Can a LatticeData image show more than a unit cell?

Question

I have an interactive code for showing the crystallographic in Mathematica.

LatticeData[
 Which[
   Type == "FCC", "FaceCenteredCubic", 
   Type == "BCC", "BodyCenteredCubic", 
   Type == "SC", "SimpleCubic"], "Image"]`

However, this shows only one unit, which means the planes that don't fit in the cell won't be shown. Is there a easy way to show more than one unit cell in the LatticeData image?

Edit: ok so apparently I had to post the full code. Note that it's written by Danyel Cavazos and can be found with this link: https://demonstrations.wolfram.com/CrystallographicPlanesForCubicLattices/ Basically what I wanted to originally do was to add another control function to the interactive graphic for showing how many lattice planes to show.

Manipulate[ 
 Show[
  LatticeData[ 
   Which[Type == "FCC", "FaceCenteredCubic", Type == "BCC", 
    "BodyCenteredCubic", Type == "SC", "SimpleCubic"], "Image"],
Table[ ContourPlot3D[  hx + ky + l*z - r == 0,
    {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, 
    ColorFunction -> Function[{x, y, z, f}, {Blue, Opacity[Op]}],
    Mesh -> False, BoundaryStyle -> Black], {r, -(h + k + l), 
    h + k + l - 2, 2}],
Graphics3D[{Thick,
     Red, Arrow[{{-1.01, -1.01, -1.01}, {1.6, -1.01, -1.01}}],
    Blue, Arrow[{{-1.01, -1.01, -1.01}, {-1.01, 1.6, -1.01}}],
    Darker@Green, Arrow[{{-1.01, -1.01, -1.01}, {-1.01, -1.01, 1.6}}]
    }],
  SphericalRegion -> True, ImageSize -> 1.1 {500, 400}
  ],
  {{Type, "SC", "lattice"}, {"SC", "BCC", "FCC"}},
 Row[{"Miller indices", Spacer[20],
   Control[{{h, 1, Style["h", Italic]}, Range[0, 10, 1]}], Spacer[20],
   Control[{{k, 1, Style["k", Italic]}, Range[0, 10, 1]}], Spacer[20],
   Control[{{l, 1, Style["l", Italic]}, Range[0, 10, 1]}]}],
 {{Op, 0.5, "plane opacity"}, 0, 1}]

You say you have interactive code that illustrates your probllem, but that's not what you post. You would have a much better chance of getting an answer if you wete to post working code. As it is, this question is likely to be closed without being answered. — m_goldberg, Jan 16 '21 at 17:59

score 7 · Answer 1 · answered Jan 16 '21 at 18:06

We first create a single cell:

Type = "FCC";
gr = LatticeData[
   Which[Type == "FCC", "FaceCenteredCubic", Type == "BCC", 
    "BodyCenteredCubic", Type == "SC", "SimpleCubic"], "Image"];

We now need to translate this cell to several locations. The problem is, that MMA can only translate graphics primitives and not whole graphics. We therefore need to dig out the graphic primitive by gr[[1]]. Now we can translate them. In the end we have to change them back to a Graphic3D. As the width of a cell is from -1 to 1, we must translate it by 2. Let n be the number of cell in any dimension:

n = 3;
Show[Graphics3D[
  Table[Translate[gr[[1]], 2 {i, j, k}], {i, n}, {j, n}, {k, n}]], 
 Boxed -> False]

Can a LatticeData image show more than a unit cell?

1 Answers1