Squirt. End Mass and Cue Flexibility.

[dr_dave]

FYI, to those interested, I did some measurement and calculations. The analysis is here:

TP B.19 - Comparison of cue ball deflection (squirt) "endmass" and stiffness effects

For the example numbers I used, the direct stiffness effect is only about 1.5% of the total cue ball deflection (squirt) effect. That is indeed very small.

The effect of stiffness on squirt appears to be even smaller than I thought - inconsequential at the level you suggest.

I honestly thought it would be even smaller, but it is still extremely small in relation to what many people seem to think causes squirt.

Again, it is the acceleration of (or momentum transfer to) the "endmass" that causes the vast majority of squirt ... not the force required to flex the shaft.

Thanks for keeping us as real as possible under the circumstances.

I aim to squerve.

P.S. Does 1.5% include the increase of involved end mass attributable to stiffness?

No. The analysis just looks at the independent effect of flex force (due to shaft stiffness) as a percentage of the total squirt force.

Regards,
Dave

 

[dr_dave]

For me, I never understood ball spin until I got that there is only one axis (and the torque acted at right angles to applied force). What really did it for me was trying figure out what a natural rolling left english ball would do after hitting cushion. Now 'top' isn't top any longer. What is it? Can you quickly intuit what the new direction of travel will do to the current spin? It is pretty easy if you consider the existing spin axis and the new force applying torque perpendicular to the new (friction) force. For me anyway.

Have you watched the following video yet?

NV B.10 - Drag spin loss and sidespin persistence, with spin-axis "flip"

People in physics and engineering often refer to the individual spin components about different axes. For example, when you hit a shot with top-right english, the CB has both topspin about the horizontal axis (which makes the ball roll forward) and sidespin about the vertical axis (which makes the ball spin like a top). The net effect of the combined rotations can be described by an effective axis of rotation (as demonstrated in the video) which results from a vector sum of the two individual components. This axis can change direction during motion (as demonstrated in the video) as the individual component of spin change independently.

Both ways of describing, visualizing, and analyzing the rotation (either with separate spin components, or with a combined-spin effective axis) are valid and useful in different contexts.

I hope that helps,
Dave

 

[dr_dave]

Rod Cross, Physics Department, University of Sydney invested a lot of time resources and science to conclude that a thinner shaft reduces squirt. I hope that he is informed that his study was not for naught and contributes 1.5% according to your calculation. Even scientist can be punked.

It is true that a "thinner shaft reduces squirt," and for several reasons. The main reason is that the actual shaft mass close to the tip will be much less (unless the tip is very heavy). Also, the transverse stiffness will be less. This creates two important effects. This will result in less sideways force as the shaft flexes (as the tip gets pushed sideways slightly as it rides the CB during the incredibly brief tip contact time). This is the effect that I show in my TP B.19 analysis to be extremely small. The other stiffness-related effect is that with a less stiff shaft, the speed of the elastic wave that travels down the shaft during tip contact will be slower, and this will reduce the "effective endmass" of the shaft. For those interested, these effects are explained and demonstrated in more detail on the endmass and stiffness resource page.

Regards,
Dave

 

[LAMas]

It is true that a "thinner shaft reduces squirt," and for several reasons. The main reason is that the actual shaft mass close to the tip will be much less (unless the tip is very heavy). Also, the transverse stiffness will be less. This creates two important effects. This will result in less sideways force as the shaft flexes (as the tip gets pushed sideways slightly as it rides the CB during the incredibly brief tip contact time). This is the effect that I show in my TP B.19 analysis to be extremely small. The other stiffness-related effect is that with a less stiff shaft, the speed of the elastic wave that travels down the shaft during tip contact will be slower, and this will reduce the "effective endmass" of the shaft. For those interested, these effects are explained and demonstrated in more detail on the endmass and stiffness resource page.

Regards,
Dave

Rod Cross is wrong?

Cue and ball deflection (or “squirt”) in billiards
Rod Cross
Physics Department of Sydney, Sydney NSW 2006, Australia

VII Conclusion

....The experimental data shows that elasticity of the cue tip plays a dominant role in the collision process and suggests that cues with thin shafts might generate lower squirt angles as a result of their greater flexibility rather than their lower mass...

Acknowledgements
This article was inspired by previous theoretical efforts by Ray Higley and by Professors David Alciatore and Ron Shepard, concerning the origin of squirt in billiards,

http://www.physics.usyd.edu.au/~cros.... squirt.pdf

 

[dr_dave]

From the OP by Rod Cross:

III. EXPERIMENTAL METHODS
When a ball is struck with side spin on a billiard table, the amount of sidespin is difficult to measure experimentally because the ball initially slides and then rolls forward. The ball therefore rotates about two different axes simultaneously.

I know some people might have an issue with this statement (as pool players); but to a physicist or engineering, the statement is perfectly fine. A ball struck with pure sidespin initially rotates about the vertical axis only; but as drag causes the CB to gradually pick up topspin, the CB now has two components: sidespin rotation about the vertical axis and topspin rotation about the horizontal axis. These rotations, when combined, are equivalent to a single rotation about a different axis, but both descriptions are valid. When doing analysis, it is actually helpful and easier to deal with the rotations as separate components; but when visualizing the actual ball motion, it helps to understand the combined-rotation axis.

Regards,
Dave

 

[dr_dave]

Dr Dave
Wouldn't it depend on the end mass vs flexibility?

I think I just answered this in a recent post. Check it out.

What if you had a 14 mm tip with brass ferrule and the last 10 inches shaped like an hour glass with the mid point 9 mm?
Verses same 14 mm shaft with pro taper?

I think the experiments I did in the past answer these questions. See the following articles that document the results of the experiments:

"Squirt - Part VII: cue test machine results" (BD, February, 2008)
"Return of the squirt robot" (BD, August, 2008)

And for more info and additional supporting resources, see the endmass and stiffness resource page.

Here's the short answer, in case you don't want to do the "homework:"

If you make the tip end of a shaft (the last 5-8 inches) lighter, squirt is reduced. And if you change the mass or stiffness of the shaft beyond 5-8 inches, it has absolutely no effect on squirt.

Now, one must be careful to not confuse "squirt" with "the combined effects of squirt and swerve" (AKA "net CB deflection" or "squerve"). Swerve is affected by many factors including the speed of the stroke, the weight of the cue, the efficiency of the tip, the elevation of the cue, the properties of the ball and cloth, etc. Swerve (and not squirt) is what really makes aiming with sidespin over a wide range of shots challenging (and fun).

Regards,
Dave

 

[Corwyn_8]

Have you watched the following video yet?

NV B.10 - Drag spin loss and sidespin persistence, with spin-axis "flip"

People in physics and engineering often refer to the individual spin components about different axes. For example, when you hit a shot with top-right english, the CB has both topspin about the horizontal axis (which makes the ball roll forward) and sidespin about the vertical axis (which makes the ball spin like a top). The net effect of the combined rotations can be described by an effective axis of rotation (as demonstrated in the video) which results from a vector sum of the two individual components. This axis can change direction during motion (as demonstrated in the video) as the individual component of spin change independently.

Both ways of describing, visualizing, and analyzing the rotation (either with separate spin components, or with a combined-spin effective axis) are valid and useful in different contexts.

I have seen the video, in fact, I looked for it (badly) to illustrate this discussion.

The key, for me, to understanding what is going on, is shown in that video. The axis of spin, in the balls reference frame, never changes. The mark you added is always on the axis of spin, the tilt of the axis is what changes. It is not the case that the side spin axis is rotating about another axis (in my mental picture of that description, anyway) Perplexingly, to me, others find it other ways of looking at more intuitive, such is learning.

Thank you kindly.

 

[dr_dave]

I doubt in about 5 hours or less that Dr. Dave could compared every type of shaft taper & all of the various degrees of flex to all of the various amounts of front end masses to arrive at his definitive figure

You are correct, but you obviously missed the entire point of the analysis.

Many people seem to think squirt (CB deflection) is caused mostly by force resulting from flexing the shaft while the tip is in contact with the CB. This is what I have convincingly proved wrong. I did this only for a single shaft and for a shot of one speed, but all of the important values in the analysis would be of the same order of magnitude for other equipment and shots. In other words, the final result will be very similar for any typical pool equipment. Again, it ain't force from shaft flex that is responsible for the majority of squirt.

a 1.5% variation can mean the difference between a make or a miss

I don't think you understand the implications of the 1.5% result.

Think for a minute what the numbers in the analysis imply. The 1.5% applies to the 1.8 degrees of total squirt, so the amount of squirt due to the flex-force effect is a whopping 0.027 degrees (1.5% of 1.8 degrees)!!! Do you realize how small this is? No human could possibly detect or create an angle change of that magnitude.

And even if a cue had 10-times as much squirt as the example in my analysis, the flex-force effect would probably still be too small to distinguish (not to mention that the cue would be useless at a pool table).

Regards,
Dave

PS: Please don't take any offense with my post. If I seem defensive or frustrated, it is because you often wrongly-extrapolate results, misinterpret conclusions, or use misleading analogies in reference to posts that are carefully written and backed up by solid principles, experimental results, and video demonstrations (which took a lot of time and effort to create).

 

[dr_dave]

Dr. Dave, in view of the conservation of momentum (transverse), I don't understand the dichotomy between stiffness and endmass. In other words, I humbly suggest that all of the lateral force on the ball has to be attributed to accelerating mass. (The endmass itself, though, is a function of stiffness and mass.)

Jim,

I think I've answered this in recent posts. If I haven't, please let me know.

Best regards,
Dave

 

[dr_dave]

Is there an easy way to retain multiple and/or nested quotes?

I do it by clicking on the link to the previous post, copying what I want from a quote of that post, then returning to the future post and pasting the quote (or quotes) into my new quoted reply. It isn't easy at first, but now I can do it with ease. I hope that makes sense.

Regards,
Dave

 

[Cornerman]

For all these years, It has been held here that only end mass determines squirt

If you would like to talk privately, I'm good with that, but NOBODY with any scientific sense says the above quote.

I (and others of course) say what causes squirt, but in the end, reducing mass at the end of the shaft is by far the easiest and most prevalent way to reduce squirt

But in the end, there is a lateral force in play, and that force is dependent on th amount of shaft involved while the tip and ball is in contact. It's the transverse wave concept, which is the deterring factor of squirt. That's been the truth for a long time. It's most definitely a materials thing. But, the math equation is about mass.

Pm me if you want.

 

[dr_dave]

It is true that a "thinner shaft reduces squirt," and for several reasons. The main reason is that the actual shaft mass close to the tip will be much less (unless the tip is very heavy). Also, the transverse stiffness will be less. This creates two important effects. This will result in less sideways force as the shaft flexes (as the tip gets pushed sideways slightly as it rides the CB during the incredibly brief tip contact time). This is the effect that I show in my TP B.19 analysis to be extremely small. The other stiffness-related effect is that with a less stiff shaft, the speed of the elastic wave that travels down the shaft during tip contact will be slower, and this will reduce the "effective endmass" of the shaft. For those interested, these effects are explained and demonstrated in more detail on the endmass and stiffness resource page.

Rod Cross is wrong?

Cue and ball deflection (or “squirt”) in billiards
Rod Cross
Physics Department of Sydney, Sydney NSW 2006, Australia

VII Conclusion

....The experimental data shows that elasticity of the cue tip plays a dominant role in the collision process and suggests that cues with thin shafts might generate lower squirt angles as a result of their greater flexibility rather than their lower mass...

Acknowledgements
This article was inspired by previous theoretical efforts by Ray Higley and by Professors David Alciatore and Ron Shepard, concerning the origin of squirt in billiards,

http://www.physics.usyd.edu.au/~cros.... squirt.pdf

It has been a long time since I carefully and completely read that article. I will read it again soon and comment if my opinion changes. However, I do think the statement in his conclusion might be inaccurate (i.e., not totally correct).

Regards,
Dave

 

[dr_dave]

I (and others of course) say what causes squirt, but in the end, reducing mass at the end of the shaft is by far the easiest and most prevalent way to reduce squirt.

Agreed. Here's a pertinent quote from the endmass resource page:

The characteristic that makes an LD shaft have less squirt (cue ball deflection) is reduced "endmass." See Diagram 4 in "Squirt - Part VII: cue test machine results" (BD, February, 2008). People who think extra stiffness is required to produce more squirt are incorrect. Added endmass alone (without added stiffness) produces significant increases in squirt. This supports the theory in TP A.31. The squirt of a shaft can be lowered by reducing the weight of the last 5-8 inches. This can be done by reducing the shaft's diameter, drilling out the core of the end of the shaft, using a lighter and/or harder tip (for more info, see cue tip hardness effects), and/or using a lighter (or no) ferrule. As demonstrated with the experiment in the article, mass closer to the tip has a greater effect on "effective endmass" than mass farther from the tip because it is moving more during tip contact (see what causes squirt), and beyond a certain distance, added mass has no effect at all.

