5 ms·

> We shouldn't round off internally because we might use it in as an intermediate result in a calculation. Only if the precision is meaningful. The author's p

> We shouldn't round off internally because we might use it in as an intermediate result in a calculation. Only if the precision is meaningful. The author's point is that it isn't for geographical coordinates.

whatshisface 7y ago
Error tends to grow as the calculation proceeds. 6th-decimal "insignificant rounding" done over and over as the calculation proceeds can grow to become arbitrarily large.
- Luc 7y ago
  Which calculation on lat/lon coordinates would make an atom-scale error grow, say, a million times to millimeter-size?
  - deleted 7y ago
    [deleted]
  - caf 7y ago
    "Build the gravitational wave observatory between the Thai restaurant and the railway station"
    - krapht 7y ago
      You joke but a lot of radar/radio observation is done with ECEF coordinates, which is a (lat, lon, z-axis) triple.
      - jhayward 7y ago
        ECEF is a Cartesian (x,y,z) not spherical (𝞺, θ, 𝜑) coordinate system. Units in ECEF are in meters from (0,0,0).
        krapht 7y ago
        Blah, shows how much I remember about it. We always converted to ENU locally to start with, hah.
- Majromax 7y ago
  What numerically unstable algorithms are ever performed with geolocations? Moreover, given the author's point that real measurement errors exceed the false precision of published data, if such a calculation were performed and did provide "arbitrarily large" error, it would indicate that the result should in fact be nonsense.
  - whatshisface 7y ago
    I'm thinking about the case where a calculation extends across many roundings.
    - Majromax 7y ago
      That doesn't necessarily make the algorithm unstable. See for example: >>> import numpy as np >>> x0 = np.random.rand(1) >>> x_rnd = np.round(100*x)/100.0 >>> for i in range(1000): ... x = np.mod(x+np.pi,1) ... x_rnd = np.mod(x_rnd + np.pi,1) ... x_rnd = np.round(100*x)/100 ... >>> x_real = np.mod(x0+1000*np.pi,1) >>> print(x_real,x,x_rnd) (array([0.51013166]), array([0.51013166]), array([0.51])) Note no loss of accuracy for all of the intermediate roundings. The accuracy is preserved because there's no mechanism here to amplify the error. That's why I ask about the kind of algorithms typically applied to geolocated data. Off the top of my head, I can't think of anything that would be both useful and error-amplifying.
  - jhayward 7y ago
    Vincenty's formula has some instabilities/failure to converge at certain areas of the globe. [1] https://en.wikipedia.org/wiki/Vincenty%27s_formulae https://en.wikipedia.org/wiki/Vincenty%27s_formulae
- caf 7y ago
  If you're doing calculations with geographical co-ordinates that are so unstable that a nanometre position difference in the original makes an arbitrarily large difference in the result, your results aren't sensible anyway.

whatshisface 7y ago

Error tends to grow as the calculation proceeds. 6th-decimal "insignificant rounding" done over and over as the calculation proceeds can grow to become arbitrarily large.

Luc 7y ago
Which calculation on lat/lon coordinates would make an atom-scale error grow, say, a million times to millimeter-size?
- deleted 7y ago
  [deleted]
- caf 7y ago
  "Build the gravitational wave observatory between the Thai restaurant and the railway station"
  - krapht 7y ago
    You joke but a lot of radar/radio observation is done with ECEF coordinates, which is a (lat, lon, z-axis) triple.
    - jhayward 7y ago
      ECEF is a Cartesian (x,y,z) not spherical (𝞺, θ, 𝜑) coordinate system. Units in ECEF are in meters from (0,0,0).
      - krapht 7y ago
        Blah, shows how much I remember about it. We always converted to ENU locally to start with, hah.
Majromax 7y ago
What numerically unstable algorithms are ever performed with geolocations? Moreover, given the author's point that real measurement errors exceed the false precision of published data, if such a calculation were performed and did provide "arbitrarily large" error, it would indicate that the result should in fact be nonsense.
- whatshisface 7y ago
  I'm thinking about the case where a calculation extends across many roundings.
  - Majromax 7y ago
    That doesn't necessarily make the algorithm unstable. See for example: >>> import numpy as np >>> x0 = np.random.rand(1) >>> x_rnd = np.round(100*x)/100.0 >>> for i in range(1000): ... x = np.mod(x+np.pi,1) ... x_rnd = np.mod(x_rnd + np.pi,1) ... x_rnd = np.round(100*x)/100 ... >>> x_real = np.mod(x0+1000*np.pi,1) >>> print(x_real,x,x_rnd) (array([0.51013166]), array([0.51013166]), array([0.51])) Note no loss of accuracy for all of the intermediate roundings. The accuracy is preserved because there's no mechanism here to amplify the error. That's why I ask about the kind of algorithms typically applied to geolocated data. Off the top of my head, I can't think of anything that would be both useful and error-amplifying.
- jhayward 7y ago
  Vincenty's formula has some instabilities/failure to converge at certain areas of the globe. [1] https://en.wikipedia.org/wiki/Vincenty%27s_formulae https://en.wikipedia.org/wiki/Vincenty%27s_formulae
caf 7y ago
If you're doing calculations with geographical co-ordinates that are so unstable that a nanometre position difference in the original makes an arbitrarily large difference in the result, your results aren't sensible anyway.