Break accuracy v. speed - please quantify

Bob Jewett said:
To look at the original question slightly differently, just consider the energy transfer from the cue ball to a single ball. The nature of that is not going to be much different from hitting a full rack. This simpler problem was solved a long time ago:

The velocity of the object ball is proportional to the cosine of the cut angle and the velocity remaining on the cue ball is proportional to the sine of the cut angle. What we are mostly interested in is energy transfer, and energy is proportional to the square of the velocity. Here is a table for various cases:

View attachment 209055
So for a 15-degree cut (about a 3/4-full hit) 93% of the energy is transferred into the object ball leaving 7% on the cue ball.

In fact the transfer of energy into a single ball is more efficient than into the rack as you can see by the fact that the cue ball bounces back after hitting a full rack -- it takes some energy away when it bounces back.
Click to expand...
If I understand the meaning of "fullness" in the table, then it seems to approximately agree with the other two: .9 fullness ~= .99 energy transfer, .8 fullness ~= .95 energy transfer. That's about a quarter inch and half inch, right?

On the last note, yes, I am told that the transfer is less efficient due to the difference in masses of the cue ball v. the total mass of the rack. But I guess the interesting thing is the % of potential transfer given the masses involved.
 
derangedhermit said:
Leave the rack on and break from the head spot. You show how far the entire rack will move towards the foot rail. Jal shows the total distance the entire rack will move in any direction. Right? My question is, which of these is more useful when looking at the chances of pocketing a random ball and getting a nice spread. It's not clear to me that your method is more enlightening. I suspect the best model might be somewhere between the two.
Click to expand...
I think I understand what you're saying.

Jal's formula is actually technically sound. But the reason it's misleading is because the scalar result doesn't give an indication of how much the balls would spread, but rather just how much of the CB's energy gets transferred to the first ball of the rack. The spread of the balls really depends on the force component going through the center of the rack, which is what my formula attempts to approximate.

To illustrate the issue better, take a standard 9-ball rack and put the 10 ball ball directly in front of the one ball in the rack. So now the rack has 10 balls instead of 9. Instead of breaking the rack facing the 10 ball head on, break from the complete side of the 10 ball such that the CB travels perpendicular to the foot/head rails. So you're hitting the 10 ball completely full on its side.

So what happens with this break? Because you hit the 10 ball full, you are technically transferring 100% of the CB's energy into the "rack", correct? Will the give you a good spread of the balls? Absolutely not because the rack is made up of more than only one ball. Only the 10 ball would move while the remaining 9 balls wouldn't budge after the initial collision. That's because there is no force component that is going straight through the center of the rack.
 
Bob Jewett said:
To look at the original question slightly differently, just consider the energy transfer from the cue ball to a single ball. The nature of that is not going to be much different from hitting a full rack. This simpler problem was solved a long time ago:

The velocity of the object ball is proportional to the cosine of the cut angle and the velocity remaining on the cue ball is proportional to the sine of the cut angle. What we are mostly interested in is energy transfer, and energy is proportional to the square of the velocity. Here is a table for various cases:

View attachment 209055
So for a 15-degree cut (about a 3/4-full hit) 93% of the energy is transferred into the object ball leaving 7% on the cue ball.

In fact the transfer of energy into a single ball is more efficient than into the rack as you can see by the fact that the cue ball bounces back after hitting a full rack -- it takes some energy away when it bounces back.
Click to expand...
Thanks for the data Bob. If people want to see various plots for single-ball collisions, I have some here:

Concerning whether or not the results would be similar with a full rack, I'm not so sure. The effective mass of the rack might vary significantly with the squareness of hit. Also, there is additional energy loss within the rack due to the complicated physics of simultaneous collisions, and the percentage energy loss might also depend on the squareness of hit.

This would be interesting to study some day with some controlled experiments ... maybe the next time we get together for such things.

