ShootingArts said:
Dave,
I tried to use the normal stroke I would use to make a bank and have to admit I couldn't see how fast the object ball spun. No clue. I guess we will go with your findings or grant the effect which we both agree on without knowing the exact mechanism.
Hu
Dave was right all along.
The amount of spin the object ball acquires is directly related to how much it's thrown. Normal throw can range from zero to about five degrees. The largest values occur at very slow cueball speeds. Let's take, say, two degrees as typical, and consider a full hit with some sidespin on the cueball. If the cueball is travelling at 5 mph (lag speed) and throws the object ball two degrees, the OB will spin at about one revolution per second. Since it takes less than a third of a second for it to travel two feet (two diamonds), it will only rotate about a third of a revolution over this distance, ignoring what it picks up as topspin.
At 10 mph, with two degrees of throw, it will spin twice as fast but only take about half as long to travel the two feet. So again, it will only complete around a third of a revolution.
It doesn't have to be spinning fast to affect the rebound off the cushion, as is obvious from experience. In the process of removing this spin, or adding to it, the cushion will slow the ball down in the direction parallel to the rail. It will slow it down more if it has to remove this spin first, but not so much if it merely adds to it.
To get maximum spin on the object ball, the required english, surprisingly, is nowhere near maximum. For a very slow cueball, about 3 mph, the optimal tip offset (contact point) is about 1/5 of its radius. For a very fast cueball, say 15 mph, it's one-half this, about 1/10 its radius. This is much less than the offset at which maximum english can be applied at 1/2 its radius.
When I first read this thread, I thought the object ball would spin much faster too. But theory and Dave's observations seem to be in accord.
Jim