A catenary with scale *a* is the graph of the function

*f*(*x*; *a*) = *a* cosh(*x*/*a*) − *a*.

The *x* and the *a* are separated by a semicolon rather than a comma to imply that we think of *x* as the variable and *a* as a parameter.

This graph passes through the origin, i.e. for any *a*, *f*(0, *a*) = 0. To find the scaling parameter *a* we need to specify the value of *f* at one more point and solve *f*(*x*, *a*) = *y*. This is the equation I alluded to recently as not having a closed-form solution.

Without loss of generality, we can assume *x* = 1. Why is that?

Define

*g*(*a*) = *f*(1; *a*).

Then

*f*(*x*‘; *a*‘) = *y*‘

if and only if

*g*(*a*‘/*x*‘) = *y*‘/*x*‘.

So will assume *x* = 1 for now and focus on solving for a value of *a* such that *g*(*a*) = *y*. But we will include Python code shortly that goes back to *f*, i.e. does not assume *x* = 1.

The Taylor series for *g* looks like

*g*(*a*) = 1 / 2*a* + 1 / 24*a*³ + …

and so for large *a*,

*g*(*a*) ≈ 1 / 2*a.*

Here are a couple plots showing how good this approximation is, even for *a* not that large. First, a plot of *g* and its approximation.

And here’s a plot of the relative error in approximating *g*(*a*) by 1/2*a*.

This means that for small *y*,

*g*(1 / 2*y*) ≈ *y.*

It’s fortunate that we have a convenient approximation when *y* is small, because in practice *y* is usually small: catenaries are usually wider than deep, or at least not much deeper than wide.

Since the terms in the Taylor series that we discarded are all positive, we also have a bound

*g*(1 / 2*y*) > *y.*

If we want to solve for *a* numerically, 1 / 2*y* makes a good starting guess, and it also makes a left bracket for root-finding methods that require a bracket around the root.

Here’s Python code to solve *f*(*x*, *a*)* = y* for

*, given*

*a**and*

*x*

*y*.from numpy import cosh from scipy.optimize import root def g(a): return a*cosh(1/a) - a def solve_g(y): assert(y > 0) lower = 0.5/y return root(lambda a: g(a) - y, lower).x def solve_catenary(x, y): "Solve for a such that a cosh(x/a) - a == y." return x*solve_g(y/abs(x))

Now that we can solve *g*(*a*) = *y* numerically, we can go back and see how far 1 / 2*y* is from the actual solution for varying values of *y*.

Recall that the second plot above showed that the relative error in approximating *g*(*a*) by 1 / 2*a* is large when *a* is small (and thus *y* is big). But the plot immediately above shows that nevertheless, 1/ 2*y* is a good guess at a solution to *g*(*a*) = *y*, never off by more than 0.175.

Recall also that we said 1 / 2*y* is a lower bound on the solution, and could be used as a left bracket in a root-finding method that requires a bracket. The plot above suggests that 0.18 + 1 / 2*y* would work as a right bracket

For “eluded” read “alluded”.