ensuing vibra tions. In order to develop an analytical expression for the torsional natural frequency as a function of th e mass moment of inertia, first of all, the following two linear equivalent systems are considered and their equations of motion are derived in order to establish an expression for the horizonta l equivalent stiffness of a pendulum. 92 Fig. 33 Equivalent linear vibrating systems. It can be seen from figure 33 that the two systems are equivalent and given the frequency of vibration of one, the frequency of vibration of the other can be deduced. The governing equations of motion and th e kinematics for the above systems are presented as follows: ∑•• =θO OI M 02= +•• θ θmgL mL 0= +•• θ θLg (76) x θ km L m O 93 For small amplitudes ‘ θ’, the kinematics of two systems are related as: θLx= Substituting this relation in the differential equation for ‘ θ’, the following equation of motion for an equivalent spring-mass system is obtained: 0= +•• xLgx 0= +•• xLmgxm If the equivalent system is considered se parately, its governing di fferential equation of motion is(neglecting damping): 0= +•• kxxm Thus the equivalent stiffness coefficient for the system is: Lmgk= (77) This expression provides the stiffness of an equivalent spring-mass system to that of a mass hanging by a rod or a taut and an inflexible string. Based on the same principle of an equivalent system, consider now a disk shaped object suspended by two strings, which is subj ected to a twisting test (torsional vibration about its center of mass) and an equivalent system supported tangentially by equivalent springs. This is illustrated in figure 34 below: 94 Fig. 34 Equivalent torsiona l vibrating systems. In this set of vibrating systems, the above system is subjected to an angular twist about the center of mass of the object supported by two strings of equal length. As a result, the system starts vibrating about its vertical centroidal axis which is shown as a dotted line in figure 34. An equivalent system is obtained if, inst ead of being supported on strings, the object is supported by the tw o equivalent springs connected tangentially to it. When the object is given a small twist about its vertical centr oidal axis, it will start vibrating. The following equations of moti on are derived to determine the mass moment L mR θ R O k k 95 of inertia of the object about its mass center in terms of the other system parameters with the help of the equivalent system: ∑•• =θO OI M Now the external moment acting on the vibr ating object (neglecting damping) can be expressed as follows: ∑•• = −= θO O I FR M 2 where ‘F’ is the tangential force exerted on the object by a spri ng and the two supporting springs are assumed to be identical. Furt hermore, ‘F’ can be expressed as follows: θ θ RLmgkRF2= = where the value of the equivalent stiffness coe fficient is obtained from the consideration of the previous equivalent systems. The fact or of 2 comes from the presence of the two strings to support the object in stead of a single string. Thus the differential equation of motion of the equivalent system is obtained by combining the above equations as follows: 02= +•• θ θ RLmgIO Equivalently, this equation can be written as: 96 02 = +•• θ θLImgR O (78) Comparing equation (78) above to the standard second-orde r linear differential equation describing a vibrating system, the natural frequency can be expressed as follows: LImgR On2 2=ω From above equation, the mass moment of iner tia of the object about its centroidal axis can be obtained as: 22 nOLmgRI ω= (79) Thus in the experimental setup for a twisting test, if the torsional period of vibration ‘T’ of the object is measured, its mass moment of inertia can be determined, since Tnπω2= Experimental setup A simple experimental setup based on th e idea of a twisting test is used to determine the mass moment of inertia of the tennis ball. The setup consists of a tennis ball hung from the ceiling by two strings, each a bout 1m (39.37 in.) long. A meter stick and a foot-scale are required for the measur ements of string length. A stopwatch is needed to measure the rotational period of vibration of the tennis ball. This is depicted in figure 35. 97 Fig. 35 Experimental setup to measure the mass moment of inertia. The strings are taped to the ball at the points on the surface so that it becomes very nearly tangent to the surface when th e upper ends are joined to the ceiling by the tape as well. To make the two supporting strings parallel and of equal length is not very easy and it took several trials and measuremen ts of the string lengths until they became equal and parallel. Then the twisting test was performed on the ba ll and the readings were obtained. Strings Tennis ball 98 Results of the experiment Several tests, with each one recording the time for 10 complete vibrations were performed and then the average was taken to determine the time period of torsional vibrations. Results are listed in the following table: Table 8. Experimental results of the twisting test on the tennis ball Test Number Number of Cycles Time for Cycles(seconds) 1 10 15.2 2 10 15.1 3 10 15.2 4 10 15.1 5 10 15.3 6 10 15.1 7 10 15.1 8 10 15.2 9 10 15.1 10 10 15.2 Total 100 151.6 The average time period of the torsional osc illations is calculated from the above experimental results as: 516.11006.151= =T seconds Based on this value of the time period, the circular natural frequency of the torsional oscillations is calculated as: 99 144.42= =Tnπω rad/s The values of the other physical parameters of the system are as follows: m = 0.127 lb = 0.000329 lb-s 2/in g = 386.4 in/s2 R = 1.25 in. L = 101.6 cm = 39.99 in. Substituting these values in equation (79), the mass moment of inertia of the tennis ball about its centroida l axis is calculated as: 00028.0 )144.4)(99.39()25.1)(4.386)( 000329.0( 22 = =OI lb-s2-in This value of the mass moment of inertia will be used as a benchmark for the simulations of the tennis ball (Chapter V). 100 CHAPTER V BEST RESULTS COMPARISONS WITH THE MEASUREMENTS From the simulation results as presented in Appendix C, there are some parameters of the two-mass elastic systems th at consistently seem to produce reasonable results for the rebound motion of the tennis ba ll as compared against the experimental measurements of the same parameters. These results seem to produce reasonable results for the incident zero spin, top spin and back spin cases for varying angles of incidences. Therefore, the results of the rebound kinematic s developed by those selected parameters of the two-mass elastic systems are described in the following pages. In order to get the damping coefficient in th e vertical direction, c y, for each of the cases of the zero spin, the top spin and the back spin, the average vertical coefficient of restitution for each case is calculated as follows: 77 1∑ ==ii COR COR Based on the value obtained above for that particular case, and having the spring constant and the mass ratios fixed, equation (6 ) is utilized to vary the coefficient for different values of the damping coefficients until the theoretical value of the coefficient of restitution, as given by equation (6), matches with the experimental value above. Three cases for the parameters that yield reas onable agreements with the experiments are described in the following table: 101 Table 9. Dynamic ratios for all cases of incident angles and spins giving best results For the zero spin impact, it can be seen from the simulation results and from the simulated kinematic graphs during the cont