Ask Uncle Colin: A dangling rope

A Hanging Rope

Dear Uncle Colin,

I’m designing a small cathedral and have an 80-metre long rope I want to hang between two vertical poles. The poles are both 50 metres high, and I want the lowest point on the rope to be 20 metres above the ground. How far apart should I place the poles?

Geometry And Undefined Definite Integration

Hi, GAUDI, and thanks for your message!

A free-hanging rope generally forms a catenary curve, and can be modelled with a function of the form $y = a \cosh\br{\frac{x}{a}} + b$, as long as the turning point is placed at $x=0$.

What we need to do is find an expression for the arclength of the curve between two arbitrary limits and solve for where that is equal to 80.

Let’s get rid of the $b$ first

First of all, we know that when $x=0$, $y=20$ and when $x=\pm K$, then $y= 50$.

This points us at $20 = a + b$ and $50 = a \cosh\br{\frac{K}{a}} + b$, leaving us with $a\br{\cosh\br{\frac{K}{a}} – 1} = 30$.

So how about that arclength?

The arclength of a curve can be expressed as $L = \int_a^b \sqrt{ 1 + \br{ \dydx }^2} \dx$. In our case, $\dydx = \sinh\br{\frac{x}{a}}$, so the integral becomes $\int_{-K}^K \sqrt{1 + \sinh^2 \br{\frac{x}{a}}} \dx$. Because $1+\sinh^2(z) = \cosh^2(z)$, this becomes $\int{-K}^{K} \cosh\br{\frac{x}{a}} \dx$, or $2a \sinh \br{\frac{K}{a}}$.

We now have $a\br{\cosh\br{\frac{K}{a}} – 1} = 30$ and $a \sinh \br{\frac{K}{a}} = 40$.

Rearranging gives $a \cosh \br{\frac{K}{a}} = 30 + a$ and $a \sinh \br{\frac{K}{a}} = 40$; squaring both gives $a^2 \cosh^2\br{\frac{K}{a}} = 900 + 60a + a^2$ and $a^2 \sinh^2\br{\frac{K}{a}} = 1600$.

Because $\cosh^2(z) – \sinh^2(z) = 1$, we can subtract these to get $a^2 = -700 + 60a + a^2$, or $a = \frac{35}{3}$.

But it’s $K$ we want!

What’s the simplest equation we’ve got with a $k$ in? I’d say $a \sinh\br{\frac{K}{a}} = 40$; and now we know that $a = \frac{35}{3}$ this becomes $\sinh\br{\frac{3K}{35}} = \frac{24}{7}$.

Now, $\arsinh(z) = \ln \br{z +\sqrt{z^2 + 1}}$, so we now have $\frac{3}{35}K = \ln \br{ \frac{24}{7} + \sqrt{ \br{\frac{24}{7}}^2 + 1} }$.

That last part under the square root is, reassuringly, two legs of a Pythagorean triple, so $ \sqrt{ \br{\frac{24}{7}}^2 + 1} = \frac{25}{7}$ – and neatly, that makes the argument to the logarithm… just 7!1

We end up with $K = \frac{3}{35} \ln(7)$, so the posts must be $\frac{6}{35}\ln(7)$ metres apart.

Hope that helps — good luck with the cathedral. Let me know when it’s finished!

– Uncle Colin

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. I mean, just 7, not 7! []

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