Siz said:
....The conjecture says that the physical basis of a good hit is all to do with maximizing the contact time
Since you've been trying to put numbers on it, maybe you have some physics background. Consider a center-ball hit to make things as simple as possible. The only major assumption is that both the stick and cueball travel the same distance during the collision. Barring gross permanent deformation, this is essentially true.
First of all, suppose you compared two cues, one that produced a contact time of 1/1000 second, the other 1 second. Both have the same mass and are equally energy efficient. The result of a collision is then completely specified by the initial velocity of the cue. The cueball will take off with exactly the same speed using either one. This is because the force, averaged over time, for the second cue is 1/1000'th as great as for the first cue.
But suppose a constant force is applied by the shooter's hand during impact in order to increase the contact time. Does it? It sounds like a simple physics question, but frankly, I don't know the answer to it. However, a related question can be answered easily: how much of that force will the cueball see? Its time averaged acceleration, Aav, will be:
Aav = (2/B)Vcm/T + (1/B)Acm
where Vcm is the pre-impact speed of the center of mass of the system (stick+ball), T is the length of the contact period, Acm is the acceleration of the center of mass of the system, and B is a number which represents the symmetry of the stick/ball force over time. In an elastic collision, B=1; in a completely inelastic collision (the stick and ball never separate), B=2.
The first term of the equation represents the time averaged acceleration of the cueball when no extra force is applied. Since the cueball's final speed is T times Aav, multiplying the first term by T yields a number which is only dependent on the pre-impact speed of the center of mass of the system. The contact time is irrelevant. (By the way, the speed of the center of mass is Vs/(1 + Mb/Ms), where Mb and Ms are the masses of the ball and stick.)
The second term comes from the extra force applied by the shooter during impact. If this force is F, then:
Acm = F/(Ms + Mb)
and the extra force, F', that the cueball sees is:
F' = (1/B)Mb(Acm) = (1/B)F[Mb/(Ms + Mb)]
So for an 18 oz cue, the cueball sees only about 1/(3 + 1) or 1/4'th of the applied force, assuming B is approximately 1.
How significant can this force be? On a hard hit, the cue is accelerated during the stroke by something like 10 pounds on average, peaking at something like 15 pounds. Typically, it's reduced to a small value by the time the tip reaches the ball, or it's actually negative (decelerating). But if, say, 10 pounds were still being applied, the cueball will see only about 2.5 pounds of this by the above logic.
The amount of force represented by the first term is very large compared to it, ranging up to 400 pounds or so on a hard center-ball hit. You would have to have a huge increase in contact time to make any noticeable difference. But how are you going to get it? After all, the added acceleration of the stick compared to the cueball is:
(10 - 2.5) pounds/18oz = .42 versus 2.5 pounds/6oz = .42 (in strange units)
The "added acceleration" of the stick just reduces the much larger deceleration from the first term of the equation (equal and opposite forces). So it appears that both will be accelerated by this extra force by the same amount. (I'm not exactly sure how that symmetry factor "B" is affected.)
An off-center hit is more complicated as you know, and the nominal impact force is reduced to something like 100 pounds (averaged over time). But it still certainly appears that the extra force will be insignificant by comparison. I think the reduction factor of Mb/(Mb + Ms) is even smaller from other considerations, such as the fact the cueball's time averaged speed is less than the stick's averaged speed during impact (the stick travels farther forward than the cueball). On a center-ball hit, they're the same and equal to the center of mass velocity.
Jim