this post was submitted on 21 Apr 2024
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cross-posted from: https://lemm.ee/post/29980757

Let π = 5

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[–] [email protected] 2 points 6 months ago (1 children)

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) .

[–] [email protected] 2 points 6 months ago (1 children)

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).

[–] [email protected] 2 points 6 months ago