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View Full Version : Cue Stiffness? (Technical)

Jal
01-21-2006, 11:03 PM
The model for squirt that's most often cited uses the concept of "endmass" at the tip portion of the shaft. The ball must push it aside as it begins to rotate, and the tip pushes the ball back with an equal amount of force. One reason cited that the entire mass of the cue is not involved is the relatively slow propagation of the transverse (shear) wave down the shaft.

But the cue also acts like a cantilever beam, providing some spring force resistance. If you look up "cantilever beam stiffness" on Google, you'll get endless references to the subject of Atomic Force Microscopy. Here can be found the formula:

k = (3/4)E[r^4/L^3]

where E is Young's modulus, r is the radius of the rod, and L its length. This is for a rod fixed at one end and of uniform radius, and for the equilibrium case.

While I think a cue can be treated as if it is fixed at one end during the brief contact period, and the formula can be modified for taper, it's not clear what its effective length is in relation to the distance the shear wave has traveled down the shaft.

Without a background in continuum physics, it's a tough nut to crack, so I'm looking for a relatively easy way out (although an extended discussion would be most welcome too).

If you're not well versed but have some thoughts, I'd be interested in hearing them. They cannot be any more bizarre than the ones I've entertained (such as thinking its effective length is the distance the shear wave has traveled).

Jim

Cornerman
01-22-2006, 06:29 AM
<blockquote><font class="small">Quote Jal:</font><hr> The model for squirt that's most often cited uses the concept of "endmass" at the tip portion of the shaft. The ball must push it aside as it begins to rotate, and the tip pushes the ball back with an equal amount of force. One reason cited that the entire mass of the cue is not involved is the relatively slow propagation of the transverse (shear) wave down the shaft.<hr /></blockquote> Right. I think... that's it's material dependent, the slow transverse wave propagation. So, the length of mass involved is determined by the contact time x speed of tranverse wave propagation.

That's been my working theory. The value ends up being from 3-6" using the measured values from the Jacksonville project.

If the speed of propagation can actually be changed, then I would think that a mechanical engineer involved in acoustic and vibration should be able to tell us what are the parameters that can change this propagation speed. Similar to tuning a tuning fork, I would think.

[ QUOTE ]
But the cue also acts like a cantilever beam, providing some spring force resistance. ...
While I think a cue can be treated as if it is fixed at one end during the brief contact period,<hr /></blockquote>I don't think this a good assumption. It was the fact that the robot was initially causing a fixed end point by its too-rigid grip that caused a squirt haywire.

I think the cue has to be treated as a free moving body with no fixed points. The beam in question is moving, so there is a momentum issue to consider. That is, this isn't a static equation. It's dynamics.

Fred

SpiderMan
01-22-2006, 09:02 PM
Jal,

I believe the key to accepting the "endmass" explanation is considering the extremely short dwell time of the tip on the ball. If the acceleration must occur in only a couple of milliseconds, the corresponding force could be much greater than the force you'd measure just by bending the stick. So, even though the entire beam is somewhat stiffly joined, the lion's share of resistance to sideways motion may still be due to inertia of the localized "endmass".

SpiderMan

cushioncrawler
01-22-2006, 10:28 PM
G'day jal.
If its bizarre u want............
I'm not sure where u (we) want to go, or why, but while i am impressed with much of the klever &amp; novel thinking that i kum akross re squirt, i tend to disagree with som.