Regards,
Dave

 

[Jal]

How could it not? If the combined spin axis is a combination of two components, how could changing one of those components NOT affect the combination?

Precession is NOT associated only with fixed spin rate. See the definition. The Earth precesses, its spin rate is not fixed. A top precesses, its spin rate is not fixed. Check out the video I linked. The spin of the bicycle wheel is slowing due to friction, and the torque from hanging from only one side is causing precession.

Thank you kindly.

Corwyn, after my posts challenging your use of the term "precession," I had second thoughts. I think the term is applicable, though I might do some reading up later.

My apology for jumping the gun.

Jim

 

[Jal]

Jim,

I think I've answered this in recent posts. If I haven't, please let me know.

Hi Dr. Dave,

Sorry if I missed it, but a perusal of your posts didn't reveal an answer.

The point I was trying to make is that squirt has to be totally attributed to the endmass concept and not to any force arising directly from the static resistance of the shaft to bending, however small. Otherwise, in the lateral direction we would have:

MballVball = -MendmassVendmass + FT

where F is the force arising from the resistance to bending and T is the contact time.

Thus, the conservation of momentum in the transverse direction would be violated (MballVball + MendmassVendmass would not be equal to zero.)

Again my apology if I'm not getting the gist of your analysis.

Jim

 

[dr_dave]

Hi Dr. Dave,

Sorry if I missed it, but a perusal of your posts didn't reveal an answer.

The point I was trying to make is that squirt has to be totally attributed to the endmass concept and not to any force arising directly from the static resistance of the shaft to bending, however small. Otherwise, in the lateral direction we would have:

MballVball = -MendmassVendmass + FT

where F is the force arising from the resistance to bending and T is the contact time.

Thus, the conservation of momentum in the transverse direction would be violated (MballVball + MendmassVendmass would not be equal to zero.)

Again my apology if I'm not getting the gist of your analysis.

Jim

Jim,

The forward impulse on the CB is:

F_imp = m_ball * v_fwd

where v_fwd is the forward speed of the CB.

For a given squirt angle "a," the sideways impulse acting on the CB, which acts equal and opposite on the tip, is:

S_imp = F_imp * tan(a) = m_ball * v_side

where v_side is the sideways speed of the CB.

The "effective endmass" feels this impulse, but it also feels an impulse resulting from the force resisting shaft bending because the flexed shaft is pushing back on the endmass, so I think the proper momentum equation for the endmass is:

S_imp - F_flex*T = m_end*v_end

where F_flex is the average force associated with the flex of the shaft and T is the tip contact time.

In my analysis, I'm comparing the peak force involved with S_imp to the peak force involved with F_flex.

There is a little "smoke and mirrors" going on here due to the awkward definition of "effective endmass" and because the flex force is actually a dynamic and distributed force acting along the "endmass." Also, my static measurement of flex force and deflection doesn't perfectly model the shaft flex involved with the sideways endmass motion, but I think the analysis is reasonably sound to get good ball-park numbers.

Please let me know if you still have concerns or think I am grossly misrepresenting something.

Thanks,
Dave

PS: We should probably continue this discussion privately since few if any users reading this thread probably aren't interested in less level of math and physics details.

 

[Patrick Johnson]

Hi Dr. Dave,

Sorry if I missed it, but a perusal of your posts didn't reveal an answer.

The point I was trying to make is that squirt has to be totally attributed to the endmass concept and not to any force arising directly from the static resistance of the shaft to bending, however small. Otherwise, in the lateral direction we would have:

MballVball = -MendmassVendmass + FT

where F is the force arising from the resistance to bending and T is the contact time.

Thus, the conservation of momentum in the transverse direction would be violated (MballVball + MendmassVendmass would not be equal to zero.)

Again my apology if I'm not getting the gist of your analysis.

Jim

Jim,

The forward impulse on the CB is:

F_imp = m_ball * v_fwd

where v_fwd is the forward speed of the CB.

