The Science of Break Cue Weights

J

justtapitin

AzB Silver Member
Silver Member
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
 
justtapitin said:
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
Click to expand...

Actually that was quite a fun post. Now Duckie will come on here and lecture us about the "one inch punch" made famous by martial arts movies and Bruce Lee in exhibitions around the world.

This is the infamous punch that breaks boards and sends grown men flying backwards from a distance of only one inch away from the target.

How would you account for that in the equation?
 
I think you still need to do the last part which is calculate the speed of the cue ball. The amount of energy transferred is not a constant fraction of the stick energy. There is a simple formula for ball velocity for a given stick velocity and ball/stick masses. Typically, there is some energy lost and the cue ball ends up going only about 85% as fast as expected for the lossless situation.
 
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.
 
justtapitin said:
.... I still claim it comes down to the player delivering the highest kinetic energy they can into the cue. ...
Click to expand...
That would be for the heaviest possible cue stick since it is easier to maintain the full force over the whole distance with a larger mass. Like 1000 tons. The problem with a 1000 ton cue stick is that while it would have maximum energy at impact, it would not be moving very fast and the cue ball would only go twice as fast as that (neglecting losses). You still need to do the rest of the problem.

You may want to get Ron Shepard's paper which talks about this stuff and lots of other stuff. http://www.sfbilliards.com/Shepard_apapp.pdf

Dr. Dave also has a lot of physics-related stuff on his site.
http://billiards.colostate.edu/
 
JB Cases said:
Actually that was quite a fun post. Now Duckie will come on here and lecture us about the "one inch punch" made famous by martial arts movies and Bruce Lee in exhibitions around the world.

This is the infamous punch that breaks boards and sends grown men flying backwards from a distance of only one inch away from the target.

How would you account for that in the equation?
Click to expand...

Uh...I do the one-inch punch for demos.

It's uses the same principles of biomechanics as a baseball pitcher's windup--just on a smaller, more compact scale. I used to do something similar when I thought breaking hard was the right thing to do.

Colin Colenso has a good YouTube video that explains power breaking.
 
Bob Jewett said:
That would be for the heaviest possible cue stick ....

That is incorrect.

Kinetic energy is proportional to the mass of the cue, but it is also proportional to the square of the velocity.

You will give up too much velocity with a 1000 ton cue stick, as you realize intuitively. If you had read and understood the documents you cite, you would know this.

My physics are correct. Nothing in these two sources conflict with what I have stated. I know about the losses addressed in the documents - frictional, some of the energy becoming rotational instead of translational, etc. Rotational losses become larger as one chooses follow, draw, or English for breaking. But those don't affect what cue weight should be chosen, the only problem I am addressing.

I am not trying to calculate the exact losses and come up with the final energy actually transferred to the rack. This is academic to the player. He or she just wants to smash the rack efficiently, and wants to know if they should choose a certain cue weight. They can't do anything about the losses, real as they are, which occur AFTER they have hit the cue ball.

I am merely trying to show what the effect of cue weight is on the ability to generate kinetic energy in layman's terms. And for the arbitrary force I chose (30 lb) and the cues weights (18 oz and 20 oz) it is astonishingly small.

I stand by my conclusions. Cue weight makes virtually no difference to the player. If you want to smash the rack, try to accelerate for a longer period of time, which means use a longer bridge. Your arm strength and fast-twitch muscles are what they are and you're basically stuck with those. And you can't do squat about the energy losses which occur after you hit the cue ball, although there is some really nice calculus and physics in Ron's paper in particular that addresses those if you need to know the exact kinetic energy you impart to the rack. I enjoyed reading it immensely because I'm a bit of a physics junky. And physics and billiards together? Pool-playing geek nirvana.

I don't have any "rest of the problem" to finish because my problem is cue weight.
Click to expand...
 
And the one-inch punch effect is basically follow-through where body tissue continues to be compressed (ouch) while he hits "through" you.

If the strong guy gives you a 6-inch punch, you're going to enjoy it even less. At a certain distance of course he will be unable to continue gaining punch velocity. Pool player breaking too. I doubt if a 1000 ft. bridge will let you go to warp speed.

