Hello there. Have something that I am working on that is leaning towards longer winded maths. Think I could get there with a page or two of calcs, but feel I am missing something really simple/obvious. This is not a trick or test that I know the answer to. Just looking to bounce it off some geometry minded people. Take a standard circle. Draw a square round it. If you know the diameter of the circle it is easy to figure out the dimensions of the square. Scale up, just as easy to figure the size of the box that fits round a sphere. My project involves a sphere. Around it is a truncated icosahedron. I am trying to figure out a simple rule to decipher the largest sphere that can fit inside a standard truncated icosahedron. I am good with both shapes. Figuring dimensions/volumes etc. But the way I am combining them, I keep going down a rabbit hole of long winded equations. What I am trying to achieve is a simple rule/calc/ratio, where I have the dimensions of a sphere and can do a quick calculation to take me to the dimensions of the truncated icosahedron. For example, if my sphere has a diameter of 300cm how would I quick calc the truncated icosahedron to fit around it. I plan to play around with multiple sphere sizes and want to quickly gauge the corresponding truncated icosahedron. If there is no quick answer, don't feel you have to do lots of work to build me an equation. Happy to try myself. But I feel it maybe much simpler than I am making it and thought someone here may point out the obvious solution. Many thanks for any response on my new little personal puzzle.
truncated icosahedron and sphere
- Thread starter Stewart17
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Otis
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- Apr 22, 2015
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Hi Stewart. I'm way out of my comfort zone here, but (at the risk of embarrassing myself) I'll share some info that I found in a Google Books preview.
On page 111 of Eric W. Weistein's "CRC Concise Encyclopedia of Mathematics" (2nd Edition), there's a chart that lists the inradius (r) of a truncated icosahedron having edges of unit length.
r = 9/872 × (21 + √5) × √[58 + 18√5]
I'm hoping that can be easily scaled.
?
I want to write a long reply, but working from home today and need to press on with things that pay the bills. Will respond properly later tonight. Short gist is that I kinda been stealing that equation already and it is very helpful. Also wanted to say a warm thank you for your help on it.