Leaning Chairs

Tom White, April 2015

While on holiday, I noticed a stack of folding chairs in the apartment we were staying in. The chairs were all folded flat, and they were leaning against each other in such a way that the top of each chair was a little lower than the one to its left that it was leaning on.

To illustrate for two chairs, the leftmost chair was vertical, while the one to its right was leaning on it like this:

The third chair was leaning on the second like this:

This made me wonder: what is the shape of the curve described by the tops of the chairs (the red dots)? Assuming each chair touches the one to its left at a point that is a fixed distance from the top, the curve looks something like this:

For simplicity, assume that the chairs are infinitely thin and all of length 1, and that \(\epsilon\) is the fixed distance of the point from the top of the chair that the next chair rests on. To compute the top of the next chair from the previous one, consider the following diagram.

Looking at similar triangles \(OAB\) and \(OA'B'\), $$x_{n+1} = x_n + \epsilon a_n$$ $$y_{n+1} = y_n - \epsilon y_n$$

Then using Pythagoras' theorem, \(y_n^2 + a_n^2 = 1\), we obtain the following recurrence relation, $$x_{n+1} = x_n + \epsilon \sqrt{1 - y_n^2}$$ $$y_{n+1} = y_n - \epsilon y_n$$

with the initial condition (for the first chair): $$x_0=0$$ $$y_0=1$$

Rewriting, we get $$\begin{align}\frac{y_{n+1} - y_n}{x_{n+1} - x_n} = -\frac{\epsilon y_n}{\epsilon \sqrt{1 - y_n^2}} \\ = -\frac{y_n}{\sqrt{1 - y_n^2}}\end{align}$$

In the limit as \(\epsilon \to 0\), this becomes $$\frac{dy}{dx} = -\frac{y}{\sqrt{1 - y^2}}$$

with $$y(0)=1$$

Using Wolfram Alpha to help solve the differential equation, we find that the closed-form equation for the curve is $$x = \log{\left(1 + \sqrt{1 - y^2}\right)} - \sqrt{1 - y^2} - \log y$$