Nappy cloth = incredible draw?

[Bob Jewett]

Isn't this ignoring dissipated energy? Any heat produced by the friction is energy from the backspin that will not contribute to draw distance. Any "wear" of the cloth or the ball also uses energy from the rotational energy of the ball, which is therefore unavailable to contribute to draw. I don't think these are negligible, and I don't think they're equal on different types of surface.

-Andrew

No, the dissipated energy is included, or in a way, ignored. For this particular problem you look at momentum and not energy. The angular momentum and the linear momentum are connected by the force of friction on the bottom of the ball. That force reduces the angular momentum and in a fixed proportion it increases the linear momentum. It doesn't matter what the force is at any instant; more friction just increases how quickly the angular is turned into linear.

 

[Andrew Manning]

No, the dissipated energy is included, or in a way, ignored. For this particular problem you look at momentum and not energy. The angular momentum and the linear momentum are connected by the force of friction on the bottom of the ball. That force reduces the angular momentum and in a fixed proportion it increases the linear momentum. It doesn't matter what the force is at any instant; more friction just increases how quickly the angular is turned into linear.

But some of the angular is turned into heat or otherwise dissipated instead of turning into linear, is it not?

-Andrew

 

[Jal]

So, to summarize:

1. Cloth has both "sliding resistance" (slick vs. sticky) and "rolling resistance" (smooth vs. nappy), and both independently affect how far the CB will draw. Sliding resistance and rolling resistance can be present in different combinations: smooth/slick, smooth/sticky, nappy/slick, nappy/sticky. Sliding resistance affects the CB before and after contact (while it's skidding); rolling resistance affects the CB only after contact (while it's rolling).

2. For the same tip height and speed, the CB will reach the OB with more backspin on slick cloth, and will draw farther.

3. If the CB reaches the OB with the same backspin (because it's hit differently or from a different distance), it will draw farther on slick cloth before achieving roll and will roll farther on smooth cloth after achieving roll. Therefore, maximum draw is achieved on cloth that is both slick and smooth.

Does that cover it?

pj
chgo

Perfect summary. I think madMac (CCB forum) would point out that rolling resistance acts while a ball is sliding as well, but it's a fairly negligible addition to the sliding friction force (about 20X smaller). And as a practical matter (no ice), and ignoring rolling distance while comparing the two sliding friction phases, one before and one after CB-OB impact, it's the one before impact (item #2) that has the most effect on draw distance, by far.

Jim

 

[Jal]

But some of the angular is turned into heat or otherwise dissipated instead of turning into linear, is it not?

-Andrew

Andrew, that's definitely true. There is a net loss of kinetic energy. But as Bob pointed out, the loss of rotational energy and concomitant gain in translational energy is the same regardless of how fast it happens. This is because the torque (loss of spin) and force (gain in speed) are in a fixed ratio to each other.

Jim

 

[Andrew Manning]

Andrew, that's definitely true. There is a net loss of kinetic energy. But as Bob pointed out, the loss of rotational energy and concomitant gain in translational energy is the same regardless of how fast it happens. This is because the torque (loss of spin) and force (gain in speed) are in a fixed ratio to each other.

Jim

Imagine this scenario, far-fetched though it may be:

There's a round indentation just behind the object ball, of the same radius as the cue ball. After the CB strikes the OB, it falls into this indentation, and spins in place since the spin doesn't create enough torque for the CB to climb out. ALL of the backward energy of the CB will be dissipated. None of it will become backward roll.

Extend that concept to hitting a draw shot on a deep-pile shag carpet. The CB will spin in place, and due to the high rolling resistance of the shag, most of the energy will be dissipated, with very little of it becoming backward roll.

So now extend that concept to a high-nap cloth vs. new simonis. I'd hypothesize that you'd have a lot more dissipated energy on the nap cloth, since it offers much more resistance to the roll of the ball.

So while I understand that the rotational momentum becomes linear momentum regardless of how long it takes, I still think differences in cloth will mean difference in dissipation of energy, meaning differences in final linear momentum.

