The Science of Break Cue Weights

......So this is a tiny bit more kinetic energy, 1.0008527 times as much, or 0.08% more. (I did less rounding than in the other thread).........
In the other thread it seemed as if you were making too much of the difference in kinetic energy for the two cues. But here you indicate it may be the result of rounding off the numbers. That's of course correct since the same force applied over the same distance will yield exactly the same kinetic energy, regardless of the mass being accelerated.

In order to figure optimal cue weight - and there is an optimum for each individual - you have to include the mass of the player's arm (and whatever else gets put into motion) during the stroke. This is because the forces being generated by the muscles are moving more than the cue. Unfortunately, that's very hard to determine. You also have to include the collision dynamics and the fact that once the tip makes contact, body mass (arm plus whatever) is no longer relevant because of "soft tissue coupling," as Bob Jewett puts it. But it is relevant during the acceleration phase.

If you do the above and assume "reasonable" values for arm mass based upon typical weight preferences for normal playing cues (under the assumption that they're at least near optimal), and then plot cueball speed vs. cue mass, you'll probably find that weight is indeed not very critical, as per your overall conclusion. That is, the curve tends to be relatively flat in the area of the maxima (optimal weight). (As an aside, optimal weight does vary with tip offset, decreasing considerably at large offsets.)

There are, however, other things to consider, such as how the generated forces may vary with different cue weights because of the relative proportions of fast and slow twitch muscles, per Dr. Dave's remarks in the other thread.

The only easy way to find out is to try out different cues with the same tip or similar characteristics.

Jim
 
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So I guess my question would be which way would contain the most power and why.

Moving an 18 oz. cue with an arm speed of let say 20 mph, or moving a 22 oz cue with an arm speed of let say 18 mph.

Im not sure exactly how fast your arm travels during the break, im just throwing numbers out as an example.

It probably depends on the individual and might not matter. While maybe a baseball player can't swing a 33 oz. bat as fast as a 27 oz. bat, a 2 oz. difference in cue weight might not change how much speed you can generate. It is possibl I suppose that the fastest speed you are capable of you can attain with either a 20 oz. or 22 oz. cue. Now, given constant speed my understanding is you would generate more force with the heavier cue.
 
...
That is incorrect.

Kinetic energy is proportional to the mass of the cue, but it is also proportional to the square of the velocity. ...
In the situation you describe, the energy in the cue stick will be force applied -- which you assume to be constant -- times the distance traveled before impact -- which you also assume to be constant. That's why you got identical energy results for the two cases you calculated (within rounding error).

Energy = force times distance

The problem I pointed out is that if the mass is very large, it will be going very slowly with "only" that much energy.

I think that the real answer here is that the plot of cue ball speed versus stick weight has a very broad maximum and that an ounce or two of stick weight is far less important than finding an efficient tip -- if speed is your primary concern.
 
Irrelevent

All this is based on the assumption that the harder you break the better your results.

This of course is not true.
 
I think that the real answer here is that the plot of cue ball speed versus stick weight has a very broad maximum and that an ounce or two of stick weight is far less important than finding an efficient tip -- if speed is your primary concern.

Can you elaborate upon this hidden nugget of information that using a efficient cue tip is far more important than cue weight?

What are the most efficient cue tips to use and how much of a difference does the tip make to spreading the balls of a rack?
 
All this is based on the assumption that the harder you break the better your results.

This of course is not true.

I am not making this assumption at all. I fully acknowledge many players get better results with softer breaks. I've seen many pro games lately where someone used a softer break on purpose and got much better results (balls made, good spread, cue ball in position, conducive to a run-out) than their opponent using a harder break.

I am merely trying to answer the question whether a heavier or lighter cue is capable of delivering more energy to the cue ball for players who choose to break as hard as they can.

I have never made any claim this maximizes results.
 
