Squirt. End Mass and Cue Flexibility.

john coloccia

AzB Silver Member
Silver Member
Jim,

I know you don't agree with me on this, but I will try once more:

1.) Shaft endmass can be increased significantly by adding mass close to the tip (e.g., by using a heavier tip or ferrule, by inserting something heavy but not stiff into a cored-out shaft, or by physically adding external mass to the tip end of the shaft). This has been clearly demonstrated with numerous experiments by me, Mike Page, and others. In these cases, the endmass is increased dramatically with no increase in shaft stiffness.

2.) Removing endmass from a shaft without significantly changing the stiffness of the shaft (e.g., by using a lighter ferrule and/or by drilling out the core of the shaft end), can significantly reduce the amount of effective endmass and squirt. This has also been demonstrated with the design of Predator shafts. Drilling out the core does not have a huge effect on shaft-end stiffness, but it does dramatically reduce endmass and squirt (as does the lighter ferrule).

3.) As my TP B.19 analysis shows, the total sideways force acting between the tip and CB is due to two physical effects. Part of the force contributes impulse which imparts momentum to the endmass of the shaft. The other part of the force (much smaller) is required to flex the shaft. The end of the shaft effectively looks like a mass supported by a spring. Think of a simple diagram of a linear spring-mass system with an applied force pushing on a mass supported by a spring. Some of the force applied to the mass goes into accelerating the mass (imparting momentum), and some goes into compressing the spring (F_total = F_applied - kx = ma). The resultant force experienced by the mass is not the force applied to the mass (F_applied); rather, it is the amount of excess force not being resisted by the spring (F_applied - kx). I think the answer to the question "In your TP B.19 analysis, how does the flex-force impulse relate to the sideways impulse between the tip and CB?" near the bottom of the squirt endmass and stiffness effects resource page shows how the same logic applies to the squirt-endmass-stiffness problem. The difference between the shaft-endmass-lateral-stiffness problem and the simple linear-spring-mass problem is that the total endmass of the shaft depends on shaft stiffness (in addition to the weight of the tip, ferrule, and anything else on or in the end of the shaft), but the spring force is still there.

Sorry Jim, but that's the best I can do. If these 3 things don't convince you, then we will need to agree to disagree on this one.

Best regards,
Dave

Are you saying that if you have a shaft with no mass that is perfectly stiff, you will get no squirt ever? It seems like that should be wrong.
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
Are you saying that if you have a shaft with no mass that is perfectly stiff, you will get no squirt ever? It seems like that should be wrong.

Of course it should seem wrong. You've put up the impossible situation , so of course it can't happen.
 

john coloccia

AzB Silver Member
Silver Member
Of course it should seem wrong. You've put up the impossible situation , so of course it can't happen.

You've never heard of a thought experiment? Sometimes, you examine an impossible edge case to get insight into the real situations in between. It brings clarity your thinking. Just because you can't build it doesn't mean you can't analyze it. I'm sure Dr Dave will agree.

edit:
Just to be clear, I don't know the answer. I haven't thought about it or analyzed it, and honestly none of this interests me very much since none of it really has much to do with actually playing pool IMHO, but I do have something of a physics background, and just thinking about it for two seconds my intuition says that a massless shaft will still squirt and I'd be interested in hearing Dave's analysis since he's spent an awful lot of time looking at it.
 
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Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
You've never heard of a thought experiment? Sometimes, you examine an impossible edge case to get insight into the real situations in between. It brings clarity your thinking. Just because you can't build it doesn't mean you can't analyze it. I'm sure Dr Dave will agree.

edit:
Just to be clear, I don't know the answer. I haven't thought about it or analyzed it, and honestly none of this interests me very much since none of it really has much to do with actually playing pool IMHO, but I do have something of a physics background, and just thinking about it for two seconds my intuition says that a massless shaft will still squirt and I'd be interested in hearing Dave's analysis since he's spent an awful lot of time looking at it.
Seriously?
 

ENGLISH!

