...I'm trying to find the surface speed of the revolving CB, and for that I need to know RPMs. I know it will vary depending on how hard I hit the ball, and how far the tip is offset from center. I'm just looking for a ballpark range for various speeds and offsets. I guess I'll just have to go with what Bob mentioned and figure it out on my own on a case-by-case basis. I'm just being lazy and was hoping there were charts somewhere.
If you want to calculate it yourself, the formula is very easy.
Let:
R == radius of the ball,
W == spin rate in radians per unit time (1 revolution per second = 2pi rad/sec = 6.28 rad/sec),
V == speed of the ball.
b == tip offset (same units as the radius - the ratio b/R doesn't need units, of course)
The surface speed of the spinning ball (the product of R and W) at the ball's spin equator is then given by:
RW = (5/2)(b/R)V
Example:
If struck at the miscue limit (b/R = 1/2), the formula becomes:
RW = (5/4)V
If the ball is traveling at 16 mph, the rotational surface speed is then 20 mph. To convert to feet/sec, multiply that by 1.467. To convert to inches/sec multiply by 1.467*12 = 17.6.
From the first formula, the spin rate in radians per unit time is:
W = (5/2)(b/R)V/R
So if the ball's speed and the subsequently calculated surface speed are in units of mph, to convert to revolutions per second, first multiply by 17.6 to convert to inches/sec, then divide by the radius (1.125 inch) to convert to rad/sec, then divide by 6.28 to convert to revolutions/sec. Putting it together, if the ball's speed is in mph, its rotational surface speed at the spin equator in inches per second is:
RW = 17.6(5/2)(b/R)V = 44(b/R)V
and its spin rate in revolutions per second is:
W = (17.6/6.28)(5/2)(b/R)(V/1.125) = 6.23(b/R)V
If you want its spin rate in radians per second, leave out the division by 6.28 in the above:
W = 39.1(b/R)V
Hope that's clear enough.
Jim