The Archimedeans can be described, physically, as truncations of the vertices of the five regular solids. These truncations, however, are based on specific proportions of the figures that compose the regular solid. For example, if we trisect each side of a triangle into three equal parts and connect the points of the division in a certain way, because of the nature of the way the triangle is proportioned this division will produce a hexagon (see figure).
However, if we tried dividing the four sides of the square into a similar manner, that is trisecting, will we produce an equilateral octagon? (see figure)
It seems that we have to slightly change our division if we are to produce an equilateral octagon. (see figure)
Now if we take this idea to the tetrahedron, what solid figure will be produced? (see figure)
We create a body which has two hexagons and one triangle at every vertex. In doing this we have now generated one of our Archimedean congruences. It's called the truncated tetrahedron.
Now, for the remainder of our examples, we will deal solely with the cubic family, that is the truncations of the cube and octahedron. Kepler states in Book V that the volumes of the dodecahedric family are "inexpressible," i.e., do not have relationships to each other that can be expressed in rational numbers. So we'll leave that investigation for another time.
Let's take a look at an interesting property of the duals. If we take the midpoints of the edges of both the octahedron and the cube, respectively, and connect them, we'll produce another truncation, and thus another solid, or congruence (see figure).
However, they both yield the same congruence, that is, at every vertex there are two squares and two triangles, which is the congruence called the cuboctahedron. One question is why do they produce the same result? Aren't the cube and octahedron two completely different types of bodies? The only thing they share in common is that they're both regular solids. Does it perhaps have something to do with their dualship? The cube has three four-sided shapes at every vertex and the octahedron has four three-sided shapes at every vertex.
So let's look at it physically as one multifaceted motion! (see animations)
So, when the cube's vertices are truncated, equilateral triangles are produced and among the square faces of the cube, distorted octagons change their shape until they culminate into a square at the halfway mark of the edge, which is the cuboctahedron. But when the octahedron's vertices are truncated, instead of triangles, squares are produced and instead of distorted octagons, distorted hexagons are produced on the triangular faces and culminate into a triangle at the point of the cuboctahedron. One interesting question is, since both the cube and octahedron truncate to the cuboctahedron, what will be the proportion of the edges of the cube to the octahedron when the cuboctahedron is of the same volume? (see figure)
Once this is known, then you can form the idea of an interval, such that the volumes of both the cube and octahedron will be specific before they both truncate to the same cuboctahedron. And from this idea of the interval, the idea of the relationships between them is knitted.
So the following question remains: how do we knowably construct an Archimedean congruence? In other words, how do we know at what point in the truncation do we produce a solid which is an Archimedean? Well, to start with, let's perform on the octahedron what we did with the tetrahedron, that is, trisect the edges. We know for sure that we'll produce a hexagon on every triangular face and a square for each vertex.
And judging from what we did before with the square, we probably can guess how we can truncate the cube to get a congruence made up of equilateral octagons and triangles. The whole process of truncation can be seen in the following figure, where the third figure for each set are the truncated octahedron and the truncated cube, as discussed above (see figures).
So you see there are constructive methods in finding the different truncations which produce the Archimedean congruences. The three we haven't touched upon yet are the rhombicuboctahedron, truncated cuboctahedron, and snub cube. The snub cube will be elaborated on later, for it's constructiblity can be dealt with by building a solid in which every vertex has four triangles and a square, but it's knowability is much more elusive.
We'll start with the rhombicuboctahedron. This congruence can be produced in the octahedron by the following way. If we take the midpoints of certain edges of the solid in the first truncation, which formed the other congruences, we will produce a series of solids that have two rectangles, a square, and a triangle at every vertex (see animation).
The square and the triangle will be at varying proportion to each other, since the one will share the long side of the rectangle and the other will share the short side. However, you'll notice a moment where it appears that the two rectangles become squares, and in that case, all of the edges of the solid become equal, hence producing the rhombicuboctahedron (see animation).
It is, so to speak, a truncation of a truncation, though the intermediary solid that is truncated to produce the rhombicuboctahedron is not an Archimedean congruence. The reader will be left with the task of geometrically calculating at what exact moment this double truncation occurs.
Also, if we perform this same experiment on the cube, you'll notice that the triangles and squares will not get the chance to become equal, since the truncation stops at the cuboctahedron and the rectangles do not get the chance to become squares (see animation).
Next is what Kepler calls the truncated cuboctahedron, which has a square, hexagon, and an octagon at every vertex. Some say that Kepler is being deceptive, or even making a mistake, by calling this solid a truncated cuboctahedron, since if you truncate the cuboctahedron by dividing the edges into thirds, then not only will you produce rectangles instead of squares, but depending on what proportion you trisect, you will produce regular hexagons and irregular octagons, or vice versa (see figures).
However, to those challenged by thoroughly working through Kepler's writings, will find in Book II, Kepler says, "...which I call a truncated cuboctahedron: not because it can be formed by truncation but because it is like a cuboctahedron that has been truncated." In fact, this Archimedean congruence can be found through a specific truncation of the cube, which, following the animation, the reader is also encouraged to figure out the geometric proportions for such a truncation to exist (see animation).
Upon further investigation, as Kepler would put it, the speculative mind reaches further into degrees of knowability of such congruences or harmonies. The speculative mind would then ask, what is the relationship of these congruences and their volumes to one another? Since the congruencies happen at, what seems like non-arithmetic and changing intervals, then perhaps their volumes to one another express a certain unseen or unheard relationship, by which finding their volumes will establish why these congruencies happen at these specific moments.
The experiment that follows on the next page was inspired by Kepler though he doesn't talk much about it directly. It does however show Kepler to be most correct about the nature of Harmony and the physical universe, and will not only perhaps shed some light on Book V, but open doors to a whole new investigation concerning the periodic table of elements and the nature of isotopes in our self-developing universe. But before we take up the die that Kepler has cast, some light must be shed upon the nature of volumes.