A ball struck by a cue in billiards with English goes straight at first....

[sophiecentaur]

The 1/10 factor is for the table normal force. The max. table friction is just a fraction of that (~0.2). And the lateral friction component is some fraction of that. This limits the lateral table friction to ~0.02 of the horizontal cue impact force (in the specific case I was considering).

That seems the wrong way round to me. The limiting friction force is less than the normal force but the normal component of the cue's force will increase the normal force (usually just the weight) considerably. 1/10 of the net, brief force from the cue will be greater (or at least as great) than the weight of the ball, which is easy to show from the fact that a struck ball can be lifted to a significant height above the table in some trick shots. So the net friction force, which is what will cause the ball to rotate, will be significantly higher with a slightly downwards stroke. It will be approximately proportional to the angle, aamof, because sin and tan are about the same as the angle for small angles (radians of course). Even if there is slippage, the tangential force against the table will still be greater than with a horizontal impact.
I have played very little pool but I have heard comments about 'controlling the ball' by digging in a bit, which justifies my view.

 

[A.T.]

So the net friction force, ..., will be significantly higher with a slightly downwards stroke.

Higher than from gravity alone. But still just ~2% of the horizontal force from a 5° downwards stroke. So at most ~1° horizontal direction change.

 

[David VH]

Higher than from gravity alone. But still just ~2% of the horizontal force from a 5° downwards stroke. So at most ~1° horizontal direction change.

How did you come to the 1 degree (could be correct, I don't see all the relevant math in my head)? 1 degree is enormous for a snooker player by the way. For a not-that-hard half table shot of ~1.7m a 1 degree deviation is more than the radius of a snooker ball, in other words the difference between a full ball contact and a miss.-

 

[A.T.]

How did you come to the 1 degree (could be correct, I don't see all the relevant math in my head)?

If the friction is 0.02 of the impact force, then it can change the direction of the net force by max. atan(0.02) ~= 1°. But that assumes that it acts perpendicularly to the impact force, so it's actually even less than 1°.

1 degree is enormous for a snooker player by the way.

Not for the question of the OP, which is to explain the ~15° deviation from the contact normal in his video.

 

[David VH]

Not for the question of the OP, which is to explain the ~15° deviation from the contact normal in his video.

I am lost. I thought this was the discussion we were having:
- if there were no friction between cue tip and cueball and both were undeformable, the cueball would go 15 degrees left of the cue direction;
- there is friction in the contact, causing the force to be mostly imparted in the longitudinal direction of the cue, keyword mostly;
- the residual part causes the cue to flex away after the contact, and should cause the cueball to veer off to the left noticeably;
- pool and snooker players do not cue significantly out of line with their intended cueball path even when applying significant side however;
- therefore we are looking for any forces which explain the straighter-than-expected cueball trajectory.

You could even ask "is the majority of the cue deflection force converted into extra spin instead of cueball deflection?" and it would be the same analysis.

Did I misinterpret the problem? I thought we did away with most of the 15 degrees from the start assuming friction contact.

 

[A.T.]

I thought we did away with most of the 15 degrees from the start assuming friction contact.

If you agree that cue friction explains most of the ~15°, and table friction accounts for merely ~1° at best, then we have no disagreement.

 

[sophiecentaur]

If you agree that cue friction explains most of the ~15°, and table friction accounts for merely ~1° at best, then we have no disagreement.

I thought the point of the top of this thread was that the ball goes more or less along the line of the cue and that needed explanation. No one would be surprised if it went as far as 15° off. The videos seem to show very nearly a parallel course and that's what I've been working to. No wonder you (A.T.) and I have been arguing the toss so much.

 

[A.T.]

No one would be surprised if it went as far as 15° off.

It goes ~15° off the contact normal. That was what the OP asked about.

 

[sophiecentaur]

It goes ~15° off the contact normal. That was what the OP asked about.

In the video I am referring to (Post #24) the ball path is parallel with the cue. That was what the question needed an explanation for, as far as I can see. Have you seen it?
I am not sure how your 15° angle is relevant here when what counts is surely the direction relative to the line of the cue - which is how the player views the event.

 

[poolplayer]

It goes ~15° off the contact normal. That was what the OP asked about.

Exactly. I don't think the cue ball goes dead parallel with the cue, and I don't care 1° difference for now.

Even if there is slippage, the tangential force against the table will still be greater than with a horizontal impact.

