cue to ball energy transfer

[DynoDan]

FYI, this topic is covered fairly well here:

cue efficiency

Enjoy,
Dave

I believe firearm ballistics science has proved the same powder charge projecting a light bullet at higher speed delivers more energy to the target than a heavy one at lower speed. But, practical application is only within a middle range. Take (for instance), baseball bats: Hit a fastball with a super-heavy bat, it’s a home run. But, good luck swinging it up to speed before the ball crosses the plate! A super-light alloy bat can be swung quickly, and at high speed, but then, good luck hitting out of the infield. Likely lighter (middle range) bats are favored since getting only to first base still beats striking out.

 

[dr_dave]

I believe firearm ballistics science has proved the same powder charge projecting a light bullet at higher speed delivers more energy to the target than a heavy one at lower speed. But, practical application is only within a middle range. Take (for instance), baseball bats: Hit a fastball with a super-heavy bat, it’s a home run. But, good luck swinging it up to speed before the ball crosses the plate! A super-light alloy bat can be swung quickly, and at high speed, but then, good luck hitting out of the infield. Likely lighter (middle range) bats are favored since getting only to first base still beats striking out.

Determining optimal cue weight for a player (for power shots like the break), is similar to the determining the best choice for baseball bat weight. It depends on the person's anatomy and muscle physiology (fast twitch vs. slow twitch muscle fibers). Some people (fast twitch) will be more effective with a lighter cue (bat), and some (slow twitch) will be more effective with a heavier cue (bat). For more info, see:

optimal cue (and baseball bat) weight

Regards,
Dave

 

[DynoDan]

Determining optimal cue weight for a player (for power shots like the break), is similar to the determining the best choice for baseball bat weight. It depends on the person's anatomy and muscle physiology (fast twitch vs. slow twitch muscle fibers). Some people (fast twitch) will be more effective with a lighter cue (bat), and some (slow twitch) will be more effective with a heavier cue (bat). For more info, see:

optimal cue (and baseball bat) weight

Regards,
Dave

Interesting. Don’t follow baseball, but I assume hitters still swing multiple/weighted bats on their way to the plate (?). Makes me wonder if 9 ballers ordinarily using a heavy cue would benefit similarly if switching to a lighter one for the break?

 

[Patrick Johnson]

...the basic result is that the cue ball will ideally be going 150% of the cue stick speed just before contact and the cue stick will slow to 50% of its previous speed.

So the CB only accelerates to 1.5 x the stick speed because the stick slowing to half its speed during contact is as if a stick half its weight (1.5 x CB weight) hit the CB?

pj <- dumb it down for me
chgo

 

[Bob Jewett]

So the CB only accelerates to 1.5 x the stick speed because the stick slowing to half its speed during contact is as if a stick half its weight (1.5 x CB weight) hit the CB?

pj <- dumb it down for me
chgo

The easy way to get to roughly the answer without going into actually solving equations is to change the frame of reference of the stick and ball to one where they are coming at each other to start with and the total momentum is zero. After the collision, the total momentum must still be zero, and if no energy is lost the two objects must have perfectly reflected at the contact point. That makes the analysis really simple, and gives the 2:1 speed multiplier for very light targets that I mentioned above.

I'm afraid that changing the frame of reference loses a lot of people. And, you have to get them to agree to conservation of momentum and energy. Also, you have to introduce the idea that momentum has a direction, and that momentum is the product of mass and velocity, with velocity having a direction. So, it's only a slightly dumber way to look at things.

The actual equation of the speed of the ball relative to the speed of the stick, which has been posted here before and is in Byrne's book and is on Dr. Dave's website and is in every high school physics text book, is:

VelocityBall/VelocityStick = 2 MassStick/(MassStick+MassBall)

This can be checked for reasonableness by looking at extreme cases. If the ball is very, very heavy, it's speed will almost be zero because its huge mass dominates the denominator on the right side. If the masses are equal (as with direct ball-ball collisions) the speeds are the same. If the ball is very light, like the ping pong ball and the locomotive, the ball speed is twice the locomotive speed.

