 | | Re: icosahedron in soccer-ball style —Alan Findlay
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| | | Well, that's weird. My original message said: (...) But you're copy of it said: (...) I wonder how that got mixed up? < shrug > Anyway, back to my original idea. The reason I had suggested 4/3/4 was based on Phillippe's formula of 4/3.5/3.5 -- (...) (21 years ago, 16-Sep-03, to lugnet.build.sculpture, FTX)
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| | | | | Re: icosahedron in soccer-ball style —John Gerlach
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| | | | | (...) I changed it, because I was thinking you meant that each of the black triangles would have one side of length 4 (that connects to the white hexagon), and two sides of length 3 (that connect to the two ajoining black triangles within the (...) (21 years ago, 16-Sep-03, to lugnet.build.sculpture, FTX)
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| | | | | | | Re: icosahedron in soccer-ball style —Alan Findlay
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| | | | | | (...) Hoo, boy. How to explain without pictures? Hmmm... okay here goes... Suppose the following represented two black triangles set side by side, touching at one vertices... _ _ \/ \/ The sides represented by "_" are 1x4's connecting to white 1x4's (...) (21 years ago, 16-Sep-03, to lugnet.build.sculpture)
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| | | | | | | | Re: icosahedron in soccer-ball style —John Gerlach
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| | | | | | Ok, now I understand what you're going for. Interesting idea! And I can imagine why you refered to it as a "pinwheel"... :-) I was stuck on using the equalateral triangles due to me having a large collection of (URL) Polydron> (gasp! A different (...) (21 years ago, 16-Sep-03, to lugnet.build.sculpture, FTX)
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| | | | | Re: icosahedron in soccer-ball style —Philippe Hurbain
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| | | | (...) Or you can go simpler: if you suppress the triangles inside the pentagon, the problem disappear! But the structure may look somewhat hollow... Philo (21 years ago, 16-Sep-03, to lugnet.build.sculpture, FTX)
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