For a given squirt angle "a," the sideways impulse acting on the CB, which acts equal and opposite on the tip, is:

S_imp = F_imp * tan(a) = m_ball * v_side

where v_side is the sideways speed of the CB.

The "effective endmass" feels this impulse, but it also feels an impulse resulting from the force resisting shaft bending because the flexed shaft is pushing back on the endmass, so I think the proper momentum equation for the endmass is:

S_imp - F_flex*T = m_end*v_end

where F_flex is the average force associated with the flex of the shaft and T is the tip contact time.

In my analysis, I'm comparing the peak force involved with S_imp to the peak force involved with F_flex.

There is a little "smoke and mirrors" going on here due to the awkward definition of "effective endmass" and because the flex force is actually a dynamic and distributed force acting along the "endmass." Also, my static measurement of flex force and deflection doesn't perfectly model the shaft flex involved with the sideways endmass motion, but I think the analysis is reasonably sound to get good ball-park numbers.

Please let me know if you still have concerns or think I am grossly misrepresenting something.

Thanks,
Dave

PS: We should probably continue this discussion privately since few if any users reading this thread probably aren't interested in less level of math and physics details.

I think it's totally entertaining - especially the parts where your cats run across your keyboards.

pj
chgo

 

[Corwyn_8]

Some differences of note between the two analyses (noted without prejudice):

1) Cross is using a small ball 50.3mm = 1.98" weighing 120g = 4.2 ounces versus DrDave using a standard ball 2.25" and 6 oz.

2) Cross is using a short cue 94cm = 37" long weighing 194g = 6.8 ounces. DrDave is presumably talking about 58" cues closer to 19 ounces.

3) Cross has mass ratio (stick to ball) of 1.6 versus DrDave (assumed) closer to 3.

4) Cross stick was not chosen for low deflection, and we are left to speculate about the end-mass contribution of the screw-on tip. DrDave was specifically using a (known) low-deflection cue

5) Cross is seeing squirt angles of around 10° while DrDave is seeing 1.8°, more than a 5 fold difference.

6) Cross took video at 25 fps (and some at 100 fps), while DrDave uses 1000 fps. This could be important when determining contact time, and whether the cue stick has time to return energy to the cue ball due to elasticity of the wood portions of the stick (as claimed by Cross).

7) Cross used an aluminum bar (20mm x 6mm) with a rubber tip to determine tip acceleration.


I tried to replicate DrDave's equations with Cross's numbers but got lost at the calculation for Δx (and a slight different value for ΔΘ
than he got (1.17 rather than 1.12)).

Thank you kindly.

 

[Corwyn_8]

PS: We should probably continue this discussion privately since few if any users reading this thread probably aren't interested in less level of math and physics details.

Please don't. Those who don't want to see it can go elsewhere on the forum, and some of us are enjoying it, and don't want to be left out.

Thank you kindly.

 

[LAMas]

Some differences of note between the two analyses (noted without prejudice):

1) Cross is using a small ball 50.3mm = 1.98" weighing 120g = 4.2 ounces versus DrDave using a standard ball 2.25" and 6 oz.

2) Cross is using a short cue 94cm = 37" long weighing 194g = 6.8 ounces. DrDave is presumably talking about 58" cues closer to 19 ounces.

3) Cross has mass ratio (stick to ball) of 1.6 versus DrDave (assumed) closer to 3.

4) Cross stick was not chosen for low deflection, and we are left to speculate about the end-mass contribution of the screw-on tip. DrDave was specifically using a (known) low-deflection cue

5) Cross is seeing squirt angles of around 10° while DrDave is seeing 1.8°, more than a 5 fold difference.

6) Cross took video at 25 fps (and some at 100 fps), while DrDave uses 1000 fps. This could be important when determining contact time, and whether the cue stick has time to return energy to the cue ball due to elasticity of the wood portions of the stick (as claimed by Cross).

7) Cross used an aluminum bar (20mm x 6mm) with a rubber tip to determine tip acceleration.


I tried to replicate DrDave's equations with Cross's numbers but got lost at the calculation for Δx (and a slight different value for ΔΘ
than he got (1.17 rather than 1.12)).

Thank you kindly.

Thanks for the comparison.
Transparency is a good thing.:thumbup:

Be well