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I would like to create a diamond lattice structure or similar -- close to this: hexagon or octagon lattice structure in 3D.

I do not have any code I´m working at the moment. I'm very new to Mathematica, and I usually use a CAD program to do this kind of task. However, it is very (extremly).

I would like to have cylinder or similar shape connected to each other in a big 3D structure. That can be varied in size and shape. I know this might be a big task, but anything to get me started would help.

Any suggestions?

m_goldberg
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pblomstrom
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  • Graphics3D[Table[Sphere[{x + y/2 + z/2, y (3/4)^(1/2) + z (3/4)^(1/2)/2, 3/4 z},1/2], {x, 10}, {y, 10}, {z, 5}] hexagonal close packing of spheres as a starter – LLlAMnYP Mar 02 '16 at 13:12
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    I would encourage you to spend at least as much time/diligence on composing your question (spelling, content, formatting) as you expect of others spending on answering. This is not your first question, and so far each of them was reformatted and corrected by users here. – Yves Klett Mar 02 '16 at 13:38
  • i will do that , sorry about that @Yves Klett – pblomstrom Mar 02 '16 at 13:39
  • Thanks for taking your time (also, a nicely formatted question has higher chance of motivating good answers)! Please add more info, optimally some sample code to show what you are after. You can edit your question at any time to optimize it. Formatting advice etc. is available e.g. here: http://mathematica.stackexchange.com/editing-help. – Yves Klett Mar 02 '16 at 13:43
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    Related: http://mathematica.stackexchange.com/questions/35515/finding-111-plane-of-carbon-diamond – Michael E2 Mar 02 '16 at 13:45
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    Are you looking for LatticeData["TetrahedralPacking", "Image"]? – Szabolcs Mar 02 '16 at 14:00
  • Using bobthechemist's answer in the linke Michael gave, I was able to create this image in short order by changing the {i, 1} to {i, 5}. From reading the question here, I'd say that is what you are looking for. – Jason B. Mar 02 '16 at 14:24
  • thank you all for the answers ! – pblomstrom Mar 02 '16 at 14:50

1 Answers1

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coords = {{0, 0, Sqrt[2/3] - 1/(2 Sqrt[6])}, 
          {1/Sqrt[3], 0, -(1/(2 Sqrt[6]))}, 
          {-(1/(2 Sqrt[3])), -(1/2), -(1/(2 Sqrt[6]))}, 
          {-(1/(2 Sqrt[3])), 1/2, -(1/(2 Sqrt[6]))}, 
          {1/Sqrt[3], 0, -(1/(2 Sqrt[6]))}};
g = Polygon@coords;
Graphics3D[
 Table[Translate[g, {i, j, k}], 
  {i, 1, 5},
  {j, 1, 5},
  {k, 1, 5}],
 Axes -> True]
David G. Stork
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  • A nice approach. Thank you, I will use also. I only would put spheres instead of the polygon statements, just to make it more resemble an atomic structure in question. Where did you take the coordinates? – Alexei Boulbitch Mar 02 '16 at 14:45