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Date: Fri, 14 Dec 90 13:38 EST
From: Jeannine Mosely
Subject: Peter Beck's construction project
To: cube-lovers@life.ai.mit.edu
Message-Id: <19901214183850.8.J9@MOE.ICAD.COM>
I have made something along the lines that Peter Beck describes in
his "construction project", but it does not quite fit his
description, so I don't know if it is the same thing. It uses only
50 modules and I can't for the life of me imagine where the other 70
should go.
My object looks like this. Imagine a regular icosahedron (20
equilateral triangular faces, with 5 coming together at each
vertex). Erect on each of these faces a triangular prism (20
modules). At each edge of the icosahedron, two square faces of
adjacent prisms rise up from the surface of the icosahedron. Band
each such pair together with a module (30 modules). The reulting
form resembles the Archmidean solid most conveniently designated
(3,4,5,4), which means that each vertex contains a triangle, square,
pentagon, square, in that order.
I say "resembles" this solid, in part, because only the squares are
actually present, the triangular and pentagonal "faces" are voids.
But a more compelling reason for saying "resembles" is that the
geometry is only approximate. If one uses the modules you describe
for the triangular prisms (that is, the height of the prism equals
the edge of the triangle) then the quadrilateral faces on the outer
surface connecting the triangular and pentagonal voids are not
squares, but rectangles whose side are in the ratio of (sqrt 5)-1 to
(sqrt 3). This discrepancy can be fudged, by allowing the squares
to bulge outward slightly. On the other hand, a figure could be
constructed where the outer quadrilaterals were in fact square, but
this would require the prisms to be shorter, and that cannot be
fudged.
Better results can be achieved if you do not fudge the geometry (or
at least not much). It turns out that
(/ (- (sqrt 5) 1) (sqrt 3)) = 5/7
(pardon my lisp) to within one tenth of one percent. Hence I make my
modules as diagrammed below. Dimensions given assume paper in the
ratio of 2 to 1.
This module is used to make the triangular prisms:
_______________________________________________
| : : : | 5/24
|.........:.............:.............:.........|
| : : : |
| : : : | 7/12
| : : : |
|.........:.............:.............:.........|
| : : : | 5/24
|_________:_____________:_____________:_________|
1/2 1/2 1/2 1/2
This module is used to band the triangular prisms together:
_______________________________________________
| : : : | 1/4
|.........:.............:.............:.........|
| : : : |
| : : : | 1/2
| : : : |
|.........:.............:.............:.........|
| : : : | 1/4
|_________:_____________:_____________:_________|
5/12 7/12 7/12 5/12
Natually, you might ask, how do I fold 5/12? There is a trick.
First fold the the long edge in half, and then in quarters at one
end, but don't make the second crease go all the way across--just
nick one edge of the paper, as a marker (point B). Now fold point B
to touch the upper left-hand corner (point A). This would make a
diagonal crease across the strip, but again, don't make the crease
go all the way across--just nick the lower edge (point C). The line
AC is the hypoteneuse of the old 5,12,13 right triangle, and point C is
at 5/12, as desired. (Pretty neat, huh?)
A _______________________________________________
| : |
| : |
| : |
12/12 | : |
| : |
| : |
| : |
|_______:______________:_____________:__________|
5/12 C 7/12 6/12 B 6/12
A similar technique is used to make the other module.
I did not need any staples.
-- jeannine mosely