Catch you later,
Dave
 
derangedhermit said:
If I understand the meaning of "fullness" in the table, then it seems to approximately agree with the other two: .9 fullness ~= .99 energy transfer, .8 fullness ~= .95 energy transfer. That's about a quarter inch and half inch, right?

On the last note, yes, I am told that the transfer is less efficient due to the difference in masses of the cue ball v. the total mass of the rack. But I guess the interesting thing is the % of potential transfer given the masses involved.
Click to expand...
0.5 fullness means a half-ball hit. 0 fullness is barely missing the ball. A ball is 2.25 inches across, so 0.9 fullness would be off-center by 0.225 inches.
 
OK, so we have had a nice long discussion about putting power into the CB but almost nothing about what the margin of error is in hitting the rack.

I suspect that if you miss your aim point on the ball in the rack by 1mm you will probably not pot the wing ball or pot the head ball in the side pocket.

What data do we have on accuracy requirements as the CB makes contact with the rack?

What data do we have on what spin (top/bottom, left/right) have on the balls in the rack separating after contact from CB?
 
derangedhermit said:
Well, I took them from Jal's table:
.9 ball hit = .99 of full head on force
.7 = .95
.5 = .87
.3 = .71
.1 = .44
and jsp's table:
.9 => .99
.7 => .91
.5 => .75
.3 => .51
.1 => .19

If .1 ball = .225 inches:
99% is .9 balls, about a quarter inch
95% is between .7 and .8 balls, or between 0.45 and 0.675 inches.

jsp pointed to the logic and math behind. It doesn't match my intuition either, but there is enough description to say where they are going off track, if indeed they are.

I gotta go practice. Later!
Click to expand...
I haven't seen Jal's or jsp's analysis or assumptions, but maybe any discrepancy between the numerical results and anecdotal experience is related to the comments in my reply to Bob.

Regards,
Dave
 
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derangedhermit said:
If I understand the meaning of "fullness" in the table, then it seems to approximately agree with the other two: .9 fullness ~= .99 energy transfer, .8 fullness ~= .95 energy transfer. That's about a quarter inch and half inch, right?

On the last note, yes, I am told that the transfer is less efficient due to the difference in masses of the cue ball v. the total mass of the rack. But I guess the interesting thing is the % of potential transfer given the masses involved.
Click to expand...
Just for the heck of it, here's a plot of a single object ball's velocity and kinetic energy relative to the cueball's as a function of ball fraction:

cos_cossq_fract.JPG

And here's one of the kinetic energy given over to a rack of balls. This assumes an effective rack mass of twice the cueball's mass, which is (roughly) what's been measured by Bob J. and, I think, Dr. Dave, and myself. (I forget the exact values, offhand).

Kinetic_Energy_of_Rack.JPG

This isn't to be taken too seriously since the effective mass of the rack almost certainly varies with the direction the apex ball is driven with respect to the remaining balls in the rack (as Dr. Dave pointed out earlier). But I just thought I'd throw it out there since I think you were looking for graphs at one point.

Of course, these ignore friction and the inelasticity of the collision(s) too.

Jim
 
Jal said:
Just for the heck of it, here's a plot of a single object ball's velocity and kinetic energy relative to the cueball's as a function of ball fraction:

View attachment 209071

And here's one of the kinetic energy given over to a rack of balls. This assumes an effective rack mass of twice the cueball's mass, which is (roughly) what's been measured by Bob J. and, I think, Dr. Dave, and myself. (I forget the exact values, offhand).

View attachment 209072

This isn't to be taken too seriously since the effective mass of the rack almost certainly varies with the direction the apex ball is driven with respect to the remaining balls in the rack (as Dr. Dave pointed out earlier). But I just thought I'd throw it out there since I think you were looking for graphs at one point.

Of course, these ignore friction and the inelasticity of the collision(s) too.

Jim
Click to expand...
While I truly appreciate the attempts at the quantitative analysis why not just go with a qualitative look... break 10 racks, if the cue ball goes left or right to the rail then you missed the head ball. Look at the impact & spread.

