xkcd: Coordinate Precision but pi (π)?

I tried looking for some answer but found mostly

  • People reciting pi
  • People teaching how to memorize pi
  • How to calculate pi using different formula
  • How many digits NASA uses

Update question to be more specific

In case someone see this later, what is the most advanced object you can build or perform its task, with different length of pi?

0, 3 => you can’t make a full circle

1, 3.1 => very wobbly circle

2, 3.14 => perfect hole on a beach

3, 3.142 => ??

4, 3.1416 => ??

5, 3.14159 => ??

Old question below

In practice, the majority of people will never require any extra digit past 3.14. Some engineering may go to 3.1416. And unless you are doing space stuff 3.14159 is probably more than sufficient.

But at which point do a situation require extra digit?
From 3 to 3.1 to 3.14 and so on.

My non-existing rubber duck told me I can just plug these into a graphing calculator. facepalm

y=(2πx−(2·3.14x))

y=abs(2πx−(2·3.142x))

y=abs(2πx−(2·3.1416x))

y=(2πx−(2·3.14159x))

Got adequate answer from @dual_sport_dork and @howrar
Any extra example of big object and its minimum pi approximation still welcome.

  • @Magister
    23•1 year ago

    Not a lot, and this is why to speedup thing on some architecture, when working with (unsigned)integer you multiply by 355 then divide by 113 (it’s like 3.14159292035)

      • Kogasa
        3•1 year ago

        There are infinitely many. Any sequence of rational numbers converging to pi contains infinitely many. 22/7 and 355/113 are just particularly good ones for their small denominators. You can find such good approximations (“the best rational approximations of a given size”) by truncating the continued fraction representation of pi:

        pi = [3;7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,2,2,2,…]
pi = 3  + 1/(7 + 1/(15 + 1/(1 + ... )))

        These approximations yield:

        pi ~ 3
pi ~ 3 + 1/7 = 22/7
pi ~ 3 + 1/(7 + 1/15) = 333/106
pi ~ 3 + 1/(7 + 1/(15 + 1)) = 355/113
pi ~ 3 + 1/(7 + 1/(15 + 1/(1 + 1/292))) = 97591/31065

        and so on. Note that by pure coincidence one of the first few terms is quite large (292), so the difference between the corresponding partial expansions is small (1 + 1/292 is close to 1). That’s why 355/113 is an unusually good approximation for such a small denominator.

    • @[email protected]
      6•1 year ago

      I’m not sure that’s a great trick. You have to remember 6 digits to calculate an approximation accurate to 8 digits.

      How many architectures in 2023 still lack a FPU? They were getting pretty rare when I last worked with this stuff 15 years ago.

      • @Magister
        3•1 year ago

        you have to remember 11 33 55 and put the bar in the middle. It is mainly small MCU like ATMEGA and co. lacking FPU, and yes old stuff like Z80 ,6502, etc.