Golf Ball Flight Dynamics - Union College

1 Golf Ball Flight DynamicsBrett BurglundRyan Street5-13-20111 AbstractProjectile motion is a field of study that has many practical appli-cations. In this case the projection of a golf ball will be analyzed. Thegolf balls projection is effected by many different forces, here the spinand drag will be the main areas of concern. A system of differentialequations will be interpreted from aerodynamics and drag to aid inthe discussion. Specifically, the magnus effect and resistance due toaerodynamic drag are IntroductionGolf balls didn t all ways have dimples on them. The dimpled golf ballwas first introduce in 1905 by Wiliam Taylor, an engineer. He found thatdimpled balls were more aerodynamic , allowing golfers to have better controland further drives. The number of dimples on a typical golf ball ranges from250-450. Each manufacture has a different theory on the number of dimples,depth of dimple and in the pattern in which the dimples are placed.

2 All theseattributes contribute to the balls overall aerodynamic , golf ball manufactures are trying to engineer the perfect golf United States Golf Association (USGA) strictly regulates the design andmaterial used in golf ball manufacturing. With the use of modern material tocreate different core a designer has the ability to create a ball that has differ-ent properties such as greater spin, soft feel at impact and lower these reasons the manufacturing of golf balls is highly Our ModelA golf balls Flight pattern is effected directly by the presence of air resistanceand the rotation of the ball. Looking a both air resistance and rotationflight paths can be predicted. In this model both air resistance and rotationare used to set up a system of differential equations that can be used topredict several different Flight paths. Matlab was used to solve the systemof equations using a numerical solver. The calculated trajectories were thenplotted to show the effects of angular acceleration and aerodynamic system is used to model a smooth ball while another is used model adimpled ball.

3 Throughout the models the initial velocity and angle of attackaffect the trajectories. Initial velocity is the velocity the ball leaves thex yplane. Angle of attack is the angle between thex yplane and initial Classic Parabolic MotionFirst, a discussion in classic parabolic motion is in order to acquire the knowl-edge needed to understand the golf balls Flight . classic parabolic motion canbe derived from a system of two differential equations, in thexandydi-rections. In the classical model, every force, excluding gravity, is assumes no wind, drag , magnus, change in altitude, or changes in hu-midity. From Newton s second law of motion,and the fact that gravity is theonly force acting on the ball, Fy=my my = mgy = gy = gt+C1y= 12gt2+C1t+C2 Solving forC1andC2yieldsC1=V0yC2=y0andy= 12gt2+V0yt+y0In thexdirection, no forces act on the ball. Fx=mx mx = 0x = 0x =C3x=C3t+C4solving forC3andC4yieldsC3=V0xC4=X0andx=V0xt+X04 Figure 1: Classic Parabolic MotionFrom this equation it is easy to see that the velocity of the ball in thexdirection stays constant throughout the duration of the Flight .

4 Note thatthroughout these systems, the initial coordinates of the ball will be (0,0)4 General EquationsFrom Newtons second law of motion the Flight equations were derived for thetrajectories of the ball. The equations were derived using air resistance andthe Magnus Force. the Flight equations were in three directions. Fy=my = Weight AirResistance+MagnusForce Fx=mx = AirResistance+MagnusForce Fz=mz = AirResistance+MagnusForce55 Air ResistanceThe force put on the golf ball due to its interaction with air molecules is calledaerodynamic drag or air resistance. The air resistance can be calculated usingthe equation shown below,R=12CD AV2where,CD= drag Coefficient =Air densityA=Surface AreaV=VelocityThe air resistance must be considered in all three directions:Rx=12Cd AV2xRy=12Cd AV2yRz=12Cd AV2z66 Smooth BallLets think of a smooth ball such as a ping pong ball for this model. When aball travels though the air there is resistance due to the contact between thesurface of the ball and the air molecules.

5 The air surrounding the ball as itflies is called the boundary layer. In this case, the air molecules tend to passby the ball and around the ball in a parallel fashion this is laminar flow. Thelaminar flow creates separation around the ball creating a large wake of lowpressure behind the ball, decelerating it. Only air resistance is considered inthis case. Figure 2: Laminar AssumptionsThe following assumptions were made to simplify the smooth ball No Spin2. No wind3. No changes in air conditions due to altitude or humidity4. No change in Barometric pressureR=Air Resistancem=MassVy=Velocity inYVx=Velocity inXVz=Velocity Flight Equations Fy=my = mg R(Vy) Fx=mx = R(Vx) Fz=mz = R(Vz)Figure 3: Smooth Ball Projection87 Magnus ForceHave you ever notice that then you throw a baseball or hit a golf ball theflight pattern is much different when you place spin on the ball. When spinis placed on a ball the pressure around the ball is altered, this is called themagnus effect.

6 Lets look at a backspin and how it effects the Flight of a golfball. When backspin is placed on a golf ball it will allow the ball to gainloft and have a longer Flight . When the ball spins it directs air upwards infront and downward behind it, creating a pressure difference. The pressuredifference pushes the ball upward allowing the ball to stay in Flight for greatertime, this is called the Magnus Magnus FormulaThe Magnus force lifting the ball is caused by the rotation and the velocity ofthe ball. The rotation causes air disturbances perpendicular to the directionof the angular velocity causing the direction of Flight is altered perpendicularto the linear velocity. This relationship shows that the Magnus force iscalculated using a cross ( V)V=Velocity =Angular VelocityS=Magnus CoefficientThe cross-product providesMagnus Forcein the three dimensionsS ijk i j kViVjVk =S[( jVk iVj)i ( iVk kVi)j+ ( iVj jVi)k]FMx=S( jVk kVj)FMy=S( kVi iVk)FMz=S( iVj jVi)98 Dimpled BallAs discussed in the smooth ball model, when a ball travels through the aira disturbance is produced.

