Question for Bob Jewett ...

[iusedtoberich]

I think the term used is "blocking" but I did a search and found nothing. I learned about it awhile back. Basically the object used to propel an object ( arm,hockey stick, etc, flex perpendicular to intended path of movement. Think of a fishing rod, if it did not flex you could only throw a lure however fast you can swing your arms. But since it does you can now add the energy stored in the flex of the shaft on top of the speed you swing your arms. In throwing a baseball they break it down like this I think it was three blocks. Block 1 the speed which you can twist your body, Block 2 the speed which you can rotate your arm at your shoulder and Block 3 the speed at which you can snap your wrist.

Block 1 = 20 MPH
Block 2 = 30 MPh
Block 3 = 30 MPH

If you tried throwing a baseball with out using your wrist and holding your arm straight out only twisting your torso or upper body you would only achieve 20 MPH. But the energy builds on itself in the complex movement of throwing allowing us to achieve better results.
Applying this to the break in pool doesn't work since its only one block (the bicep or arm swing however you look at it).

Interesting. A forum member here Colin Colenso made a video on power breaking about 5 years ago. He describes the same thing you have, but don't believe he used the same vocabulary. He showed how to move all parts of your body during the break to get the most power. It is a very good video, definelty worth watching if you want to be a power breaker.

*Edit: Here is the video:
http://www.youtube.com/watch?v=xW1tsONEI_U

 

[iusedtoberich]

Hey PJ, do you know if the cue stick mass increases or decreases the momentum the cue ball has at impact? I know the cue stick is roughly 3 times the mass of the cue ball, but I never fully understood how the distribution of that mass should be applied to conservation of momentum equations.

In other words, does the cue ball move at a faster or slower velocity than the break hand (all velocities with respect to the plane of the table)?

Thanks.

 

[Bob Jewett]

The question has mostly been answered. The basic reason is that the normal stance is not designed to get maximum speed into the cue stick. An interesting side point is that the ball leaves the tip traveling about 130% of the speed of the stick at the instant the stick hits the ball. A player that gets a 26MPH break shot speed only moves the stick at 20MPH.

 

[RRfireblade]

The question has mostly been answered. The basic reason is that the normal stance is not designed to get maximum speed into the cue stick. An interesting side point is that the ball leaves the tip traveling about 130% of the speed of the stick at the instant the stick hits the ball. A player that gets a 26MPH break shot speed only moves the stick at 20MPH.

I think I may be on to something....



Phenolic tip on a javelin. I'll be right back......

 

[iusedtoberich]

The question has mostly been answered. The basic reason is that the normal stance is not designed to get maximum speed into the cue stick. An interesting side point is that the ball leaves the tip traveling about 130% of the speed of the stick at the instant the stick hits the ball. A player that gets a 26MPH break shot speed only moves the stick at 20MPH.

Bob,

How do you think the distribution of mass of the stick affects the cue ball speed? If you inserted a small steel rod inside the cue, at x distance from the tip, would the change in x be proportional to the change in velocity of the cue ball?

 

[Mitchxout]

Bob,

How do you think the distribution of mass of the stick affects the cue ball speed? If you inserted a small steel rod inside the cue, at x distance from the tip, would the change in x be proportional to the change in velocity of the cue ball?

I hope it doesn't work. We have enough gimmicks polluting our great game already.

 

[Patrick Johnson]

Hey PJ, do you know if the cue stick mass increases or decreases the momentum the cue ball has at impact?

A heavier cue at the same speed has more momentum, as does the same weight cue at higher speed. Momentum increases linearly with greater speed and/or mass.

However, it's kinetic energy that really matters, and kinetic energy increases linearly with greater mass but quadratically with greater speed, so increasing speed has a much greater effect than increasing weight. Doubling the weight of your cue and swinging it at the same speed doubles your break force, but doubling the speed without increasing the weight quadruples the break force.

This is why conventional wisdom says a lighter break stick can be more effective. Of course, that depends on being able to swing the lighter stick faster, which ain't necessarily so.

I know the cue stick is roughly 3 times the mass of the cue ball, but I never fully understood how the distribution of that mass should be applied to conservation of momentum equations.

I don't know that either, but I suspect the overall mass matters much more than the distribution.

In other words, does the cue ball move at a faster or slower velocity than the break hand?

Bob says the CB leaves the tip at 130% of the speed of the tip - I believe Bob until shown otherwise.

pj
chgo

[EDIT: Based on comments below, I've changed the term "exponentially" to "quadratically".]

 

[jsp]

However, it's kinetic energy that really matters, and kinetic energy increases linearly with greater mass but exponentially with greater speed...

Technically, kinetic energy increases quadratically with greater speed. "Exponentially" has a different meaning mathematically (if velocity v is the exponent in the energy equation, then what you said would be true).

