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Railroad Track Problem


Given a straight segment of track of length $l$, add a small segment $\Delta l$ so that the track bows into a circular Arc. Find the maximum displacement $d$ of the bowed track. The Pythagorean Theorem gives

R^2=x^2+({\textstyle{1\over 2}}l)^2,
\end{displaymath} (1)

But $R$ is simply $x+d$, so
\end{displaymath} (2)

Solving (1) and (2) for $x$ gives
x={{\textstyle{1\over 4}}l^2-d^2\over 2d}.
\end{displaymath} (3)

Expressing the length of the Arc in terms of the central angle,

$\displaystyle {\textstyle{1\over 2}}(l+\Delta l)$ $\textstyle =$ $\displaystyle \theta (d+x)=\theta \left({d+{{\textstyle{1\over 4}}l^2-d^2\over 2d}}\right)= \theta \left({2d^2+{\textstyle{1\over 4}}l^2-d^2\over 2d}\right)$  
  $\textstyle =$ $\displaystyle \theta\left({d^2+{\textstyle{1\over 4}}l^2\over 2d}\right).$ (4)

But $\theta$ is given by
\tan\theta ={{\textstyle{1\over 2}}l\over x} = {{\textstyle{...
...{1\over 4}}l^2-d^2} = {dl\over {\textstyle{1\over 4}}l^2-d^2},
\end{displaymath} (5)

so plugging $\theta$ in gives
{\textstyle{1\over 2}}(l+\Delta l)=\left({d^2+{\textstyle{1\...
...)\tan^{-1}\left({dl\over {\textstyle{1\over 4}}l^2-d^2}\right)
\end{displaymath} (6)

d(l+\Delta l)=(d^2+{\textstyle{1\over 4}}l^2)\tan^{-1}\left({dl\over {\textstyle{1\over 4}}l^2-d^2}\right).
\end{displaymath} (7)

For $l\gg d$,
{dl\over {\textstyle{1\over 4}}l^2\left({1-{d^2\over 4l^2}}\...
...right)^{-1} \approx {4d\over l}\left({1+{4d\over l^2}}\right).
\end{displaymath} (8)


$\displaystyle d(l+\Delta l)$ $\textstyle \approx$ $\displaystyle (d^2+{\textstyle{1\over 4}}l^2)\left\{{{4d\over l}\left({1+{4d^2\...
...{1\over 3}\left[{{4d\over l}\left({1+{4d^2\over l^2}}\right)}\right]^3}\right\}$  
  $\textstyle \approx$ $\displaystyle (d^2+{\textstyle{1\over 4}}l^2)\left[{{4d\over l}+{16d^3\over l^3}-{1\over 3}\left({4d\over l}\right)^3 \left({1+3 {4d^2\over l^2}}\right)}\right].$ (9)

Keeping only terms to order $(d/l)^3$,
dl+\Delta l\approx {4d^3\over l}+dl+{4d^3\over l}-{16\over 3} {d^3\over l}
\end{displaymath} (10)

\Delta l \approx \left({8-{\textstyle{16\over 3}}}\right){d^3\over l} = {24-16\over 3} {d^3\over l} = {8\over3} {d^3\over l},
\end{displaymath} (11)

d^2={\textstyle{3\over 8}} l\Delta l
\end{displaymath} (12)

d\approx {\textstyle{1\over 2}}\sqrt{{\textstyle{3\over 2}} l\Delta l} = {\textstyle{1\over 4}}\sqrt{6l\Delta l}.
\end{displaymath} (13)

If we take $l=1{\rm\ mile}=5280{\rm\ feet}$ and $\Delta l=$ 1 foot, then $d\approx 44.450$ feet.

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© 1996-9 Eric W. Weisstein