### Perfectly centered break of a perfectly aligned pool ball rack

Imagine the beginning of a game of pool, you have 16 balls, 15 of them in a triangle <| and 1 of them being the cue ball off to the left of that triangle. Imagine that the rack (the 15 balls in a triangle) has every ball equally spaced apart and all balls touching all other appropriate balls. All balls are perfectly round. Now, imagine that the cue ball was hit along a friction free surface on the center axis for this triangle O-------<| and hits the far left ball of the rack dead center on this axis. How would the rack react? I would imagine this would be an extension of newtons cradle and only the 5 balls on the far end would move at all. But in what way would they move? Thanks

For many particles in contact, conservation of energy and momentum do not determine the outcome, so I don't think your problem has a unique solution, unless only the far left ball moves. Physically, there are fractional forces between the balls and between the balls and surface, that depend on how the arrangement was set up.

I would appreciate it if someone else could give the correct tags for this problem. I'm not a complete genius at physics and I could not create the tags I thought were relevant, nor find any tags that I believed to be more relevant. Vector spaces was a guess as it had vector in it. I am writing a pool simulation program, and am looking for a qualitative answer to base my program around, rather than a specific quantitative result.

I re-tagged the question (for approval; I don't have enough rep). However, I wonder whether this question would be better at the physics site?

At the moment of impact, force balance would imply that the forces propagate only along the sides of the triangle. Therefore, after the impact the corner balls and the cue ball would be moving. Solving for energy and momentum conservation we get that the cue ball moves back at 1/5 its original speed and the corner balls move at $2\sqrt{3}/5$ that speed.

Hmm, I guess you are right. The problem seems underdetermined, with my solution being only one possibility.

@GlenTheUdderboat: the "reasonable" solution is the one I gave, since the impact on the top rack ball is divided between the next two as given by the cosine, but the impact on those balls cannot be balanced by the inside balls, only by the next ball down along the edge.

Those simulations do not take into account quantum perturbations, aka low level randomness which cant be ignored. Once the cue ball gets hit then its wave function for vector motion collapses to one solution (the system get measured) forcing its trajectory down an unpredictable path, roughly known but not exactly. ie its impossible to hit the first ball perfectly, the question is flawed.

I protected the question as there were *two* "answers" by new users in a short time that did not really answer the questions. PLease comment-notify me if you want this undone.

@tacorama I think physically the roughness of the balls (and hence friction) is probably more important than quantum effects. The fact that the outcome depends on the chosen potential means that it is important to get the physics right to develop an accurate simulation tool.

100k views in 12 days for such a particular question, unbelievable!

Anyone care to model a bowling bowl hitting square in the 1-3 pocket?

Jim Belk Correct answer

7 years agoThis question was cross-posted on Math Stack Exchange. Here is a copy of my answer for it there.

This is it. The perfectly centered billiards break. Behold.

## Setup

This break was computed in

*Mathematica*using a numerical differential equations model. Here are a few details of the model:- All balls are assumed to be perfectly elastic and almost perfectly rigid.
- Each ball has a mass of 1 unit and a radius of 1 unit.
- The cue ball has a initial speed of 10 units/sec.
- The force between two balls is given by the formula $$ F \;=\; \begin{cases}0 & \text{if }d \geq 2, \\ 10^{11}(2-d)^{3/2} & \text{if }d<2,\end{cases} $$ where $d$ is the distance between the centers of the balls. Note that the balls overlap if and only if $d < 2$. The power of $3/2$ was suggested by Yoav Kallus in the comments, because it follows Hertz's theory of non-adhesive elastic contact.

The initial speed of the cue ball is immaterial -- slowing down the cue ball is the same as slowing down time. The force constant $10^{11}$ has no real effect as long as it's large enough, although it does change the speed at which the initial collision takes place.

## The Collision

For this model, the entire collision takes place in the first 0.2 milliseconds, and none of the balls overlap by more than 0.025% of their radius during the collision. (These figures are model dependent -- real billiard balls may collide faster or slower than this.)

The following animation shows the forces between the balls during the collision, with the force proportional to the area of each yellow circle. Note that the balls themselves hardly move at all

*during*the collision, although they do accelerate quite a bit.## The Trajectories

The following picture shows the trajectories of the billiard balls after the collision.

After the collision, some of the balls are travelling considerably faster than others. The following table shows the magnitude and direction of the velocity of each ball, where $0^\circ$ indicates straight up.

