power vs. speed on the break

[asn130]

i'm not sure which cue will generate more mph on the cueball, but one thing is for sure, there is no such thing as more power or more speed.

The cueball weighs 6oz period. The only thing you can change is how fast the cueball is moving when it leaves your cue, so if you hit it with a match stick & can make it travel @ 30mph it will apply the same force to the rack as if you hit it with a baseball bat making it travel @ 30mph.

 

[Jason Robichaud]

The weight of the ball, and the velocity of the ball is all that matters. A pea at 5 mph wont hurt you, while a train will. Most people misunderstand-the cueball has a fixed weight, and therefore it is weight and velocity that compute. A 20 mph cueball, whether struck with a 21 oz dufferin or with a 7 oz jump cue, has the same energy.

The topic is cue not CB. They want to know what will get the most MPH into the CB. You are right, 20 mph CB is 20 mph. The poster wanted to know, is it easier to get a CB to 20 mph with a x oz or y oz cue or do x and y produce the same.

If we look at golf, people are hitting longer with clubs that weigh less. The club weigh less and the faces are bigger to hit the ball better, so maybe we should be looking at tip size on breakers! Maybe a 15mm tip would better than a 12.25...

 

[dave sutton]

i dont know that tip size would matter. for the fraction of a sec its in contact with the cueball. if it did it would be bigger not smaller. a golf club head is bigger for a more consistant hit.

altho im sure you can throw a smaller cue faster also. yet another piece of the pie

i guess the only true answer is to take a fundamentally sound break like larry nevel. give him 5 different cues and measure speed at the point of contact. maybe do the same experiment a few times. like 5 or so times on different days. not looking for if he liked it but just the numbers

personally for me arm speed tops mass. not that i cant break with a heavy cue. i feel like i crush the ball with a lighter cue

 

[cmssuits]

It's all about "club head speed." You can accelerate the lighter cue faster. Just say'n. :groucho:

 

[Colin Colenso]

Some Simple Math

If your cue has infinite weight, the CB will bounce off it at about 2 times the speed of the cue.

Obviously a player cannot move a 1 tonne cue very fast.

Also a cue that weights around 6oz? (weight of the CB??) will send the CB off at about the same speed of the cue.

So the ultimate CB zone is somewhere in the middle.

I guess for a powerful player the optimum for break speed is near 21oz and for the whippy thin player, perhaps closer to 18 oz.

Colin

 

[Jason Robichaud]

i dont know that tip size would matter. for the fraction of a sec its in contact with the cueball. if it did it would be bigger not smaller. a golf club head is bigger for a more consistant hit.

altho im sure you can throw a smaller cue faster also. yet another piece of the pie

i guess the only true answer is to take a fundamentally sound break like larry nevel. give him 5 different cues and measure speed at the point of contact. maybe do the same experiment a few times. like 5 or so times on different days. not looking for if he liked it but just the numbers

personally for me arm speed tops mass. not that i cant break with a heavy cue. i feel like i crush the ball with a lighter cue

The bigger tip would help with consistant hits. If you are hitting the CB fuller more of the cue speed would be transfered to the CB, thats my guess.

Anyone have bigger tips on the break cue?

 

[cmssuits]

If your cue has infinite weight, the CB will bounce off it at about 2 times the speed of the cue.

Obviously a player cannot move a 1 tonne cue very fast.

Colin

I knew this would come up, (I mean, I am after all on the interwebz ) to which I must plead:

1. Common sense

2. Law of diminishing returns

 

[gpeezy]

IMO. When dealing with something the weight difference of a cueball and a cue stick I don't believe 3-4 oz will make a difference in causing the cueball to move any faster. As far as making MPH, I think a lighter cue would be easier. But if your looking for the best break, its in what YOU can swing the best. If your not able to have a good delivery on a cueball including angle of cue, cueball hit, and able to hit the rack square, MPH to a certain point will make no difference. Meaning a guy/girl with a good swing can have a much more effective break at 18-20mph than someone with a bad 30mph break. I have to best swing with a cue 19-19.5 which is also what I like to play with.

 

[gpeezy]

The bigger tip would help with consistant hits. If you are hitting the CB fuller more of the cue speed would be transfered to the CB, thats my guess.

Anyone have bigger tips on the break cue?

