<p>In northern hemisphere, when looking at <a href="https://en.wikipedia.org/wiki/Cassiopeia_(constellation)" rel="nofollow noreferrer">Cassiopeia</a> (sometimes referred to as  <a href="https://en.wikipedia.org/wiki/Cassiopeia_(constellation)#%27W%27_asterism" rel="nofollow noreferrer">the &quot;W&quot;</a>), it’s rightmost brightest star &quot;Caph&quot; (<a href="https://en.wikipedia.org/wiki/Beta_Cassiopeiae" rel="nofollow noreferrer">Beta Cassiopeiae</a>) is almost at 0hr RA.</p>
<p><a href="https://i.sstatic.net/oQzi9.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oQzi9.png" alt="File:Cassiopeia_IAU" /></a></p>
<p><a href="https://en.wikipedia.org/wiki/File:Cassiopeia_IAU.svg" rel="nofollow noreferrer">Source</a></p>

CC BY-SA 4.0
2020-08-23T11:22:38.447Z
`+1` I added some goodies, maybe too many; feel free to edit further, and *Welcome to Astronomy!*
uhoh
CC BY-SA 4.0
2020-08-23T08:47:24.277Z
2020-08-23T11:21:31.583Z
2020-08-23T11:21:31.583Z
Shital Shah
<p>A brief overview of RA is the angular distance measured eastward from the vernal equinox along the celestial equator. Another way to say that refers to where the celestial equator intersects the horizon at a right angle.</p>

<p>If you are able to follow the celestial equator to the point known as the vernal equinox, 0Deg RA is at the intersection point, with RA increasing directly east. The vernal equinox can be found at the first point of Aries, on a celestial sphere it is the point where the Sun crosses the celestial equator from south to north during the March equinox. The point during the March equinox is located in Pisces.</p>

<p>There are methods to determine the vernal equinox at other times, which can be dependent on location and the time. A great way that I personally use, is the same method as 'star hopping'. Using reference points, you can generally determine a line of sight between two or more points, for example if you are aware of a point and it's approximate RA, you can follow the path through another known RA point and evaluate the distance between the points. Following the line of sight you can them determine how far along the line the point of 0 would be, using this method takes much practice and you could even use visual guides such as a thumb as a measuring tool.</p>

<p>Using my thumb, I can say held at .5 meters is equivalent to 2.5 degrees, using stars as reference points I could approximate the line of sight and have a reliant measuring tool to judge distance.</p>

<p>The method you may require, could be entirely dependent on the reason you need to determine the point. If using electronic devices you can follow the celestial equator west along the celestial equator to find the point etc. I can explain other methods in more detail if required, others may have developed their own methods also.</p>

CC BY-SA 3.0
2015-01-28T00:29:55.437Z
Thanks James, I can work with your suggestions. Just starting out and have lots to learn. -lige
gaintwist
CC BY-SA 3.0
2015-01-24T02:24:15.690Z
2015-01-24T02:24:15.690Z
Ashley James
<p>I'd suggest a good star atlas and look for stars that are close to that meridian. Depending on how accurate you want to be you can use a telescope to get closer (ie to observe stars that are not visible to the naked eye).</p>

<p>But what is it that you are trying to do? Set setting circles on a telescope? You don't need to look for RA=0h0'00" then - simply look for an object with a known RA and Dec.</p>

CC BY-SA 3.0
2015-02-16T19:43:49.387Z
2015-02-16T19:43:49.387Z
adrianmcmenamin
<p>In the Southern Hemisphere, the Southern Cross (<a href="https://en.wikipedia.org/wiki/Crux" rel="nofollow noreferrer">Crux</a>) lies at approximately 12.5h RA. You can <a href="https://www.abc.net.au/science/articles/2009/07/27/3169018.htm" rel="nofollow noreferrer">use the pointer stars (yellow) to locate the south celestial pole</a>, and then continue the line from Crux (blue), perhaps bending it clockwise a little, as a rough guide to the location of 0h RA (green).</p>
<p><a href="https://i.sstatic.net/6dZYX.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6dZYX.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-03-14T13:25:51.303Z
2021-03-14T13:25:51.303Z
sjy
<p>It depends on the your current date, time of day and your location since, from your point of view, Right Ascension (in hours) is always progressing in the same direction as the stars are moving. Remember that RA "0" is the same as RA "24" and is a line that runs from the north celestial pole to the south celestial pole. If we are trying to find the actual "line" in the sky that corresponds to RA "0" we have a couple of options. One is to obtain a star chart, find any star that is on that line, locate it in the sky and you are there. If you can't find a star chart, figure out the "Local Apparent Sidereal Time" (LAST) which will be the same as whatever RA is directly overhead (at your location) at that moment and you can estimate from that or wait until your "LAST" time is "0", at which time it will be directly overhead. Star charts as well as your current "LAST" time can be found directly on many astronomy smartphone apps or can be calculated (another question).</p>

