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I'm high school student, and I am doing research about buildings using Mathematica. I want to plot the polyhedron shown below:

the desired polyhedron

I have tried using Table, but it doesn't work. Please help me.

v = Table[ { {-2, -2, h} , {-2, 2, h} , {2, 2, h} , {2, -2, h} , {-3, -3, h + 3} , {-3, 3, h + 3} , {3, 3, h + 3} , {3, -3, h + 3} } , {h, 0, 1, 3} ]; i = { {1, 2, 3, 4} , {1, 2, 6, 5} , {2, 3, 7, 6} , {3, 4, 8, 7} , {4, 1, 5, 8} , {5, 6, 7, 8} }; Graphics3D[GraphicsComplex[v, Polygon[i]], Axes -> True]
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    $\begingroup$ Welcome to Mathematica.SE. Could you be more specific in your question? . Only good questions are likely to get great answers. Please edit your question to improve it and make more clear what you are asking. Also consider taking the tour. $\endgroup$ Commented Aug 21, 2015 at 14:57
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    $\begingroup$ Aren't you doing exactly that? $\endgroup$ Commented Aug 21, 2015 at 15:04
  • $\begingroup$ Can you clarify what you need, because you are using a Table, do you just want help to simplifiy your code? $\endgroup$ Commented Aug 21, 2015 at 15:51
  • $\begingroup$ sorry. I want to regulate height using table $\endgroup$ Commented Aug 22, 2015 at 1:05
  • $\begingroup$ In case this was contributing to your confusion, please note that the iteration specifier {h, 0, 1, 3} makes h run from 0 to 1 in steps of 3. So in fact h only ever takes the value 0. To run from 0 to 3 in steps of 1, use {h, 0, 3, 1} or just {h, 0, 3}. (The final value and step are given in the opposite order compared to, e.g., Matlab.) $\endgroup$ Commented Aug 23, 2015 at 10:00

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I am glad that you are using Mathematica in your high school project.

I think you forgot to mention in your question that the code you posted doesn't actually produce the image you showed; you may also want to mention where you obtained that image.

Anyway, since your figure is made up of repeating units, I generated one unit, then translated it multiple times to generate the rest of the figure. You could use the following code as an inspiration to tweak to your desires: 

Graphics3D[ Table[ GeometricTransformation[ Pyramid[{{0, 0, 0}, {2, 0, 0}, {2, 2, 0}, {0, 2, 0}, {1, 1, -2}}], TranslationTransform[{0, 0, n}] ], {n, 0, 5, 1} ], Axes -> True ]

pyramids

UPDATE

As @J.M. mentioned in his comment, this is a pretty versatile approach. You can swap out different 3D repeating units, and the rest of the code will do the stacking for you.

In fact, a more faithful representation of what you showed in the picture can be obtained using a pyramidal frustum (i.e. a pyramid with the tip chopped off) as the repeating unit:

repeatingUnit = Hexahedron[{ (*bottom face*) {-2, -2, 0}, {2, -2, 0}, {2, 2, 0}, {-2, 2, 0}, (*top face*) {-3, -3, 3}, {3, -3, 3}, {3, 3, 3}, {-3, 3, 3} }]; Graphics3D[ Table[ GeometricTransformation[ repeatingUnit, TranslationTransform[{0, 0, n}] ], {n, 0, 15, 3} ], Axes -> True ]

frusta

You can explore further by considering e.g. less regular pyramids, pyramids with non-horizontal bases, etc.

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  • $\begingroup$ Hopefully, OP realizes that the Pyramid[] in your code is replaceable, if, say, he wants frustums instead. $\endgroup$ Commented Aug 21, 2015 at 16:58
  • $\begingroup$ @J. M. Good point. I updated the code to include that, in specific to separate the construction of an explicit repeating unit from the generation of the 3D graphic itself. $\endgroup$ Commented Aug 21, 2015 at 17:19
  • $\begingroup$ +1 for the effort put into interpreting a weird question $\endgroup$ Commented Aug 21, 2015 at 17:32

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