A nice thinker from Futility Closet:

A rail one mile long is lying on the ground. If you push its ends closer together by a single foot, so that the distance between them is 5279 feet rather than 5280, how high an arc will the rail make?

Feel free to have a go yourself! Spoilers below the line.

Assuming the rail forms the arc of a circle, I’m not certain it’s possible to solve this analytically (it has $\theta$s and functions thereof involved), but it’s possible to come up with a reasonable estimate.


Suppose the track forms an arc of a circle, with a radius of $R$. Let the angle subtended by the arc at the centre be $2\theta$.

Then the length of the rail is $2R\theta$, and the height of its middle above the ground is $R(1-\cos(\theta))$.

The distance between the ends of the rails is $2R\sin(\theta)$.

So, working in feet ((Ugh. Don’t tell the Ninja)), and calling $L= 5280$, we have:

  • $2R\theta = L$; and
  • $2R\sin(\theta) = L-1$.

And we want $R(1-\cos(\theta))$


I think we can assume that $\theta$ is small, certainly small enough that we can say $\sin(\theta) \approx \theta - \frac{1}{6}\theta^3$.

If we divide the two equations, we get $\frac{\sin(\theta)}{\theta} = 1 - \frac{1}{L}$.

However, $\frac{\sin(\theta)}{\theta} \approx 1 - \frac{1}{6}\theta^2$, so we can say $\frac{1}{L}\approx \frac{1}{6}\theta^2$, so $\theta \approx \sqrt{\frac{6}{L}}$.

We’re aiming for $R(1 - \cos(\theta))$. We can find $R$ from the first equation: $2R \sqrt\frac{6}{L} = L$, so $24R^2 = L^3$. We can work that out later if we need to.

How about $\cos(\theta)$? That’s not too tricky: we know $\cos(\theta) \approx 1 - \frac{1}{2}\theta^2$, which is $1 - \frac{3}{L}$ - which makes $1 - \cos(\theta) \approx \frac{3}{L}$.

Finishing up

Now we’re getting somewhere!

All we need is to work out $h = R(1 - \cos(\theta))$. I’d be tempted to square everything to begin with: $h^2 = R^2 (1 - \cos(\theta))^2$, which is $\frac{L^3}{24} \times \frac{9}{L^2}$, or $\frac{9}{24}L$.

For once, the imperial system’s fabled divisibility works in our favour: 5280 ÷ 24 is 220, and nine times that is 1980.

We still have to square root that…


“I might have known you’d show up once all the hard work was done.”

“1980 is 44 × 45, so its square root is 44.5, less $\frac{1}{89}$. Which, as we all know, is 0.011235…”

“How very Fibonacci.”

“Quite. So I’d estimate 44.49.”

“The Futilitarians just say ‘more than 44.’”

With something between a tut and a pshaw, the Mathematical Ninja was off.