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Math Help - Parabola for Golf Ball Roll on Green

  1. #1
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    Parabola for Golf Ball Roll on Green

    Hello,

    I'm having a hard time coming up with a formula for showing how a golf ball will "roll" on a sloped green.

    For example, assume there is a right to left break of 1 degree and the ball is hit straight. The ball will end up on a path that is exactly 10 degrees to the left of the hole.

    I need a formula so that the line will start at the center of the golf ball and pass through a point 10 degrees left of the hole (with a curve, not a straight line)

    Any help is appreciated.

    Thanks
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  2. #2
    Junior Member slider142's Avatar
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    Brooklyn, NY
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    Do you wish to include friction? A frictionless analysis gives a parabola whose curvature depends on the ball's initial velocity and the degree of the slope.
    Specifically, if we let the Cartesian plane be the plane that is sloped 1 degree (pi/180 radians) from the positive y-axis to the negative y-axis, and we hit the ball along the negative x-axis from the origin, the ball will describe the path
    y = -\left(\frac{g\sin\left(\frac{\pi}{180}\right)}{2v^  2}\right)x^2
    g is the acceleration due to gravity where the green is located. At sea level, it is usually 9.8 meters/(second^2). v is the initial velocity with which the ball is hit along the negative x-axis. You will need to find the correct v to deviate from the hole in the correct way:
    If the hole is located at (-R, 0), then you must solve the equation
    R = \frac{2v^2\tan\left(\frac{\pi}{18}\right)}{g\sin\l  eft(\frac{\pi}{180}\right)}
    to get the initial velocity v necessary to accomplish your deviation of 10 degrees (pi/18 radians).
    If you simply want the curve and you don't actually care about v, simply replace v with the relevant function of R in the original equation:
    y = -\left(\frac{\tan\left(\frac{\pi}{180}\right)}{R}\r  ight)x^2
    If you want to include non-zero friction, the result is a solution of a differential equation that is no longer a parabola.
    Last edited by slider142; June 24th 2010 at 05:01 PM.
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  3. #3
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    Thank you very much...I'm going to try this in my graphing tool.

    I really appreciate the help!
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