-Andrew

 

[jsp]

Jsp,

Actually, I remember Mike Page once made some sort of a wager as to how far a ball might be drawn on level ice. The general argument goes like this. Assuming the cueball comes to essentially a complete stop (full hit), and has a spin of W (radians/sec), initially the bottom of the ball is moving over the surface at a speed of R*W (radius X spin rate). Sliding friction at the surface produces a force and a torque, the latter being equal to R*F, and both act until natural roll takes over. The force propels the cueball as a whole, while the torque reduces its spin. This takes place until its acquired speed equals and therefore exactly cancels the spin component of surface speed (V'=R*W', which is natural roll, by definition).

The relevant point is that the ratio of force to torque is fixed when a force (friction at the surface in this case) acts at a fixed distance from the center of a sphere. Therefore, a ball always acquires a certain increment of speed for every unit of spin it loses to friction. This continues until they "meet" (V'=R*W' ). The moment of inertia of a sphere (= (2/5)*M*R^2) dictates that this will occur when V'=(2/7)*R*W, where W was its initial spin rate immediately after OB impact. All of this is independent of how fast it reaches natural roll, i.e., whether it happens on a slick cloth or a sticky one (or ice), due to the fixed ratio of force to torque. Once natural roll takes over, rolling resistance comes into play, which is a different dynamic as you point out.

The longer (time-wise) it takes to reach natural roll, the more distance a ball will have traveled up to that point. That's because its time averaged speed is the same in any case (V'/2), and the distance traveled is (V'/2)*T, where T is the time it took to reach the rolling state. So even though a more "grabby" surface would seem to have the edge, you will nevertheless see more total draw distance with a slicker one, all else equal (e.g., rolling resistance). In the case of ice, which has a very small rolling resistance to boot, you should be able to draw much farther than on rubber, assuming the friction doesn't generate enough heat such that the cueball digs itself into a pit.

Does that make any sense?

Jim

Thanks for the explanation Jal. After I posted my last response last night, I slept on it and independently came to your above explanation. But I'm still not satisfied with it. I still think there is missing energy.

Take the case where there is very high static friction between CB and cloth. The CB initially has backspin, and once it drops and gets in contact with the cloth (neglect bouncing effects), no slippage occurs and the CB immediately achieves forward roll with a velocity Vo.

Now take the case with a slick cloth. The CB has the same amount of initial backspin. When it contacts the cloth, it momentarily slides along the cloth has it builds up its linear velocity. You argue that once the CB stops sliding and achieves natural roll, then this velocity is exactly equal to the rolling velocity of the first case Vo. The difference is that the CB in the second case would have traveled a distance D from its starting point before it reaches Vo, while in the first case the CB immediately reaches Vo (or at least travels very little before achieving Vo).

Do you agree so far? If so, why does the second case have more total energy than the first case? In both cases you start off with exactly the same amount of energy (same amount of backspin). And in both cases the CB ends up having the same amount of total energy (same linear and rotational velocities). But the key difference is that in the second case, the CB gets displaced a distance D from its starting point before achieving natural roll. Throughout that distance, there is force/torque acting on the CB, so that means a certain amount of work (force times distance) is involved displacing the CB.

So because of this extra energy/work in the second case, the total amount of energy in the system doesn't add up. What happened to this extra energy in the first case? If both CBs reach the same rolling velocity, then this extra work is simply missing. Since energy must be conserved, doesn't this mean that the rolling velocity in the second must be slower than the first case, to make up for the energy dissipated by displacing the CB a distance D?

Make sense?

 

[jsp]

So while I understand that the rotational momentum becomes linear momentum regardless of how long it takes, I still think differences in cloth will mean difference in dissipation of energy, meaning differences in final linear momentum

I'm with you on this Andrew. Though, I have been proven wrong multiple times by Jal in the past. We'll see how this discussion plays out, because I think we're correct on this.

 

[Jal]

Imagine this scenario, far-fetched though it may be:

There's a round indentation just behind the object ball, of the same radius as the cue ball. After the CB strikes the OB, it falls into this indentation, and spins in place since the spin doesn't create enough torque for the CB to climb out. ALL of the backward energy of the CB will be dissipated. None of it will become backward roll.

Extend that concept to hitting a draw shot on a deep-pile shag carpet. The CB will spin in place, and due to the high rolling resistance of the shag, most of the energy will be dissipated, with very little of it becoming backward roll.

