RE: Coordinates of vertices
- To: mathgroup at smc.vnet.net
- Subject: [mg87774] RE: [mg87720] Coordinates of vertices
- From: "Maarten van der Burgt" <maarten.vanderburgt at icos.be>
- Date: Wed, 16 Apr 2008 22:31:34 -0400 (EDT)
Steve,
For version 5.2 I could do the following.
I assume this will also work for 5.1.
<<Graphics`Polyhedra`
<<graphics`graphics3d`
trc =Show[Truncate[Polyhedron[Cube]]]
vtc =(Flatten[trc[[1]]/.Polygon->List,2]//Union)
ScatterPlot3D[vtc]
Maarten
"King, Peter R"
<peter.king@imper To: "Maarten van der Burgt" <maarten.vanderburgt at icos.be>
ial.ac.uk> cc: <mathgroup at smc.vnet.net>
Subject: RE: [mg87720] Coordinates of vertices
16/04/2008 16:00
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
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> "King, Peter R"
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> <peter.king@imper To:
> mathgroup at smc.vnet.net
>
> ial.ac.uk> cc:
>
>
> Subject:
> [mg87720] Coordinates of vertices
>
> 16/04/2008 11:02
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> 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)
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> Thanks,
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> Peter
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