<p>Motion of a Golf Ball on a Putting Green</p><p>David J. Murrow 2/8/2014</p><p>1 Motion of a Golf Ball on a Putting Green—D.J. Murrow</p><p>The surface of a putting green can be described by a function u=z-f(x,y)=0 , which gives the height of the green above some reference horizontal plane. The motion of the ball on this surface may then be derived by a constrained minimization of the "action" of the ball on the surface, where the total action is the time-integral of the difference between in it's kinetic and potential energy. This is as in Hamilton's Principle of Least Action. The constraint in the minimization is that the ball must stay on the surface of the green. This leads to the equations of motion of the golf ball on the green in terms of the forces and resultant accelerations, but does not take into account the forces of drag and "stiction", which is a static force keeping the ball from moving until a certain minimum force is applied. These additional forces can be accounted for after the minimization of the action.</p><p>Hamilton's Principle of Least Action</p><p>Objects subjected to conservative forces move in such a way as to minimize the total action in the system.</p><p>Let A be the total action in the system, defined by</p><p>T A  (KE  PE)dt 0</p><p>Where KE is the total system kinetic energy and PE is the system total potential energy</p><p>2 2 2 2 KE  1/ 2 mv  1/ 2m(vx  vy  vz ) and PE  mgz .</p><p>The surface constraint is added with the help of a Lagrange multiplier, giving</p><p>T A  (KE  PE  (t)u)dt . 0 or</p><p>T A  (1/ 2m(v2  v2  v2 )  mgz  (t)(z  f (x, y))dt  x y z . 0</p><p>2 To minimize A, the calculus of variations is used. Setting the variation of A to zero determines the minimum value of A, i.e.,</p><p>T A  (m(v v  v v  v v )  mgz  (t)(z  f (x, y))dt  0  x x y y z z 0</p><p>Note that vx = dx/dt = dx/dt, and </p><p>f (x, y)  (f / x)x  (f / y)y , so after integrating the first three terms by parts</p><p>T    T T T 0  mv x | mv y | mv z |  (m(v x x  v y y  v z z)  mgz  (t)(z  (f / x)x  (f / y)y)dt x 0 y 0 z 0  0 Recognizing that x and y are independent and arbitrary functions, it is concluded that</p><p> mv x  (t)(f / x)  mv y  (t)(f / y)  mv z  mg  (t)</p><p>Now it is clear that these are the three components of forces acting on the golf ball. They do not include the non-conservative drag forces. Assuming these are directly proportional to the component velocities, and act in the opposite direction, </p><p> mv x  (t)(f / x)  kvx  mv y  (t)(f / y)  kvy  mv z  mg  (t)  kvz</p><p>Recall that the gradient of u is normal to the surface u =0. Thus</p><p> n  u  (f / x)ux  (f / y)uy  uz</p><p>Thus, if r is any vector to a point on the surface of the green, the vector n above is normal to it, i.e., nr  0 .</p><p>Taking first and second time derivatives of this constraint reveals</p><p>3  nv  0 , nv  0</p><p>In vector form, the three equations of motion above can be written as</p><p> . mv  n  mguz  kv</p><p>Dotting the normal vector n into this vector equation gives</p><p>0   || n ||2 mg , from which</p><p>  mg / || n ||2 .