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&quot;
which was obtained based on my location at +50°54'13.9&quot; -2°34'19.1&quot; 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.