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Let π = 5 (files.catbox.moe)

submitted 7 months ago by [email protected] to c/[email protected]

39 comments fedilink hide all child comments

cross-posted from: https://lemm.ee/post/29980757

Let π = 5

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[–] [email protected] 2 points 7 months ago* (last edited 7 months ago) (1 children)

Better question: What curvature of space is necessary for the apparent value of π to be 5?

~~honestly I don't know if there is any way to measure curvature of space , but its slightly more curved than the surface of a ball (where π=~4.712)~~

edit : its more complex than that and topology of non euclidean spaces hurts

[–] [email protected] 2 points 7 months ago (1 children)

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.

[–] [email protected] 2 points 7 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 7 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 7 months ago

oopsie :3