RE: Coordinates of vertices
- To: mathgroup at smc.vnet.net
- Subject: [mg87772] RE: [mg87720] Coordinates of vertices
- From: "King, Peter R" <peter.king at imperial.ac.uk>
- Date: Wed, 16 Apr 2008 22:31:12 -0400 (EDT)
- References: <OFD482F44A.E0878572-ONC125742D.00375EA0@icos.be>
Many thanks to all those suggesting using PolyhedronData. Unfortuantely I forgot to mention that I was using version 5.1 so I don't think I have access to that. My fault. Any other suggestions? > -----Original Message----- > From: Maarten van der Burgt [mailto:maarten.vanderburgt at icos.be] > Sent: 16 April 2008 11:08 > To: King, Peter R > Cc: mathgroup at smc.vnet.net > Subject: Re: [mg87720] Coordinates of vertices > > > Peter, > > The vertices are in the graphics object. > Just use [[ ]] to get them out. > > tc=Truncate[PolyhedronData["Cube"]] > vertices=ph[[1,1]]//N > ListPointPlot3D[vertices ,BoxRatios->Automatic] > > Hope this helps, > > maarten > > > > > > > "King, Peter R" > > > <peter.king@imper To: > mathgroup at smc.vnet.net > > ial.ac.uk> cc: > > > Subject: > [mg87720] Coordinates of vertices > > 16/04/2008 11:02 > > > > > > > > > > > > > I am sure I have done this before but I can't for the life of me > remember how, and I can't find it in the manual. Some help > would be much > appreciated. I'd like to find the coordinates of the vertices of a > truncated cube. > > Whilst Vertices[cube] correctly gives me the coordinates of a cube > Vertices[Truncate[Polyhedron[Cube]]] doesn't do what I want because > Truncate[Polyhedron[... is a graphics object not a polyhedron > object. So > can I convert the graphics object into a polyhedron and use > vertices, or > work out the vertices some other way (for a cube this is trivial to do > but for more complicated shapes this is much more tedious) > > Thanks, > > Peter > > > > > > >