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Wall Thickness

 
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hamp856



Joined: 21 Jul 2011
Posts: 16

PostPosted: Wed Jul 27, 2011 11:17 am    Post subject: Wall Thickness Reply with quote

I am making plutonic objects out of wood. I have had resonalble success. I want to take the objects I make, put them in a wood lathe and turn them into spheres. Does anyone know how to calculate the thickness of the wood before it is turned to achieve a final wall thickness of 1/2"

Hamp
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guy



Joined: 11 Feb 2008
Posts: 75
Location: England

PostPosted: Thu Jul 28, 2011 8:30 pm    Post subject: Reply with quote

Well, the wood will be thinnest in the corners and thickest at the centre of each piece. Assuming you want at least 1/2" everywhere, you need to:

Take the difference between the inradius (radius of the sphere tangent to the faces at their centres) and the out- or exradius (of the sphere tanget to the vertices). That tells you how much fatter each piece is in the middle.

As a rough approximation, add 1/2" to get the starting thickness. You could calculate more accurately but it would depend on some complicated angles of each Platonic, so personally I'd just round the number off to a handy thickness - if it's a tetrahedron I'd round down.

HTH
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hamp856



Joined: 21 Jul 2011
Posts: 16

PostPosted: Mon Aug 01, 2011 6:00 pm    Post subject: Reply with quote

Hey, Thanks a bunch for the info. I see in Stella 4-D under the Scale Tab an item for "inradius" but I don't see the "out Radius" There are several tabs underscale. Which would you use ?

This week end I was able to build a "Rhombicosidodecahedron". I had good success for a first shot. I made it out of MDF plywood since it was prototype. It is about 13" in dia. I'll post a picture as soon as I can get my hands on a digital camera.

thanks for your help.
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robertw
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Joined: 10 Jan 2008
Posts: 394
Location: Melbourne, Australia

PostPosted: Tue Aug 02, 2011 10:42 am    Post subject: Reply with quote

Circumradius (or out-radius) is simply referred to as "Radius". Technically speaking, a polyhedron may not have a circumradius, which would have to go through all points, so I just use the more generic term radius.
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hamp856



Joined: 21 Jul 2011
Posts: 16

PostPosted: Tue Aug 02, 2011 11:19 am    Post subject: Reply with quote

Thanks, I see the objects you build. What material do you use. How do you cut the angles necessary?

I am experimenting with a product called "Sign Foam" which is "high density polyunrethane. It is perfect for these objects.

Thanks,

Hamp
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robertw
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Joined: 10 Jan 2008
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Location: Melbourne, Australia

PostPosted: Tue Aug 02, 2011 11:27 am    Post subject: Reply with quote

I presume that question's for Guy? I only make models from paper. Just wanted to answer the part of the question I could Smile
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hamp856



Joined: 21 Jul 2011
Posts: 16

PostPosted: Tue Aug 02, 2011 7:45 pm    Post subject: Reply with quote

I made an "Archimedan Rhombicodidodechedron" how do I post a picture of it?[/img]
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robertw
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PostPosted: Wed Aug 03, 2011 10:30 am    Post subject: Reply with quote

See this thread, about posting images: http://software3d.com/Forums/viewtopic.php?t=23
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guy



Joined: 11 Feb 2008
Posts: 75
Location: England

PostPosted: Sun Aug 07, 2011 7:09 pm    Post subject: Reply with quote

robertw wrote:
I presume that question's for Guy? I only make models from paper. Just wanted to answer the part of the question I could Smile

No, I don't post images of real models (except the odd icosahedron in a museum that I think may have been made by H.T. Flather of Coxeter, Du Val, Flather and Petrie fame).

And yes, circumradius is the usual term. I knew that. Sad
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hamp856



Joined: 21 Jul 2011
Posts: 16

PostPosted: Sun Aug 07, 2011 7:33 pm    Post subject: archimedan rhombicosidodecahedron Reply with quote

If am building a "archimedan rhombicosidodecahedron" how do I calculate the # of vertices?

Thanks,

Hamp
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robertw
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PostPosted: Mon Aug 08, 2011 11:34 am    Post subject: Reply with quote

Stella shows number of vertices in the Info window.

Otherwise, you could note that each vertex touches exactly one pentagon. There are 12 pentagons, and so 5 x 12 = 60 vertices.
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