act that above an in cident angle of 20 degrees, the ball enters into the ro lling mode. The time of contact, t c, predicted by the simulations agrees very well with the experimental determinations of the contact times of tennis balls [4]. The contact duration usually spans from 4.5 ms to 5.5 ms. The predicted contact times vary in the same range for the different angles of incidence. Thus the contact duration is in close agre ement with the experimental values. For the topspin impacts, the simulations s how that the rolling mode occurs during contact as angle of incidence is increased fr om 17 degrees to 23 de grees. Due to rolling, the horizontal velocity at th e rebound is higher as compared to what it would be if it were sliding throughout the contact. This cau ses the simulated angles of rebound to be smaller than the angles of incidence. This agrees well with the e xperimental observations (Figs.30-32) in that the rebounds from the tops pin impact are usually at smaller angles than incident and relatively less loss of hor izontal component of incident velocity. The simulated rebound angles for the topspin imp acts agree well with this observation. The results achieved for the rebound spins for the three cases of the zero spin, the topspin and the backspin are encouraging and agreement with the experiments seems reasonable (Figs.36-44). In almost all case, the spin values are slightly higher than the experimental values. This might be attributed to some complicated effects of stick-slip Cases M 1 M2 Ky I1 I2 kθ Lb-s2/in Lb-s2/in Lb/in Lb-in-s2 Lb-in-s2 Lb-in/rad 1 2M/3 M/3 80 0.532I 0.468I 1000 2 2M/3 M/3 90 0.532I 0.468I 1000 3 M/2 M/2 72 0.353I 0.647I 1000 102 occurring at the inte rface of the tennis ba ll and the ground, which have not been taken into account in this model. For the backspin impacts, the simulations show that the rolling mode occurs later than for the corresponding cases of the zer o spin or the topspin impacts. Due to occurrence of rolling at later inst ant of contact in case of backspin as compared to either zero spin or topspin, the rebound horizontal velocity is smaller as compared to either of zero spin or topspin incidence (since most of the contact time of the ball is in sliding motion for the backspin impact, and a small percentage of the contact time is in rolling mode). This causes the simulated angles of rebound to be higher than the angles of incidence. This simulation prediction agrees very well with the experimental observations (Figs.40-42) which reveal that the backspin impacts on the tennis court surfaces generate large angles of reb ound, which are always higher than the corresponding angles of incidences. The simulated horizontal velocities at th e rebound are generally higher than the experimental values (Figs.36-44) and the re ason again might be the presence of some tangential flexibility, or in other words, elas ticity in the X direction, that has not been accommodated in this model. 103 Incident vs. rebound angles Incident vs. rebound horizontal velocities Incident vs. rebound vertical velocity Fig. 36 Incident vs rebound parameters for the zero spin (average COR = 0.765), case 1. (c y = 0.0225 lb-s/in) 10 20 30 40 50 60 70100806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees6 8 10 12 14 16051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s 4 6 8 1 01 21 41 61 81412108642 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s+θ -θ X Y 10410 20 30 40 50 60 70100806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees6 8 10 12 14 16051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocities 4 6 8 1 01 21 41 61 81412108642 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s Incident vs. rebound vertical velocity Fig. 37 Incident vs rebound parameters for the zero spin (average COR = 0.765), case 2. (c y = 0.0239 lb-s/in) +θ -θ X Y 10510 20 30 40 50 60 70100806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees6 8 10 12 14 16051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocities 4 6 8 10 12 14 16 181412108642 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s Incident vs. rebound vertical velocity Fig. 38 Incident vs rebound parameters for the zero spin (average COR = 0.765), case 3. (c y = 0.0185 lb-s/in) +θ -θ X Y 10610 20 30 40 50 60 70 80806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees4 6 8 1 01 21 41 61 82468101214 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocities 4 6 8 1 01 21 41 61 8141210864 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s135 140 145 150 155 160100200300400500 Experimental SimulatedIncident vs rebound spinRebound spin, rad/s Incident spin, rad/s Incident vs. reb ound vertical velocity Incident vs. rebound angular spin Fig. 39 Incident vs rebound parameters for the t op spin (average COR = 0.778), case 1. (c y = 0.0211 lb-s/in) 10710 20 30 40 50 60 70 80806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees468 1 0 1 2 1 4 1 6 1 82468101214 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizon tal velocities 4 6 8 10 12 14 16 18141210864 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s135 140 145 150 155 160100200300400500 Experimental SimulatedIncident vs rebound spinRebound spin, rad/s Incident spin, rad/s Incident vs. rebound vertical velocity Incident vs. rebound angular spin Fig. 40 Incident vs rebound parameters for the top spin (average COR = 0.778), case 2. (c y = 0.0224 lb-s/in) 10810 20 30 40 50 60 70 80806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees468 1 0 1 2 1 4 1 6 1 8051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocit ies 4 6 8 10 12 14 16 18141210864 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s135 140 145 150 155 160100200300400500 Experimental SimulatedIncident vs rebound spinRebound spin, rad/s Incident spin, rad/s Incident vs. rebound vertical velocity Incident vs. rebound angular spin Fig. 41 Incident vs rebound parameters for the t op spin (average COR = 0.778), case 3. (c y = 0.0173 lb-s/in) 10910 20 30 40 50 60 70100806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees6 8 10 12 14 16 18051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocit ies 4 6 8 10 12 14 1612108642 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s 190 180 170 160 150 14050100150200250300 Experimental SimulatedIncident vs rebound spinRebound spin, rad/s Incident spin, rad/s Incident vs . rebound vertical velocity Incident vs. rebound angular spin Fig. 42 Incident vs. rebound parameters for the back spin (average COR = 0.732), case 1. (c y = 0.0262 lb-s/in) 11010 20 30 40 50 60 70100806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees6 8 10 12 14 16 18051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocit ies 468 1 0 1 2 1 4 1 612108642 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s190 180 170 160 150 14050100150200250300 Experimental SimulatedIncident vs rebound spinRebound spin, rad/s Incident spin, rad/s Incident vs. rebound vertical velocity Incident vs. rebound angular spin Fig. 43 Incident vs. rebound parameters for the back spin (average COR = 0.732), case 2. (c y = 0.0277 lb-s/in) 11110 20 30 40 50 60 70100806040200 Experimental SimulatedIncident vs Rebound AnglesRebound Angle, degrees Incidence Angle, degrees6 8 10 12 14 16 18051015 Experimental SimulatedIncident vs Rebound horizontal velocityRebound horizontal velocity, m/s Incident horizontal velocity, m/s Incident vs. rebound angles Incident vs. rebound horizontal velocit ies 468 1 0 1 2 1 4 1 612108642 Experimental SimulatedIncident vs Rebound vertical velocityRebound vertical