Let π = 5 - eviltoast
  • MinekPo1@lemmy.ml
    link
    fedilink
    English
    arrow-up
    2
    ·
    7 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) .

    • MBM@lemmings.world
      link
      fedilink
      English
      arrow-up
      2
      ·
      7 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).