CTE Sweep: a short study

mohrt

Student of the Game
Silver Member
This is something I was playing with recently, and thought I'd share. Note that I have no proof of accuracy or correctness of these findings, this is only something found through pure observation.

First I setup a straight-in shot, and marked a dot on the shot line (the dot in the middle.) Ignore the pink ring on the table, that just happened to be there.

7EbG0PV.jpg


Then I lined up my eyes on visual CTEL-A. Once eye position was locked, I took note of where CCB was. It was the dot I marked just left of the middle dot.

Then I lined up my eyes on visual CTEL-C. Once eye position was locked, I took note of where CCB was. It was the dot I marked just right of the middle dot.

The distance between the dots is right about 2mm.

What does this tell us? What it should tell us, is that any time we are executing a LEFT sweep, our eyes move RIGHT on the CB (to the center dot). Likewise, any time we are executing a RIGHT sweep, our eyes move LEFT on the CB (to the center dot). Of course when we move onto this new center, we must align our cue exactly perpendicular (or STRAIGHT through the core) on the new CB center. That means, the movement is a slight rotation of the eyes around the CB. As a right-hander, the right sweep is a very short movement since our cue is already very close to this position. The left sweep requires a bit more of a movement to get the cue into position, and thus, the "rotation of the shoulder" that many have stated previously. The entire cue/body is rotating around the CB to the new center.

I tried this with varying shots and it seemed pretty consistent... and that makes sense if sweeps are truly consistent. This might help someone understand what to look for when executing a sweep. ie. they can focus on the CB and memorize what this movement looks like from the surface of the CB, and make that your target. The CB is the target after all, right?

Now, I have no claim on this 2mm distance thing, this is only what I came up with on my own experiments. However, I thought I'd share as this may open doors to someone trying to get a grasp on sweeps.
 
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mohrt, not to derail your thread, but I believe you mean mm, not cm. 1 cm is just slightly less than 0.4 inches, so 2 cm would be about 1/3 the width of the ball.
 
mohrt, not to derail your thread, but I believe you mean mm, not cm. 1 cm is just slightly less than 0.4 inches, so 2 cm would be about 1/3 the width of the ball.

You are right my mistake. It is 2mm, or 1/5 of a cm. I changed the post. And to be clear, the distance between each dot is 2mm.
 
Interesting analysis, seems to be in line with what I experienced as well when using Pro One. Specifically, that the sweeps are typically more subtle then newbies make them out to be, and therefore can be at once more consistent or more elusive, depending on which way you want to look at it... :)

Thanks for continuing to analyze your own implementation of Pro One and continuing to add details, as Stan as, for the people that are trying to work with the system, much appreciated!
Scott
 
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