I'm going to address a couple comments here, sorry I don't know how to quote from multiple posts

1-inch punch:

The point of the 1-inch punch wasn't that it was a more powerful strike than what could be delivered from farther away, it was that Bruce Lee could coordinate all the other moveable parts of his body to deliver a great amount of force over a distance of one inch. If you watch video of him demonstrating the punch, his body begins to move well before the strike is delivered, but his fist is never more than an inch away from the target prior to the strike. It's a question of body coordination and acceleration.

length of bridge:

It's true that constant acceleration over a longer distance produces a higher speed, but the blanket statement that a longer bridge produces a more powerful break is questionable at best. At some point, the player will reach terminal velocity, i.e., their fastest possible break, and after that point any increase in distance is counterproductive. Based on the same concept as the one-inch punch, the idea with a short bridge on the break is to achieve that terminal velocity in a shorter distance. The key is your acceleration to terminal velocity, and the distance you require to do that, not constant acceleration over a longer bridge.

equal speed = more force with heavier cue:

True, but a lighter cue will have a higher speed, and a heavier cue will have a lower speed. This applies to the acceleration in the stroke as well as to the terminal velocity, so it's a wash. If I'm able to deliver a certain amount of force, then that force is conserved regardless of the weight of the cue. If I'm only able to maintain a certain amount of momentum, then that momentum, again, is conserved regardless of the weight of the cue. At that point, the question becomes accuracy and comfort, with accuracy being the most important.

weight of cue:

The weight you need on your break cue depends entirely on what you're comfortable breaking with. I'm equally comfortable breaking with 18, 19, 21, 23, and 25 oz, and currently I'm using a 23 because I like the feel of it. I get about the same results regardless of the weight of the cue.

Incidentally, I bridge about 10 inches when breaking, because I'm more comfortable breaking with my hand on the cloth than on the rail. I start to shift my body weight forward as I begin the accelerating stroke with my arm, so I can accelerate as much as possible in the 10 inches and use as much of my body weight as possible to increase the momentum on the cue ball. My follow through tends to be about a foot, because the impact with the cue ball actually arrests the forward motion of the cue. I play with a guy who follows through all the way to the rack, and we get about as much action on both of our breaks.
 
That's all well and good but the lighter the cue the harder it is to stroke in a straight line.

I agree with that. I did state in another weight thread an argument could be made a slower accelerating cue would be easier to control. I'd much rather use anything in the 17 to 24 oz range than a 6 oz cue. I'm sure a 6 oz. cue would be a squib fest for me personally.

I also believe a light cue will feel to the player like he really smashed the rack with greater power. The lighter cue will be going faster, and the feedback from the collision will be greater, hence I think they may feel like they broke "better".
 
very simple

If a lighter break cue gives you more cue speed (for most but not all people it does) then you will get a higher break speed with a lighter cue.

The same cue speed applied to the cue ball with a heavier cue will create more force.

You will not see a baseball player hit more home runs with a heavy bat unless they are built like a weight lifter with no fast twitch muscle.All major league baseball players have fast twitch muscle that allows them good bat speed thus they all search for the lightest bat that wont break on impact.

Tennis with a 10.LB racket anyone?
 
If a lighter break cue gives you more cue speed (for most but not all people it does) then you will get a higher break speed with a lighter cue.

The same cue speed applied to the cue ball with a heavier cue will create more force.

You will not see a baseball player hit more home runs with a heavy bat unless they are built like a weight lifter with no fast twitch muscle.All major league baseball players have fast twitch muscle that allows them good bat speed thus they all search for the lightest bat that wont break on impact.

Tennis with a 10.LB racket anyone?

Speed is a function of momentum. Velocity = Momentum/Mass
Force is a function of acceleration. Force = Mass*Acceleration

In the case of a collision between a moving object and a stationary object, the acceleration referred to is the instantaneous acceleration of the stationary object. More force correlates to higher mass and/or higher acceleration.

So you can compare the momentum of the cue prior to contact to the cue ball after contact, or you can measure the force applied to the cue ball at contact, but you can't directly compare speed to force the way you did earlier.

If I could perfectly hit a pitched baseball with a 10 lb bat, I'd hit homerun speed almost every time due to the transfer of momentum to the ball. The question with a heavy bat is, can you time it so that the sweet spot of the bat travels through the ball's path when the ball is in the strike zone? It's a question of timing, not power.

For the ease of the math, let's say you're hitting a 6 oz cue ball with an 18 oz cue. Momentum is conserved, so

Momentum(ball) = Momentum(stick), ergo
Mass(ball)*Velocity(ball)=Mass(stick)*Velocity(stick)

The mass of the stick is three times the mass of the ball, so the only way this equation balances is if the speed of the ball is three times the speed of the cue.

