4
$\begingroup$

I'm currently using JPL DE440 ephemerides. Following the documentation here, verified with the header file (specially the records in GROUP 1050) for the DE440 ephemerides (file header.440 available here), and using the Clenshaw algorithm to calculate the sum of the Chebyshev approximation, as described in section 3.3.4 of Fundamentals of Astrodynamics and Applications by David Vallado, I was able to obtain my own calculations.

I then aimed to compare with the values produced by Horizons online app. However, I am getting a strange discrepancy, where the X coordinate for each planet is exactly correct but Y and Z deviate substantially.

As a test case, I am using the position of Mercury (in ICRF frame, with the center at the Solar System barycenter, i.e., as produced directly after calculating the sum of the Chebyshev polynomials). In particular, for Julian date $JD=2469776.5$ (corresponding to 2049-Dec-01 00:00:00.0000 TDB), equivalently Modified Julian date $MJD=69776$, the coefficients are found in the first data block of file named ascp02050.440 here. This is confirmed by the fact that the first value in that data block (i.e., the first float in the second line of the file, since the first one indicated the block number and the number of coefficients) is 2469776.5. The second float is 2469808.5, indicating that this block contains the data for Julian dates comprised in that interval. Furthermore, as indicated by the GROUP 1050 section of the header file, DE440 contains 4 time subintervals for Mercury (1st value of line 3 of GROUP 1050), each one with 14 Chebyshev coefficients (1st value of line 2 of GROUP 1050). Having 4 time subintervals, each one corresponds to a period of 8 days. But since we are calculating the position at the beginning date of the interval, we are in subinterval 1.

Furthermore, there are Chebyshev coefficients for the X, Y and Z position. Overall, this means we need in total 42 coefficients, which would be the first 42 floats in the mentioned ascp02050.440 after skipping the first two that indicate the Julian dates for which the coefficients apply. Of that set, the first 14 would be for the X component, the next 14 for the Y component, and the final 14 for the Z component.

I then obtain the following values from the Clenshaw algorithm (values in km):

Component Value
X 46410910
Y -33842102
Z -22826642

On the other hand, from Horizons Web app, for Mercury, for the same JD, and with coordinates also in ICRF and referenced to the Solar System barycenter, I obtain:

Component Value
X 46410910
Y -40129438
Z -7481419

Values are truncated in both cases to the units of km, but for the X component they match up to at least 0.01 mm, which is the last digit provided by Horizons app.

I have verified the same occurs for other planets and dates.

The only difference I see with Horizons app is that they use the DE441 ephemeris instead of DE440 even in the range of application of DE440, but it seems unlikely that this would lead to such a large difference (and also no difference in the X component).

Furthermore, I checked the example provided here for Mercury, where they state explicitly the X, Y and Z coefficients for Mercury for $JD=2458850.5$. I compared these coefficients with those extracted by my implementation and, while some of them have slight differences due to the fact that the coefficients in the linked example are from DE405, they agree very well, which indicates that I seem to be extracting the correct coefficients.

Does anybody have any idea what I could be possibly missing here?

$\endgroup$
7
  • 1
    $\begingroup$ Could you post a screenshot of the HORIZONS settings you are using? I'm guessing you're using J2000 in one place and J(current) in another, or some other minor frame reference issue. $\endgroup$ Commented Jan 20, 2022 at 12:38
  • $\begingroup$ @barrycarter You were spot on! I just went to the settings and in addition to the Reference frame setting (which was set to ICRF), there was a "Reference plane" setting. The default value was "ecliptic x-y plane derived from reference frame", but upon changing it to "x-y axes of reference frame (equatorial or equatorial-aligned, inertial)", the output values match exactly what I calculated! Do you have by any chance a good understanding of what these exactly mean? Intuitively, I guess the Y-Z plane is somehow different between the 2 settings, but it would be nice to properly understand how! $\endgroup$
    – Rafa
    Commented Jan 20, 2022 at 15:37
  • 1
    $\begingroup$ The X axis in both frames points towards the first point of Aries on J2000, if that helps. In equatorial reference frame, the XY plane is the celestial equator, and in the ecliptic reference frame the XY plane is the ecliptic. The difference in YZ coordinates should be small near the first point Aries (and its antisol, the first point of Libra) and highest near the solstices. $\endgroup$ Commented Jan 20, 2022 at 17:56
  • 1
    $\begingroup$ The Horizons web app provides a link that gives all the parameters in batch-file format. "the batch-file corresponding to the settings above can be viewed by using this link". That may be more useful / convenient than a screenshot. You can run the batch file via the Horizons APIs. The following comment contains a link to a basic Sage script which runs such batch files; you can also find it on gist.github.com/PM2Ring/b1fec75e78cc08f6fc28c6f6c43529c3 $\endgroup$
    – PM 2Ring
    Commented Jan 21, 2022 at 3:56
  • 1
    $\begingroup$ Horizons batch-file runner $\endgroup$
    – PM 2Ring
    Commented Jan 21, 2022 at 3:56

1 Answer 1

2
$\begingroup$

I'm almost certain this is essentially a 23 degree shift to a standard coordinate system. See https://ssd.jpl.nasa.gov/planets/approx_pos.html for the details. Essentially you need to convert the x and y as follows.

$e = 23.43928^\circ$

$y_{eq} = \cos(e) y_{ecl} - sin(e)z_{ecl}$

$z_{eq} = \sin(e) y_{ecl} + cos(e)z_{ecl}$

JPL Horizons uses the J2000 frame, which isn't the same as the frame planets are measured in for reasons I don't fully understand.

$\endgroup$
2
  • 2
    $\begingroup$ That was indeed it! By applying the conversion, I get matching values. It seems a bit confusing. They say they are using the ICRF frame (almost equivalent as J2000), but have two different possible positions for the X-Y plane, aligned with the ecliptic (which is, as far as I understand, not the real ICRF frame, but nevertheless is the default in JPL Horizons!), and aligned with the X-Y plane of the reference frame ("real" ICRF frame). This setting can be changed in the Table settings section, to obtain values of Y and Z without rotation to the ecliptic plane directly. Thanks! $\endgroup$
    – Rafa
    Commented Jan 21, 2022 at 2:29
  • 2
    $\begingroup$ Glad that my pain could help someone else as well! $\endgroup$ Commented Jan 21, 2022 at 3:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .