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dpapavas
11-24-2007, 12:38 PM
Hello all,

most analyses of friction on a billiard ball I've seen begin with a ball being hit without sidespin or follow/draw. Sliding friction brings it then to natural roll at which point rolling friction kicks in to decelerate it to rest.

What happens if you use sidespin though? If I'm not mistaken the initial sliding phase will still terminate in natural roll but the spin axis of the ball will then not be in the XY plane and perpendicular to the velocity vector. It will still be at right angles to the velocity vector but at some angle to the z axis which depends on the linear velocity. I haven't come to this conclusion analytically. I'm developing a billiards simulator and this is the behavior I get with only sliding friction: The spin vector starts off horizontal and rises to some angle theta relative to the z axis. It also rotates on the xy plane. Once it has reached this state it stays there and the ball rolls and spins at the same time with the velocity of the contact point with the table being zero. Again this is with only sliding friction.

If I implement rolling friction as a force acting at the center of mass of the ball with magnitude mu_r * N and direction against the linear velocity vector and also a matching torque to keep the ball rolling then this happens: The initial phase is the same as before but once the spin and velocity vector reach their "steady" positions the spin vector slowly starts to turn towards the z axis and the velocity vector diminishes. That is the rolling friction decelerates the ball but leaves the z spin component intact. At the end the ball remains spinning like a top with the contact point still having zero velocity.

This seems reasonable to me. So a third friction force or rather torque is needed. A "spinning friction" torque. Assuming a torque of magnitude mu_s * N where mu_s a suitable coefficient an N the normal force and direction opposite the spin direction the behavior is this: After the initial sliding phase the spin vector "falls" from its initial direction to the XY plane and the velocity vector diminishes. That is the extra friction torque brings the ball to a "textbook" natural roll state.

Now to my questions to the physics-savvy members of this forum are:
1) Does this behavior seem accurate to you?
2) Is the coulomb-like model for "spinning friction" correct or do you know of anything better?
3) Would you know of any experimental data regarding the coefficient of spinning friction? If not would it be easy to measure it experimentally?

I've googled for answers quite a bit (including papers on spinning tops etc.) but haven't found anything interesting so far. Any help is greatly appreciated.

PS1: It's a pity I can't show you videos of all this in action but I haven't succeeded in capturing video of my simulator. I'll keep trying though.
PS2: Do you btw know of any experimental data regarding the coefficient of restitution between ball and table? This is one of the easiest coefficients to measure and I haven't found anything on this. I need it to make jumpshots and elevated cue stick shots in general work correctly.

Ralph_Kramden
11-24-2007, 03:55 PM
Spinning friction would probably differ with two different grades of cloth. It almost seems that a ball will never stop spinning if you use your fingers to spin a ball on a glass plate.

DeadCrab
11-24-2007, 05:07 PM
For more on billiard ball physics than anyone should know, see this:

http://www.sfbilliards.com/Shepard_apapp.pdf


My guess is that your answer is in there somewhere.

Jal
11-24-2007, 06:12 PM
For the most part, your model seems to be right. But I'm not sure what you mean at a couple of places.

<blockquote><font class="small">Quote dpapavas:</font><hr>...What happens if you use sidespin though? If I'm not mistaken the initial sliding phase will still terminate in natural roll but the spin axis of the ball will then not be in the XY plane and perpendicular to the velocity vector. It will still be at right angles to the velocity vector but at some angle to the z axis which depends on the linear velocity.<hr /></blockquote>Agreed.

<blockquote><font class="small">Quote dpapavas:</font><hr>I haven't come to this conclusion analytically. I'm developing a billiards simulator and this is the behavior I get with only sliding friction: The spin vector starts off horizontal and rises to some angle theta relative to the z axis.<hr /></blockquote>Not sure about this. If a ball starts off with pure sidespin (a non-zero z-axis component and no other spin), the spin vector is vertical and gradually becomes more horizontal.

<blockquote><font class="small">Quote dpapavas:</font><hr>It also rotates on the xy plane. Once it has reached this state it stays there and the ball rolls and spins at the same time with the velocity of the contact point with the table being zero. Again this is with only sliding friction.<hr /></blockquote>Yes, if it has a non-zero component of spin in the direction of motion (or anti-direction), the spin vector will rotate in the x-y plane.