Now hit the same cue ball with a 24 oz cue. The speed of the cue ball will be four times the speed of the stick.

Now for the fun math: If the speed of your 18 oz cue is 1, then the speed of your 24 oz cue only needs to be 0.75 to match momentum.

The question here is the ratio of your cue speed with the lighter cue to your cue speed with the heavier cue. At some point as you continue to lighten your cue, you'll get to the point where you can't generate enough speed with the cue to get the momentum you need on the cue ball. If you were hitting a 6 oz ball with a 6 oz cue, you'd actually need to generate break speed with your arm to get break speed on the cue ball. To get to the other extreme, where the cue is too heavy to move, you'd need to measure it in pounds, not ounces. Tap a 6 oz ball with a 40 lb cue, and that ball will fly off the tip.

What it boils down to is finding a cue weight where you don't need to kill yourself chucking your cue at the ball to get your max break speed, so when you dial back some of the speed to increase your accuracy, you still can.

You can probably break reasonably with anywhere from 16 oz to 25 oz, and it still boils down to whatever you personally are most comfortable with.
 
By all means, let's nit pick a little :)

Conservation of energy applies to isolated systems only. A billiard table with balls on it is not one. Some energy is converted to other forms like heat, but the total energy (Kinetic energy + potential energy) before and after the collision is not conserved in an open system, such as the case in your example.


What is your point?

An open system allows for the exchange of energy and matter between the system and its surroundings.

A closed system allows for the exchange of energy but not matter. This energy can be exchanged as energy itself or as work.

An isolated system allows for neither the exchange of energy or mass.

We are at liberty to define our system to suit analysis of a motion or process. Everything else other than system is “surrounding”.

Explain where you believe energy is lost or destroyed on a break shot.
 
In the situation you describe, the energy in the cue stick will be force applied -- which you assume to be constant -- times the distance traveled before impact -- which you also assume to be constant. That's why you got identical energy results for the two cases you calculated (within rounding error).

Energy = force times distance

The problem I pointed out is that if the mass is very large, it will be going very slowly with "only" that much energy.

I think that the real answer here is that the plot of cue ball speed versus stick weight has a very broad maximum and that an ounce or two of stick weight is far less important than finding an efficient tip -- if speed is your primary concern.

As far as I can tell, I think we are in agreement. I had not originally approached the problem as a conservation of momentum, which should work equally well. In fact, as you or someone pointed out, that makes it a short problem.

Again, I thought a given player could generate a maximum force that was basically constant based on strength. Now others want to bring up arm length, etc., but I don't care to try to figure out exactly what their maximum force is.

I only wanted to use the same force for a lighter or heavier cue to see if one struck the cue ball with more kinetic energy. I chose 30 lbs. in an entirely arbitrary manner. It seemed clear to me a given player can generate only so much force, and that would be the force acting on each cue. Assuming they want to use their maximum effort.

From there I calculated what the cue velocity would be at the end of acceleration through the same distance, figuring again that a given player would use the same bridge length regardless of cue weight, and how much time would elapse during the stroke.

But in the end it was entirely obvious (as it is much sooner as a conservation of momentum problem) the best way to increase energy going into the rack was to learn to use a longer bridge without sacrificing accuracy or the ability to position the cue ball as wanted.

But Jesus, many people (not you) want to argue every minor phrase in my analysis which only consisted of "don't worry about your break cue weight when it comes to obtaining a "better" break".

At times I wish I had just kept it to myself.
 
... the plot of cue ball speed versus stick weight has a very broad maximum
FYI, a good article dealing with baseball bat weights can be found here:

Example curves like you describe for break cues can be found about 3/4 down the page. The plots are for baseball bats, but the physics and arguments are very similar for a break cue.

Also, for people who want more info on the topic of optimal cue weight, I have some info and related resources here:

Regards,
Dave
 
Can you elaborate upon this hidden nugget of information that using a efficient cue tip is far more important than cue weight?

