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

[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?

 

[poolplayer]

'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?

Some players including professional players use that kind of shots when they have to hit with right/left English. I think it is not like stroking or scratching on the ball, but rather an attempt not to push against the impact of cue-to-ball collision, which results in a sudden change of the cue direction to the side during/after the collision. They believe that it makes the ball go more straight (more parallel to the initial cue direction). Massive 'stroking' might be used in some masse shots, but this is not related to this thread.

This kind of shots may affect the ball direction. However, many people do not use that shot, but still can make a ball go almost straight with right/left English. So, I guess its effect would be small, maybe ~1-2° at most. I don't use that shot because it hurts consistency of shots.

 

[poolplayer]

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.

You are absolutely right.. but this is a kind of tautology. When I was asking about applied force F by the cue, I was talking about a simple case in theory where some unknown parameters can be ignored.

When force F (whose direction is off the normal) is applied to a ball, is the tangential acceleration of the ball F/m sinθ if there is enough friction between the force and ball (enough friction means Fsinθ < μFcosθ , where μ is static friction coefficient)?

Is this an ill-posed problem? If not and if it is right, maybe this was the simplest answer that I could get. My problem was that I thought that all of the tangential force Fsinθ cannot be used for linear acceleration (I even thought that the tangential force only spins a ball...). But, it seems that all the force can be used for linear acceleration of center-of-mass after a little research. (if there is enough friction on the contact surface)
1) Application: pull on two hockey pucks
https://books.google.co.jp/books?id=8oyNPd5QbYgC&pg=PA350&lpg=PA350&dq=puck+string+center+of+mass+velocity+google+book&source=bl&ots=61cEZe4Zaj&sig=blaDAk_ZGOhw0PbgPsm-6To0X7U&hl=en&sa=X&ved=0ahUKEwisivLSvLPRAhWr1IMKHVcaC68Q6AEIIzAB#v=onepage&q=puck string center of mass velocity google book&f=false
2) Puck-to-stick collision and fusion on frictionless ice
http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W11D2-4.pdf

Of course, we now know that some factors such as cue flex or player's arm movement do not guarantee that the force is always in the direction of the cue during the collision, which makes the case complex. But, I would say the applied force is almost always near to the initial direction of the cue. Because players usually try to push a cue straight forward.

 

[A.T.]

When force F (whose direction is off the normal) is applied to a ball, is the tangential acceleration of the ball F/m sinθ ...

Yes.

But, it seems that all the force can be used for linear acceleration of center-of-mass after a little research.

Yes.

 

[poolplayer]

I think I am done here if no one points to a flaw in the logic. Thank you.

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.

This video is still mysterious, though... maybe some unidentified leftwards force makes the ball go significantly off from parallel to the cue direction, but I will think about this in other occasions.

 

[sophiecentaur]

Because players usually try to push a cue straight forward.

I believe that is true but that doesn't imply that there is no lateral force from the cue (the fact that it requires some effort must imply there is some lateral force). So how could one measure the actual impulse? Answer has to be by observing the actual direction of the resulting motion of the ball when hung on a string. You have already done that on a few (or one) occasions and you would really need to do several runs (>10) with the same contact point etc. and analyse the result by taking the mean after discarding any that don't fit the pattern. It may need a mechanical actuator, rather than a human. Also, it would be interesting to find out the coeff of friction, which could be dominated by the cue tip, so a flat ceramic plate could be used with a cue tip under a known mass and a spring balance used to pull it along the surface of the plate. There have been various values suggested but I wonder how well justified they are. Varying the applied weight could show how linear the friction is - which could really spoil your day if you wanted to make real sense of it.