Previously, the ball has started rolling. There is nothing more we humans can do. Let’s see if it will make it into the cup or come close to an OK putt. So, this time, we will calculate the rolling of the ball.
In this case, we will consider putting on a level surface parallel to a level surface with no slope on the green. The force on the ball is the acceleration of gravity \(g\), which we all love, the force derived from the ball’s initial velocity \(v_0\), and the friction on the green. The figure below illustrates this.
The newly emerged power💪 is the friction force \(\mu{}mg\). Among these, \(\mu\) is the coefficient of friction, which is the force acting in the opposite direction as the ball rolls forward. The force obtained from the putter head is gradually reduced by the frictional force and eventually comes to a stop. Friction force, friction force… What is the coefficient of friction, \(\mu\)? I don’t know.
It needs to be replaced by a coefficient that is understandable to the golfer. The force that restrains the ball from rolling is the turf on the green. You generally describe it as the “speed of the green. Did you know that this speed index is quantified? Of course, I had no idea. I learned about it for the first time when I created “Putt Physics”. The name of the device is “Stimpmeter. It is a device to measure the speed of greens, and it is sold like this.
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It is quite expensive. It seems to be the de facto standard for green speed measurement, and I wanted to buy one and measure it, but I could not afford it. Some golf courses seem to post the stimpmeter values for the day, so I would like to make use of it for “Putt Physics” calculations when I hit a course that is grateful for such a service.
This is off topic, but the distance rolled using the stimpmeter translates to the speed of the green. The unit is feet. The average golf course is 8 to 9 feet. Let’s calculate the speed of the green using the stimpmeter value as \(s\). All we need is the three equations for iso-accelerative linear motion.
$$F=ma$$
$$x=\frac{1}{2}at^2+v_0t$$
$$v=at+v_0$$
The frictional force is \(\mu{}mg\), which is the gravitational acceleration on the ball multiplied by the coefficient of friction. If the direction of rolling is +, \(ma=F=-\mu{}mg\), and \(a=-\mu{}g\).
From equation 3, since \(v=0\) when the ball stops.
$$0=at+v_0=-\mu{}gt+v_0$$
$$t=\frac{v_0}{\mu{}g}$$
When the stimpmeter \(s=9\), the coefficient of friction \(\mu=0.1\). Assuming initial ball speed \(v_0=2m/s\), the time \(t\) until the ball comes to a stop is
$$t=\frac{2}{0.1*9.8}=1.96秒$$
and stops after 1.96 seconds. From equation 2, it follows that
$$\frac{1}{2}(-\mu{}g)t^2+v_0t=-0.5*0.1*9.8*1.96^2+2*1.96=2.0m$$
The ball’s initial speed is 2m/s on a green with a stimpmeter value of s=9. On a green with a stimpmeter value of \(s=9\), if the ball is launched at an initial speed of 2m/s, it will roll 2m.
Finally, what is not being calculated in the customary “snapshot”? As I mentioned before, we ignore wind effects and air resistance. Then we consider the ball to be a perfect sphere, so we ignore ball mumbo jumbo. The other thing I didn’t include in the formula is the force consumed by rolling, but it may be included empirically in the conversion to friction coefficient and stimpmeter value. I have no proof of this, so if the rolling distance deviates significantly, I would appreciate it if you could let me know the conditions.
This completes the calculation of the motion of a ball rolling on a plane.