What are the most efficient cue tips to use and how much of a difference does the tip make to spreading the balls of a rack?
See the following video:

Phenolic tips can be 10-15% more efficient than typical leather tips. That basically results in 3-4% more cue ball speed (for a given cue weight and speed), which basically results in the object balls traveling 10-15% farther, slightly increasing the likelihood of pocketing a ball. Those aren't big numbers, but even a slight edge can be a big factor at the professional level.

Regards,
Dave
 
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Cue weight keeps coming up with a lot of speculation whether a heavier or lighter cue is better for breaking. Let's do a couple of real-world examples with some simple equations.

Newton's second law, F=MA, reads "force equals mass times acceleration". So dividing both side by M gives F/M=A.

An 18 oz. cue is 510.3 grams, or 0.5103 kg. (1 oz = 28.35 grams)

If your arm were able to supply a constant 30 lb force, or 133.4466 Newtons (1 lb = 4.44822 Newtons), then the acceleration of your cue stick is 133.4466 Newtons divided by 0.5103 kg equals 261.506 meters per second per second, aka meters per second squared.

Now we need to know for how much time does this acceleration take place, and then we can calculate the velocity at the end of that time. From there we can calculate how much kinetic energy (energy due to motion) the cue possesses when it strikes the cue ball.

For initial position and velocity at zero, s (distance) equals 1/2 times A times T squared, where A is acceleration and T is time.

Multiplying both sides by 2 and dividing both sides by A gives 2 times S divided by A equals T squared.

If your cue tip is 1/2 meter away from the cue when you start, then 2 times 1/2 meter divided by 261.506 meters per second squared equals T squared. Notice that dividing meters by meters per second squared gives units of seconds squared. So T is the square root of 1/261.506 equals 0.0618385 seconds. A little over 6 hundredth of a second - pretty quick.

Now for initial velocity zero, velocity equals accelertion times time. So velocity at cue ball strike (again, starting 1/2 meter away) is 261.506 meters per second squared times 0.0618385 seconds is 16.1711 meters per second. This is about 36.17 miles per hour - a helluva break. The kinetic energy the cue possesses at impact is KE=1/2 times M times V squared, where M is the mass of the cue (0.5103 kg) and V is the velocity at impact (16.1711 meters per second). So the kinetic energy is 66.7229 kg meters squared per second squared, or 66.7229 Joules.

Quickly doing the same calculation for a 20 oz. cue (0.567 kg) with the same constant 30 pounds force by your arm gives:
Acceleration = 133.4466 Newtons / 0.567 kg = 235.556 meters per second squared (less than before since we are accelerating a heavier object with the same force).
Time of acceleration (again through 1/2 meter) equals the square root of 1/235.556 equals 0.0651558 seconds. (more time than last time, again because we are accelerating a heavier object through the same distance with an equal force).
Velocity = accel x time = 235.556 times 0.0651558 = 15.3478 meters per second (slower than before as expected).
So the kinetic energy = 1/2 x M (0.567 kg) x V squared = 66.7798 Joules.

So this is a tiny bit more kinetic energy, 1.0008527 times as much, or 0.08% more. (I did less rounding than in the other thread).

Now I'm going to do a few more examples for myself to see if for less force (I know I'm not reaching 36 mph on the break) to see if the kinetic energy advantage (if you can call it that) flips to the lighter cue.

The answer appears to be don't worry AT ALL about your break cue weight. The difference in kinetic energy is absolutely negligible.

However, the farther away from the cue ball you can start your break while using the maximum force you can generate will result in a boatload more kinetic energy. I claim this is the "secret" of a killer break. Use a short bridge for the break and it will suck big time.

So start practicing those long bridges for breaking if you really want to smash the rack.

Ain't science fun? Mr. Geek out.:p


Dude ! You have way too much time on your hands,LOl:smile: My brain hurts now! :grin:



H.P.
 
nice stick

This is all mental masturbation.

If you don't know what weight break stick to use, get one the same weight as your playing cue.

You need a break cue that is stiff and has a hard tip.

Learn to hit the cue ball directly on center and hit the head ball in the rack directly on center. Shoot as hard as you can but not so hard that you miss the center on the cue ball or the center of the head ball.

This will transfer the most energy to the balls in the rack and give the best break.

Change positions for the cue ball until balls start to fall in the pockets on the break.