 

[gregcantrall]

“Simplified” and “perfect world”. My statement is actually true, much to your dismay. Sixty years ago does not apply to today. There is not one public pool hall that uses ivory cue balls, so why even bring up a statement so blatantly misleading? Ivory has a lower CoR than phenolic, which kind of proves my napkin match there. Will we ever see a cue that imparts 3x speed to a cue ball, no. But it’s easier to start there and then explain the rest, than starting at the right answer and explain it incorrectly. I don’t think there’s a single cue out there that performs at a 1.35x level anymore due to proliferation of the use of phenolics.

Ok Mr Hard, I weigh 185 lbs. If you were stationary and I hit you head on traveling at 15 mph. What would be our speeds after the collision? Just kidding. I feel safe in accepting the conclusions of Bob Jewett and Dr. Dave.

 

[Hits 'em Hard]

If what you said was true, a ping pong ball struck by a locomotive going five miles per hour would break the sound barrier. That doesn't happen. The maximum speed a light object can go when set in motion by being struck by a much heavier object is twice the speed of the heavy object.

Please stop trying to twist what I’m saying. I’m not even close to saying what you’re implying. Due to the CoR of a ping pong ball, it is incapable of transferring energy at 100% efficiency. Same thing I’m implying. But I guess since the OP was having trouble understanding the why, I decided to not explain it in collegiate level but a more simplistic basic way. Not my fault you want to take my explanation out of context.

Ok Mr Hard, I weigh 185 lbs. If you were stationary and I hit you head on traveling at 15 mph. What would be our speeds after the collision? Just kidding. &#55357;&#56860; I feel safe in accepting the conclusions of Bob Jewett and Dr. Dave.

If you and I both had a CoR of 100%, and I weighed the same as you. Then I would end up going 15mph, and you would be going 0mph. But alas we are squishy and don’t have a 100% a CoR on a collision. Give me the CoR of the human body and I could tell you. But the CoR of a phenolic ball is different than that of an ivory ball, so to imply that it’s only 1.35x because of an old experiment is a scientific failure.

 

[Bob Jewett]

... I decided to not explain it in collegiate level but a more simplistic basic way. ... .

Your explanation was also wrong. You do not understand the physics of simple collisions. Try asking a local high school physics teacher if you don't trust people whose day jobs require them to understand physics.

 

[duckie]

Just remember, the cue is controlled by muscle movement of the arm and is not hung by wires, nor is it dropped as shown in one video.

There’s is a difference between testing to prove a theory and testing to see what is happening in real situations. In real situations, the cue is controlled by the arm and not gravity.

 

[Patrick Johnson]

Just remember, the cue is controlled by muscle movement of the arm and is not hung by wires, nor is it dropped as shown in one video.

The arm is disconnected from the cue by the soft flesh of the hand flexing during impact.

There’s is a difference between testing to prove a theory and testing to see what is happening in real situations.

You haven’t tested this either way.

pj
chgo

 

[iusedtoberich]

Physics aside, from a purely practical pool player standpoint, I've never noticed a meaningful difference. In 25 years of playing, I've played with hundreds of random house cues, dozens of cue's I've owned, dozens of tips, borrowed friend's cues, etc. When I strike a shot with any of the above, the shot generally works the same.

I'd wager any one of us could take any random cue in the pool room, make a lag shot, and be about as close to the end rail as our personal daily player.

 

[duckie]

Let’s look at this from another view point.......moving balls.

There is a cut shot where the OB is x distance from the pocket and the CB needs to go x distance for the next shot.

There is only one amount of force that can do this. Less or more force and the balls won’t be placed where you want. With less force, the OB won’t be made cause of the distance from the pocket nor will the CB go where it needs for the next shot.

More force, the OB rattles out of the pocket, the CB goes way beyond where it needs to be for the next shot.

A 19 oz cue is used to apply the proper amount of force to make the OB and placed the CB where it needs to be for the next.

A 20 oz must apply the same amount of force as the 19 oz in order to obtain the same results.

This can only be done by stroke control.

