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Physics of Billiard

  1. May 20, 2009 #1
    First of, PhysicsForum is the best thing ever XD

    Hello everyone, I'm trying to write a research paper about billiard. I've already ask a question on another thread about angular velocity and quaternions and everything worked out well. Now, I have another problem.

    Let's start : There's a cue ball on the table. When I hit it with the stick, it first slides on the cloth, then, after a some time, the ball starts rolling. My problem is with the sliding part. Here's what we know :

    [tex]v_r(t) = v(t) + Re_3\times\omega(t)[/tex]

    That's the non-vanishing relative velocity of the contact point of the ball with the cloth (sliding movement)

    [tex]v(t) = v(0) - g\mu_s \hat{v}_r(0)t[/tex]

    This is the equation for linear velocity when the ball is sliding. [tex]\mu_s[/tex] is the sliding friction coefficient (if it was [tex]\mu_r[/tex], it would be the rolling friction coefficient). -g is the the well-known gravitational acceleration and v hat-r at time zero is the relative initial velocity of the ball.

    [tex]\omega(t) = \omega(0) - \frac{Re_3 \times f}{I}t[/tex]

    This is the angular velocity equation. R is the raduis of the ball, so [tex]Re_3[/tex] is the position of the center of mass of the ball. f is the frictional force ([tex]f = -mg\mu_s\hat{v}_r(0)[/tex] where m is the mass of the ball) and I is the inertia tensor ([tex]I = \frac{2mR^2}{5}[/tex]).

    Easily, I can obtain the linear position :

    [tex]p(t) = p(0) + v(0)t - \frac{g\mu_s\hat{v}_r}{2}t^2[/tex]

    and the angular position of the sliding ball :

    [tex]a(t) = a(0) + \omega(0)t - \frac{Re_3\times f}{2\cdot I}t^2[/tex]

    These formulas work well, but for one thing : when I want to visualize those using a program, there's a problem with angular position. Here's why : if I hit the ball with a certain "top" angle with the stick, it should spin with a certain velocity in x, in y and in z (so there's a resulting axis that passes through the center of mass around which the ball should spin, let's call that axis u). The u-axis should be fixed, but, unfortunately, it appears that it is not.

    I'll compare this situation to the Earth's rotation : As we know, the Earth is a pseudo-spherical rigid-body (haha). I takes one day to rotate around what we'll call the Earth's u-axis. But we also know the the u-axis also rotates. Do you see what I mean? I think this is called the "precession", it may have something to do with torque-free motion (correct me if I'm wrong).

    So this is the same thing that happens to the cue ball. Instead of having a fixed u-axis, we have an axis that rotates with time : [tex]u(t)[/tex]. I'd like to find a way to relate [tex]\omega(t)[/tex] and/or [tex]p(t)[/tex] to the rotating u-axis [tex]u(t)[/tex].

    Also : Do not forget that every equation works fine, the only problem is if I want to visualize in a 3D program that my colleague made. For example, if I hit the ball number 5 with the stick (with a certain "top" angle), after sliding, the "5" doesn't seem to have to right orientation and the ball doesn't even seem to spin correctly. That's why I was told by a Ph.D. that the "u-axis" might rotate too, but that it shall not affect the results, but it would be interesting to be able to visualize the motion of the ball with the program.

    Finally, you may have noticed that English isn't my mother tongue. Also, I am not really advanced in physics. I know a bit more about maths, but still... not that much.

    Thank you for your help, if something I said wasn't clear enough, please just tell me and I'll try to explain what I meant.
     
  2. jcsd
  3. May 20, 2009 #2

    Danger

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    Gold Member

    I'm afraid that I have no knowledge of math, and not much of physics. I do, however, play a fair bit of pool. There are a lot of practical factors that you aren't considering. First and foremost, probably, is the amount of follow-through on the stroke. Also, how hard the stroke is administered is important. The composition of the cuestick tip matters a great deal, as does the type of surface. 'Mercury felt', for instance, does not interact with the ball the same way that a typical 'bar table' does.
    While this isn't strictly scientific, I recommend that you go out and shoot at least a dozen games per day for a while. You'll start to see things that the theories don't readily reveal. And, you'll have some fun while learning.
     
  4. May 20, 2009 #3
    Thank you for your reply. I understand that there might other factors to inckude into my equations, but this only an "approximated" program. My colleague included some of the factors you described, because those were important enough to have a MAJOR modification on the velocity/position of the ball.

    The only important thing here is the "precession" of the ball, that's what I'm having a problem here. Thank you for you advice, I'll surely play some games form time to time : it's so much fun!
     
  5. May 21, 2009 #4
    Here guys, I found this to explain what I meant :

    http://imtuoradea.ro/auo.fmte/MECANICA_files/DELIMAN%20TITUS%201.pdf [Broken]

    The only difference is that my ball's precession isn't caused by electrostatic effect.

    Could someone help me understanding please? =)
     
    Last edited by a moderator: May 4, 2017
  6. May 21, 2009 #5
    I just noticed that the article was written in BAD English :( unfortunately
     
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