<p>In <em>Astronomical Algorithms</em>, Second Edition, Jean Meeus shows us how to convert local coordinates to equatorial and vice versa.</p>
<p>Formulas 13.5 and 13.6, p. 93:</p>
<p><span class="math-container">$$ \mathrm{tan}\ A = \frac {\mathrm{sin}\ H}{\mathrm{cos}\ H\ \mathrm{sin}\ \phi - \mathrm{tan}\ \delta\ \mathrm{cos}\ \phi} $$</span></p>
<p><span class="math-container">$$ \mathrm{sin}\ h = \mathrm{sin}\ \phi\ \mathrm{sin}\ \delta + \mathrm{cos}\ \phi\ \mathrm{cos}\ \delta\ \mathrm{cos} H $$</span></p>
<p>Unnumbered formulas, p. 94:</p>
<p><span class="math-container">$$ \mathrm{tan}\ H = \frac {\mathrm{sin}\ A}{\mathrm{cos}\ A\ \mathrm{sin}\ \phi + \mathrm{tan}\ h\ \mathrm{cos}\ \phi } $$</span></p>
<p><span class="math-container">$$ \mathrm{sin}\ \delta = \mathrm{sin}\ \phi\ \mathrm{sin}\ h - \mathrm{cos}\ \phi\ \mathrm{cos}\ h\ \mathrm{cos}\ A $$</span></p>
<p>where <span class="math-container">$ \alpha $</span> is the right ascension in <em>degrees</em>, <span class="math-container">$ \delta $</span> is the declination in degrees, <span class="math-container">$ \phi $</span> is the latitude (+ North, <span class="math-container">$ - $</span> South) in degrees, <span class="math-container">$ H $</span> is the hour angle measured westwards from South in <em>degrees</em>, <span class="math-container">$ A $</span> is the azimuth in degrees, and <span class="math-container">$ h $</span> is the height of the object in degrees. Please note that Meeus measures azimuths from the <em>South</em> heading East (90°) then North (180°) and West (270°).</p>
<p>With these formulas, you can find the required position of the point you want to know about.</p>
<p>In order to get the nadir, I would just put <span class="math-container">$ h = -90° $</span>. 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.</p>
<p>Hope this helps!</p>
<p>Clear skies.</p>

CC BY-SA 4.0
2021-02-03T23:01:55.437Z
2022-07-19T17:19:27.750Z
2022-07-19T17:19:27.750Z
Pierre Paquette
<p>What would be the easiest way to translate a zenith position to a nadir on the celestial sphere?</p>
<p>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.</p>

<positional-astronomy>
Translating a zenith position to the nadir
user36093
# Translating a zenith position to the nadir

(Asked by: username_0 on 2021-02-03)

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.

**Tags:** positional-astronomy

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## 1 answers

### Answer by username_1 on 2021-02-03. Score: 3 | Accepted answer

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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