Page 1 of 1

How to calculate planet rotation/orientation

Posted: 24.09.2018, 01:50
by posfan12
I am creating a POV-Ray scene to model the Solar System:

http://lib.povray.org/searchcollection/index2.php ... emOrrery&contributorTag=SharkD

Using this PDF I was able to plot the planet positions:

https://ssd.jpl.nasa.gov/txt/aprx_pos_planets.pdf

However, now I want to rotate/orient each planet properly using rotational elements found here:

https://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2009reprint.pdf

I *think* this is how it works:

1. Position the planet so that the pole axis is aligned with the Z (up) coordinate axis.
2. Position the planet so that the the prime meridian points toward the negative X axis.
3. Rotate around the Z axis by W.
4. Rotate around the X axis by (90-delta_0) degrees.
5. Rotate around the Z axis by (90+alpha_0) degrees.
6. Rotate around the X axis by 23.43928 to get out of the ICRF frame and into the ecliptic frame.
7. Rotate around the Z axis by 90 degrees for reasons unknown to me.

These steps seem to work reliably for Earth and Mars (give or take one hour). But for Saturn they are off by about ~90 degrees around the Z axis (as far as I can tell). Am I following the wrong steps? Where am I making a mistake? Thanks!

Posted: 27.09.2018, 22:50
by posfan12
Here are some comparisons showing that my math is somehow wrong.

https://imgur.com/a/iJcVme7

I skipped step 7, above, however.

Posted: 02.11.2018, 13:38
by onetwothree
Hi posfan12,

Have you asked this question on http://forum.celestialmatters.org? There some original Celestia authors can be found.

Posted: 10.06.2019, 18:53
by amzounslide
I'd just make one small point. In your first diagram, it appears as if you are showing the Moon's orbit as being on the same plane as the Celestial plane/equator, where, in reality, it is inclined by ~5 degrees to the ecliptic (~18 degrees to the Celestial equator).

Posted: 11.06.2019, 21:16
by posfan12
There are no moons in any of my images.