1...... Re effektiv length of cue -- if some one made a flexible joint, then L would be fixed. This might make som of the theory redundant (not important) -- but shood reduce squirt allso (not important either).
2.......Re reducing squirt -- i dont see this az being helpfull either. I reckon that konsistency iz paramount -- there iznt much konsistency in having a flexible lightweight cue (tip) that sometimez pushez (squirts) the ball ok -- but next shot bendz like mad &amp; puts on heaps of side etc, &amp; the qball brakes &amp; swervz like mad &amp; allmost missez the objekt ball kompletely (at long range). Here i am mainly talking about when u hit the qball below the equator (in addition to the sidespin).
3.......Re break cuez -- not having ever seen one, much less uzed one -- surely a big fat cue &amp; tip, with a big flat tip, would be best (&amp; az light or heavy az u like). This would be very forgiving if u suffer from misshitting the qball off center i think.
4.......The same advice might help for one'z main cue. I think a big flat tip would help beginnerz a lot, &amp; good playerz even more so. Som of the threadz in the forum (eg where duz the qball go???) are i think due to thin whippy cuez &amp; nickel shaped tips. Kontrol beats world rekord spin. Try skrew shots etc with a big flat tip -- u will be amazed at the power &amp; kontrol -- &amp; just for fun, play about 6 skrew shots without chalking up (hey, what happened to misscuez????)
5......When i see an interesting cue, i hold it near the butt (az uzual) &amp; i bang it against my other hand (about a foot from the tip) -- the vibrational frequency tellz me how it playz. Perhaps this leedz to an equation (but i hate equationz).
6....... When i buy a new cue, i hollow out the butt (to looz mass), &amp; i add mass nearer the join. I know that this kontradikts Shepard -- but i suspekt that this reducez squirt. No, this duznt kontradikt what i sed earlyr -- i reckon it iz ok to reduce squirt, but not by taking the strength out of the cue.
7.......Anyhow, i reckon that every cue in the world iz better than every other cue, for some sort of shot, somewhere, sometime.
.......Good luck with yor quest &amp; equationz.

dr_dave
01-23-2006, 09:35 AM
I agree with Spiderman 100% on this one.

The cue tip impact forces acting over the extremely brief contact time are huge compared to any elastic bending forces. The sideways (squirt) impact forces can be provided only by the inertial "resistance" of the end mass of the shaft.

Regards,
Dave

<blockquote><font class="small">Quote SpiderMan:</font><hr> Jal,

I believe the key to accepting the "endmass" explanation is considering the extremely short dwell time of the tip on the ball. If the acceleration must occur in only a couple of milliseconds, the corresponding force could be much greater than the force you'd measure just by bending the stick. So, even though the entire beam is somewhat stiffly joined, the lion's share of resistance to sideways motion may still be due to inertia of the localized "endmass".

SpiderMan <hr /></blockquote>

dr_dave
01-23-2006, 09:39 AM
Jim,

FYI, there has been some discussion on this and related topics in the past. You might find some of it interesting and/or insightful. Links can be found under "cue stick" in the threads summary area of my website (http://www.engr.colostate.edu/~dga/pool/threads.html).

Regards,
Dave

<blockquote><font class="small">Quote Jal:</font><hr> The model for squirt that's most often cited uses the concept of "endmass" at the tip portion of the shaft. The ball must push it aside as it begins to rotate, and the tip pushes the ball back with an equal amount of force. One reason cited that the entire mass of the cue is not involved is the relatively slow propagation of the transverse (shear) wave down the shaft.

But the cue also acts like a cantilever beam, providing some spring force resistance. If you look up "cantilever beam stiffness" on Google, you'll get endless references to the subject of Atomic Force Microscopy. Here can be found the formula:

k = (3/4)E[r^4/L^3]

where E is Young's modulus, r is the radius of the rod, and L its length. This is for a rod fixed at one end and of uniform radius, and for the equilibrium case.

While I think a cue can be treated as if it is fixed at one end during the brief contact period, and the formula can be modified for taper, it's not clear what its effective length is in relation to the distance the shear wave has traveled down the shaft.

Without a background in continuum physics, it's a tough nut to crack, so I'm looking for a relatively easy way out (although an extended discussion would be most welcome too).

If you're not well versed but have some thoughts, I'd be interested in hearing them. They cannot be any more bizarre than the ones I've entertained (such as thinking its effective length is the distance the shear wave has traveled).

Jim<hr /></blockquote>

Deeman3
01-23-2006, 09:41 AM
As a mechanical engineer and pool player, I really think you guys need to get out more, date a very troubled women, experiment with a few class "A" narcodics, exceed the legal speed limits on occasion and dump those pocket protectors... /ccboard/images/graemlins/cool.gif

Deeman

dr_dave
01-23-2006, 09:49 AM
<blockquote><font class="small">Quote Deeman3:</font><hr> As a mechanical engineer and pool player, I really think you guys need to get out more, date a very troubled women, experiment with a few class "A" narcodics, exceed the legal speed limits on occasion and dump those pocket protectors... /ccboard/images/graemlins/cool.gif<hr /></blockquote>

Deeman,

FYI, I just came off a relationship with a "troubled woman," and I think I need a break. I'm not into narcotics (class A or otherwise), but beer has been a good friend lately (in Fort Collins, we have the highest per capita of microbreweries than any place in the country ... we like our beer in Colorado). Our speed limits are also quite high Colorado (75 mph on the interstates, even through most towns), and we usually drive much faster than that, so I don't feel much of a need for speed. I'm also sorry to report that I've never once worn a pocket protector; although, maybe that's the "chick magnet" I need to catch a "rebound relationship" woman.

