974.00 Initial Frequency
974.01 The initially potential-only frequency (F = +1, -1) vector equilibrium has a volume of 20 regular tetrahedra, each of which consists of 24 A modules. 20 \times 24 = 480 modules = initial vector equilibrium.
974.02 The initial-frequency vector equilibrium has alternatively either a radiant rhombic dodecahedron core or a gravitational rhombic dodecahedron core, either of these alternates being of identical overall size and shape. Both consist of 144 modules. From the 480 modules of the vector equilibrium, we subtract the 144 modules of the rhombic dodecahedron, which leaves 366 modules surrounding either the radiant or gravitational rhombic dodecahedron nuclear-sphere-enveloping core. Each module = \sfrac{1}{144}th of our spherical domain. $336/144 = 2\sfrac
{1}; and 480/144 = 3\sfrac; therefore, one nuclear-sphere domain surrounded by the parts of exactly 27s additional spherical-domain-producing modules, distributed symmetrically around the nucleus in exactly 12 groups of 28 modules each. 27s spheres divided by 12 = 73sths of one spherical domain. 273 = 73 = ^84^/36 spheres = 73eths of a nuclear sphere. We do not produce any complete regular polyhedron by adding 28 modules to each of the 12 rhombic dodecahedron faces. While 28 modules, i.e., 73sths of one spherical domain, may be added to each corner vertex of the vector equilibrium, they do not produce any complete regular polyhedron at *initial* *frequency* where F$ = both + 1 and. Each nucleated vector equilibrium = 3 V3 spheres exactly, and not 13 whole spheres as do 12 closest packed around one. They have the centers of 12 fractional spheres (7/36 each) close packed around one whole sphere.
974.03 Each cube = 3 tetrahedra x 24 modules = 72 modules. 144 modules = 1 sphere. Each initial-frequency cube = V2 a spherical domain. Eight cubes in aF2 cube=four spherical domains. Eight cubes have one whole central nuclear sphere and eight Vsth spheres on the eight outer corners. The eight F = 1 cubes combined to = IF = 2 cubes have 8 x 72 = 576 modules.
974.04 One octahedron has 4\times 24 = 96 modules = 1/144 = 2/s spherical domain. One vector equilibrium = 3–7s spheres. Therefore, one vector equilibrium plus one octahedron = four spheres = one tetrahedron of four closest-packed spheres. One Eighth-Octahedron = 96Is = 12 modules. If we add eight Eighth-Octahedra to each of the vector equilibrium’s eight triangular faces, we produce a cube of 480 + 96 modules = 576 modules, which is the same as the eight-cube F = 2 cube.