velocity, m/s Incident vertical velocity, m/s190 180 170 160 150 1400100200300400 Experimental SimulatedIncident vs rebound spinRebound spin, rad/s Incident spin, rad/s Incident vs. rebound vertical veloc ity Incident vs. rebound angular spin Fig. 44 Incident vs. rebound parameters for the back spin (average COR = 0.732), case 3. (c y = 0.0215 lb-s/in) 112 CHAPTER VI CONCLUSIONS Simulation results of the e xperimental measurements on the tennis balls with selected parameters were presented in Chapter V. The results show that in general, the two mass model gives a fairly reasonable approximation of the rebound kinematics of the tennis ball. From the simulations, it can be c oncluded that the offset parameter ‘ ε’ (rolling friction) plays an important and a crucial part for the success of most of the simulations where there is a transition from the sliding to the rolling mode. This will always occur when the angle of incidence gets larger, whic h in turn implies a large vertical velocity component. The larger the vertical velocity component, the greater is the effect of the moment about the center of mass of the system produced by the normal contact force. The application of linear vi brations theory and impact dynamics to the simulation of tennis ball using a two-mass model predicts the kinematics of the rebound, especially the rather challengi ng parameter of the spin of rebound reasonably well. The model is efficient from the point of view of comput ation time and effort, and it is simple to understand, since it is a linear model. The resu lts indicate that the linear theory can be used to predict the impact phenomena with good success. Incorporating the vertical stiffness and the torsional stiffness as well as the damping in the model, which can be deduced by the simple bounce height experiments, the model can be used to predict the rebound velocities, the spins and the angles of rebounds. Also the model can be used to succes sfully predict the time of contact of the 113 tennis ball with the ground, the onset of rolli ng or no-slip motion and the contact forces acting on the ball. The simulations of the experimental results as presented in tables C-7 to tables C- 27 present the time of contact (duration of c ontact). As can be found from references [3] and [5], the average time of contact that ha s been measured for the tennis balls is somewhere around 4.5 ms to 5 ms. From the simulation results, the average contact time varies very much in the same range, with 4. 5ms to 5ms as the most frequent occurrence. This indicates that the model indeed predicts the time of contact very well. Finally, the selected three cases of the dynamic parameters that gave the best agreements with the measurements indicate th at the inner core mass is always slightly higher in both the mass and the mass moment of inertia to give the best fit with the measurements. 114 REFERENCES [1] Wang, J.C., 1989, The Impact Dynamics of a Tenni s Ball Striking a Hard Surface, PhD Dissertation, Oregon State University [2] Smith III, J.F., 1987, The Effect of Angle, Velocity and Rotation of Incidence on the Angle Deviation of Rebounding Tennis Balls, PhD Dissertation, Texas A&M University [3] Cross, R., 1999, “The Bounce of a Ball,” American Journal of Physics, 67 (3), pp. 222-227 [4] Cross, R., 2002, “Measurements of the Ho rizontal Coefficient of Restitution for a Superball and a Tennis Ba ll”, American Journal of Physics, 70 (5), pp. 482-489 [5] Brody, H., 1984, “That’s How the Ball Bounces,” The Physics Teacher, pp. 494-497 [6] Hubbard, M., and Stronge, W.J., 2001, “Bounce of Hollow Balls on Flat Surfaces,” Sports Engineering, 1, pp.94-102 [7] Stronge, W.J., 2000, Impact Mechanics , Cambridge University Press [8] Stevenson, B., Bacon, M.E., and Baines, C.G.S., 1998, “Impulse and Momentum Experiments Using Piezo Disks,” American Journal of Physics, 66(5), pp. 445- 448 [9] Malcolm, R.E., 1989, “A Measurement Usi ng the Piezoelectric Effect,” The Physics Teacher, pp. 637-639 [10] Minnix, R.B., and Carpenter,D.R., 1985, “Piezoelectric Film Reveals F vs t of Bounce Ball,” The Physics Teacher, pp. 180-181 [11] International Tennis Federation Rules and Regulations, ITF, UK, Year 2002 [12] Lankarani,H.M., and Nikravesh, P.E ., 1990, “ A Contact Force Model With Hysteresis Damping for Impact Analysis of Multibody Systems,” Journal of Mechanical Design, 112, pp.369-376 [13] Stronge, W.J., 1992, “Energy Dissipated in Planar Collisions,” ASME Journal of Applied Mechanics, 59, pp. 681-683 115 [14] Sondergaard, R., Chaney,K., and Bre nnen,C.E., 1990, “Measurements of Solid Spheres Bouncing Off Flat Plates ,” ASME Journal of Applied Mechanics, 57, pp. 694-699 [15] Keller, J.B., 1986, “Impact With Fricti on,” ASME Journal of Applied Mechanics, 53, pp. 1-4 [16] Hudnut, K., and Flansburg, L., 1979, “Dynamic Solutions for Linear Elastic Collisions,” American Journal of Physics, 47(10), pp. 911-914 [17] Bayman, B.F., 1976, “Model of the Beha viour of Solid Objects During Collision,” American Journal of Physics, 44(7), pp. 671-676 [18] Johnson, K.L., 1982, “The Bounce of a Superball,” International Journal of Engineering Education, pp. 57-63 [19] Hunt, K.H., and Crossley, F.R.E., 1975, “Coefficient of Restitution Interpreted as Damping in Vibroimpact,” 42, pp.440-445 [20] Spradley, J.L., 1987, “Velocity Amplifi cation in Vertical Collisions,” American Journal of Physics, 55(2), pp. 183-184 116 APPENDIX A PARTICULAR SOLUTION OF SECOND-ORDER, NON- HOMOGENOUS, ORDINARY LINE AR DIFFERENTIAL EQUATION Consider the following differential equation: ) cos( ) sin(2 1 22 t e t e cdtdb dtdat tβ γ β γθθ θα α+ = + + (A1) In the above equation, the coefficients of the variables on the left hand side are all constants. In this equation, only θ and t are the variables involved, hence the remaining symbols on the right hand side are all constants as well. The constants γ1 and γ2 are non-zero, so equation (A1) is a non-homogenous, second- order, ordinary differential equation with consta nt coefficients. It is desired to determine the particular solution of the above differential equation. In order to determine the solution of equation (A1), the method of undetermined coefficients is used. Accordingly, the assume d particular solution of equation (A1) is of the form: ) cos( ) sin( )( t Bet Aett tβ β θα α+ = (A2) where ‘A’ and ‘B’ are the constants w hose values have to be determined. The first and second derivatives of th e assumed function in equation (A2) are readily evaluated as follows: 117 ) )( cos( ) )( sin( B At e B At edtdt tα ββ β αβθα α+ + − = ( A 3 ) )] ( 2)[ cos( ] 2) ()[ sin(2 2 2 2 22 β α αβ β αβ β α βθα α− + + − − = BA t e B At e dtdt t (A4) Substituting equations (A2), (A3) and (A4) in equation (A1), the following algebraic equation is obtained: ) cos( ) sin( ) cos( ) sin( ) cos() () sin() () cos(]) ( 2[) sin(] 2) ([ 2 12 2 2 2 t e t e t cBet cAet eB Abt eB Abt eB A at eB Aa t t t t tt t t β γ β γ β β β α ββ β α β β α αβ β αβ β α α α α α αα α α + = + + + +− + − + + − − (A5) The coefficients for the time functions on the left and the right hand side in equation (A5) can be compared to obtain th e following linear simultaneous equations in ‘A’ and ‘B’: 12 2) (] 2) ([ γ β α αβ β α = + − + − − cA B AbB Aa (A6) 22 2) (]) ( 2[ γ α β β α αβ = + + + − + cB B AbB A a (A7) Separating the terms involving ‘A’ and ‘B’, the above equations can be re-written as: 12 2) 2( ] ) ([ γ β αβ α β α = + −+ + − b aBc b aA (A8) 22 2] ) ([ ) 2( γ α β α β αβ =+ + − + + c b aB b aA (A9) 118 Solving equations (A8) and (A9) simulta neously, the values of ‘A’ and ‘B’ are determined as: ]) 2() ([ ]) 2(} ) ([{) 2(2 2 1 2 2 2 2 2β αβα β αγγ β αβ α β αβ αβ b ac b a a b a c b ab aA++ + −+ + + + + −+= (A10) ) 2() (12 2 β αβγ α β α b ac b a aAB+−+ + −= (A11) (‘B’ is expressed in terms of ‘A’. Once ‘A’ is known from equati on (A10), ‘B’ is known from equation (A11)). From equations (A10), (A11) and (A2), it can be seen that the particular solution of equation (A1) is completely established in terms of the constants involved in the differential equation (A1). Thus substituting the values of ‘A’ and ‘B’ from equations (A10) and (A11) into equation (A2), which is the assumed particular solution, the particular integral of equation (A1) is determined. 119 APPENDIX B MATHCAD CODES FOR VARIOUS KINEMATICS Vertical displacement during contact Dn M 1, cy,ky,Vy1, ( ) Vy1i Vy1 39.37 ⋅ ← W 0.127← g 386.4← R 1.25← ωyky M1← ξycy 2M 1⋅ω y⋅← ωdy ωy1 ξy2−⋅ ← Vy1i Rωdy⋅en−ξ y⋅π⋅ω y⋅ ωdy⋅ sin n π⋅()⋅:= The code above generates the vertical displacement of the inner core mass M 1, given the stiffness of the spring and the dampi ng coefficient of the damper in the vertical direction, as well as the incident vertical velocity. The code requires, as its arguments, the numerical values of the inner core mass M 1, the damping coefficient, the stiffness coefficient and the incident vertical veloci ty. The ‘n’ inside the brackets on the right hand side of the first line of th e code indicates that the dime nsionless contact time ‘n’ is also present in the equations. As the equations inside the code indicate, the displacement is plotted as a function of dimensionless cont act time ‘n’. Given the physical parameters, it calculates the damping ratio, the undamped natural frequency and the damped natural 120 frequency inside the code and then on the ba sis of these quantities, the vertical velocity during the contact is determined. Code for rebound velocity in Y direction The following code generates the velocity of rebound (in m/s) in Y direction during any instant of contact and at any value of th e dynamic parameters (to be inputted in FPS units). Vy M1 ky ,cy,n,Vy1, ( ) Vyi Vy1 39.37 ⋅ ← ωyky M1← ξycy 2M 1⋅ω y⋅← ωdy ωy1 ξy2−⋅ ← Vyi−ξ y⋅ω y⋅eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ ωdysin n π⋅()⋅ Vyi eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos n π⋅()⋅ +⎛⎜ ⎜ ⎜⎝⎞ ⎟ ⎠ 39.37:= The code for the vertical velocity duri ng the contact is shown above. With the incident vertical velocity input in mete rs per second, and the remaining physical parameters in FPS units, it gives the vertical velocity during the cont act as a function of dimensionless contact time, ‘n’. It gi ves the final answers in SI units. 121 Code for X velocity of the ball during sliding VxsµM1, cy,ky,n,Vx1, Vy1, () Vyi Vy1 39.37 ⋅ ← Vxi Vx139.37 ⋅ ← ωyky M1← ξycy 2M 1⋅ω y⋅← ωdy ωy1 ξy2−⋅ ← yVyi ωdyen−ξ y⋅π⋅ω y⋅ ωdy⋅ sin n π⋅()⋅ ← VyVyi−ξ y⋅ω y⋅eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ ωdysin n π⋅()⋅ Vyi eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos n π⋅()⋅ +⎛⎜ ⎜ ⎜⎝⎞ ⎟ ⎠← axµ− Mcy Vy⋅ ky y⋅ + ()⋅ ← Vx1 cyVyi ωdy⋅ eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sin n π⋅()⋅ ← Vx2ky Vyi⋅ ωdyξy−ω y⋅ ξy2ωy2⋅ω dy2+eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sin n π⋅()⋅⎛⎜ ⎜ ⎜⎝⎞ ⎟ ⎠⋅ ← Vx3ky Vyi⋅ ωdyωdy− ξy2ωy2⋅ω dy2+eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos n π⋅()⋅ωdy ξy2ωy2⋅ω dy2++⎛⎜ ⎜ ⎜⎝⎞ ⎟ ⎠⋅ ← Vxiµ MVx1 Vx2+ Vx3+ ()⋅ −:= 122 The above code evaluates the horizontal component of velocity of the two-mass system during the sliding contac t as a function of the dimensionless contact time, ‘n’. As can be seen form the code, it is a function that is dependent on the ve rtical parameters of the two-mass system. The results that combin e to produce the horizontal velocity during the contact are the integrations of the accel eration function. The code indicates the results for the case in which the friction for ce goes opposite to the X velocity. In case it goes in the same direction as the X velocity (topspin, in which surface velocity due to spin is higher than incident horizontal velocity), simply the sign if the term following Vxi in the last line should be put positive inst ead of the negative to give the correct velocity variation. Code for the angular velocity of the ball during sliding ω1:= (Incident angular spin) Vy1 := (Incident vertical velocity in m/s) Vx1 ::= (Incident horizontal velocity in m/s) Vyi Vy1 39.37 ⋅ := Vy1 (Incident vertical velocity in in/s) Vxi Vx1 39.37 ⋅ := Vx1 (Incident horizontal velocity in in/s) I 0.000342 := (Mass moment if inertia of ball (FPS units)) IG := (Mass moment of inertia of inner core) I2:= (Mass moment of inertia of outer shell) k θ:= (Torsional stiffness) 123 a11 1 IG1 I2+⎛⎜⎝⎞ ⎠← ωyky M1← ξycy 2M 1⋅ω y⋅← ωθ kθ1 IG1 I2+⎛⎜⎝⎞ ⎠⋅ ← b1 c θ := c θ c1 k θ := k θ ω d θ ωθ 1 ξθ 2 − ⋅ := ξθ ω dy ω y 1 ξ y 2 − ⋅ := ξ y R 1.25 := (Outer radius of the ball(FPS units)) cy:= (Damping coefficient(FPS units)) ky:= (Stiffness coefficient(FPS units))M 0.000329 := (Mass of the ball(FPS units)) M1:= (Mass of inner core) M2:= (Mass of outer shell) 124 γ1µR⋅ I2ky Vyi⋅ ωdycyξy⋅Vyi⋅ω y⋅ ωdy−⎛⎜⎝⎞ ⎠⋅1 IG1 I2+⎛⎜⎝⎞ ⎠1− ⋅ ← γ2µR⋅ I2cy⋅Vyi⋅1 IG1 I2+⎛⎜⎝⎞ ⎠1− ⋅ ← A12a 1⋅α 1⋅β 1⋅ b1β1⋅ + () a1α12⋅ a1β12⋅− b1α1⋅ + c1+ ()2 2a 1⋅α 1⋅β 1⋅ b1β1⋅ + ()2+⎡ ⎣⎤ ⎦γ2γ1a 1 α12⋅ a1β12⋅− b1α1⋅ + c1+ ( )⋅ 2a 1⋅α 1⋅β 1⋅ b1β1⋅ + ()+⎡ ⎢ ⎣⎤ ⎥ ⎦⋅ ← B1A1 a1 α12⋅ a1β12⋅− b1α1⋅ + c1+ ( )⋅γ 1− 2a 1⋅α 1⋅β 1⋅ b1β1⋅ + ()← vθos1cθ I2A1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sin n π⋅()⋅ B1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos n π⋅()⋅ + B1−⎛ ⎜ ⎝⎞ ⎠ ⋅ := vθos2kθ I2 0n n A1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sin n π⋅()⋅ B1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos n π⋅()⋅ +⎛ ⎜ ⎝⎞ ⎠π ωdy⋅⌠ ⎮ ⎮⎮ ⌡d ⋅ := vθos3 0n n γ1eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sinωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅γ 2eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos ωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅ +⎛⎜ ⎜⎝⎞ ⎠π ωdy⋅⌠ ⎮ ⎮⎮ ⌡d ω1+ := α 1 ξ y − ω y ⋅ := ω y β 1 ω dy := ω dy 125 vθos v θos1 v θos2 + vθos3 + := (Angular velocity of outer shell) Code for the angular velocity of the ball during rolling Vy1:= (Incident vertical velocity in m/s) Vx1:= (Incident horizontal velocity in m/s) Vyi:=Vy1.39.37 (Incident vertical velocity in in/s) Vxi:=Vx1.39.37 (Incident horizontal velocity in in/s) ω1:= (Incident angular spin) ε := (Eccentricity of normal force)R 1.25 := (Outer radius of the ball(FPS units)) cy:= (Damping coefficient(FPS units)) ky:= (Stiffness coefficient(FPS units))M 0.000329 := (Mass of the ball(FPS units)) M1:= (Mass of inner core) M2:= (Mass of outer shell)I 0.000342 := (Mass moment if inertia of ball (FPS units)) IG:= (Mass moment of inertia of inner core) I2:= (Mass moment of inertia of outer shell) k θ:= (Torsional stiffness) 126 ωθ kθ1 IG1 I2 M R2⋅++⎛ ⎜ ⎝⎞ ⎠⋅ ← ξθ 1ωy ωθ⎛⎜⎝⎞ ⎠2 1ξy2−()⋅ − ← ωdθω θ 1ξθ2−⋅ ← a11 1 IG1 I2 M R2⋅++⎛ ⎜ ⎝⎞ ⎠← cθ 2ξθ⋅kθ 1 IG1 I2 M R2⋅++⎛ ⎜ ⎝⎞ ⎠⋅ ← b1 c θ← c1 k θ← γ1ε− I2 M R2⋅+ky Vyi⋅ ωdycyξy⋅Vyi⋅ω y⋅ ωdy−⎛⎜⎝⎞ ⎠⋅1 IG1 I2 M R2⋅++⎛ ⎜ ⎝⎞ ⎠1− ⋅ ← γ2ε− I2 M R2⋅+cy⋅Vyi⋅1 IG1 I2 M R2⋅++⎛ ⎜ ⎝⎞ ⎠1− ⋅ ← ω y ky M1 := M1 ξ y cy 2 M1 ⋅ ω y ⋅ := ω y ω dy ω y 1 ξ y 2 − ⋅ := ξ y 127 A12a 1⋅α 1⋅β 1⋅ b1β1⋅ + () a1α12⋅ a1β12⋅− b1α1⋅ + c1+ ()2 2a 1⋅α 1⋅β 1⋅ b1β1⋅ + ()2+⎡ ⎣⎤ ⎦γ2γ1a 1 α12⋅ a1β12⋅− b1α1⋅ + c1+ ( )⋅ 2a 1⋅α 1⋅β 1⋅ b1β1⋅ + ()+⎡ ⎢ ⎣⎤ ⎥ ⎦⋅ ← B1A1 a1 α12⋅ a1β12⋅− b1α1⋅ + c1+ ( )⋅γ 1− 2a 1⋅α 1⋅β 1⋅ b1β1⋅ + ()← θiM Vxi2()⋅ Iω12⋅+ IGω12⋅− I2 M R2⋅+← n:=0,0.001…1 vθor1 n()kθ I2 M R2⋅+0n nω1θi−() ωdθeξθ−ω