Sidespin on a stopshot

[Jal]

I'm going out on a limb here, but I don't believe anybody ever mentioned how much force was or could be used for this hit. At a slower hit, or using far left/right english, there is going to be much more grab between the cb and ob. I believe the answer is both, depending on how the cb is struck. Practicing banks with spin from the kitchen I think I've seen both happen.

It has been considered and mentioned.

My weak theory on the physics would be.. if the cb is struck harder, the contact is very brief and therefor has much less time to grab the effects of the spin, leaving the cb to spin helplessly in place.

It takes a very large increase in speed to significantly reduce the contact time (see Hertz Law). But the coefficient of friction between the surfaces is quite sensitive to it, diminishing with more speed. Offsetting this is the increase in the normal (compression) force between the surfaces. The net effect is that the friction force itself doesn't change too much. Neither does the contact time. So the cueball is thrown to the side by about the same amount over a wide range of speeds (an inch as far as I can tell). This is for tip offsets near maximum.

It's true that there is much less object ball throw at higher speeds, but this doesn't necessarily mean that the magnitude of the friction is reduced much, for the reasons just given. (The object ball doesn't so much lose sideways velocity as it gains forward speed. Thus, the throw angle is reduced). What you're seeing is likely the effect of more squirt coupled with less swerve.

If you're not very far over left/right, this makes it even more easily done.

If you're just slightly left/right, yes. But you actually get more throw, according to theory, at moderate offsets. Maximum throw occurs at roughly 1/3 of maximum offset, but this varies with speed and surface conditions.

Jim