<p>No, it's not enough. The coordinates you give are the position in the sky. To plot a location in the galaxy, you'd need also to know the distance from Earth.</p>
<p>However, nearly all naked eye visible stars are almost next to the sun at the scale of that picture.</p>
<p>Your coordinates are, I think, those of M13, the globular cluster.  It is more distant, about 22000-25000 light years distant. It has a galactic longitude of 59 degrees (you can convert to galactic longitude using the equations below).  This places M13 above the Perseus arm.  It is also quite a long way above the disc of the galaxy, but you can't show that in a 2d map.</p>
<hr />
<p>These equations are copied directly from Wikipedia <a href="https://en.wikipedia.org/wiki/Galactic_coordinate_system" rel="nofollow noreferrer">Galactic Coordinate system</a>
<span class="math-container">$$\begin{align}
\sin(b) &amp;= \sin(\delta_\text{NGP})\sin(\delta)+\cos(\delta_\text{NGP})\cos(\delta)\cos(\alpha-\alpha_\text{NGP}) \\
\cos(b) \sin(l_\text{NCP}-l) &amp;= \cos(\delta)\sin(\alpha-\alpha_\text{NGP}) \\
\cos(b) \cos(l_\text{NCP}-l) &amp;= \cos(\delta_\text{NGP}) \sin(\delta) - \sin(\delta_\text{NGP}) \cos(\delta) \cos(\alpha - \alpha_\text{NGP})
\end{align}
$$</span>
where
<span class="math-container">$$\alpha_\text{NGP}=12^h 51.4^m \qquad \delta_\text{NGP}=27.13^{\circ} \qquad l_\text{NCP}=122.93314^{\circ}$$</span>
and <span class="math-container">$\alpha, \delta$</span> are the sky coordinates that you give in your question, and <span class="math-container">$l, b$</span> are the galactic longitude and latitude.</p>
<p>Remember to convert angles to degrees or radians as appropriate to your software.</p>
<p>To project down to the plane, the project distance / actual distance = cos(latitude)</p>

CC BY-SA 4.0
2023-09-01T09:23:35.793Z
If I understood correctly, I just need to calculate the longitude and find out the the distance from Earth to the object (eg for m13 it says about 22,180 ly)? Then I can use the numbers to plot approximate the position on the map using simple trigonometry with origin at Sun?
Eugene
CC BY-SA 4.0
2023-09-01T09:40:39.110Z
Yes, that's right.
James K
CC BY-SA 4.0
2023-09-01T06:19:46.263Z
2023-09-01T09:42:41.470Z
2023-09-01T09:42:41.470Z
James K
<p>I am wondering how is it possible to plot the celestial object on the map I have attached if I only know following data:</p>
<ol>
<li>16h 41m 41,24s</li>
<li>36° 27′ 35,5″</li>
</ol>
<p><a href="https://i.sstatic.net/ZapLi.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZapLi.jpg" alt="galaxy coordinates system drawing" /></a></p>

CC BY-SA 4.0
2023-09-01T09:13:35.170Z
It may be worth noting that the image is an "artist's impression" of what the milky way may look like.  Our actual knowledge of the structures and appearance is rather limited. It's hard to know the appearance of a cloud while you are inside it.
James K
<coordinate><galaxy-center><map>
How to plot object on the map correctly
Eugene
# How to plot object on the map correctly

(Asked by: username_0 on 2023-08-31)

I am wondering how is it possible to plot the celestial object on the map I have attached if I only know following data:

1. 16h 41m 41,24s
2. 36° 27′ 35,5″

[![galaxy coordinates system drawing](https://i.sstatic.net/ZapLi.jpg)](https://i.sstatic.net/ZapLi.jpg)

**Tags:** coordinate, galaxy-center, map

> **Comments:**
> **username_1**: It may be worth noting that the image is an "artist's impression" of what the milky way may look like.  Our actual knowledge of the structures and appearance is rather limited. It's hard to know the appearance of a cloud while you are inside it.  

---

## 1 answers

### Answer by username_1 on 2023-09-01. Score: 4

No, it's not enough. The coordinates you give are the position in the sky. To plot a location in the galaxy, you'd need also to know the distance from Earth.

However, nearly all naked eye visible stars are almost next to the sun at the scale of that picture.

Your coordinates are, I think, those of M13, the globular cluster. It is more distant, about 22000-25000 light years distant. It has a galactic longitude of 59 degrees (you can convert to galactic longitude using the equations below). This places M13 above the Perseus arm. It is also quite a long way above the disc of the galaxy, but you can't show that in a 2d map.

---

These equations are copied directly from Wikipedia [Galactic Coordinate system](https://en.wikipedia.org/wiki/Galactic_coordinate_system)
$$\begin{align}
\sin(b) &= \sin(\delta\_\text{NGP})\sin(\delta)+\cos(\delta\_\text{NGP})\cos(\delta)\cos(\alpha-\alpha\_\text{NGP}) \\
\cos(b) \sin(l\_\text{NCP}-l) &= \cos(\delta)\sin(\alpha-\alpha\_\text{NGP}) \\
\cos(b) \cos(l\_\text{NCP}-l) &= \cos(\delta\_\text{NGP}) \sin(\delta) - \sin(\delta\_\text{NGP}) \cos(\delta) \cos(\alpha - \alpha\_\text{NGP})
\end{align}
$$
where
$$\alpha\_\text{NGP}=12^h 51.4^m \qquad \delta\_\text{NGP}=27.13^{\circ} \qquad l\_\text{NCP}=122.93314^{\circ}$$
and $\alpha, \delta$ are the sky coordinates that you give in your question, and $l, b$ are the galactic longitude and latitude.

Remember to convert angles to degrees or radians as appropriate to your software.

To project down to the plane, the project distance / actual distance = cos(latitude)

> **Comments:**
> **username_0**: If I understood correctly, I just need to calculate the longitude and find out the the distance from Earth to the object (eg for m13 it says about 22,180 ly)? Then I can use the numbers to plot approximate the position on the map using simple trigonometry with origin at Sun?  
> **username_1**: Yes, that's right.  

---

