modifying the tangent line

[ShootingArts]

how is it possible to transfer spin?

when you put it that way...hard not to agree.

I can't claim to know the small physics involved in the collision between two objects of equal elasticity and different mass. I don't know if they actually are in contact for more time or not. If they are, then Randy's distinction between contact point and release point is meaningful and necessary. If they are not in contact for any longer, then hmm...tough to say, but it would seem that contact point and release point are two words for the same thing.

KMRUNOUT

When we shoot the cue ball with stun, the object ball takes on spin according to the angle it is hit. This is easily demonstrated when we see the object ball come off of the rail. We can put spin on the cue ball and offset some or all of the collision induced spin by transferring some spin from the cue ball to the object ball. That last is a key point because it demonstrates that we have more going on than the assumed static collision of two objects. If the collision and release were instantaneous or a very smooth slide this wouldn't be possible.

If we were to write out a list of the properties or possible properties of a collision between two balls and then have to explain each of these things we would find that existing accepted theory simply doesn't explain everything. Any hypothesis that doesn't cover all known empirically proven facts is either wrong or incomplete.

How does the collision between the cue ball and the object ball transfer spin between the cue ball and the object ball? (addressing transferred spin here, not collision induced spin) Remember we can actually reverse the effects of collision induced spin on some shots.

Hu

 

[Bob Jewett]

My guess is we don't need one.

I think "release point" is a byproduct of an incorrect mental picture of the interaction.

The concept is perhaps more useful for the tip-ball collision.
...
The balls don't move together like the tip and cueball do. Instead they rub across one another while they are compressing and decompressing. So the contact patch starts as a point, grows to whatever it grows to, and ends as a point. If the contact time is 0.2 ms and the sideways relative speed of the balls is 4 mph (like it would be for a 6 mph cueball doing a quarter-ball hit), then the sliding distance (distance the cueball moves while in contact with the object ball) is perhaps 0.5 mm, half the thickness of a dime.

I don't think this matters for anything. I don't think it affects the tangent line or anything else we need to think about to understand pool.

Maybe Bob J. or Dave A. disagree with me?

I agree, mostly. I think we had a discussion about the effect of finite contact time over in RSB (or maybe it was here -- it was a while ago), and the conclusion was that it could, in fact, change the cut angle but that change in cut angle was typically smaller than the change in angle due to throw.

Briefly, because the cue ball moves partly "into" the object ball, the correct position to use for the cut angle is the position of the cue ball at the instant of maximum compression and largest contact area. But, as Mike points out, this is only a tiny bit away from the cue ball position at the start of contact.
The cut angle is expected to be very slightly increased due ball-ball penetration. I vaguely recall the change was about 1/4 degree for high-speed shots. Can anyone find that discussion?

 

[Bob Jewett]

... How does the collision between the cue ball and the object ball transfer spin between the cue ball and the object ball? ...

Physics predicts that is it from a tangential frictional force due to the sliding of the surfaces of the two balls against each other. Do you think the spin transfer is from something else? So far as I know, there is no experiment that disproves this simple explanation from physics.

 

[ShootingArts]

agree completely

Physics predicts that is it from a tangential frictional force due to the sliding of the surfaces of the two balls against each other. Do you think the spin transfer is from something else? So far as I know, there is no experiment that disproves this simple explanation from physics.

Bob,

I agree completely. Now that we have agreed that there is enough friction for a long enough time to transfer spin how can we say that it is too short and insignificant a contact to do other things?

Hu

 

[dr_dave]

How does the collision between the cue ball and the object ball transfer spin between the cue ball and the object ball? (addressing transferred spin here, not collision induced spin) Remember we can actually reverse the effects of collision induced spin on some shots.

Physics predicts that is it from a tangential frictional force due to the sliding of the surfaces of the two balls against each other. Do you think the spin transfer is from something else? So far as I know, there is no experiment that disproves this simple explanation from physics.

FYI, video demonstrations and physics proofs of spin transfer effects can be found here:

http://billiards.colostate.edu/threads/throw.html#spin​

Regards,
Dave

 

[Bob Jewett]

... Now that we have agreed that there is enough friction for a long enough time to transfer spin how can we say that it is too short and insignificant a contact to do other things? ...

What other thing do you have in mind? What phenomenon are you hoping to explain that has not been explained so far?

 

[mikepage]

Mike,

I think I'm with you here. The tangent line may not require a circle, but by definition it definitely requires a curve. It is entirely possible to hit a perfect stop shot on a cut shot in which case the cue ball will travel along what for all intents and purposes is a straight line.

We might be saying the same thing: at the moment of contact between cue ball and object ball, the cue ball will initially begin travelling parallel to a line tangent to the object ball at the point of contact. How it moves after that is entirely dependent on the spin on the cue ball.

Are we saying the same thing? I think we are.

KMRUNOUT

Yes, I think so too.

 

[Island Drive]

what other thing do you have in mind? What phenomenon are you hoping to explain that has not been explained so far?

ivory balls......................

 

[mikepage]

I agree, mostly. I think we had a discussion about the effect of finite contact time over in RSB (or maybe it was here -- it was a while ago), and the conclusion was that it could, in fact, change the cut angle but that change in cut angle was typically smaller than the change in angle due to throw.

Briefly, because the cue ball moves partly "into" the object ball, the correct position to use for the cut angle is the position of the cue ball at the instant of maximum compression and largest contact area. But, as Mike points out, this is only a tiny bit away from the cue ball position at the start of contact.
The cut angle is expected to be very slightly increased due ball-ball penetration. I vaguely recall the change was about 1/4 degree for high-speed shots. Can anyone find that discussion?

For the object ball,

throw: decrease the cut angle
compression: increase the cut angle
inelasticity: no effect? on the cut angle
mass differential: no effect? on the cut angle

For the cueball,

throw: no effect? on the carom angle
compression: decrease the carom angle
inelasticity: decrease the carom angle
mass differential: increase or decrease the carom angle

Does this sound about right?

I'm thinking the big effects are throw for the object ball and inelasticity for the cueball

 

[peteypooldude]

I would have thought this would be a one or two post reply,amazing lol

 

[Bob Jewett]

For the object ball,

throw: decrease the cut angle
compression: increase the cut angle
inelasticity: no effect? on the cut angle
mass differential: no effect? on the cut angle

For the cueball,

throw: no effect? on the carom angle
compression: decrease the carom angle
inelasticity: decrease the carom angle
mass differential: increase or decrease the carom angle

Does this sound about right?

I'm thinking the big effects are throw for the object ball and inelasticity for the cueball

I think that covers it. One other factor that enters is that friction decreases for faster surface speed at the point of contact. In general, the effect is similar to compression, but causes larger angle changes for the object ball.