Compare that to a break where the cue ball dies off the front of the rack and stays center table. Again compare the impact & spread, but, more importantly understand that with the cue ball in the center of the table if you have a clear look at an object ball it will be within half a table length which ought to help set up your run out as you will have an initial shot you are confident in.

I only see down side to losing the cue ball off the side of the rack due to a non-center ball hit.

There are too many variables for me to really buy in to the tables and graphs but they are fun to look at.
 
joelpope said:
... There are too many variables for me to really buy in to the tables and graphs but they are fun to look at.
Click to expand...
Well, yes, but, I think the OP asked:
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?
​
Of course what most nine ball players are interested in is "how can I make a ball more often?" Some people simplify this to: "how can I hit the rack harder?" but that has been shown to be the wrong question for a tight rack.
 
Bob Jewett said:
Well, yes, but, I think the OP asked:
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?
​
Of course what most nine ball players are interested in is "how can I make a ball more often?" Some people simplify this to: "how can I hit the rack harder?" but that has been shown to be the wrong question for a tight rack.
Click to expand...

What about moving balls up table and getting a good spread?
 
Bob Jewett said:
Well, yes, but, I think the OP asked:
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?
​
Of course what most nine ball players are interested in is "how can I make a ball more often?" Some people simplify this to: "how can I hit the rack harder?" but that has been shown to be the wrong question for a tight rack.
Click to expand...
I found a satisfactory practical answer to my original question: By keeping the cue ball under control in the middle of the table, you ensure you are accurate enough to be transfering cue ball energy to the rack with maximum efficiency.

I asked the question the way I did since I was looking for a general answer, not one for nine ball (or any other specific game).
C.Milian said:
What about moving balls up table and getting a good spread?
Click to expand...

I'll posit that the goal of making a ball on the break (in 8-ball or 10-ball or any "ideal" game where that is an objective) and the goal of getting a good spread are best done by the same action: putting maximum energy into the rack. This is based on the assumptions that the best spread you can hope for is a random distribution of balls on the table, and that there is a direct relationship between the number of collisions a ball has and how likely it is to drop in a pocket. Obvious, but it's hard to imagine something better that I wouldn't classify as a trick shot and cause me to lose some interest in the game.

People can usually mix the balls well enough in 9-ball that there is no visual pattern left from the break. It's less common to see that good a break in 8-ball - often you can see patterns left because the mixing is incomplete. Which makes me think more energy going in would improve the spread.
 
Jal,

Thanks for the very illustrative plots and for explaining how you derived them.

I think it would be very interesting to experimentally find out how much effective rack mass and percentage energy loss vary with ball-hit-fraction, and to see the effects of these changes on the rack energy vs. accuracy results.

Anecdotally, it seems like the rack energy might fall off sooner and faster than the simple model predicts; but as we all know, anecdotes can often be terribly wrong.

Regards,
Dave

Jal said:
Just for the heck of it, here's a plot of a single object ball's velocity and kinetic energy relative to the cueball's as a function of ball fraction:

View attachment 209071

And here's one of the kinetic energy given over to a rack of balls. This assumes an effective rack mass of twice the cueball's mass, which is (roughly) what's been measured by Bob J. and, I think, Dr. Dave, and myself. (I forget the exact values, offhand).

View attachment 209072

This isn't to be taken too seriously since the effective mass of the rack almost certainly varies with the direction the apex ball is driven with respect to the remaining balls in the rack (as Dr. Dave pointed out earlier). But I just thought I'd throw it out there since I think you were looking for graphs at one point.

Of course, these ignore friction and the inelasticity of the collision(s) too.

Jim
Click to expand...
 
dr_dave said:
...I think it would be very interesting to experimentally find out how much effective rack mass and percentage energy loss vary with ball-hit-fraction, and to see the effects of these changes on the rack energy vs. accuracy results.
Click to expand...
It certainly would be, Dr. Dave, but who would be the person to do it? (There might be a gentleman and scholar at one of our universities (in the southwest) willing to take it on, but I can't think of anybody else.)