CC BY-SA 3.0
2015-02-17T18:14:48.293Z
2015-02-17T18:14:48.293Z
TerryDH
<p>Planisphere. Batteries not required.</p>
<ul>
<li><a href="https://www.amazon.com/s?k=planesphere&amp;ref=nb_sb_noss" rel="nofollow noreferrer">https://www.amazon.com/s?k=planesphere&amp;ref=nb_sb_noss</a></li>
<li>make your own: <a href="https://in-the-sky.org/planisphere/index.php" rel="nofollow noreferrer">https://in-the-sky.org/planisphere/index.php</a></li>
<li><a href="https://astronomy.stackexchange.com/q/24141/7982">Star map - finding stars</a></li>
</ul>

CC BY-SA 4.0
2020-08-23T22:57:44.633Z
`+1` I hope you don't mind my additions
uhoh
CC BY-SA 4.0
2020-08-24T03:56:20.747Z
It would be helpful if the planisphere's had some lines of Right Ascension (RA) on them.
JohnHoltz
CC BY-SA 4.0
2020-08-24T19:24:22.653Z
https://d2yol3sfogi69i.cloudfront.net/wp-content/uploads/2016/09/planisphere-set-20-nov.jpg         Are all Planespheres created equal???? The hour angles might not be marked on the map but what is there on the outside of the ring is enough for the general purpose of the Planesphere.
L d Bonnie
CC BY-SA 4.0
2020-08-23T19:05:17.823Z
2020-08-23T22:54:24.507Z
2020-08-23T22:54:24.507Z
L d Bonnie
<p>I am standing in an open area and can see the sky for a full 360 degrees. I have the coordinates of an object (<a href="https://en.wikipedia.org/wiki/Right_ascension" rel="nofollow noreferrer">RA</a> and <a href="https://en.wikipedia.org/wiki/Declination" rel="nofollow noreferrer">DEC</a>). Where do I look to find RA 0?</p>

CC BY-SA 3.0
2015-02-17T10:33:41.997Z
@barrycarter OP has not indicated which hemisphere they are in. Pegasus is not currently visible from the South.
andy256
<amateur-observing>
How to find RA 0
gaintwist
# How to find RA 0

(Asked by: username_0 on 2015-01-24)

I am standing in an open area and can see the sky for a full 360 degrees. I have the coordinates of an object ([RA](https://en.wikipedia.org/wiki/Right_ascension) and [DEC](https://en.wikipedia.org/wiki/Declination)). Where do I look to find RA 0?

**Tags:** amateur-observing
> **Comments:**
> **username_1**: @barrycarter OP has not indicated which hemisphere they are in. Pegasus is not currently visible from the South.  

---

## 6 answers

### Answer by username_2 on 2020-08-23. Score: 4

In northern hemisphere, when looking at [Cassiopeia](https://en.wikipedia.org/wiki/Cassiopeia_(constellation)) (sometimes referred to as [the "W"](https://en.wikipedia.org/wiki/Cassiopeia_(constellation)#%27W%27_asterism)), it’s rightmost brightest star "Caph" ([Beta Cassiopeiae](https://en.wikipedia.org/wiki/Beta_Cassiopeiae)) is almost at 0hr RA.

