<p>The <a href="https://www.aanda.org/articles/aa/pdf/2003/48/aa4068.pdf" rel="nofollow noreferrer">model used by the IAU</a> has (in arcseconds) the obliquity, <span class="math-container">$\epsilon_A$</span> as</p>
<p><span class="math-container">$$
\epsilon_A = \epsilon_0 − 46''.84024 t − 0''.00059 t^2 + 0''.001813 t^3$$</span></p>
<p>Where <span class="math-container">$t$</span> is measured in Julian centuries of 36525 days since J2000 (midday on Jan 1 2000) and <span class="math-container">$\epsilon_0=84381''.448 = 23^{\circ}.439167$</span></p>
<p>So to find when it was 23.44, or 84384'' you <a href="https://www.wolframalpha.com/input/?i=84384%20%3D84381.448%20-46.84024%20t-%200.0059t%5E2%2B0.001813t%5E3" rel="nofollow noreferrer">solve</a></p>
<p><span class="math-container">$$84384 =84381.448 -46.84024 t- 0.0059t^2+0.001813t^3$$</span></p>
<p>to give <span class="math-container">$t≈-0.0544831$</span> or <a href="https://www.timeanddate.com/date/dateadded.html?d1=1&amp;m1=1&amp;y1=2000&amp;type=sub&amp;ay=&amp;am=&amp;aw=&amp;ad=1990&amp;rec=" rel="nofollow noreferrer">1990 days before Jan 1 2000 Terrestrial Time</a> which was Thursday, 21 July 1994.</p>

CC BY-SA 4.0
2022-01-23T21:34:08.493Z
Thanks a lot, James K. Just a little question: how do you convert, in the last step, the −0.0545128 to 1991 days?
Unix
CC BY-SA 4.0
2022-01-23T21:49:56.280Z
multiply by 36525 (days in a Julian century)
James K
CC BY-SA 4.0
2022-01-23T21:52:43.407Z
Oh I see! Thanks again!
Unix
CC BY-SA 4.0
2022-01-24T10:36:04.523Z
The expression you wrote from the cited paper is IAU 1977 model. It would probably be better to use the IAU 2000 model, equation (14) rather than (6) from the cited paper. This is $$\epsilon_A = \epsilon_0 − 46″.84024 t − 0″.00059 t^2 + 0″.001813 t^3$$
David Hammen
CC BY-SA 4.0
2022-01-24T10:52:50.580Z
The above yields a slightly different value for $t$: than does your answer: $t\approx -0.0544831$. This results in 12:06:52 21 July 1994 TT.  It would also be useful to note that your "midday on Jan 1 2000" is midday Terrestrial Time (TT). TT was ahead of UTC (Coordinated Universal Time) by 64.184 seconds on 1 Jan 2000, and ahead of UTC by 61.184 seconds on 21 July 1994.
David Hammen
CC BY-SA 4.0
2022-01-24T11:05:36.027Z
You're right:  that'll teach me not to scan read.
James K
CC BY-SA 4.0
2022-01-27T22:25:19.687Z
If you want a precise answer, you will also need to take nutation into account.  The effect is less than an arcminute, so only ignore it if that's precise enough for your purposes.  The search term to find an algorithm including nutation is "true obliquity of the ecliptic".
Greg Miller
CC BY-SA 4.0
2022-01-23T16:39:09.200Z
2022-01-24T11:10:37.000Z
2022-01-24T11:10:37.000Z
James K
<p>How can I calculate the Earth's axial tilt in degrees. I tried Skyfield, Poliastro and PyEphem. I know it's changing <code>-0.47&quot;/year</code>, and it was <code>23° 27' 00&quot;</code> in the year 1917; so I created a spreadsheet (Calc) with this information. However, I think it's not exact enough.</p>
<p>In decimal degrees, in 1917 it was 23,45°, so I wanted to know when it was 23,44° and when it will be 23,43°. According to my Calc file, it was in 1993 and it will be in 2069 (-0,01 every ~76 years). According to an online tool, it was in 1992 and it will be in 2067 (-0,01 every ~75 years). Probably it's not a linear function. That's the reason I'm searching for an astronomy tool.</p>

<earth><skyfield><obliquity>
How to calculate the Earth's axial tilt (obliquity)
Unix
# How to calculate the Earth's axial tilt (obliquity)

(Asked by: username_0 on 2022-01-23)

How can I calculate the Earth's axial tilt in degrees. I tried Skyfield, Poliastro and PyEphem. I know it's changing `-0.47"/year`, and it was `23° 27' 00"` in the year 1917; so I created a spreadsheet (Calc) with this information. However, I think it's not exact enough.

In decimal degrees, in 1917 it was 23,45°, so I wanted to know when it was 23,44° and when it will be 23,43°. According to my Calc file, it was in 1993 and it will be in 2069 (-0,01 every ~76 years). According to an online tool, it was in 1992 and it will be in 2067 (-0,01 every ~75 years). Probably it's not a linear function. That's the reason I'm searching for an astronomy tool.

**Tags:** earth, skyfield, obliquity


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## 1 answers

### Answer by username_1 on 2022-01-23. Score: 5 | Accepted answer

The [model used by the IAU](https://www.aanda.org/articles/aa/pdf/2003/48/aa4068.pdf) has (in arcseconds) the obliquity, $\epsilon\_A$ as

$$
\epsilon\_A = \epsilon\_0 − 46''.84024 t − 0''.00059 t^2 + 0''.001813 t^3$$

Where $t$ is measured in Julian centuries of 36525 days since J2000 (midday on Jan 1 2000) and $\epsilon\_0=84381''.448 = 23^{\circ}.439167$

So to find when it was 23.44, or 84384'' you [solve](https://www.wolframalpha.com/input/?i=84384%20%3D84381.448%20-46.84024%20t-%200.0059t%5E2%2B0.001813t%5E3)

$$84384 =84381.448 -46.84024 t- 0.0059t^2+0.001813t^3$$

to give $t≈-0.0544831$ or [1990 days before Jan 1 2000 Terrestrial Time](https://www.timeanddate.com/date/dateadded.html?d1=1&m1=1&y1=2000&type=sub&ay=&am=&aw=&ad=1990&rec=) which was Thursday, 21 July 1994.

> **Comments:**
> **username_0**: Thanks a lot, username_1. Just a little question: how do you convert, in the last step, the −0.0545128 to 1991 days?  
> **username_1**: multiply by 36525 (days in a Julian century)  
> **username_0**: Oh I see! Thanks again!  
> **username_2**: The expression you wrote from the cited paper is IAU 1977 model. It would probably be better to use the IAU 2000 model, equation (14) rather than (6) from the cited paper. This is $$\epsilon_A = \epsilon_0 − 46″.84024 t − 0″.00059 t^2 + 0″.001813 t^3$$  
> **username_2**: The above yields a slightly different value for $t$: than does your answer: $t\approx -0.0544831$. This results in 12:06:52 21 July 1994 TT.  It would also be useful to note that your "midday on Jan 1 2000" is midday Terrestrial Time (TT). TT was ahead of UTC (Coordinated Universal Time) by 64.184 seconds on 1 Jan 2000, and ahead of UTC by 61.184 seconds on 21 July 1994.  
> **username_1**: You're right:  that'll teach me not to scan read.  
> **username_3**: If you want a precise answer, you will also need to take nutation into account.  The effect is less than an arcminute, so only ignore it if that's precise enough for your purposes.  The search term to find an algorithm including nutation is "true obliquity of the ecliptic".  

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