# Re: OT hypercube (was: Con-other)

From: | John Vertical <johnvertical@...> |

Date: | Friday, May 30, 2008, 21:23 |

On Fri, 30 May 2008 15:07:50 -0400, Mark J. Reed wrote:
>In neither case does it matter if the object is solid or hollow. But
>if it's not transparent then the drawing degenerates to a square.
>Which isn't that exciting. :)

Not necessarily… This is, again, the difference between a direct 2D
projection - which could also be a hexagon or an octagon, but indeed flat -
and a 2D picture of a 3D projection; which could be eg. a rhombic
dodecahedron. You can construct one of those from four skew'd,
3D-Penrose-tile-like pieces. This is the projection equivalent to a cube,
viewed corner-on, looking like a hexagon made from three rhombuses (yeah,
I'm not even going to try fancy-pluralize that right).
…These rhombic parallelepipeds aren't the actual 3D Penrose tiles however;
those will produce a rhombic triacontahedron (aka d30) insted. Supposedly,
such a construction is a 3D projection of a penteract, but as said, my
visualization skills don't go that far. Still, compare with the
tesseract-in-octagon construction I described in my previous post; subtract
edges that cross one another, and you'll be left with an octagon constructed
from squares and rhombuses. Intuition would say that this is an analogical case.
John Vertical

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