Many analytical and mathematical properties of the snub cube can be found anywhere there is an anal professor trying to extract sun-beams from cucumbers or trying to reduce human excrement to its original food. Attempting to understand the nature of the snub cube through mathematical formalisms is like trying to get to the moon by unicycle. Though many of the analyses on the snub cube remain in obscurity, the goal here is to make intelligible to anyone the paradoxes of it's geometrical character.
As was alluded to before, anyone with poster board, compass, scissors, and tape could build what Kepler describes as the snub cube. All you need is six squares and thirty two triangles, and build a solid angle of four triangles and a square. Build it yourself.
You may think you have built the snub cube, but a little more is to be known about it before we can say we've found it. You may notice that when you stand it up on its square base, that the six square faces are corresponding to the six square faces of the cube with an exception; they're not quite aligned with being a cube. They're sort of twisted. There's also the question of whether or not these twisted squares are at right angles to each other like the faces of the cube. Now some may describe the formation of the snub cube as simply taking the six square faces of a cube and, at the same time pulling out, and twisting them until you produce thirty two triangles in between them. However, this may be physically plausible, it is not very accurate. For how would you know, a) how far to pull out and b) at what angle to twist? It appears we may need the equivalent of a sextant to be able to see relationships to try to determine the proportions of this thing. What's our measure?
One initial attempt is to take what we found in the rhombicuboctahedron, inside of the intermediary solid, and twist the triangular faces, which will in turn twist the squares, until twelve of the squares each become two triangular faces (see animation).
However, we run into the same problem as before. How much do we twist the triangles and squares before they form equilateral triangles between them? Not only that, but a much bigger problem arises: the square's and triangle's edges, when twisted on their respective faces of the intermediary, are actually growing at two different rates (see animation).
So you see, that the square's edge, when twisted until maximum, becomes the edge of the square of the intermediary. And the triangle's edge, when twisted until maximum, becomes a diagonal of the irregular hexagon, an edge which is obviously longer than the square's edge at maximum. So we know the two lengths' proportion to each other at minimum and maximum, but how do we determine the rate of change of that proportion as the two shapes are twisting? Is this perhaps an elliptical problem?
To add onto our problems, when the proportions of our intermediary solid change, then so do the square's and triangle's edges to each other. Not only at minimum and maximum, but anywhere, which means that the proportions of the edges of the intermediary solid are going to have to be specific in order for this twist to produce the snub cube!
What is it about these twisting triangles and squares that makes it so difficult to find the moment of congruence? One unique aspect about the snub cube is that this so called "twist" can occur in a left handed and right handed form, and as a result of which way it twists, will be two distinctly different forms (see figure).
This cannot occur with any other Platonic or Archimedean solid, except for the snub dodecahedron of course.
Upon further investigation of the Archimedeans, you will find, according to what types and how many regular figures come to a vertex, that the Archimedeans have classes of their own (see table).
According to each of the six classes, there is the name of the solid, and the figures that meet at each vertex of the solid; the figures written in black show the changing aspect of the vertices and the figures in red show the constant aspect. Two of the classes, in order for the aspects of the vertices to be consistent, each have to begin with a platonic solid. However, take note that the snub cube exists in a class with two other solids that have five-fold symmetry, i.e. the pentagon. However, you don't really see any pentagons in the snub cube. The only thing five-fold about the snub cube is that it has five figures (four triangles and one square) at each vertex and thus five edges at each vertex. Interestingly enough, Kepler in his snowflake paper raises the question why most trees and bushes grow flowers of five-fold symmetry. As you saw in Book I, there are many interesting properties of the pentagon, which Kepler describes as the "flag of the faculty of propagation" which is flown by living processes.
But what does the pentagon have to do with the snub cube? Is their a relationship or are we being tricked to think that the rock in the water appears to be the head of an alligator? Let's return to the Archimedean class of the icosahedron, snub cube, and snub dodecahedron. Luca Pacioli, in his "Divine Proportion," describes a construction of the icosahedron inside the octahedron, in such a manner as dividing the edges of the octahedron into the golden section and connecting them. The following animation shows this relationship according to the truncations we've discussed before, starting with the tetrahedron (see animation).
So in the animation, the truncation of what Kepler considers one of the primary solids, the tetrahedron, occurs at half way along the edges, forming the octahedron. Then, the triangles of the octahedron are twisted until they divide the edges of the octahedron into the golden section. In this case, the icosahedron can be called a snub tetrahedron.
So, the next question to ask is: if this type of truncation works on one primary solid, the tetrahedron, to produce the first solid of our sixth class from the table above, then what kind of truncation will have to be taken on another primary solid, the cube, to produce the second solid of that same class, the snub cube? If you perform this experiment, you'll see that obviously, truncating the cube until half way of its edges to produce the cuboctahedron will not be enough to generate the type of "twist" necessary to form an Archimedean congruence. You will actually have to extend the truncation past the cuboctahedron, so that the triangles will have to intersect each other. How far past? Well, why don't we try the golden section? And then twist the squares and triangles to the golden section of the side of our truncating triangles, similarly what happened in the icosahedron/octahedron construction (see animation).
Is this a valid construction, or is it a fool's snub? It perceptibly seems plausible, even somewhat conceptually. Once again, the reader must decide, after they've gone through the construction more thoroughly.
As Kepler proved many times; in a universe which is self-developing, your models certainly seem plausible to your senses, and may even seem structurally sound, but are only on the surface of a universally principled reality.