Re: Question of function for hexahedron

*To*: mathgroup at smc.vnet.net*Subject*: [mg23580] Re: [mg23562] Question of function for hexahedron*From*: BobHanlon at aol.com*Date*: Sun, 21 May 2000 18:12:50 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

In my initial reply I stated: > >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)]]; > Needs["Graphics`Polyhedra`"] Needs["DiscreteMath`Permutations`"] This function allows the vertices of the Hexahedron to be given in any sequence drawHexahedron[vertices_List, opts___] := Module[{origin = Chop[(Plus @@ #) & /@ Transpose[vertices]], seq1, seq2, remap}, seq1 = (FromDigits[#, 2] & /@ ((1 + Sign[#])/2 & /@ ((# - origin) & /@ vertices))); seq2 = FromDigits[#, 2] & /@ ((1 + Sign[#])/2 & /@ Vertices[Hexahedron]); remap = Inner[Rule, seq2, Range[8], List]; orderedVertices = Transpose[ Rest[Transpose[ Sort[Transpose[Join[{seq1 /. remap}, Transpose[vertices]]]]]]]; Show[Graphics3D[ Polygon[#] & /@ (orderedVertices[[#]] & /@ Faces[Hexahedron])], opts] ] 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}}; Demonstrating that the vertices can be entered in any order drawHexahedron[vert[[#]] & /@ RandomPermutation[8], ViewPoint -> {1.300, -2.400, 2.000}]; Bob BobHanlon at aol.com