So now extend that concept to a high-nap cloth vs. new simonis. I'd hypothesize that you'd have a lot more dissipated energy on the nap cloth, since it offers much more resistance to the roll of the ball.

So while I understand that the rotational momentum becomes linear momentum regardless of how long it takes, I still think differences in cloth will mean difference in dissipation of energy, meaning differences in final linear momentum.

-Andrew

Andrew, that's a valid point and relates to madMac's contention that rolling resistance is in effect even while a ball is sliding. It's relatively small compared to sliding friction, but it is there and will be different with different cloths. I kind of glossed over that--maybe because I didn't think of it --to try and make the point about the effect of sliding friction alone on distance to natural roll. It shouldn't be too hard to figure how much extra rolling resistance it would take to completely eliminate any gain in distance to natural roll from reduced sliding friction, but I haven't done the math.

Jim

 

[spoons]

I'm not sure if these questions have simple answers, but it's worth a shot...

If I'm reading this correctly, it seems that most everyone who's posted here agrees that the friction prior to the collision has a greater effect on draw distance than the friction after contact.

So, assuming that initial backspin on the cue ball immediately after the collision is equal in all cases: How big is the difference in draw distance likely to be between the following playing surfaces? Are we talking about a couple of inches? a couple of feet?


New Simonis vs. a run of the mill, dirty, nappy felt

New Simonis vs. newly installed, slick, nappy cloth


I'm just trying to get a feel for the range of differences. It would seem to me that those are the extremes in the comparison between worsted and nappy cloth, so any other scenario would likely fall somewhere between the two. Right?

Seriously, folks. I really love geeking out on this stuff, so I appreciate you all taking the time to answer these questions.

Thanks!

 

[Jal]

...Take the case where there is very high static friction between CB and cloth. The CB initially has backspin, and once it drops and gets in contact with the cloth (neglect bouncing effects), no slippage occurs and the CB immediately achieves forward roll with a velocity Vo.

I agree Jsp, that this would result in essentially no loss of cueball energy, except that tiny, almost infinitesimal amount imparted to the cloth-table-room-earth half of the system. You're essentially describing a sort of tangential (surface-to-surface) shear collision between a massive sphere (earth) and a smaller one (cueball). If this were elastic (your premise), your conclusion follows.

Now take the case with a slick cloth. The CB has the same amount of initial backspin. When it contacts the cloth, it momentarily slides along the cloth has it builds up its linear velocity. You argue that once the CB stops sliding and achieves natural roll, then this velocity is exactly equal to the rolling velocity of the first case Vo. The difference is that the CB in the second case would have traveled a distance D from its starting point before it reaches Vo, while in the first case the CB immediately reaches Vo (or at least travels very little before achieving Vo).

Do you agree so far? If so, why does the second case have more total energy than the first case? In both cases you start off with exactly the same amount of energy (same amount of backspin). And in both cases the CB ends up having the same amount of total energy (same linear and rotational velocities). But the key difference is that in the second case, the CB gets displaced a distance D from its starting point before achieving natural roll. Throughout that distance, there is force/torque acting on the CB, so that means a certain amount of work (force times distance) is involved displacing the CB.

So because of this extra energy/work in the second case, the total amount of energy in the system doesn't add up. What happened to this extra energy in the first case? If both CBs reach the same rolling velocity, then this extra work is simply missing. Since energy must be conserved, doesn't this mean that the rolling velocity in the second must be slower than the first case, to make up for the energy dissipated by displacing the CB a distance D?

Make sense?

Yes it does. Of course, the bone of contention is that the two described cases are very different. Static friction can conserve system wide kinetic energy, in principle (in the ideal case), while sliding friction doesn't. Let me get back to you later on that, if no one else provides a explanation of why that's so. Frankly, I have thought about this before, but I've either forgotten, or more likely, never have come up with a very clear reason (other than that energy is lost as heat, for instance). It's not all that obvious why assuming some "friction force" acts between two surfaces, inevitably leads to deficit in total kinetic energy. But that nevertheless shows up when you do the math. What is it in the original force/torque assumptions that lead to this imbalance?