Sorry to put a wet blanket on all of your advice,
Dave /ccboard/images/graemlins/wink.gif

dr_dave
01-23-2006, 09:59 AM
Jal,

To get a better feel for the stiffness effect, you might want to check out some of the high-speed video clips I have posted. Two good examples are HSV A.5 (http://www.engr.colostate.edu/~dga/pool/high_speed_videos/new/HSVA-5.htm) and HSV A.16 (http://www.engr.colostate.edu/~dga/pool/high_speed_videos/new/HSVA-16.htm). The cue stick does not deflect much during cue tip contact (although, it does deflect quite a bit after contact as the cue stick vibrates). Therefore, the sideways elastic forces during contact are extremely small (and probably negligible).

What do you think?
Dave

<blockquote><font class="small">Quote Jal:</font><hr> The model for squirt that's most often cited uses the concept of "endmass" at the tip portion of the shaft. The ball must push it aside as it begins to rotate, and the tip pushes the ball back with an equal amount of force. One reason cited that the entire mass of the cue is not involved is the relatively slow propagation of the transverse (shear) wave down the shaft.

But the cue also acts like a cantilever beam, providing some spring force resistance. If you look up "cantilever beam stiffness" on Google, you'll get endless references to the subject of Atomic Force Microscopy. Here can be found the formula:

k = (3/4)E[r^4/L^3]

where E is Young's modulus, r is the radius of the rod, and L its length. This is for a rod fixed at one end and of uniform radius, and for the equilibrium case.

While I think a cue can be treated as if it is fixed at one end during the brief contact period, and the formula can be modified for taper, it's not clear what its effective length is in relation to the distance the shear wave has traveled down the shaft.

Without a background in continuum physics, it's a tough nut to crack, so I'm looking for a relatively easy way out (although an extended discussion would be most welcome too).

If you're not well versed but have some thoughts, I'd be interested in hearing them. They cannot be any more bizarre than the ones I've entertained (such as thinking its effective length is the distance the shear wave has traveled).

Jim<hr /></blockquote>

SpiderMan
01-23-2006, 11:28 AM
<blockquote><font class="small">Quote Deeman3:</font><hr> As a mechanical engineer and pool player, I really think you guys need to get out more, date a very troubled women, experiment with a few class "A" narcodics, exceed the legal speed limits on occasion and dump those pocket protectors... /ccboard/images/graemlins/cool.gif
Deeman <hr /></blockquote>

Dave,

I'm not certain, but I think this BBQ-eating Memphis redneck just called us a couple of nerds /ccboard/images/graemlins/grin.gif

SpiderMan

dr_dave
01-23-2006, 11:45 AM
<blockquote><font class="small">Quote SpiderMan:</font><hr> <blockquote><font class="small">Quote Deeman3:</font><hr> As a mechanical engineer and pool player, I really think you guys need to get out more, date a very troubled women, experiment with a few class "A" narcodics, exceed the legal speed limits on occasion and dump those pocket protectors... /ccboard/images/graemlins/cool.gif
Deeman <hr /></blockquote>

Dave,

I'm not certain, but I think this BBQ-eating Memphis redneck just called us a couple of nerds /ccboard/images/graemlins/grin.gif

SpiderMan <hr /></blockquote>
Spiderman,

Very astute observation. Although, I took no offense because I like the D-man. But I think he at least owes us a couple of beers (and maybe some of that great BBQ) if we ever hook up with him again.

Dave

Deeman3
01-23-2006, 11:50 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote SpiderMan:</font><hr> <blockquote><font class="small">Quote Deeman3:</font><hr> As a mechanical engineer and pool player, I really think you guys need to get out more, date a very troubled women, experiment with a few class "A" narcodics, exceed the legal speed limits on occasion and dump those pocket protectors... /ccboard/images/graemlins/cool.gif
Deeman <hr /></blockquote>

Dave,

I'm not certain, but I think this BBQ-eating Memphis redneck just called us a couple of nerds /ccboard/images/graemlins/grin.gif <font color="blue">Only if you feel uncomfortable about it, or guilty. /ccboard/images/graemlins/grin.gif </font color>

SpiderMan <hr /></blockquote>
Spiderman,

Very astute observation. Although, I took no offense because I like the D-man. <font color="blue"> I like both of you. </font color> But I think he at least owes us a couple of beers (and maybe some of that great BBQ) if we ever hook up with him again. <font color="blue"> I'll stand good for the beers, Spiderman can bring the bar-b-que. Don't worry, once SPetty gets a group of us together, we're bonded for life. I'm sure we will draw each other at the first tournament we all attend. /ccboard/images/graemlins/confused.gif

Deeman
I was thinking Dave was giving the peace sign with his 30 degree finger thingy....</font color>

Dave <hr /></blockquote>

dr_dave
01-23-2006, 11:52 AM
Jim and others,

FYI, Ron Sheppard wrote a great paper describing the "end mass" effect. It can be found under "technical articles" in the pool physics section of my website (http://www.engr.colostate.edu/~dga/pool/physics.html). He doesn't get into the elastic wave propagation physics you guys are discussing, but his analysis and results are still insightful and useful.

Regards,
Dave

<blockquote><font class="small">Quote Jal:</font><hr> The model for squirt that's most often cited uses the concept of "endmass" at the tip portion of the shaft. The ball must push it aside as it begins to rotate, and the tip pushes the ball back with an equal amount of force. One reason cited that the entire mass of the cue is not involved is the relatively slow propagation of the transverse (shear) wave down the shaft.

But the cue also acts like a cantilever beam, providing some spring force resistance. If you look up "cantilever beam stiffness" on Google, you'll get endless references to the subject of Atomic Force Microscopy. Here can be found the formula:

k = (3/4)E[r^4/L^3]

where E is Young's modulus, r is the radius of the rod, and L its length. This is for a rod fixed at one end and of uniform radius, and for the equilibrium case.

While I think a cue can be treated as if it is fixed at one end during the brief contact period, and the formula can be modified for taper, it's not clear what its effective length is in relation to the distance the shear wave has traveled down the shaft.

Without a background in continuum physics, it's a tough nut to crack, so I'm looking for a relatively easy way out (although an extended discussion would be most welcome too).

If you're not well versed but have some thoughts, I'd be interested in hearing them. They cannot be any more bizarre than the ones I've entertained (such as thinking its effective length is the distance the shear wave has traveled).

Jim<hr /></blockquote>

dr_dave
01-23-2006, 11:55 AM
I look forward to the next opportunity to hook up with you guys at PettyPoint in Texas. If you are nice (and let me win some games), I might even show you some more of my hand tricks.

Regards,
Dave

<blockquote><font class="small">Quote Deeman3:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote SpiderMan:</font><hr> <blockquote><font class="small">Quote Deeman3:</font><hr> As a mechanical engineer and pool player, I really think you guys need to get out more, date a very troubled women, experiment with a few class "A" narcodics, exceed the legal speed limits on occasion and dump those pocket protectors... /ccboard/images/graemlins/cool.gif
Deeman <hr /></blockquote>

Dave,

I'm not certain, but I think this BBQ-eating Memphis redneck just called us a couple of nerds /ccboard/images/graemlins/grin.gif <font color="blue">Only if you feel uncomfortable about it, or guilty. /ccboard/images/graemlins/grin.gif </font color>

SpiderMan <hr /></blockquote>
Spiderman,

Very astute observation. Although, I took no offense because I like the D-man. <font color="blue"> I like both of you. </font color> But I think he at least owes us a couple of beers (and maybe some of that great BBQ) if we ever hook up with him again. <font color="blue"> I'll stand good for the beers, Spiderman can bring the bar-b-que. Don't worry, once SPetty gets a group of us together, we're bonded for life. I'm sure we will draw each other at the first tournament we all attend. /ccboard/images/graemlins/confused.gif

Deeman
I was thinking Dave was giving the peace sign with his 30 degree finger thingy....</font color>

Dave <hr /></blockquote> <hr /></blockquote>

Jal
01-23-2006, 12:10 PM
Thanks for all of your responses. You've given me a few things to think about.

