Tennis serving machine--- How does a spinning ball bounce? I have an idea of making a tennis serving machine. I will briefly describe what it is:


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*The machine is configured to serve the ball at a fixed speed to the center of the left (or right) service court

*Based on the rotation in 3 directions (X,Y,Z) of the ball, the machine would let the ball land in various positions in the left service court.


How can I determine the relationship between the angular velocity of the ball and the location where it lands?
 A: If you want to describe the dynamics of the ball, you need to use the SO(3) matrix which describes the ball's orientation. This is a 3 by 3 matrix whose transpose is its inverse. The orientation may be parametrized by Euler angles, and most of the literature on rigid rotating bodies uses this convention. But I think it is best just to use the matrix entries themselves because Euler angles are hard to work with. The orientation matrix itself has a velocity vector which is described by an anti-symmetric matrix.
The center of mass motion of the ball, ignoring air resistance, will be a parabolic arc in the direction of the original motion, and the spin has no effect. The ball will not go off-center just by spinning it unless it collides with some part of the machine going out.
There is asymmetric air drag on the ball when it is rotating, which will lead the ball to arc its trajectory a little bit. This is a small effect. You shouldn't use spin to do this, because the main effect of the spin will be on the bounce when landing, which is going to be impossibly wild given the enormous spin you intend to put on the ball. Human servers can't put too much spin on the ball, because they are limited by the racket design. You are better off just aiming the ball in a different direction for different serves, and if you want to disguise the direction, do it with a reflector plate which the ball bounces off. You should match to experiment, not theory because it will be easier to fit the experimental data with a curve than to predict it from mechanics.
Given your current spin design, you will make inhuman and unanswerable serves.

Reflection of spinning tennis balls
Since you keep asking for this, I worked it out. It is somewhat interesting because there are several issues that one has to keep straight. Most of the answer is determined independent of tennis-ball details, but one quantity, the change in rotation in the plane of reflection, is impossible to predict well.
Consider a tennis ball impacting the z=0 plane, travelling freely with velocity $v_x,v_z$, and with rotation vector $(\omega_x, \omega_y, \omega_z)$. This is the general case


since you choose the z axis perpendicular to the wall, and the x-axis parallel to the velocity in the plane of the wall.
At the moment of impact, the friction force of the ball will quickly and irreversibly enforce the no-slip condition, before any significant deformation of the ball. The reason is that the total impulse (momentum transfer) from the impact is about $2Mv_x$, so that the available friction impulse is of the order of $2\mu Mv_x$, where $\mu$ is the coefficient of friction, which is of order 1, and this is much bigger than the impulse required to enforce no slip or a reasonable range of $\omega$, say 10-500.
No-slip-on-contact means that the friction impulse imparted to the ball in the x-y plane $P_x,P_y$ must satisfy the following:
$$ v_x + {P_x\over M} = R\omega_y - {\alpha P_x\over M}$$
$$ {P_y\over M} = R\omega_x - {\alpha P_y\over M}$$
These conditions enforce that the velocity and the angular velocity after the impulse are those required for no-slip. This gives the impulse. The constant $\alpha$ is the coefficient of the moment of inertia,
$$ I = {2MR^2\over \alpha}$$
For a shell, like a tennis ball, $\alpha=3$.
These conditions determine the outgoing velocity in the x,y directions and the outgoing angular velocity in the x,y directions.
$$\Delta V_x = {R\omega_y - v_x\over 1+\alpha}$$
$$\Delta V_y = {R\omega_x \over 1+\alpha}$$
The z direction is executing a reflection independent of the x and y, because once no-slip is established, the elastic process happens as it would for a non-rotating ball. The final z velocity is $\kappa v_z$ in the opposite direction, so that
$$\Delta v_z = -(1+\kappa)v_z$$
Here $\kappa$ is a phenomenological bounce-loss parameter, for a tennis ball, I would guess about .8 (from bouncing tennis balls, they go back to about 64% of their original height each bounce). 
The final values of $\omega_x$ and $\omega_y$ are determined by the no-slip condition :
$$\omega_y^f = {v_x\over R}$$
$$\omega_x^f = {v_y\over R}$$
The undetermined quantity is the final value of $\omega_z$, the rotation in the plane of the impact wall. This rotation is reduced by the friction force as the ball elastically deforms, and bounces off. I will write this as
$$\omega_z^f = q(v_z,\mu) \omega_z$$
Where $q(v_z,\mu)$ is a phenomenological function which you need to parametrize. The friction torque is reduced by the impact area from the friction force, and this is a factor of maybe 1% for a ball at 60 m/s, but the total friction impact available is $2\mu Mv_z$, which is about 100 times the amount needed to stop a reasonable rotation. So these are comparable, and it is not easy to predict how the impact will affect this component of rotation, except that it will reduce it, perhaps by as little as 1%, perhaps as much as 90% per bounce. This needs to be measured for a real tennis ball at real impact speeds.
