I recently became curious to know how to calculate the volume of a regular spherical tetrahedron — that is, a subset of the unit 3-sphere S^3 bounded by 4 spherical triangles, such that all 6 dihedral angles are equal. The 2D case is just a "regular spherical triangle" with its 3 angles equal. It's well known that if the common angle is t, then its area is A(t) = 3t - π. I've perused a few papers on the volume of a spherical tetrahedron, and this is astonishingly more complicated than the 2D case. We can, however use the limiting cases and the regular 4-polytopes to assert a few of its values. Let V(t) denote the volume for dihedral angle = t, and let a = arccos(1/3). Note that 2π^2 is the volume of the entire 3-sphere S^3. Then: V(a) = 0 V(2π/5) = π^2/300 V(π/2) = π^2/8 V(2π/3) = 2π^2 V(π) = π^2 V(2π-a) = 2π^2 If we only knew the *form* of V(t), this might be (more than) enough to determine it for all t. But I don't. —Dan