I've heard it suggested before that the "average offset" of the tip while it's in contact with the CB determines the final spin-speed ratio. But I wonder why it wouldn't be the maximum tip offset (tip offset just before contact ceases) that determines final spin-speed ratio. Is there a specific reason?
pj
chgo
Patrick, my two cents.
Take the simple case of a constant force applied through the center of mass of an object over a finite time T, (i.e., no spin involved). At time T the object will have moved a certain distance X. Its average distance (averaged over time), however, is exactly 1/3 of X (Xav = 1/3 X). It turns out that the range of of Xav is very small for a force which is symmetric about T/2 (which is the case for a constant force)...at least the ones I've looked at. Some examples:
Sawtooth: Xav=.292X
Sinusoidal: Xav=.297X
Thin spike at T/2: Xav=.25X
The symmetry about T/2 represents a perfectly elastic collision. The numbers don't change all that much with less than perfect elasticity, and fall into a similarly small range.
In the case of tip offset, things are more complicated, but the principle is the same. The average angle of rotation during impact, which is integral to figuring the average moment arm (offset) of the applied force, is a somewhat smallish fraction of the total rotation, as in the linear cases above. I was in the process of working this out a year or two ago, but was distracted (easily done!). From a few of the preliminary and incomplete results (sawtooth and sinusoidal - both elastic and inelastic) it appears that the numbers are very nearly the same within these cases, and as a guess, somewhere between 10% and 25% of the total rotation. This should be dependent on the initial offset itself. (I'm a bit fuzzy on this stuff now, so don't quote me

).
The moral is that the total amount of ball rotation is misleading since the force function, along with the moment arm (via ball rotation), and thus the
torque (=force X moment arm), vary with time. You have to average the torque over the contact period, and then divide by the average force (also averaged over time) to isolate the average or effective offset. But from the linear cases, you get the intuitive sense that ball rotation doesn't add that much. And its effect is to just partially reclaim some (maybe all?) of the offset lost to squirt (i.e., the direction of the force is not straight ahead because of squirt, as I know you know).
Welcome back Dr. Dave!
Jim