El corte de Leones

The Alhambra, in Granada, Spain, is one of the most precious treasures of architecture, and rightfully a UNESCO World Heritage Site. I had the chance of visiting it a few years ago, and I had the chance of taking tons of photos. Recently, and that’s why I’m telling you about the Alhambra, I stumble upon a couple of texts that say that the proportions of the Alhambra are somewhat all built around ratios of $\sqrt{2}$. But, well, is it true?

In this week’s entry, let us first see what geometry $\sqrt{2}$ is supposed to give us in the Alhambra, then we will look for it in the actual building, and finally, we will see that we can arrive to the same results with a much easier method.

The proportions of the façades of the different buildings of the Alhambra are supposed to be constructed around the $\sqrt{2}$: squares for the columns, rectangles of (relative) height $\sqrt{2}$ and width $1$ for the architrave and frieze, and then $\sqrt{1+\sqrt{2}}$ for the cornice, etc. OK, let’s see how we construct that with nothing more than a ruler and a compass. First, we start with a square and, centering the compass on the lower-left corner, we draw a circle:

Let the compass stay centered on the same corner, but let it stretch to the upper-right corner, with diameter $\sqrt{2}$, as shown by the dotted line, and let draw an arc of a circle from the upper-right corner to directly above the lower-left corner. We complete with a new rectangle that touches the arc as follows:

The new side has height $\sqrt{2}$. We can repeat the process:

The new side has height $\sqrt{1+\sqrt{2}^2}$. We repeat:

The new side has height $\sqrt{1+\sqrt{1+\sqrt{2}^2}^2}$.

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Now, are those proportions really found in the architecture of the Alhambra? Well. Some. Kind of. No. Not really.

I sifted through my pictures of the Alhambra and found a couple of candidates, but none that matched very exactly the theory. Let’s look at the puerta del vino (the door of wine, where wine merchants entered the fortress) and superimpose the rectangles $1:1$, $\sqrt{2}:1$, $\sqrt{1+\sqrt{2}^2}:1$:

The square doesn’t match anything special. The $\sqrt{2}:1$ rectangle matches almost perfectly, but the $\sqrt{1+\sqrt{2}^2}:1$ doesn’t match anything really. I also went through my pictures of the Court of the Lions (the corte de leones), probably the most famous court of the palace, where we find a splendid fountain surrounded by ten lions. Unfortunately, it was (mostly) closed for restoration, but I got a couple of good enough shots anyway.

The red line is the floor level. Again, the square doesn’t match anything, or maybe the top of the three arches. The $\sqrt{2}:1$ fits the width of the façade and its total height, up to the roof, but that doesn’t match the main architectural element. The other rectangles again do not match anything. So maybe we don’t have a very complicated hierarchy. Maybe we only have $\sqrt{2}:1$?

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Now that we have kind of established that $\sqrt{2}:1$ is used (and not always, not everywhere), we can still ask ourselves 1) what would it look like if we used the hypothesized system to build $n$ layers in façades, and 2) what is the general formula, if any.

Let’s start by answering 2), because it will make answering 1) much easier. If we use the ruler-and-compass construction outlined above, we will have the following recurrence:

$r(n)=\begin{cases}1&\text{if }n\leqslant{1}\\ \sqrt{1+r(n-1)^2}&\text{otherwise}\end{cases}$

The first few values are given by:

$r(1)=1$

$r(2)=\sqrt{1+\sqrt{1}^2}=\sqrt{2}$

$r(3)=\sqrt{1+\sqrt{1+\sqrt{1}^2}^2}=\sqrt{3}$

$r(4)=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}^2}^2}^2}=\sqrt{4}=2$

$r(5)=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}^2}^2}^2}^2}=\sqrt{5}$

Simplifying correctly the recurrence always gives $r(n)=\sqrt{n}$, so it’s basically a complicated way of computing square roots.

Therefore, if we want a 10-level building, it will have a height of $\sqrt{10}\approx{}3.16228$ times its width:

(click to embiggen)

And since $r(n)=\sqrt{n}$, the series diverges, so we can’t build a finite-height building with an infinite number of layer, for that, we would need to use another process, maybe like the wild golden ratio of last week.

One Response to El corte de Leones

1. […] method is related to the Court of Lions method I presented a good while ago to compute square roots. This method, based on a simple […]