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01-28-2016, 03:01 PM

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Originally Posted by LAMas View Post
I shot with a hollow aluminum 19 oz house stick and it was high squirt/deflection.
That's probably because the lateral or transverse stiffness is much higher than a maple cue, causing the effective endmass to be much larger.

Regards,
Dave
  
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01-29-2016, 08:29 PM

This instructor looks like he is swiping the CB or is his stroke straight but the shaft is bending or both?

https://www.youtube.com/watch?v=G1C_O8AO8yY

Be well


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01-30-2016, 11:35 AM

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Originally Posted by Jal View Post
Dr. Dave's measurement was with the cue supported back in the butt region, so most of the cue was flexing, whereas the calculation I did was based on a cantilever length of only six inches (because of the slow propagation of the shear wave). The "spring constant" is inversely proportional to the cube of the cantilever's length, so that's where the major difference comes in. Also, since Dr. Dave's measurement involved the tapered portion of the cue, differences arise from that as well, but would tend to bring the numbers closer together since that portion is much stiffer than the untapered section. The chance of these static measurements/calculations reflecting reality for the dynamic case is not all that great, imo.
Jim,

FYI, I did another set of stiffness measurements with only the 8 inches closest to the tip allowed to flex (which is closer to the situation occurring during tip contact). The measurements and new calculations are included in the latest version of TP B.19. This changed the bottom-line result from 1.6% to 14.1% (the percentage contribution of flex to squirt, compared to the endmass-momentum effect). As you point out, this is a big difference, but I think the conclusion is still accurate:

As has been shown with experiments dealing with adding and removing mass from the end of the shaft, endmass is much more important than shaft flex concerning how much CB deflection (squirt) a shaft creates (although, as pointed out on the endmass resource page, lateral or transverse stiffness does affect "effective endmass").

Thanks again for making me reconsider some of the assumptions and implications of the analysis.

Catch you later,
Dave

PS: I certainly agree with you that a static measurement doesn't perfectly model what is going on during the dynamic flexing of the cue during tip contact (nor is "effective endmass" a perfect model of the complex dynamic momentum transfer occurring during impact), but I am still confident with the "ball-park" results and conclusion of the analysis.
  
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01-30-2016, 12:07 PM

Big E,

Did you see that?

From 1.5% to 14.1% just because of a different method.

Now I wonder what if a shaft's end flexibility can be increased say with a long flexible ferrule &/or an hour glass taper at just the right location?

Maybe it can climb to 25%.

Then there is making the 'whole' shaft more flexible with a much longer then 'normal' pro parallel taper.

What do you think?

Best 2 Ya,
Rick

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01-30-2016, 12:51 PM

Quote:
Originally Posted by ENGLISH! View Post
Big E,

Did you see that?

From 1.6% to 14.1%[/COLOR][/SIZE][/B] just because of a different method.

Now I wonder what if a shaft's end flexibility can be increased say with a long flexible ferrule &/or an hour glass taper at just the right location?

Maybe it can climb to 25%.

Then there is making the 'whole' shaft more flexible with a much longer then 'normal' pro parallel taper.

What do you think?

Best 2 Ya,
Rick
ENGLISH,
After 4,000+ hits we and others get a better answer to the question in this thread.
Thanks to intuition and tenacity, you and others helped to keep it alive and more real.

Thanks all and be well.


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01-30-2016, 12:59 PM

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Originally Posted by ENGLISH! View Post
Now I wonder what if a shaft's end flexibility can be increased say with a long flexible ferrule &/or an hour glass taper at just the right location?

Maybe it can climb to 25%.

Then there is making the 'whole' shaft more flexible with a much longer then 'normal' pro parallel taper.
Making a shaft more flexible (less stiff) would actually reduce the percentage contribution of flex force to CB deflection (squirt), not increase it.

And if the endmass is also reduced, the total amount of CB deflection (squirt) would also be lower.

Regards,
Dave
  
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01-30-2016, 01:08 PM

Quote:
Originally Posted by dr_dave View Post
Making a shaft more flexible (less stiff) would actually reduce the percentage contribution of flex force to CB deflection (squirt), not increase it.

