6 ms·

You know, even if I know plenty of digits, I can't think of any real world reason in my personal life for needing more decimals than 3.14 . If I add one mor

You know, even if I know plenty of digits, I can't think of any real world reason in my personal life for needing more decimals than 3.14 . If I add one more, that's a correction of less than 1/1000 and no measurement I do is that precise. NASA's rocket trajectories are less precise than that. Of course, extremely precise devices need better, but they are removed far from my everyday life. Has anybody needed more than 3.14?

CydeWeys 6y ago
> correction of less than 1/1000 and no measurement I do is that precise. NASA's rocket trajectories are less precise than that. Source? If you make that big of an error you're gonna fail your orbital insertion at the other end entirely when going to places like Mars, let alone any farther.
- vidanay 6y ago
  I don't have a source to support the OP, but I would think there is a significant allowance for course correction that reduces the required precision. It's not a purely ballistic trajectory.
  - CydeWeys 6y ago
    The mid-course correction is not because you used an imprecise value of pi. If that's all it were about, you'd just use a precise pi from the beginning and then not even need the mid-course correction. Missions have failed because the rocket engines failed to start up for the mid-course correction. Unnecessarily risking that because for some reason you insist on using far fewer digits of pi than your computer is capable of would be insanity.
    - vidanay 6y ago
      The context of my statement/question was definitely "given the precision that can be achieved with available precision" and not with an arbitrary imprecise constant in the calculations. As you say, the mechanical aspect of the launch and transfer orbit insertion is going to be several orders of magnitude more imprecise than the mathematical goal - that is what the mid-course corrections are for (aka the difference between theory and reality)
      - CydeWeys 6y ago
        The comment I was originally responding to said "Has anybody needed more [digits in pi] than 3.14?" and "NASA's rocket trajectories are less precise than [one part in a thousand]", both of which I take issue with. You then stepped in to defend those statements against my objections. I'm glad it turns out you don't actually agree with those statements, but hopefully you can see my confusion.
- dmurray 6y ago
  How precisely do you have to get it right up front? Maybe it's OK to get your initial trajectory off by a 1/1000 part, and make an adjustment when you're 3/4 of the way to Mars and your relative error is now more like 1/250.
  - CydeWeys 6y ago
    There's no world in which it makes sense to save a few digits on pi and then risk running out of fuel while making your mid-course correction because your initial burn was too far off. The mid-course correction is because you can only be so precise when you're firing rocket engines, not because you're foolishly using imprecise constants for no good reason at all. So I'll reiterate, I want to see a source that NASA uses only 3.14 as the value for pi in making their trajectory calculations. And I want to point out the absurdity of even so much as having this debate in the very comment thread for an article on nasa.gov where NASA itself is saying that they use 3.141592653589793 for pi. We already have as good as a source as we're going to find, and there's a lot more sig figs in it than 3!
    - dmurray 6y ago
