<p>I'm converting the <em>W</em> for each planet in this <a href="https://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2015reprint.pdf" rel="nofollow noreferrer">IAU report</a> into <a href="https://en.wikipedia.org/wiki/Celestia" rel="nofollow noreferrer">Celestia</a>'s MeridianAngle, which is converted like this:
<span class="math-container">$$
\textrm{MeridianAngle}=W-Q
$$</span>
Where Q is the point where the planet's equator intersects with the ICRF equator (aka Earth's equator), and W is the position of the planet's prime meridian in the J2000 epoch.
I'm using ecliptic coordinates for the MeridianAngle, so setting it to 0 is the equivalent to Q.</p>
<p>I think Q is calculated like this, according to the report:
<span class="math-container">$$Q=90^{\circ}+\alpha_{0}$$</span>
Where <span class="math-container">$\alpha_{0}$</span> is the right ascension of the north pole.</p>
<p>Earth's <em>Q</em> is 90°, which is correct because I checked it by setting the MeridianAngle to 0 in the equatorial reference frame, and then going to 90° East in geographic coordinates, where it intersected the ecliptic. (Red line is the ecliptic plane.)</p>
<p><a href="https://i.sstatic.net/YI1BB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YI1BB.png" alt="" /></a></p>
<p>I also checked it by setting the MeridianAngle to 100.147 (Earth's W subtracted by 90°) and converting its rotational elements to the ecliptic reference frame. After converting them, the Sun was now shining near Earth's prime meridian in the J2000 epoch.</p>
<p><a href="https://i.sstatic.net/RWlVd.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RWlVd.png" alt="" /></a></p>
<p>For Mars, I added 90° to the right ascension of its north pole, which equals around 47°.</p>
<p>The problem is, 47° doesn't intersect the ecliptic.</p>
<p><a href="https://i.sstatic.net/kr5fx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kr5fx.png" alt="" /></a></p>
<p>But around 40° does.</p>
<p><a href="https://i.sstatic.net/I5tau.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/I5tau.png" alt="" /></a></p>
<p>I'm not even sure if this diagram helps me calculate the Q for other planets:</p>
<p><a href="https://i.sstatic.net/OEDEi.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OEDEi.png" alt="" /></a></p>
<p>I think the pole's declination also affects <em>Q</em>. When I change it, the <em>Q</em> also changes.</p>
<p>So, how do you calculate <em>Q</em>, the point where the equator of the planet and the ecliptic intersect, for other planets, when the RA and Dec of the planet's pole affect it? Do you have to convert those 2 to ecliptic coordinates first in order to calculate it?</p>

CC BY-SA 4.0
2022-10-20T13:06:13.933Z
it's a good idea if you can avoid formula symbols in questions but give a name to the thing you are asking about. Symbols are exchangable and used in many different contexts and widely varying meaning. (E.g. my first guess for Q is critical impact energy, could be latent heat, could be...)
planetmaker
CC BY-SA 4.0
2022-10-20T13:35:27.513Z
In the report, Q is where the the equator of the planet and the ICRF equator (or the ecliptic plane) intersect. W is the position of the prime meridian at epoch.
ILikeSaturn
CC BY-SA 4.0
2022-10-20T13:39:34.860Z
@ILikeSaturn, the ICRF equator is not the ecliptic plane.  The ICRF equator is the same as Earth's, it's inclined about 23.4 degrees to the ecliptic plane.
Greg Miller
CC BY-SA 4.0
2022-10-20T14:05:25.047Z
Oh, I see. I thought it was the same as the ecliptic plane because both look inclined by 23.4 degrees, and I thought the diagram was using Earth as an example. So, does this mean I have to convert the RA and Dec of the north pole to ecliptic coordinates before calculating Q?
ILikeSaturn
CC BY-SA 4.0
2022-10-20T16:54:51.143Z
@ILikeSaturn From https://ssd.jpl.nasa.gov/horizons/manual.html#frames "The ICRF was constructed to closely align with the older FK5/J2000 dynamic reference frame [...] the ICRF is thought to differ from the previous FK5/J2000 dynamical system by at most 0.02 arcseconds". So the XY plane of the ICRF is very close to the J2000 equatorial plane.
PM 2Ring
<rotation><iau>
Calculation of intersection point Q of planetary equator with ICRF equator for other planets
ILikeSaturn
# Calculation of intersection point Q of planetary equator with ICRF equator for other planets

