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Consider the leaves from a bunch of trees in a terraced plaza in the Autumn. It may well happen that the tiles of the terrace are squares whose length easily exceeds the length of the stem of the leaves (assuming leaves all of the same kind). Cannot this be regarded as a Buffon process? That is, if you took a photograph of it and counted the line crossing of the leaf-stems with the lines determined by the tiles, you would get Buffon's calculation for pi. Furthermore, you could do this even in the absence of the tiles, simply by imposing an appropriate tiling over the photograph later, right? And if we can impose a tiling later, what about imposing it on a TV program, with respect to some recurring object? Would that, properly done, not constitute a Buffon process as well?

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I couldn't resist leaving this hilarious interpretation of floor "pi" –  amWhy May 26 '13 at 0:37

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up vote 15 down vote accepted

Let's do an experiment! Here's a deck with leaves on it:

enter image description here

Counting the leaves, I see about 100 leaves. Now lets assume all the stems are the same size, even for the larger and smaller leaves. Also let's try to visualize stems on some leaves which don't seem to have stems. Interpolating, I see about 30 crossings.

Now the Buffon needle problem says that the probability of a crossing is

$$P(\mbox{crossing})=\frac{2L_{stem}}{L_{deck}\pi}$$

Now the rough ratio I'm seeing from the image is $L_{stem}/L_{deck}\approx 1/2$, which gives:

$$P(\mbox{crossing})\approx \frac{1}{\pi}$$

or with around 100, leaves, we'd expect to see $100/\pi\approx 32$ crossings. Not bad, since we counted 30, which gives our experimental approximation of $\pi$ as 3.33, and since I probably counted pretty poorly, I probably have a standard error of around three leaves, which puts $\pi$ within our experimental error. By the way, it can be shown that we used an unbiased estimator for $\pi$, whose mean-square error scales something like $1/n$, where $n$ is the number of leaves, so we should get a somewhat accurate guess of $\pi\approx 3.14$ past around 1000 leaves. I leave that experiment to you.

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