I've always accepted the endmass explanation, which is a good one since conservation of momentum must be obeyed, and pretty obviously the entire mass of the cue cannot be involved or you would get a whole bunch more squirt.

But, I've been looking at Dr. Dave's videos and from them it appears that the contact time for the observed shots is about 1.3 msec, and that it takes about 3 msec for the tip to cease its sideways motion. On some of the harder shots it appears to be more like 4 msecs. From one sequence of the Australian film (linked to at his site), although no clock is present the corrsponding times can be seen to be about 5.5 frames of contact and 13 frames until the tip stops moving. This is about the same ratio as 1.3 msec / 3 msec.

Using this, and assuming a sine like function for the lateral spring force (Hookes Law), you can figure out how much force the shaft was exerting during contact due to its stiffness alone, as a fraction of the total force exerted on the cueball. The figure I get is 22% (0.22).

Okay, this is not very much but sufficiently borderline so that if there are reasons to believe that it may be greater during contact, the spring force might very well be significant (remember it varies with the fourth power of the shaft's radius while mass varies with the second power). Whether it is or isn't hinges on how its stiffness varies with time (if at all), and how much of the cue as a whole or in part has moved during that 3 msec period. The latter would tend to make the actual time at which the tip reaches its maximum sideways displacement during its sinusoidal movement shorter than it appears in the videos, and the derived spring force greater.

But I think more importantly, a 3 msec quarter period implies a vibration frequency of 1/[4x3] or 83 Hz. Reported frequencies are around 40 Hz, suggesting that a cue's stiffness may decrease after the collision. I haven't done much in trying to measure the after collision frequency, but I did look at the length of the first half-period in HSVA-5 after it completed its first sideways excursion (which took 3-4 msec), and this is indeed about 12 msec, implying a frequency close to 40 Hz.

So the evidence is suggesting (to me) that its stiffness may be a function of time, perhaps due to the shear propagation. It may also be true that most of the squirt is developed during the early part of the contact period, but I haven't worked this out yet and am only going on some idle thoughts here.

Since I'm ignorant of all the gory details that take place when something bends, such as what does the shear force have to do with anything, I'll take another look at Ron Shepard's treatment, but any further ideas or criticisms are certainly welcome (expert, speculator, anyone). A formula for K(t) would be even nicer. /ccboard/images/graemlins/smile.gif

Jim

dr_dave
01-23-2006, 03:24 PM
Jim,

I am impressed with the level of thought you have put into this. I have also put a lot of thought into understanding squirt in the past. But after a while, I decided that it wasn't worth it (to me personally). To fully understand and model all of the important impact, friction, shock wave, and cue tip physics would take a monumental effort (experiments to characterize material properties [especially cue tip deformation and friction], finite element analyses to accurately model all of the dynamic effects and material interactions, etc.). To me, there is enough evidence and understanding to support the important conclusion that "end mass" is the most important predictor of squirt. Beyond that, I'm not sure what additional conclusions a more complete analysis would yield. Do you have any thoughts (or hopes) on this?

I think cue tip deformation and friction might be important, and maybe some work in this area could yield some useful results and conclusions (e.g., alternative cue tip and ferrule materials and geometries that result in less squirt), but I'm not sure.

Concerning trying to predict the sideways force on the cue stick during tip contact based on the maximum deflection amplitude well after impact, I think there are potential problems with making such conclusions given some of the complexities at work (e.g., the shock waves through the cue stick), but I could be wrong.

Let us know if you come up with some more squirt insights. I think squirt is still a very interested effect for which the true fundamentals are not totally understood.

Regards,
Dave

<blockquote><font class="small">Quote Jal:</font><hr> Thanks for all of your responses. You've given me a few things to think about.

I've always accepted the endmass explanation, which is a good one since conservation of momentum must be obeyed, and pretty obviously the entire mass of the cue cannot be involved or you would get a whole bunch more squirt.

But, I've been looking at Dr. Dave's videos and from them it appears that the contact time for the observed shots is about 1.3 msec, and that it takes about 3 msec for the tip to cease its sideways motion. On some of the harder shots it appears to be more like 4 msecs. From one sequence of the Australian film (linked to at his site), although no clock is present the corrsponding times can be seen to be about 5.5 frames of contact and 13 frames until the tip stops moving. This is about the same ratio as 1.3 msec / 3 msec.

