Squirt. End Mass and Cue Flexibility.

[john coloccia]

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.

Yes, thanks. I missed that. That's more what I would have expected.

 

[Cornerman]

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.

Jeezus. This one is the shaft/tip and ball collision. Are you talking about something else?

There is one basic equation. Conservation of momentum. Use (john's) his theoretical mass-lass shaft. Done.

Let's not try to confuse this just for confusion sake. They all follow the same equation. The Zero mass immediately goes to zero.

Even if you add a flex impulse piece, that impulse has a mass component.

 

[row21097]

E=MC^2

In the words of the famous boxer / physicist Roberto Duran:

No mass / No energy / No squirt

Bert

 

[Corwyn_8]

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.

The consensus seems to be, it doesn't matter when we are talking about hands and arms, but does if we start talking about rigid supports and robots.

Thank you kindly.

 

[Cornerman]

Yes, thanks. I missed that. That's more what I would have expected.

F_Flex* T = Force*Time=Mass * delta V=Impulse (Change in momentum of a body)

A mass-less body with a change in velocity would still yield zero impulse.

Do you read this differently? I may have certainly misunderstood Dr. Dave's equation, but he used impulse convention.

 

[Corwyn_8]

Jeezus. This one is the shaft/tip and ball collision. Are you talking about something else?

Yes we are. If the thought experiment doesn't intrigue you, just ignore it, it has already been resolved.

Thank you kindly.

 

[Cornerman]

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.

It's not a free body, but that's why we consider wave analysis: so that the masses in question can be understood.

It's a good question, and one that's got good answers. The current transverse wave theory absolutely considers the support structure, specifically the grip of the cue. An overly rigid group will yield different results. But, a human with skin can't grip any tighter (significantly) than the pliable skin is capable. So how the skin affects the overall collision is important. That's why we have high speed video of the tip ball contact, so that we can just use that empirically-observed time to then do equations on the rest of the collision.

Rigid supports like solid blocks and overly tight grips will affect squirt, and the transverse wave model is in line with that. The reality is that human hands and skin, when taken "as a whole" in the equation, results in an insignificant contribution to the squirt analysis. Rigid supports can have a significant contribution (Bob Meucci's Myth Destroyer Robot for example).

IF anyone doesn't think that the analysis hasn't done a good job at viewing the colliision in whole, you need to read closer. Lots of great information in there.

 

[ENGLISH!]

Jeezus. This one is the shaft/tip and ball collision. Are you talking about something else?

There is one basic equation. Conservation of momentum. Use (john's) his theoretical mass-lass shaft. Done.

Let's not try to confuse this just for confusion sake. They all follow the same equation. The Zero mass immediately goes to zero.

Even if you add a flex impulse piece, that impulse has a mass component.

Hi Freddie,

Even Gravity is the attraction of bodies of MASS one to the other.

No mass equals no gravitational pull.

Best 2 Ya,
Rick

PS It seems that confusion is desirable by some. It's seems we may soon be moving the ball by a different temperature of the air near a certain point of the ball relative to the surrounding air. Does not even atmospheric air molecules & atoms have MASS?

 

[Cornerman]

Yes we are. If the thought experiment doesn't intrigue you, just ignore it, it has already been resolved.

Thank you kindly.

Resolve? Really? Maybe I should just talk to John privately.

 

[LAMas]

For those that didn't take physics:

Cornerman suggested an "I" beam shaped shaft at the front of the shaft rather than a hourglass shape.

Imagine a wooden yardstick with a tip on the end. With the "I" pointed up or perpendicular to the slate, the stick could just glance off of the CB because of its flexibility.

Now turn the stick 90 degrees and parallel to the slate and you can move the CB forward with squirt.

In both cases, the mass id the same.

Be well

 

[Cornerman]

For those that didn't take physics:

Cornerman suggested an "I" beam shaped shaft at the front of the shaft rather than a hourglass shape.

Imagine a wooden yardstick with a tip on the end. With the "I" pointed up or perpendicular to the slate, the stick could just glance off of the CB because of its flexibility.

Now turn the stick 90 degrees and parallel to the slate and you can move the CB forward with squirt.

In both cases, the mass id the same.

Be well

seems like something we could actually try.

 

[ENGLISH!]

For those that didn't take physics:

Cornerman suggested an "I" beam shaped shaft at the front of the shaft rather than a hourglass shape.

Imagine a wooden yardstick with a tip on the end. With the "I" pointed up or perpendicular to the slate, the stick could just glance off of the CB because of its flexibility.

Now turn the stick 90 degrees and parallel to the slate and you can move the CB forward with squirt.

In both cases, the mass id the same.

Be well

Hi E,

That is what I suggested with my middle lamination of the the Black Dot shaft.

The same 'end mass' with one rigid application & one flexible application.

There would be two different effective end masses with the same static mass.

All that would need be done is to apply the same velocity.

