  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.
