From j9@icad.com Fri Dec 14 19:26:43 1990 Return-Path: Received: from BU.EDU by life.ai.mit.edu (4.1/AI-4.10) id AA01678; Fri, 14 Dec 90 19:26:43 EST Received: by BU.EDU (1.99) Fri, 14 Dec 90 13:43:49 EST Received: from MOE.ICAD.COM by icad.COM (4.1/SMI-4.0) id AA21059; Fri, 14 Dec 90 13:34:43 EST 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