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Explain the 90 degree rule in Billiards

  1. Nov 12, 2007 #1
    Every pool shot, that does not have a spin or anything crazy done to it, always hits and moves away at a 90 degree angle. How can I prove this mathematically?
  2. jcsd
  3. Nov 12, 2007 #2


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    I assume you mean "hits another ball".

    The collisions are elastic. All kinetic energy directed along the line between the two balls is transferred to the target ball - with no residual - since the two balls are identical in mass - leaving the cue ball with zero momentum in that direction. What's left is the residual component. With all momentum in one line at zero, the remaining component will have to be perpendicular.

    Or so it seems to me...

    Not that that helps you prove it mathematically.
  4. Nov 12, 2007 #3
    yes I meant "hits another ball"

    I understand what you were saying except for the "residual component". I am unsure of what that means. And is there an equation I can use to show this idea?
  5. Nov 12, 2007 #4

    The angle of deflection always matches 90 degrees less the angle of incedence. Like bouncing light off a mirror, or through a prism. If it always left at 90 degrees some really wierd things would happen in a pool hall. Not to say that really wierd things don't happen in a pool hall, just that most of them have nothing to do with the small collection of orbs on a felt covered slate surface.

    Imagine if you lined up right behind the ball and hit it and it suddenly moved at right angles. You can see my point. It doesn't happen in the real world. But imagine a line from the queue ball through your target ball as a line. Hit the target at an incendence of 30 degrees to the tangent of it's surface and it will always move at 60 degrees from parallel, or 30 degrees from the right angle.
  6. Nov 12, 2007 #5

    Ben Niehoff

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    Scroll down the page to find the two-dimensional case. The ball bounces off of the other ball at 90 degrees because of geometry.
  7. Nov 13, 2007 #6
    Thanks for the link Ben.

    I guess I need to clarify. From the Balls frame of reference it is always 90 degrees from the tangent of contact. From the shooters perspective (The big red arrow in Ben's link) it is always 180 degrees less the angle of incidence.

    Personally I've always had an easier time visualizing the problem looking straight down the stick. But whatever works. The perspective is different, but the math is still comes out to the same answer.

    Go Math!
    Last edited: Nov 13, 2007
  8. Nov 13, 2007 #7


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    A 90 degree angle requires a slight amount of backspin when struck by the cue so it has no spin when it collides with the other ball. If there's backspin at the point of contact, the angle is greater than 90 degrees, and if there's topspin, the angle is less than 90 degrees. This, combined with speed of the shot, and side spin allow a good player to pocket a ball and control where the cue ball ends up.
  9. Nov 13, 2007 #8

    D H

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  10. Nov 13, 2007 #9

  11. Jul 28, 2009 #10
  12. Jul 31, 2009 #11


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    There are a couple of things to keep in mind in the real world. First, pool ball collisions are not elastic. Some of the energy is released as sound, and a wee bit as heat.
    Secondly, the object ball always travels in the same direction as the cue ball for a very brief distance before it diverts to its new angle.
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