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@@ -2,6 +2,179 @@ pagetitle: yummers lang: en --- +# 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 straight +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 angle between two neighboring points where the circle and + polygon intersect. +- $h$: 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 $r$. + +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: + +$$ +\begin{align*} +\frac{e}{r} &= 2 \frac{\sin{\frac{2 \pi}{n} - \frac{\pi}{2}} + 1}{\frac{\pi (n-2)}{n}} && \text{Plug in definitions of } \phi \text{ and } \theta \text{.} \\ + &= 2 \frac{1 - \cos{\frac{2\pi}{n}}}{\dots} && \text{In general, } \sin{x-\frac{\pi}{2}} = -\cos{x} \\ + &= 2 \frac{2 \sin^2{\frac{\pi}{n}}}{\dots} && \text{Double angle formula.} \\ + &= 2 \frac{\dots}{\sin{\pi - \frac{2 \pi}{n}}} && \text{Simplify.} \\ + &= 2 \frac{\dots}{\sin{\frac{2\pi}{n}}} && \text{In general, } \sin{\pi-x} = \sin{x} \\ + &= 2 \frac{\dots}{2 \sin{\frac{\pi}{n}} \cos{\frac{\pi}{n}}} && \text{Double angle formula.} \\ + &= 2 \frac{2 \sin^2{\frac{\pi}{n}}}{2 \sin{\frac{\pi}{n}} \cos{\frac{\pi}{n}}} && \text{Write explicitly.} \\ + &= 2 \frac{\sin{\frac{\pi}{n}}}{\cos{\frac{\pi}{n}}} && \text{Cancel terms.} \\ + &= 2 \tan{\frac{\pi}{n}} && \text{Definition of tangent.} +\end{align*} +$$ + +We're in the final stretch! + +Let $P = e \cdot n$, $C = 2 \pi r$. Then: + +$$ +\begin{align*} +\frac{P}{C} &= \frac{e \cdot n}{2 \pi r} && \text{Plug in definitions.} \\ + &= \frac{n}{2 \pi} \frac{e}{r} && \text{Group terms.} \\ + &= \frac{n}{2 \pi} 2 \tan{\frac{\pi}{n}} && \text{Plug in equation from before.} \\ + &= \frac{n}{\pi} \tan{\frac{\pi}{n}} && \text{Simplify.} \quad \square +\end{align*} +$$ + +Here is what this equation looks like: + + + +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 $\frac{1}{n}$). + + + +Here are the values of $P/C$ and $(P/C-1) \cdot n$ for up to 30 tiles: + +|# of tiles | P/C | (P/C-1)*n | +|------------|-----|----------| +|3 |1.653986686 |1.961960059| +|4 |1.273239545 |1.092958179| +|5 |1.156328347 |0.7816417349| +|6 |1.102657791 |0.6159467451| +|7 |1.073029735 |0.511208143| +|8 |1.054786175 |0.4382894013| +|9 |1.042697915 |0.3842812313| +|10 |1.034251515 |0.3425151527| +|11 |1.028106371 |0.3091700813| +|12 |1.023490523 |0.2818862802| +|13 |1.019932427 |0.2591215493| +|14 |1.017130161 |0.2398222536| +|15 |1.014882824 |0.2232423644| +|16 |1.013052368 |0.2088378934| +|17 |1.011541311 |0.1962022837| +|18 |1.010279181 |0.185025256| +|19 |1.009213984 |0.1750656961| +|20 |1.008306663 |0.1661332692| +|21 |1.007527411 |0.1580756349| +|22 |1.006853153 |0.1507693603| +|23 |1.006265797 |0.1441133352| +|24 |1.005750997 |0.1380239172| +|25 |1.005297252 |0.1324312968| +|26 |1.004895259 |0.1272767379| +|27 |1.004537424 |0.1225104564| +|28 |1.004217499 |0.1180899697| +|29 |1.003930303 |0.1139788006| +|30 |1.003671515 |0.1101454484| + +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: + +$$ +e = 2r \tan{\frac{\pi}{n}} +$$ + # histogram-preserving tri-planar projection 31 March 2026 |