Regular arrangement of equal spheres in a plane changing to an irregular arrangement of unequal spheres (bubbles).
hcp and fcc closepacking of spheres
In geometry, closepacking of equal spheres is a dense arrangement of congruent spheres in an infinite, regular arrangement (or lattice). Carl Friedrich Gauss proved that the highest average density – that is, the greatest fraction of space occupied by spheres – that can be achieved by a lattice packing is

\frac{\pi}{3\sqrt 2} \simeq 0.74048.
The same packing density can also be achieved by alternate stackings of the same closepacked planes of spheres, including structures that are aperiodic in the stacking direction. The Kepler conjecture states that this is the highest density that can be achieved by any arrangement of spheres, either regular or irregular. This conjecture is now widely considered proven by T. C. Hales.^{[1]}^{[2]}
Many crystal structures are based on a closepacking of a single kind of atom, or a closepacking of large ions with smaller ions filling the spaces between them. The cubic and hexagonal arrangements are very close to one another in energy, and it may be difficult to predict which form will be preferred from first principles.
Contents

fcc and hcp lattices 1

Cannonball problem 1.1

Positioning and spacing 1.2

Lattice generation 2

Miller indices 3

Filling the remaining space 4

See also 5

Notes 6

External links 7
fcc and hcp lattices
There are two simple regular lattices that achieve this highest average density. They are called facecentered cubic (fcc) (also called cubic close packed) and hexagonal closepacked (hcp), based on their symmetry. Both are based upon sheets of spheres arranged at the vertices of a triangular tiling; they differ in how the sheets are stacked upon one another. The fcc lattice is also known to mathematicians as that generated by the A_{3} root system.^{[3]}
Cannonball problem
Cannonballs piled on a triangular
(front) and rectangular
(back) base, both
fcc lattices.
The problem of closepacking of spheres was first mathematically analyzed by Thomas Harriot around 1587, after a question on piling cannonballs on ships was posed to him by Sir Walter Raleigh on their expedition to America.^{[4]} Cannonballs were usually piled in a rectangular or triangular wooden frame, forming a threesided or foursided pyramid. Both arrangements produce a facecentered cubic lattice – with different orientation to the ground.
Positioning and spacing
In both the fcc and hcp arrangements each sphere has twelve neighbors. For every sphere there is one gap surrounded by six spheres (octahedral) and two smaller gaps surrounded by four spheres (tetrahedral). The distances to the centers of these gaps from the centers of the surrounding spheres is \scriptstyle \sqrt{\frac{3}{2}} for the tetrahedral, and \scriptstyle \sqrt2 for the octahedral, when the sphere radius is 1.
Relative to a reference layer with positioning A, two more positionings B and C are possible. Every sequence of A, B, and C without immediate repetition of the same one is possible and gives an equally dense packing for spheres of a given radius.
The most regular ones are:

fcc = ABCABCA (every third layer is the same)

hcp = ABABABA (every other layer is the same).
In closepacking, the centertocenter spacing of spheres in the x–y plane is a simple honeycomblike tessellation with a pitch (distance between sphere centers) of one sphere diameter. The distance between sphere centers, projected on the z (vertical) axis, is:

\text{pitch}_Z = \sqrt{6} \cdot {d\over 3}\approx0.81649658 d,
where d is the diameter of a sphere; this follows from the tetrahedral arrangement of closepacked spheres.
The coordination number of hcp and fcc is 12 and its atomic packing factor (APF) is the number mentioned above, 0.74.
Comparison between hcp and fcc


Figure 1 – The hcp lattice (left) and the fcc lattice (right). The outline of each respective Bravais lattice is shown in red. The letters indicate which layers are the same. There are two "A" layers in the hcp matrix, where all the spheres are in the same position. All three layers in the fcc stack are different. Note the fcc stacking may be converted to the hcp stacking by translation of the uppermost sphere, as shown by the dashed outline.



Figure 2 – Shown here is a stack of eleven spheres of the hcp lattice illustrated in Figure 1. The hcp stack differs from the top 3 tiers of the fcc stack shown in Figure 3 only in the lowest tier; it can be modified to fcc by an appropriate rotation or translation.

Figure 3 – Thomas Harriot, circa 1585, first pondered the mathematics of the cannonball arrangement or cannonball stack, which has an fcc lattice. Note how adjacent balls along each edge of the regular tetrahedron enclosing the stack are all in direct contact with one another. This does not occur in an hcp lattice, as shown in Figure 2.

Lattice generation
When forming any spherepacking lattice, the first fact to notice is that whenever two spheres touch a straight line may be drawn from the center of one sphere to the center of the other intersecting the point of contact. The distance between the centers along the shortest path namely that straight line will therefore be r_{1} + r_{2} where r_{1} is the radius of the first sphere and r_{2} is the radius of the second. In close packing all of the spheres share a common radius, r. Therefore two centers would simply have a distance 2r.
Simple hcp lattice
An animation of closepacking lattice generation. Note: If a third layer (not shown) is directly over the first layer then the HCP lattice is built, if the third layer is placed over holes in the first layer then the FCC lattice is created
To form an ABAB... hexagonal close packing of spheres, the coordinate points of the lattice will be the spheres' centers. Suppose, the goal is to fill a box with spheres according to hcp. The box would be placed on the xyz coordinate space.
First form a row of spheres. The centers will all lie on a straight line. Their xcoordinate will vary by 2r since the distance between each center if the spheres are touching is 2r. The ycoordinate and zcoordinate will be the same. For simplicity, say that the balls are the first row and that their y and zcoordinates are simply r, so that their surfaces rest on the zeroplanes. Coordinates of the centers of the first row will look like (2r, r, r), (4r, r, r), (6r ,r, r), (8r ,r, r), ... .
Now, form the next row of spheres. Again, the centers will all lie on a straight line with xcoordinate differences of 2r, but there will be a shift of distance r in the xdirection so that the center of every sphere in this row aligns with the xcoordinate of where two spheres touch in the first row. This allows the spheres of the new row to slide in closer to the first row until all spheres in the new row are touching two spheres of the first row. Since the new spheres touch two spheres, their centers form an equilateral triangle with those two neighbors' centers. The side lengths are all 2r, so the height or ycoordinate difference between the rows is \scriptstyle\sqrt{3}r. Thus, this row will have coordinates like this:

\left(r, r + \sqrt{3}r, r\right),\ \left(3r, r + \sqrt{3}r, r\right),\ \left(5r, r + \sqrt{3}r, r\right),\ \left(7r, r + \sqrt{3}r, r\right), \dots.
The first sphere of this row only touches one sphere in the original row, but its location follows suit with the rest of the row.
The next row follows this pattern of shifting the xcoordinate by r and the ycoordinate by \scriptstyle\sqrt{3}. Add rows until reaching the x and y maximum borders of the box.
In an ABAB... stacking pattern, the odd numbered planes of spheres will have exactly the same coordinates save for a pitch difference in the zcoordinates and the even numbered planes of spheres will share the same x and ycoordinates. Both types of planes are formed using the pattern mentioned above, but the starting place for the first row's first sphere will be different.
Using the plane described precisely above as plane #1, the A plane, place a sphere on top of this plane so that it lies touching three spheres in the Aplane. The three spheres are all already touching each other, forming an equilateral triangle, and since they all touch the new sphere, the four centers form a regular tetrahedron.^{[5]} All of the sides are equal to 2r because all of the sides are formed by two spheres touching. The height of which or the zcoordinate difference between the two "planes" is \scriptstyle\sqrt{6}r2/3. This, combined with the offsets in the x and ycoordinates gives the centers of the first row in the B plane:

\left(r, r + \frac{\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\ \left(3r, r + \frac{\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\ \left(5r, r + \frac{\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\ \left(7r, r + \frac{\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right), \dots.
The second row's coordinates follow the pattern first described above and are:

\left(2r, r + \frac{4\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\ \left(4r, r + \frac{4\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\ \left(6r, r + \frac{4\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\ \left(8r,r + \frac{4\sqrt{3}r}{3}, r + \frac{\sqrt{6}r2}{3}\right),\dots.
The difference to the next plane, the A plane, is again \scriptstyle \sqrt{6}r2/3 in the zdirection and a shift in the x and y to match those x and ycoordinates of the first A plane.^{[6]}
In general, the coordinates of sphere centers can be written as:
\begin{bmatrix} 2i + ((j\ +\ k)\ \bmod{2})\\ \sqrt{3}\left[j + \frac{1}{3}(k\ \bmod{2})\right]\\ \frac{2\sqrt{6}}{3}k\\ \end{bmatrix}r
where i, j and k are indices starting at 0 for the x, y and z coordinates.
Miller indices
Miller–Bravais index for hcp lattice
Crystallographic features of hcp systems, such as vectors and atomic plane families can be described using a fourvalue Miller index notation ( hkil ) in which the third index i denotes a convenient but degenerate component which is equal to −h − k. The h, i and k index directions are separated by 120°, and are thus not orthogonal; the l component is mutually perpendicular to the h, i and k index directions.
Filling the remaining space
The FCC and HCP packings are the densest known packings of equal spheres. Denser sphere packings are known, but they involve unequal sphere packing. A packing density of 1, filling space completely, requires nonspherical shapes, such as honeycombs.
Replacing each contact point between two spheres with an edge connecting the centers of the touching spheres produces tetrahedrons and octahedrons of equal edge lengths. The FCC arrangement produces the tetrahedraloctahedral honeycomb. The HCP arrangement produces the gyrated tetrahedraloctahedral honeycomb. If, instead, every sphere is augmented with the points in space that are closer to it than to any other sphere, the duals of these honeycombs are produced: the rhombic dodecahedral honeycomb for FCC, and the trapezorhombic dodecahedral honeycomb for HCP.
Spherical bubbles in soapy water in a FCC or HCP arrangement, when the water in the gaps between the bubbles drains out, also approach the rhombic dodecahedral honeycomb or trapezorhombic dodecahedral honeycomb. However, such FCC of HCP foams of very small liquid content are unstable, as they do not satisfy Plateau's laws. The Kelvin foam and the Weaire Phelan foam are more stable, having smaller interfacial energy in the limit of a very small liquid content. ^{[7]}
See also
Notes

^

^ "Mathematics: Does the proof stack up?".

^ Conway, John Horton; Sloane, Neil James Alexander; & Bannai, Eiichi. Sphere packings, lattices, and groups. Springer, 1999, Section 6.3.

^ David Darling. "Cannonball Problem". The Internet Encyclopedia of Science.

^ "on Sphere Packing". Grunch.net. Retrieved 20140612.

^ Weisstein, Eric W., "Hexagonal Close Packing", MathWorld.

^ Foams, Structure and Dynamics, I. Cantat et al, Oxford University Press 2013
External links

P. Krishna & D. Pandey, "ClosePacked Structures" International Union of Crystallography by University College Cardiff Press. Cardiff, Wales. PDF

"3D Sphere Packing Applet" ClosePacking of Spheres java applet
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