Banned
Silver Member
You've never heard of a thought experiment? Sometimes, you examine an impossible edge case to get insight into the real situations in between. It brings clarity your thinking. Just because you can't build it doesn't mean you can't analyze it. I'm sure Dr Dave will agree.

edit:
Just to be clear, I don't know the answer. I haven't thought about it or analyzed it, and honestly none of this interests me very much since none of it really has much to do with actually playing pool IMHO, but I do have something of a physics background, and just thinking about it for two seconds my intuition says that a massless shaft will still squirt and I'd be interested in hearing Dave's analysis since he's spent an awful lot of time looking at it.

What would your shaft of NO MASS be composed of...

your will?

What would make up it's 'stiffness'?

Please remember, we're moving a solid phenollic ball that has mass, weight, that is being held in place by gravitational forces to a another rather very large 'sphere', the planet Earth.

Is your shaft going to be an anti gravity shaft of no mass?

What type of force with no mass is going to move an object that has mass in an environment of gravitational forces.

Does not velocity of ZERO mass, no mass, equal zero force?

Are we going to change the temperature drastically on one said of the ball & at one point on the ball? Does not the atoms of air have mass?

Sorry for the sarcasm, but we're talking about forces that involve mass.

Best Wishes.
 
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Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
What would your shaft of NO MASS be composed of...

your will?

What would make up it's 'stiffness'?

Please remember, we're moving a solid phenollic ball that has mass, weight, that is being held in place by gravitational forces to a another rather very large 'sphere', the planet Earth.

Is your shaft going to be an anti gravity shaft of no mass?

If you swipe the directional application of your
"shaft with no mass" across the face of the ball, will the ball spin away with no movement in the opposite direction of the swipe?

What type of force with no mass is going to move an object that has mass in an environment of gravitational forces.

Sorry for the sarcasm, but we're talking about forces that involve mass.

Does not velocity of ZERO mass, no mass, equal zero force?

Best Wishes.
Correct

Even if someone were to do "analysis," the standard momentum and conservation of energy equations for this collision would yield zero rather quickly.

Freddie <~~~ studied physics once
 

john coloccia

AzB Silver Member
Silver Member
This is complete nonsense. Any first year physics student can analyze what happens to a ball when you apply a constant force. That's equivalent to saying that the shaft is rigid with a constant acceleration, and the mass is completely irrelevant...it can be zero...it can be infinite...it can be 19oz...whatever. I choose it to be zero to make the point, but choose whatever you think is convenient. Since the shaft is rigid it CAN'T move offline regardless, and since acceleration is constant, the mass of the cue is irrelevant. That's the same exact situation.

Again, I'm pretty sure Dave will agree with this, but we'll see. I think perhaps the point is that all shafts are sufficiently flexible and/or all bridges are sufficiently sloppy that the flexibility becomes irrelevant and the entire thing becomes dominated by the weight at the tip, i.e. the part that needs to move the most.
 

Corwyn_8

Energy Curmudgeon
Silver Member
Without the wall (the support), there is no compression.

Why not? Isn't the compression felt even before the speed of sound in the spring has reached the wall? Isn't that in fact what we model when we say that the transverse wave hasn't reached the bridge hand yet?

Thank you kindly.

p.s. to SEE the non-intuitiveness of this see:
https://www.youtube.com/watch?v=wGIZKETKKdw
[featuring the fellow who started this discussion]
 

Corwyn_8

Energy Curmudgeon
Silver Member
Even if someone were to do "analysis," the standard momentum and conservation of energy equations for this collision would yield zero rather quickly.

Gravity is a force with infinite stiffness and no mass.

Thank you kindly.
 

john coloccia

AzB Silver Member
Silver Member
No, that is NOT what he is saying.

Thank you kindly.

I didn't think that could be right, but that's the impression I got reading through some of this thread.