<blockquote><font class="small">Quote dpapavas:</font><hr>If I implement rolling friction as a force acting at the center of mass of the ball with magnitude mu_r * N and direction against the linear velocity vector and also a matching torque to keep the ball rolling then this happens: The initial phase is the same as before but once the spin and velocity vector reach their "steady" positions the spin vector slowly starts to turn towards the z axis and the velocity vector diminishes. That is the rolling friction decelerates the ball but leaves the z spin component intact. At the end the ball remains spinning like a top with the contact point still having zero velocity.<hr /></blockquote>Yes, agreed.

<blockquote><font class="small">Quote dpapavas:</font><hr>This seems reasonable to me. So a third friction force or rather torque is needed. A "spinning friction" torque. Assuming a torque of magnitude mu_s * N where mu_s a suitable coefficient an N the normal force and direction opposite the spin direction the behavior is this: After the initial sliding phase the spin vector "falls" from its initial direction to the XY plane and the velocity vector diminishes. That is the extra friction torque brings the ball to a "textbook" natural roll state.<hr /></blockquote>If I'm reading it correctly, it sounds right to me. But I'm not sure if you're allowing for some or most of the z-axis component to remain after the ball reaches natural roll. A rough estimate (by me) is that the ball's z-axis spin undergoes an angular acceleration of -12 radians/sec/sec. This is based on how long it takes a ball spinning in place to lose the spin and is only a very crude estimate.


<blockquote><font class="small">Quote dpapavas:</font><hr>1) Does this behavior seem accurate to you?<hr /></blockquote>Overall, yes.

<blockquote><font class="small">Quote dpapavas:</font><hr>2) Is the coulomb-like model for "spinning friction" correct or do you know of anything better?<hr /></blockquote>If you'd like a sure test, you can check that your model predicts the ball's final speed and direction at natural roll correctly. This is given by:

V' = (5/7)[V - (2/5)W X R]

where V' is its final velocity, V its initial velocity, W its initial spin, and R is the displacement (radius) vector from the center of the ball to the point of contact with the cloth. This holds for any spin W and for any value of the sliding coefficient of friction mu (and it doesn't matter if mu changes from point to point on the cloth).

<blockquote><font class="small">Quote dpapavas:</font><hr>3) Would you know of any experimental data regarding the coefficient of spinning friction? If not would it be easy to measure it experimentally?<hr /></blockquote>A figure of 0.2-0.22 for the sliding coefficient is good. There is an article linked to at Dr. Dave Alciatore's website (a frequent poster) here:

http://billiards.colostate.edu/

Click on "Pool/Billiards Physics Resources" and see the last entry under technical articles, titled "Rolling and sliding resistive forces on balls moving on a flat surface", by Witters and Duymelinck. Bob Jewett (another frequent poster) has a test to measure it, I believe, but I've forgotten the details. I've measured it off of the high-speed videos at Dr. Dave's website using spin loss off a cushion impact, but something between 0.2 and 0.22 should be accurate.

The rolling coefficient for a 100 speed cloth is 0.01. In general:

mu_r = a/g = 4d/sg

where a is the horizontal acceleration from the rolling resistance, g the gravitational acceleration, d is the length of a 9' table's playing surface (97.75"), and s the cloth speed. The cloth speed s is calculated as twice the square of the time it takes a perfectly lagged ball to rebound from the foot cushion and come to rest on the head cushion (ie, s = 2t^2).

<blockquote><font class="small">Quote dpapavas:</font><hr>I've googled for answers quite a bit (including papers on spinning tops etc.) but haven't found anything interesting so far. Any help is greatly appreciated.<hr /></blockquote>You'll find plenty in the technical proofs section of his website. Also, another poster was doing some modeling too last year and some info can be found in the thread "Equations of Motion" in the middle of the page here (http://www.billiardsdigest.com/ccboard/postlist.php?Cat=&amp;Board=ccb&amp;page=56&amp;view=collapsed &amp;sb=5&amp;o=).