θ⋅nπ⋅ ωdy⋅ ⋅ sinωdθnπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅⎡⎢ ⎢⎣⎤⎥ ⎥⎦π ωdy⋅⌠ ⎮ ⎮⎮ ⌡d ⋅ := vθor2 n()kθ I2 M R2⋅+0n n A1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sinωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅ B1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos ωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅ + B1−⎛⎜ ⎜⎝⎞ ⎠π ωdy⋅⌠ ⎮ ⎮ ⎮ ⌡d ⋅ := vθor3 n()cθ I2 M R2⋅+ω1θi−() ωdθeξθ−ω θ⋅nπ⋅ ωdy⋅ ⋅ sinωdθnπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅⎡⎢ ⎢⎣⎤⎥ ⎥⎦⋅ := vθor4 n()cθ I2 M R2⋅+A1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sinωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅ B1 eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos ωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅ + B1−⎛⎜ ⎜⎝⎞ ⎠⋅ := α 1 ξ y − ω y ⋅ := ω y β 1 ω d θ := ω d θ 128 vθor5 n() 0n n γ1eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ sinωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅γ 2eξy−ω y⋅nπ⋅ ωdy⋅ ⋅ cos ωdynπ⋅ ωdy⋅⎛⎜⎝⎞ ⎠⋅ +⎛⎜ ⎜⎝⎞ ⎠π ωdy⋅⌠ ⎮ ⎮⎮ ⌡d θ i+ := vθor n() v θor1 n() v θor2 n() + vθor3 n() + vθor4 n() + vθor5 n() + Vxr n() R v θor n() ⋅ Code for the transition point during co ntact (sliding to rolling transition) ω 2 v θ os sol ( ) := sol n= sol root Vxs n ( ) R v θ os n ( ) ⋅ − n , ( ) := n sol = sol Vx sol ( ) = sol Vx (sol R = v θ os sol ( ) = sol 129 APPENDIX C COMPARISON OF THEORY WITH EXPERIMENT SIMULATION RESULTS Some simulation results for the tennis ball dynamics are shown on the following pages. For each incident angle and for the major cases of zero spin, topspin and backspin, there are several possibilities of varying the mass, the stiffness, the damping, the mass moment of inertias and the torsional stiffness parameters and then obtaining the rebound kinematics. Accordingly, first of all, three cases for the impact have been categorized as zero spin, topspin and backspin . Then, for each of those major categories, sub-categories have been formed based on the angles of incidence of the ball with the ground. For the experimental data available [1], the angles of incidence vary from 17 degrees to 70 degrees. For each angle of incidence, the cases column on the extreme left of the Tables indicates that the dynamic i nput parameters of the tennis ball model are varied i.e., the inner core mass, the torsional stiffness, the moments of inertia etc. and these variations produce the simu lated values in the column s for the rebound velocities, the rebound angle and the time of contact of the ball with th e ground, which are the last five columns. The fixed inputs are the parameters that are constant for each incidence angle. These include the horizontal and the vertical components of the incident velocity, with the values being input the same as those found out from the experimental results for each of these angles. Another important fixed parame ter to be used in order to generate the simulated values of the kinematics includes the vertical coefficient of restitution, with the value being input same as that obtaine d from the corresponding experimental result. The first four columns in the table are ones that directly affect the value of the vertical coefficient of restitution (equation (6), Chapter 1) and these input parameter values are 130 adjusted for each simulation such that the theoretical value of the coefficient of restitution is equal to the one input, which in turn is equal to the experimental value for that particular case. Another fixed input parameter is coefficient of sliding friction. All simulations are based on an assumption that the damped period of vibration in relative rotation is equal to the period of vibration in vertical motion. This ensures that the rotational vibrations complete their one cycle during the impact and that the cycle is not incomplete when the contact finally ends. E quating these two periods also provides with an estimation of the damping ratio in the ro tation. The value of 1000 lb-in/rad has been used for torsional stiffness, since it produces the best agreements with the experiments, and also that it fits the descrip tion of a tennis ball as a relatively stiff torsional spring [1]. The solutions of the differential equations deri ved in chapter II have been coded in the MathCAD software and all the results have been obtained from there. The actual experiments [1] were performe d on an acrylic surface, for which the average sliding coefficient of friction is 0. 55. This same value is used throughout the simulations for this parameter. It is helpful to ascertain some fixed phys ical parameters of the whole system. Mass of tennis ball According to [1], the tennis ball weight is around 0.0576 kg or 0.127 lb. From the International Tennis Federation (ITF) standa rds, the weight should be between 56.0 grams and 59.4 grams. For the simulations, th e weight of 0.127 lb has been selected. The mass can be calculated as follows: 4.386127.0= =gWM =inslb2 000329.0 − 131 SELECTED CASES The selected cases for comparison with the experiment are presented in tables 7 to 27 and figures 35 to 51. 132ZERO SPIN IMPACT 17 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =15.62 m/s Incident vertical velocity = V y1 = 4.81 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.877 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C1. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s2 Lb-in-s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -4.217 13.399 73.846 -15.356 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -4.217 13.403 73.901 -15.352 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -4.217 13.394 74.527 -15.362 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -4.217 12.308 244.57 -16.645 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -4.22 12.309 245.44 -16.645 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -4.219 12.307 245.86 -16.646 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -4.219 11.691 298.83 -17.471 5.202 1332. 23 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =15.62 m/s Incident vertical velocity = V y1 = 6.71 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.858 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C2. Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in-s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -5.757 12.521 104.24 -22.292 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -5.757 12.528 103.09 -24.669 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -5.754 12.54 104.57 -24.671 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -5.76 11.08 335.56 -27.468 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -5.76 11.087 335.76 -27.441 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -5.757 11.095 336.01 -27.444 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -5.762 11.47 361.26 -26.649 5.202 1343. 35 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =14.49 m/s Incident vertical velocity = V y1 = 9.95 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.7698 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C3. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -7.654 9.895 152.47 -37.569 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -7.654 9.904 152.87 -37.730 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -7.663 9.923 152.74 -37.622 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -7.648 9.97 301.94 -37.714 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -7.654 10.144 307.20 -37.014 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -7.648 10.148 307.32 -37.018 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -7.652 11.089 335.82 -34.608 5.202 1354. 