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

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

But $R$ is simply $x+d$, so

$$R^2=(x+d)^2=x^2+2xd+d^2.$$ (2)

Solving (1) and (2) for $x$ gives

$$x={{\textstyle{1\over 4}}l^2-d^2\over 2d}.$$ (3)

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

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

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

But $\theta$ is given by

$$\tan\theta ={{\textstyle{1\over 2}}l\over x} = {{\textstyle{... ...{1\over 4}}l^2-d^2} = {dl\over {\textstyle{1\over 4}}l^2-d^2},$$ (5)

so plugging $\theta$ in gives

$${\textstyle{1\over 2}}(l+\Delta l)=\left({d^2+{\textstyle{1\... ...)\tan^{-1}\left({dl\over {\textstyle{1\over 4}}l^2-d^2}\right)$$ (6)

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

For $l\gg d$,

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

Therefore,

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

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

Keeping only terms to order $(d/l)^3$,

$$dl+\Delta l\approx {4d^3\over l}+dl+{4d^3\over l}-{16\over 3} {d^3\over l}$$ (10)

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

so

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

and

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

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