# How to calculate the Earth's axial tilt (obliquity)

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.

The model used by the IAU 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

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

to give $$t≈-0.0544831$$ or 1990 days before Jan 1 2000 Terrestrial Time which was Thursday, 21 July 1994.

• Thanks a lot, James K. Just a little question: how do you convert, in the last step, the −0.0545128 to 1991 days?
– Unix
Jan 23 at 21:34
• multiply by 36525 (days in a Julian century) Jan 23 at 21:49
• 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 $$\epsilon_A = \epsilon_0 − 46″.84024 t − 0″.00059 t^2 + 0″.001813 t^3$$ Jan 24 at 10:36
• 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. Jan 24 at 10:52
• 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". Jan 27 at 22:25