Hi Colin,
Thanks for the detailed info. I think it's pretty clear where our point of departure is: the amount throw. But first:
Colin Colenso said:
...To me, it would seem to make more sense to define a pivot point as the length of bridge at which the CB will travel along the exact same line after being pivoted. This I believe is significantly different from the resultant line of an OB placed at one diamond distance travelling the same line.
I think that I and everyone else agrees with your definition, if you mean by traveling along the same line that its initial line of travel is the same, ie, sans later swerve. I think you do mean this but correct me if I'm wrong.
Colin Colenso said:
...From my tests, the OB when hit at firm speed (travel 2.5 table lengths) with 1-2 tips of english will turn off line by about 1" in every 20" travelled.
This is where our differences lay. The figure of 1" in 20" is almost three degrees. Dr. Dave's equations show a throw value of less than one degree at this speed, which is, I believe, about 8 mph for the 2.5 table lengths of travel. (The values shown in his graph for spin induced throw are for a cueball speed of 3 mph). Obviously, this makes for a large difference in the estimate of the error. It is of course possible that the balls you've been using are different than the ones his theory is based upon (from Wayland Marlow's experiments). A factor of three seems unlikely, but who knows?
Colin Colenso said:
That means to hit it in the intended direction, the CB must hit it about 1mm from the center contact point. Meaning that such a test actually measures the pivot point for a squirt angle of a couple of degrees.
I don't believe this is right if I'm reading it correctly. 1mm in one diamond's worth of travel is an angle of arctan (1/25.4/12) = .19 degree. This is the error in the squirt angle and is even smaller, by a third, if the throw is actually as the aformentioned theory predicts.
Here is an error calculation for a cue with a true pivot point of 15", a tip offset of (2/5)R (= .45"), and a throw value of one degree. (Dr. Dave's theory predicts .70 degree whereas my version of it predicts .63 degree at a cueball speed of 8 mph. But let's say it's 1.0 degree.)
To have the cueball stop dead, the object ball has to be hit off-center by R*sin(1 deg) = 1.125"*sin(1 deg) = .0196".
With 12" of cueball travel, this introduces an error of arctan(.0196"/12") = .094 deg.
For a cue with a 15" pivot point, the true squirt angle is arctan(.45"/15") = 1.718 deg.
With the above error of .094 deg, the pivot point will instead be measured at .45"/tan(1.718 deg - .094 deg) = 15.9".
The above is not quite exact geometry, but pretty close. Do you agree with it, if not the throw part?
Colin Colenso said:
It's worth noting that many still seem to think that speed does not significantly increase squirt, and that the differences are mainly a result of reduced swerve when hitting hard. I've trialled this quite a lot recently and I'm convinced that increasing speed significantly increases squirt. Two aspects confuse the matter in opposing directions, swerve and reduced spin induced throw at speed.
I think that increasing speed probably does increase squirt a bit because of the increased tip offset. Is there another speed related mechanism that might affect it?
Jim