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Amy C. Edmondson
A Fuller Explanation
Chapter 8, Tales Told by the Spheres: Closest Packing
pages 114 through 116

Vector Equilibrium

In cubic packing, twelve spheres surround one sphere, with each sphere tangent to every neighbor and without any gaps. This perfect geometric fit of the thirteen omnisymmetrical forms provides a basis for understanding the fundamental directions inherent in space. But we saw that the configuration is more specific than just a numerical consistency; the spheres outline the vertices of the cuboctahedron, or VE. This shape seems to appear out of nowhere., Created by the cluster of cornerless spheres shoved together, this result is as counter-intuitive as it is reliable. The more we learn about the shape of space, however, the more natural the appearance of the VE—or any manifestation of "twelve degrees of freedom"—becomes.


One two and three frequency progression of vector equilibrium formed from clustered spheres
Fig. 8-13
Click on thumbnail for larger image.

The six squares and eight triangles outlined by the closepacked spheres, although unmistakable even in the simplest case, become more and more distinct as the frequency increases. As Fuller's convention is to refer to the number of spaces (rather than spheres) along the "edge" of the cuboctahedral cluster as the frequency of the system, the first twelve-around-one group is "one-frequency." The next layer—in the VE case, a surrounding envelope rather than just another layer added to the bottom—is two-frequency, having three balls along each edge. The next is three-frequency, with four balls per edge, and so on. It follows that the higher the frequency, the smaller an individual sphere is in relation to its polyhedral face, and so these polygonal faces look progressively less bumpy, or more sharply defined (Fig. 8-13). The precise planar organization of the clustered spheres becomes more obvious as the frequency increases. The appearance of eight triangles and six squares was not an accidental property of the first layer: the VE is here to stay. This lesson is continually reinforced by additional layers.

      Having established the shape of symmetrical nuclear sphere-packing we proceed to investigate numbers. Twelve balls fit tightly around one; how many does it take to completely surround the twelve with a second layer? By carefully placing balls in the "nests" on the cluster's surface, we generate a two-frequency VE shell of exactly forty-two balls. We might begin by placing one ball at the center of each of the six squares and then in each nest along the twenty-four VE "edges", thereby superimposing six two-frequency squares (nine balls each) over the simple four-ball squares (Fig. 8-13b). The edges of the six two-frequency squares supply the spheres for the adjacent two-frequency triangles (without adding any more balls). These second-layer triangles cover the eight three-ball triangles of the first layer. The twelve vertices, or corner spheres, are each shared by two squares, and so twelve must be subtracted from fifty-four (nine spheres per square times six squares) to get the total number for the second VE layer: forty-two spheres. (See Fig. 8-13.)

      Envelop the whole package with a third layer, a three-frequency shell with four balls per edge. Following the above procedure, we count sixteen balls in each of the six squares, for a total of 16 × 6 = 96, from which we must subtract the twelve vertices counted twice due to overlap between squares. The edges of the eight triangles are again already in place—provided by the edges of the squares—but on this shell, a central sphere which belongs in each three-frequency triangle must still be added; eight more spheres are therefore needed to complete the enveloping layer. 96 minus 12 plus 8 yields a total of 92 balls.

      The next shell (four-frequency) requires 162 balls; the five-frequency layer consists of 252, six of 362, and so on. We are now able to detect a pattern by looking carefully at these numbers: 12,42, 92, 162, 252, 362,... . It will come as no surprise to the observant student of numbers to learn that the next shell consists of 492 balls.

      What exactly is going on? To begin with, we notice the consistent last digit: every single number ends with 2, reminiscent of Euler's law and its "constant 2." Fuller interprets this persistent "excess of 2" in radial sphere packing as further affirmation of the inevitable "poles of spinnability," inherent in the topology of singly closed systems. And indeed, the temptation to embrace a single explanation is strong. The fact that the number of spheres per shell always ends with the digit 2—even though those numbers increase drastically with each successive layer—seems too strange to ignore; we want an explanation for nature's behavior. But at this stage, speculation as to significance is a sidetrack: our task is to fully describe the configurations. As soon as we fully understand the patterns and are thus armed with the facts, such speculation will be appropriate and indeed inevitable.

      After observing the reliable last digit, we can simplify our sequence—following Fuller's procedure—by subtracting the 2 from each term, removing the distraction to assist further analysis. We are left with 10, 40, 90, 160, 250, 360,490,..., all divisible by 10. So let's divide by 10. This leaves 1,4,9, 16,25, 36,49,..., and now the pattern is clear.

      The latter sequence is generated by f ², for f = 1,2,3,4,5,6,7 We choose "f " in this case, to represent frequency, for it turns out that the relationship between frequency and number of units per shell can be directly specified. Nature thus reveals yet another "generalized principle." This equation actually describes a straight-forward edge-length-to-surface-area relationship, exactly what we expect from geometry—in a slightly different format.

      The next question is how to specify the relationship in precise terms. We work in reverse from our final sequence (f = 1,2, 3,4,...) to generate the original sequence. First we must raise the frequency f to the second power, then multiply each term by 10, and then add 2: l0f ² + 2 therefore gives us the total number of spheres for any shell (specified by frequency) in nuclear sphere packings.

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