Catenary Curve

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The catenary curve is the shape assumed by a cable hanging under its own weight when supported at its endpoints.

Naively, you might think it’s a parabola, and in fact the catenary and parabola are remarkably similar near their vertices. But the catenary is not a parabola, nor any part of a conic section.

The curve assumed by a freely hanging cable can be found using one important physical principle: systems tend towards the configuration that minimises their potential energy.

For example, the tendency of an object to fall can be explained by saying there is a potential energy associated with the object’s height above the ground. The higher the object is, the greater its potential energy. Hence the object moves downwards so as to minimise its potential energy.

The same principle can be used to explain the form of the catenary. A freely hanging cable will assume a shape that minimises its gravitational potential energy. There are several ways of calculating the functional form of the catenary; in the following post we’ll use the calculus of variations.

Let’s set up the problem. The x axis sits in the plane of the Earth’s surface and the y axis is parallel to the vertical direction. The shape of the catenary is described by y(x).

The gravitational potential energy of a point mass m at a height y above the ground is

U=mgy

where g is the strength of gravity. We can model the cable as a continuous chain of infinitesimal elements of differential length d\gamma. If the linear mass density of the cable is \rho,  the mass m of a tiny element of the chain is

m=\rho\,d\gamma

So the gravitational potential energy of a differential element of the cable is

dU=\rho\,d\gamma\,gy

Using Pythagoras’ theorem, the length d\gamma is related to small changes in x and y by

(d\gamma)^2=(dx)^2+(dy)^2

We manipulate this:element

d\gamma=\sqrt{(dx)^2+(dy)^2}

d\gamma=\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}dx

Hence

dU=\rho gy\sqrt{1+y'^2}\,dx

So, assuming the mass density of the cable is uniform, the total gravitational potential energy of the cable is

\displaystyle U=\rho g\int_a^b y\sqrt{1+y'^2}\,dx

where a and b are constant limits of integration. Using the minimum potential energy principle, the cable assumes a particular shape y which minimises its potential energy U. So we want to find the function y that minimises an integral … this is a job for the Euler-Lagrange equation!

The Lagrangian for this system is

L(y,y')=y\sqrt{1+y'^2}

ignoring the constant factor outside the integral. Note that this Lagrangian does not depend explicitly on x, so we can use a trick derived from the Euler-Lagrange equation which says that

\displaystyle L-y'\frac{\partial L}{\partial y'}=\text{constant}

Written out explicitly, this is

\displaystyle y\sqrt{1+y'^2}-y' \frac{\partial}{\partial y'}(y\sqrt{1+y'^2})=\text{constant}

\displaystyle y\sqrt{1+y'^2}-\frac{yy'^2}{\sqrt{1+y'^2}}=\text{constant}

We’ll call the constant a and multiply through by the square root:

y(1+y'^2)-yy'^2=a\sqrt{1+y'^2}

y=a\sqrt{1+y'^2}

\displaystyle \Big(\frac{y}{a}\Big)^2-(y')^2=1

How do we solve this differential equation? It looks a little like the identity

\cosh^2 x-\sinh^2 x=1

So let’s try the trial solution

y=a\cosh bx

The derivative is

y'=ab\sinh bx

Substituting this ansatz into the differential equation gives

\cosh^2 bx - a^2 b^2\sin^2 bx = 1

So our guess works provided

\displaystyle b=\frac{1}{a}

Hence the catenary curve is described by

\displaystyle y(x)=a\cosh \Big(\frac{x}{a}\Big)

\

The diagram below shows a catenary curve in gold and a parabola with the same endpoints and vertex in brown. As you can see, the parabola has a slightly greater curvature near the vertex, and approaches its endpoints with a smaller gradient.

parcat

The catenary has some other interesting properties. An arch in the form of an upside-down catenary is the most structurally sound of all arches. A square wheel can roll smoothly over a series of catenary bumps. The shape of the soap film formed between two rings is the surface of revolution of a catenary, called a catenoid. This is derived from a similar minimisation problem.

This is just one way of deriving the form of the catenary curve. If you don’t trust Euler-Lagrangian witchcraft, Wikipedia presents an explanation using the concept of forces, rather than of energy.

I hope this slapdash derivation has given you a glimpse of the power of the Euler-Lagrange equations. If you want a challenge, you could cook up a potential field of your own, and see how the shape of the cable changes. A catenary is formed when the potential energy of an element of the cable varies linearly with height. What happens if the potential energy varied (say) quadratically with height? To what physical situation might it correspond?

Further applications of the Euler-Lagrange equations will follow at some point this summer.

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