Students using Quadray Coordinates to better grasp XYZ (comparing and contrasting is an old technique), may wonder where best to locate (0,0,0,0) relative to an IVM ball packing.
Given we're planning to use the 12 combinations of {2, 1, 1, 0} as the centers of IVM spheres, 12 around the nuclear ball at (0, 0, 0, 0), we know we want the origin in a ball center. If that's the case, how do we orient the four basis vectors (each one free to grow and shrink independently of the others, with these four sufficient to span our vector space or "room" through addition)?
Remember how the rhombic dodecahedron (volume 6) shrink-wraps every ball (voronoi cell concept)? Its corners occur smack in the voids between spheres, where IVM ball centers are not. We find two types of void however, and not in equal number.
The rhombic dodecahedron is a combination of two Platonics, duals of one another, the cube and octahedron. The octahedron termini define a complementary IVM i.e. balls inflated in these voids will grow to another IVM should the current balls shrink correspondingly to give them room. The cube termini reduce to two cases: two tetrahedrons. Each of these is likewise an IVM waiting to happen, for a total of Four IVMs, a main focus of Russell Chu's visualizations.
The rhombic dodecahedron's long diagonals define the octahedron of the paired IVM, while the short diagonals define the cube connecting the two remaining alternative IVMs.
Remember the negative basis rays of any Quadray Tetrahedron are its dual complement, at the center of our cube, and therefore our rhombic dodecahedron which embraces an IVM ball. That's how to picture your origin then. Imagine two Quadray Tetrahedrons making the Stella Octangula with its tips reaching to the voids of alternate IVMs. The octahedral voids are further away, at the dual octahedron's tips, at six locations (that's six in addition to this four-and-four of the Quadray star, for the 14 corners of the rhombic dodecahedron, dual of the cuboctahedron).
Now linear combinations of {2, 1, 1, 0} i.e. vector additions of however many such vectors, grabbing any of the 12 at random, with as many iterations as we like, keeps us to the IVM ball centers of a single IVM. What's maybe strange here is the length of those basis vectors, which use for unit some length we mostly don't need as such. R=1 or R=0.5 vis-a-vis the IVM balls will suit our needs more often. The basis vectors simply delimit the four quadrants of the home base tetrahedron, each rated at one fourth of unit volume.
Given we're planning to use the 12 combinations of {2, 1, 1, 0} as the centers of IVM spheres, 12 around the nuclear ball at (0, 0, 0, 0), we know we want the origin in a ball center. If that's the case, how do we orient the four basis vectors (each one free to grow and shrink independently of the others, with these four sufficient to span our vector space or "room" through addition)?
Remember how the rhombic dodecahedron (volume 6) shrink-wraps every ball (voronoi cell concept)? Its corners occur smack in the voids between spheres, where IVM ball centers are not. We find two types of void however, and not in equal number.
The rhombic dodecahedron is a combination of two Platonics, duals of one another, the cube and octahedron. The octahedron termini define a complementary IVM i.e. balls inflated in these voids will grow to another IVM should the current balls shrink correspondingly to give them room. The cube termini reduce to two cases: two tetrahedrons. Each of these is likewise an IVM waiting to happen, for a total of Four IVMs, a main focus of Russell Chu's visualizations.
The rhombic dodecahedron's long diagonals define the octahedron of the paired IVM, while the short diagonals define the cube connecting the two remaining alternative IVMs.
Remember the negative basis rays of any Quadray Tetrahedron are its dual complement, at the center of our cube, and therefore our rhombic dodecahedron which embraces an IVM ball. That's how to picture your origin then. Imagine two Quadray Tetrahedrons making the Stella Octangula with its tips reaching to the voids of alternate IVMs. The octahedral voids are further away, at the dual octahedron's tips, at six locations (that's six in addition to this four-and-four of the Quadray star, for the 14 corners of the rhombic dodecahedron, dual of the cuboctahedron).
Now linear combinations of {2, 1, 1, 0} i.e. vector additions of however many such vectors, grabbing any of the 12 at random, with as many iterations as we like, keeps us to the IVM ball centers of a single IVM. What's maybe strange here is the length of those basis vectors, which use for unit some length we mostly don't need as such. R=1 or R=0.5 vis-a-vis the IVM balls will suit our needs more often. The basis vectors simply delimit the four quadrants of the home base tetrahedron, each rated at one fourth of unit volume.
