Paper Thin Stone Walls

The last time I visited King's College Chapel Cambridge I noticed a small poster which talked about the construction.  It turns out that just before they built it they discovered an innovation which allowed them to use much less masonry.  The poster said that if you build an arch such that it is possible to draw a catenary between the inside and outside wall, then the arch will be stable.  This discovery meant that they could make the walls of King's College Chapel as thin as they wanted, subject to their ability to measure accurately.

This got me thinking.  Why should that be?  Then I realised that it all has to do with the name: catenary.  A catenary is a hyperbolic cosine like $cosh(x)$.  (Obviously if you stretch or translate along the x or y axes it is still a catenary.)  The name comes from catena or "chain", because it is the shape a chain makes if held in two places at the same height.

Now imagine a chain being held in that manner.  Each link has a weight and is being pulled in two different directions by the links either side of it.  The sum of these forces must be zero because the link remains stationary.  You can draw the diagram of this, in which the weight is a vector pointing downwards and the other two forces are tensions that point along the chain.  If you turn that picture upside down and reverse all the arrows you have a diagram that shows a different structure - an arch. The tensile forces become compressive forces but - critically - it is still a diagram of a structure where the forces on each element are in balance.

The new diagram is of an arch made from paper thin walls.  However if the walls are made thicker, but still allow a catenary to be drawn within their cross section, and the density (defined as mass per length of catenary) is constant, then the diagram still works!  (You just have to imagine a single point of contact between each block - the key point is that no frictional forces are needed between the blocks, only compressive ones along the catenary.)

So, why is a catenary the shape a chain makes?  Let $\rho$ be the density of the chain and let $\textbf{T} = (T_x, T_y)$ be the tension in the chain.  Also, let $x=0$ be the lowest point of the chain. Then since links are not accelerating along the x-axis we must have

$$
T_x = constant
$$

And since $T_y$ counteracts the force of gravity
$$
\begin{align}
T_y &= \int_0^x dweight \\
 &= \rho g \int_0^x ds \\
 &= \rho g \int_0^x \sqrt{d{x'}^2+dy^2} \\
 &= \rho g \int_0^x \sqrt{1 + \big(\frac{dy}{dx'}\big)^2}dx'
\end{align}
$$

So
$$
\begin{align}
\frac{dy}{dx} &= \frac{T_y}{T_x} \\
 &= \frac{\rho g}{T_x} \int_0^x \sqrt{1 + \big(\frac{dy}{dx'}\big)^2}dx'
\end{align}
$$

Applying the Fundamental Theorem of Calculus gives us

$$
\frac{d^2y}{dx^2} = \frac{\rho g}{T_x} \sqrt{1 + \big(\frac{dy}{dx}\big)^2}
$$
which gives us the differential equation
$$
\alpha^2 {y''}^2 - {y'}^2 = 1
$$
where $\alpha = \frac{T_x}{\rho g}$.   Finally, using the identities
$$
\begin{align}
cosh(x) &= \frac{1}{2}(e^x + e^{-x}) \\
sinh(x) &= \frac{1}{2}(e^x - e^{-x})
\end{align}$$

It is easy to show that $y = \alpha cosh(\frac{x}{\alpha})$ is a solution to the differential equation.  Lovely.

POSTSCRIPT:

The other thing I found interesting about King's College Chapel - mainly because it fits in with my ideas about both monarchy and religion - was the depiction of Christ on the cross in the stain glass of the east window.  Apparently Henry VIII demanded that the face be replaced with one more resembling his own.