def crystal_surface_family(Mil_ind):
'''
determine all the equivalent crystal planes of the crystal plane family in the 32 point group.
input:
Mil_ind: Miller indices of crystallographic planes.
return:
f: crystal plane family.
'''
cc1 = CC.CrystalClass32()
f = []
for index in Mil_ind:
family = cc1.plane_family(index)
f.append(family)
return f
def crystal_plane(hkl):
'''
return the geometric planes of the polyhedron, where the 32 point group is considered,
and the lattice parameter is taken as 0.4nm.
hkl: Miller indices of crystallographic planes.
'''
b1 = basis.cubic(0.4) # unit: nm
cc1 = CC.CrystalClass32()
latt = CrystalLattice(b1, cc1)
planes = []
number_planes = []
for index in hkl:
pf = latt.geometric_plane_family(index)
number_planes.append(len(pf))
planes.extend(pf)
return planes, number_planes
number_planes1 = []
for index in [[1,2,3], [1,1,0], [2,1,0], [3,3,1], [1,0,0], [1,1,1]]:
pf = latt.geometric_plane_family(index)
number_planes1.append(len(pf))
planes1.extend(pf)
#set distances of the planes
ps_setd_mingled(number_planes1, planes1, [2.0, 2.0, 2.0, 2.0, 2.0, 2.0])
diameter = 10.0 # unit: nm
volume = 4.0/3.0*np.pi* (diameter*0.5)**3
qfp = enclose_polyhedron(number_planes1, planes1, volume)
print "q=", qfp[0]
print "f=", qfp[1]
print "numbers of points are", [len(p) for p in qfp[2]]
latt2 = CrystalLattice(b1, CC.CrystalClass32())
f = shape_by_f([0.25, 0.75], latt2, [[0,0,1], [1,1,1]], 10.0, [1.0, 1.0])
print "shape by f", f
ps = ps_by_family([[0,0,1], [1,1,1]], latt2)
ps_setd_mingled(ps[0], ps[1], f)
qfp2 = enclose_polyhedron(ps[0], ps[1], 10.0)
print "q2=", qfp2[0]
print "f2=", qfp2[1]