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This new pi value should land us on the precise nanometer on a planetary rock of a sun located some 18 trillion light years away. More than good enough for a S

This new pi value should land us on the precise nanometer on a planetary rock of a sun located some 18 trillion light years away. More than good enough for a Star Trek transporter targeting system, provided that sufficient power can reach it and able to compensate for planetary orbital speed, orbital curvature, surface axial rate, as well same value set for its solar system pathway around its galaxy, and its galaxy pathway thru its eyewatery cornucopia of galaxies. But it may not be good enough for precise calculation of field interaction within a large group of elementary particles of quantum physics. Thanks to Heisenburg’s Indeterminacy Principle (aka Uncertainty Principle).

slyall 2y ago
> This new pi value should land us on the precise nanometer on a planetary rock of a sun located about 18 trillion light years away. 40 digits or so will get you that...
- hughesjj 2y ago
  Nah, observable universe is only 93B light years in "diameter" at the current "moment"
  - theandrewbailey 2y ago
    93 billion light years is 8.798×10^26 meters[0], so about 35 digits ought to suffice for any given nanometer. [0] https://www.wolframalpha.com/input?i=93+billion+light+years+in+meters https://www.wolframalpha.com/input?i=93+billion+light+years+...
    - egberts1 2y ago
      Now that's a solid win for us math nerds.
- constantcrying 2y ago
  In floating point arithmetic two consecutive operations can have an unbounded error. Just because the precision is good enough for one computation doesn't mean it is good enough for all computations.
- egberts1 2y ago
  Very good. Here's the WolframAlpha equation to your assertion of 40-digit ... or so. https://www.wolframalpha.com/input?i=%28180+x+17+trillion+light+years%29%2F%28pi+x1nm%29 https://www.wolframalpha.com/input?i=%28180+x+17+trillion+li...
  - SushiHippie 2y ago
    Pardon my ignorance, but what does this calculate?
    - egberts1 2y ago
      Central angle, of which you can use that exponential power to determine how many digits of π to minimize error tolerance. Also, I err'd. I swapped L and R. https://www.wolframalpha.com/input?i=%28180+x+1+nanometer%29%2F%28pi+x+17+trillion+ly%29 https://www.wolframalpha.com/input?i=%28180+x+1+nanometer%29...
LeoPanthera 2y ago
This is why Star Trek transporters have “Heisenburg compensators”. Everyone knows that. And also that you have to disable them if you want to beam holographic objects off a holodeck. It’s just good science.
- batch12 2y ago
  Only if you need to trick a sentient AI into thinking it's part of the real world and not in a Russian doll cube of hypervisors.
  - _joel 2y ago
    We just need to crack out a Tommy Gun for that.
  - 0cf8612b2e1e 2y ago
    I know it was a monster-of-the-week format, but this episode really stuck with me. Created sentient life to never be discussed again. Data is only special in that he has a physical body.
    - batch12 2y ago
      Stuck on a shelf somewhere, oblivious to the fact that it's in a simulated environment. It'd be an interesting Star Trek II type followup when someone finds the cube and plugs in a cable only to have Moriarty escape and find a mobile emitter prototype somewhere on the network.. but I digress..
    - LeoPanthera 2y ago
      For what it’s worth, Moriarity does show up again in Star Trek Picard, but only as a brief cameo.
- loloquwowndueo 2y ago
  Heisenberg, not Heisenburg.
NKosmatos 2y ago
Perhaps the following statement from NASA will help ;-) "For JPL's highest accuracy calculations, which are for interplanetary navigation, we use 3.141592653589793" (15 digits). How Many Decimals of Pi Do We Really Need? : https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimals-of-pi-do-we-really-need/ https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimal...
- zamadatix 2y ago
  It's a shame they don't mention why they use specifically 15 digits (because of doubles?). Would give some satisfaction to why the specific amount after the explanation.
  - tomtom1337 2y ago
    Agreed! "What would happen if we were off by some tiny fraction" is a really interesting question to me!
    - gnramires 2y ago
      Something like spaceflight is subject to chaotic forces and unpredictable interactions, so having that many digits (more than 15 decimal) becomes negligible. (For example, a planetary force will probably vary by that much in ways models don't capture, like subtle tidal forces and subtle variations in its orbit, etc.). Navigation usually involves methods of measuring your position and adjusting course in theory needing much less precision. Simulating physical systems to extremely high precision (e.g. more than double precision) in general seems pointless in most situations because of those effects.
      - fwip 2y ago
        Adding on to this - the physical characteristics of the spacecraft itself are also not machined to nearly 15 digits of tolerance, so these feedback systems are necessary even if the rest of the universe were perfectly modeled.
  - 2y ago
kens 2y ago
That's a nice visual, but completely wrong. You're underestimating the accuracy by the absurd amount of roughly 10^160000000000000.
- onlyrealcuzzo 2y ago
  That's 10^(1.6*10^14) for anyone who can't read that many 0s.
- aaron695 2y ago
  > amount of roughly 10^160000000000000. You're also underestimating the accuracy by the absurd amount of roughly 10^202000000000000 ;) You need ~ zero of the digits of the calculated pi to do OPs calculation. [edit] My brains melting, I think I'm wrong and you are underestimating the underestimation of the accuracy by the absurd amount of roughly 10^42000000000000. OP is underestimating by 10^202000000000000.
  - kens 2y ago
    Yes, your edit is correct.
    - egberts1 2y ago
      To be fair, he did say "roughly". And boy, that was a shaffing error.
0x1ceb00da 2y ago
> This new pi value should land us on the precise nanometer on a planetary rock of a sun located some 18 trillion light years away. What does this mean?
- greycol 2y ago
  Imagine I'm using pi = 3 (accuracy of 1 significant figure) that's an error of about 4.5% of π(pi), 3.1 is only 1.3% 3.14 only 0.05% with the error decreasing with each additional significant figure. Imagine there's a circle with radius 1m and you've got a calculated bearing to it calculated using pi = 3 . In the worst case in 2 dimensions for every meter you walked you could be walking off to the side ~0.0225 meters (napkin maths) from where the circle really is, so it would only take the circle being ~45m away for you to walk right by it rather than through it. With pi =3.1 you're diverging ~0.0066m per meter so the circle would need to be ~152m away before there was a chance you'd miss it. 11 digits of pi gets you about 3 light years of walking before you had a chance of missing. They were discussing (with a large degree of understatement,as discussed by others above) that this value of pi gives great precision in these kinds of calculations.

