<p>You can use this fundamental formula in spherical astronomy<sup><a href="https://en.wikipedia.org/wiki/Astronomical_coordinate_systems#Equatorial_%E2%86%94_horizontal" rel="noreferrer">[1]</a></sup>
<span class="math-container">$$\sin a=\sin \phi \sin \delta + \cos \phi \cos \delta \cos H$$</span>
where</p>
<ul>
<li><span class="math-container">$a$</span> is the wanted altitude,</li>
<li><span class="math-container">$\phi$</span> is your latitude,</li>
<li><span class="math-container">$\delta$</span> is the declination of the star, and</li>
<li><span class="math-container">$H$</span> is the hour angle, measured in the clockwise direction.</li>
</ul>
<p>Pay attention to the units! (Don't mix degrees, radians and grads. Common cause of error!)</p>
<p>Since I don't know how you are familiar with the trigonometric functions (I believe pretty well), you only get <span class="math-container">$\sin a$</span> using that formula. You need to get the <span class="math-container">$\arcsin$</span> of that value in order to get wanted altitude <span class="math-container">$a$</span>.</p>
<hr />
<p>The solution above is perfectly correct in theory (on competitions, exams, and for personal use), but if you are writing a program on computer, you might find the following useful:</p>
<p>The factor we haven't yet addressed is the atmospheric refraction<sup><a href="https://en.wikipedia.org/wiki/Atmospheric_refraction" rel="noreferrer">[2]</a></sup> It causes the star to look higher than in reality. The effect is pretty small, on range of few arc minutes.</p>
<p>First you need to calculate the factor <span class="math-container">$R$</span> by the formula<sup><a href="http://star-www.st-and.ac.uk/%7Efv/webnotes/chapt11.htm" rel="noreferrer">[3]</a></sup>
<span class="math-container">$$R=\frac{16.27''\cdot P}{273 + T}$$</span>
where <span class="math-container">$P$</span> is the pressure <strong>in millibars</strong> and <span class="math-container">$T$</span> is the temperature <strong>in degrees of Celsius</strong>. You are perfectly fine using just <span class="math-container">$R=60''=1'$</span>. Then the apparent altitude of the star is given by <span class="math-container">$a'=R+a$</span>. Again, pay attention to the units (everything in degrees or everything in minutes ...)</p>
<hr />
<p>If you are interested in learning about <em><a href="https://www.britannica.com/science/positional-astronomy" rel="noreferrer">positional or spherical astronomy</a></em>, then I advise you to visit <a href="https://astronomy.stackexchange.com/questions/44368/best-books-on-spherical-astronomy">another Stack Exchange question about this topic</a>. Personaly, I have had great fun with <a href="http://observatory.znu.ac.ir/files/uploaded/editor_files/observatory/files/Fundamental%2BAstronomy%2B5th%2BEdition.pdf" rel="noreferrer">Fundamental Astronomy</a>. But in general, the formula is derived using the <a href="https://en.wikipedia.org/wiki/Spherical_law_of_cosines" rel="noreferrer">spherical law of cosines</a>.</p>

CC BY-SA 4.0
2021-09-19T18:43:56.180Z
Thank you so much! I looked at the link [1] you shared (which is great), but I don't seem to see how that was derived. Would you be able to link me to where I could learn more about htat?
altitudeFinder
CC BY-SA 4.0
2021-09-19T18:52:15.307Z
Included in the answer. [Fundamental Astronomy](http://observatory.znu.ac.ir/files/uploaded/editor_files/observatory/files/Fundamental%2BAstronomy%2B5th%2BEdition.pdf) has the derivation. You can read more about it in my answer [here](https://astronomy.stackexchange.com/a/44376/25994).
User123
CC BY-SA 4.0
2021-09-19T18:29:34.050Z
2021-09-19T18:44:19.113Z
2021-09-19T18:44:19.113Z
User123
<p>I'm trying to find the altitude of a star for observing, but all I have is the hour angle and declination of the star, along with latitude of the location I'm observing from. How can I find the altitude?</p>

<celestial-mechanics>
How to calculate the altitude of a star given the hour angle, declination, and latitude?
altitudeFinder
# How to calculate the altitude of a star given the hour angle, declination, and latitude?

(Asked by: username_0 on 2021-09-19)

I'm trying to find the altitude of a star for observing, but all I have is the hour angle and declination of the star, along with latitude of the location I'm observing from. How can I find the altitude?

**Tags:** celestial-mechanics

---

## 1 answers

### Answer by username_1 on 2021-09-19. Score: 9 | Accepted answer

You can use this fundamental formula in spherical astronomy[[1]](https://en.wikipedia.org/wiki/Astronomical_coordinate_systems#Equatorial_%E2%86%94_horizontal)
$$\sin a=\sin \phi \sin \delta + \cos \phi \cos \delta \cos H$$
where

* $a$ is the wanted altitude,
* $\phi$ is your latitude,
* $\delta$ is the declination of the star, and
* $H$ is the hour angle, measured in the clockwise direction.

Pay attention to the units! (Don't mix degrees, radians and grads. Common cause of error!)

Since I don't know how you are familiar with the trigonometric functions (I believe pretty well), you only get $\sin a$ using that formula. You need to get the $\arcsin$ of that value in order to get wanted altitude $a$.

---

The solution above is perfectly correct in theory (on competitions, exams, and for personal use), but if you are writing a program on computer, you might find the following useful:

The factor we haven't yet addressed is the atmospheric refraction[[2]](https://en.wikipedia.org/wiki/Atmospheric_refraction) It causes the star to look higher than in reality. The effect is pretty small, on range of few arc minutes.

First you need to calculate the factor $R$ by the formula[[3]](http://star-www.st-and.ac.uk/%7Efv/webnotes/chapt11.htm)
$$R=\frac{16.27''\cdot P}{273 + T}$$
where $P$ is the pressure **in millibars** and $T$ is the temperature **in degrees of Celsius**. You are perfectly fine using just $R=60''=1'$. Then the apparent altitude of the star is given by $a'=R+a$. Again, pay attention to the units (everything in degrees or everything in minutes ...)

---

If you are interested in learning about *[positional or spherical astronomy](https://www.britannica.com/science/positional-astronomy)*, then I advise you to visit [another Stack Exchange question about this topic](https://astronomy.stackexchange.com/questions/44368/best-books-on-spherical-astronomy). Personaly, I have had great fun with [Fundamental Astronomy](http://observatory.znu.ac.ir/files/uploaded/editor_files/observatory/files/Fundamental%2BAstronomy%2B5th%2BEdition.pdf). But in general, the formula is derived using the [spherical law of cosines](https://en.wikipedia.org/wiki/Spherical_law_of_cosines).

> **Comments:**
> **username_0**: Thank you so much! I looked at the link [1] you shared (which is great), but I don't seem to see how that was derived. Would you be able to link me to where I could learn more about htat?  
> **username_1**: Included in the answer. [Fundamental Astronomy](http://observatory.znu.ac.ir/files/uploaded/editor_files/observatory/files/Fundamental%2BAstronomy%2B5th%2BEdition.pdf) has the derivation. You can read more about it in my answer [here](https://astronomy.stackexchange.com/a/44376/25994).  

---

