<p>According to the <strong>old</strong> nomenclature, the azimuth <span class="math-container">$A_s$</span> was defined starting from the <strong>South</strong> towards the West, from 0º to 360º (SWNE direction).</p>
<p>If <span class="math-container">$a$</span> = altitude, (not zenith angular distance = <span class="math-container">$z$</span>) and <span class="math-container">$H$</span> = hour angle, then the relationships were:</p>
<p><span class="math-container">$\left.
\begin{aligned}\
sin \delta &amp;= \sin a \sin \phi - \cos a \cos \phi \cos A_s \\
\sin H \cos \delta &amp;= \cos a \sin A_s \\
\cos H \cos \delta &amp;=\sin a \cos \phi + \cos a \sin \phi \cos A_s
\end{aligned}
\right \}$</span></p>
<p><span class="math-container">$\phi$</span> is the observer's latitude</p>
<p><strong>Subsequently</strong>, the International Astronomical Union (<strong>IAU</strong>) <strong>changed the definition</strong>: <strong>Azimuth</strong> <span class="math-container">$A_n$</span> is <strong>now</strong> defined from the <strong>North</strong> towards the Est, from 0º to 360º, (NESW direction)</p>
<p><span class="math-container">$A_s=A_n+\pi$</span></p>
<p>Now, the current relationships are:</p>
<p><span class="math-container">$$\left.
\begin{aligned}
\sin \delta &amp;= \sin a \sin \phi + \cos a \cos \phi \cos A_n \\
\sin H \cos \delta &amp;= -\cos a \sin A_n \\
\cos H \cos \delta &amp;=\sin a \cos \phi - \cos a \sin \phi \cos A_n
\end{aligned}
\right \}$$</span></p>
<p>If you prefer the latter expressions as a function of zenith angle <span class="math-container">$z$</span> instead of altitude <span class="math-container">$a$</span> :</p>
<p><span class="math-container">$a=\dfrac{\pi}2 - z$</span></p>
<p><span class="math-container">$\left.
\begin{aligned}
\sin \delta &amp;= \cos z \sin \phi + \sin z \cos \phi \cos A_n \\
\sin H \cos \delta &amp;= -\sin z \sin A_n \\
\cos H \cos \delta &amp;=\cos z \cos \phi - \sin z \sin \phi \cos A_n
\end{aligned}
\right \}$</span></p>
<p>This can be read on the IAU page:</p>
<p><em>Description: In a horizontal coordinate system, azimuth refers to the direction (along the horizon) at which the object is found. It is measured in degrees <strong>starting from the north and towards the east</strong>. Azimuth values cover a full circle from 0 deg. to 360 deg. In other words, if you draw an imaginary arc on the celestial sphere from the object to the horizon and perpendicular to the horizon, the azimuth will tell you the location of the point where this arc meets the horizon. An object located directly north would have 0 deg. azimuth, an object directly east would have 90 deg. azimuth and so on. <strong>In older textbooks used in multiple countries, the convention was to start measuring the azimuth from the south towards the west</strong>. Thus, azimuth values in those textbooks would be shifted by 180 deg.</em></p>
<p>The link is: <a href="https://www.astro4edu.org/resources/glossary/term/36/" rel="nofollow noreferrer"><strong>GLOSSARY TERM: AZIMUTH</strong></a></p>
<p>Best regards.</p>

CC BY-SA 4.0
2023-05-01T09:04:57.997Z
2023-05-03T06:19:06.510Z
2023-05-03T06:19:06.510Z
Albert
<p>Hello I am trying to convert my Horizontal coordinate system dataset into Equatorial coordinate system. However, it seems that all the information from the internet is different.</p>
<p>I will use symbols as following</p>
<p>θ : (Observer's) Longitude (I am not going to use it now, but required to get LST)</p>
<p>φ : (Observer's) Latitude</p>
<p>A : azimuth</p>
<p>a : altitude (or zenith)</p>
<p>α : R.A.</p>
<p>δ : declination</p>
<p>h : hour angle (Not going to use it now, but required to calculate R.A.)</p>
<p>t : LST (Local Sidereal Time, required to calculate R.A.)</p>
<p>Also, I will just write down formula for the declination because I think I am stuck with this.</p>
<p>First, from Wikipedia,</p>
<p><a href="https://en.wikipedia.org/wiki/Astronomical_coordinate_systems" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Astronomical_coordinate_systems</a></p>
<pre><code>sin(δ)=sin(φ)sin(a)-cos(φ)cos(a)cos(A)
</code></pre>
<p>Next, from the University of St. Andrews (I guess this is one of the British institution?)</p>
<p><a href="http://star-www.st-and.ac.uk/%7Efv/webnotes/chapter7.htm" rel="nofollow noreferrer">http://star-www.st-and.ac.uk/~fv/webnotes/chapter7.htm</a></p>
<pre><code>sin(δ)=sin(a)sin(φ)+cos(a)cos(φ)cos(A)
</code></pre>
<p>From an YouTube page, (Watch from 5:45)</p>
<p><div class="youtube-embed"><div>
                <iframe width="640px" height="395px" src="https://www.youtube.com/embed/5UY1WBKQtNQ?start=0"></iframe>
            </div></div></p>
<pre><code>sin(δ)=sin(φ)cos(a)+cos(φ)cos(a)cos(A)
</code></pre>
<p>I think I just have to use arcsine to get declination, however, they all suggests me different formula. The more frustrating is that they don't seem to match with example dataset which I have. Can anyone tell me what would be the correct formula? Or am I missing something like definition in domain?</p>