How can you be sure that? You really push the friction between the ball and table... The cue angle would be always downwards in the game because there are rails, but we always try to strike horizontally as possible. So, the angle should be less than 5° in most cases. In my videos, I would say shots are very horizontal because there is no rail on my table.

Anyway, I am going to post probably decisive evidence against the friction between the ball and table. I hang a ball with a rope in the air and stroke the ball with right English.

I am sorry for the low quality experiment, but you would agree that the ball still goes straight without contact with the table.

So, now we can focus on the real force that makes the ball goes rightwards from the normal direction.

 

[A.T.]

In the video I am referring to (Post #24) the ball path is parallel with the cue.

Nope. Watch the video again.

That was what the question needed an explanation for, as far as I can see.

Nope. Read the OP again. It asks about the difference to ball-on-ball, where the initially resting ball goes off aprox. along the contact normal. That difference is ~15° in that video.

 

[sophiecentaur]

Nope. Watch the video again.Nope. Read the OP again. It asks about the difference to ball-on-ball, where the initially resting ball goes off aprox. along the contact normal. That difference is ~15° in that video.

This is in the top post and is what I have been addressing.

I heard that it is because the ball and cue can be thought as unity so the force only propagates to the cue direction.

There have been a lot of statements and you have obviously been looking at others; the thread is almost endless.

 

[A.T.]

Anyway, I am going to post probably decisive evidence against the friction between the ball and table. I hang a ball with a rope in the air and stroke the ball with right English.

Thanks for the experiment. A very good idea - the vertical string doesn't apply horizontal forces to the ball, so it's all cue contact force.

So, now we can focus on the real force that makes the ball goes rightwards from the normal direction.

Well it's obviously friction, and the required friction coefficient is below 0.3, so entirely feasible

 

[poolplayer]

Well it's obviously friction, and the required friction coefficient is below 0.3, so entirely feasible

Glad to hear that. Here, 0.3 comes from tan(15°) because the tangential force is [tan(θ) * normal force] , right? Some say that friction coefficient between the ball and cue tip is around 0.6, so it would be much greater than 0.3.

But, how can I know how much of the tangential force affects the movement of the ball? I thought I could use energy conservation to know speed of the ball to tangential direction, but it seems not good because I don't know how much energy I applied to the ball...

 

[sophiecentaur]

Anyway, I am going to post probably decisive evidence against the friction between the ball and table. I hang a ball with a rope in the air and stroke the ball with right English.

That's the best evidence so far and it suggests that I am wrong in some respects. (Great to see the experiment is continuing, btw)
I am definitely more interested in getting a right answer to this than 'not being proved wrong'.

 

[A.T.]

Glad to hear that. Here, 0.3 comes from tan(15°) because the tangential force is [tan(θ) * normal force] , right?

Yes.

Some say that friction coefficient between the ball and cue tip is around 0.6, so it would be much greater than 0.3.

The static friction coefficient is a limit. The actual ratio can be anything less than that.

But, how can I know how much of the tangential force affects the movement of the ball?

Not sure what you mean. All of it affects the movement via F = ma.

 

[poolplayer]

Not sure what you mean. All of it affects the movement via F = ma.

Maybe I should not say that the tangential force induces torque... but the tangential force somehow relates to the torque, right? When I think of a ball on a conveyor belt, I guess that the ball rolls if its moment of inertia is small and the ball moves with the belt without rolling if its inertia is large. Is it correct?

What I want to do is to estimate the tangential force if possible and confirm that it is near tanθ of the normal force.

 

[A.T.]

Maybe I should not say that the tangential force induces torque... but the tangential force somehow relates to the torque, right? When I think of a ball on a conveyor belt, I guess that the ball rolls if its moment of inertia is small and the ball moves with the belt without rolling if its inertia is large. Is it correct?

To confirm that linear and angular velocity changes are consistent with the assumption of a single impulse from the cue, check if the following relation holds:

w/v = 5/2 * r/R²

or:

w/v = 5/2 * sin(θ)/R

where:

w : angular speed
v : speed
r : lever arm of the force around the center
R : radius of the sphere
θ : angle between contact normal and the ball velocity / net force.

What I want to do is to estimate the tangential force if possible and confirm that it is near tanθ of the normal force.

It is per definition, if the above assumption holds.

