# hypervolume of truncated tesseract?

Don Reble djr at nk.ca
Wed Mar 28 03:37:29 CEST 2007

```> In geometry, a truncated tesseract is a uniform polychoron
> (4-dimensional uniform polytope) which is bounded by 24 cells: 8
> truncated cubes, and 16 tetrahedra. ...
> ...The truncated tesseract may be constructed by
> truncating the vertices of the tesseract at 1 / (2 + sqrt(2)) of the
> edge length."

In other words, the truncated squares are regular octagons.

> the hypervolume of a truncated tesseract is ... ?

In N dimensions, a rectangular simplex is a simplex where N
of the edges are mutually orthogonal line segments of the same
length L. It's size is L^N / N! .

From the unit hypercube, each truncated corner loses one such
simplex, of hypervolume 1 / (1632 + 1152 sqrt2). There are 16
corners, so the total loss is 1 / (102 + 72 sqrt2) = 0.00491...

That may be surprisingly tiny, but remember, those corners are
small to the fourth power, divided by 4!.

--
Don Reble  djr at nk.ca

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