      Sure. But the question asked was not about what value NASA use, but what precision you need for space travel. The rule of thumb I use when looking at noisy estimates is that you should shoot for twice the precision (one extra bit of information) that you have in your noisiest other value (or of your total noise - they're usually about the same thing). Any less precision than that and your estimate gets worse, but more precision doesn't materially improve your estimate. I don't know a rigorous proof for this but you can hand wave it from the Nyquist-Shannon sampling theorem [0]. So how much uncertainty comes from other sources when deciding on an initial burn for your trip to Mars? Contributing factors could be your fine control over the thrust from the rocket engine, but also solar radiation, gravity or - I suspect this is biggest - attitude control. I'd guess NASA can achieve a precision of better than 1/500 but probably not 1/50,000, so they'd need about 5 or 6 digits of pi. But I'm interested in hearing a more educated guess! [0] https://en.m.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem https://en.m.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_samp...
- hyperman1 6y ago
  Wow, calm down, CydeWeys, you're scaring me. To be clear: No source, imprecise sentence, should have added a ton of buts and ifs, but that would make the question twice as long and thrice as boring. I meant it more in a philosophical way. For example: I don't think rockets start on course with a precision of 1e-3, and even if they do, small atmospheric changes will probably cause errors greater than that. I assume the feedback loop of the guidance systems easily smooth over errors this small. So while the theoretical trajectory calculations need more precision, the real world guidance might do not. Another consideration: Is e.g. the length of the rocket know this precise? If temperature on a day changes over a range of e.g. 10 degrees, the materials might expand and contract more than 1e-3. Now that's NASA. There are clearly things requiring much smaller errors. My CPU litography being off a factor 1e-3 will not end well. I meant the quest in my everyday life. If I build a round table for my home, 3.14 will probably serve me well enough on the drawing table. At what point will random Joe Shmoe notice when pi is not exactly 3.14
  - CydeWeys 6y ago
    Do you have any sources you can point to from research you've done on this subject? This is all coming off as uninformed speculation.
tsomctl 6y ago
That's a thousandth of an inch. Lots of things are machined to a thousandth of an inch tolerance, and some things are machined to a significantly finer tolerance. And if you're drilling bolt holes on a diameter that's greater than an inch, you'll need to use a lot more than 3.14.
GloriousKoji 6y ago
For work I needed the accuracy out to 15 decimal places. It was for calculating random jitter based on integrating and averaging a large number of phase noise measurements.
reaperducer 6y ago
I can't think of any real world reason in my personal life for needing more decimals than 3.14 If you keep your feet on the ground, that's probably enough for most things. But your grandchildren will probably need more, as spaceflight becomes commoditized. A couple of decimal places might mean the difference between landing on Mars, and icy death in the vacuum of space.
stochastician 6y ago
This is an incredibly interesting question, and the answer is an emphatic YES! Many systems involve iterative schemes, where the output of one step is used as the input to the next step. Here, these precision errors can accumulate, and if there's a multiplicative term in your equations, they can explode! These sorts of problems are actually very common in a lot of scientific computing and simulation contexts, which is why many in the scientific computing community look aghast at the rise of FP16 (and even fp32) in machine learning applications. Of course, those algorithms are often of a _very_ different nature from (say) the large-scale linear algebra or PDE solvers we're using, but still it's pretty shocking if you're used to worrying about machine precision!
- mjevans 6y ago
  Machine learning might also prefer focusing on the magnitude rather than an exact value. (With the lower precision number part more about being nudged between magnitude bins.) E.G. bfloat16 https://en.wikipedia.org/wiki/Bfloat16_floating-point_format https://en.wikipedia.org/wiki/Bfloat16_floating-point_format

