hypervolume of truncated tesseract?

[Don Reble]

> 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


-- 
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.