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Numerically determine area and circumference

  1. Nov 6, 2004 #1
    Hey guys

    I'm currently making a small game (simple space game) and have a problem with numerically determine area and circumference of a ball that is being thrown.

    The problem is presented here:


    Thanks in advance
  2. jcsd
  3. Nov 6, 2004 #2
    That looks like a cardioid. It's general equation in polar coords is: r=a(1-cost), but it could also be r=a(1+cost), i.e. flipped around. I think the maximum distance from the point where it has an 'indent' is 2a, but I'm not entirely sure. a could possibly be either the length of the rope or the radius of the circle.

    The path of motion of the ball looks like a distored ellipse/circle. Perhaps someone who's into cosmology can help you out.

    Sorry I couldn't be of much help.
  4. Nov 6, 2004 #3
    I think you meant topology, not cosmology. :)
  5. Nov 7, 2004 #4
    Topology? I don't know much about topology, but I thought some of the astrophysicists/cosmologists could help because the path of the ball looks like several elliptic and circular orbits. :shy:
  6. Nov 7, 2004 #5
    Oh, I see.
  7. Nov 7, 2004 #6
    Assuming devious is right. Then it would just be a matter of integrating in polar coordinates. Just get a function r in terms of [itex]\theta[/itex] that describes the distance of the ball from the center of the circle and integrate:
    [tex]A = \frac{1}{2}\int_0^{2\pi}r^2(\theta)d\theta[/tex]
    edit - looked up a cardioid, a = the radius of the circle. So:
    [tex]A = \frac{1}{2}\int_0^{2\pi}a^2(1-cos(\theta))^2d\theta[/tex]
    [tex] = \frac{1}{2}a^2\int_0^{2\pi}sin(\theta)^2d\theta[/tex]
    [tex] = \frac{1}{4}a^2\int_0^{2\pi}(1-cos(2\theta))d\theta[/tex]
    [tex] = 3a^2\pi[/tex]
    Last edited: Nov 7, 2004
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