I was trying to optimize the cells in a honeycomb structure and I eventually came across the rhombic dodecahedron (I am aware other figures such as the bitruncated octahedron entail greater efficiency). Bees make (let us assume it is their intention) hexagonal prisms with an apex enclosure as the one in the image to minimize the wax used. I wanted to utilize a Lagrangian function that took into consideration the surface area and volume (equated to a constraint of 0.35) in a multivariable system of equations, but need an actual formula for the volume to be able to carry out the task. The surface area I found was 3s(2h+(s√2)/2) (calculating already for the angle to be 54.7º). I have tried to use integrals of volume and have tried to estimate the irregularities of this prism with regular prisms, but none of these seem to be the correct approach (or at least I have not figured out how to use them appropiately). To sum up, my question is: ¿How can I derive an expression for the volume of a rhombic dodecahedron that has got an hexagonal base as the one shown in the figure? Any help would be appreciated.
Edit: I am now looking to optimize the sides of the shape as previously suggested. The area is 3s(2h+(s√2)/2) and the volume is ((3√3)/2 s^2 )(h-s/(2√2))=0.35 (0.35 is the constraint). Utilizing Lagrange I get sides s and h to be approximately 0.4567 and 0.8073 respectively. This yields a surface area of approximately 2.6547. The thing is that I did it as well with a regular hexagonal prism and the surface area was surprisingly less; approximately 2.254. The area used in that case was 3sa+6sh (becuase the hexagonal aperture is open) and the volume 3ash=0.35. I thought this might be because in the long run the rhombic dodecahedron might end up tessellating space more efficiently, but I am not sure. Did I go wrong in my operations, or is there a suitable explanation?
Best Answer
Your solid is the union of three truncated prisms with rhombic bases. The lower basis has sides of length $s$ and area ${\sqrt3\over2}s^2$. The longest of the four lateral edges has length $h$, the two middle lateral edges have length $h-{1\over2\sqrt2}s$ (check this please) and the shortest lateral edge has length $h-{1\over\sqrt2}s$.
By symmetry, the volume of each truncated prism is then $$ V={\sqrt3\over2}s^2\left(h-{s\over2\sqrt2}\right). $$ Muliply that by $3$ to get the volume of the whole solid.