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?