DV-cell of the fcc-lattice

The DV-cell of the face centered cubic lattice has 12 facetts, 14 vertices and 24 edges (this fulfills the euler-formula). This polytope is called a rhombic dodecahedron.
The fcc-lattice can be described by a basis of integer vectors, eg. ‹1,1,0›, ‹1,0,1›, ‹0,1,1›. In this coordinate system the normal vectors of the DV-cell are the following twelve vectors (which can be connected to the edges of the unit-cube):

‹1,1,0› ‹1,-1,0› ‹-1,1,0› ‹-1,-1,0›
‹1,0,1› ‹1,0,-1› ‹-1,0,1› ‹-1,0,-1›
‹0,1,1› ‹0,1,-1› ‹0,-1,1› ‹0,-1,-1›

The vertices can be described by 14 vectors. 6 (type 1) coresponding to the normalvectors of the cubefacets (e.g. ‹0,-1,0›) , and 8 (type 2) coresponding to the vertices of the cube (e.g. ‹1,-1,-1›). These have different distance to the origin. Suppose the vertices coresponding to the vertices of the cube have distance 1 to the origin, than the other 6 vertices have distance 2/sqrt(3). As you can see in the picture, a vertice coresponding to a cubevertice has 3 adjancent facets, the other 6 vertices have 4 adjacent facets. Cutting the cell by a hyperplane in the origin with a normalvector pointing in direction of a vertice of the cube, you get a sixgon.

With the purpose to give concrete values for the distance,area and volume, we choose the fcc-lattice with minum distance 2. In this case the facets are described using the above mentioned normalvectors n in the following way:
1/sqrt(2) * n = 1
The volume of the DV-cell results in 4sqrt(2). The packing of spheres with radius 1 an centers at the points of the fcc-lattice, has the density pi/sqrt(18). Now its simple to obtain the values for the distances:
distance value
Vertice type 1 from the origin sqrt(2)
Vertice type 2 from the origin sqrt(3/2)
facet from the origin 1
edgelength of the cube
fitting in the cell 		sqrt(2)
second diagonal of the rhombus 2

The density of the above mentioned spherepacking can now also be obtained by the observation that the euclidian space can be filled with cubes of edgelength sqrt(2) at every point of the lattice and filling the gaps with the same number of cubes (obviousliy the cube has half the volume of the dv-cell):
delta = 4pi/3 / (2*sqrt(2)^3) = pi/sqrt(18)
The DV-cell of the hexagonal closest packing (hcp) looks similar. One half of the cell is turned 60 degrees (see the following picture). In this case not all the DV-cells are translates of one another. In adjacent layers the cells are rotated 60 degrees.

The next picture shows the dv-cell of the fcc-lattice intersected with a sphere of radius 2/sqrt(3).

Peter Scholl
Last modified: Wed Dec 6 11:24:52 CET 2000