Using this, and assuming a sine like function for the lateral spring force (Hookes Law), you can figure out how much force the shaft was exerting during contact due to its stiffness alone, as a fraction of the total force exerted on the cueball. The figure I get is 22% (0.22).

Okay, this is not very much but sufficiently borderline so that if there are reasons to believe that it may be greater during contact, the spring force might very well be significant (remember it varies with the fourth power of the shaft's radius while mass varies with the second power). Whether it is or isn't hinges on how its stiffness varies with time (if at all), and how much of the cue as a whole or in part has moved during that 3 msec period. The latter would tend to make the actual time at which the tip reaches its maximum sideways displacement during its sinusoidal movement shorter than it appears in the videos, and the derived spring force greater.

But I think more importantly, a 3 msec quarter period implies a vibration frequency of 1/[4x3] or 83 Hz. Reported frequencies are around 40 Hz, suggesting that a cue's stiffness may decrease after the collision. I haven't done much in trying to measure the after collision frequency, but I did look at the length of the first half-period in HSVA-5 after it completed its first sideways excursion (which took 3-4 msec), and this is indeed about 12 msec, implying a frequency close to 40 Hz.

So the evidence is suggesting (to me) that its stiffness may be a function of time, perhaps due to the shear propagation. It may also be true that most of the squirt is developed during the early part of the contact period, but I haven't worked this out yet and am only going on some idle thoughts here.

Since I'm ignorant of all the gory details that take place when something bends, such as what does the shear force have to do with anything, I'll take another look at Ron Shepard's treatment, but any further ideas or criticisms are certainly welcome (expert, speculator, anyone). A formula for K(t) would be even nicer. /ccboard/images/graemlins/smile.gif

Jim <hr /></blockquote>

Cornerman
01-24-2006, 10:16 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> To me, there is enough evidence and understanding to support the important conclusion that "end mass" is the most important predictor of squirt. Beyond that, I'm not sure what additional conclusions a more complete analysis would yield. <hr /></blockquote>I would suggest to print this out before considering a response.

Nobody is saying that the tip end mass isn't the major contributor. That's far from what is being said.

The conclusions won't mean squat to anyone except someone who might want to be able to change the squirt characteristic of a cue, but doesn't want to bore holes or decouple the ferrule.

The idea of "end mass" is fine for a "what causes squirt" question. And it's become too much Standard Fare Answer to the point that no matter what the question, "tip end mass" is the answer for those who like to answer with SFAs. But, that's not always the answer to the question being asked.

When we as an industry find out what parameters control how much end mass is in effect during the collision, then maybe cuemakers and cue buyers who care about such things will have a better direction on shaft construction such that if you actually want a low squirt shaft, you don't have to give up feel.

I personally would love to have a definitive answer of:

1. at what distance down the shaft starting from the tip is no longer in effect given a .001 contact time, a .002 contact time, and a .0026 contact time and why?

2. what parameters affect that length? And don't say tip end mass. That's not the answer to this question. Tip end mass is the output to the equation. What are the inputs aside from the cueball mass?

3. what paramaters can a cuemaker control?

4. Is it material dependent?

As of today, Clawson/Predator simply bore a hole. I'm sure they tested different lengths of holes, and found a point of no change. That is, they empirically deduced a distance by shooting a cannon with shot and reducing the shot size until they came to stop point. Aside from empirical testing, is there a reasonable equation that gets this distance without shooting cannon?

IMO, it has to involve the speed of the transverse wave propagation down the length of the shaft during contact. If it is, how do we change that speed? If it isn't, what other theories are on the table to describe how much mass is in effect during the tip/ball contact?

Fred

dr_dave
01-24-2006, 01:54 PM
Fred,

Your points are very well taken ... you ask some good questions. I agree with you that the speed of the shock wave is probably directly related to the mass (and stiffness ?) distribution in the shaft. Here are a few more questions that might also be pertinent:

a) How important are cue tip deformation and friction characteristics to squirt?

b) Could different tip materials and treatments result in less squirt?

c) How much does hollowing out a cue affect its long-term structural integrity?

d) How should the mass and stiffness be distributed in the shaft to minimize squirt?

I guess that's all we know for now is that if cue makers make the end of the shaft lighter (and less stiff?), there will be less squirt.

Regards,
Dave

PS: Are you Fred Agnir by chance?

