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 Post subject: Question about polyhedra related to cubes.
PostPosted: Fri Mar 29, 2013 8:55 am 
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Joined: Fri Nov 05, 2010 2:20 am
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There are 3 'fundamental' ways to make a puzzle turn, faces, corners and edges.

A face turning cube->face turning cube
Corner turning cube -> face turning octahedron,
Edge turning cube-> Face turning rhombic dodecahedron.

When moving on to hybrid turning axes,

Face+ edge turning cube -> Face turning (what solid?)
Face+ Corner->???
edge+corner->???

I have found the solution for face+corner+edge which is the rhombicuboctahedron, but i cannot deduce the rest. Anyone?

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 Post subject: Re: Question about polyhedra related to cubes.
PostPosted: Fri Mar 29, 2013 9:16 am 
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Truncating the corners of a cube yields a Cuboctahedron.

I think FTO versions of non-regular polyhedra are no more interesting than other forms of the puzzle. Things like the FT-Rhombic Dodecahedron are interesting because you can gain some insight into the geometry and why it jumbles. When there is more than one face type that turns there isn't any shape that is going to make that look natural.

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 Post subject: Re: Question about polyhedra related to cubes.
PostPosted: Fri Mar 29, 2013 11:13 am 
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See figure 4a on this page to see an edge-truncated cube and edge-truncated octahedron.

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 Post subject: Re: Question about polyhedra related to cubes.
PostPosted: Fri Mar 29, 2013 11:39 am 
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jaap wrote:
See figure 4a on this page to see an edge-truncated cube and edge-truncated octahedron.

That link has some really cool potential new puzzle shapes. Thanks for posting. :D

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 Post subject: Re: Question about polyhedra related to cubes.
PostPosted: Fri Mar 29, 2013 11:50 am 
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Face + edge = 4-fold truncated rhombic dodecahedron:

Face + corner = cuboctahedron:

Edge + Corner = 3-fold truncated rhombic dodecahedron:
Attachment:
3-fold truncated rhombic dodecahedron.jpg
3-fold truncated rhombic dodecahedron.jpg [ 168.67 KiB | Viewed 597 times ]

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