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?
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.