<p>The <a href="https://astrogeology.usgs.gov/search/map/Docs/WGCCRE/WGCCRE2015reprint" rel="nofollow noreferrer">2015 Report</a> of the IAU Working Group on Cartographic Coordinates and Rotational Elements (WGCCRE) gives the same ICRF (equatorial, J2000) coordinates for the north pole of the Sun:</p>
<p>α<sub>0</sub> = 286.13°
<br>δ<sub>0</sub> = 63.87°</p>
<p>But the Earth orbits the Sun in the ecliptic plane, so let's <a href="https://ned.ipac.caltech.edu/coordinate_calculator" rel="nofollow noreferrer">convert</a> to ecliptic coordinates (J2000):</p>
<p>λ<sub>0</sub> = 345.77°
<br>β<sub>0</sub> = 82.75°</p>
<p>When the Sun appears at that longitude, its north pole points away from Earth.
At the opposite longitude, its north pole points toward Earth.
Using the <a href="https://www.imo.net/resources/solar-longitude-tables/" rel="nofollow noreferrer">meteor observing</a> utility <a href="https://kikimoreau.shinyapps.io/SollongCalc/" rel="nofollow noreferrer">SollongCalc</a>:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Sun orientation</th>
<th style="text-align: center;">Geocentric ecliptic longitude of Sun (J2000)</th>
<th style="text-align: center;">2024 Date (UT)</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">N far, S near</td>
<td style="text-align: center;">345.77°</td>
<td style="text-align: center;">Mar 6.2</td>
</tr>
<tr>
<td style="text-align: center;">equator facing</td>
<td style="text-align: center;">75.77°</td>
<td style="text-align: center;">Jun 6.3</td>
</tr>
<tr>
<td style="text-align: center;">N near, S far</td>
<td style="text-align: center;">165.77°</td>
<td style="text-align: center;">Sep 8.3</td>
</tr>
<tr>
<td style="text-align: center;">equator facing</td>
<td style="text-align: center;">255.77°</td>
<td style="text-align: center;">Dec 7.7</td>
</tr>
</tbody>
</table>
</div>
<p>As you expect, these recur at intervals of a sidereal year, about 20 minutes longer than a tropical year.
If the figures in question are 140 years old, the dates then should be about 2 days earlier.</p>
<p>Some formulas to compute the Sun's ecliptic longitude produce its longitude of date, which increases 0.014° per year relative to J2000 longitude.
If you use HORIZONS, a geocentric <code>ObsEcLon</code> is a longitude of date but a heliocentric <code>hEcl-Lon</code> is a J2000 longitude.</p>

CC BY-SA 4.0
2024-01-10T18:40:29.633Z
That answers the question perfectly; it also both explains what was incorrect with the first comment on the question (not converting to ecliptic coordinates) and shows that the dates have drifted a couple of days from whatever year the diagram was made in (which would indeed mean it's roughly 140 years old, with some margin due to rounding). Thanks!
Outis Nemo
CC BY-SA 4.0
2024-03-26T21:07:13.817Z
I'm revisiting this answer again, and wondering about something: you mentioned that the Solar ecliptic longitude increases 0.014°/yr relative to J2000; shouldn't that mean that the date where the Solar ecliptic longitude is e.g. 345.77° should slightly decrease rather than increase? The latter is of course what I see happens, so I'm just wondering what I've gotten wrong about that.
Outis Nemo
CC BY-SA 4.0
2024-01-08T22:45:46.283Z
2024-01-08T22:45:46.283Z
Mike G
<p>In case the title is a bit confusing, what I'd like to know is essentially respectively on which dates the Solar axial tilt is pointing straight towards Earth, straight away from Earth, and perpendicular to us (which will be two dates).</p>
<p>I'd also like to know how I could calculate this for myself.</p>
<p>The two things I've found about it so far are from a couple of different answers. One mentions the north pole of rotation as being follows:</p>
<blockquote>
<p><strong>North Pole of Rotation</strong></p>
<p>Right Ascension: 286.13</p>
<p>Declination : 63.87</p>
</blockquote>
<p>I am however not sure how to calculate it from that.</p>
<p>The other is an image I found, which seemingly does provide the dates, but I believe it's quite outdated:</p>
<p><a href="https://i.sstatic.net/0M7fs.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0M7fs.png" alt="Solar equator" /></a></p>
<p>This puts the dates for the minimum and maximum on March 4 and September 6 respectively, with the two dates of zero (perpendicular, i.e. where we see exactly half of both the northern and southern hemisphere of Sol) being June 4 and December 6 respectively.</p>
<p>However, from observation this doesn't quite seem to match up anymore; is that just me doing something wrong, or have the dates indeed drifted since then? When I thought about it I concluded that the only reason that I can see that the dates would move would be due to defining our dates according to Earth's axial tilt, and thus it being subject to axial precession (&quot;precession of the equinoctes&quot;), but this effect would only lead to ~1 day of &quot;slippage&quot; every ~70 years or so; I guess that diagram could be quite old and that it has slipped a day or two since it was made, but that's not very much.</p>
<p>I guess with an accurate calculation of the dates for today based on the coordinates of the rotation I could compare the two and see. Hopefully someone can help.</p>

CC BY-SA 4.0
2024-01-08T15:38:03.427Z
What are the sources of the quote and the figure?
Mike G
CC BY-SA 4.0
2024-01-08T17:08:12.820Z
Since there are 365 days in a year and 360 degrees in a circle, a rough approximation is 286 days from the vernal equinox the North pole is pointed towards Earth. Add +-90 for the perpedicular dates, and 180 for the pole pointed away.  A more precise method is to use an ephemeris (eg JPL Horizons) to get the RA of the Sun.  When the RA of the sun is 180-RA of the pole, the North pole is pointing towards Earth.
Greg Miller
CC BY-SA 4.0
2024-01-08T17:59:30.353Z
@GregMiller: that sounds strange to me; by that logic, wouldn't it be when the right ascension is 360 - the RA of the pole for when it's pointing away from us? That would be 73.87 degrees of right ascension, but it points away from us roughly two weeks ahead of the March equinox, which can't possibly correspond to that many degrees of right ascension given how the equinox is defined at 0 degrees. Am I missing something?
Outis Nemo
<the-sun><solar-system><positional-astronomy><axial-tilt>
On what four dates dates do we observe the minimal, maximal, and two days of zero Solar axial tilt relative to us?
Outis Nemo
# On what four dates dates do we observe the minimal, maximal, and two days of zero Solar axial tilt relative to us?