And if the endmass is also reduced, the total amount of CB deflection (squirt) would also be lower.

Regards,
Dave
Dave,

I think you misinterpreted my post to Lamas.

He seems to have understood it.

Best Wishes.
  
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01-30-2016, 03:59 PM

Quote:
Originally Posted by dr_dave View Post
FYI, I did another set of stiffness measurements with only the 8 inches closest to the tip allowed to flex (which is closer to the situation occurring during tip contact). The measurements and new calculations are included in the latest version of TP B.19. This changed the bottom-line result from 1.6% to 14.1% (the percentage contribution of flex to squirt, compared to the endmass-momentum effect). As you point out, this is a big difference, but I think the conclusion is still accurate:
Hi Dr. Dave,

Thanks for the additional work and info!

I'm in the middle of "refurbishing" the math (E-B beam treatment for the whole cue as well as the tip end) and as a matter of curiosity, looking forward to comparing your measurements to its predictions. (I may do a few myself.)

Quote:
Originally Posted by dr_dave View Post
As has been shown with experiments dealing with adding and removing mass from the end of the shaft, endmass is much more important than shaft flex concerning how much CB deflection (squirt) a shaft creates (although, as pointed out on the endmass resource page, lateral or transverse stiffness does affect "effective endmass").
Forgive me. and perhaps as a petulant side note, I have to lodge a word of protest as to the notion that shaft flex/stiffness contributes anything to squirt apart from its role in creating (and delivering) endmass.

Jim
  
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01-30-2016, 05:23 PM

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Originally Posted by Jal View Post
Forgive me. and perhaps as a petulant side note, I have to lodge a word of protest as to the notion that shaft flex/stiffness contributes anything to squirt apart from its role in creating (and delivering) endmass.
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

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01-30-2016, 11:32 PM

Quote:
Originally Posted by dr_dave View Post
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.
Dr. Dave,

Thanks for taking the time to explain.

I'm sure of two things. One is that given the fact that the forces involved are all internal to the cue-cueball system, lateral momentum must be conserved during every moment of the collision (i.e., = 0 for a straight stroke). Second is that you know that (and know it far better than me, given everything).

If you can construct a scenario where a force is acting on the cueball that is not acting equally and opposite on the endmass, I think, well, I won't know what to think. It's true that a flex force does act on the cueball, but in acting equally and opposite on the endmass, it acts in the opposite direction of what our intuition suggests, which would be to push the endmass in the same direction as the cueball.

If this is the situation described by your example:

]///////////(mass 1)<----- Force----->(mass 2)

then the force is pushing on the wall behind the spring besides compressing it. Without the wall (the support), there is no compression. So the analogous "endmass" here is mass1 plus the wall plus whatever it's attached to. If the momentum of the combined system: m2, m1, wall.. was zero to begin with, it's zero throughout the interaction. That is to say, whatever force acts against m2, acts equally (and oppositely) on m1-wall... There is no force acting on m2 that's not acting on m1-wall... It's surely true, as you say, that the force on m1 is less than that on m1-wall.. But m1 isn't the entire "endmass," by my argument. Obviously, with a cue's endmass, there is no wall, just m1:

(mass 1)///////////<----- Force----->(mass 2)

in which case m1 does feel whatever m2 feels (but opposite).

That's probably my best shot too, so we just may have to agree to disagree. (It was you and Patrick that convinced me that it was all endmass, so to speak. So I think one thing we can agree on for sure, is that we must be trapped in some sort of diabolical space-time loop!)

Jim

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01-31-2016, 04:26 AM

Quote:
Originally Posted by dr_dave View Post
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.
  
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01-31-2016, 05:40 AM

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Originally Posted by john coloccia View Post
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.


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01-31-2016, 06:48 AM

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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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01-31-2016, 08:35 AM

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Originally Posted by john coloccia View Post
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?


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01-31-2016, 08:56 AM

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Originally Posted by john coloccia View Post
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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