The Catenary

The problem

A uniform chain of length 2.6 m hangs freely between two points 2.0 m apart at the same height. What curve does it take, and how far does it sag?

Galileo said a parabola. He was wrong, and the discrepancy is measurable.

Setting up the functional

The chain hangs so as to minimise its potential energy, subject to having a fixed length. With \(y\) measured upward and mass per unit length \(\rho\):

\[U[y] = \rho g\int y\,ds = \rho g\int_{-d}^{d}y\sqrt{1+y'^2}\,dx\]

subject to the constraint

\[S[y] = \int_{-d}^{d}\sqrt{1+y'^2}\,dx = S \quad\text{(fixed)}\]

This is a constrained variational problem, so introduce a Lagrange multiplier \(\lambda\) and extremise \(U - \lambda S\), i.e. take the integrand

\[f = (y - \lambda)\sqrt{1+y'^2}\]

(absorbing \(\rho g\)). The multiplier is not a bookkeeping device here — it has a physical identity, as we will see.

Solving it

\(f\) has no explicit \(x\), so the Beltrami identity applies:

\[f - y'\frac{\partial f}{\partial y'} = \text{const} = C\]

\[(y-\lambda)\sqrt{1+y'^2} - \frac{(y-\lambda)y'^2}{\sqrt{1+y'^2}} = \frac{y-\lambda}{\sqrt{1+y'^2}} = C\]

Rearranging and shifting the origin so \(\lambda = 0\):

\[y' = \sqrt{\frac{y^2}{C^2} - 1} \qquad\Longrightarrow\qquad \boxed{\;y = a\cosh\frac{x}{a}\;}\]

with \(a = C\). The chain hangs in a hyperbolic cosine — the catenary, from Latin catena, chain.

Fitting it to our chain

Two conditions. Half-span \(d = 1.0\) m; total length \(S = 2.6\) m. Arc length of \(a\cosh(x/a)\):

\[S = \int_{-d}^{d}\sqrt{1+\sinh^2(x/a)}\,dx = 2a\sinh\frac{d}{a}\]

This is transcendental in \(a\), so solve numerically. With \(d = 1\), \(S = 2.6\):

\[a = 0.77682\ \text{m}\]

(Check: \(2(0.77682)\sinh(1/0.77682) = 2.60000\) ✓)

The sag — the drop from the endpoints to the lowest point:

\[\text{sag} = a\left(\cosh\frac{d}{a} - 1\right) = 0.7376\ \text{m}\]

Why not a parabola?

Compare the catenary with the parabola through the same three points (the two supports and the lowest point), at the quarter-span \(x = d/2\):

	height above the low point
Catenary	0.1665 m
Parabola	0.1844 m

About 11% apart at the quarter-span — comfortably visible.

The physical difference is what the weight is distributed along. A parabola is the shape of a cable carrying a load uniform in horizontal distance — which is exactly a suspension bridge, where the deck weighs far more than the cable. A catenary carries a load uniform along its own arc length — a chain supporting only itself.

So the Golden Gate's main cables really are close to parabolic, and Galileo's error was to picture a bridge rather than a chain. In the limit of a shallow chain, \(\cosh\) expands as \(1 + x^2/2a^2 + \dots\), so the two agree to leading order — which is why the mistake survived for decades.

The multiplier's identity

\(\lambda\) shifted the origin, and \(a = C\) turns out to be

\[a = \frac{H}{\rho g}\]

where \(H\) is the horizontal tension, which is constant along the whole chain (nothing supplies a horizontal force). So the Lagrange multiplier is not an abstraction: it is a tension, and the catenary's shape parameter is the ratio of horizontal tension to weight per unit length. A taut chain has large \(a\) and is shallow; a slack one has small \(a\) and plunges.

This is the general pattern — Lagrange multipliers are usually the physical quantity conjugate to the constraint. Constrain volume and the multiplier is a pressure; constrain length and it is a tension; constrain particle number and it is a chemical potential.

Where it turns up

The inverted catenary is the optimal arch. Turn the chain upside down and its pure tension becomes pure compression, with no bending moment anywhere. Gaudí famously designed the Sagrada Família by hanging weighted strings and photographing them upside down.
Overhead power lines and mooring cables are catenaries; the sag is how installers set tension.
The catenoid — a catenary revolved about the \(x\)-axis — is the minimal surface a soap film forms between two rings, which is the same variational problem with area in place of potential energy.

Common mistakes

Forgetting the constraint. Without fixed length, minimising potential energy sends the chain to \(y\to-\infty\). The multiplier is what makes the problem well posed.
Assuming a parabola. Correct for a bridge deck, wrong for a chain.
Trying to fit \(a\) algebraically. \(S = 2a\sinh(d/a)\) has no closed-form inverse; solve it numerically and say so.
Ignoring that the catenary equation still holds for unequal support heights. The curve is the same \(\cosh\); only the position of its minimum shifts, and it may lie outside the span.

Calculus of variations · Euler–Lagrange equation · Lagrange multipliers · Brachistochrone — the other Beltrami classic · Minimal surfaces — the catenoid · Geodesics