## Friday, June 10, 2016

### Exploding Diagram

Clearly when it comes to sharing the experience of a rhombic triacontahedron exploding into 120  E-modules (were they named something else before?), a JavaScript library such as three.js is a goldmine.

I'm awed by Ricardo Cabello's skill-set here.  He has picked a steep trail and climbed steadily, with a little help from collaborators.

The E-modules are so named within the canonical context of nested polyhedrons, with a focus on the Jitterbug Transformation, so named by R. Buckminster Fuller, and cited officially by some mathematicians.

The cuboctahedron (we can call it that, no problem (Fuller uses other names too)) "torques down" (rotate-contracts) into an icosahedron with the same edge-lengths, yet with more faces (20, not 14).  The VE (as Fuller calls it) has 24 radial and 24 circumferential edges (hence "vector equilibrium"), as the eight tetrahedrons comprising it are "hinge-bonded".

Let the aforesaid cuboctahedron (VE) be defined by 12 uni-radius balls (r=R) around a nuclear ball of diameter D (2R = D). In Synergetics this thing has volume 20 in being comprised of 8 tetrahedrons and 6 half-octahedrons, of volume 1 and 2 respectively (8 x 1 + 6 x 2 = 8 + 12 = 20).

By jitterbugging the cuboctahedron, making it torque, we get this icosahedron of volume ~18.51 (there's a phi-based expression) that still needs to be phi-reduced, meaning all edges multiplied by 0.618... to get us the rhombic triacontahedron (RT, 30-diamond-faced) of diameter, you guessed it, D.

The E-modules are R-deep to the center.  The aforementioned nuclear ball, and all the others, are "shrink wrapped" by the E-modules' rhombic triacontahedron (faces inter-tangent to contained ball).  Each of the 30 diamonds contributes four "pizza slices" or "wedges" pointing inward to a common center.

Inversely (in the sense of reciprocally), the E-module RT scales up by phi (1.618...) to give what Koski and I call the "super-RT" (because bigger), the long diagonals of which are this 18.51... volumed icosahedron (the short diagonals form its dual, the pentagonal dodecahedron).

Readers attempting to follow all this may be stuck back at the cuboctahedron of volume 20 and its constituent tetrahedrons of volume 1.  "All math is ethno-math" and in this neck of the woods, a cube of edges 2nd root of 2, diagonal two (2R), has a volume of exactly 3, not 2nd root of 2 to the 3rd power as from school we'd expect.  How could this be?

Choosing a cube as the model of 3rd powering was a choice and our mathematics today derives from making that choice.  However we know mathematics to be a branching tree-like structure, or maybe a graph, in that where there's a road not taken, we may sometimes go back and take it, exploring an alternative branch.

Our model of 3rd powering along this other road is not the cube, but the tetrahedron.

Picture starting with the corner of a regular tetrahedron, perhaps starting with XYZ's 90-90-90 and simply changing the angles to 60-60-60.  Instead of building out in a parallelepiped (such as a cube) when multiplying, just connect the tips of the three rods of lengths A, B, C directly ("close the lid") to trap the corresponding volume ABC (or A to the 3rd if all three lengths equal).

A regular tetrahedron of edges D has volume one.  Reduce edges to 0.5 (or R) and the volume drops by 0.5 to the 3rd power, to 1/8.

Likewise when starting with the icosahedron of volume ~18.51 and multiplying all edges by 0.618, volume goes down by 0.618 to the 3rd power.  So-called phi-scaling, of E-modules but also S-modules, other modules (U, V, W) is where David Koski's "playing with blocks" fits in.

For example, not directly mentioned in Synergetics (the original two volumes) is that the cuboctahedron to icosahedron volume ratio, is the same as the S:E module volume ratio.  How to morph an E into an S is another "Koski cartoon" or scenario.

Back to the cube of exactly volume three, edges 2nd root of two:  assume in XYZ the cube of edges R is the unit of volume (R=1; 1x1x1 = cube) whereas in the IVM (alternative scaffolding), the tetrahedron of edges D is the unit of volume (D=2R; 1x1x1 = tetrahedron).

We have a ratio then, between these two unit volumes, the 2nd root of 9/8.  That's how much bigger the R-edged cube is, in terms of volume. It'd take fewer XYZ cubes than IVM tetrahedrons to fill a cuboctahedron of edges D.  20 x (2nd root of 8/9) would be the answer.

T-mods and E-mods are angularly identical, and so one might think distinguishing by scale is immaterial, given the 4D sculpture under discussion is not pegged to any particular size or time.  However, given surface:volume is not a constant, T and E are indeed distinguishable.

As pointed out in the above video, not everyone relates to "pure geometry".  Starting the video with a sports car, maybe with a basket ball in the back seat, could kick things off.  Every so often, we could dissolve to something more "real".