Break accuracy v. speed - please quantify

D

derangedhermit

AzB Gold Member
Gold Member
Silver Member
If you consider only the total energy transfer between cue ball and the rack of balls (9, 10, or 15 - I think it makes no difference), what does the trade-off look like between cue ball speed and off-center hits?

I'm trying to get a feel for how accurate you need to be. For example, will being off center 1/16" (1.6mm) cost the equivalent energy of 1 mph lower cue ball speed? How about 1/4"? I guess the relevant speed range is about 18-25mph.

Assume a tight rack (all balls touching). I think, based on intuition, the energy drop off increases non-linearly as the contact point moves further away from a center hit.

A graph would be outstanding!:grin:
 
I don't see how it matters all that much. The objective is to be in line to the head ball even if you are not trying to concentrate on hitting it dead on. Once you are in line there is very little you can do to mess up a good break. About nine practice strokes is all you need to send the cue ball 4-5 feet midair. If I'm in for a short set, I'll try to crush all the balls every time, otherwise I'll pop all the breaks. Break hard enough and you can send the cue ball flying and still land good.
 
derangedhermit said:
If you consider only the total energy transfer between cue ball and the rack of balls (9, 10, or 15 - I think it makes no difference), what does the trade-off look like between cue ball speed and off-center hits?

I'm trying to get a feel for how accurate you need to be. For example, will being off center 1/16" (1.6mm) cost the equivalent energy of 1 mph lower cue ball speed? How about 1/4"? I guess the relevant speed range is about 18-25mph.

Assume a tight rack (all balls touching). I think, based on intuition, the energy drop off increases non-linearly as the contact point moves further away from a center hit.

A graph would be outstanding!:grin:
Click to expand...
I have sent hours on end with numerous people including a top 3 US pro playing around with break speed & results. I have used radar guns, iPhone break Speed App, video replay and charting and believe that there is a direct correlation between where your cue tip hits the cue ball and your breaking success.

2 main reasons...

1) most people do not hit the cue ball where they think they are hitting it during a break. The extreme force throws their aim and cue delivery way off. Maximum energy transfer from cue to cue ball is a dead center hit, over the course of hundreds of breaks I found that for every 1/4 tip you miss center ball by there is a 5% drop in speed... easy math is that a 20 MPH break goes to 19 with a 1/4 tip miss, 18 with a half tip miss, etc...

try marking a cue ball with a sharpie & see if you can deliver your tip to center ball at break speed... you will be surprised.

2) just as maximum energy transfer between cue and cue ball is achieved with a center ball hit, maximum energy transfer between cue ball and rack is achieved with a "center ball hit" on the head ball. if you are not delivering the cue tip to the cue ball precisely on the vertical axis (center ball) then the cue ball is not going to go where you are aiming. Deflection from a non center ball hit is massive due to the extreme force of the break shot. Personally, I only used LD shafts on my break cue just for this reason, even though I don't use an LD shaft on my playing cue.

have a friend track the path of your cue ball after it hits the rack. if it veers left or right, you did not hit the head ball head on. The cue ball should either rebound back or up slightly upon impact and die... that is the sign of a 100% energy transfer from the cue ball to the rack.

run a few tests and check it out
 
Last edited:
derangedhermit said:
If you consider only the total energy transfer between cue ball and the rack of balls (9, 10, or 15 - I think it makes no difference), what does the trade-off look like between cue ball speed and off-center hits?

I'm trying to get a feel for how accurate you need to be. For example, will being off center 1/16" (1.6mm) cost the equivalent energy of 1 mph lower cue ball speed? How about 1/4"? I guess the relevant speed range is about 18-25mph.

Assume a tight rack (all balls touching). I think, based on intuition, the energy drop off increases non-linearly as the contact point moves further away from a center hit.

A graph would be outstanding!:grin:
Click to expand...


Its almost never worth it to hit the pack off center because you would lose some energy. The mph would remain the same, but less energy is transferred to the balls(the cue ball retains some after contact). If there was such a graph, I dont think it'd show show an off center sweet spot.
 
