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Re: Question of function for hexahedron

  • To: mathgroup at smc.vnet.net
  • Subject: [mg23577] Re: [mg23562] Question of function for hexahedron
  • From: BobHanlon at aol.com
  • Date: Sat, 20 May 2000 17:44:32 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 5/20/2000 3:35:53 AM, khs at procd.sogang.ac.kr writes:

>I want to find out the function that will generate hexahedron,however I
>heard there are some founctions related to 'hexahedron' like 'Cubid' or
>'Ployhedron'. But I want to generate hexahedron for given eight numbers.
>
>for example
>
>  (0.0057    0.0052    0.0052)
>  (-0.0057    0.0063    0.0051)
> ( 0.0057   -0.0052    0.0062)
> ( 0.0057   -0.0063    0.0063)
> ( 0.0057   -0.0063   -0.0051)
> (-0.0057   -0.0052   -0.0052)
> (-0.0057    0.0063   -0.0063)
>  (0.0057    0.0052   -0.0062)
>

 vert = {{0.0057,    0.0052,    0.0052}, {-0.0057,    0.0063,    0.0051}, { 
        0.0057,   -0.0052,    0.0062}, { 0.0057,   -0.0063,    0.0063}, { 
        0.0057,   -0.0063,   -0.0051}, {-0.0057,   -0.0052,   -0.0052}, \
{-0.0057,    0.0063,   -0.0063}, {0.0057,    0.0052,   -0.0062}};

Checking the location of the vertices

Show[Graphics3D[{RGBColor[1, 0, 0], Line[vert]}]];

There appears to be a mistake in defining the vertices. The third point 
should have a negative x value.

 vert = {{0.0057,    0.0052,    0.0052}, {-0.0057,    0.0063,    
        0.0051}, { -0.0057,   -0.0052,    0.0062}, { 0.0057,   -0.0063,    
        0.0063}, { 
        0.0057,   -0.0063,   -0.0051}, {-0.0057,   -0.0052,   -0.0052}, \
{-0.0057,    0.0063,   -0.0063}, {0.0057,    0.0052,   -0.0062}};

Rechecking

Show[Graphics3D[
      Join[{RGBColor[1, 0, 0], Line[vert], RGBColor[0, 0, 1]}, 
        Table[Text[ToString[k], vert[[k]]], {k, Length[vert]}]]]];

Each of the six faces will be a polygon formed by four vertices

faces = {{1, 2, 3, 4}, {5, 6, 7, 8}, {3, 4, 5, 6}, {1, 2, 7, 8}, {2, 3, 6, 
        7}, {1, 4, 5, 8}};

Drawing the hexahedron

Show[Graphics3D[Polygon[#] & /@ (vert[[#]] & /@ faces)]];


Bob

BobHanlon at aol.com


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