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Break shot in pool

  1. Nov 17, 2009 #1
    15 balls are racked in a triangle. They are all alike and each one is touching its neighbors. Ignore friction and assume all collisions are elastic. You break by hitting the apex ball hard with the cue ball, as for a game of 8-ball. Assuming you know the velocity of the cue ball and where it contacts the apex ball, what else do you need to know to predict the resulting motions of the other balls? Conservation of momentum and energy are not enough by themselves, as would be the case if there were only one ball in the rack. My experience is that, provided the balls are well racked and I hit the apex ball full-on, the cue ball bounces back as if it had collided with a heavier object. But I'm not sure that the "effective mass" of this object is the mass of all 15 balls, because they do not all move together as a whole. I'm pretty sure that simplifying the problem to a triangle of 3 balls would capture most of the relevant aspects of the problem.
     
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  3. Nov 17, 2009 #2

    rcgldr

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    One issue is that not all 15 balls will be in contact with each other, mostly due to the irregulaties in the surface of the pool table. There are enough balls for the cue ball to bounce back a bit. Youtube has a few videos like this one:

    http://www.youtube.com/watch?v=_lYmiTjWHLE&fmt=18
     
  4. Nov 17, 2009 #3
    That's right. In real life, the balls will never be touching all their neighbors perfectly. Not even with a new table and a contraption like the Sardo Tight Rack (which comes close enough to perfection for practical purposes). Another factor is that, owing to the popularity of the game 9-ball, balls 1-9 usually will have more wear, and will thus be slightly smaller than the rest. Since the question of how a real-life rack of balls will behave is so difficult, I'm considering the simpler question of an idealized rack with all the balls perfectly touching on a flat, level, frictionless table. And I'm assuming the collisions between balls are perfectly elastic...no energy loss to sound, heat, deformation, etc. These are the assumptions used in freshman physics problems, but I'm not sure if freshman physics is enough to tackle even this idealized scenario.

    Thanks for the vid link. The slow motion is very helpful. Johnny Archer has one of the best breaks. I wish someone would do a break analysis of the Pyramid (Russian Pool) player Evgeny Stalev. He is a freak of nature:

    (playing 8-ball in this video)
     
    Last edited by a moderator: Sep 25, 2014
  5. Nov 19, 2009 #4
    so you have 16 balls, each with a centre of mass and velocity, so 64 degrees of freedom (no friction=no angular momentum).


    are you trying to model how the initial collision works? I think the white will hit the first red, which will pass the impulse down each side of the triangle and to the two end balls, so I think you might be able to work this out purely by energy and momentum.

    the middle balls wouldn't be affected since there is no friction and no net impulse in their direction from the outer balls, as it is all channelled down the sides by the first ball. I don't know if the white will bounce back... I just have this image of those metal swinging balls in my head.
     
  6. Nov 19, 2009 #5
    Right, by ignoring friction I'm ignoring the possibility that balls will be spinning or rolling after the collision. In this simplified scenario, the white ball slides into the rack and the resulting motions of the balls are sliding motions. It's not terribly realistic, but I think it's a first step to figuring out what is going on. And yes, I would be happy just to how the initial collision works. In particular, I'd like to know how fast the white bounces back off the apex ball (if it indeed bounces back). That would tell me what the "effective mass" of the rack is. By "effective mass," I mean the mass a solid object would need to have for the white ball to bounce back from it at the observed velocity. This can be calculated from conservation of momentum and energy.

    I suspect the effective mass depends on what angle you are breaking from. I think you are describing a break shot right down the center line of the table, hitting the first red head on. What you say about the impulse being channeled down the outer balls makes a lot of sense, as the break shot is perfectly symmetric with respect to left and right in that case. I have observed this in 9-ball, where nine balls are racked in a diamond shape. If the balls are racked well, a break from the middle leaves the center ball (the 9) almost motionless while all the others fly around. The white bounces straight back. However, a break from the side gets all the balls moving, and the white does not bounce back as much. It's like the white feels less resistance from the rack at that angle.

    Also, I'm glad you mentioned the metal balls on strings (Newton's Cradle). An understanding of how that works should shed some light on the pool problem. BTW, do you play UK-style 8 ball, with reds and yellows and two turns after a foul? If so, you probably recognize names like Darren Appleton, Chris Melling, etc.
     
  7. Nov 22, 2009 #6
    If you totally ignore friction then it's never going to look like real pool is it.
    In your sceneario the cue ball slides into the pack without rolling so it has a lot less kinetic energy than a real cue ball would have. I don't know the exact figure, (you have to add the rotationional energy of a sphere) but it probably works out something like twice as much energy as a sliding ball. Maybe answers why you noticed the cue ball seems to carry a lot of more 'weight' with it than you'd expect it to.
     
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