I know some cue makers make their break cue a little bigger to keep a stiff hit. My thoughts on tips are the should be hard! A soft tip would make it tougher to hit the cueball consistently if you ask me. My thoughts are the soft (when thinking of breaking any leather tip would be soft with any break above 21mph) is gonna compress quite a bit and could cause a change in the flight pattern and even direction. I like the predator BK2. It has a phenolic tip with a pad below it (think mezz does the same) and it is easiest to control. Solid phenolic tips seem to be harder to control than a phenolic/pab/ferrule break cue. The low deflection shaft on the bk2 helps also.
Just what I THINK I've learned since I've been playing.

 

[maha]

both weight and velocity determine the speed of the cueball. velocity rules though. so a lighter cue up to a point will make the cue ball go faster. as you get too light your mass slows down the cue ball and also makes it harder to propel the stick in a straight line.
for most players your stick as close to your playing stick in size and weight is optimum for all the factors combined. also after breaking with a stick of vastly different weight than your playing cue it could affect the feel, for the next few shots. which could be critical and negate any advantage of a few ounces of extra breaking force.

 

[Patrick Johnson]

You can accelerate the lighter cue faster.

This is certainly true for a 1 oz cue vs. a 10 lb cue, but is it necessarily true for an 18 oz cue vs. a 20 oz cue? Those weights might be so close together that our arm can't capitalize on the difference.

pj
chgo

 

[Jude Rosenstock]

This is certainly true for a 1 oz cue vs. a 10 lb cue, but is it necessarily true for an 18 oz cue vs. a 20 oz cue? Those weights might be so close together that our arm can't capitalize on the difference.

pj
chgo

Well, I know it's true for a 32 oz. baseball bat versus a 34 oz. Those two weights were the difference between making contact on a 90mph pitch (in the cage) versus wiffing 20 times in a row.

 

[Shaft]

If your cue has infinite weight, the CB will bounce off it at about 2 times the speed of the cue.

....

Colin

Whoa, Colin... I don't follow you here. In my momentum/energy balancing spreadsheet I used a 21,000,000 oz cue stick and got the cue ball off at roughly 9x the initial stick speed, (but both were VERY slow.) (In my spreadsheet I ignore arm mass, which changes the numbers but not the comparison between sticks).

A lot of the posters have stated opinions here based on true-but-only-part-of-the-story physics formulas.

The only energy on the table is what is supplied by your arm. When comparing light and heavy cues, you have to start with arm energy as a constant, and conserve pre-collision and post-collision energy and momentum.

If your calculations assume the same arm energy (that is, supplying a consistent force over a consistent distance like a robot does), you will see that heavier cues retain more of their pre-impact energy and transfer less energy to the CB.

For the same swing energy, a light cue results in higher CB speed and higher CB energy. The difference is not much, maybe too little to argue over, but there is no question that lighter cues "win."

I may have made an error, I have before. I posted all my calculations in Post #17 so anyone can check me. If you disagree, I would love to talk about it, but please show your math using arm energy as a constant.

If you argue that arm energy is not constant between sticks in this experiment, then there can be no answer to this question.

 

[asn130]

Whoa, Colin... I don't follow you here. In my momentum/energy balancing spreadsheet I used a 21,000,000 oz cue stick and got the cue ball off at roughly 9x the initial stick speed, (but both were VERY slow.) (In my spreadsheet I ignore arm mass, which changes the numbers but not the comparison between sticks).

A lot of the posters have stated opinions here based on true-but-only-part-of-the-story physics formulas.

The only energy on the table is what is supplied by your arm. When comparing light and heavy cues, you have to start with arm energy as a constant, and conserve pre-collision and post-collision energy and momentum.

If your calculations assume the same arm energy (that is, supplying a consistent force over a consistent distance like a robot does), you will see that heavier cues retain more of their pre-impact energy and transfer less energy to the CB.

For the same swing energy, a light cue results in higher CB speed and higher CB energy. The difference is not much, maybe too little to argue over, but there is no question that lighter cues "win."

I may have made an error, I have before. I posted all my calculations in Post #17 so anyone can check me. If you disagree, I would love to talk about it, but please show your math using arm energy as a constant.

If you argue that arm energy is not constant between sticks in this experiment, then there can be no answer to this question.

can anyone explain in layman terms how it is possible that the cueball can be traveling faster than the stick after impact? Or am i missunderstanding somthing here?

 

[Shaft]

First, Colin was right, I made a mistake on my spreadsheet. Infinite mass will result in a 2x CB speed after impact. My apologies, Colin, for speaking too soon. I was going to edit my post, but someone had already quoted it.

I know this all sounds counter-intuitive and I don't know how to explain it in layman's terms. If the collision is considered elastic, the combined energy and momentum of the stick and the ball must be the same after the collision as it was before the collision.

