How can I go about creating a 3D model / 3D image of a Kelvin’s Tetrakaidecahedron Cell / Tetrakaidecahedron. I planned on using Octave to 3D model an image it mathematically then convert that into a Blender 3D file for animation. But how does one begin to model it?
Some background on it:
https://ardentmetallurgist.wordpress.com/2020/07/26/tetradecahedron/
"Lord Kelvin is one of them and his proposed structure referred to as ‘Kelvin Cell’. Kelvin cell possesses alternative quadrilateral and hexagonal faces, i.e., there are six planar quadrilateral faces and eight equiangular non-planar hexagonal faces. OK, I know it’s crucial to imagine such structure. Let us start with an Octahedral structure. If we cut regular octahedron from all meeting points by using quadrilateral face, this will automatically create alternative hexagonal and quadrilateral structure. This structure holds 36 edges and 24 corners."
http://soft-matter.seas.harvard.edu/index.php/Tetrakaidecahedron_(Kelvin_Cell)
Assuming the polyhedron you are talking about is the one with the $24$ vertices at $(0, \pm 1, \pm 2)$, $(0, \pm 2, \pm 1)$, $(\pm 1, 0, \pm 2)$, $(\pm 2, 0, \pm 1)$, $(\pm 1, \pm 2, 0)$, and $(\pm 2, \pm 1, 0)$, then, if we use notation $(n_x, n_y, n_z; d)$ for an outwards facing halfspace, so that point $(x, y, z)$ is in the halfspace if and only if $x n_x + y n_y + z n_z \le d$, the $14$ halfspaces defining this convex polyhedron are $(0, 0, \pm 1; 2)$, $(0, \pm 1, 0; 2)$, $(\pm 1, 0, 0; 2)$, and $(\pm 1, \pm 1, \pm 1; 3)$.
The six square faces are at distance $1$ from origin with unit normals along the coordinate axes ($(\pm 1, 0, 0)$, $(0, \pm 1, 0)$, or $(0, 0, \pm 1)$).
The eight hexagonal faces are at distance $\sqrt{3}$ from origin with diagonal unit normals ($(\pm\sqrt{1/3}, \pm\sqrt{1/3}, \pm\sqrt{1/3})$).
There are multiple ways in which these can be used in Blender to construct a solid primitive; I recommend you ask Blender Stack Exchange about the details.