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Comment on Let π = 5

MinekPo1@lemmy.ml ⁨6⁩ ⁨months⁩ ago

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)

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MBM@lemmings.world ⁨6⁩ ⁨months⁩ ago
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.

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- MinekPo1@lemmy.ml ⁨6⁩ ⁨months⁩ ago
  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) .
  
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  - MBM@lemmings.world ⁨6⁩ ⁨months⁩ ago
    
    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).
    
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    MinekPo1@lemmy.ml ⁨6⁩ ⁨months⁩ ago
    oopsie :3
    
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