The Solution for the Railrod Track Problem

In the figure above $c$ is the chord of the circle, $s$ is the arc length $ADB$, $r$ is the radius, $\theta$ is the central angle determined by the chord $c$, $h$ is the distance from the chord to the circle and $d$ is the distance from the center of the circle to the chord. The following relationships hold: $r = d + h$, $c=2r\sin \frac{\theta}{2}$, $d=r\cos \frac{\theta}{2}$ and $s=r\theta$. From these we get

$$\frac{s}{c}=\frac{r\theta}{2r\sin\frac{\theta}{2}}=\frac{\theta}{2\sin\frac{\theta}{2}}.$$ (1)

We can use the approximation

$$\sin\frac{\theta}{2}\doteq \frac{\theta}{2}-\frac{\theta^3}{48}$$ (2)

to get a very good approximation for $\theta$. Hence substituting (2) into (1), we get

$$\frac{s}{c} \doteq \frac{\theta}{2\left( \frac{\theta}{2} -\frac{\theta^3}{48}\right) }.$$ (3)

From this we can get

$$\theta^2\doteq 24\left( 1- \frac{c}{s}\right)$$ (4)

Thus,

$$\theta \doteq \sqrt{24\left( 1- \frac{c}{s}\right) }.$$

We can now use this value for $\theta$ to compute $h$. Now

$$\begin{eqnarray*} r &=& \frac{s}{\theta}\doteq \frac{s}{\sqrt{24\left( 1- \frac{... ...( 1- \frac{c}{s}\right) }}\left( 1-\cos \frac{\theta}{2}\right) \end{eqnarray*}$$

Thus, for our prolbem, if $c=5280$ and $s=5281$, we have

$$\theta \doteq \sqrt{24\left( 1- \frac{5280}{5281}\right) }\doteq 0.067413603$$

$$r \doteq \frac{5281}{0.067413603}=78337.305306171 \mbox{ feet}.$$

$$\begin{eqnarray*} h&\doteq& \frac{5281}{\sqrt{24\left( 1- \frac{5280}{5281}\righ... ...71\left( 1-\cos0.033706802\right)\\ &\doteq& 44.5 \mbox{ feet}. \end{eqnarray*}$$

Is that surprising, or not?

About this document ...

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Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.

The translation was initiated by Bill Richardson on 2007-04-18