(The collision is not perfectly elastic, but the assumption simplifies the calculations and does not change the comparison).

My family is joking about the time I spend on these posts, so Merry Christmas!

 

[Banger]

I posted all my calculations in Post #17 so anyone can check me.

Hey Shaft, I haven't really checked all of your calculations, but on first glance, they look good. I do have to take exception with one thing, though.

The coefficient of restitution is not actually the main culprit at work here. The coefficient of restitution, by definition, is a ratio of the velocity of two objects involved in the collision prior to, and after impact (more about that later). According to the Handbook of Chemistry and Physics, the actual formula for the coefficient of restitution is as follows:

C= (v4-v3)/(v2-v1)

where two bodies moving in a straight line with velocities v1 and v2 respectively, collide, and after the impact have velocities v3 and v4.


Now, because we are assuming the collision is perfectly elastic (that is, no energy is lost to heat, sound, etc.), the coefficient of restitution is going to be "1" for both collisions. If we plug the actual velocities of the cues and cue balls that you calculated into the formula, we will indeed get a C.O.R. ratio of "1" for each collision. And since the coefficient of restitution is the same in both cases, it is safe to say that it is not responsible for the difference in energy transfer, because both collisions are 100% efficient (only for the purpose of these calculations, of course).

As you pointed out, both cues have the same energy prior to the collision with the cue ball, but the heavier cue has more momentum. Also, the lighter cue has a greater velocity prior to the collision. Because of the heavier cue's momentum, an interesting happens after the collision. Things turn around, and after the impact, and the heavier cue now has a greater velocity than the lighter cue. Since the heavier cue has a greater velocity after impact, it obviously has retained more of it's kinetic energy than the light cue. This, of course, means that the heavier cue transfered less energy to the cue ball. And since less energy is transfered to the cue ball, it will have a lower velocity. The bottom line, as you said, is that the lighter cue will impart more velocity to the cue ball due to a greater transfer of energy, assuming all other things are equal.

Shaft's calculations appear to be dead nuts on, and I thank him for taking the time to grind them out for us.



One final note on the coefficient of restitution. The golf world has perverted the definition of the the coefficient of restitution. The C.O.R. is actually the property of two colliding objects, and IS NOT a property that can be assigned to one particular object or material. Golfers have been fooled into thinking a club or ball can be assigned a C.O.R. and that is simply wrong. For instance, if you drop a golf ball onto a concrete walkway, it will bounce back up with an initial velocity that is very close to the final velocity it had when it hit the walkway. Therefore, the collision of those two objects would result in a C.O.R of close to "1". But if you drop the same golf ball into a sand trap, it will not bounce back up much, if at all, and thus the C.O.R would be near "0". So you must take both objects into account when calculating the actual coefficient of restitution. In other words, what did you bounce the golf ball off of? It makes a difference.

 

[Shaft]

Hey Shaft, I haven't really checked all of your calculations, but on first glance, they look good. I do have to take exception with one thing, though.

The coefficient of restitution is not actually the main culprit at work here. The coefficient of restitution, by definition, is a ratio of the velocity of two objects involved in the collision prior to, and after impact (more about that later). According to the Handbook of Chemistry and Physics, the actual formula for the coefficient of restitution is as follows:

C= (v4-v3)/(v2-v1)

where two bodies moving in a straight line with velocities v1 and v2 respectively, collide, and after the impact have velocities v3 and v4.


Now, because we are assuming the collision is perfectly elastic (that is, no energy is lost to heat, sound, etc.), the coefficient of restitution is going to be "1" for both collisions. If we plug the actual velocities of the cues and cue balls that you calculated into the formula, we will indeed get a C.O.R. ratio of "1" for each collision. And since the coefficient of restitution is the same in both cases, it is safe to say that it is not responsible for the difference in energy transfer, because both collisions are 100% efficient (only for the purpose of these calculations, of course).

As you pointed out, both cues have the same energy prior to the collision with the cue ball, but the heavier cue has more momentum. Also, the lighter cue has a greater velocity prior to the collision. Because of the heavier cue's momentum, an interesting happens after the collision. Things turn around, and after the impact, and the heavier cue now has a greater velocity than the lighter cue. Since the heavier cue has a greater velocity after impact, it obviously has retained more of it's kinetic energy than the light cue. This, of course, means that the heavier cue transfered less energy to the cue ball. And since less energy is transfered to the cue ball, it will have a lower velocity. The bottom line, as you said, is that the lighter cue will impart more velocity to the cue ball due to a greater transfer of energy, assuming all other things are equal.

