nataddrho said:

After I said this, I exported 10 minutes of DigiBall data and calculated that my SPF was an average of 1.326, with a standard deviation of 0.506.

I think this is useless, lol.

It would only have meaning if you could compare it to, say Jayson Shaw, running 500+ balls, and maybe noting you aren't using as much spin as he is. Maybe using less could help, for 14.1.

It would have to be very specific for it to matter.

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lfigueroa said:

Well, that sure cleared it up.

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jay helfert said:

Shipping has become quite expensive these days. I do have a batch of posters left. I'd send you a bunch of them (ten, twelve or more good ones) for $100. I'll pay the shipping.

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dr_dave said:

RPF definitely increases linearly with tip offset. The math and physics is here:

TP A.12 – The relationship between cue ball spin and cue tip offset

SRF = omega / (v / R) = (5/2) * (b/R)​

where SRF is the spin-rate factor (spin-to-speed ratio), omega is the angular speed in radians/sec, v is the ball speed, R is the ball radius, and b is the tip offset from center.

Speed is related to distance (d) and time (t):

v = d / t​

So:

SRF = (omega * t) * R / d​

The first term is related to revolutions (rev) according to:

(omega * t) = rev * (2 * pi) [there are 2*pi radians per revolution]​

So:

SRF = (2 * pi * rev) * R / d​

So revolutions per foot (RPF) is:

RPF = rev / d = SRF / (2 * pi * R)​

with d and R measured in feet.

The ball radius (in feet) is:

R = (2.25" / 2) * (1ft / 12") = 1.125/12 feet​

For an assumed maximum spin at the standard miscue limit of 0.5*R,

SPF = (5/2) * (b/R) = (2.5) * (0.5) = 1.25​

which gives:

RPF = 1.25 / (2 * pi * 1.125/12) = 2.1 rev/ft
​

If you roll a ball (SRF = 1) and see how many revolutions it makes in one foot, you would observe:

RPF = 1 / (2 * pi * R) = 1.7 rev/ft​

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