I don't know if we need to go there, but the conservation of energy has been shown to be an instance of a principle known as Noether's theorem. It states that if something is shown to be invariant under continuous symmetry transformations, it leads to a conservation law. This is a deep underlying principle that shows itself in all levels of physics. I don't yet understand it, but apparently kinetic friction violates this in some way, or verifies it with the generation of heat?

At any rate, if you assume some sliding friction force--any sliding friction force--the math leads you to what we've been asserting about the natural roll speed, and its independence of the magnitude of that force. In essence, for every unit of surface speed reduced by the gain in velocity, 5/2's is reduced by a change in spin rate. (The 5/2 factor comes from a solid sphere's moment of inertia.) If V' is the natural roll speed, and R*W the initial surface speed, then:

V' + (5/2)V' = R*W

(7/2)V' = R*W

V' = (2/7) R*W

I don't think I'm addressing your concerns in a satisfactory way, but that's all I can offer at the moment.

Jim

 

[Bob Jewett]

... How big is the difference in draw distance likely to be between the following playing surfaces? Are we talking about a couple of inches? a couple of feet? ...

Different cloths have different "speeds." The higher the "speed" the lower the rolling resistance. It will primarily be the speed of the cloth that determines the distance the draw shot will cover. See http://www.sfbilliards.com/articles/1995-04.pdf for the details of how you can easily measure the speed of a table for the purposes of comparison. By that method, thick, cruddy cloth might come in at 60, while Simonis 860 might be 100 and Simonis Super Rapide on a heated table might be 200. A ball will roll twice as far on a 200 table as on a 100 table (if no cushion is encountered).

 

[spoons]

Different cloths have different "speeds." The higher the "speed" the lower the rolling resistance. It will primarily be the speed of the cloth that determines the distance the draw shot will cover. See http://www.sfbilliards.com/articles/1995-04.pdf for the details of how you can easily measure the speed of a table for the purposes of comparison. By that method, thick, cruddy cloth might come in at 60, while Simonis 860 might be 100 and Simonis Super Rapide on a heated table might be 200. A ball will roll twice as far on a 200 table as on a 100 table (if no cushion is encountered).

Thanks again, Bob!

And thanks everyone else. I think I'm almost there on understanding some of the science behind this. If I could, I'd like to try to state it back to you folks who are smarter than I. I'm not sure whether my terminology is correct here, but hopefully it's clearly written enough to overcome that. Here goes...


The whole problem is counter-intuitive to what I had believed up until this point. I had been operating under the assumption that, the longer a ball spins in place, the more rotational momentum that is lost due to friction between the cloth and ball, and therefore the shorter the draw distance should be.

Instead, it sounds like the rotational momentum from the spinning cue ball is converted to the same amount of backward roll momentum regardless of how long it takes for the conversion to take place. And, just because a cue ball is spinning in place, doesn't mean it's necessarily losing momentum. But rather, because there is less friction, the rotational momentum of the cue ball is maintained longer before being converted to rolling momentum.

Therefore, for a given spin rate after contact with the object ball, the cue ball will essentially attain the same amount of rolling momentum regardless of the playing surface.

Having established that, it makes sense that the ball will draw back farther on faster cloth (say 200 speed) than on slower cloth (say 60 speed).

And, to bring it full circle to my original question... The nappy cloth in question, is very likely made out of a slicker material, so the cue ball retains more spin before contact with the object ball. So, for a given initial spin rate, the cue ball will generally be spinning faster when it makes contact with the object ball than it would be on a "stickier" cloth.

Therefore, on this slick cloth, after contact with the object ball, there is more rotational momentum to be converted into rolling momentum. This in turn makes it possible to achieve farther draw distances for a given initial spin rate than would occur on say, a stickier (dirty) worsted cloth might.

Is that close? Perhaps an oversimplification?

Thanks again!

 

[jsp]

Yes it does. Of course, the bone of contention is that the two described cases are very different.

Thinking about it some more, I think you're right that my two cases are very different.

At any rate, if you assume some sliding friction force--any sliding friction force--the math leads you to what we've been asserting about the natural roll speed, and its independence of the magnitude of that force.

Yup, I can't find any holes in the math or your reasonings. So I guess I'll admit my wrongness now.