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{ball} & \text{cue} & 1 & 2,3 & 4,6 & 5 & 7,10 & 8,9 & 11,15 & 12,14 & 13 \\ \hline \text{angle} & 0^\circ & 0^\circ & 40.1^\circ & 43.9^\circ & 0^\circ & 82.1^\circ & 161.8^\circ & 150^\circ & 178.2^\circ & 180^\circ \\ \hline \text{speed} & 1.79 & 1.20 & 1.57 & 1.42 & 0.12 & 1.31 & 0.25 & 5.60 & 2.57 & 2.63 \\ \hline \end{array} $$

For comparison, remember that the initial speed of the cue ball was 10 units/sec. Thus, balls 11 and 15 (the back corner balls) shoot out at more than half the speed of the original cue ball, whereas ball 5 slowly rolls upwards at less than 2% of the speed of the original cue ball.

By the way, if you add up the sum of the squares of the speeds of the balls, you get 100, since kinetic energy is conserved.

## Linear and Quadratic Responses

The results of this model are dependent on the power of $3/2$ in the force law -- other force laws give other breaks. For example, we could try making the force a linear function of the overlap distance (in analogy with springs and Hooke's law), or we could try making the force proportional to the

*square*of the overlap distance. The results are noticeably different## Stiff Response

Glenn the Udderboat points out that "stiff" balls might be best approximated by a force response involving a higher power of the distance (although this isn't the usual definition of "stiffness"). Unfortunately, the calculation time in

*Mathematica*becomes longer when the power is increased, presumably because it needs to use a smaller time step to be sufficiently accurate.Here is a simulation involving a reasonably "stiff" force law $$ F \;=\; \begin{cases}0 & \text{if }d \geq 2, \\ 10^{54}(2-d)^{10} & \text{if }d<2.\end{cases} $$

As you can see, the result is very similar to my initial answer on Math Stack Exchange. This seems like good evidence that the behavior discussed in my initial answer is indeed the limiting behavior in the case where this notion of "stiffness" goes to infinity.

As you might expect, most of the energy in this case is transferred very quickly at the beginning of the collision. Almost all of the energy has moves to the back corner balls in the first 0.02 milliseconds. Here is an animation of the forces:

After that, the corner balls and the cue ball shoot out, and the remaining balls continue to collide gently for the next millisecond or so.

While the simplicity of this behavior is appealing, I would guess that "real" billard balls do not have such a force response. Of the models listed here, the intial Hertz-based model is probably the most accurate. Qualitatively, it certainly seems the closest to an "actual" break.

**Note:**I have now posted the*Mathematica*code on my web page.Beautifully effective animations! :-)

"Of the models listed here, I think the initial, linear model is probably the most accurate." Hertz's theory of non-adhesive elastic contact (http://en.wikipedia.org/wiki/Contact_mechanics) gives $F\sim(2-d)^{3/2}$.

@YoavKallus Thanks for the suggestion! I have incorporated this into the model.

As a novice programmer who wanted to check that his pool break simulation that he'd programmed had the correct end result, I have been absolutely blown away by this answer. My simulation currently represents the incredibly stiff result, though I feel spurred on now to attempt to incorporate the maths of the linear answer into my design. I'm going to be spending the next week just trying to wrap my head around the maths that went into this. Thanks @JimBelk for an incredibly extensive answer!

@user2376055 From a programming perspective, everything difficult here was done by *Mathematica*, which has a built-in numerical differential equations solver. I don't know what algorithm(s) it uses to solve a problem like this, but it's potentially very complicated, and I'm not sure I would have been capable of solving this problem in, say, Java (or at least, not without finding a good numerical mathematics library). If you like, I'd be happy to share my *Mathematica* code with you.

Using the trick suggested in this answer, you can probably figure out which numerical method Mathematica is using.

How stable is all this to small random errors of the initial position of the balls in the triangle?

That's not a billiards break, it's a pool break (as was requested). Billiards only uses three balls, and doesn't have breaks.

@MikeScott Interesting. According to Wikipedia, "billiards" can be used as an umbrella term for all cue sports. However, it *is* more precise to say that this is a pool break, or specifically a break of an eight-ball rack.

@JimBelk Must be an example of different usage in the US and UK. In the UK, billiards would only ever refer to the game played with two white balls and one red ball. Which I see Wikipedia calls "English billiards", so I guess that explains it.

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Alex Degtyarev 7 years ago

This seems like a nice question, but what does it have to do with the vector spaces. Did you try $3+1$ balls to begin with?