<blockquote><font class="small">Quote dpapavas:</font><hr>PS2: Do you btw know of any experimental data regarding the coefficient of restitution between ball and table? This is one of the easiest coefficients to measure and I haven't found anything on this. I need it to make jumpshots and elevated cue stick shots in general work correctly. <hr /></blockquote>I've bounced balls on the surface of the Gold Crowns (Simonis 860 cloth) where I play (Security wasn't happy about it), and it rebounds to about a third of the dropping height, implying a coefficient of about 0.6. I've measured roughly the same from some of the high speed videos at Dr. Dave's site, but I would double-check that. I think that modeling of a jump shot is going to be tough. But you can check it against some of the videos.

Some of the physics is not easy (jump shots, cushion rebound angles), so you may have to set your goals to a less than perfect simulation. Dr. Dave has stated that he's planning to look at the cushion thing in the future. I also have a model, but it needs work. Another poster (Cushioncrawler) has a model as well, and he tends to look at things in extreme detail.

Jim

dr_dave
11-24-2007, 07:40 PM
A list of many useful pool physics resources, with many links, can be found here (http://billiards.colostate.edu/physics/index.html). The Marlow book is probably the single best resource for what you are doing.

You can also find a list of some property ranges under "technical information" here (http://billiards.colostate.edu/threads.html).

Good luck with your simulator.

Regards,
Dave

cushioncrawler
11-24-2007, 11:16 PM
<blockquote><font class="small">Quote dpapavas:</font><hr> Hello all, most analyses of friction on a billiard ball I've seen begin with a ball being hit without sidespin or follow/draw. Sliding friction brings it then to natural roll at which point rolling friction kicks in to decelerate it to rest. What happens if you use sidespin though? If I'm not mistaken the initial sliding phase will still terminate in natural roll but the spin axis of the ball will then not be in the XY plane and perpendicular to the velocity vector. It will still be at right angles to the velocity vector but at some angle to the z axis which depends on the linear velocity. I haven't come to this conclusion analytically. I'm developing a billiards simulator and this is the behavior I get with only sliding friction: The spin vector starts off horizontal and rises to some angle theta relative to the z axis. It also rotates on the xy plane. Once it has reached this state it stays there and the ball rolls and spins at the same time with the velocity of the contact point with the table being zero. Again this is with only sliding friction. If I implement rolling friction as a force acting at the center of mass of the ball with magnitude mu_r * N and direction against the linear velocity vector and also a matching torque to keep the ball rolling then this happens: The initial phase is the same as before but once the spin and velocity vector reach their "steady" positions the spin vector slowly starts to turn towards the z axis and the velocity vector diminishes. That is the rolling friction decelerates the ball but leaves the z spin component intact. At the end the ball remains spinning like a top with the contact point still having zero velocity. This seems reasonable to me. So a third friction force or rather torque is needed. A "spinning friction" torque. Assuming a torque of magnitude mu_s * N where mu_s a suitable coefficient an N the normal force and direction opposite the spin direction the behavior is this: After the initial sliding phase the spin vector "falls" from its initial direction to the XY plane and the velocity vector diminishes. That is the extra friction torque brings the ball to a "textbook" natural roll state. Now to my questions to the physics-savvy members of this forum are:
1) Does this behavior seem accurate to you?
2) Is the coulomb-like model for "spinning friction" correct or do you know of anything better?
3) Would you know of any experimental data regarding the coefficient of spinning friction? If not would it be easy to measure it experimentally? I've googled for answers quite a bit (including papers on spinning tops etc.) but haven't found anything interesting so far. Any help is greatly appreciated. PS1: It's a pity I can't show you videos of all this in action but I haven't succeeded in capturing video of my simulator. I'll keep trying though.
PS2: Do you btw know of any experimental data regarding the coefficient of restitution between ball and table? This is one of the easiest coefficients to measure and I haven't found anything on this. I need it to make jumpshots and elevated cue stick shots in general work correctly.<hr /></blockquote>Dimi -- The way i see it iz....

1.... A sliding ball iz slowed by ball-to-bed friktion (ie sliding friktion), plus (ie at the same time) rolling rezistance.

2.... I dont like the term "rolling friktion", i prefer "rolling rezistance".