(Asked by: username_0 on 2022-10-20)

I'm converting the *W* for each planet in this [IAU report](https://astropedia.astrogeology.usgs.gov/download/Docs/WGCCRE/WGCCRE2015reprint.pdf) into [Celestia](https://en.wikipedia.org/wiki/Celestia)'s MeridianAngle, which is converted like this:
$$
\textrm{MeridianAngle}=W-Q
$$
Where Q is the point where the planet's equator intersects with the ICRF equator (aka Earth's equator), and W is the position of the planet's prime meridian in the J2000 epoch.
I'm using ecliptic coordinates for the MeridianAngle, so setting it to 0 is the equivalent to Q.

I think Q is calculated like this, according to the report:
$$Q=90^{\circ}+\alpha\_{0}$$
Where $\alpha\_{0}$ is the right ascension of the north pole.

Earth's *Q* is 90°, which is correct because I checked it by setting the MeridianAngle to 0 in the equatorial reference frame, and then going to 90° East in geographic coordinates, where it intersected the ecliptic. (Red line is the ecliptic plane.)

[![](https://i.sstatic.net/YI1BB.png)](https://i.sstatic.net/YI1BB.png)

I also checked it by setting the MeridianAngle to 100.147 (Earth's W subtracted by 90°) and converting its rotational elements to the ecliptic reference frame. After converting them, the Sun was now shining near Earth's prime meridian in the J2000 epoch.

[![](https://i.sstatic.net/RWlVd.png)](https://i.sstatic.net/RWlVd.png)

For Mars, I added 90° to the right ascension of its north pole, which equals around 47°.

The problem is, 47° doesn't intersect the ecliptic.

[![](https://i.sstatic.net/kr5fx.png)](https://i.sstatic.net/kr5fx.png)

But around 40° does.

[![](https://i.sstatic.net/I5tau.png)](https://i.sstatic.net/I5tau.png)

I'm not even sure if this diagram helps me calculate the Q for other planets:

[![](https://i.sstatic.net/OEDEi.png)](https://i.sstatic.net/OEDEi.png)

I think the pole's declination also affects *Q*. When I change it, the *Q* also changes.

So, how do you calculate *Q*, the point where the equator of the planet and the ecliptic intersect, for other planets, when the RA and Dec of the planet's pole affect it? Do you have to convert those 2 to ecliptic coordinates first in order to calculate it?

**Tags:** rotation, iau
> **Comments:**
> **username_1**: it's a good idea if you can avoid formula symbols in questions but give a name to the thing you are asking about. Symbols are exchangable and used in many different contexts and widely varying meaning. (E.g. my first guess for Q is critical impact energy, could be latent heat, could be...)  
> **username_0**: In the report, Q is where the the equator of the planet and the ICRF equator (or the ecliptic plane) intersect. W is the position of the prime meridian at epoch.  
> **username_2**: @ILikeSaturn, the ICRF equator is not the ecliptic plane.  The ICRF equator is the same as Earth's, it's inclined about 23.4 degrees to the ecliptic plane.  
> **username_0**: Oh, I see. I thought it was the same as the ecliptic plane because both look inclined by 23.4 degrees, and I thought the diagram was using Earth as an example. So, does this mean I have to convert the RA and Dec of the north pole to ecliptic coordinates before calculating Q?  
> **username_3**: @ILikeSaturn From https://ssd.jpl.nasa.gov/horizons/manual.html#frames "The ICRF was constructed to closely align with the older FK5/J2000 dynamic reference frame [...] the ICRF is thought to differ from the previous FK5/J2000 dynamical system by at most 0.02 arcseconds". So the XY plane of the ICRF is very close to the J2000 equatorial plane.  

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## 0 answers

