Can your gravitational pull affect your billiards game?

Think of this plot in terms of area. The area of ​​the graph covered by the blue data (to hit the ball 2) is much larger than the area of ​​the graph showing the speeds needed to hit the ball 3. It becomes a lot harder to achieve a collision that involves all four balls.

Let’s do one more. What if I add 4 balls to the chain of collisions?

Illustration: Rhett Allen

Just to be clear, this is a comparison of the range of the starting velocities of the cue ball, which leads to the ball 3 hitting the ball 4. Let’s go through some rough ranges for the initial velocities of the cue ball.

To make the ball 1 hit the ball 2, the x-speed can be from close to 0 m / s to 1 m / s. (I have not calculated velocities greater than 1 m / s.) Y-velocities can be from about 0.02 to 0.18 m / s. This is a velocity range x of 1 m / s and a velocity range y of about 0.16 m / s.

In order for the ball 2 to hit the ball 3, the speed x can be from 0.39 to 1 m / s with the speed y from 0.07 to 0.15 m / s. Note that the velocity range on x has dropped to 0.61 m / s, and the velocity range on y is now 0.08 m / s.

Finally, to hit the ball 3 on the ball 4, the x-speed can be from 0.42 to 1 m / s and the y-speed from 0.08 to 0.14 m / s. This gives an x-range of 0.58 m / s and a y-range of 0.06 m / s.

I think you can see the trend: more collisions means a smaller range of initial values ​​that will result in a hit on the last ball.

Now we need to test the final case: nine balls. Here’s what it looks like:

Video: Rhett Allain

Okay, that works. But will the last ball still be hit if we take into account the extra gravitational force caused by the interaction between the cue ball and the player?

This is relatively easy to test. All I have to do is add some type of person. I will use the approximation of a spherical person. I know, people aren’t really spheres. But if you want to calculate the gravitational force due to a real player, you will have to make some seriously complex calculations. Each part of the face has a different mass and would be at a different distance (and direction) from the ball. But if we assume that man is a sphere, then it would be the same as if the whole mass were concentrated in one point. This is a calculation we can do. And in the end, the difference in gravitational force between a real and a spherical person probably won’t matter much.

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