3.... Rolling Friktion can be considered to be the ball-to-bed friktion force (for a rolling ball) that pushes the ball along, making it roll further.

4.... On a friktionless bed the ball would roll perhaps only 5/7ths (say) az far az it would on a friktional bed.

5.... When a ball iz rolling happyly, with sidespin allso, the spin-roll-axis will not be in a plane perpindikular to the direktion of travel, the axis will be very nearly perpindikular to the "grade of the hill" which itself iz perhaps 7/5ths of the "effektiv hill" suggested by the "rolling rezistance".

6.... Whether the ball ends up spinning (with zero rolling) would depend largely on whether rolling endz before spinning endz (or vice-versa). If u had a very slow but very slippery bedcloth then this would act differently to a very fast but hi-friktion bedcloth. Some simple hand-tests will show which initial axis-angle iz "kritikal" -- i wouldnt be surprized if 35dg to 40dg off vertikal gave a "dead-heat".

7.... I have dunn a few drop-bounce tests for ball off bed -- its diffikult to mezure accurately -- i dont remember the rezults, about 0.7 i think.

8.... I dont see any need to mezure the spinning friktion -- it wouldnt be much good to u unless u tested and found the (a) diameter and depth of the ball'z footprint, and (b) a relationship between depth of footprint and bed-reaktion, and (c) set up an excel or similar computer program calculating and integrating the incremental (annular) contributions to reaktion and friktion and torq, all of which i have dunn a long time ago.

9.... Simpler to find the kritikal axis angle and calculate the spin-rezistance-torq uzing skoolkid logik, assuming say 1 in 100 for the roll-rezistance. madMac.

dpapavas
11-25-2007, 04:00 AM
Jal,

[ QUOTE ]

Not sure about this. If a ball starts off with pure sidespin (a non-zero z-axis component and no other spin), the spin vector is vertical and gradually becomes more horizontal.
<hr /></blockquote>

Ooops, sorry about this. I got a little mixed up there. That is indeed the case. It starts of vertical and "falls" a little towards the XY plane and also rotates around Z.

[ QUOTE ]

If you'd like a sure test, you can check that your model predicts the ball's final speed and direction at natural roll correctly. This is given by:

V' = (5/7)[V - (2/5)W X R]

where V' is its final velocity, V its initial velocity, W its initial spin, and R is the displacement (radius) vector from the center of the ball to the point of contact with the cloth. This holds for any spin W and for any value of the sliding coefficient of friction mu (and it doesn't matter if mu changes from point to point on the cloth).
<hr /></blockquote>

-12 rad / sec^2 corresponds to a mu of about 0.0005 for spinning friction and with this figure (which by coincidence is the last figure I tried) my model aggrees with the above formula to 3-4 decimal places. I'm quite happy with that!

[ QUOTE ]

Some of the physics is not easy (jump shots, cushion rebound angles), so you may have to set your goals to a less than perfect simulation. Dr. Dave has stated that he's planning to look at the cushion thing in the future. I also have a model, but it needs work. Another poster (Cushioncrawler) has a model as well, and he tends to look at things in extreme detail.
<hr /></blockquote>

This is true I'm afraid. Still what I aim for is a simulator that will be playable by an experienced player. You needn't necessarily be able to do all those fancy trick shots but you should get the result you'd expect on a normal shot (with english,follow/draw and some elevation maybe). That said jump shots do work actually. I can make the ball jump high or low, far ot close depending on elevation and english. What I can't get are those curved shots you get with high elevation. That is I can get the ball to curve but it also jumps initially. Oh well...

Cushioncrawler,

[ QUOTE ]

8.... I dont see any need to mezure the spinning friktion -- it wouldnt be much good to u unless u tested and found the (a) diameter and depth of the ball'z footprint, and (b) a relationship between depth of footprint and bed-reaktion, and (c) set up an excel or similar computer program calculating and integrating the incremental (annular) contributions to reaktion and friktion and torq, all of which i have dunn a long time ago.
<hr /></blockquote>

What you describe would be required in order to analytically calculate the coefficient I'm after. I assume a simple mu * N model for this "spinning" friction (which is reasonable since it is in fact just sliding friction of th points in the contact patch of the ball and the bed). So in order to measure it one would need to spin a ball with dots on a table and film it, measure its angular velocity at some point from the velocity of the dots in the footage and then measure the time it takes from that point for the ball to stand still.

Bob_Jewett
11-25-2007, 05:52 PM
<blockquote><font class="small">Quote dpapavas:</font><hr>.... It's a pity I can't show you videos of all this in action but I haven't succeeded in capturing video of my simulator. ... <hr /></blockquote>
You may want to take a look at Virtual Pool which has fairly good physics models (including of balls bouncing on the bed of the table), although some of the phenomena such as cushion rebound are implemented by look-up tables or curve fits. I think the VP programmers have already dealt with many of the things you have questions about.

Ralph_Kramden
11-25-2007, 10:09 PM
If I am reading correctly you are saying that a ball struck low with off center contact, will both slide and spin until friction slows it down, but will keep on spinning after it comes to rest. If you are saying something otherwise I am completely misunderstanding what you are trying to say.

That can't happen unless the cueball collides with another ball. A cueball will not keep spinning in place unless it comes to an abrupt stop while it is still spinning horizontaly.

Try hitting a drawshot down the table with a striped ball. Place the stripe tilted about 15 degrees from vertical. Hit the stripe on the centerline as if you were hitting a drawshot with a little side english. Hit this shot to the end rail with no other ball on the table.

Mark one side of the ball opposite the stripe. Place the striped ball, with the striped tilted, on the headspot and hit it with low off center draw just hard enough so it is sliding as it goes over the footspot. The ball will not come up to a horizontal spin and keep sliding as it moves forward.

The ball will actually flip it's axis completely over 90 degrees and keep rolling as if it had been hit with high off center follow. The spot that was marked on one side of the ball will be facing in the complete opposite direction.

If the ball would collide with another ball so it would stop suddenly moving forward at the same time it is spinning horizontaly, it will spin in place.

If the ball does not hit anything it will flip over and keep rolling on the opposite tilted axis until it slows down enough to roll on a horizontal axis before it comes to a stop.

This is why it is important to hit a long drawshot on dead center. If the axis is flipping over from a off center hit it will throw the OB offline if it is spinning horizontaly.

Bob_Jewett
11-25-2007, 10:51 PM
<blockquote><font class="small">Quote Ralph_Kramden:</font><hr> If I am reading correctly you are saying that a ball struck low with off center contact, will both slide and spin until friction slows it down, but will keep on spinning after it comes to rest. ... That can't happen unless the cueball collides with another ball. A cueball will not keep spinning in place unless it comes to an abrupt stop while it is still spinning horizontaly. ... <hr /></blockquote>
While I agree that such a thing is never seen in practice, I wonder if there is some combination of rolling and "boring" friction that could produce a spinning-in-place cue ball without any other ball contact. For this purpose, I don't think masse can help.

cushioncrawler
11-26-2007, 01:46 AM
<blockquote><font class="small">Quote Ralph_Kramden:</font><hr> If I am reading correctly you are saying that a ball struck low with off center contact, will both slide and spin until friction slows it down, but will keep on spinning after it comes to rest. If you are saying something otherwise I am completely misunderstanding what you are trying to say. That can't happen unless the cueball collides with another ball. A cueball will not keep spinning in place unless it comes to an abrupt stop while it is still spinning horizontaly. Try hitting a drawshot down the table with a striped ball. Place the stripe tilted about 15 degrees from vertical. Hit the stripe on the centerline as if you were hitting a drawshot with a little side english. Hit this shot to the end rail with no other ball on the table. Mark one side of the ball opposite the stripe. Place the striped ball, with the striped tilted, on the headspot and hit it with low off center draw just hard enough so it is sliding as it goes over the footspot. The ball will not come up to a horizontal spin and keep sliding as it moves forward. The ball will actually flip it's axis completely over 90 degrees and keep rolling as if it had been hit with high off center follow. The spot that was marked on one side of the ball will be facing in the complete opposite direction. If the ball would collide with another ball so it would stop suddenly moving forward at the same time it is spinning horizontaly, it will spin in place. If the ball does not hit anything it will flip over and keep rolling on the opposite tilted axis until it slows down enough to roll on a horizontal axis before it comes to a stop. This is why it is important to hit a long drawshot on dead center. If the axis is flipping over from a off center hit it will throw the OB offline if it is spinning horizontaly.<hr /></blockquote>Yes, i agree that there iz no bedcloth in existance that would provide enuff rolling rezistance plus enuff slipperyness to allow a qball to stop with rezidual spin, unless the qball hits another ball. That bizness about a marked ball flipping over az sliding stops iz certainly one of the weird n wonderfull thingz that iz ever seen on a billiardz table. madMac.

