https://www.youtube.com/watch?v=ia45cZ-LEPA&list=UUW8lTFYIYGN2AjHKN23M-RQ&index=10&t=0s
<....sensational demonstration of nine balls sent into a corner pocket which is shimmed up to 3 1/2 inches. All using the same 15degree perception
from nine different cue ball locations. No looking for the 5/8 fraction, the 13/16 fraction, the 7/8 fraction, the 21/32 fraction, or any of that complex arithmetic stuff that gives marginal results. And not depending on any pocket 'slop'. Now THIS is top level pool playing, in my opinion.
GREAT post, SpiderMan.
Cool video. And a little math proves how he is able to pocket the ob from all 9 different cb spots. Math is awesome.
From the footspot, with a 3.5" pocket, the ob has a 2° window to go cleanly into the pocket. That's a margin of error of +/- 1° from a perfect center pocket shot.
Here's the math for that, the "why":
The ob is 2.25" wide, so a 3.5" opening would only allow 0.625" left or right of center pocket in order for the ball to fit in cleanly. And the ball is 35.4" away from the pocket
Margin of Error = +/- atan (0.625/35.4) = +/-
1.0°
This means the ob can only be off by 1° left or right of center pocket, which is only a 2° window. That's tight!
Now, this same 2° window applies to the cb also, meaning the cb has to be within a 2° window from the ob aim point or ob reference. Here's the math for that:
The cb is about 28" from the ob, based on what it looks like in the video. So let's calculate how much the cb's location can be displaced (left or right of the shot line) and still be within the required 2° window:
CB margin for displacement = tan(1°) x 28 x 2 = 0.98", or basically
1 inch
This means at 28" away from the ob, the cb can be anywhere within a 1" wide window and the shot works. In other words, by looking at all those little white hole reinforcements, they appear to be within the required 1" (2°) window that allows one aiming reference to work for each cb placement. Stan uses a 15° perception with an inside sweep, and he does an excellent job with consistency.
The same thing could be done using one fractional aiming reference also, but I understand this thread is geared to helping understand or learn CTE. I just thought showing a little math to explain why it works would be beneficial, a little proof that it's very possible to use one aming reference to cover a range of shot angles