Talk:Lattice:Hexagonal diamond

Absolute

• $4\,\mathrm {bottom\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
  • $\left(0,0,0\right),(a,0,0),\left({\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},0\right),\left(a+{\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},0\right)$
• $4\,\mathrm {mid\,\,layer} :\,{\frac {1}{6}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{3}}=1$
  • $\left(0,0,{\frac {5c}{8}}\right),\left(a,0,{\frac {5c}{8}}\right),\left({\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},{\frac {5c}{8}}\right),\left(a+{\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},{\frac {5c}{8}}\right)$
• $2\,\mathrm {internal\,\,strut} :\,1+1=2$
  • $\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {c}{8}}\right),\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {4c}{8}}\right)$
• $4\,\mathrm {top\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
  • $\left(0,0,c\right),(a,0,c),\left({\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},c\right),\left(a+{\frac {b}{2}},{\frac {{\sqrt {3}}b}{2}},c\right)$

Distances

For a particle-particle bond-length of $l$:

$a={\frac {2{\sqrt {6}}}{3}}l$

$b=a={\frac {2{\sqrt {6}}}{3}}l$

$c={\frac {8}{3}}l$

${\frac {a}{c}}={\frac {\sqrt {6}}{4}}\approx 0.61237$

${\frac {c}{a}}={\frac {4}{\sqrt {6}}}\approx 1.63299$

Absolute (in terms of particle-particle distance)

• $4\,\mathrm {bottom\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
  • $\left(0,0,0\right),\left({\frac {2{\sqrt {6}}}{3}}l,0,0\right),\left({\frac {\sqrt {6}}{3}}l,{\sqrt {2}}l,0\right),\left({\sqrt {6}}l,{\sqrt {2}}l,0\right)$
• $4\,\mathrm {mid\,\,layer} :\,{\frac {1}{6}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{3}}=1$
  • $\left(0,0,{\frac {5}{3}}l\right),\left({\frac {2{\sqrt {6}}}{3}}l,0,{\frac {5}{3}}l\right),\left({\frac {\sqrt {6}}{3}}l,{\sqrt {2}}l,{\frac {5}{3}}l\right),\left({\sqrt {6}}l,{\sqrt {2}}l,{\frac {5}{3}}l\right)$
• $2\,\mathrm {internal\,\,strut} :\,1+1=2$
  • $\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {c}{8}}\right),\left({\frac {a}{2}},{\frac {b}{2{\sqrt {3}}}},{\frac {4c}{8}}\right)$
• $4\,\mathrm {top\,\,layer} :\,{\frac {1}{12}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{6}}={\frac {1}{2}}$
  • $\left(0,0,{\frac {8}{3}}l\right),\left({\frac {2{\sqrt {6}}}{3}}l,0,{\frac {8}{3}}l\right),\left({\frac {\sqrt {6}}{3}}l,{\sqrt {2}}l,{\frac {8}{3}}l\right),\left({\sqrt {6}}l,{\sqrt {2}}l,{\frac {8}{3}}l\right)$