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The "truncated octahedron," or tetrakaidecahedron, is also a known space filler (Fig. 12-4a). A model of this semiregular polyhedron can be constructed by taping together eight cardboard hexagons and six squares of the same edge length. But Fuller is wary of that approach, a geometry which supports the illusion of "solids." Instead, synergetics prescribes that we view the system in context. Therefore, to generate this shape we start with a "three-frequency octahedron." With its edges divided into three equal segments, this octahedron submits quite naturally to vertex truncation, as shown in Chapter 9, Figure 9-9b. A (single-frequency) half octahedron can be chopped off from each comer, creating six square faces and converting the three-frequency triangles into equilateral hexagons.
How do they fit together? Hexagonal faces of two truncated octahedra come together after rotating 60 degrees with respect to each other, so that square faces alternate (rather than landing next to each other) and begin to frame a cavity in the exact tetrakaidecahedral shape (Fig. 12-4b).
Four tetrakaidecahedra fit together around one IVM vertex. To ascertain the exact relationship of this packing to the IVM will require some investigation: how remote are the vertices involved, and how many cells are incorporated in between? We will answer these questions below, but first we go back to the simpler cases, for a sense of the whole progression.
Two to One: A Review
We can now appreciate Loeb's development of the requirement that a space-filling polyhedron can be broken down into twice as many octahedra as tetrahedra. (3)
The cube consists of one tetrahedron plus four eighth octahedraor a total of one half octahedron. This ratio of 1 to ½ certainly qualifies. Eight quarter tetrahedra (for a total of two) embrace one octahedron to create the rhombic dodecahedron. With this confirmation, we start a list of results, displayed in Table VI.
The VE consists of eight tetrahedra and six half octahedra: 8 to 3. Lacking one octahedron, the VE must not be a space filler, a conclusion that is consistent with our earlier observation of the IVM. However, pair a VE with the missing octahedron and all space can be filled. The two-to-one ratio also serves as a prescription for complementary pairing; if an IVM system is not a space filler, the ratio tells us what's missing.
The rest? All-space-filling rhombohedra are utterly straightforward; with two tetrahedra on either side of an octa, this shape gave us our starting point for the space-filling ratio. Icosahedra cannot be broken down into tetrahedra or octahedra; the icosahedron is eternally out of phase. That brings us to the truncations, which can be analyzed in terms of the IVM. As a representative example, we explore the truncated octahedron.
The three-frequency octahedron incorporates nineteen single-frequency octahedra and thirty-two tetrahedra. It can be a frustrating experience to count these individual cells, but it can be done! Skeptics are encouraged to get out a box of toothpicks and some mini-marshmallows. Figure 12-5 illustrates the dissection of one half of a three-frequency octahedron, to make it easier to count the unit cells in individual layers. Sixteen tetrahedra and nineteen half octahedra are clearly visible in the three layers shown. These results are simply doubled to verify the totals in the first sentence of this paragraph.
The ratio of 32:19 does not qualify. Not surprisingly, the three-frequency octahedronlike any regular octahedron, for size does not affect shapeis not a space filler. The truncation process removes half of a small octahedron from each of the six comers, or three octahedra altogether. Nineteen minus three leaves sixteen, while the total of thirty-two tetrahedra does not change. 32:16 is indeed two to one, and the rule holds true once again.
The truncated tetrahedron can be dissected in the same manner as its octahedral counterpart. The IVM starting point, a three-frequency tetrahedron, consists of eleven unit tetrahedra and four octahedra, and is clearly not a space filleras we well know. Figure 12-6 enables the unit cells to be counted, by separating the layers of a four-frequency tetrahedron. In the process of counting, we can also finally verify the claim introduced in Chapter 2 that third-power numerical values (2³, 3³, 4³, etc.) are represented by the volumes of tetrahedra of increasing frequency. Recall that the expression "x cubed" is derived from the fact that cubes of progressively higher frequency consist of 1, 8, 27, 64,... unit cubes. Fuller points out that exactly the same volumetric increase is exhibited by tetrahedra, as portrayed in Figure 2-lb. We now take advantage of the disected four-frequency tetrahedron to verify these values. Adding up the volumes of the unit tetrahedra (each with a volume of one) and unit octahedra (each with a volume of four) in successive layers, we quickly find that the "cubic" numbers are confirmed, as shown in Figure 12-6. A two-frequency tetrahedron has a volume of 8; a three-frequency, 27; a four-frequency, 64; and so on. "Nature is not 'cubing,' she is 'tetrahedroning,'" announces the triumphant Bucky.
Back to the space-filling investigation, our next candidate is the (semiregular) truncated tetrahedron, which in synergetics is derived from a three-frequency tetrahedron. Subtracting a unit tetrahedron from each of the four corners yields a truncated tetrahedron consisting of seven small tetrahedra and four octahedra (Fig. 9-9a). Seven-to-four does not correspond to the desired ratio, and we thus learn that this shape will not fill space either; it lacks one unit tetrahedron. As a final illustration of the disection method, we note the new totals when a three-frequency tetrahedron is paired with a three-frequency octahedron. The resulting inventory consists of 43 tetrahedra and 23 octahedra, which does not quite reach the desired double ratio. What if we add a second three-frequency tetrahedron? The final total is then 54 tetrahedra and 27 octahedra, and that's it. Two 3v tetrahedra paired with one 3v octahedron are able to fill space. Once again, shape proves independent of size.
The above examples provide first-hand confirmation of the space-filling hypothesis, allowing us to feel confident that a new truth has been revealed.
Out of all the accumulated data, a consistent finding emerges. With this generalized principle, we are equipped to determine whether any polyhedron that submits to analysis in terms of octahedral and tetrahedral components is a space filler. We just have to break it down into these constituent parts and check the ratio for the requisite two-to-one. The ratio replaces the arduous task of trying to push shapes together and twist them around to see if they pack; it is another shortcut. IVM provides the framework; two-to-one is the ratio; all criteria supplied.
Coincidence? Magic? Or rules of spatial order?
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