## What shape is the Golden Gate Bridge?

I was visiting San Francisco last week and took a boat across the bay. Looking at the Golden Gate Bridge made me wonder what shape the supporting cables should make.

Traditionally, two kinds of answers are given to this question. If a cable is suspended by two points and holds nothing but its own weight, then it assumes the shape of a hyperbolic cosine. If on the other hand it is holding a constant weight along its span and the weight of the cable is negligible, it is a parabola. But what happens when the cable is holding the weight of a bridge but we do not want to ignore the weight of the cable itself entirely? Wikipedia states that the shape is a parabola – but is it??? There is a great tool in a mathematician’s toolbox: asymptotic analysis.

Let’s begin by drawing a force diagram for a segment of bridge.

Balance of forces means that all the vector forces must add to 0. The key property of cables we exploit is the fact that they can transmit force only tangent to rope itself. Therefore we know that the sum of the vertical components must be equal to the weight supported. The weight supported has two components: $F_r$ and $F_b$, the forces exerted by the mass of the rope and of the bridge respectively. The former is proportional to the line density of the rope $\rho$ multiplied by the arc length:

$F_r~=g~dx ~\rho~\sqrt{(1+y'^2)}$

whereas the force exerted by the bridge is only proportional to $dx$ and to the line density of the bridge $P$:

$F_b~= g~dx~P$

Invoking balance of forces leads to:
$f\frac{ y'(x+dx)-y'(x)}{dx}=g~P+g~\rho~\sqrt{(1+y'^2)}$

Taking the limit as $dx\rightarrow~0$ leads to a second order ordinary differential equation:

$f y''=g~P+g~\rho~\sqrt{(1+y'^2)}$

with two boundary conditions for y:

$y(-a)=y(a)=0$

If the first term of the equation was the only one that mattered (the weight of the cable is ignored) the solution is a quadratic. If the second term is the only one that matters (the weight of the bridge is ignored) then the solution is a hyperbolic cosine. Let us non-dimensionalise the problem choosing $\bar{x}a=x$ and $\frac{g~P}{f}a^2 \bar{y}=y$. Then dropping the bars and defining two new non dimensional constants we end up with the following problem:

$y''=1+\epsilon~\sqrt{(1+\gamma~y'^2)}~~~~y(-1)=y(1)=0~~~~\epsilon=\frac{\rho}{P}~~~\gamma=\left(\frac{a~g~P}{f}\right)^2$

I think that it may be possible to solve this (WolframAlpha gives this solution to the definite integral. However a far better approach is to realise that we are interested in the problem where $\epsilon$ is small (the bridge is “denser” than the cable) and $\gamma<1$ (the reason for this will become apparent). Therefore expand the solution in the following power expansion:

$y=y_0+\epsilon~y_1+\epsilon~\gamma~y_2+O(\epsilon~\gamma^2)$

Inserting this into the field equation allows us to write down the following problem:

$y_0''=1~~~~y_1''=1~~~~y_2''=\frac{y_0'^2}{2}$

Solving these equations with the correct boundary conditions leads to the following solution to order $O(\epsilon\gamma)$:

$y=(x^2-1)+\epsilon(x^2-1)~+~\epsilon~\gamma\frac{x^4-1}{12}$

The first term is the quadratic corresponding to the weight of the bridge, the second term corrects the parabola for the weight of the cable and third corrects the parabola for the fact that the weight of the cable is exerted not over the length of the bridge, but over the arc-length of the cable. Now we can see why $\gamma$ must be small. It corresponds to the square of the ratio of the droop of the bridge to half its span. Patently the bridge does not droop more than the half its width, therefore $\gamma$ must be smaller than one.

Pheeew! So it looks like the solution is indeed a quadratic with some correction terms that are fourth order in space.

I am a researcher in solar energy at the University of Oxford. I am interested in mathematics, programming and trying to understand why things work. I also like the great outdoors and riding my bike.
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### 8 Responses to What shape is the Golden Gate Bridge?

1. Matt says:

Cool, I really enjoyed that! Watch though I think by ‘hyperbolic sign’ you meant cosh?
Also the justification for the left hand side of the DE was kinda glossed over. Would there be some constant for the tension of the rope?
Matt

2. Thanks for reading and finding the typo: you are correct.
I believe the tension in the rope is wrapped into the parameter that determines the constant horizontal force on the cable $f$. In principle this could be determine knowing – for example – the length of the rope. I guess that knowing the desired tension would tell you what value of the horizontal force you need.

3. Matt says:

Cheers.
Is there a little f missing from the first DE?

4. Thara says:

I would have thought that it would be cosh.

• that is only if the bridge has no weight . In the limit that the cable has no weight it would be a parabula. In the limit where the line density of the cable is small compared to that of the bridge it is a quadratic plus a quartic!

5. Daniel says:

What about the weight of the support (vertical) cables? They are not constant across the span.