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Let π = 5
(files.catbox.moe)
cross-posted from: https://lemm.ee/post/29980757
Let π = 5
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this post was submitted on 21 Apr 2024
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Kind of curious how you got that value. I think the ratio of circumference to diameter ("pi") is actually smaller in spherical geometry, in the most extreme case (the equator) it's just 1. You could say "pi = 5" for circles of a specific radius in hyperbolic geometry, I guess.
my mistake was using the sum of angles in a triangle which was kinda dum but whatever . I also tried calculating via the circumstance of a circle placed at a pole where π was 20x smaller for the case I was using but its not linear so I looked deeper which was a big mistake .
BTW the ratio of circumstance to radius for a circle which is also an equator of the space is ¼ not 1 (r=½π₀ , C=2π₀) .
I think you mean 4, which makes the ratio of circumference to diameter 2 (either way, no idea how I messed up that one).
oopsie :3