Accuracy of Cue Ball hit on the Rack & accuracy of Cue Stick hit on the Cue Ball are real important, for dispersion of energy into the Rack & Cue Ball control after impact.
 
joelpope said:
I have sent hours on end with numerous people including a top 3 US pro playing around with break speed & results. I have used radar guns, iPhone break Speed App, video replay and charting and believe that there is a direct correlation between where your cue tip hits the cue ball and your breaking success.

2 main reasons...

1) most people do not hit the cue ball where they think they are hitting it during a break. The extreme force throws their aim and cue delivery way off. Maximum energy transfer from cue to cue ball is a dead center hit, over the course of hundreds of breaks I found that for every 1/4 tip you miss center ball by there is a 5% drop in speed... easy math is that a 20 MPH break goes to 19 with a 1/4 tip miss, 18 with a half tip miss, etc...

try marking a cue ball with a sharpie & see if you can deliver your tip to center ball at break speed... you will be surprised.

2) just as maximum energy transfer between cue and cue ball is achieved with a center ball hit, maximum energy transfer between cue ball and rack is achieved with a "center ball hit" on the head ball. if you are not delivering the cue tip to the cue ball precisely on the vertical axis (center ball) then the cue ball is not going to go where you are aiming. Deflection from a non center ball hit is massive due to the extreme force of the break shot. Personally, I only used LD shafts on my break cue just for this reason, even though I don't use an LD shaft on my playing cue.

have a friend track the path of your cue ball after it hits the rack. if it veers left or right, you did not hit the head ball head on. The cue ball should either rebound back or up slightly upon impact and die... that is the sign of a 100% energy transfer from the cue ball to the rack.

run a few tests and check it out
Click to expand...



Nice post, well done.
randyg
 
I thought he meant an off center hit on the pack.

Off topic here, but didnt Dr. Dave run deflection tests and determine that speed does not effect deflection?
 
C.Milian said:
I don't see how it matters all that much. The objective is to be in line to the head ball even if you are not trying to concentrate on hitting it dead on. Once you are in line there is very little you can do to mess up a good break. About nine practice strokes is all you need to send the cue ball 4-5 feet midair. If I'm in for a short set, I'll try to crush all the balls every time, otherwise I'll pop all the breaks. Break hard enough and you can send the cue ball flying and still land good.
Click to expand...

Maybe I am misunderstanding you, but there are allot of things you can do wrong once you are "in line". I think one of the biggest issues is one that has already been mentioned here. Most people don't hit the cue ball exactly where they think they do which can cause a variety of problems. When you crank up the power you typically loose accuracy.
 
Most of the time when I see pros using a cut break it is only because they are breaking from the box. Otherwise they are going for a good solid hit on the head ball.

I did notice however, that this was not the case at the last Mosconi Cup. Most of the guys were using a cut break from the side rail, the Europeans to great effect. I'm still a little stumped by it, but I can only suppose they simply solved the riddle of how that table was breaking.
 
Cdryden said:
Maybe I am misunderstanding you, but there are allot of things you can do wrong once you are "in line". I think one of the biggest issues is one that has already been mentioned here. Most people don't hit the cue ball exactly where they think they do which can cause a variety of problems. When you crank up the power you typically loose accuracy.
Click to expand...

I think it is very hard to miss the one ball or the stack for that matter if you follow all the rules to a clean hit, especially the break. Guys practice 100s of breaks daily and are used to being in the right spot so they can crank up the power without losing accuracy. I personally do not worry about accuracy or at least stroking accurately for the break shot. It is a fun shot and I like hammering it every chance I get. I'll try to get it into the kitchen all day until my arm falls off, if someone will rack for me.
 
Scott Lee said:
Just curious about how you arrived at that number?

Scott Lee
www.poolknowledge.com
Click to expand...

It's actually a low number. I can go 12 practice strokes for the break shot. Cue ball...one ball.....cueball then I'm in line. Wait...I think its oneball...cueball....no Well whatever pattern you want that works but I'm sure as you zome in on the one, you should close in on the cueball or at least that's how I do it until I have the entire stack locked in before putting everything into it. Then sometimes Its a few stroke and its gone.

There has been times when I know the cue ball will stay on the table no matter how much of my body I throw at the ball. I guess I generate more power that way.
 
joelpope said:
I [...] believe that there is a direct correlation between where your cue tip hits the cue ball and your breaking success.

