The following is an example of a truncated octahedron, which has 24 vertices, 36 equal edges, and 14 faces:
One way to layout this 3-dimensional object onto a flat surface is as follows:
Given an edge length a, the volume V and surface area S are:
- V = (8√2)a3
- S = (6 + 12√3)a2
Since a truncated octahedral pot is missing a hexagon:
It follows that a pot's surface area Spot = (6 + 12√3)a2 - ½(3√3)a2 = 1.5(4 + 7√3)a2
Lastly, truncated octahedra can tessellate (bitruncated cubic honeycomb) in 3-space, as shown:
