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
Perfectly Centered Break in Pool Ball Rack – Physics and Dynamics
billiardsds.dynamical-systemsphysics
Related Solutions
How about the following? It suggests the answer to Q1 is no.
The 2,15,6,9 and cueball are in a horizontal row and touching (or very nearly touching). The 14,cueball and 4 are in a vertical row and (very nearly) touching.
When the cueball is hit, one or both of the following will happen (depending on which direction the cueball is hit).
- The cueball moves horizontally to the right.
- The 4 moves vertically downwards, hitting the 8-ball, which travels down the table, off the bottom cushion and back up the table, off the left cushion and hits the 15 head on and slightly from the left (the angle in my illustration is probably not accurate though). The 8 and 15 will come back down the table, while the cueball is knocked horizontally to the right along with the 9 and, possibly, the 6.
In any case, the white should go in the corner pocket off the 3.
[Edit: The layout has been modified since the initial revision, but it is still the same idea.]
We need to be careful here and check that something unintended doesn't happen, like the 8,15 or 4 going in a pocket before the white. This seems unlikely and, in any case, we can put extra balls on the table to block their path if necessary.
There is still one problem though. If the cueball is hit at a specific angle, the white will start moving horizontally towards the 3, and the 9 will follow it shortly afterwards. What happens if, just after the cueball deflects off the 3, the 9 catches up and hits it? Maybe this is enough to deflect it away from the pocket? I'm not sure about this. Maybe the setup can be simplified a bit.
[The pool table above was copied from Wikimedia commons]
Here's a simpler example. The white ball is hemmed in by the top-right corner pocket. All you can do is to either hit it straight into the pocket, or against the 7 or 14. If the 7 is hit, it rolls along the top cushion into the 3, which bounces off the left cushion into the 12, which rolls towards towards the top right corner, pocketing the white. Similarly, hitting the 14 will knock the 9 towards the bottom cushion and back up into the 8, which again pockets the white.
Edited 1. Some suggestions are added at the end concerning Q2.
Edited 2. An "explanation" of spikes at $17^0$ is added at the very end...
Q1 I think that the answer to Q1 is positive provided the boundary of the tube is smooth. I'll consider the case of dimension $2$.
So, by our assumptions the light is propagating in the strip bounded by two smooth curves $L_+$ and $L_-$ that are both equidistant from the central curve $L$ on distance $\frac{1}{2}$. It is important that the whole strip is foliated by unit intervals orthogonal to all three curves, we call this foliation $F$.
Now consider our ray of light $R(t)$ propagating in the strip and introduce a function $\angle(t)$ that equals the angle between $R(t)$ and the orthogonal to $F$ in the direction of the tube. On the entrance of the tube the angle equals $0$.
Claim 1. For any moment $t$ we have $\angle(t)<\frac{\pi}{2}$.
Proof. Indeed, suppose that at some time $\angle(t)=\frac{\pi}{2}$. This means that $R(t)$ at this time goes in the direction of the foliation $F$. But since any segment of $F$ is a periodic ray in the strip, $R(t)$ must coincide with the segment, which is an absurd. END.
So, we see now that $R(t)$ will always propagate in the strip in one direction. So the only possibility for the ray to stay forever in the strip is to accumulate at some point to a segment $F_0$ of the foliation $F$. Let me explain why this is impossible. The main idea is that this is impossible in the case the curve $L$ is a circle of radius $r>1$. In this case it is easy to check the statement. The statement for general $L$ roughly follows from the fact that $L$ can be approximated well by the circle at any point.
To spell out the above in more details we can reduce the question to a question of billiards. Indeed, on the two-dimensional set of straight directed segments that join $L_+$ with $L_-$ there is a (partially defined) self map, consisting of two consequent reflection of the segment (first with respect $L_-$ then with respect to $L_+$). All the segments of $F$ are fixed points of the map. We need to show that for $F_0$ there is no a point that tends to it under the infinite iterations of the map. This self-map have three properties: 1) it preserves an area form 2) it fixes a segment (parametrizing the segments of $F$) 3) its linearisation is never identical on the fixed segment.
These 3 properties are sufficient to deduce that everything roughly boils down to the following exercise:
Exercise. Consider a sequence $a_n$, such that $a_{n+1}=a_n(1-a_n)$, with $a_0$ positive and less than one . Then $\sum_i a_i=\infty$.
PS. I think, that we can ask the curves $L_+$, $L_-$ and $L$ to be only $C^3$-smooth, but the proof uses the fact that the curvature of $L$ is strictly larger than $1$. It is not obvious if this condition can be relaxed.
Q2 This is more of a suggestion rather than an answer. But this suggestion might help to get some clues to the answer. I would suggest you to make one more picture, namely, the picture of the Phase portrait - standard thing one does when dealing with a billiard. So, one only needs to consider the trajectory and for each reflection of the trajectory from the upper curve plot the point with two coordinates:
(angle of the ray; $x$-coordinate modulo $2$)
If you plot 1500 points, a certain shape will appear. Probably the points will fill a two-dimensional domain, but according to the histogram, the trajectory will avoid a large part of the phase portrait. This just reflects the fact that this billiard is not ergodic. I think, that to understand why there are no rays with angles in $[19^0, 111^0]$ one should analyse the boundary of the shape that will appear. This boundary might correspond to some "quasi-periodic" trajectory(ies) of the billiard.
Further on Q2. I want to add a couple of remarks on Q2, that are rather superficial. So, from the experiment of Joseph we see that with some probability it turns out that the original trajectory is quasi-periodic. I.e. the segments constituting the trajectory land on a one-dimensional curve in the 2-dimensional space of all segments. This at least explains the appearance of spikes in the first histogram. Indeed, when you project a measure evenly distributed on a curve on a plane to the $x$ axes - the projected measure will have singularities at points where the vertical lines $x=const$ are tangent to the curve.
Now, I guess, that in order to really answer the question one can indeed try to prove that the initial trajectory is quasi-periodic. The billiard is rather simple of course, but I don't know how hard it will be. And before you prove this, you can not be sure that the trajectory is really quasi periodic...
Best Answer
This 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:
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.