What is the volume of the largest truncated octahedron that can be inscribed in the unit sphere?

Question

If you were to maximize the volume of a truncated octahedron while keeping it in completely inside a given sphere, what percentage of the sphere's volume would it take up?

This question is an extension of a larger one I've been wondering. What is the largest space-filling polyhedron of any kind that can be inscribed in a sphere? Space-filling meaning, can perfectly tile the 3D plane in Euclidean space. Would, for example, a rhombic dodecahedron maximized in a sphere take up more space than the maximized truncated octahedron?

More concretely, imagine you have spheres of a valuable material you need to pack. You can pack them more densely by cutting them in a more packable shape, thus saving more of the material, but you lose whatever you cut off. Is there a space-filling polyhedron you could cut each sphere into that contains more than ~74.048% (maximum sphere packing density) of the original sphere?

Answer 1

I think the thing that you're interested in here is the covering density of the sphere, rather than its packing density - this is the lowest-density arrangement of overlapping unit spheres needed to cover every point in $\mathbb{R}^3$.

In the plane, the hexagonal lattice is optimal for both packing and covering. In three dimensions, though, they differ - the FCC lattice is best for packing, but the BCC lattice (a cubic lattice along with the centers of each cube) is the lattice offering the thinnest covering density of $\frac{5\sqrt{5}\pi}{24}\approx1.4635$. (Whether a non-lattice packing could do better is an open problem, I believe - see here.)

If you take the Voronoi cells of the BCC lattice with side length $1$, you get truncated octahedra with edge length $\sqrt{2}/4$, volume $1/2$, and circumradius $\sqrt{5}/4$, thus taking up $\frac{24}{5\sqrt5\pi} \approx 0.6833$ of the sphere, which of course ends up being the reciprocal of the covering density.

This correspondence between covering density of a sphere arrangement and relative volume of the voronoi cell means that the truncated octahedron is optimal, at least among space-filling bodies which tile by translation, and that any improvement would lead to a previously-unknown thinnest sphere covering. (The converse is not necessarily true, since the Voronoi cells of a non-lattice packing don't have to be congruent.)

For reference, though, a rhombic dodecahedron with coordinates $(\pm1,\pm1,\pm1)$ and all permutations of $(\pm2,0,0)$ has volume $16$ within a sphere of radius $2$, so occupies only $\frac{3}{2\pi}\approx0.4775$ of the circumsphere.

Answer 2

The regular truncated octahedron is an Archimedean solid; eight faces are regular hexagons and six faces are squares with common edge length $a$. For example:

All permutations of $(0, ±1, ±2)$ are Cartesian coordinates of the vertices of a truncated octahedron of edge length $a = \sqrt 2$ centered at the origin.

The volume of this truncated octahedron of edge length $a$ is $V=8\sqrt 2 a^3 = 32$. Its circumscribing sphere has radius $\sqrt 5$, so volume $(4/3)\pi \cdot(5\sqrt 5)$. Consequently the sphere is filled in the ratio:

$$ \frac{32}{(4/3)\pi \cdot(5\sqrt 5)} = 0.68329204... $$

For comparison the regular octahedron inscribed in the unit sphere has a volume of $4/3$, giving a filled ratio of $1/\pi \approx 0.31831$, so truncation makes quite an improvement. Indeed the best filled ratio of a Platonic solid inscribed in a sphere is for the dodecahedron, at about $0.6649$. See the blog post: Which Platonic Solid is Most-Spherical? which references an 1841 "Penny Cyclopaedia" printed by the Society for the Diffusion of Useful Knowledge.