</p><p>Sustituting this result back into the vector equation gives</p><p> 2 . mv  mgn/ || n || mguz  kv or, dividing through by m and letting  = k/m,</p><p> 2 . v  gn / || n || guz  v</p><p>In general, n / ||n||2 is a non-linear function of x and y, which in turn are time functions describing the x an y coordinates of the golf ball trajectory. This makes the differential equation, in general, non-linear and intractable in closed form except for special cases. However, once f is specified, it can be numerically integrated using the difference equations</p><p>2 an  (gn / || n || guz  v)n</p><p> vn1  vn  ant</p><p>2 rn1  rn  vnt 1/ 2ant</p><p> tn1  tn  t</p><p>4 There is an additional force called the "stiction" force, short for static friction. This force prevents the ball from rolling unless it is overcome by the accelerating forces. Since the accelerating force is an above, the ball will not move until ||a|| > asf .</p><p>Planar Surface</p><p>A special case where the equations can be solved in closed form is when the green surface is a tilted plane. Without loss of generality, the axis of rotation of the tilt is taken to be the y-axis, so that the plane is tilted in elevation. In this case, u  z  f  z  tan()x  0 where  is the elevation tilt angle. </p><p>Then the normal to the plane is</p><p> n  u  uz  tan()ux .</p><p>Thus ||n||2 = 1+tan2() = 1/cos2(), so the vector equation of motion becomes</p><p> 2 v  g(sin ()uz  sin()cos()ux )  v</p><p>2 Let ax = gsin()cos() , ay=0, az = -gsin ( .</p><p>Then the vector equation becomes</p><p> v  a  v .</p><p>The solution to the homogeneous equation is </p><p>t v  v0e</p><p>The general solution is </p><p>t t v  v0e  a(1 e )/ .</p><p>Integrating to find the position vector gives</p><p>5  t  t r  r0  v0 (1 e )/  a(t  (1 e )/ ) / .</p><p>Note that this equation has all forces present except that of stiction.</p><p>The foregoing equations of motion have been programmed on the computer using the Matlab programming language to give some following sample results. Trajectories for the integrated difference equation are compared to the closed form equation for a few cases. The Matlab source code is attached. </p><p>Figure DJM-1 Motion of a Golf Ball on a Putting Green</p><p>6 % Pgm by D.J. Murrow to calculate path of golf ball on tilted green % Green is tilted in elevation by specified amt, % i.e., rotated about y-axis % Hole is at origin, initial ball x,y location % and velocity vx,vy specified clear;clf; pi=4*atan(1);dtr=pi/180; g=9.8; % gravitational accel(m/ssq)</p><p>% Green Tilt Angle: phid=2.5; phi=phid*dtr;</p><p>% Initial Ball x,y Position %x0=-3;y0=3; x0=3;y0=3; % Initial Ball xy-Velocities %vx0=2.78;vy0=-2.0; % vx0=3;vy0=-2.5; vx0=-.5;vy0=-1.75;</p><p>% Drag to mass Coefficient (k/m) gam=0.5;</p><p>% Static Friction Force/m and min vel afs=0.43;vs=0.5; tc=1/gam; % Time-constant figure (1) % Draw Hole dh=3.5*.0254;rh=0.5*dh; ad=0:360;a=ad*dtr;xh=rh*cos(a)*cos(phi);yh=rh*sin(a); fill(xh',yh','b') axis([-3 3 -3 3]) grid on hold on</p><p>% Set up time steps and duration tmax=10*tc;dt=.005*tmax; t=0:dt:tmax;N=length(t); a=0.5*g*sin(2*phi); as=(a<afs); % Calculate trajectory tphi=tan(phi);dt2=dt^2; x=zeros(1,N);y=zeros(1,N);z=zeros(1,N); vx=zeros(1,N);vy=zeros(1,N);vz=zeros(1,N); afx=zeros(1,N);afy=zeros(1,N);afz=zeros(1,N); vmag=zeros(1,N);af=zeros(1,N); x(1)=x0;y(1)=y0;z(1)=tphi*x(1); vx(1)=vx0;vy(1)=vy0; vz(1)=tan(phi)*vx(1); vmag(1)=sqrt(vx(1)^2+vy(1)^2+vz(1)^2);</p><p>7 for n=1:N-1 vmag(n)=sqrt(vx(n)^2+vy(n)^2+vz(n)^2); afx(n)=-a-gam*vx(n); afy(n)=-gam*vy(n); afz(n)=tphi*afx(n); af(n)=sqrt(afx(n)^2+afy(n)^2+afz(n)^2); plot(x(n),y(n),'.r') if ((vmag(n)<vs).*as) af(n)=0;afx(n)=0;afy(n)=0;afz(n)=0;a=0; vx(n)=0;vy(n)=0;vz(n)=0; end vx(n+1)=vx(n)+afx(n)*dt; vy(n+1)=vy(n)+afy(n)*dt; vz(n+1)=tphi*vx(n+1); x(n+1)=x(n)+vx(n)*dt+0.5*afx(n)*dt2; y(n+1)=y(n)+vy(n)*dt+0.5*afy(n)*dt2; z(n+1)=tphi*x(n+1); end</p><p>%pause</p><p>%figure(2) %plot(t,vx,'r',t,vy,'b',t,vz,'g') %grid on</p><p>8</p>