41 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =13.46 m/s Incident vertical velocity = V y1 = 11.74 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.7436 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C4. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -8.733 8.038 180.24 -48.172 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -8.733 8.049 180.37 -47.334 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -8.733 8.071 180.21 -47.259 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -8.734 9.269 280.70 -43.278 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -8.728 9.169 277.67 -43.582 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -8.726 9.273 280.83 -43.263 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -8.727 9.219 279.18 -42.429 5.202 1365. 48 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =11.7 m/s Incident vertical velocity = V y1 = 13.25 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.7072 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C5. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 k θ Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -9.363 6.14 185.95 -58.795 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -9.363 6.149 186.21 -56.703 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -9.362 6.16 186.55 -56.667 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -9.366 7.891 238.98 -49.876 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -9.363 7.894 239.06 -49.896 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -9.373 7.894 239.07 -49.884 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -9.369 7.795 245.5 -50.239 5.202 1376. 59 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =9.16 m/s Incident vertical velocity = V y1 = 15.19 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.7156 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C6. Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -10.87 4.702 142.39 -67.970 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -10.87 4.672 141.49 -66.766 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -10.88 4.68 141.47 -66.723 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -10.88 5.747 174.05 -62.136 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -10.87 5.745 173.99 -62.155 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -10.87 5.742 173.90 -62.168 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -10.87 5.185 163.30 -64.511 5.202 1387. 68 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =6.57 m/s Incident vertical velocity = V y1 = 16.14 m/s Incident spin velocity = ω1 = 0 rad/s Vertical coefficient of restitution = 0.6871 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C7. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.029 0.352I 0.647I 0.00028 -11.14 3.179 96.283 -75.562 4.791 2 M/2 M/2 80 0.031 0.352I 0.647I 0.00028 -11.14 3.183 96.389 -74.049 4.546 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -11.13 3.188 96.55 -74.032 4.288 4 2M/3 M/3 72 0.0107 0.532I 0.468I 0.00026 -11.14 3.701 112.09 -71.543 5.488 5 2M/3 M/3 80 0.0113 0.532I 0.468I 0.00026 -11.09 3.698 111.99 -71.561 5.206 6 2M/3 M/3 90 0.0119 0.532I 0.468I 0.00026 -11.13 3.693 111.84 -71.644 4.908 7 3M/4 M/4 90 0.00187 0.634I 0.366I 0.00025 -11.13 0.554 16.78 -87.151 5.202 1390 0.2 0.4 0.6 0.8 134567 Sliding RollingX Velocity transition Sliding to RollingHorizontal velocity, m/s Dimensionless contact time,n0 0.2 0.4 0.6 0.8 1201001020Vertical Velocity During ContactVertical velocity, m/s Dimensionless contact time, n 0 0.2 0.4 0.6 0.8 1050100150 Sliding RollingAngular Vel.transition Sliding toRollingAngular spin, rad/s Dimensionless contact time, n Fig.C1 Kinematic parameters and their vari ation with time during contact, case 1 140TOPSPIN IMPACT 18 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =16.13 m/s Incident vertical velocity = V y1 = 5.33 m/s Incident spin velocity = ω1 = 138.20 rad/s Vertical coefficient of restitution = 0.876 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C8. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -4.67 13.681 219.55 -18.847 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -4.67 13.683 217.01 -18.841 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -4.669 13.683 220.02 -18.852 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -4.672 12.766 386.62 -20.085 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -4.668 12.771 386.76 -20.082 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -4.669 12.777 386.93 -20.077 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -4.67 12.762 386.48 -20.099 5.517 1412. 23 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =16.19 m/s Incident vertical velocity = V y1 = 6.75 m/s Incident spin velocity = ω1 = 158.06 rad/s Vertical coefficient of restitution = 0.905 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad T a b l e C 9 . Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -6.11 13.088 261.08 -25.025 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -6.11 13.091 260.02 -25.009 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -6.107 13.092 259.25 -25.022 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -6.111 12.968 392.72 -25.217 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -6.107 12.971 392.83 -25.248 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -6.117 12.975 392.95 -25.216 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -6.11 13.821 418.55 -23.849 5.517 1423. 34 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =14.83m/s Incident vertical velocity = V y1 = 9.98 m/s Incident spin velocity = ω1 = 152.53 rad/s Vertical coefficient of restitution = 0.827 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C10. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 k θ Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -8.257 10.244 304.86 -38.87 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -8.257 10.248 305.18 -38.862 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -8.258 10.249 305.73 -38.836 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -8.251 11.78 356.76 -35.002 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -8.249 11.737 355.44 -35.110 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -8.252 11.783 356.86 -34.995 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -8.249 12.414 375.96 -33.604 5.517 1434. 42 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =13.76m/s Incident vertical velocity = V y1 = 12.55 m/s Incident spin velocity = ω1 = 136.32 rad/s Vertical coefficient of restitution = 0.726 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C11. Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -9.119 9.318 282.18 -44.382 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -9.119 9.324 282.36 -44.319 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -9.105 9.33 282.56 -44.307 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -9.107 10.803 327.16 -40.134 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -9.108 9.813 297.18 -42.859 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -9.106 10.638 322.16 -40.573 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -9.109 11.096 336.04 -39.384 5.517 1445. 48 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =11.98m/s Incident vertical velocity = V y1 = 13.17 m/s Incident spin velocity = ω1 = 147.58 rad/s Vertical coefficient of restitution = 0.750 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C12. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -9.883 8.487 257.04 -49.346 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -9.883 8.491 257.15 -49.321 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -9.879 8.496 257.28 -49.293 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -9.875 9.373 283.85 -46.503 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -9.878 9.371 283.80 -46.523 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -9.883 9.369 283.75 -46.549 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -9.89 9.622 291.38 -45.755 5.517 1456. 61 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =8.45 m/s Incident vertical velocity = V y1 = 15.29 m/s Incident spin velocity = ω1 = 146.74 rad/s Vertical coefficient of restitution = 0.688 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C13. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -10.52 6.425 194.56 -58.591 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -10.52 6.425 194.59 -58.586 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -10.52 6.798 205.88 -57.112 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -10.51 6.500 196.84 -58.287 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -10.52 6.496 196.72 -58.303 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -10.52 6.491 196.58 -58.322 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -10.52 6.328 191.65 -58.69 5.517 1467. 71 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =5.67 m/s Incident vertical velocity = V y1 = 16.39 m/s Incident spin velocity = ω1 = 145.88 rad/s Vertical coefficient of restitution = 0.674 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad T a b l e C 1 4 . Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.03 0.352I 0.647I 0.00028 -11.05 4.795 145.21 -66.538 4.794 2 M/2 M/2 80 0.0318 0.352I 0.647I 0.00028 -11.05 4.794 145.17 -66.533 4.548 3 M/2 M/2 90 0.0338 0.352I 0.647I 0.00028 -11.04 4.791 145.11 -66.556 4.288 4 2M/3 M/3 72 0.0123 0.532I 0.468I 0.00026 -11.05 4.27 129.32 -68.867 5.489 5 2M/3 M/3 80 0.0129 0.532I 0.468I 0.00026 -11.05 4.265 129.16 -68.888 5.208 6 2M/3 M/3 90 0.0137 0.532I 0.468I 0.00026 -11.05 4.259 128.99 -68.915 4.910 7 3M/4 M/4 90 0.00345 0.634I 0.366I 0.00025 -11.05 3.799 115.06 -71.021 5.517 1470 0.2 0.4 0.6 0.8 13.544.555.56 Sliding RollingX Velocity transition Sliding to RollingHorizontal velocity, m/s Dimensionless contact time, n0 0.2 0.4 0.6 0.8 1201001020Vertical Velocity During ContactVertical velocity, m/s Dimensionless contact time, n 0 0.2 0.4 0.6 0.8 1120140160180 Sliding RollingAngular Vel.transition Sliding toRollingAngular spin, rad/s Dimensionless contact time, n Fig.C2 Kinematic parameters and their varia tion with time during contact, case 6. 148BACKSPIN IMPACT 1. 17 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =16.18 m/s Incident vertical velocity = V y1 = 5.08 m/s Incident spin velocity = ω1 = -168.32 rad/s Vertical coefficient of restitution = 0.769 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C15. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 k θ Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in/rad m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -5.265 13.798 -88.94 -15.814 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -5.265 13.799 -88.74 -15.808 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -5.268 13.494 -100.6 -16.139 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -5.267 12.614 95.18 -17.214 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -5.265 12.622 95.76 -17.191 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -5.263 12.614 96.46 -17.209 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -5.266 5.302 166.99 -36.386 5.517 1492. 22 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =15.95 m/s Incident vertical velocity = V y1 = 6.95 m/s Incident spin velocity = ω1 = -148.05 rad/s Vertical coefficient of restitution = 0.799 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C16. Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 k θ Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in/rad m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -3.908 12.86 -45.07 -22.264 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -3.908 12.861 -44.81 -22.275 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -3.907 12.466 -30.09 -22.905 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -3.905 11.324 194.01 -24.936 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -3.908 11.335 193.25 -24.906 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -3.905 11.324 195.44 -24.939 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -3.907 11.558 350.03 -24.495 5.517 1503. 35 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =14.35 m/s Incident vertical velocity = V y1 = 9.97 m/s Incident spin velocity = ω1 = -157.58 rad/s Vertical coefficient of restitution = 0.711 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C17. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 k θ Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in- s2 Lb- in/rad m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -7.087 9.675 -1.783 -36.223 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -7.087 9.677 -1.386 -36.206 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -7.084 9.611 20.875 -36.432 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -7.094 8.594 260.26 -39.523 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -7.09 8.595 260.29 -39.507 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -7.087 8.607 260.68 -39.48 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -7.09 10.027 303.65 -35.264 5.517 1514. 40.5 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =13.8 m/s Incident vertical velocity = V y1 = 11.78 m/s Incident spin velocity = ω1 = -179.12 rad/s Vertical coefficient of restitution = 0.733 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C18. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -8.635 8.276 4.962 -46.216 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -8.635 8.279 5.43 -46.189 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -8.63 7.572 31.65 -48.763 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -8.638 7.893 239.05 -47.577 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -8.637 7.895 239.09 -47.579 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -8.64 7.838 237.37 -47.779 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -8.638 9.073 274.76 -43.593 5.517 1525. 45 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =12.02 m/s Incident vertical velocity = V y1 = 12.11 m/s Incident spin velocity = ω1 = -164.57 rad/s Vertical coefficient of restitution = 0.749 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C19. Dynamic parameters and simulation results Cases M 1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -9.074 6.342 24.668 -55.049 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -9.074 6.344 25.15 -55.032 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -9.071 5.617 52.104 -58.225 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -9.068 6.701 202.94 -53.548 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -9.072 6.704 203.04 -53.518 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -9.066 6.708 203.16 -53.523 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -9.073 7.853 237.82 -49.123 5.517 1536. 59 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =9.12 m/s Incident vertical velocity = V y1 = 15.04 m/s Incident spin velocity = ω1 = -184.04 rad/s Vertical coefficient of restitution = 0.684 Coefficient of sliding friction = µ = 0.55 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C20. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 I Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in- s2 m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -10.29 2.068 50.984 -78.635 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -10.29 2.071 51.583 -78.622 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -10.29 1.708 53.797 -80.572 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -10.29 4.111 124.51 -68.205 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -10.28 4.112 124.52 -68.208 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -10.29 4.112 124.52 -68.214 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -10.29 4.891 148.11 -64.573 5.517 1547. 68 degrees angle of incidence Fixed Inputs: Incident horizontal velocity = V x1 =6.57 m/s Incident vertical velocity = V y1 = 15.95 m/s Incident spin velocity = ω1 = -165.53 rad/s Vertical coefficient of restitution = 0.679 Coefficient of sliding friction = µ = 0 Torsional stiffness coefficient = k θ = 1000 lb-in/rad Table C21. Dynamic parameters and simulation results Cases M1 M2 k y cy I 1 I2 k θ Vy2 Vx2 ω2 θ2 t c Lb- s2/in Lb- s2/in Lb/in Lb-s/in Lb-in- s2 Lb-in-s 2 Lb-in/rad m/s m/s Rad/s Degrees msec 1 M/2 M/2 72 0.026 0.352I 0.647I 0.00028 -10.83 0.76 23.02 -85.986 4.783 2 M/2 M/2 80 0.0275 0.352I 0.647I 0.00028 -10.83 0.767 23.226 -85.952 4.537 3 M/2 M/2 90 0.00633 0.352I 0.647I 0.00028 -10.84 0.685 20.748 -86.378 4.248 4 2M/3 M/3 72 0.00728 0.532I 0.468I 0.00026 -10.82 2.273 68.844 -78.150 5.485 5 2M/3 M/3 80 0.00768 0.532I 0.468I 0.00026 -10.83 2.272 68.8 -78.149 5.204 6 2M/3 M/3 90 0.00815 0.532I 0.468I 0.00026 -10.83 2.269 68.728 -78.161 4.906 7 3M/4 M/4 90 0.00230 0.634I 0.366I 0.00025 -10.82 2.734 82.803 -75.824 5.517 1550 0.2 0.4 0.6 0.8 12468 Sliding RollingX Velocity transition Sliding to RollingHorizontal velocity, m/s Dimensionless contact time, n 0 0.2 0.4 0.6 0.8 1201001020Vertical Velocity During ContactVertical velocity, m/s Dimensionless contact time, n 0 0.2 0.4 0.6 0.8 12001000100 Sliding RollingAngular Vel.transition Sliding toRollingAngular spin, rad/s Dimensionless contact time, n Fig.C3 Kinematic parameters and their vari ation with time during contact, case 6+θ -θ X Y 156 APPENDIX D ERROR ANALYSIS A simple error analysis can be performed for each of the cases of zero spin, top spin and back spin whereby the percentage difference between the theoretically predicted rebound parameters and experimentally determined values of these parameters are obtained. These percentage differences, w ith experimental values as reference, are obtained for all three sets of dynamic paramete rs used to perform the simulations of the tennis ball, namely Case 1, Case 2, and Case 3, as presented in Chapter V. The comparison of percentage differences will also give an idea of how the selection of varying dynamic parameters in Cases 1, 2 and 3 affects the rebound parameters. 1. Zero Spin Case 1 Table D1 . Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 10.992 19.476 2.655 25.418 23.300 7.205 32.428 18.107 4.736 27.876 12.398 16.992 20.108 5.141 10.404 48.449 9.554 20.254 34.963 3.669 16.357 157 Case 2 Table D2. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 10.974 19.464 2.829 24.423 22.762 9.509 32.480 18.132 4.778 27.903 12.411 17.017 22.009 5.927 12.151 48.372 9.537 20.189 34.781 3.642 16.196 Case 3 Table D3. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 20.821 25.699 -69.115 41.380 31.064 -66.717 29.177 16.585 -48.017 10.869 4.288 -24.897 -5.100 6.184 -12.767 21.498 3.479 -1.583 14.635 0.516 0.034 158 2. Top Spin Case 1 Table D4. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 16.348 22.116 0.202 25.203 27.828 -1.327 21.627 17.481 9.791 10.010 1.807 12.860 20.916 8.337 19.980 17.895 1.679 20.044 -6.264 5.778 8.457 Case 2 Table D5. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 16.348 22.116 -0.026 25.241 27.848 -1.295 22.104 17.735 10.229 10.010 1.807 12.857 20.890 8.325 19.958 17.804 1.649 19.959 -6.396 5.815 8.312 159 Case 3 Table D6. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 24.918 27.147 -43.135 26.100 28.293 -34.418 6.155 8.426 -5.832 -3.640 6.944 -1.148 9.509 2.920 8.664 16.606 1.256 18.733 5.385 2.651 21.932 3. Back Spin Case 1 Table D7. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 19.526 19.175 -10.165 15.427 18.395 13.443 43.25 18.974 -0.474 47.295 18.331 14.416 62.718 19.334 15.943 46.335 7.035 16.601 44.713 3.343 -3.682 160 Case 2 Table D8. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 19.451 19.127 -8.566 15.315 18.324 14.726 43.450 19.054 -0.331 46.231 17.975 13.590 62.815 19.359 16.015 46.335 7.035 16.605 44.522 3.326 -3.783 Case 3 Table D9. Percentage error for different incident angles Rebound Horizontal Velocity Rebound Angle Rebound Angular Spin 30.663 25.736 -184.300 30.957 27.216 -126.457 61.250 25.616 -100.682 54.403 20.646 -97.625 53.932 17.019 -85.914 -26.406 6.126 -52.257 -51.592 5.574 -67.773 These results are presented in graphical form on next pages for each of the three incidences of zero spin, top spin and back spin. 161 Zero Spin Comparison of three cases for rebound horizontal velocity in Zero spin -100102030405060 02468 1 0 1 2 Rebound Horizontal Velocity (m/s)Percentage difference Case 1 Case 2 Case 3 Comparison of three cases for rebound angle in Zero spin 05101520253035 -80 -60 -40 -20 0 Rebound angle(Degrees)Percentage differenceCase 1 Case 2 Case 3 162 Comparison of rebound an gluar spin for three cases with Zero incident spin -80-70-60-50-40-30-20-100102030 0 50 100 150 200 250 300 350 Rebound spin,rad/sPercentage differenceCase 1 Case 2 Case 3 Top Spin Comparison of rebound horizontal velocity for three cases in Top spin -10-5051015202530 02468 1 0 1 2 Rebound horizontal velcoty(m/s)Percentage differenceCase 1 Case 2 Case 3 163 Comparison of rebound angle for three cases in Top spin 051015202530 -80 -60 -40 -20 0 Rebound angle (Degrees)Percentage differenceCase 1 Case 2 Case 3 Comparison of rebound angular spin for three cases with Top spin -50-40-30-20-100102030 0 100 200 300 400 500 Rebound spin(rad/s)Percentage differenceCase 1 Case 2 Case 3 164 Back Spin Comparison of rebound horizontal velocit y for three cases in Back spin -60-40-20020406080 02468 1 0 1 2 Rebound horizontal velcoty(m/s)Percentage differenceCase 1 Case 2 Case 3 Comparison of rebound angle for three cases in Back spin 051015202530 -100 -80 -60 -40 -20 0 Rebound angle (Degrees)Percentage differenceCase 1 Case 2 Case 3 165 Comparison of rebound an gular spin for three cases with Back spin -200-150-100-50050 0 50 100 150 200 250 300 Rebound spin(rad/s)Percentage differenceCase 1 Case 2 Case 3 166 VITA Syed Muhammad Mohsin Jafri B-311, Block-10, F.B. Area, Karachi, Pakistan EDUCATION Master of Science in Mechanical Engineer ing from Texas A&M University, College Station,USA(August 2000-May 2004) Area of interest Vibrations, Dynamics and Rotordynamics Relevant courses Engineering dynamics, Mechanical vibrati ons, Vibration measurements in rotating machinery and machine structures, Dynamics of rotating machinery, Methods of partial differential equations Bachelor of Mechanical Engineering fr om NED University of Engineering & Technology, Karachi (February 1995-May 1999) Final year design project Design of liquified petroleum storage facility Relevant courses Engineering mechanics, Solid mechanics, Fl uid mechanics, Stress analysis, Vibrations PROFESSIONAL AFFILIATIONS Member of Pakistan Engineering Council CONTACT INFORMATION E-mail: [email protected] Phone: (979)739-2220 Local Address: 4302 College Main, Apt.114 Bryan, TX 77801