7 The difference between the smooth ball and thedimpled ball is this disturbance. The boundary layer surrounding a dimpledball is turbulent. This turbulent flow is able to stay very close to the surfaceof the ball for a longer period of time than a laminar flow. The turbulentflow allows for a smaller separation of air flow around the ball creating asmaller area of low pressure behind the ball, causing deceleration at a slowerrate than the smooth ball. The magnitude of the magnus force is drasticallyincreased with the dimples, therefor both the Magnus Force and air resistanceare considered in this case. Figure 4: Turbulent AssumptionsThe following assumptions were made to simplify the dimpled ball No wind2. No change in air conditions due to altitude or humidity3. No changes in Barometric pressureR=Air ResistanceFM=Magus Force =Omegam=MassVy=Velocity inYVx=Velocity inXVz=Velocity Flight EquationsThe Flight equations make up the system of equations that govern the tra-jectory of the golf ball.

8 The equations account for air resistance and magnusforce. Fy=my = mg R(Vy) +FM( ,vx,vz,)y = g Rm(Vy) +FMm( ,vx,vz) Fx=mx = R(Vx) +FM( ,vy,vz,)x = Rm(Vx) +FMm( ,vy,vz) Fz=mz = R(Vz) +FM( ,vx,vy,)z = Rm(Vz) +FMm( ,vx,vy) BackspinThese figures use the same initial velocity and angle of 5: Backspin Hook and SliceFigure 6: Hook, Pull, Push, and Slice, from left to right13 Figure 7: 3-dimensional view. These figures use the same initial velocity andangle of ComparisonThese figures use the same initial velocity and angle of attack. There isbackspin ( k) of 130 radians per second on the dimpled ball trajectory. Noticehow the dimpled golf ball flies more than twice the distance of the smoothball. This is caused by the reduced coefficient of drag due to the turbulentflow behind the dimpled 8: Comparing the Models9 ConclusionIt is quit clear to see the effect dimples have on a golf ball trajectory andcontrol. The increase in control and trajectory is directly related to theturbulent boundary layer that is created by the dimples.

9 The smooth ballstrajectory is about half a far as the trajectory of a golf ball. The systemsof equations that were used are very simple, only taking into account ballrotation and drag . The dimpled ball system accounted for the effects ofdimples but for further analysis a greater knowledge in aerodynamic theoryis References1. Xu, L. (2006). Analysis of baseball pitches. Informally publishedmanuscript, Carnegie Mellon University, Pittsburgh, Pensilvania. Re-trieved from Healey, J. ( ). The golf ball. Retrieved Elert, G. (1998). aerodynamic drag . The physics May 5, 2011, from Barber III, J. ( ). Golf ball Flight Dynamics . Informally publishedmanuscript, A and EP, Cornell University, New York, New York. Re-trieved Labeau, et al., K. (2004). Golf ball projectile motion. Informallypublished manuscript, School of Engineering, Grand Valley State Uni-versity, Allendale, from lait/312 Original image obtained from: Accessed 5/5/20111611 Classic Parabolic Motion m-filefunction Classicball%initial velocity = feet/sec = m/h @ Clasical Model )xlabel( X )ylabel( Y )grid onendfunction xprime=flights(t,x)xprime=zeros(2,1);%Xx prime(1)=x(2);xprime(2)=0;%Yxprime(3)=x( 4);xprime(4)= ;%Zxprime(5)=x(6);xprime(6)=0;end$ Smoothball$function velocity = feet/sec = m/h @ degreesplot(x(:,1),x(:,3));title( Smooth Ball Model )xlabel( X )ylabel( Y )grid onendfunction xprime=flights(t,x)17%ParametersC_d=.

10 25;r=.002378;A=.25*pi*( )^2;m=( (16* ));%D =((1/2)*C_d*r*A*V^2%Square of the velocity is in the equation, thereforD =((1/2)*C_d*r*A);xprime=zeros(2,1);%Xxpr ime(1)=x(2);xprime(2)=-(D/m)*x(2)^2;%Yxp rime(3)=x(4);xprime(4)= (D/m)*x(4)^2;%Zxprime(5)=x(6);xprime(6)= -(D/m)*x(6)^2;end$ 3-dimensional Dimpled Ball Models$function -ob );title( Golf Ball Model )xlabel( X (feet) )ylabel( Z (feet) )zlabel( Y (feet) )grid onhold onendfunction xprime=flightg(t,x)%Parameters%DragC_d=. 15;r=.002378;A=.25*pi*( )^2;18m=( (16* ));%D =((1/2)*C_d*r*A*V^2%Square of the velocity is in the equation, thereforD =((1/2)*C_d*r*A);%Magnuss=.000005;m=( (16* ));M=(s/m);W_I=-100;W_J=0;W_K=110;xprime =zeros(2,1);%Xxprime(1)=x(2);xprime(2)=- (D/m)*x(2)^2+M*(W_J*x(6)-W_K*x(4));%Yxpr ime(3)=x(4);xprime(4)= (D/m)*x(4)^2+M*(W_K*x(2)-W_I*x(6));%Zxpr ime(5)=x(6);xprime(6)=-(D/m)*x(6)^2+M*(W _I*x(4)-W_J*x(2));end$19)