 

[sophiecentaur]

What I want to do is to estimate the tangential force if possible and confirm that it is near tanθ of the normal force.

To get the direction out of it, you only need to know the ratio of the forces. The Impulse is what counts so, as long as F_radial/ F_tan doesn't vary over the impact due to some time slipping and some without slipping during contact, the actual time profile of the forces is not too important. Its could be that the initial contact would not have any slipping as the angle will be steeper and the ball will be rotating a bit during the impact.

 

[poolplayer]

To confirm that linear and angular velocity changes are consistent with the assumption of a single impulse from the cue, check if the following relation holds:

w/v = 5/2 * r/R²

or:

w/v = 5/2 * sin(θ)/R

where:

w : angular speed
v : speed
r : lever arm of the force around the center
R : radius of the sphere
θ : angle between contact normal and the ball velocity / net force.

Thank you for posting this equation again. I calculated it and the result is pretty good. My understanding is that this means that we don't need to think other force than the impulse, which is awesome. But, we cannot say that the direction of the ball is near straight from this, right?

Just for your information, I will note the values used for the calculation. I measured these values in the video in which I hit a ball hung in the air (240fps).

w : 0.113 rad/ms (I used a red dot on the ball to measure this)
v : 4.55 mm/ms (linear speed of the ball just after collision)
R : 28.575 mm (known radius of the pocket ball)
θ : 15° (it is hard to know the actual angle, but it is in the range of 15-25deg)

The result is 0.024835 = 0.022644. If values w and v are happen to be correct, the actual hit angle would be ~16.4°.

 

[poolplayer]

I was not familiar with the equation A.T. posted, but I took that the equation indicates that torque is just a byproduct of the force for linear acceleration. This is good to know. Then, maybe can I calculate the tangential force for linear acceleration from the angular speed (applied torque) and the length of the lever arm? maybe I am just confused...

 

[poolplayer]

Wait... if the cue applies force F to the ball, is the normal force Fcosθ and tangential force Fsinθ? And if all the tangential force Fsinθ can be used for linear acceleration to the ball due to the friction, is it obvious that the ball goes straight?

 

[A.T.]

But, we cannot say that the direction of the ball is near straight from this, right?

What do you mean by "goes straight"? Parallel to the cue? Then no, the cue orientation and velocity don't come into this, just the cue contact position and the resulting ball velocities. It confirms that no other forces than the cue contact are relevant, but it doesn't go into how cue velocity / orientation lead to that force.

 

[poolplayer]

What do you mean by "goes straight"? Parallel to the cue? Then no, the cue orientation and velocity don't come into this, just the cue contact position and the resulting ball velocities. It confirms that no other forces than the cue contact are relevant, but it doesn't go into how cue velocity / orientation lead to that force.

Yes, that was what I meant.

I am quite confused now, but if my last post is one way to understand why the ball goes straight (parallel to the cue), it was really easy question to answer. And I don't know why my experiments suggested that cue flex is important for determining the ball direction.

 

[David VH]

I don't know why my experiments suggested that cue flex is important for determining the ball direction.

Think of it as adding gradually less important factors (or terms...) to the model. The basic model is a cue and cueball made of frictionless adamantium, in a vacuum, without bridge hand. This is the 15 degree scenario from the first page, and the cue would bounce off to the side uninhibited. But the contact is not frictionless, so it's much less than 15 degrees, let's say 0.8 degrees. The bridge hand and back hand inhibit the cue bouncing off sideways, this transfers more lateral force to the ball, so we're at maybe 1.1 degrees. The cue flexes however, which is akin to letting the cue bounce off sideways just a little. We're back at 0.9 degrees because of that. The resistance of the ball against the cloth, amplified by the slightly downward cue stroke, means that the cueball resists the sideways force somewhat compared to the vacuum. That brings it back lower, say 0.7 degrees. The longitudinal friction of the cloth and the downward angle tilt the rotation axis forward, which curves the path slightly, also due to cloth friction. By the time the cueball hits the object ball, maybe it's only 0.5 degrees anymore.

Any more marginal factors we've discovered over 9 pages that I missed?

 

[poolplayer]

Any more marginal factors we've discovered over 9 pages that I missed?

Thank you for summarizing it. I think that is quite much, although I have been at a loss at the most basic level 2) (please see the quote below)...

