Talk:Lattice:Hexagonal diamond - Revision history

KevinYager at 15:07, 9 January 2018

2018-01-09T15:07:58Z

KevinYager at 15:03, 9 January 2018

2018-01-09T15:03:23Z

KevinYager: Created page with "====Absolute==== * $4 \, \mathrm{bottom\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}$ ** $\left(0,0,0\right),(a,0,0),\..."$

2018-01-09T14:58:26Z

Created page with "====Absolute==== * <math> 4 \, \mathrm{bottom\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math> ** <math>\left(0,0,0\right),(a,0,0),\..."

New page

====Absolute====
* <math> 4 \, \mathrm{bottom\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math>
** <math>\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)</math>
* <math> 4 \, \mathrm{mid\,\, layer}: \, \frac{1}{6} + \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = 1</math>
** <math>\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)</math>
* <math> 2 \, \mathrm{internal\,\, strut}: \, 1 + 1 = 2</math>
** <math>\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)</math>
* <math> 4 \, \mathrm{top\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math>
** <math>\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)</math>

====Distances====
For a particle-particle bond-length of <math>l</math>:
:<math>a=\frac{2\sqrt{6}}{3}l</math>
:<math>b=a=\frac{2\sqrt{6}}{3}l</math>
:<math>c=\frac{8}{3}l</math>
:<math>\frac{a}{c}=\frac{\sqrt{6}}{4}\approx 0.61237</math>
:<math>\frac{c}{a}=\frac{4}{\sqrt{6}}\approx 1.63299</math>

====Absolute (in terms of particle-particle distance)====
* <math> 4 \, \mathrm{bottom\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math>
** <math>\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)</math>
* <math> 4 \, \mathrm{mid\,\, layer}: \, \frac{1}{6} + \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = 1</math>
** <math>\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)</math>
* <math> 2 \, \mathrm{internal\,\, strut}: \, 1 + 1 = 2</math>
** <math>\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)</math>
* <math> 4 \, \mathrm{top\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math>
** <math>\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)</math>

@@ Line 23: / Line 23: @@
 ** <math>\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)</math>
 * <math> 2 \, \mathrm{internal\,\, strut}: \, 1 + 1 = 2</math>
-** <math>\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)</math>
+** <math>\left(\frac{\sqrt{6}}{3}l,\frac{\sqrt{2}}{3}l,\frac{1}{3}l\right),\left(\frac{\sqrt{6}}{3}l,\frac{\sqrt{2}}{3}l,\frac{4}{3}l\right)</math>
 * <math> 4 \, \mathrm{top\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}</math>
 ** <math>\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)</math>

Talk:Lattice:Hexagonal diamond - Revision history

KevinYager at 15:07, 9 January 2018

KevinYager at 15:03, 9 January 2018

KevinYager: Created page with "====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),\..."

KevinYager: Created page with "====Absolute==== * $4 \, \mathrm{bottom\,\, layer}: \, \frac{1}{12} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{1}{2}$ ** $\left(0,0,0\right),(a,0,0),\..."$