Proof that the cable forms a parabola (optional)

We will show here that the cable of a suspension bridge actually hangs in the shape of a parabola: that is, a curve with equation y = Ax2. The mathematics is a little difficult, so if you are happy to accept this result, then feel free to skip this page.

We can think of the cable as a weightless string from which equal weights w are suspended at equal intervals a. We shall assume that one of the weights hangs at the centre of the string. We label this point of the cable A0, and succeeding points A1, A2, ... as in the diagram. The symbols T1, T2, ... denote the tension in the cable. At each of the points Ai, there are three forces acting: if i > 0 these are Ti, Ti+1 and w. We can resolve these forces in the horizontal and vertical directions. Since the system is in equilibrium, these forces must balance out.

Resolving horizontally at A1, A2, ... gives:

     T1 cos 1 = T2 cos 2,
     T2 cos 2 = T3 cos 3, etc.

So
     T1 cos 1 = T2 cos 2 = T3 cos 3 = ... = H say. (1)

Resolving vertically, we find

     T2 sin 2 – T1 sin 1 = w,
     T3 sin 3 – T2 sin 2 = w, etc.

Eliminating the Ti by substituting from (1) gives

     tan 2 – tan 1 = w/H = tan 3 – tan 2 = ... = tan n+1 – tan n.    (2)

This actually tells us that tan 1, tan 2, tan 3, ... form an arithmetic progression: each term after the first is obtained from the one before by adding on the constant w/H.

In particular, at the point A0,

    T1 cos 1 = H, and T1 sin 1 = 1/2.w, so tan 1 = w/2H . (3)

Now let the coordinates of A0 to be (0, 0), and determine the coordinates of A1, A2, ... .
We find:

A1 (a, a tan 1), A2 (2a, a tan 1 + a tan 2), ... ,
                             An (na, a tan 1 + a tan 2 + ... + a tan n).

We note that

a(tan 1 + tan 2 + ... + tan n)
     = a(tan n – tan n-1) + 2a(tan n-1 – tan n-2) + ...
                             + (n – 1)a(tan 2 – tan 1) + na tan 1
     = aw/H .(1 + 2 + 3 + ... + (n–1)) + n. aw/2H (substituting from (2) and (3))
     = aw/H . [ (n–1)n/2 + n/2 ] = awn2 /2H.

Thus for each n, An (na, awn2 /2H) lies on the parabola with equation y = w /2Ha . x2. This parabola touches the x-axis at the origin, and is symmetric about the y-axis.

It follows that the cable of a suspension bridge assumes the shape of a parabola.