# Number of sphere sections on the surface of a hexagonally packed cube

• Jun 22nd 2010, 08:27 AM
grastgirl
Number of sphere sections on the surface of a hexagonally packed cube
I hope you can follow the way I've written the formulae, I don't know how to use LaTEX.

I am trying to calculate the number of sphere sections that would be present on the faces of some hexagonally close packed or face centered cubic packed spheres where the dimensions of the cube are not a tidy multiple of the sphere radius/diameter. Each sphere section should only be counted once. I'm actually looking at this as a model system for plant cells and their contents i.e. once a cell is broken it will empty so I don't need to count it again.

I've got some formulae that someone else derived, but I can't see how they arrived at the formulae for the number of sphere sections (broken cells) per face. I think it must be related in some way to the total number of spheres in a cubic particle, but I'm not sure how.

I'm also not sure how the packing fraction in a volume is applicable to what appears to be an area problem.

Sphere diameter (d), cube edge length (p), Packing fraction of spheres in a cube (P = pi/(3root2)), Cube has faces a,b,c,d,e and f
Total no. of spheres in a cubic particle = p^3/((4/3).pi.(d/2)^2)*P
Na = Number of sectioned spheres on face a etc..
Total number of sectioned spheres (N) = Na+Nb+Nc+Nd+Ne+Nf
Na = Nb, Nc = Nd, Ne = Nf
Na = (64p^2)/(pi^3.d^2)*P
Nc = (64p^2)/(pi^3.d^2)*P - 2(p/((pi/4).d)*P)
Ne = (64p^2)/(pi^3.d^2)*P - 4*P(p/((pi/4).d) - 1)

Any help gratefully appreciated.