There is nothing else you need to know.

Kim
 
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...I had not originally approached the problem as a conservation of momentum, which should work equally well. In fact, as you or someone pointed out, that makes it a short problem.
How does the conservation of momentum lead you to any conclusion about cue weight?

Again, I thought a given player could generate a maximum force that was basically constant based on strength. ....

I only wanted to use the same force for a lighter or heavier cue to see if one struck the cue ball with more kinetic energy. I chose 30 lbs. in an entirely arbitrary manner. It seemed clear to me a given player can generate only so much force, and that would be the force acting on each cue. Assuming they want to use their maximum effort.
But there's a problem with the premise: using an equal amount of force on cues of different weights. If a force F is applied to a 2-mass system, m1 and m2, which are stuck together (e.g. arm and cue), the forces acting on them, respectively, are:

F1 = [m1/(m1+m2)]F = [1/(1+m2/m1)]F

F2 = [m2/(m1+m2)]F = [1/(1+m1/m2)]F

So while the muscles generate some maximum force F (or safely controllable force F), the force that m2 experiences depends on the ratio m1/m2. You can equalize F2 for different m2's by adjusting F, but that would be cheating. The nature of the problem is to determine optimum cue mass (m2) based on some amount of force a player is able or willing to produce. That determination cannot be done unless you include the player's physical characteristic (m1).

Of course, even for a simple pendulum stroke, the above equations are an oversimplification. The lower arm and cue is actually rotating about the elbow as a result of an applied torque by the muscles. It's not a linear motion, and because the cue is moving in a quasi-straight line, the moment of inertia of the arm-cue system is constantly changing throughout the stroke.

For a relatively small arc, you can transform this otherwise complicated problem into a simpler one by treating the arm as if it were moving linearly with equivalent mass m1. m1 is equal to the arm's moment of inertia about the elbow divided by the square of the distance from the elbow to the axis of the cue. Now, the problem is to find the arm's moment of inertia for a given player? If you have any ideas on how to do that, I'm all ears.

At times I wish I had just kept it to myself.
Such are the hazards of posting something without, perhaps, thinking it through. Most, if not all of us, have done it. (It's one of my favorite activities.)

Jim
 
What is your point?

An open system allows for the exchange of energy and matter between the system and its surroundings.

A closed system allows for the exchange of energy but not matter. This energy can be exchanged as energy itself or as work.

An isolated system allows for neither the exchange of energy or mass.

We are at liberty to define our system to suit analysis of a motion or process. Everything else other than system is “surrounding”.

Explain where you believe energy is lost or destroyed on a break shot.

My point is you can't use energy conservation to explain the importance of the weight of a break stick, simply because you don't know exactly how much of your arm strength is transferred to the cue ball upon impact, you don't know how much of the linear energy is converted to the rotational energy, you don't know how much heat is generated at the impact, etc... You are simply assuming a perfect %100 linear kinetic energy transfer, and all these might affect the end result.

However, using momentum conservation, you don't need to know all this. All that is required here is the assumption of a perfect center ball hit.

Using the latter, it seems that our conclusions are similar, so all those things that you disregarded at the beginning of your analysis might be negligible after all. Yet, the ends don't justify the means...

What I am claiming though is that in order to control the cue ball better in a break shot, there will be a certain break cue weight requirement for a certain player with which he'll feel more comfortable.

Even though 18oz or 20oz break sticks might generate similar break speeds, the cue ball might end up dead center of the table with one, while with the other it might go break somebody's nose at a nearby table...

:)
 
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Yes, you lose energy in the transfer. Some will go into compressing the tip, into compressing and flexing the cue, and into heat. But the player can't do anything about it. (OK, stiff cue, hard tip that resists compression, the obvious)

That's why I left it at generating the highest kinetic energy of the cue at impact. It's something the player can do something about.

I also didn't address a non-center contact with the cue ball where you have components acting in different directions.

I still claim it comes down to the player delivering the highest kinetic energy they can into the cue.

And I believe the best way to do that is to become proficient at using a longer bridge.



Good post. Now I need some advice about the longer bridge.

If I normally use about an 18 inch bridge. Should I go to 20 or 30 inches????? What would you suggest?

Thanks
randyg
 
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