Why does the cue ball do that?

[Tommy-D]

> Follow me here. Go find a gumball machine or go to Wal-Mart and buy one of those "super" balls. Find a spot to bounce it,and twist your wrist towards you as you just let it drop. Watch it from above,hopefully you got a ball with some wild stripes in it so you can see the effect. On an even numbered bounce,the spin is in effect,on an odd bounce the spin is negated. It will also bounce straight up when it has no spin,and will want to move slightly in the direction of the spin when it is in effect. I have no idea why but it is the real deal,kind of like how running english takes off the first rail,but when contacting the second rail it becomes reverse. The reason the cue ball squats is because after contacting the rack,it bounces twice,then maybe a little one that only get 1/16 off the cloth. When it shoots forward,if you watch it under well-lit conditions and are extremely observant,you will see and probably hear that the cue ball bounces an odd number of times. Hope your head doesn't hurt now,mine did for a long time when I noticed it. Tommy D.

 

[asn130]

My guess would be that the cue ball picks up some "follow" or top spin when coming into contact with the rack.

The cueball when hit so hard actually leaves the surface of the table slightly, which causes it to jump off the headball when you break.

We all know that when two balls come into contact at an angle (when cutting a ball), they impart english on one another in a gear-like effect.

I imagine this is the same thing that happens on the break. But instead of english imparted along the vertical axis (i.e. left/right engligh) you get spin imparted along the horizontal axis (i.e. top/bottom). Since the cueball strikes the rack while its still in the air, it contacts the headball above the head ball's horizontal axis. The gear-effect (one would assume) would impart "top" english on the cueball from the perspective of the breaker.

Hadn't thought of that one as a possibility. Good answer.

 

[Jal]

...The effective mass is nowhere near 9/7 the mass of one ball.

I believe it is. If you can slow the following video down, you'll see that the cueball rebounds with about 1/5 - 1/6 of its initial speed.

http://www.engr.colostate.edu/~dga/pool/high_speed_videos/HSV7-6.htm

This is about the same as it hitting two balls which are frozen to each other simultaneously (ie, cutting them at 30 degree angles). In turn, this is the same as it colliding with one ball having 3/2's of its mass.

If you would like, I could waste like 3 hours of my life proving this by going through the physics of the collision, but I assume everyone can imagine that 15 balls acting like the mass of ~1 ball is counterintuitive.

Being counterintuitive does not make it wrong. But I would like to hear any further thoughts you may have on this (if you don't consider it a waste of time!).

Jim

 

[jsp]

Not only can you not assume that all collisions happen at different times (if the rack is tight), but you cannot also assume that the collisions are perfectly elastic (when using 15 balls; two ball collisions are different). The effective mass is nowhere near 9/7 the mass of one ball.

If you would like, I could waste like 3 hours of my life proving this by going through the physics of the collision, but I assume everyone can imagine that 15 balls acting like the mass of ~1 ball is counterintuitive.

Two words for you...Newton Balls.

If all the balls in the rack were perfectly solid spheres, all the collisions were completely elastic and happened exactly at the same time, and you had an absolutely perfect rack (all balls frozen to each other), then the effective mass of the rack would look exactly like one ball. But because the balls aren't perfectly solid spheres, the collisions aren't completely elastic and don't occur all at the same time, and you don't get a perfect rack all the time...then the 9/7 number looks like a pretty good estimate.

 

[jsp]

Two words for you...Newton Balls.

If all the balls in the rack were perfectly solid spheres, all the collisions were completely elastic and happened exactly at the same time, and you had an absolutely perfect rack (all balls frozen to each other), then the effective mass of the rack would look exactly like one ball. But because the balls aren't perfectly solid spheres, the collisions aren't completely elastic and don't occur all at the same time, and you don't get a perfect rack all the time...then the 9/7 number looks like a pretty good estimate.

I think my post begs the question if, in the ideal world, a 15-ball rack in a triangle configuration would have the same effective mass (that the incoming CB sees) as a perfectly straight line configuration of 15 balls, as in the Newton balls case.

If you had 15 balls all frozen together in a straight line and you hit the CB directly at the head ball, then this scenario exactly resembles the Newton balls. In a perfect world with perfectly hard balls, totally elastic collisions, and absolutely no friction, you would expect the CB to stop exactly frozen to the first ball in the string and the last ball in the line would depart the string with the same initial velocity and direction as the CB. Therefore, the CB effectively would see only one ball's worth of mass, regardless of how many balls are configured in the line. Simple conservation of momentum.

The question now is if it matters if the 15-balls are configured in a triangle rack formation. Would the CB still only see one ball's worth of mass, provided it hits the head ball perfectly straight on? Again, assume an ideal world with perfectly hard spheres, etc. I've thought about this exact situation long and hard. Although it may seem counterintuitive, I cannot think of a reason why the CB would NOT see only one ball's worth of mass. Therefore, in this ideal situation, the CB would sit like a rock right by next to the head ball, the head ball wouldn't move a lick, and the rest of the balls in the rack would be traveling backwards (towards the end rail, none would be moving towards the head rail).

Of course, you would never see that happening in real life...simply because the balls compress, all the collisions don't happen at exactly the same time, blah, blah, blah. Also, the cloth friction plays a role as well. But I've always thought the extremely ideal case is interesting to think about. Please let me know if my reasoning is at all whacked-up.

 

[Bob Jewett]

... If you would like, I could waste like 3 hours of my life proving this by going through the physics of the collision, but I assume everyone can imagine that 15 balls acting like the mass of ~1 ball is counterintuitive....

But the experiment is done often on the table, and the effective mass is clearly not as large as 2 in those experiments. I think your detailed analysis will show you where your intuition went wrong. I offered an analysis that gives a result of 9/7 with some assumptions that are realistic if not perfect. If you have a better analysis, please present it.

 

[Colin Colenso]

Bob's analysis seems right to me.

I should add that I think the bounce-back can vary considerably depending on which balls are touching and the line the the CB takes into them. Hence the importance of consistant rack.

As for the topspin that stops the CB after the bounce-back, and sometimes brings it forward a little, I imagine most of this is coming from a slightly above center hit on the CB, rather than via the friction on the cloth.

Some high speed videos would come in handy to better analyze this though.

Colin

 

[Bob Jewett]

... one of those "super" balls. ...

I don't think the spin on the cue ball on the break shot reverses. The cloth-ball friction is too small.