Calculating a formula to express the volume of a rhombic dodecahedron (BEE PRISM WITH HEXAGONAL BASE)

Question

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?

Answer 1

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.

Answer 2

The body has 6 lateral edges with alternating lengths. Different from your choice it is more convenient to choose the length of the longer edge as the parameter $H$, since with this choice the volume of the body will be simply $V=A\cdot H$, where $A=\frac{3\sqrt3}2s^2$ is the area of the base. To see this draw a plane parallel to the base through the ends of the longer edges (two of them are $J,L$ in the figure) and prolong the shorter lateral edges till the intersection with the plane, so that the prism with parallel bases appears. Consider also the point of intersection of the central height ($h$) with the plane. Then you will find two congruent pyramids with triangle base, the volume of one of them being added to and that of the other being subtracted from the volume of the prism. Observe that the result does not depend on the value of the angle $\phi$.