You don't need a formula. Remember that there are precisely $5$ regular convex polyhedrons:
- The tetrahedron has $4$ vertices, $6$ edges, $4$ faces
- The octahedron has $6$ vertices, $12$ edges, $8$ faces
- The cube has $8$ vertices, $12$ edges, $6$ faces
- The icosahedron has $12$ vertices, $30$ edges, $20$ faces
- And the dodecahedron has $20$ vertices, $30$ edges, $12$ faces
So the most faces that a regular convex polyhedron can have is $20$, which is the icosahedron. If you want to find an estimate of the number of faces you can divide a sphere in to, given the number of points (vertices), you can make your own imaginary polyhedron using Euler's polyhedra formula. Which states that the number of vertices minus the number of edges plus the number of faces equals $2$.
$$
V - E + F = 2
$$
Notice that the number of edges equals one half of the number of sides of the face multiplied by the number of faces. This is true because when the faces are connected to form a polyhedron, each side is shared with another face, so each side must only be counted once as an edge. So if we let s equal the number of sides of whatever face you decide to use (remember you can only choose from a triangle, square, or pentagon) then we can write:
$$
\frac12 sF = E
$$
Plug this in to Euler's formula to get:
$$
V - \frac12 sF + F = 2\\
V + F \left(-\frac12 s + 1\right) = 2\\
F = \frac{2 - V}{-\frac12 s+1}
$$
Plug in your number of vertices (points) and the number of sides of the face you decide to use to obtain an estimated number of faces you can divide your surface area in to. Note that the larger the number of points on the sphere, the more accurate your answer will be. This is true because this solution was obtained from assuming that whatever face you chose to use is planar. So the more points you have, the more faces you have, so the faces take up a smaller area of the sphere, in turn making each face less hyperbolic and more planar.
I struggled with this one, but it finally became clear. I hope this answer helps someone else along the way.
The comment around the original MathForum question that suggested that one could calculate the volume of the kinds of solids described above by taking the hyperarea of the base and multiplying it by the average of the heights, while technically on the mark, its wording sent me down the wrong road.
To find the volume of these kinds of solids one needs to multiply the volume formula for a regular simplex (also shown in the original post):
times the average of the heights of the cap.
Calculating a solid with 3 heights to the cap will need a 2 dimensional regular simplex as a base (an equilateral triangle). In this case the formula for the volume of a regular simplex will give the triangle's area, hence the confusion about hyperarea.
When calculating for heights of n dimensions, the base simplex will have n-1 dimensions.
So, a little Mathematica function will do the trick:
volSimplexWithHyperCap[heights_List] := Module[{n, simplexVolume},
n = Length[heights] - 1;
simplexVolume = Sqrt[n + 1]/(n! * Sqrt[2^(n)]);
N[simplexVolume * Mean[heights]]]
volSimplexWithHyperCap[3, {1, 1, 1}]
0.433013
This corresponds correctly to the prism volume in the question.
It also, readily extends to higher dimensions:
volSimplexWithHyperCap[{1, 3, 2, 3, 5}]
0.0652186
This has the advantage over integration of being relatively simple and very fast to calculate. Not a Cholesky decomposition solution like I imagined at first, but it looks serviceable.
Maybe @MvG will think of a proof.
Thanks to everyone for their patients with all the edits.
Best Answer
(Converting a comment to an answer, as requested.)
The appearance of $\pi$ suggests that this is not the formula for the volume of a (flat-sided, straight-edged) polyhedron, whether inscribed or circumscribed about a sphere. Rather, there must be a circular component.
@LeeDavidChungLin's commented suggestion ---a cone of radius $r$ and height $\sqrt{3}\,r$--- seems to be the "most natural" one: the target volume decomposes as $$\frac13\cdot \pi r^2 \cdot \sqrt{3}\,r \;=\; \frac13\cdot(\text{area of base})\cdot(\text{height})$$
I'll note that such a cone arises from revolving an equilateral triangle of side $2r$ about an axis of symmetry.