In fact the strong guy could give a zero-inch punch that could do substantial damage as well. Push hard and fast and let the guy's body inertia keep him standing there for a bit.
 
Without complicating matters due to the non-conservation of energy in the process, using conservation of linear momentum (which pretty much applies here unlike energy conservation), one has:

Break speed = (cue mass/ball mass) * (cue speed)

Even though this formula dictates that your break speed depends only on the initial momentum (simply put, how strong you are), there has to be a certain combination of cue mass and cue speed such that a particular player feels more comfortable with controlling the cue ball after the impact.

That is why choosing a correct weight for the break stick is important: not because heavier or lighter sticks will make you break any faster.
 
Last edited:
I break better with a lighter cue.
My best break is with a 13 oz cue with a hard tip.
 
SeanC said:
Without complicating matters due to the non-conservation of energy in the process, using conservation of linear momentum (which pretty much applies here unlike energy conservation)

Not to pick a nit, but one applies exactly as much as the other. Energy is most definitely conserved unless you know how to break the laws of physics. A portion of both energy and linear momentum are converted to other forms during the process of striking the cue ball and having it contact the rack.

I do agree completely with the "be comfortable" side of your position. That has been my position all along. You will obtain a virtually identical energy OR linear momentum with a heavier or lighter break cue in the reasonable range. It would be meaningless for example to discover a 1 oz. cue or a 100 lb. cue delivers a little more energy or linear momentum as both are impractical on several levels.

The only reason to change break cue weight is for comfort and feel.
Click to expand...
 
JB Cases said:
Actually that was quite a fun post. Now Duckie will come on here and lecture us about the "one inch punch" made famous by martial arts movies and Bruce Lee in exhibitions around the world.

This is the infamous punch that breaks boards and sends grown men flying backwards from a distance of only one inch away from the target.
Click to expand...


You are referring of course, to the Five Point Palm Exploding Heart technique developed by Pai Mei and taught only to a select few.
 
Last edited:
DogsPlayingPool said:
You are referring of course, to the Five Point Palm Exploding Heart technique developed by Pai Mei and taught only to a select few.
Click to expand...

Yes and in the world of pool only Shane Van Boening knows how to apply it. When he does it it's call the 10 Ball Exploding Rack Technique.
 
justtapitin said:
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
Click to expand...

That's all well and good but the lighter the cue the harder it is to stroke in a straight line.
 
***********************
Your arm strength and fast-twitch muscles are what they are and you're basically stuck with those.
**********************
Not really. Try flipping a coin by "smoothly" accelerating your thumb. Then try it the standard way by wedging your thumb under the index finger and building tension prior to rapid release. Different result?


***********************
And physics and billiards together? Pool-playing geek nirvana.
*************************
Beware getting married to a "mature" technology.
 
I've been using a 18.3 oz break cue. My playing cues are about an once heavier. I want to try a 16 oz. break cue. Anyone here ever used one that light?
 
Not Dead Ted said:
***********************
Your arm strength and fast-twitch muscles are what they are and you're basically stuck with those.
**********************
Not really. Try flipping a coin by "smoothly" accelerating your thumb. Then try it the standard way by wedging your thumb under the index finger and building tension prior to rapid release. Different result?


***********************
And physics and billiards together? Pool-playing geek nirvana.
*************************
Beware getting married to a "mature" technology.
Click to expand...


iit's got more to do with timing and getting your weight behind the cue than it does arm strength
 
Not to pick a nit, but one applies exactly as much as the other. Energy is most definitely conserved unless you know how to break the laws of physics. A portion of both energy and linear momentum are converted to other forms during the process of striking the cue ball and having it contact the rack.
Click to expand...

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.
 
poolplayer2093 said:
That's all well and good but the lighter the cue the harder it is to stroke in a straight line.
Click to expand...

While this is somewhat true for me too, I'm not sure everyone would agree with this. But in any event the point the OP was making is that cue weight does not make a significant difference anyway, but that bridge length is what matters most. So use whatever weight cue you get best results from and don't worry about it.
 
Which has more power

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.
 
You must log in or register to reply here.
Back
Top