For what it's worth, the slinky thing is cool, but I personally think it's confusing to analyze it as a wave, information going here and there, etc. It's simpler to analyze it considering that a spring has a constant k for small displacements, so when you let go of the top it's still pulling on both the top and bottom parts with the same force it was before you let go of it. End result is that the bottom stays put, and the top accelerates faster than it would with just gravity, and because k is constant the bottom stays dead still until the spring displacement returns to 0. Not trying to argue - just offering a different point of view that's maybe a bit more intuitive.
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
This is complete nonsense. Any first year physics student can analyze what happens to a ball when you apply a constant force. That's equivalent to saying that the shaft is rigid with a constant acceleration, and the mass is completely irrelevant...it can be zero...it can be infinite...it can be 19oz...whatever. I choose it to be zero to make the point, but choose whatever you think is convenient. Since the shaft is rigid it CAN'T move offline regardless, and since acceleration is constant, the mass of the cue is irrelevant. That's the same exact situation.

Again, I'm pretty sure Dave will agree with this, but we'll see. I think perhaps the point is that all shafts are sufficiently flexible and/or all bridges are sufficiently sloppy that the flexibility becomes irrelevant and the entire thing becomes dominated by the weight at the tip, i.e. the part that needs to move the most.

Please then do the very basic momentum equation for a collision with the mass-less object at whatever velocity striking an object at rest. As you know your Physics, you should have your answer rather quickly.

I know that you're asking the good Dr. Dave to answer, but you seem to not be giving anyone else a shake at this? Some of us having been dissecting squirt about 10 years longer than Dave hit the internet forums. Surely, you're not dismissing that body of work. Dave even got answers from me. I know, it must come as a shock.

Freddie <~~~ going back in time is apparently possible
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
Gravity is a force with infinite stiffness and no mass.

Thank you kindly.

You opened the door. Are you suggesting a shaft made of gravity?

Are you suggesting gravity has something to do with "this collision" and the my quoted statement?

I"m all ears. Throw a gravitational device in a shaft, and surely you can reduce squirt.

Is that really what we're entertaining here?
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
This is complete nonsense. Any first year physics student can analyze what happens to a ball when you apply a constant force. That's equivalent to saying that the shaft is rigid with a constant acceleration, and the mass is completely irrelevant...it can be zero...it can be infinite...it can be 19oz...whatever. I choose it to be zero to make the point, but choose whatever you think is convenient. Since the shaft is rigid it CAN'T move offline regardless, and since acceleration is constant, the mass of the cue is irrelevant. That's the same exact situation.

Again, I'm pretty sure Dave will agree with this, but we'll see. I think perhaps the point is that all shafts are sufficiently flexible and/or all bridges are sufficiently sloppy that the flexibility becomes irrelevant and the entire thing becomes dominated by the weight at the tip, i.e. the part that needs to move the most.
I'm going to try this again, against my better judgement.

Stiffness IS important. It is the stiffness of the material that determines the speed of the transverse wave during the tip ball contact. The speed of the transverse wave determines the LENGTH of shaft involved in the collision. The LENGTH of shaft involved determines the MASS involved in the collision. How??? By average DENSITY of the material, in simple terms.

If the DENSITY is zero, by your magical mass-less shaft, then the amount of mass involved in the collision regardless of stiffness is ZERO (ignoring the tip mass, why not?)

As also has been said now for about 15 years, the stiffness isn't significantly changed if we're using a 58" (or 29") shaft with dimensions of a pool cue. Some other significant material change would have to happen without changing other things.

If you can come up with a great material that changes the transverse wave properties significantly without changing the mass AND have it so that you can actually play, then there you have it. But within the realm of profiles that we play pool with and the range of woods, the range of stiffnesses for cues, whether conical or delayed pro cylindrical taper) are too close in a tight window to significantly change that transverse wave.

But it reads that your "intuition" :

"that all shafts are sufficiently flexible and/or all bridges are sufficiently sloppy that the flexibility becomes irrelevant and the entire thing becomes dominated by the weight at the tip,"

That has been the very basics of squirt and "stiffness is not significant" discussions. The VARIATION of "flexibility" or "stiffness" is not enough to cause significant difference in the materials and profiles used in today's cues. But stiffness is absolutely important in the overall equation of determining the actual "mass in effect."