Jal
11-26-2007, 02:14 AM
<blockquote><font class="small">Quote dpapavas:</font><hr>-12 rad / sec^2 corresponds to a mu of about 0.0005 for spinning friction and with this figure (which by coincidence is the last figure I tried) my model aggrees with the above formula to 3-4 decimal places. I'm quite happy with that!<hr /></blockquote>I'm glad it's working out, but I'm afraid I don't see what the spinning friction has to do with the cited formula (V' = (5/7)...). In that formula, the cross product of any z-axis component of W and R will be zero, and thus contribute nothing to V'. In reality, it will, for the reasons Mac indicated: the interaction of W with the rolling resistance. Is your program calculating this?

<blockquote><font class="small">Quote dpapavas:</font><hr>...That said jump shots do work actually. I can make the ball jump high or low, far ot close depending on elevation and english. What I can't get are those curved shots you get with high elevation. That is I can get the ball to curve but it also jumps initially.<hr /></blockquote>If the ball is jumping, why don't you consider that realistic?

<blockquote><font class="small">Quote dpapavas:</font><hr><blockquote><font class="small">Quote Cushioncrawler:</font><hr>I dont see any need to mezure the spinning friktion -- it wouldnt be much good to u unless u tested and found the (a) diameter and depth of the ball'z footprint, and (b) a relationship between depth of footprint and bed-reaktion, and (c) set up an excel or similar computer program calculating and integrating the incremental (annular) contributions to reaktion and friktion and torq, all of which i have dunn a long time ago.
<hr /></blockquote>

What you describe would be required in order to analytically calculate the coefficient I'm after. I assume a simple mu * N model for this "spinning" friction (which is reasonable since it is in fact just sliding friction of th points in the contact patch of the ball and the bed). So in order to measure it one would need to spin a ball with dots on a table and film it, measure its angular velocity at some point from the velocity of the dots in the footage and then measure the time it takes from that point for the ball to stand still. <hr /></blockquote>I don't think it's that important, but you might want to treat it along the lines Mac is suggesting. Instead of an effective mu for the spinning friction torque, you might want to use an effective moment arm r for the torque. Playing with the numbers a bit, the effective moment arm can be estimated using:

Ia_ = (mu)rN

where I is the balls moment of inertia (= (2/5)mR^2), a_ is the angular acceleration, mu the sliding cof (= 0.2 say), and N the normal force (= mg). Thus:

r = a_(2/5)R^2/[(mu)g]

Using a_ = 12 rad/sec^2, and g = 386 inches/sec^2, we get r = .07869 inch.

In a recent thread on measuring the cut angle, DeadCrab noted that the bottom of a ball sitting in the cloth dimple emerges from the dimple at about 5 mm from the vertical center axis of the ball (ie, the dimple is about 5mm in apparent radius). Allowing for cloth fuzz, the shape of the dimple, and the fact that a moving ball is going to be riding a little higher, we'll estimate the dimple's "working" radius at 3mm.

Treating the downward pressure the ball exerts on the dimple as uniform, and the dimple as a flat disk to a reasonable approximation, the average moment arm can be shown to be 2/3'rds of this, or 2mm. We get:

r = 2mm/25.4mm/inch = .07874 inch

Comparing this with the number derived above, we can say that physics has been berry, berry good to us. But seriously, it doesn't seem to be too unreasonable. Maybe Mac has a more solid figure.

