| Popis: |
The average distance of the equal hard spheres is introduced to evaluate the density of a given arrangement. The absolute smallest value is two radii because the spheres can not be closer to each other than their diameter. The absolute densest arrangement of two, three and four spheres is defined, which gives the absolute highest density in one, two and three dimensions. The absolute highest density of equal spheres in three dimensions is the tetrahedron formed by the centers of four spheres touching each other with density of 0.7796. The density of this tetrahedron unit can be maintained only locally because the tetrahedron units can not be expanded to form a tightly packed arrangement in three dimensions. The maximum number of tetrahedron units that one sphere is able to accommodate is twenty which corresponds to the density of 0.684. The only compatible formation of equal spheres which can be mixed with tetrahedron is octahedron. In order to mix the tetrahedron and octahedron units certain geometrical constrains must be satisfied. It is shown that the only possible mixture of tetrahedrons and octahedrons units is the one which accommodates eight tetrahedron and six octahedron vertexes which is identical to FCC and an alternative proof for the Kepler conjecture. It is suggested that there is a density gap between the FCC density and the highest density of disordered arrangements and that the icosahedrons configuration with its 0.684 density represents the upper bound on the disordered arrangements. |