Is large friction between the ball and cue sufficient that the ball goes parallel to the cue? What still confuses me is some videos that I posted before. I hit the ball with a metal cue (so small cue flex. tip is normal, so there should be the same friction as a normal cue) and it seems that the ball went rather to the normal direction. Do you think this is just a sort of double hit? It seems that the ball already started to go oblique immediately after the first contact.

1) The basic model is a cue and cueball made of frictionless adamantium, in a vacuum, without bridge hand. This is the 15 degree scenario from the first page
2) But the contact is not frictionless, so it's much less than 15 degrees, let's say 0.8 degrees.
3) The bridge hand and back hand inhibit the cue bouncing off sideways, this transfers more lateral force to the ball, so we're at maybe 1.1 degrees.
4) The cue flexes however, which is akin to letting the cue bounce off sideways just a little. We're back at 0.9 degrees because of that.
5) The resistance of the ball against the cloth, amplified by the slightly downward cue stroke, means that the cueball resists the sideways force somewhat compared to the vacuum. That brings it back lower, say 0.7 degrees.
6) The longitudinal friction of the cloth and the downward angle tilt the rotation axis forward, which curves the path slightly, also due to cloth friction. By the time the cueball hits the object ball, maybe it's only 0.5 degrees anymore.

I don't know if these values are reasonable, but my intuition says changes of the ball direction due to these factors would be correct.

 

[A.T.]

And if all the tangential force Fsinθ can be used for linear acceleration to the ball due to the friction, is it obvious that the ball goes straight?

Friction explains why the ball doesn't move along the contact normal. But nothing implies that the ball must move parallel to the cue (and as far I can see it doesn't).

 

[poolplayer]

Friction explains why the ball doesn't move along the contact normal. But nothing implies that the ball must move parallel to the cue (and as far I can see it doesn't).

OK. Let me confirm this. When the cue applies net force (F) to the ball, is the tangential force Fsinθ? If so, do you think tangential acceleration of the ball is F/m sinθ if there is enough friction? (m: mass of the ball)

 

[A.T.]

When the cue applies net force (F) to the ball, is the tangential force Fsinθ?

If θ is the angle between F and contact normal, then Fsinθ is the tangential force per definition.

If so, do you think tangential acceleration of the ball is F/m sinθ if there is enough friction? (m: mass of the ball)

If Fsinθ is the tangential force then F/m sinθ is the tangential acceleration (assuming no other relevant forces than the cue contact).

 

[poolplayer]

If Fsinθ is the tangential force then F/m sinθ is the tangential acceleration (assuming no other relevant forces than the cue contact).

And then, in theory the ball goes parallel to the cue direction, doesn't it? Because the normal acceleration of the ball is F/m cosθ.

 

[A.T.]

If θ is the angle between F and contact normal, then Fsinθ is the tangential force per definition.

If Fsinθ is the tangential force then F/m sinθ is the tangential acceleration (assuming no other relevant forces than the cue contact).

And then, in theory the ball goes parallel to the cue direction, doesn't it?

I have no idea how you come to this conclusion, since nothing I wrote above references the cue direction in any way.

 

[poolplayer]

I have no idea how you come to this conclusion, since nothing I wrote above references the cue direction in any way.

I think cue direction only changes θ.

 

[poolplayer]

I have no idea how you come to this conclusion, since nothing I wrote above references the cue direction in any way.

I think I got what you said. I meant that the directions of the cue and F are the same, but you meant that the direction of F is the direction of the ball acceleration and not necessarily the cue direction. I think you are right... anyway, it seems that you don't think that the ball always goes parallel to the cue direction even if there is enough friction between the cue and ball and there is no other force than collision.

 

[A.T.]

I meant that the directions of the cue and F are the same, but you meant that the direction of F is the direction of the ball acceleration and not necessarily the cue direction.

Yes, exactly. If F is the only relevant force, then the velocity will be parallel to it's average value during impact. This is the simple part.

I think you are right... anyway, it seems that you don't think that the ball always goes parallel to the cue direction even if there is enough friction between the cue and ball and there is no other force than collision.

At least I don't see an obvious reason why this should always be true. But maybe there is a combo of friction coefficients and other parameters where it goes parallel or even to the other side. This is the complex part.

 

[sophiecentaur]

we don't need to think other force than the impulse, which is awesome.

That assumption is fine as long as the contact is brief. 'Stroking' along the side may well affect the accuracy of that approximation. Would I be right unsaying that stroking is sometimes used for some shots?