Subject:
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Re: icosahedron in soccer-ball style
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Newsgroups:
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lugnet.build.sculpture
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Date:
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Tue, 16 Sep 2003 19:51:44 GMT
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Viewed:
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989 times
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In lugnet.build.sculpture, Alan Findlay wrote:
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Well, that’s weird. My original message said:
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How about making each black triangle a 4/3/4?
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But you’re copy of it said:
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How about making each black triangle a 4/3/3?
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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 -- 4+3.5+3.5=11 and 4+3+4=11 also. It may
be a too simplistic approach to this type of problem, but the arithmetic
caught my attention.
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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
pentagon). So I pictured each black triangle being 4/3/3. I tried this last
week, and it didn’t work - the five triangles couldn’t connect.
Now you’ve got me confused - how would you have a black triangle of 4/3/4? One
‘4’ would be attached to the white hexagon, then what?
John confused... :-)
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Message has 1 Reply:
 | | Re: icosahedron in soccer-ball style
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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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Message is in Reply To:
| | Re: icosahedron in soccer-ball style
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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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