Parabola for Golf Ball Roll on Green

Jun 2010

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.

May 2009
Brooklyn, NY
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
\(\displaystyle y = -\left(\frac{g\sin\left(\frac{\pi}{180}\right)}{2v^2}\right)x^2\)
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
\(\displaystyle R = \frac{2v^2\tan\left(\frac{\pi}{18}\right)}{g\sin\left(\frac{\pi}{180}\right)}\)
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:
\(\displaystyle y = -\left(\frac{\tan\left(\frac{\pi}{180}\right)}{R}\right)x^2\)
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:
Jun 2010
Thank you very much...I'm going to try this in my graphing tool.

I really appreciate the help!