I'd love to see some actual experimental results regarding this, because part of me still feels the empirical data won't exactly match the theory. I think there might be some second/third order effects that come into play that might make reality fit closer to what my intuition thinks, but then again maybe not. Do you know of any stores that sells this so I can try out some experiments?

But thanks for taking the time to school me once again.

 

[Jal]

The whole problem is counter-intuitive to what I had believed up until this point. I had been operating under the assumption that, the longer a ball spins in place, the more rotational momentum that is lost due to friction between the cloth and ball, and therefore the shorter the draw distance should be.

Instead, it sounds like the rotational momentum from the spinning cue ball is converted to the same amount of backward roll momentum regardless of how long it takes for the conversion to take place. And, just because a cue ball is spinning in place, doesn't mean it's necessarily losing momentum. But rather, because there is less friction, the rotational momentum of the cue ball is maintained longer before being converted to rolling momentum.

Therefore, for a given spin rate after contact with the object ball, the cue ball will essentially attain the same amount of rolling momentum regardless of the playing surface.

Having established that, it makes sense that the ball will draw back farther on faster cloth (say 200 speed) than on slower cloth (say 60 speed).

And, to bring it full circle to my original question... The nappy cloth in question, is very likely made out of a slicker material, so the cue ball retains more spin before contact with the object ball. So, for a given initial spin rate, the cue ball will generally be spinning faster when it makes contact with the object ball than it would be on a "stickier" cloth.

Therefore, on this slick cloth, after contact with the object ball, there is more rotational momentum to be converted into rolling momentum. This in turn makes it possible to achieve farther draw distances for a given initial spin rate than would occur on say, a stickier (dirty) worsted cloth might.

Is that close? Perhaps an oversimplification?

Thanks again!

The only thing someone might take issue with--and it really isn't very important--is that while the cueball might appear to spin in place, especially on slick cloth, it's actually accelerating backward and picking up speed. It'll do this in fits and starts as it bounces slightly, but it is moving backward, albeit slowly at first. That aside, I think you've summarized it very well.

And just in case we've left the wrong impression, it probably should be noted that the question of the ball traveling farther on slicker versus stickier cloth during this brief acceleration period is really a minor issue. It's been a "principle of the thing" debate--the difference in distance is very small during this phase compared to subsequent roll. At least that's true for the range of cloth conditions we're likely to encounter on pool tables--the exception being the one Jsp managed to locate.

Jim

 

[Jal]

...I'd love to see some actual experimental results regarding this, because part of me still feels the empirical data won't exactly match the theory. I think there might be some second/third order effects that come into play that might make reality fit closer to what my intuition thinks, but then again maybe not. Do you know of any stores that sells this so I can try out some experiments?

Jsp, thanks for the generous (and greatly exaggerated!) remarks.

I'm sure you're right about there being additional effects; and that they should bring actual results in the direction of your intuition.

Rolling resistance and air drag, for instance, will have longer to act during a lengthened acceleration period. These should, I think, reduce natural roll speed accordingly, along with the distance over which the ball is accelerating. In fact, there's reason to believe that rolling resistance is increased by a factor of 7/5'ths for a sliding ball, since static friction at the ball/cloth interface of a rolling ball reduces the effect of a force trying to slow it down by a factor of 5/7'ths. But I'm still betting that they're smallish perturbations. That's just a gut estimate. Are you referring to something else, though?

I looked at one of Dr. Dave's videos (kick losses, HSV B-15), which uses both normal and slicked up balls, but I don't think we can tell much from this particular one. There might be another more suited to the issue. If not, and if you really want to cook up a test, maybe we can figure something out. In the meantime, I'll keep my fingers crossed that you'll never have access to one of those tables shown in your link. (I can't imagine how you found that.)

Jim

 

[Patrick Johnson]

jsp, JAL, Andrew, etc.:

I'm really enjoying your exchanges, even though it takes several readings to understand them even as poorly as I eventually do (all attempts to translate math to English are appreciated). Many of the effects and countereffects you describe may be negligible at the scale of my ability to control things on a pool table, but I'm confident that a better understanding at all levels is worth the effort (well, maybe quantum pool would be a little overboard) - not to mention just plain fun to learn.

Thanks from another fan of geek pool discussion (that makes two!),

pj
chgo