Shaft's calculations appear to be dead nuts on, and I thank him for taking the time to grind them out for us.



One final note on the coefficient of restitution. The golf world has perverted the definition of the the coefficient of restitution. The C.O.R. is actually the property of two colliding objects, and IS NOT a property that can be assigned to one particular object or material. Golfers have been fooled into thinking a club or ball can be assigned a C.O.R. and that is simply wrong. For instance, if you drop a golf ball onto a concrete walkway, it will bounce back up with an initial velocity that is very close to the final velocity it had when it hit the walkway. Therefore, the collision of those two objects would result in a C.O.R of close to "1". But if you drop the same golf ball into a sand trap, it will not bounce back up much, if at all, and thus the C.O.R would be near "0". So you must take both objects into account when calculating the actual coefficient of restitution. In other words, what did you bounce the golf ball off of? It makes a difference.

Yes Banger, we agree, I was being sloppy with my writing. It had been a year since I did the calculations and was mis-remembering my research. COR is more a description of how elastic or inelastic a collision is, not a property of a massive object. Cheers.

 

[Island Drive]

can anyone explain in layman terms how it is possible that the cueball can be traveling faster than the stick after impact? Or am i missunderstanding somthing here?

Possibly from the whip action of the shaft bending and then releasing??????

 

[Big Perm]

Dave, this is a tough question with so many variables....how good is the hit on the cue ball, what type of english, how solid is the hit on the object ball, how well are the balls racked, felt, cushions.....I would think the physics would be rough on this one

In practice, I have seen many more good breaks with a 20+ ounce cue in regards to ball action or force....but good players can break with an 18 and accomplish what they want.....personally, I seem to get better results from a controlled aggressive strock with a heavy cue....

Back in the day, I had a 14 and 16 ounce among my house cues....breaking with them was a total joke (yeah, I tried), but I would think that mph is really all that matters in the end....

 

[Jal]

....If you argue that arm energy is not constant between sticks in this experiment, then there can be no answer to this question.

Shaft,

Arm energy is not constant, but nevertheless, there is an answer. The entity that should be set constant is the time averaged force applied to the cue. This may not be strictly true - you might be able to generate a little more force with a heavier cue - but it's a good starting point that produces an approximate answer.

To get it, use the conservations laws of energy and momentum to arrive at an expression for the cueball's speed (Vb) in terms of the mass of the stick and cueball (Ms and Mb), and the speed of the stick (Vs). If you like, you can include the tip offset (b) as a variable as well. Assuming an elastic collision for simplicity sake, and that a player's arm mass does not participate in the collision (this is a very good assumption!), you should get:

Vb = 2Vs/(u + Mb/Ms)

where

u = (5/2)(b/R)^2 + 1.

and R is a ball's radius. At zero offset (b = 0), u = 1.

For Vs, you can use the standard kinematic formula. For some time averaged force, F, which is symmetric over time, and bridge distance X.

Vs = sqrt(2FX/M).

where M is the combined inertia of the stick and player's arm:

M = Ms + Ma

Since the player's arm mass is distributed and different parts of it are moving at different speeds (it's rotating about some axis - going through the elbow for instance), using Ia as its moment of inertia about that axis, and p for the distance from the axis to the grip hand, an approximation to Ma is:

Ma = Ia/p^2

Plugging all of this into the formula for the cueball's speed:

Vb = [2/((5/2)(b/R)^2 + 1 + Mb/Ms)]sqrt[2FX/(Ms + Ia/p^2)]

Now, to find the stick mass, Ms, that'll produce the most cueball speed for some given (fixed) time averaged force (F), there is a standard technique used in calculus. Differentiate the above expression with respect to Ms, set that equal to zero, and then solve for Ms. I should have it sitting around somewhere, but not at hand. Another way is to graph Vb as a function of Ms and see where it peaks.

The main problem here is to assign some value to Ia. It's different for every player, a little or a lot, and how do you go about measuring it? Since players have found that cue weights of (very) roughly 20 oz work well, you can back calculate it, but then you've already assumed some answer for Ms (20 oz).

The main thing this exercise shows is that there is an optimal weight which is at least a little different for each player. It also shows that this varies with tip offset. And, if you assume some value for Ia and plug in different numbers for Ms, it reveals that Ms is not critical; you'll produce about the same cueball speed as Ms is varied over several ounces.

Jim