# Cue the Queue Solution

The following is a solution to Puzzle #14 – Cue the Queue. Take a look if you haven’t already.

Let’s consider just a single collision in the chain. The diagram above should help you visualise it.

The ingoing ball’s trajectory makes an angle $\theta$ with the axis. The outgoing ball’s path makes an angle $\varphi$ with the axis. It is important to note that, when the two balls collide, the outgoing ball travels in a direction parallel to the line joining the centres of the balls at the moment of collision. This is how we justify the fact that the outgoing angle $\varphi$ and the angle $\varphi$ shown in the diagram are one and the same.

The objective is to find $\varphi$ in terms of $D,d$ and $\theta$. There are probably a hundred ways of doing this; the following is the method that sprung to my mind.

The law of sines says the ratio of any side of a triangle to the sine of the angle ‘opposite’ it is a constant. This means that for the triangle in the diagram above,

$\displaystyle \frac{\sin\theta}{2r}=\frac{\sin(\pi-(\theta+\varphi))}{D}$

This can be simplified:

$\displaystyle \frac{\sin\theta}{d}=\frac{\sin(\theta+\varphi)}{D}$

Here we’ve used the factor that twice the radius of a circle is its diameter $d$, and that $\sin(\pi-x)=\sin x$.

Now in theory, $\sin\varphi$ can be solved for by using a compound angle formula, rearranging, squaring, rearranging, and finding the relevant root of a quadratic equation. But we’ll use an approximation, exploiting the fact that the balls are much further apart than they are wide. In this case, all angles are small:

$\sin\theta\approx\theta,\ \sin(\theta+\varphi)\approx\theta+\varphi$

Then the equation above is easily rearranged to give

$\displaystyle \varphi\approx\Big(\frac{D}{d}-1\Big)\theta$

In fact, you could go further and say

$\displaystyle \varphi\approx\frac{D}{d}\theta$

assuming $\frac{D}{d}\gg 1$. You might like to think about why $D$ and $d$ have the positions they do in this equation.

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This is really the key point: we’ve learnt that the outgoing angle $\varphi$ is some large multiple of the ingoing angle $\theta$. Every time a collision happens, a relatively small divergence from the axis is amplified to a much greater one. And remember, this happens at every collision. If we label the angle at which the $n$th ball travels $\theta_n$ (taking the very first ball to be the zeroth ball), then

$\theta_1=\frac{D}{d}\theta_0$

$\theta_2=\frac{D}{d}\theta_1$

$\theta_3=\frac{D}{d}\theta_2$

which is to say that

$\displaystyle \theta_n=\Big(\frac{D}{d}\Big)^n\theta_0$

Now we ask: at what value of $n$ will the ingoing ball miss entirely? Some thought shows that this occurs when

$\displaystyle \sin\theta_n\ge\frac{d}{D}$

or approximately,

$\displaystyle \theta_n\ge\frac{d}{D}$

Using our expression for $\theta_n$ yields

$\displaystyle \Big(\frac{D}{d}\Big)^n\theta_n\ge\frac{d}{D}$

Rearrangement gives

$\displaystyle n\ge\frac{\ln(\frac{1}{\theta_0}\frac{d}{D})}{\ln(\frac{D}{d})}$

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Let’s plug in some numbers to give an idea of the size of $n$. Suppose the balls are roughly ten times further apart than they are wide. That is,

$\displaystyle \frac{D}{d}\approx 10$

Let’s take the angle at which the very first ball travels relative to the axis is something small – say

$\theta_0=10^{-9}$

Then

$\displaystyle n\ge\frac{\ln(10^9)}{\ln(10)}$

$n\ge 9$

Wow! So if you’re off axis by just one billionth of a radian, the greatest number of balls you’re ever going to hit is a tiddly 9. Errors propagate so rapidly through this system that the chain reaction will break down almost immediately, no matter how accurate you think you are.

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A remark: the post about the motivation for statistical mechanics cited rapid error propagation as a major source of problems. It so happens that, at standard temperature and pressure, gas molecules in the atmosphere are roughly ten times further apart than they are wide, so the example above provides a fair analogy with interparticular collisions. It gives an idea of how rapidly a deterministic model breaks down if your data is truncated.

What’s more, the mean free path of a molecule in the atmosphere is a few tens of nanometers, but the average speed at which it travels is a few hundred of meters per second. This means a particle undergoes ten collisions in a nanosecond or so. If your data is wrong by one part in a billion, your simulated system will be completely incomparable to the true system in the blink of an eye.