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I want to draw a Cuboid accroding to the coordinates of its geometric center and its dimensions, rather than its diagnoal coordiantes. And How can I draw a triangular prism?

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  • $\begingroup$ Possible duplicate, or at least strongly related: mathematica.stackexchange.com/q/10572/121 $\endgroup$ – Mr.Wizard Apr 8 '13 at 4:01
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    $\begingroup$ Dear @user5463: you asked eleven questions and voted only twice. Is there something wrong with the answers you're getting in the site? Also, please change your userid for something more "human" :) $\endgroup$ – Dr. belisarius Apr 8 '13 at 4:35
  • $\begingroup$ Sorry about that, I did't realise that I had to vote that much. I voted for questions asked by others, does it have anything to do with my answers? $\endgroup$ – novice Apr 8 '13 at 5:53
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cuboid[center_, dim_ ] := Cuboid[center - dim/2, center + dim/2]

Graphics3D[cuboid[{6, 7, 8}, {1, 2, 3}], Axes -> True]

enter image description here

For the triangular prism see my answer here.

Edit

Please note that (by design)

cuboid[{c, c, c}, {xd, yd, zd}] == cuboid[ c, {xd, yd, zd}] 

and

cuboid[{cx, cy, cz}, {d, d, d}] == cuboid[ {cx, cy, cz}, d]

If you want

cuboid[ 1, 3 ]

to represent a cube of size 3 centered at {1, 1, 1}, you can modify the definition as follows:

cuboid[center_, dim_ ] := Cuboid[center - dim/2 + {0,0,0}, center + dim/2 + {0,0,0}]
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  • $\begingroup$ belisarius Thanks´╝îBig favor! $\endgroup$ – novice Apr 8 '13 at 7:04
  • $\begingroup$ center + dim/2 + {0, 0, 0} ?? $\endgroup$ – andandandand Apr 14 '16 at 17:26

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