<observational-astronomy><coordinate><positional-astronomy><spherical-astronomy><coordinate-systems>
Discrepancies in the equations for converting Horizontal coordinates to Equatorial coordinates
Kyle
# Discrepancies in the equations for converting Horizontal coordinates to Equatorial coordinates

(Asked by: username_0 on 2023-05-01)

Hello I am trying to convert my Horizontal coordinate system dataset into Equatorial coordinate system. However, it seems that all the information from the internet is different.
I will use symbols as following
θ : (Observer's) Longitude (I am not going to use it now, but required to get LST)
φ : (Observer's) Latitude
A : azimuth
a : altitude (or zenith)
α : R.A.
δ : declination
h : hour angle (Not going to use it now, but required to calculate R.A.)
t : LST (Local Sidereal Time, required to calculate R.A.)
Also, I will just write down formula for the declination because I think I am stuck with this.
First, from Wikipedia,
https://en.wikipedia.org/wiki/Astronomical_coordinate_systems
sin(δ)=sin(φ)sin(a)-cos(φ)cos(a)cos(A)

Next, from the University of St. Andrews (I guess this is one of the British institution?)
http://star-www.st-and.ac.uk/~fv/webnotes/chapter7.htm
sin(δ)=sin(a)sin(φ)+cos(a)cos(φ)cos(A)

From an YouTube page, (Watch from 5:45)



sin(δ)=sin(φ)cos(a)+cos(φ)cos(a)cos(A)

I think I just have to use arcsine to get declination, however, they all suggests me different formula. The more frustrating is that they don't seem to match with example dataset which I have. Can anyone tell me what would be the correct formula? Or am I missing something like definition in domain?

**Tags:** observational-astronomy, coordinate, positional-astronomy, spherical-astronomy, coordinate-systems

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## 1 answers

### Answer by username_1 on 2023-05-01. Score: 4 | Accepted answer

According to the **old** nomenclature, the azimuth $A\_s$ was defined starting from the **South** towards the West, from 0º to 360º (SWNE direction).

If $a$ = altitude, (not zenith angular distance = $z$) and $H$ = hour angle, then the relationships were:

$\left.
\begin{aligned}\
sin \delta &= \sin a \sin \phi - \cos a \cos \phi \cos A\_s \\
\sin H \cos \delta &= \cos a \sin A\_s \\
\cos H \cos \delta &=\sin a \cos \phi + \cos a \sin \phi \cos A\_s
\end{aligned}
\right \}$

$\phi$ is the observer's latitude

**Subsequently**, the International Astronomical Union (**IAU**) **changed the definition**: **Azimuth** $A\_n$ is **now** defined from the **North** towards the Est, from 0º to 360º, (NESW direction)

$A\_s=A\_n+\pi$

Now, the current relationships are:

$$\left.
\begin{aligned}
\sin \delta &= \sin a \sin \phi + \cos a \cos \phi \cos A\_n \\
\sin H \cos \delta &= -\cos a \sin A\_n \\
\cos H \cos \delta &=\sin a \cos \phi - \cos a \sin \phi \cos A\_n
\end{aligned}
\right \}$$

If you prefer the latter expressions as a function of zenith angle $z$ instead of altitude $a$ :

$a=\dfrac{\pi}2 - z$

$\left.
\begin{aligned}
\sin \delta &= \cos z \sin \phi + \sin z \cos \phi \cos A\_n \\
\sin H \cos \delta &= -\sin z \sin A\_n \\
\cos H \cos \delta &=\cos z \cos \phi - \sin z \sin \phi \cos A\_n
\end{aligned}
\right \}$

This can be read on the IAU page:

*Description: In a horizontal coordinate system, azimuth refers to the direction (along the horizon) at which the object is found. It is measured in degrees **starting from the north and towards the east**. Azimuth values cover a full circle from 0 deg. to 360 deg. In other words, if you draw an imaginary arc on the celestial sphere from the object to the horizon and perpendicular to the horizon, the azimuth will tell you the location of the point where this arc meets the horizon. An object located directly north would have 0 deg. azimuth, an object directly east would have 90 deg. azimuth and so on. **In older textbooks used in multiple countries, the convention was to start measuring the azimuth from the south towards the west**. Thus, azimuth values in those textbooks would be shifted by 180 deg.*

The link is: [**GLOSSARY TERM: AZIMUTH**](https://www.astro4edu.org/resources/glossary/term/36/)

Best regards.


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