slyall 2y ago

> This new pi value should land us on the precise nanometer on a planetary rock of a sun located about 18 trillion light years away. 40 digits or so will get you that...

hughesjj 2y ago
Nah, observable universe is only 93B light years in "diameter" at the current "moment"
- theandrewbailey 2y ago
  93 billion light years is 8.798×10^26 meters[0], so about 35 digits ought to suffice for any given nanometer. [0] https://www.wolframalpha.com/input?i=93+billion+light+years+in+meters https://www.wolframalpha.com/input?i=93+billion+light+years+...
  - egberts1 2y ago
    Now that's a solid win for us math nerds.
constantcrying 2y ago
In floating point arithmetic two consecutive operations can have an unbounded error. Just because the precision is good enough for one computation doesn't mean it is good enough for all computations.
egberts1 2y ago
Very good. Here's the WolframAlpha equation to your assertion of 40-digit ... or so. https://www.wolframalpha.com/input?i=%28180+x+17+trillion+light+years%29%2F%28pi+x1nm%29 https://www.wolframalpha.com/input?i=%28180+x+17+trillion+li...
- SushiHippie 2y ago
  Pardon my ignorance, but what does this calculate?
  - egberts1 2y ago
    Central angle, of which you can use that exponential power to determine how many digits of π to minimize error tolerance. Also, I err'd. I swapped L and R. https://www.wolframalpha.com/input?i=%28180+x+1+nanometer%29%2F%28pi+x+17+trillion+ly%29 https://www.wolframalpha.com/input?i=%28180+x+1+nanometer%29...

