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Hexagonal Close Packing


HexagonalClosePackingClusHexagonalClosePackingSolid

In hexagonal close packing, layers of spheres are packed so that spheres in alternating layers overlie one another. As in cubic close packing, each sphere is surrounded by 12 other spheres. Taking a collection of 13 such spheres gives the cluster illustrated above. Connecting the centers of the external 12 spheres gives Johnson solid J_(27) known as the triangular orthobicupola (Steinhaus 1999, pp. 203-205; Wells 1991, p. 237).

Hexagonal close packing must give the same packing density as cubic close packing, since sliding one sheet of spheres cannot affect the volume they occupy. To verify this, construct a three-dimensional diagram containing a hexagonal unit cell with three layers (Steinhaus 1999, pp. 203-204). Both the top and the bottom contain six 1/6-spheres and one hemisphere. The total number of spheres in these two rows is therefore

 2(6·1/6+1·1/2)=3.
(1)

The volume of spheres in the middle row cannot be simply computed using geometry. However, symmetry requires that the piece of the sphere which is cut off is exactly balanced by an extra piece on the other side. There are therefore three spheres in the middle layer, for a total of six, and a total volume

 V_(spheres in unit cell)=·(4pi)/3r^3(3+3)=8pir^3.
(2)

The base of the unit cell is a regular hexagon made up of six equilateral triangles with side lengths 2r. The unit cell base area is therefore

 A_(unit cell)=6[1/2(2r)(sqrt(3)r)]=6sqrt(3)r^2.
(3)

The height is the same as that of two tetrahedra length 2r on a side, so

 h_(unit cell)=2(2rsqrt(2/3)),
(4)

giving

 eta_(HCP)=(8pir^3)/((6sqrt(3)r^2)(4rsqrt(2/3)))=pi/(3sqrt(2))
(5)

(Conway and Sloane 1993, pp. 7 and 9). Now that the Kepler conjecture has been established, hexagonal close packing and cubic close packing, both of which have the same packing density of eta=pi/(3sqrt(2))=0.74048..., are known to be the densest possible packings of equal spheres.

If we had actually wanted to compute the volume of sphere inside and outside the hexagonal prism, we could use the spherical cap equation to obtain

V_ subset =1/3pih^2(3r-h)
(6)
=1/3pir^31/3(3-1/(sqrt(3)))
(7)
=1/9pir^3(3-(sqrt(3))/3)
(8)
=1/(27)pir^3(9-sqrt(3))
(9)
V_ superset =pir^3[4/3-1/(27)(9-sqrt(3))]
(10)
=1/(27)pir^3(36-9+sqrt(3))
(11)
=1/(27)pir^3(27+sqrt(3)).
(12)
SquashedCubicSquashedHexagonal

If spheres packed in a cubic lattice, face-centered cubic lattice, and hexagonal lattice are allowed to expand uniformly until running into each other, they form cubes, hexagonal prisms, and rhombic dodecahedra, respectively. In particular, if the spheres of face-centered cubic packing are expanded until they fill up the gaps, they form a solid rhombic dodecahedron (left figure above), and if the spheres of hexagonal close packing are expanded, they form a second irregular dodecahedron consisting of six rhombi and six trapezoids (right figure above; Steinhaus 1999, p. 206) known as the trapezo-rhombic dodecahedron. The latter can be obtained from the former by slicing in half and rotating the two halves 60 degrees with respect to each other. The lengths of the short and long edges of the rotated dodecahedron have lengths 2/3 and 4/3 times the length of the rhombic faces. Both the rhombic dodecahedron and trapezo-rhombic dodecahedron are space-filling polyhedra.


See also

Circle Packing, Cubic Close Packing, Kepler Conjecture, Kepler Problem, Sphere Packing

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References

Conway, J. H. and Sloane, N. J. A. Sphere Packings, Lattices, and Groups, 2nd ed. New York: Springer-Verlag, 1993.Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, 1999.Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, pp. 53-54, 1991.

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Hexagonal Close Packing

Cite this as:

Weisstein, Eric W. "Hexagonal Close Packing." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/HexagonalClosePacking.html

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