Jim
 
Jal said:
dr_dave said:
... I think it would be very interesting to experimentally find out how much effective rack mass and percentage energy loss vary with ball-hit-fraction, and to see the effects of these changes on the rack energy vs. accuracy results.
Click to expand...
It certainly would be, Dr. Dave, but who would be the person to do it? (There might be a gentleman and scholar at one of our universities (in the southwest) willing to take it on, but I can't think of anybody else.)
Click to expand...
I'll add this to my list, but it might be a tough one. Maybe I'll be able to convince Bob to come out and help with it.

FYI, we (Coloradans) generally refer to Colorado as being in the "West" and not the "Southwest" (Texas, New Mexico, Arizona) ... or did you mean someone else?

Best regards and Happy New Year,
Dave
 
jsp said:
Jal's formula is different than mine because he doesn't square the cosine while my formula does.
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Jsp,

For what it's worth, I agree with your squaring of the cosine since, as Bob J. also pointed out later, ball travel distance is proportional to the square of its velocity. (I can't remember exactly what my rational was for not squaring it --whether it was a good rational or bad rational. :) )

Good to see you posting again!

Jim
 
Jal said:
Jsp,

For what it's worth, I agree with your squaring of the cosine since, as Bob J. also pointed out later, ball travel distance is proportional to the square of its velocity. (I can't remember exactly what my rational was for not squaring it --whether it was a good rational or bad rational. :) )

Good to see you posting again!

Jim
Click to expand...
Um, thanks...but I'm sure I'm squaring the cosine for all the wrong reasons...lol.

And for what it's worth, I'd rather quietly bow out of threads that already have you, dr_dave, and Bob Jewett posting in them. :)
 
Jal said:
[...] as Bob J. also pointed out later, ball travel distance is proportional to the square of its velocity. [...]
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I'm trying to get practical answers, and I need help. I'd like to apply the general statement above and apply it to a pool table. Ball distance traveled is determined by several things.

The distance traveled by a pool ball on a table is estimated by the following equation:

(equation goes here)

Where:
Vi is the initial velocity of the object ball
Cfr is the rolling coefficient of friction of a given felt and ball combination
Cfs is the sliding coefficient of friction of a given felt and ball combination
Da is the atmospheric drag on the ball (given the low friction of felt v. ball, I am not sure we can ignore air drag of spherical objects at 4-20 mph (2-10 m/s).)
Lc is a term that calculates the velocity loss per cushion struck at a certain speed and angle

My intuition is that the dominant factors for distance are, of course, the initial velocity, and then the number of cushions struck. (I am not sure if angle of impact is important). Obviously felt speed plays a noticeable roll. Do clean balls roll further than dirty ones? How much of a role does angular momentum play (v. translational momentum)?

Has anyone written a pool table dynamics simulator?
 
derangedhermit said:
I'm trying to get practical answers, and I need help. I'd like to apply the general statement above and apply it to a pool table. Ball distance traveled is determined by several things.

The distance traveled by a pool ball on a table is estimated by the following equation:

(equation goes here)

Where:
Vi is the initial velocity of the object ball
Cfr is the rolling coefficient of friction of a given felt and ball combination
Cfs is the sliding coefficient of friction of a given felt and ball combination
Da is the atmospheric drag on the ball (given the low friction of felt v. ball, I am not sure we can ignore air drag of spherical objects at 4-20 mph (2-10 m/s).)
Lc is a term that calculates the velocity loss per cushion struck at a certain speed and angle
Click to expand...
FYI, you can find equations like this here:
and here:

Air resistance is usually ignored or lumped in with rolling resistance.

derangedhermit said:
Has anyone written a pool table dynamics simulator?
Click to expand...
Yes. Info on some is available here:

Regards,
Dave
 
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