Will Buckner sent me an email with the following question recently. (I’m sharing this with permission.)

I am building a kiln using a catenary arch. The rear wall and front wall/door will be vertical and fill in the space under the arch, which has the dimensions of 41″W x 39.5″H. I need the area within this arch in order to calculate how many bricks I need to construct the two walls. Can you help me calculate that area?

Here’s my solution. I fit a parabola and a catenary in part to check my work and in part to see how different they are.

## Parabola

We’ll start with a parabola as our warmup because it’s easier.

Let *w* = 41 and *h* = 39.5. We want a quadratic function *y*(*x*) such that *y*(0) = *h* and *y*(±*w*/2) = 0.

Assume *y* has the form

*y*(*x*) = *b* − *a**x²*

*y*(0) = *h* leads to the conclusion that *b* = *h*, and *y*(±*w*/2) = 0. leads to

*a*(*w*/2) = *h*

And this leads to the final form.

*y* = *h*(1 – (2*x*/*w*)²)

## Catenary

Our catenary will have the form

*y*(*x*) = *b* – *a* cosh(*x*/*a*)

and again *y*(0) = *h* and *y*(±*w*/2) = 0. These two requirements lead to the equations

*b* − *a* = *h*

and

*b − *a cosh(*w*/2*a*) = 0.

Then

*b* = *h* + *a*

and

*h* + *a* (1 − cosh(*w*/2*a*) = 0.

The latter equation cannot be solved in closed form, but it’s easy enough to solve numerically. When I asked Mathematica to compute the root with

FindRoot[39.5 + a - a Cosh[20.5/a] == 0, a]

it failed. Then I plotted the function, saw that the root was around 8.5, then gave Mathematica a hint.

FindRoot[39.5 + a - a Cosh[20.5/a] == 0, {a, 8.5}]

Then Mathematica came back with

*a* = 8.475359432102444.

So the equation of our catenary is

*y* = *h* + *a*(1 – cosh(0.5 *x*/*a*))

where *h* and *a* are given above.

## Plots

Here’s a plot of both curves. The blue curve inside is the parabola, and the gold curve on the outside is the catenary.

Here’s the code that made the plot.

Plot[{h (1 - (2 x/w)^2), h + a - a Cosh[x/a] }, {x, -w/2, w/2}]

## Area

And now for the area, the thing I was asked to find. For the parabola

Integrate[h (1 - (2 x/w)^2), {x, -w/2, w/2}]

returns 1079.67. For the catenary,

Integrate[h + a (1 - Cosh[x/a]), {x, -w/2, w/2}]

returns 1166.56.

The number of bricks this will take is roughly the area of the arch divided by the area of a brick, but being more precise is a complicated question that depends on how much mortar you use, and how you use plan to use rectangular bricks to make a curved arch.