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I am researching problems relating to finding the optimal packing density of tetrahedra and I am driving myself crazy with the following very elementary calculations which do not seem to make sense.

I have a container in the shape of a rectangular prism with volume $4690$ml (I measured it with water and also computed it theoretically by measuring the length, width, and height) and I am attempting to pack tetrahedra with edge length $6.7$cm.

According to Wikipedia and other websites the volume of a regular tetrahedron with edge length $a$ is given by, $$\text{Vol}(\text{Tet})=\frac{\sqrt{2}}{12}a^3=\frac{\sqrt{2}}{12}(0.067\text{m})^3=3.545 \times 10^{-5}\text{m}^3$$

I then can convert my volume of the container in terms of ml's to m$^3$ as follows:

$$\frac{x}{4.69\text{L}} = \frac{1 \text{m}^3}{1000\text{L}}$$

So, I have my container has volume $x = 0.00469 \text{m}^3$.

For my packing density, I then have,

$$\Delta = n\frac{\text{Vol(Tet)}}{\text{Vol(Box)}}=n\left(\frac{3.545 \times 10^{-5}\text{m}^3}{0.00469 \text{m}^3}\right)=0.00756n$$ where $n$ is the number of tetrahedra I can fit in the packing. I now attempted to fill the box, and following a fairly dense packing of around $0.78$, I only was able to fit $47$ tetrahedra in the container. That gives $\Delta = 0.00756(47) = 0.36$, which is almost as bad as the Bravais lattice packing! Therefore, my calculations must be off somewhere because it looks to me like I have packed around $\Delta=0.7$, but I am coming up with a calculation of $\Delta=0.36$.

Any ideas?

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Just a note: tetrahedron packing is very hard. Ex: How many tetrahedron of edge length $1$ can you fit in the unit sphere, with one vertex each at the center? That problem is open. –  Alex Becker May 23 '12 at 17:21
    
@AlexBecker: I know Tetrahedron packing is hard, I have read tons of papers on it and am familiar with the recent work done to improve the lower bound on $\Delta$; I am actually researching improving the upper bound published by Elser in 2010 and am presenting a demonstration to a general audience where I need to have a correlated packing density to the number of tetrahedra they can fit in the container. Unfortunately, my calculations make no sense to me visually and so this is why I asked my question here. –  Samuel Reid May 23 '12 at 21:33
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1 Answer

up vote 4 down vote accepted

Your calculation is probably right. The decreased density is due to the effect of the boundary. Since your box is only about 3-4 edge lengths across, the packing densities achievable for packing all of space are pretty much meaningless for the problem of packing in your box. For example, note that the densest packing of ten circles in a square is $0.69$, whereas the densest packing of the whole plane is $0.91$. I would not at all be surprised if the boundary effects are even bigger in three dimensions and for tricky shapes like tetrahedra.

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Thank you for the response, it just seems surprising that the packing density would decrease so much! –  Samuel Reid May 23 '12 at 22:59
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By the way, you could use water again to measure the free volume in the box after you put the tetrahedra in. –  Kallus May 24 '12 at 2:06
    
I had thought of this, but unfortunately my model tetrahedra do not have a solid boundary, so the water would leak in. –  Samuel Reid May 29 '12 at 19:29
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