4 ms·

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 exp

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.
snet0 2y ago
I imagine that you'd want to use fixed-point arithmetic when it comes to these things, right? Floating-point numbers are good enough for a lot of things, but the precise thing they're bad at is dealing in high precision at multiple orders of magnitude, which feels like a thing NASA would kinda need.
- mr_mitm 2y ago
  A lot of the parameters that enter their equations are probably measurements, like the gravitational acceleration, properties of some material, and so on. The numerical solutions to their equations have an error that is at least that of the most unprecise parameter, which I can't imagine to be more than four significant digits, so doubles should provide plenty of precision. The error introduced by the numerical algorithms can be controlled. I don't see why you'd need fixed point arithmetic.
  - snet0 2y ago
    Okay yeah, what you're saying seems true. I guess the GP comment was discussing that, with this new measurement of pi, we now have enough precision (in pi) to reference a point this small on an object this far away. Once you account for all the other uncertainties in referencing that point, as you mentioned, all that precision in one dimension of the measurement is completely meaningless. It still feels weird that you'd use an arithmetic with guaranteed imprecision in a field like this, but I can definitely see that, as long as you constrain the scales, it's more than enough.
    - lanstin 2y ago
      They put in scheduled fix the course burns, as there's a lot of uncertainty outside the math - the fuel burns probably can't be controlled to 5 sig figs, for example. Also, although I have no idea if this matters, N-body orbital mechanics itself is a chaotic system, and there will be times when the math just won't tell you the answer. https://botsin.space/@ThreeBodyBot https://botsin.space/@ThreeBodyBot if you like to see a lot of examples of 3-body orbits. (maybe just in 2d, I'm not sure).
    - fwip 2y ago
      Fixed-point also has guaranteed imprecision for many operations, because you only have a finite number of digits after the decimal point. e.g, with two decimal digits: (2.83 * 0.10) = 0.283, which is stored as 0.28.
- constantcrying 2y ago
_a_a_a_ 2y ago
They do say why. By example.
- zamadatix 2y ago
  If a double had happened to give 14 significant figures the article would list the exact same "why" 14 is enough. I.e. the article explains why it's more than enough, it doesn't explain why it's the exact number of digits used.
  - _a_a_a_ 2y ago
    If they had said X digits you'd be asking "but why X digits?", for any X.
    - zamadatix 2y ago
      Precisely why I said it'd be good if the article had answered that question. The value of X isn't happenstance in this case, the article just doesn't mention that.
pclmulqdq 2y ago
16 decimal places (really about 15.9 decimal places) is what you get with double-precision floating point. Double-double would be about 31 digits, and quad precision would get you 34. Single-precision gets you a bit more than 7 digits.
- zamadatix 2y ago
  The double representation of pi just so happens to be accurate to precisely 15 digits even though the general amount of precision is slightly higher.
- zamadatix 2y ago
  Missed in my other response: Double gives about double the precision, not 4 times, that of a single: "The 53-bit significand precision gives from 15 to 17 significant decimal digits precision (2−53 ≈ 1.11 × 10−16). If a decimal string with at most 15 significant digits is converted to the IEEE 754 double-precision format, giving a normal number, and then converted back to a decimal string with the same number of digits" (from Wikipedia)

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.

snet0 2y ago

I imagine that you'd want to use fixed-point arithmetic when it comes to these things, right? Floating-point numbers are good enough for a lot of things, but the precise thing they're bad at is dealing in high precision at multiple orders of magnitude, which feels like a thing NASA would kinda need.

mr_mitm 2y ago
A lot of the parameters that enter their equations are probably measurements, like the gravitational acceleration, properties of some material, and so on. The numerical solutions to their equations have an error that is at least that of the most unprecise parameter, which I can't imagine to be more than four significant digits, so doubles should provide plenty of precision. The error introduced by the numerical algorithms can be controlled. I don't see why you'd need fixed point arithmetic.
- snet0 2y ago
  Okay yeah, what you're saying seems true. I guess the GP comment was discussing that, with this new measurement of pi, we now have enough precision (in pi) to reference a point this small on an object this far away. Once you account for all the other uncertainties in referencing that point, as you mentioned, all that precision in one dimension of the measurement is completely meaningless. It still feels weird that you'd use an arithmetic with guaranteed imprecision in a field like this, but I can definitely see that, as long as you constrain the scales, it's more than enough.
  - lanstin 2y ago
    They put in scheduled fix the course burns, as there's a lot of uncertainty outside the math - the fuel burns probably can't be controlled to 5 sig figs, for example. Also, although I have no idea if this matters, N-body orbital mechanics itself is a chaotic system, and there will be times when the math just won't tell you the answer. https://botsin.space/@ThreeBodyBot https://botsin.space/@ThreeBodyBot if you like to see a lot of examples of 3-body orbits. (maybe just in 2d, I'm not sure).
  - fwip 2y ago
    Fixed-point also has guaranteed imprecision for many operations, because you only have a finite number of digits after the decimal point. e.g, with two decimal digits: (2.83 * 0.10) = 0.283, which is stored as 0.28.
constantcrying 2y ago

_a_a_a_ 2y ago

They do say why. By example.

zamadatix 2y ago
If a double had happened to give 14 significant figures the article would list the exact same "why" 14 is enough. I.e. the article explains why it's more than enough, it doesn't explain why it's the exact number of digits used.
- _a_a_a_ 2y ago
  If they had said X digits you'd be asking "but why X digits?", for any X.
  - zamadatix 2y ago
    Precisely why I said it'd be good if the article had answered that question. The value of X isn't happenstance in this case, the article just doesn't mention that.

pclmulqdq 2y ago

16 decimal places (really about 15.9 decimal places) is what you get with double-precision floating point. Double-double would be about 31 digits, and quad precision would get you 34. Single-precision gets you a bit more than 7 digits.

zamadatix 2y ago
The double representation of pi just so happens to be accurate to precisely 15 digits even though the general amount of precision is slightly higher.
zamadatix 2y ago
Missed in my other response: Double gives about double the precision, not 4 times, that of a single: "The 53-bit significand precision gives from 15 to 17 significant decimal digits precision (2−53 ≈ 1.11 × 10−16). If a decimal string with at most 15 significant digits is converted to the IEEE 754 double-precision format, giving a normal number, and then converted back to a decimal string with the same number of digits" (from Wikipedia)