Like Freddie has been saying though, perhaps a different material other than wood might be in order.

Meucci is using what I think is a patented plastic in his Power Piston Butts.

Be & Stay Well,
Rick

PS I went to the hall the other day but I did not bring a washer & my cue has a steel pin & not a G10. Maybe a flexible pin & two of those coned shapes small end to small end would be an interesting test.:wink:

 

[dr_dave]

Sorry I missed out on all of the excitement this morning. I was out playing pool the last 4 hours. I'll read through all of the posts I missed and offer some replies.

I'll also work on a new diagram that might help explain things better in terms of how both endmass and flex contribute to squirt (CB deflection).

Regards,
Dave

 

[JoeyA]

I have a DymondWood playing cue which is VERY STIFF and it has some of the lowest cue ball squirt/deflection characteristics of any cue.

So from my personal experience in having many cues built to my specifications, I believe that the stiffness of the shaft doesn't have any significant effect on cue ball deflection/squirt.

JoeyA

 

[john coloccia]

I have a DymondWood playing cue which is VERY STIFF and it has some of the lowest cue ball squirt/deflection characteristics of any cue.

So from my personal experience in having many cues built to my specifications, I believe that the stiffness of the shaft doesn't have any significant effect on cue ball deflection/squirt.

JoeyA

Without knowing much about any of the research here, my initial guess would have been that there wouldn't be an obvious correlation between shaft stiffness and squirt, but that there would be some change if you were able to control for just one variable.

 

[dr_dave]

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

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).

Agreed. Total lateral momentum must be conserved. The sideways momentum picked up by the CB (causing squirt) must be balanced by momentum transferred to the cue. As we just discussed on the phone, and as my new diagram will show (I will create and post this soon), because there are transverse shear forces that act between the endmass and the remainder of the cue (due to cue flex), some momentum is transferred to the remainder of the cue. Therefore, the total impulse acting between the CB and tip goes into two effects: endmass momentum and flex impulse which acts equal and opposite between the endmass and the remainder of the cue.

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.

I think my explanation above (along with the new diagram I will post and explain) provides the complete picture explaining how the total sideways impulse between the CB and tip does not go entirely into endmass momentum.

More to come later,
Dave

 

[dr_dave]

Are you saying that if you have a shaft with no mass that is perfectly stiff, you will get no squirt ever?

If the shaft and butt had no mass, then not only would there be no squirt, the CB would not even move when struck with the cue (with either a center-ball or off-center hit)!!! To impart sideways momentum (squirt) to the CB, this momentum must be balanced by equal and opposite sideways momentum in the shaft and cue (for momentum to be conserved), and momentum (mass x velocity) requires mass.

Now, if the shaft were very stiff and had little or no mass, but the butt did have mass (so the CB could actually move forward), the transverse wave would travel through the shaft very fast and involve mass of the butt in the "effective endmass." In this case, there would be squirt.

It seems like that should be wrong.

Bizarre theoretical examples like this often seem wrong because they are not practically possible, so they sometimes go against our intuitive understanding.

Regards,
Dave

 

[dr_dave]

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.

As Freddie has pointed out, the transverse stiffness of all common wood shafts spans over a very small range. Therefore, stiffness differences in wood shafts really don't affect endmass that much.

You are right that mass closer to the tip contributes to effective endmass (and squirt) more than mass farther from the tip. The experiment documented in Diagram 4 of the following article illustrates this fact very clearly:

"Squirt - Part VII: cue test machine results" (BD, February, 2008)

Concerning the possible effects of the bridge on squirt: The bridge could have an effect only if it were very short (less than 6-8 inches, with less effect at larger lengths) and only if the bridge fingers were very bony and had an extremely tight (i.e., non human) grip around the cue.

Even if the bridge were perfectly rigid, it would have absolutely no effect for bridge lengths beyond about 6-8 inches. The following video (at the 2:32 point point in part 2) shows and explains why visually: NV B.96 - Grip and bridge technique and advice.

Regards,
Dave

 

[dr_dave]

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.

If there were no mass, there would also be no flex. All terms in the equation would be zero.

Regards,
Dave

 

[dr_dave]

For those that didn't take physics:

Cornerman suggested an "I" beam shaped shaft at the front of the shaft rather than a hourglass shape.

Imagine a wooden yardstick with a tip on the end. With the "I" pointed up or perpendicular to the slate, the stick could just glance off of the CB because of its flexibility.

Now turn the stick 90 degrees and parallel to the slate and you can move the CB forward with squirt.

In both cases, the mass is the same.

The effective endmass would be very different in those two situations because endmass is affected by the transverse stiffness of the "shaft."

The squirt (CB deflection) would be very different in those two cases mostly due to the effective endmass difference, but also due to cue-flex stiffness effects (which would be stronger in this extreme example as compared to a typical pool cue).

Regards,
Dave