<p>If you do not require arcminute precision, you can approximate Greenwich sidereal time as the Earth rotation angle <span class="math-container">$\theta(t)$</span>.
<a href="https://www.iers.org/IERS/EN/Publications/TechnicalNotes/tn32.html" rel="nofollow noreferrer">IERS Technical Note 32</a> &sect;5.4.4 gives
<span class="math-container">$$ \begin{align}
   \theta(t) &amp;= 2 \pi (0.77905~72732~640 + 1.00273~78119~11354~48~ t) \\
      &amp;\approx 280.46^\circ + 360.985612^\circ~ t
\end{align} $$</span>
where <span class="math-container">$t$</span> is a real number of days since JD 2451545.0 (2000-01-01 12:00 TT &approx; 11:59 UTC).</p>

<p>For an observer at north latitude <span class="math-container">$\phi$</span> and east longitude <span class="math-container">$\lambda$</span>, things should line up like this:</p>

<p>Local sidereal time <span class="math-container">$ = \mathrm{LST} \approx \theta(t) + \lambda$</span></p>

<p>Zenith (RA, Dec) <span class="math-container">$ = (\alpha, \delta) = (\mathrm{LST},~ \phi)$</span></p>

<p>North horizon <span class="math-container">$(\alpha, \delta) = \begin{cases}
   (\mathrm{LST + 12h},~ 90^\circ - \phi) &amp; \mathrm{if}~\phi &gt;= 0 \\
   (\mathrm{LST},~ 90^\circ + \phi) &amp; \mathrm{if}~\phi &lt; 0
\end{cases}$</span></p>

<p>East horizon <span class="math-container">$(\alpha, \delta) = (\mathrm{LST + 6h},~ 0^\circ)$</span></p>

<p>The transformation between equatorial and horizontal coordinates can be composed of two rotations, similar but not necessarily identical to those in <a href="https://en.wikipedia.org/wiki/Celestial_coordinate_system#Converting_coordinates" rel="nofollow noreferrer">Wikipedia: Celestial coordinate system</a>.
For example, you could:</p>

<ol>
<li>Start with the north horizon vector pointing at the north celestial pole, and the zenith vector pointing at (0h, 0&deg;).</li>
<li>Rotate the ground counterclockwise as seen from the north horizon vector by LST.</li>
<li>Rotate the ground clockwise as seen from the east horizon vector by <span class="math-container">$\phi$</span>.</li>
</ol>

CC BY-SA 4.0
2020-05-03T19:34:12.040Z
Great explanation thanks :)
PaddyDeveloper
CC BY-SA 4.0
2020-04-28T21:49:00.617Z
2020-04-28T21:49:00.617Z
Mike G
<p>How would one calculate the rotation/tilt of the earth to simulate the Night Sky in a self-written tool or app. I am trying to built an app for my telescope to show me on my phone what I am looking at. I can already simulate the exact positions of stars at the 1.1.2000 at 00:00:00 but I dont know how the "earth" should be rotated so I can simulate the sky from my current position and accurate time. </p>

<p>There are App and tools like stellarium that can achive this but I just cant get the angles right: <a href="https://i.sstatic.net/GKysC.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GKysC.png" alt="enter image description here"></a></p>

<p>If anyone could point me in the right direction and tell me what I need to change/calculate it would help me a lot.</p>

CC BY-SA 4.0
2020-04-26T22:43:03.360Z
Earth rotates wrt stars once every sidereal day, that is about 23h:56m
planetmaker
CC BY-SA 4.0
2020-04-27T02:39:26.800Z
If I understand right, you need to convert ra dec (the position of a star around the Earth) to alt az (the position of the star at given time and place).  There are some other questions about this here.  You do need your time and location.
James K
CC BY-SA 4.0
2020-04-28T11:07:45.763Z
@JamesK in that case [does this answer](https://astronomy.stackexchange.com/a/35871/7982) this question?
uhoh
<star><earth><rotation>
Calculating the orientation of the night sky
PaddyDeveloper
# Calculating the orientation of the night sky

(Asked by: username_0 on 2020-04-26)

How would one calculate the rotation/tilt of the earth to simulate the Night Sky in a self-written tool or app. I am trying to built an app for my telescope to show me on my phone what I am looking at. I can already simulate the exact positions of stars at the 1.1.2000 at 00:00:00 but I dont know how the "earth" should be rotated so I can simulate the sky from my current position and accurate time.

There are App and tools like stellarium that can achive this but I just cant get the angles right: [![enter image description here](https://i.sstatic.net/GKysC.png)](https://i.sstatic.net/GKysC.png)

If anyone could point me in the right direction and tell me what I need to change/calculate it would help me a lot.

**Tags:** star, earth, rotation

> **Comments:**
> **username_1**: Earth rotates wrt stars once every sidereal day, that is about 23h:56m  
> **username_2**: If I understand right, you need to convert ra dec (the position of a star around the Earth) to alt az (the position of the star at given time and place).  There are some other questions about this here.  You do need your time and location.  
> **username_3**: @username_2 in that case [does this answer](https://astronomy.stackexchange.com/a/35871/7982) this question?  

---

## 1 answers

### Answer by username_4 on 2020-04-28. Score: 1 | Accepted answer

If you do not require arcminute precision, you can approximate Greenwich sidereal time as the Earth rotation angle $\theta(t)$.
[IERS Technical Note 32](https://www.iers.org/IERS/EN/Publications/TechnicalNotes/tn32.html) §5.4.4 gives
$$ \begin{align}
\theta(t) &= 2 \pi (0.77905~72732~640 + 1.00273~78119~11354~48~ t) \\
&\approx 280.46^\circ + 360.985612^\circ~ t
\end{align} $$
where $t$ is a real number of days since JD 2451545.0 (2000-01-01 12:00 TT &approx; 11:59 UTC).

For an observer at north latitude $\phi$ and east longitude $\lambda$, things should line up like this:

Local sidereal time $ = \mathrm{LST} \approx \theta(t) + \lambda$

Zenith (RA, Dec) $ = (\alpha, \delta) = (\mathrm{LST},~ \phi)$

North horizon $(\alpha, \delta) = \begin{cases}
(\mathrm{LST + 12h},~ 90^\circ - \phi) & \mathrm{if}~\phi >= 0 \\
(\mathrm{LST},~ 90^\circ + \phi) & \mathrm{if}~\phi < 0
\end{cases}$

East horizon $(\alpha, \delta) = (\mathrm{LST + 6h},~ 0^\circ)$

The transformation between equatorial and horizontal coordinates can be composed of two rotations, similar but not necessarily identical to those in [Wikipedia: Celestial coordinate system](https://en.wikipedia.org/wiki/Celestial_coordinate_system#Converting_coordinates).
For example, you could:

1. Start with the north horizon vector pointing at the north celestial pole, and the zenith vector pointing at (0h, 0°).
2. Rotate the ground counterclockwise as seen from the north horizon vector by LST.
3. Rotate the ground clockwise as seen from the east horizon vector by $\phi$.

> **Comments:**
> **username_0**: Great explanation thanks :)  

---

