# Translating a zenith position to the nadir

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.

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:

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

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

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.