How to project 5D to 3D? When I have a 4D polyhedron inscribed in a sphere, I project it to 3D with the stereographic projection.
But now, if I have a 5D polyhedron inscribed in the sphere, how could I project it to the 3D space ?
Clarification
In this video (see also here), the author shows the projections of a pentaract (5 dimensional cube) in 3D. He explains that the tesseract (4 dimensional cube) has been projected to 3D by a stereographic projection. I totally understand that (I have done the same). But then, he explains that the pentaract has been projected to 3D by a double successive stereographic projection. I don't understand this point, and I'd like to. If we project the pentaract with a first stereographic projection, the result does not fall on a sphere. So how can we apply a next stereographic projection ?
 A: Having viewed the YouTube video in the question,
I think when the author says "stereographic projection" when projecting from four dimensions to three, he literally just means that the formula for the projection is
$$ P(x,y,z,w) = \frac{1}{1-w}(x,y,z), $$
as indicated in the text description of the video.
He does not mean that all the $P(x,y,z,w)$ lie on a sphere and that the projection point is also on that sphere.
For example, consider the tesseract (four-dimensional cube) when two of its cells are parallel to the projection hyperplane.
If the edge length of the tesseract is $2a$
(vertices at $(\pm a,\pm a,\pm a)$), the radius of the hypersphere on which 
those vertices lie is $\sqrt4 a = 2a.$
If the projection point is on the same hypersphere,
it is three times as far from the farthest cell as from the nearest cell.
Therefore the projection of the tesseract when two cells are parallel
to the projection hyperplane will appear as two concentric cubes,
one three times the size of the other, with edges connecting the corresponding vertices,
much like the projection that occurs at about 4:16 (and again at about 7:23) in this video.
But looking at the projections of the hypercube in 
the video in question
at about 0:37, if you pause the video just at the moment when one cube is concentric inside another, you can see that the larger cube is nowhere near three times the size of the smaller one.
This implies that the projection point is somewhere outside the hypersphere on which all the vertices of the tesseract lie.
To put it another way, the tesseract must be somewhere inside the hypersphere of radius $1$ around $(0,0,0,0),$ which is the hypersphere
that you would normally consider to be stereographically projected by
$P(x,y,z,w) = \frac{1}{1-w}(x,y,z).$
The stereographic projection from five dimensions to four is presumably
something like
$$ P(x,y,z,w,v) = \frac{1}{1-v}(x,y,z,w). $$
While I have not taken measurements of the projections to estimate the original sizes of the polytopes, I suspect that the author made
the $5$-polytopes small enough so that even with this stereographic projection to four dimensions, the projected vertices are all well inside the hypersphere $x^2 + y^2 + z^2 + w^2 = 1.$
The purpose of doing this is so that the projection from four dimensions to three will not look too distorted.