Doubling the weight of your cue and swinging it at the same speed doubles your break force, but doubling the speed without increasing the weight quadruples the break force.

This isn't technically correct either. It is true that doubling the weight doubles the kinetic energy of the cue while doubling the speed quadruples it, but the cue's kinetic energy isn't directly proportional to the "break force".

What you're missing is the transfer of energy component between the cue and the CB. In other words, how much of the cue's kinetic energy gets transferred into CB kinetic energy. This amount of transfer depends on the relative masses of the cue and CB. Because the cue is heavier than the CB, only a fraction of the cue's kinetic energy gets transferred to the CB. The rest is retained as the cue's forward momentum after impact. The heavier the cue the less percentage of energy will get transferred to the CB, whereas the lighter the cue the greater percentage of energy will get transferred. (The ideal case in terms of transfer of energy is if the cue equals the mass of the CB, which would mean fricken light cue.)

That's why a car weighing 4000 times the amount of a cue and hitting the CB at 20mph won't make the CB go 80,000mph.

 

[dr_dave]

A heavier cue at the same speed has more momentum, as does the same weight cue at higher speed. Momentum increases linearly with greater speed and/or mass.

However, it's kinetic energy that really matters, and kinetic energy increases linearly with greater mass but exponentially with greater speed, so increasing speed has a much greater effect than increasing weight. Doubling the weight of your cue and swinging it at the same speed doubles your break force, but doubling the speed without increasing the weight quadruples the break force.

Pat,

FYI, see:

TP A.30 - The effects of cue tip offset on cue ball speed and spin​

This analysis answers many questions in this thread (and more). For example, CB speed is directly proportional to the cue speed (for a given cue mass). In other words, if you double the cue speed, the CB speed will also double.

Bob says the CB leaves the tip at 130% of the speed of the tip - I believe Bob until shown otherwise.

Assuming a perfect tip and collision (i.e., elastic), and for a 19 oz cue, the CB would move 152% as fast as the cue with a center-ball hit (from Equation 7 in TP A.30, with x=0). With typical tip and collision inefficiency (e.g., e=0.75), the result is 133% (from Equation 22 in TP A.30). Therefore, your faith in Bob was well placed.

Regards,
Dave

 

[dr_dave]

What you're missing is the transfer of energy component between the cue and the CB. In other words, how much of the cue's kinetic energy gets transferred into CB kinetic energy. This amount of transfer depends on the relative masses of the cue and CB. Because the cue is heavier than the CB, only a fraction of the cue's kinetic energy gets transferred to the CB. The rest is retained as the cue's forward momentum after impact. The heavier the cue the less percentage of energy will get transferred to the CB, whereas the lighter the cue the greater percentage of energy will get transferred. (The ideal case in terms of transfer of energy is if the cue equals the mass of the CB, which would mean fricken light cue.)

Well stated.

Regards,
Dave

 

[dr_dave]

Dave,

I've always appreciated your analysis, but I have one burning question:

You are completely neglecting the mass of the person swinging the cuestick. Obviously the person's hand has a lot of flex and is not a very rigid connection. Is it your opinion (or hopefully you have researched it) that because the hit is approaching impulse conditions that differences in the person have little affect on the outcome?

That question is a bit clunky, but I am struggling with a better way to ask it, hopefully you get the gist of it and are better able to answer it.

That is an excellent question ... not "clunky" at all. My "research" on this topic can be found here:

effects of light grip resource page​

Based on the info on this page, the grip has negligible effect during tip contact for a typical pool stroke. Even for a powerful break shot, the grip force during impact is also probably negligible compared to the impulsive force between the tip and ball (although, I haven't measured this).

Regards,
Dave

 

[Patrick Johnson]

Technically, kinetic energy increases quadratically with greater speed.

Thanks. I'll edit my post with the correct term so it won't confuse readers.

It is true that doubling the weight doubles the kinetic energy of the cue while doubling the speed quadruples it, but the cue's kinetic energy isn't directly proportional to the "break force".

...Because the cue is heavier than the CB, only a fraction of the cue's kinetic energy gets transferred to the CB. The rest is retained as the cue's forward momentum after impact.

Thanks again.

Your comment about the excess kinetic energy in the cue retained as its forward momentum after impact reminds me of another possible misinterpretation of mine: that the cue essentially stops upon impact with the CB and is carried forward again (after the CB is gone) by the arm's momentum. Is my understanding of the cue/ball interaction flawed? Does the cue's retention of forward momentum affect tip/ball contact time?

pj
chgo

 

[dr_dave]

... another possible misinterpretation of mine: that the cue essentially stops upon impact with the CB and is carried forward again (after the CB is gone) by the arm's momentum. Is my understanding of the cue/ball interaction flawed?

The cue slows to about 60% of it starting value after CB impact (with typical equipment). In other words, it does not "essentially stop." For more info and a super slow-motion video explanation, see:

stroke acceleration resource page​

Regards,
Dave

 

[Patrick Johnson]

CB speed is directly proportional to the cue speed (for a given cue mass). In other words, if you double the cue speed, the CB speed will also double.

And therefore the CB's kinetic energy will quadruple, right? And does this mean that the force delivered to the rack quadruples? And does this mean that the balls in the rack will travel quadruple the combined distance (ignoring rails)?

pj
chgo

 

[dr_dave]

... CB speed is directly proportional to the cue speed (for a given cue mass). In other words, if you double the cue speed, the CB speed will also double.

And therefore the CB's kinetic energy will quadruple, right?

That's right.

And does this mean that the force delivered to the rack quadruples?

If the CB energy increases, the energy delivered to the rack would ideally also increase by the same factor.

And does this mean that the balls in the rack will travel quadruple the combined distance (ignoring rails)?

Sounds reasonable to me since ball sliding and rolling distance is directly related to ball speed (for more info, see TP B.6).

Regards,
Dave

 

[dr_dave]

Thanks for the prompt answer.

Probably a difficult thing to calculate. I have access to a 10000fps camera and I've often thought of filming some of the reactions of different things during the break.

Maybe when I get back in August I'll take it to a pool hall.

dld

That would be great. Please post links to the videos here when they are available. My camera also has 10000fps capability, but only at very limited resolution. Most of the stuff I've posted over the years was filmed at 1000fps.

Regards,
Dave

 

[Patrick Johnson]

It is true that doubling the weight doubles the kinetic energy of the cue while doubling the speed quadruples it, but the cue's kinetic energy isn't directly proportional to the "break force".

Please help me reconcile this comment with Dave's (above) that the CB's speed also doubles, quadrupling its kinetic energy and the resulting energy (ideally) delivered to the rack (what I think of as "break force"). Am I missing something?

pj
chgo

 

[jsp]

Your comment about the excess kinetic energy in the cue retained as its forward momentum after impact reminds me of another possible misinterpretation of mine: that the cue essentially stops upon impact with the CB and is carried forward again (after the CB is gone) by the arm's momentum. Is my understanding of the cue/ball interaction flawed?

Think of those heavy CBs they use on barbox tables. If you attempt to execute a stop shot such that the CB is not rolling on impact with the OB, would the heavier CB actually stop? No, it would retain forward moment because it's slightly heavier than the OB. Same holds true for the cue.

Another way to think about it is if you simply throw your cue at the CB. The cue won't stop dead in its tracks on impact. It would if the cue was just as heavy as the CB.

Does the cue's retention of forward momentum affect tip/ball contact time?

Good question. I don't know.

Please help me reconcile this comment with Dave's (above) that the CB's speed also doubles, quadrupling its kinetic energy and the resulting energy (ideally) delivered to the rack (what I think of as "break force"). Am I missing something?

Good point, and I guess I should clarify my statement. What I should have said is the cue's kinetic energy isn't necessarily proportional to the break force.

If you keep the cue mass constant (which is the assumption dr. dave made in post #29), then it is true that kinetic energy would be proportional to the break force. The reason is that fixing the mass of the cue also fixes the energy transfer percentage between cue to CB, since the ratio between the two masses are fixed. So as you increase the kinetic energy of the cue, the CB will have a proportional increase in kinetic energy.

Things are different when you change the mass of the cue for a given cue speed. Changing the cue's mass not only changes the cue's kinetic energy, but it also changes the percentage of energy transfer. So the resulting CB kinetic energy will not be proportional to the cue's kinetic energy.

 

[Patrick Johnson]

DoubleD:
The double-weight cuestick retains more of the system energy, thereby offering a less-than-doubling of the cueball speed.

jsp:
If you keep the cue mass constant (which is the assumption dr. dave made in post #29), then it is true that kinetic energy would be proportional to the break force. The reason is that fixing the mass of the cue also fixes the energy transfer percentage between cue to CB, since the ratio between the two masses are fixed. So as you increase the kinetic energy of the cue, the CB will have a proportional increase in kinetic energy.

Things are different when you change the mass of the cue for a given cue speed. Changing the cue's mass not only changes the cue's kinetic energy, but it also changes the percentage of energy transfer. So the resulting CB kinetic energy will not be proportional to the cue's kinetic energy.

Thanks to both of you. I learned something. Two things if you count the realization that I didn't already know everything.

pj
chgo

 

[pvc lou]

keep a link to this post on hand for the future N posters who will accuse you of being an arrogant, naysayer knowitall. :embarrassed2:

Thanks to both of you. I learned something. Two things if you count the realization that I didn't already know everything.

pj
chgo

keep a link to this post on hand for the future N posters who will accuse you of being an arrogant, naysayer knowitall. :embarrassed2: