how do you evenly tile a column?

12 Jul 2026

While walking through town the other day, I saw a pillar that looks a bit like this:

In other words, it was a circular vertical column decorated with flat tiles. I got to thinking: how do you make such a column? I would probably make a cylindrical base, then stick the tiles to it. But how would I know how big each tile should be so that they exactly divide the circumference of the pillar?

The problem is that, since the column is circular, and the tiles are straight, you can’t just divide the circumference of the pillar by the number of tiles. Each tile creates a tiny gap vs. the cylindrical pillar, and those gaps would add up over the circumference of the pillar. Your tiles wouldn’t exactly meet up when you get back to where you started!

There should be a simple, mathematical relation between the circumference of the pillar, and the perimeter of the regular polygon with $n$ vertices which circumscribes it.

Let’s draw a couple pictures. To keep things easy to visualize, we’ll look at a case where $n = 3$, but we’ll keep our math generalizable to any $n$.

Our circle has radius $r$, and the circumscribing polygon has edge length $e$.

Our task is to come up with some relationship between $r$ and $e$. (Or more precisely, an expression for $\frac{n e}{2 \pi r}$ solely in terms of $n$.)

Zooming in on the bottom-right corner of our circle, we can define a few more interesting quantities:

We define:

• $\sigma$: the central angle of the polygon.
• $h$: the height of the intersection point over the horizontal base of the polygon.
• $\theta$: the interior angle of the polygon.

Finally, if we focus on the region outlined by $r$, $h$ and the bottom of the polygon:

We define one final quantity, $\phi$, the interior angle of the right triangle formed by $h-r$ and $r$.

Here is a summary of the quantities defined so far:

$$\begin{align*} r & && \text{Inscribed circle radius.}\\ n & && \text{Number of vertices in circumscribing polygon.}\\ e & && \text{Edge length of circumscribing polygon.}\\ h & && \text{Height of next intersection point with respect to previous edge.}\\ \sigma & && \text{Central angle of circumscribing polygon.}\\ \theta & && \text{Interior angle of circumscribing polygon.}\\ \end{align*}$$

Let’s start defining these quantities in terms of each other - preferably exclusively in terms of $n$ where possible.

$$\begin{align*} \theta &= \frac{\pi (n-2)}{n} && \text{Interior angle of a regular polygon.}\\ \sigma &= \frac{2 \pi}{n} && \text{Central angle.} \\ \phi &= \sigma - \frac{\pi}{2} && \text{Follows from figure 3.} \\ \sin{\theta} &= \frac{2h}{e} && \text{Figure 3, definition of sine.} \\ h &= \frac{e}{2} \sin{\theta} && \text{Rearrange previous equation.} \\ \sin{\phi} &= \frac{h -r}{r} && \text{Figure 3, definition of sine.} \\ \sin{\phi} &= \frac{h}{r} - 1 && \text{Simplify previous equation.} \\ h &= r(\sin{\phi} + 1) && \text{Rearrange previous equation.} \\ \frac{e}{2} \sin{\theta} &= r(\sin{\phi} + 1) && \text{Set } h \text{ equations equal to each other.} \\ \frac{e}{r} &= 2 \frac{\sin{\phi} + 1}{\sin{\theta}} && \text{Rearrange terms.} \end{align*}$$

We have come up with an expression relating $e$ and $r$ but it’s far from the elegant solution we were searching for. Here is where I chucked it into wolframalpha and got a nice solution, then asked a clanker to derive it for me. The simplification process is:

$$\begin{align*} \frac{e}{r} &= 2 \frac{\sin{(\frac{2 \pi}{n} - \frac{\pi}{2})} + 1}{\sin{\frac{\pi (n-2)}{n}}} && \text{Plug in definitions of } \phi \text{ and } \theta \text{.} \\ &= 2 \frac{1 - \cos{\frac{2\pi}{n}}}{\dots} && \text{In general, } \sin{(x-\frac{\pi}{2})} = -\cos{x} \\ &= 2 \frac{2 \sin^2{\frac{\pi}{n}}}{\dots} && \text{Double angle formula.} \\ &= 2 \frac{\dots}{\sin{(\pi - \frac{2 \pi}{n})}} && \text{Simplify.} \\ &= 2 \frac{\dots}{\sin{\frac{2\pi}{n}}} && \text{In general, } \sin{(\pi-x)} = \sin{x} \\ &= 2 \frac{\dots}{2 \sin{\frac{\pi}{n}} \cos{\frac{\pi}{n}}} && \text{Double angle formula.} \\ &= 2 \frac{2 \sin^2{\frac{\pi}{n}}}{2 \sin{\frac{\pi}{n}} \cos{\frac{\pi}{n}}} && \text{Write explicitly.} \\ &= 2 \frac{\sin{\frac{\pi}{n}}}{\cos{\frac{\pi}{n}}} && \text{Cancel terms.} \\ &= 2 \tan{\frac{\pi}{n}} && \text{Definition of tangent.} \end{align*}$$

We’re in the final stretch!

Let $P = e \cdot n$, $C = 2 \pi r$. Then:

$$\begin{align*} \frac{P}{C} &= \frac{e \cdot n}{2 \pi r} && \text{Plug in definitions.} \\ &= \frac{n}{2 \pi} \frac{e}{r} && \text{Group terms.} \\ &= \frac{n}{2 \pi} 2 \tan{\frac{\pi}{n}} && \text{Plug in equation from before.} \\ &= \frac{n}{\pi} \tan{\frac{\pi}{n}} && \text{Simplify.} \quad \square \end{align*}$$

This represents the ratio of these two shapes’ circumferences, so we expect that at the limit of n, it should be 1. Therefore we subtract 1 to get an error function. This is the graph of $P/C-1$:

As expected, the error starts out very large with few tiles, then quickly drops towards 0 (the ratio converging to 1).

Our column-builders are more interested in the error with respect to the length of a tile. To illustrate the point: $P/C-1$ tends towards 0, but so does the length of our tiles. Which one converges faster, and by how much?

To get the error per tile, we use the formula $(P/C - 1) \cdot n$. (Intuitively: each tile is small, so the amount of error it sees is inversely proportional to its size $\frac{1}{n}$).

TODO: I think that this measure of relative error is wrong.

Here are the values of $P/C-1$ and $(P/C-1) \cdot n$ for up to 30 tiles:

As we can see, the ratio of $P/C$ quickly drops below 1% (taking only 19 tiles) but even with 30 tiles the per-tile error still doesn’t drops below 10%. Therefore in real-world conditions, you actually need to account for this source of error, or live with a narrower-than-intended tile on your column.

Finally, let’s address our problem statement directly. I have a column of radius $r$, and I want to wrap it with $n$ tiles. What should the edge length $e$ of each tile be so that the tiles wrap the column exactly?

Rearranging an equation given above:

$$e = 2r \tan{\frac{\pi}{n}}$$

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