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Let π = 5 (files.catbox.moe)

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

31 comments fedilink hide all child comments

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

Let π = 5

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[–] [email protected] 27 points 7 months ago (1 children)

Well, a r=10 and h=10 doesn't mean much without units. 10 what? Feet? Meters? Inches? CM?

Its not supposed to mean anything, it's about the equation not about the end result. We're not calculating an actual cilinder

[+] [email protected] -15 points 7 months ago (3 children)

If we're not calculating something useful, then why are we here and not in the library learing about the universe?

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

[–] [email protected] 16 points 7 months ago

If we're not calculating something useful, then why are we here and not in the library learing about the universe?

We're learning maths, which is arguably the foundation of the universe.

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

I'm afraid that that is beyond the comprehension of my human existence

[–] [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