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› |

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