Let’s Derive The Catenary Equation With Calculus

What is a catenary?

A catenary (derived from the Latin catenaria meaning “chain”) is an idealized curve in physics or maths that represents the shape that a chain (or rope) assumes under its own weight when being supported only at its ends.

Robert Hooke holding a chain, which assumes the shape of the catenary

You can see this shape everywhere in the world and in nature. From spiderwebs to overhead powerlines.

The catenary looks very much like a parabola. And it wasn’t until the 17th century that Galileo Galilei recognized that the catenary is “only an approximate parabola.” He wasn’t, however, able to derive an exact expression for the catenary.

But today we have the amazing mathematical tool of calculus that we can use to derive the catenary equation.

Derivation of the Catenary Equation
Mechanics and Statics
We can start by using basic mechanics and statics. Let’s start with a catenary and draw the forces of tension at the ends where it is supported:

These forces are obvious (I hope). The rest of the forces require a little more thinking. If you take a single chain link, for example, what are the forces acting on it? One of the forces is obvious: Gravity.

And by using Newton’s Third Law of Motion it’s easy to see that each chain link will also have a force of tension on both sides because each chain link exerts a force on every chain link next to it:

Now that we understand that, let’s draw a free body diagram:

One of the best way to solve this problem is by taking an arc of the catenary with length s and considering the forces acting on it.

The weight (distributed force of gravity) of this arc must be represented using the linear 1D density of the chain because we are not dealing with a point mass:

Where μ[kg/m] is the linear 1D density of the chain, g[m/s²] the gravitational acceleration on Earth, and s[m] the arclength.

The other forces acting on the arc (point A to P) are the tension from the chain at point A and the tension at point P.

We assume the catenary to be ideal and thus so thin and flexible that any force of tension is parallel to the chain (to the catenary curve).

At point A the force of tension T₀[N] is therefore tangent to the catenary and only horizontal. At point P the force of tension T[N] makes an angle relative to the horizontal we will call θ.

The arc (and the entire chain) is in static equilibrium and thus the net force acting on it is zero: ΣF⃗ = 0.

It’s helpful to split the net force up into its horizontal (x) and vertical (y) components:

We can rewrite sinθ as cosθtanθ to get:

Eq. 2

Analysis with Calculus
Now it’s useful to re-write θ in terms of x and y. We can do this by realizing that the tangent of theta is equal to the slope of the curve:

We can then write:

Now let’s define a constant κ = μg/T₀ to simplify the expression to:

Eq. 3

Now we have a first-order differential equation. If we solve it, we will have our catenary equation y(x). So the rest is “just maths.”

“Just Maths”
We have a first-order DE but it’s not so easy to solve because it’s written in terms of s instead of only y and x. The best way I know to solve this problem is by first increasing the order of the DE by differentiating both sides of eq. 3 with respect to x:

Eq. 4

Now we have expressed the relationship between y and x with a 2nd-order DE. So it looks more complicated now. But sometimes you first have to make things more complicated to make them more simple.

Now we need the arc length equation:

As you can see this equation is nothing more than the Pythagorean Theorem applied to an infinitesimally small arc.

We can also make things more simple by defining a new variable:

Now we can re-write our arc length equation:

We can also re-write eq. 4:

Now let’s solve this DE:

The integral on the left side of the equation is a standard form and its solution is arcsinh(z) + C. Where arcsin(z) is the inverse hyperbolic sine function. So we can write:

If we take point A to have x-coodinate 0, then C=0 because the slope at that point dy/dx=z is also zero (arcsinh 0 = 0).

Thus we have:

Now let’s isolate z:

Now we finally have a first order DE in terms of y and x.

We can easily solve it,

And eq. 6 is our catenary equation at last!