 

[Patrick Johnson]

A 19 oz cue is used to apply the proper amount of force to make the OB and placed the CB where it needs to be for the next.

A 20 oz must apply the same amount of force as the 19 oz in order to obtain the same results.

This can only be done by stroke control.

If you apply the same stroke force to both cues the CB speed will be the same.

pj
chgo

 

[Bob Jewett]

A 19 oz cue is used to apply the proper amount of force to make the OB and placed the CB where it needs to be for the next.

A 20 oz must apply the same amount of force as the 19 oz in order to obtain the same results.

If you apply the same stroke force to both cues the CB speed will be the same.
...

Technically both of these comments are slightly wrong.

The speed of the cue ball is determined by adding up the force at each instant during the tip-ball contact. If you have a soft tip, you might have 50 pounds of force for two milliseconds. If you have a hard tip, you might get 100 pounds of force for one millisecond. Both of those cases lead to the same cue ball speed.

It is not the force at a particular instant that's important, it's the combination of force and duration. (Technically, this is referred to as the "time integral of force" and what is calculated is the change in momentum of the object. And the force is not constant so you usually have to find the time profile of the force, but the right average force can give the right answer. Fortunately most collisions of two objects -- stick-to-ball -- can be calculated much more simply by using Newton's laws.)

The second statement is not quite right either and is ambiguous for the same reason as the first statement. But it can be shown easily to be probably false by taking it to the extreme. Imagine a 19 ounce cue and a 19 ton cue. If you apply the same force to the 19 ton cue that gets a good shot with the 19 ounce cue, the 19 ton cue will be barely moving when it arrives at the cue ball. And that will barely get the cue ball moving. That gives a clue that maybe there is at least a small difference between 19 and 20 ounces. It turns out to be smaller than you might think (less than the difference in weights) due to other factors.

 

[ceebee]

Please pay attention folks, all of this is intended to make you a better player...

 

[john noe]

Yes, a heavier cue takes more force to move it. I remember reading a book about 30 years ago where they tested the optimum weight of a cue to achieve the most powerful break. It was around 19 ounces if memory serves. 30 years is a long time to remember something like that but when it comes to something I'm very interested in, like pocket billiards I don't usually lose that memory.

Cheers,
John Noe

 

[Patrick Johnson]

Yes, a heavier cue takes more force to move it. I remember reading a book about 30 years ago where they tested the optimum weight of a cue to achieve the most powerful break. It was around 19 ounces if memory serves. 30 years is a long time to remember something like that but when it comes to something I'm very interested in, like pocket billiards I don't usually lose that memory.

Cheers,
John Noe

Maybe that's an average...? I think optimal break cue weight wouldn't be the same for everybody - for one obvious reason because everybody isn't equally strong.

pj
chgo

 

[Patrick Johnson]

The second statement is not quite right either

The things we say are supposed to be right? That changes everything.

...and is ambiguous for the same reason as the first statement. But it can be shown easily to be probably false by taking it to the extreme. Imagine a 19 ounce cue and a 19 ton cue. If you apply the same force to the 19 ton cue that gets a good shot with the 19 ounce cue, the 19 ton cue will be barely moving when it arrives at the cue ball. And that will barely get the cue ball moving. That gives a clue that maybe there is at least a small difference between 19 and 20 ounces. It turns out to be smaller than you might think (less than the difference in weights) due to other factors.

But is it correct to a practical degree within the normal range of cue weights (and maybe some other qualifiers I'll think of)?

pj
chgo

 

[duckie]

FWIW...

Kinetic Energy, the energy a object has in motion......K.E. = 1/2 mv2
M=mass......in this case the cue and your hand
V=velocity of cue

The get the same KE when the mass is increased, velocity needed to do so must decrease.

Which requires stroke control.

To get the same ending shot results using the exact same starting ball positions will require the same amount of force regardless of cue weight. To do so, will require a slower stroke with a heavier cue.

This is because the weight of the balls didn’t change.

A cue hanging from wires or a cue machine that uses spring are examples of potential energy, which is not used in shot making.