But you asked about a zero mass moving body. In the basic collision Physics 101 collision equation, you wouldn't get any squirt. You need a mass for that moving object for that mechanical collision.
 
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Corwyn_8

Energy Curmudgeon
Silver Member
Is that really what we're entertaining here?

I am suggesting that your claim of not being able to make sense of equations of such a collision is wrong, since we do it all the time with respect to gravity.

Thank you kindly.
 

Cornerman

Cue Author...Sometimes
Gold Member
Silver Member
I am suggesting that your claim of not being able to make sense of equations of such a collision is wrong, since we do it all the time with respect to gravity.

Thank you kindly.

Please review my posts. I specifically said "this collision" so that we didn't start talking about non-relative motions. This collision is at its most basics a mechanical collision.

The collision is simply a conservation of energy or conservation of momentum. Do you disagree with this?
 

Corwyn_8

Energy Curmudgeon
Silver Member
I didn't think that could be right, but that's the impression I got reading through some of this thread.

Go reread where DrDave gives the equation for the squirt.

m_ball * v_side = m_end*v_end + F_flex*T

If the mass_end is 0, then the squirt momentum is F_flex*T.
(and the ball velocity will be directly opposite the contact point, for infinitely stiff cues).

Thank you kindly.
 

john coloccia

AzB Silver Member
Silver Member
I'm going to try this again, against my better judgement.

Stiffness IS important. It is the stiffness of the material that determines the speed of the transverse wave during the tip ball contact. The speed of the transverse wave determines the LENGTH of shaft involved in the collision. The LENGTH of shaft involved determines the MASS involved in the collision. How??? By average DENSITY of the material, in simple terms.

If the DENSITY is zero, by your magical mass-less shaft, then the amount of mass involved in the collision regardless of stiffness is ZERO (ignoring the tip mass, why not?)

As also has been said now for about 15 years, the stiffness isn't significantly changed if we're using a 58" (or 29") shaft with dimensions of a pool cue. Some other significant material change would have to happen without changing other things.

If you can come up with a great material that changes the transverse wave properties significantly without changing the mass AND have it so that you can actually play, then there you have it. But within the realm of profiles that we play pool with and the range of woods, the range of stiffnesses for cues, whether conical or delayed pro cylindrical taper) are too close in a tight window too significantly change that transverse wave.

But it reads that your "intuition" :

"that all shafts are sufficiently flexible and/or all bridges are sufficiently sloppy that the flexibility becomes irrelevant and the entire thing becomes dominated by the weight at the tip,"

That has been the very basics of squirt and "stiffness is not significant" discussions. The VARIATION of "flexibility" or "stiffness" is not enough to cause significant difference in the materials and profiles used in today's cues. But stiffness is absolutely important in the overall equation of determining the actual "mass in effect."

But you asked about a zero mass moving body. In the basic collision Physics 101 collision equation, you wouldn't get any squirt. You need a mass for that moving object for that mechanical collision.

Thanks for your reply. That makes more sense. One thing I would add, though, is that I believe you're analyzing it as free bodies, i.e. as though the cue were simply thrown into the ball. In that case, then you absolutely need the cue to be massive or the ball doesn't move at all. It's not a free body, though. Even if the cue is massless, it's being driven by your arm and it's constrained by your bridge. It's not a free body situation. You need to consider that force as well if you want the whole picture. Maybe you have and the conclusion is still that everything is just sufficiently flexible that it never really matters.
 

Corwyn_8

Energy Curmudgeon
Silver Member
I specifically said "this collision"

Which one is "this one"? The one with a massive cue and finite stiffness, or the one with a mass-less cue, with infinite stiffness?

You can choose one, but you can't force both to follow the equations of only the one you chose.

If the thought experiment doesn't intrigue you, just ignore it. It does have a perfectly sensible answer.

Thank you kindly.
 
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