LeoPanthera 2y ago

This is why Star Trek transporters have “Heisenburg compensators”. Everyone knows that. And also that you have to disable them if you want to beam holographic objects off a holodeck. It’s just good science.

batch12 2y ago
Only if you need to trick a sentient AI into thinking it's part of the real world and not in a Russian doll cube of hypervisors.
- _joel 2y ago
  We just need to crack out a Tommy Gun for that.
- 0cf8612b2e1e 2y ago
  I know it was a monster-of-the-week format, but this episode really stuck with me. Created sentient life to never be discussed again. Data is only special in that he has a physical body.
  - batch12 2y ago
    Stuck on a shelf somewhere, oblivious to the fact that it's in a simulated environment. It'd be an interesting Star Trek II type followup when someone finds the cube and plugs in a cable only to have Moriarty escape and find a mobile emitter prototype somewhere on the network.. but I digress..
  - LeoPanthera 2y ago
    For what it’s worth, Moriarity does show up again in Star Trek Picard, but only as a brief cameo.
loloquwowndueo 2y ago
Heisenberg, not Heisenburg.

NKosmatos 2y ago

Perhaps the following statement from NASA will help ;-) "For JPL's highest accuracy calculations, which are for interplanetary navigation, we use 3.141592653589793" (15 digits). How Many Decimals of Pi Do We Really Need? : https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimals-of-pi-do-we-really-need/ https://www.jpl.nasa.gov/edu/news/2016/3/16/how-many-decimal...

zamadatix 2y ago
It's a shame they don't mention why they use specifically 15 digits (because of doubles?). Would give some satisfaction to why the specific amount after the explanation.
- tomtom1337 2y ago
  Agreed! "What would happen if we were off by some tiny fraction" is a really interesting question to me!
  - gnramires 2y ago
    Something like spaceflight is subject to chaotic forces and unpredictable interactions, so having that many digits (more than 15 decimal) becomes negligible. (For example, a planetary force will probably vary by that much in ways models don't capture, like subtle tidal forces and subtle variations in its orbit, etc.). Navigation usually involves methods of measuring your position and adjusting course in theory needing much less precision. Simulating physical systems to extremely high precision (e.g. more than double precision) in general seems pointless in most situations because of those effects.
    - fwip 2y ago
      Adding on to this - the physical characteristics of the spacecraft itself are also not machined to nearly 15 digits of tolerance, so these feedback systems are necessary even if the rest of the universe were perfectly modeled.
- 2y ago

kens 2y ago

That's a nice visual, but completely wrong. You're underestimating the accuracy by the absurd amount of roughly 10^160000000000000.

onlyrealcuzzo 2y ago
That's 10^(1.6*10^14) for anyone who can't read that many 0s.
aaron695 2y ago
> amount of roughly 10^160000000000000. You're also underestimating the accuracy by the absurd amount of roughly 10^202000000000000 ;) You need ~ zero of the digits of the calculated pi to do OPs calculation. [edit] My brains melting, I think I'm wrong and you are underestimating the underestimation of the accuracy by the absurd amount of roughly 10^42000000000000. OP is underestimating by 10^202000000000000.
- kens 2y ago
  Yes, your edit is correct.
  - egberts1 2y ago
    To be fair, he did say "roughly". And boy, that was a shaffing error.

0x1ceb00da 2y ago

> This new pi value should land us on the precise nanometer on a planetary rock of a sun located some 18 trillion light years away. What does this mean?

greycol 2y ago
Imagine I'm using pi = 3 (accuracy of 1 significant figure) that's an error of about 4.5% of π(pi), 3.1 is only 1.3% 3.14 only 0.05% with the error decreasing with each additional significant figure. Imagine there's a circle with radius 1m and you've got a calculated bearing to it calculated using pi = 3 . In the worst case in 2 dimensions for every meter you walked you could be walking off to the side ~0.0225 meters (napkin maths) from where the circle really is, so it would only take the circle being ~45m away for you to walk right by it rather than through it. With pi =3.1 you're diverging ~0.0066m per meter so the circle would need to be ~152m away before there was a chance you'd miss it. 11 digits of pi gets you about 3 light years of walking before you had a chance of missing. They were discussing (with a large degree of understatement,as discussed by others above) that this value of pi gives great precision in these kinds of calculations.