All the below is based on the paper the physics of putting from A.R. Penner. The goal is to determine the path of a golf ball.

A golf ball is launched with velocity 𝑣 and a launch angle of 𝛽 towards a hole along the y-axis. The "green" is level.

enter image description here

We know that the components of 𝑣 are

$$ v_x = v \cdot \sin{(\beta)}$$


$$ v_y = v \cdot \cos{(\beta)}$$

If $\beta$ is 0 then $v_x$ is also equal to 0.

In this case, the golf ball stops when $v_y$ is 0. In order to determine when $v_y$ is equal to 0, we need to calculate $a_y$. This is pretty straightforward if we know the coefficient of rolling friction, $\rho$.

However, I want to determine the path for a "green" where slopes are present.The surface of the green is sloped at angles, with respect to the x-axis, $\theta$ , and $\phi$ along the y-axis.

So, if it is a straight uphill put, $\theta$ will be equal to 0 and $\phi$ will be larger than 0.

enter image description here

According to the mentioned paper we have the following formulas for acceleration:

$$ a_x = -1 \cdot g \cdot sin{(\theta)} - (f \cdot sin{(\phi)}) / m$$


$$ a_y = -1 \cdot g \cdot cos{(\theta)} \cdot sin{(\phi)} - (f \cdot cos{(\phi)}) / m$$

In addition, the direction and magnitude of the tangential component of the contact force are given by:

$$\tan(\alpha) = \frac{\rho\cdot\cos(\theta) \cdot\cos(\phi)\cdot\sin(\beta)-I\cdot\sin(\theta)}{\rho \cdot\cos(\theta) \cdot\cos(\phi) \cdot\cos(\beta)- I\cdot\cos(\theta) \cdot\sin(\phi)} $$


$$f = \frac{\rho\cdot\cos(\theta) \cdot\cos(\phi)\cdot\cos(\beta)-I\cdot\cos(\theta)\cdot\sin(\phi)}{(1 + I)\cdot\cos(\alpha)} \cdot mg$$

It is assumed that

$$ \rho = 0.131 $$

and the moment of intertia of the ball is

$$ I = 0.40 $$

According to the author

"The above expressions for f and φ along with $a_x$ and $a_y$ will allow the x- and y-components of the acceleration of the golf ball to be determined for greens of various slopes. Given these accelerations, along with the initial launch conditions, the paths of the putted golf balls can be determined."

My question is when does the ball stop? With the level green it is easy; the ball stops when $v_x$ and $v_y$ are both equal to zero. However, if we have straight uphill green where $\phi$ is large the ball will stop at some point on the hill, where $v_y$ is 0, but the ball will start roll back towards the starting location because of the steepness of the hill.

What is sort of speak the "end criteria"?

  • $\begingroup$ your equation $\tan\alpha$ is wrong. The ball stop when the magnitue of the ball velocity equal zero $v=\sqrt{v_x^2+v_y^2}=0$ $\endgroup$
    – Eli
    May 10 at 13:47
  • $\begingroup$ Notice that $ v_x=\int a_x dt+v_{xo}$ analog $ v_y$ $\endgroup$
    – Eli
    May 10 at 14:00
  • $\begingroup$ I don't understand why the equation that calculates tan $\alpha$ is wrong? Please explain. I understand your other comments $\endgroup$
    – HJA24
    May 10 at 17:14
  • $\begingroup$ I compered it with the literature that you gave $~\tan(\phi)=....~$ $\endgroup$
    – Eli
    May 10 at 17:40
  • 1
    $\begingroup$ ah that way. what do you see that I don't see $\endgroup$
    – HJA24
    May 10 at 17:56

1 Answer 1


The condition for a momentary stop is, as you say $v_x = v_y = 0$. In order for the ball to remain there indefinitely and not roll off in some direction, the acceleration of the ball must also vanish, as otherwise that acceleration will increase the velocity again.

So your condition for the final position is $$ v_x = v_y = a_x = a_y = 0$$ (with a bit of abuse of notation).

  • $\begingroup$ ah that makes sense.. But how does $a_x$ and $a_y$ change? I see no variable input parameter that makes the two acceleration components change? They are all constant values right? $\endgroup$
    – HJA24
    May 10 at 11:48
  • $\begingroup$ I assumed that the angles would change and follow the current direction of the ball, but the problem remains, yes. Looking at the paper now it seems like the author simply assumed the static friction is always high enough for the ball to remain stationary once it has stopped (but doesn't reflect that in the equations because it's not practical). This is also reflected by the fact that they only talk about a critical angle downhill, when the same problem should really also arise uphill. With that assumption the condition is simply $v_x = v_y = 0$. $\endgroup$
    – noah
    May 10 at 12:23
  • $\begingroup$ gotcha, so if I want to model it like in the real world I will use your final condition. Any tips? $\endgroup$
    – HJA24
    May 10 at 17:11

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