Jim

Jal
11-26-2007, 02:17 AM
<blockquote><font class="small">Quote Ralph_Kramden:</font><hr> If I am reading correctly you are saying that a ball struck low with off center contact, will both slide and spin until friction slows it down, but will keep on spinning after it comes to rest. If you are saying something otherwise I am completely misunderstanding what you are trying to say.<hr /></blockquote>I think dpapavas does say this, but only as a step in the argument. He/she then goes on to cite the need for a torque which degrades the sidespin.

dpapavas
11-26-2007, 03:23 AM
Ralph,

[ QUOTE ]

If I am reading correctly you are saying that a ball struck low with off center contact, will both slide and spin until friction slows it down, but will keep on spinning after it comes to rest. If you are saying something otherwise I am completely misunderstanding what you are trying to say.
<hr /></blockquote>

You are, in part. This only happens if I don't apply this boring friction torque or if the CB has unnaturaly large enlish applied, which can happen since miscues are not implemented yet, and hits a cushion or two before coming to rest.

Jal,

[ QUOTE ]

I'm glad it's working out, but I'm afraid I don't see what the spinning friction has to do with the cited formula (V' = (5/7)...). In that formula, the cross product of any z-axis component of W and R will be zero, and thus contribute nothing to V'. In reality, it will, for the reasons Mac indicated: the interaction of W with the rolling resistance. Is your program calculating this?
<hr /></blockquote>

I should have mentioned this before: I didn't strike the ball with the cue but instead gave it various initial velocities (all on the XY plane of course) and spins and the let it come to natural roll on an infinitely large table. The end velocity was that predicted by the formula up to a few decimal places.

[ QUOTE ]

If the ball is jumping, why don't you consider that realistic?
<hr /></blockquote>

It's not unrealistic. I just can't get those shots with high cue elevation and a lot of sidespin where the ball swerves wildly for example but doesn't jump initially. I can get it to swerve but it also jumps a little.

[ QUOTE ]

I don't think it's that important, but you might want to treat it along the lines Mac is suggesting. Instead of an effective mu for the spinning friction torque, you might want to use an effective moment arm r for the torque. Playing with the numbers a bit, the effective moment arm can be estimated using:

Ia_ = (mu)rN
<hr /></blockquote>

Well that is the same as applying a friction torque of tau = Ia_ = (mu_s * r) * N which is what I'm doing with mu_b = mu_s * r. In fact the r you calculated results in mu_b = 0.0004 for a mu_s of 0.2 and this is very close to what I'm using (0.00047).

BTW, since you mentioned cushion deflections. Are you aware of any experimental results of CB rail deflections I could compare my model to?

Jal
11-26-2007, 12:34 PM
<blockquote><font class="small">Quote dpapavas:</font><hr>...Well that is the same as applying a friction torque of tau = Ia_ = (mu_s * r) * N which is what I'm doing with mu_b = mu_s * r. In fact the r you calculated results in mu_b = 0.0004 for a mu_s of 0.2 and this is very close to what I'm using (0.00047).<hr /></blockquote>Yes, I see. It's just that having a different mu for z-axis spinning friction seemed a little artificial, but it does the job just fine.

<blockquote><font class="small">Quote dpapavas:</font><hr>BTW, since you mentioned cushion deflections. Are you aware of any experimental results of CB rail deflections I could compare my model to? <hr /></blockquote>Not a heck of a lot. Bob Jewett has a couple of articles that cover balls stunned into the cushion without any sidespin. See entries 2004-08, 2004-09 on this page:

http://www.sfbilliards.com/articles/BD_articles.html

For the more general cases, I've set aside several of Dr. Dave's high-speed videos to do measurements on. They are HSV6-3 thru HSV6-11 and HSVA-68 thru HSVA-75. But I haven't gotten around to doing the measurements yet.

At one time I thought that cushion and bed friction alone (along with a coefficient of restitution of about .75), was sufficient to explain the "shortening" of the rebound angles for stunned balls (and no english). This seems to predict the 70% mirror system pretty well, described in one of the articles cited above. But cushion "throwback", due to the asymmetric form the cushion takes on, appears to be a necessary ingredient to explain other shots. The cushion's skewed geometry can be seen clearly on Dr. Dave's DVD, a companion to his book.

Jack Koehler's "The Science of Pocket Billiards", may have something (I've read it but forgotten).

Jim