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View Full Version : Angle limitation for this draw shot?

okinawa77
02-02-2008, 03:55 AM
Most times, I can execute this type of shot, but sometimes the CB will contact the rail and produce an undesirable outcome. I think it is because I may have too much angle at times, and I just don't know where to "draw the line" for the angle limit.

Has anyone figured out the limitations on the CB to OB angle for this shot?
I plan to create a personal project to work on this.

renard
02-02-2008, 05:21 AM
Most times, I can execute this type of shot, but sometimes the CB will contact the rail and produce an undesirable outcome. I think it is because I may have too much angle at times, and I just don't know where to "draw the line" for the angle limit.

Has anyone figured out the limitations on the CB to OB angle for this shot?
I plan to create a personal project to work on this.

The angle limitation is usually three times the angle in, on a draw. Check this out: http://www.sfbilliards.com/articles/1995-10.pdf

Dawgie
02-02-2008, 12:32 PM
Thanks for the link. It was a great article.

Jal
02-02-2008, 03:25 PM
Most times, I can execute this type of shot, but sometimes the CB will contact the rail and produce an undesirable outcome. I think it is because I may have too much angle at times, and I just don't know where to "draw the line" for the angle limit.

Has anyone figured out the limitations on the CB to OB angle for this shot?...The final direction of the cueball after it reaches natural roll can be found with some simple geometry, as shown here:

http://ww2.netnitco.net/users/gtech/CBDirection4.jpg

The method does not account for follow spin gained or draw spin lost on the way to the object ball. Nor does it or any of the other methods I've seen account for spin lost during the collision. The latter can be up to a third at small cut angles and moderate tip offsets. To figure this in, you would have to shorten the black arrow labeled "b" in the right-hand diagram by up to a third. Knowing exactly how much is not very easy as this depends on the amount of throw, and the calculation of throw is not that straightforward.

But relatively little spin is lost during the collision if a large tip offset is used. So the method becomes considerably more accurate when the arrow b approaches 1/2 the radius of the cueball, which is generally taken to be the maximum offset of the tip to cueball contact point.

On your particular shot then, let's estimate the cut angle at 5 degrees. Using maximum draw (b=1/2 R) and assuming no spin is lost before or during the collision, the cueball's direction as it reached natural roll would be 10 degrees off its original direction (the angle between the before and after lines in your diagram). If it had 80% of its initial spin after the collision, the angle would be about 12.5 degrees. If 50%, the angle would be about 20 degrees.

In algebraic terms, that angle, call it "A", is related to the cut angle "C" and the vertical tip offset "b" (of the contact point) by:

tan(A) = sin(C)cos(C)/[sin(C)sin(C) + b/R]

where R is the ball's radius. b/R is positive for offsets above center, and negative for offsets below center. As indicated, b has to be adjusted for any spin loss. It is really the effective offset after all the dirty business has been completed.

Jim