CydeWeys 6y ago

> correction of less than 1/1000 and no measurement I do is that precise. NASA's rocket trajectories are less precise than that. Source? If you make that big of an error you're gonna fail your orbital insertion at the other end entirely when going to places like Mars, let alone any farther.

vidanay 6y ago
I don't have a source to support the OP, but I would think there is a significant allowance for course correction that reduces the required precision. It's not a purely ballistic trajectory.
- CydeWeys 6y ago
  The mid-course correction is not because you used an imprecise value of pi. If that's all it were about, you'd just use a precise pi from the beginning and then not even need the mid-course correction. Missions have failed because the rocket engines failed to start up for the mid-course correction. Unnecessarily risking that because for some reason you insist on using far fewer digits of pi than your computer is capable of would be insanity.
  - vidanay 6y ago
    The context of my statement/question was definitely "given the precision that can be achieved with available precision" and not with an arbitrary imprecise constant in the calculations. As you say, the mechanical aspect of the launch and transfer orbit insertion is going to be several orders of magnitude more imprecise than the mathematical goal - that is what the mid-course corrections are for (aka the difference between theory and reality)
    - CydeWeys 6y ago
      The comment I was originally responding to said "Has anybody needed more [digits in pi] than 3.14?" and "NASA's rocket trajectories are less precise than [one part in a thousand]", both of which I take issue with. You then stepped in to defend those statements against my objections. I'm glad it turns out you don't actually agree with those statements, but hopefully you can see my confusion.
dmurray 6y ago
How precisely do you have to get it right up front? Maybe it's OK to get your initial trajectory off by a 1/1000 part, and make an adjustment when you're 3/4 of the way to Mars and your relative error is now more like 1/250.
- CydeWeys 6y ago
  There's no world in which it makes sense to save a few digits on pi and then risk running out of fuel while making your mid-course correction because your initial burn was too far off. The mid-course correction is because you can only be so precise when you're firing rocket engines, not because you're foolishly using imprecise constants for no good reason at all. So I'll reiterate, I want to see a source that NASA uses only 3.14 as the value for pi in making their trajectory calculations. And I want to point out the absurdity of even so much as having this debate in the very comment thread for an article on nasa.gov where NASA itself is saying that they use 3.141592653589793 for pi. We already have as good as a source as we're going to find, and there's a lot more sig figs in it than 3!
  - dmurray 6y ago
    Sure. But the question asked was not about what value NASA use, but what precision you need for space travel. The rule of thumb I use when looking at noisy estimates is that you should shoot for twice the precision (one extra bit of information) that you have in your noisiest other value (or of your total noise - they're usually about the same thing). Any less precision than that and your estimate gets worse, but more precision doesn't materially improve your estimate. I don't know a rigorous proof for this but you can hand wave it from the Nyquist-Shannon sampling theorem [0]. So how much uncertainty comes from other sources when deciding on an initial burn for your trip to Mars? Contributing factors could be your fine control over the thrust from the rocket engine, but also solar radiation, gravity or - I suspect this is biggest - attitude control. I'd guess NASA can achieve a precision of better than 1/500 but probably not 1/50,000, so they'd need about 5 or 6 digits of pi. But I'm interested in hearing a more educated guess! [0] https://en.m.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem https://en.m.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_samp...
hyperman1 6y ago
Wow, calm down, CydeWeys, you're scaring me. To be clear: No source, imprecise sentence, should have added a ton of buts and ifs, but that would make the question twice as long and thrice as boring. I meant it more in a philosophical way. For example: I don't think rockets start on course with a precision of 1e-3, and even if they do, small atmospheric changes will probably cause errors greater than that. I assume the feedback loop of the guidance systems easily smooth over errors this small. So while the theoretical trajectory calculations need more precision, the real world guidance might do not. Another consideration: Is e.g. the length of the rocket know this precise? If temperature on a day changes over a range of e.g. 10 degrees, the materials might expand and contract more than 1e-3. Now that's NASA. There are clearly things requiring much smaller errors. My CPU litography being off a factor 1e-3 will not end well. I meant the quest in my everyday life. If I build a round table for my home, 3.14 will probably serve me well enough on the drawing table. At what point will random Joe Shmoe notice when pi is not exactly 3.14
- CydeWeys 6y ago
  Do you have any sources you can point to from research you've done on this subject? This is all coming off as uninformed speculation.

tsomctl 6y ago

That's a thousandth of an inch. Lots of things are machined to a thousandth of an inch tolerance, and some things are machined to a significantly finer tolerance. And if you're drilling bolt holes on a diameter that's greater than an inch, you'll need to use a lot more than 3.14.

GloriousKoji 6y ago

For work I needed the accuracy out to 15 decimal places. It was for calculating random jitter based on integrating and averaging a large number of phase noise measurements.

reaperducer 6y ago

I can't think of any real world reason in my personal life for needing more decimals than 3.14 If you keep your feet on the ground, that's probably enough for most things. But your grandchildren will probably need more, as spaceflight becomes commoditized. A couple of decimal places might mean the difference between landing on Mars, and icy death in the vacuum of space.

stochastician 6y ago

This is an incredibly interesting question, and the answer is an emphatic YES! Many systems involve iterative schemes, where the output of one step is used as the input to the next step. Here, these precision errors can accumulate, and if there's a multiplicative term in your equations, they can explode! These sorts of problems are actually very common in a lot of scientific computing and simulation contexts, which is why many in the scientific computing community look aghast at the rise of FP16 (and even fp32) in machine learning applications. Of course, those algorithms are often of a _very_ different nature from (say) the large-scale linear algebra or PDE solvers we're using, but still it's pretty shocking if you're used to worrying about machine precision!

mjevans 6y ago
Machine learning might also prefer focusing on the magnitude rather than an exact value. (With the lower precision number part more about being nudged between magnitude bins.) E.G. bfloat16 https://en.wikipedia.org/wiki/Bfloat16_floating-point_format https://en.wikipedia.org/wiki/Bfloat16_floating-point_format