08-25-2010, 05:40 PM
(08-24-2010, 10:45 AM)3D Sunset Wrote: For what it's worth, as I was looking over your projection/rotation figures, it occurred to be that if you start with your 2D representation of the 4 triangles, then anchor that at the top, rather than the bottom of the pyramid, then the resulting diamond formed where the bottom of the 2D representation intersects the four faces, is at about the right height for the Queen's chamber.
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Am I describing my idea clearly enough for you to try it?
Hey 3D, I believe I interpreted you correctly, but check the image below to make sure...
The problem as I see it here is that the rotated set of triangles does not equally cover the side of the pyramid, which was a requirement that Ra stipulated in their calculation. Secondarily to this, the bottom of the triangle set also does not reach far enough down the side to meet the Queen's chamber height of 1/6th the total pyramid height (shown here in red), although it is approximate. So I think this also doesn't work, although it's a good idea.
On the flip side to this, I have had the opportunity to make various multi-colored pyramids and triangles, which is always cool.
Lavazza