What would be the easiest way to translate a zenith position to a nadir on the celestial sphere?

For example I have a position from the azimuthal grid in Stellarium; 3h44m32.64s/+50°50'16.2" which was obtained based on my location at +50°54'13.9" -2°34'19.1" and a time of ; 15/03/2019 16:24:20.


1 Answer 1


In Astronomical Algorithms, Second Edition, Jean Meeus shows us how to convert local coordinates to equatorial and vice versa.

Formulas 13.5 and 13.6, p. 93:

$$ \mathrm{tan}\ A = \frac {\mathrm{sin}\ H}{\mathrm{cos}\ H\ \mathrm{sin}\ \phi - \mathrm{tan}\ \delta\ \mathrm{cos}\ \phi} $$

$$ \mathrm{sin}\ h = \mathrm{sin}\ \phi\ \mathrm{sin}\ \delta + \mathrm{cos}\ \phi\ \mathrm{cos}\ \delta\ \mathrm{cos} H $$

Unnumbered formulas, p. 94:

$$ \mathrm{tan}\ H = \frac {\mathrm{sin}\ A}{\mathrm{cos}\ A\ \mathrm{sin}\ \phi + \mathrm{tan}\ h\ \mathrm{cos}\ \phi } $$

$$ \mathrm{sin}\ \delta = \mathrm{sin}\ \phi\ \mathrm{sin}\ h - \mathrm{cos}\ \phi\ \mathrm{cos}\ h\ \mathrm{cos}\ A $$

where $ \alpha $ is the right ascension in degrees, $ \delta $ is the declination in degrees, $ \phi $ is the latitude (+ North, $ - $ South) in degrees, $ H $ is the hour angle measured westwards from South in degrees, $ A $ is the azimuth in degrees, and $ h $ is the height of the object in degrees. Please note that Meeus measures azimuths from the South heading East (90°) then North (180°) and West (270°).

With these formulas, you can find the required position of the point you want to know about.

In order to get the nadir, I would just put $ h = -90° $. Another way would be to find the position of the object at zenith, add/remove 180° to its right ascension, and change the sign of its declination.

Hope this helps!

Clear skies.


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