(Asked by: username_0 on 2024-01-08)

In case the title is a bit confusing, what I'd like to know is essentially respectively on which dates the Solar axial tilt is pointing straight towards Earth, straight away from Earth, and perpendicular to us (which will be two dates).

I'd also like to know how I could calculate this for myself.

The two things I've found about it so far are from a couple of different answers. One mentions the north pole of rotation as being follows:

> **North Pole of Rotation**
>
> Right Ascension: 286.13
>
> Declination : 63.87

I am however not sure how to calculate it from that.

The other is an image I found, which seemingly does provide the dates, but I believe it's quite outdated:

[![Solar equator](https://i.sstatic.net/0M7fs.png)](https://i.sstatic.net/0M7fs.png)

This puts the dates for the minimum and maximum on March 4 and September 6 respectively, with the two dates of zero (perpendicular, i.e. where we see exactly half of both the northern and southern hemisphere of Sol) being June 4 and December 6 respectively.

However, from observation this doesn't quite seem to match up anymore; is that just me doing something wrong, or have the dates indeed drifted since then? When I thought about it I concluded that the only reason that I can see that the dates would move would be due to defining our dates according to Earth's axial tilt, and thus it being subject to axial precession ("precession of the equinoctes"), but this effect would only lead to ~1 day of "slippage" every ~70 years or so; I guess that diagram could be quite old and that it has slipped a day or two since it was made, but that's not very much.

I guess with an accurate calculation of the dates for today based on the coordinates of the rotation I could compare the two and see. Hopefully someone can help.

**Tags:** the-sun, solar-system, positional-astronomy, axial-tilt

> **Comments:**
> **username_1**: What are the sources of the quote and the figure?  
> **username_2**: Since there are 365 days in a year and 360 degrees in a circle, a rough approximation is 286 days from the vernal equinox the North pole is pointed towards Earth. Add +-90 for the perpedicular dates, and 180 for the pole pointed away.  A more precise method is to use an ephemeris (eg JPL Horizons) to get the RA of the Sun.  When the RA of the sun is 180-RA of the pole, the North pole is pointing towards Earth.  
> **username_0**: @GregMiller: that sounds strange to me; by that logic, wouldn't it be when the right ascension is 360 - the RA of the pole for when it's pointing away from us? That would be 73.87 degrees of right ascension, but it points away from us roughly two weeks ahead of the March equinox, which can't possibly correspond to that many degrees of right ascension given how the equinox is defined at 0 degrees. Am I missing something?  

---

## 1 answers

### Answer by username_1 on 2024-01-08. Score: 4 | Accepted answer

The [2015 Report](https://astrogeology.usgs.gov/search/map/Docs/WGCCRE/WGCCRE2015reprint) of the IAU Working Group on Cartographic Coordinates and Rotational Elements (WGCCRE) gives the same ICRF (equatorial, J2000) coordinates for the north pole of the Sun:

α0 = 286.13°
  
δ0 = 63.87°

But the Earth orbits the Sun in the ecliptic plane, so let's [convert](https://ned.ipac.caltech.edu/coordinate_calculator) to ecliptic coordinates (J2000):

λ0 = 345.77°
  
β0 = 82.75°

When the Sun appears at that longitude, its north pole points away from Earth.
At the opposite longitude, its north pole points toward Earth.
Using the [meteor observing](https://www.imo.net/resources/solar-longitude-tables/) utility [SollongCalc](https://kikimoreau.shinyapps.io/SollongCalc/):

| Sun orientation | Geocentric ecliptic longitude of Sun (J2000) | 2024 Date (UT) |
| --- | --- | --- |
| N far, S near | 345.77° | Mar 6.2 |
| equator facing | 75.77° | Jun 6.3 |
| N near, S far | 165.77° | Sep 8.3 |
| equator facing | 255.77° | Dec 7.7 |

As you expect, these recur at intervals of a sidereal year, about 20 minutes longer than a tropical year.
If the figures in question are 140 years old, the dates then should be about 2 days earlier.

Some formulas to compute the Sun's ecliptic longitude produce its longitude of date, which increases 0.014° per year relative to J2000 longitude.
If you use HORIZONS, a geocentric `ObsEcLon` is a longitude of date but a heliocentric `hEcl-Lon` is a J2000 longitude.

> **Comments:**
> **username_0**: That answers the question perfectly; it also both explains what was incorrect with the first comment on the question (not converting to ecliptic coordinates) and shows that the dates have drifted a couple of days from whatever year the diagram was made in (which would indeed mean it's roughly 140 years old, with some margin due to rounding). Thanks!  
> **username_0**: I'm revisiting this answer again, and wondering about something: you mentioned that the Solar ecliptic longitude increases 0.014°/yr relative to J2000; shouldn't that mean that the date where the Solar ecliptic longitude is e.g. 345.77° should slightly decrease rather than increase? The latter is of course what I see happens, so I'm just wondering what I've gotten wrong about that.  

---

