## calculating the interior dimensions of an altered cube

I'm working on a project to build a Star Wars Jedi Holocron based somewhat on the one you can see images of in Star Wars Rebels. I've been able to work out most of the calculations on my own, but once I hit the third dimension my brain shuts down. Here's the reference image I'm looking at:

So what we have is a cube, with the corners cut off, and a second cube inside. Here are the dimensions I have so far:

The basic ("closed") cube is 3" on a side. Each face is broken at the midpoints of all four edges, leaving behind a diamond shape. The four triangles at the four corners therefore each have a and b dimensions of 1.5" and a hypotenuse of 2.12132". This means that the equilateral triangles that connect the square sides to each other now are also 2.12132" on a side. Because I will be beveling the edges at the joins, I'm not super concerned about the material thickness (0.05") just yet.

At this point, I'm stumped. What I need to find is the interior distance between the centers of the equilateral triangles, since the inner cube touches each of its eight corners to the center points of the eight equilateral triangles. Measuring a mockup I made from card stock tells me that the outside is about 3-7/16" high from triangle to opposite triangle (3.4375") which, after taking out the material thickness top and bottom, leaves me with a interior height of ~3.3375". This would be the diagonal of the interior cube, which means that by folllowing the Pythagorean theorem again in reverse this time, the cube should be about 1.92691" on an exterior side. This MAY be close enough for my purposes, but I'd like to get it calculated out mathematically to check my work rather than rely solely on an eyeball measurement of not-entirely-rigid material.