In the previous issue, we explained the swing swing swing width calculation for the application “Putt Physics”. Also, when combined with the putter head speed calculated in the previous two articles, the swing swing width can now be calculated if the putter head speed at the lowest point is known.

Now, in the third article, the ball finally comes out. This time, we will calculate the motion of the putter head and the ball when they collide and find the initial velocity of the golf ball. From the conservation of momentum law, we get

$$m_{a}v_{a0}+m_{b}v_{b0}=m_{a}v_{a}+m_{b}v_{b}$$

formula. Each symbol means something like this.

\(m_a\): Weight of putter

\(m_b\): Weight of the ball

\(v_{a0}\): Putter head speed just before collision

\(v_{b0}\): Ball speed just before collision

\(v_a\): Putter head speed immediately after collision

\(v_b\): Ball velocity immediately after collision

Since golfers hit a stationary ball, the ball velocity vb0 just before impact is zero. Next, the coefficient of repulsion, e, should be considered.

$$e=-\frac{v_a-v_b}{v_{a0}-v_{b0}}=-\frac{v_a-v_b}{v_{a0}}$$

Actually, the number is between 0 and 1, but for simplicity, let e=1. Then the previous equation becomes.

$$v_a=v_b-v_{a0}$$

Let’s go back to the expression for the law of conservation of momentum and substitute a numerical value. The weight of a golf ball, \(m_b\), is fixed by the rules. The rule is 1.62 ounces or less, which translates to 45.93 grams. Next is the weight of the putter, \(m_a\), which I honestly don’t know how to define. Assuming a putter head, it would be around 350g, a putter club would be 500-600g, and if you include human weight, it would be, say, 50000g. Which is it?

Anyway, let’s do the math. Let \(m_a\) be 350g and 50000g as minimum and maximum values. If the velocity of the putter head was 1 m/s, the initial velocity of the ball \(v_b\) would look like this.

$$v_b=\frac{2m_{a}v_{a0}}{m_a+m_b}$$

$$m_a=350g のとき、v_b=\frac{2*350*1}{350+45.93}=1.77$$

$$m_a=50000g のとき、v_b=\frac{2*50000*1}{50000+45.93}=1.998$$

What we can see here is that no matter how large we assume \(m_a\) to be, the initial velocity of the ball will never be more than twice the head speed. Actually, this is the part of the calculation that I don’t agree with the most. When I actually putt the ball, it is more than twice as fast. I think the head speed is probably not far off, so I guess what I need to consider is how the force is transmitted.

In real-life putting, I would try to keep the head from slowing down as much as possible after making contact with the ball. I personally have been taught that the backswing and follow-through should be the same. There are various types of players who shorten their follow-through, lengthen their follow-through, or are conscious of their rhythm, but I think the common denominator is to make sure that the follow-through is firm so that the ball is not defeated. If you are not conscious of this, the ball will have a “unstable” trajectory. This means that even a person with a normal hitting style who does not make a punch shot is applying a force like a punch shot to some extent. This force is probably not reflected in the formula.

$$m_{a}v_{a0}+m_{b}v_{b0}+ ??? =m_{a}v_{a}+m_{b}v_{b}$$

I would appreciate it if you could let me know if there is a good way to add an equation in the \(???\) section or change the calculation of collision to force product. However, I believe that the head speed \(v_{a0}\) and the ball initial velocity \(v_b\) are still in a proportional relationship, so I have specified this ratio as the putter transmission ratio in the application “Putt Physics”. The default value is “3.5”, which is greater than “2”.

This value will vary depending on the characteristics of the putter and the swing method, so please find your own value and adjust it while actually putting.

This concludes the topic of collision calculation.