Bending a long bar

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.

Setup

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 feet1, and calling $L= 5280$, we have:

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

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

Solving

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…

Whoosh

“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.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Ugh. Don’t tell the Ninja []

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