2 main reasons...

1) most people do not hit the cue ball where they think they are hitting it during a break. [...]

2) just as maximum energy transfer between cue and cue ball is achieved with a center ball hit, maximum energy transfer between cue ball and rack is achieved with a "center ball hit" on the head ball. [...]

run a few tests and check it out
Click to expand...
Thank you very much.:thumbup: I do break with an LD shaft, to reduce cue ball squirt. Beyond that, I haven't focused enough on the cue <-> cue ball contact point. Missiing a dead-center hit may be caused as much or more by hitting the cue ball off center as by aim line. I'll get some impact tape and spend some practice time in discovery. I think I'll put a square on the 1-ball as well as the cue ball. I hope it's thin enough so the cue ball stays on the table.
 
jsp said:
Check out this post and thread a couple years back...

http://forums.azbilliards.com/showthread.php?p=245868#post245868

Wow, just realized that was way back in 2005. I don't know if I was smarter then or dumber...lol.
Click to expand...

If your formula is in line with Jal's table in post #8, I have trouble believing it is correct, at least in the way I have in mind. Jal's table:

.9 ball hit = .99 of full head on force
.7______ = .95
.5______ = .87
.3______ = .71
.1______ = .44

I have a bit of trouble on the "parts of a ball" thing, but 0.1 ball hit must be way past the 45-50 degrees that would have a transfer of 44%. My intuition and/or experience says that if the object ball leaves its position at 80 degrees or so different than the cue ball direction, it does not carry 44% of the mojo away with it.

For simplicity's sake, and also practically for the break case, assume the cue ball is skidding with no rotation (a "knuckleball"). (One goal for a good break must be that the cue ball have little or no angular mojo, I suppose).

I can't put numbers on it yet, but with the help of that thread, maybe I can. But here is what any result must show:

1) at a 0 degree hit (full-on dead center), the mojo transferred approaches 1. For one object ball, mojo transferred = 1 (full transfer).

2) at a 45 degree hit, half the mojo is transferred, and the balls separate travelling at equal speeds.

3) as the "ball hit" approaches 0 (the object ball direction approaches 90 degrees after contact), the mojo transferred approaches 0.
 
Last edited:
C.Milian said:
I think it is very hard to miss the one ball or the stack for that matter if you follow all the rules to a clean hit, especially the break. Guys practice 100s of breaks daily and are used to being in the right spot so they can crank up the power without losing accuracy. I personally do not worry about accuracy or at least stroking accurately for the break shot. It is a fun shot and I like hammering it every chance I get. I'll try to get it into the kitchen all day until my arm falls off, if someone will rack for me.
Click to expand...

I don't think it's a matter of hitting the one ball as much as it is hitting it in a exact spot that is the issue, ( center ball).

As long as your enjoying what you are doing that's all that really matters. You enjoy hitting them hard and I enjoy hitting them accurately so that I can keep shooting.;)
 
derangedhermit said:
If your formula is in line with Jal's table in post #8, I have trouble believing it is correct, at least in the way I have in mind. Jal's table:

.9 ball hit = .99 of full head on force
.7______ = .95
.5______ = .87
.3______ = .71
.1______ = .44

I have a bit of trouble on the "parts of a ball" thing, but 0.1 ball hit must be way past the 45-50 degrees that would have a transfer of 44%. My intuition and/or experience says that if the object ball leaves its position at 80 degrees or so different than the cue ball direction, it does not carry 44% of the mojo away with it.
Click to expand...
Jal's formula is different than mine because he doesn't square the cosine while my formula does.

Jal's formula/table is technically correct, but it's rather misleading because it calculates the magnitude of the force in the direction through the center of the one ball from the contact point. So for a 10% hit, it is true that about 44% of the CB's initial momentum (magnitude) gets transferred to the one ball, but for a 10% hit the contact point is almost completely on the side of the one ball and the direction of the force is almost completely sideways. Compare that to the force if the direction is oriented straight through the rack.

That is the reason why I square the cosine in my formula, to get the force component that is in the same direction as the CB's pre-contact direction.

So my table would be...

.9 => .99
.7 => .91
.5 => .75
.3 => .51
.1 => .19
 
Last edited:
You must log in or register to reply here.
Back
Top