<blockquote><font class="small">Quote Cornerman:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> To me, there is enough evidence and understanding to support the important conclusion that "end mass" is the most important predictor of squirt. Beyond that, I'm not sure what additional conclusions a more complete analysis would yield. <hr /></blockquote>I would suggest to print this out before considering a response.

Nobody is saying that the tip end mass isn't the major contributor. That's far from what is being said.

The conclusions won't mean squat to anyone except someone who might want to be able to change the squirt characteristic of a cue, but doesn't want to bore holes or decouple the ferrule.

The idea of "end mass" is fine for a "what causes squirt" question. And it's become too much Standard Fare Answer to the point that no matter what the question, "tip end mass" is the answer for those who like to answer with SFAs. But, that's not always the answer to the question being asked.

When we as an industry find out what parameters control how much end mass is in effect during the collision, then maybe cuemakers and cue buyers who care about such things will have a better direction on shaft construction such that if you actually want a low squirt shaft, you don't have to give up feel.

I personally would love to have a definitive answer of:

1. at what distance down the shaft starting from the tip is no longer in effect given a .001 contact time, a .002 contact time, and a .0026 contact time and why?

2. what parameters affect that length? And don't say tip end mass. That's not the answer to this question. Tip end mass is the output to the equation. What are the inputs aside from the cueball mass?

3. what paramaters can a cuemaker control?

4. Is it material dependent?

As of today, Clawson/Predator simply bore a hole. I'm sure they tested different lengths of holes, and found a point of no change. That is, they empirically deduced a distance by shooting a cannon with shot and reducing the shot size until they came to stop point. Aside from empirical testing, is there a reasonable equation that gets this distance without shooting cannon?

IMO, it has to involve the speed of the transverse wave propagation down the length of the shaft during contact. If it is, how do we change that speed? If it isn't, what other theories are on the table to describe how much mass is in effect during the tip/ball contact?

Fred<hr /></blockquote>

Cornerman
01-24-2006, 02:48 PM
<blockquote><font class="small">Quote dr_dave:</font><hr>
a) How important are cue tip deformation and friction characteristics to squirt?

b) Could different tip materials and treatments result in less squirt?
<hr /></blockquote> I think these two are at least addressed by theories that have been floated about.

These are my understandings of the theories:

The collision should be considered static, not slipping/sliding. It's not a shear effect like rubber on road.

The compression and decompression differences in soft and hard tips though different and resulting in different contact time doesn't suggest any difference in inelastic collision. I think.

[ QUOTE ]
c) How much does hollowing out a cue affect its long-term structural integrity?<hr /></blockquote> I think this is an important issue, but I don't think it's important to the squirt question. That is, unless it's suggested that a cue starts falling apart, and therefore will have less (or more) squirt as it ... splinters?

[ QUOTE ]
d) How should the mass and stiffness be distributed in the shaft to minimize squirt?<hr /></blockquote> I think this is where the most interest lies, if stiffness or any other parameter is important at all. I don't think stiffness is an important parameter in of itself, but I don't know that for sure. What I'm really interested is how to calculate how much mass will effect the collision.

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PS: Are you Fred Agnir by chance?
<hr /></blockquote>Let me check the note on my skivvies. Yup.

Fred &lt;~~~ and it's Tuesday

dr_dave
01-24-2006, 03:10 PM
<blockquote><font class="small">Quote Cornerman:</font><hr><blockquote><font class="small">Quote dr_dave:</font><hr>PS: Are you Fred Agnir by chance?<hr /></blockquote>
Let me check the note on my skivvies. Yup.<hr /></blockquote>
I was worried that you left the board. I guess you have, as "Fred Agnir." It's nice to have you back as "Cornerman." I know "Fred Agnir" and "dr_dave" didn't always "get along" in the past. I hope I can start off on a better foot with "Cornerman." Maybe I should change my user name to "SidePocketMan."

Regards,
Dave

Cornerman
01-25-2006, 07:45 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> I know "Fred Agnir" and "dr_dave" didn't always "get along" in the past. I hope I can start off on a better foot with "Cornerman." Maybe I should change my user name to "SidePocketMan."<hr /></blockquote>That would be a start.

The name change was long overdue. It stems from the fact that I write the Cue Makers Corner article. I'm not a cue or a maker. I'm just a guy who sets up the corner.

But, the name has enough double or triple entendre to mean whatever anyone wants it to me.

Fred