[![File:Cassiopeia_IAU](https://i.sstatic.net/oQzi9.png)](https://i.sstatic.net/oQzi9.png)

[Source](https://en.wikipedia.org/wiki/File:Cassiopeia_IAU.svg)

> **Comments:**
> **username_3**: `+1` I added some goodies, maybe too many; feel free to edit further, and *Welcome to Astronomy!*  

---

### Answer by username_4 on 2015-01-24. Score: 3

A brief overview of RA is the angular distance measured eastward from the vernal equinox along the celestial equator. Another way to say that refers to where the celestial equator intersects the horizon at a right angle.

If you are able to follow the celestial equator to the point known as the vernal equinox, 0Deg RA is at the intersection point, with RA increasing directly east. The vernal equinox can be found at the first point of Aries, on a celestial sphere it is the point where the Sun crosses the celestial equator from south to north during the March equinox. The point during the March equinox is located in Pisces.

There are methods to determine the vernal equinox at other times, which can be dependent on location and the time. A great way that I personally use, is the same method as 'star hopping'. Using reference points, you can generally determine a line of sight between two or more points, for example if you are aware of a point and it's approximate RA, you can follow the path through another known RA point and evaluate the distance between the points. Following the line of sight you can them determine how far along the line the point of 0 would be, using this method takes much practice and you could even use visual guides such as a thumb as a measuring tool.

Using my thumb, I can say held at .5 meters is equivalent to 2.5 degrees, using stars as reference points I could approximate the line of sight and have a reliant measuring tool to judge distance.

The method you may require, could be entirely dependent on the reason you need to determine the point. If using electronic devices you can follow the celestial equator west along the celestial equator to find the point etc. I can explain other methods in more detail if required, others may have developed their own methods also.

> **Comments:**
> **username_0**: Thanks James, I can work with your suggestions. Just starting out and have lots to learn. -lige  

---

### Answer by username_5 on 2015-02-16. Score: 2

I'd suggest a good star atlas and look for stars that are close to that meridian. Depending on how accurate you want to be you can use a telescope to get closer (ie to observe stars that are not visible to the naked eye).

But what is it that you are trying to do? Set setting circles on a telescope? You don't need to look for RA=0h0'00" then - simply look for an object with a known RA and Dec.


---

### Answer by username_6 on 2021-03-14. Score: 1

In the Southern Hemisphere, the Southern Cross ([Crux](https://en.wikipedia.org/wiki/Crux)) lies at approximately 12.5h RA. You can [use the pointer stars (yellow) to locate the south celestial pole](https://www.abc.net.au/science/articles/2009/07/27/3169018.htm), and then continue the line from Crux (blue), perhaps bending it clockwise a little, as a rough guide to the location of 0h RA (green).

[![enter image description here](https://i.sstatic.net/6dZYX.jpg)](https://i.sstatic.net/6dZYX.jpg)


---

### Answer by username_7 on 2015-02-17. Score: 1

It depends on the your current date, time of day and your location since, from your point of view, Right Ascension (in hours) is always progressing in the same direction as the stars are moving. Remember that RA "0" is the same as RA "24" and is a line that runs from the north celestial pole to the south celestial pole. If we are trying to find the actual "line" in the sky that corresponds to RA "0" we have a couple of options. One is to obtain a star chart, find any star that is on that line, locate it in the sky and you are there. If you can't find a star chart, figure out the "Local Apparent Sidereal Time" (LAST) which will be the same as whatever RA is directly overhead (at your location) at that moment and you can estimate from that or wait until your "LAST" time is "0", at which time it will be directly overhead. Star charts as well as your current "LAST" time can be found directly on many astronomy smartphone apps or can be calculated (another question).


---

### Answer by username_8 on 2020-08-23. Score: 1

Planisphere. Batteries not required.

* <https://www.amazon.com/s?k=planesphere&ref=nb_sb_noss>
* make your own: <https://in-the-sky.org/planisphere/index.php>
* [Star map - finding stars](https://astronomy.stackexchange.com/q/24141/7982)

> **Comments:**
> **username_3**: `+1` I hope you don't mind my additions  
> **username_9**: It would be helpful if the planisphere's had some lines of Right Ascension (RA) on them.  
> **username_8**: https://d2yol3sfogi69i.cloudfront.net/wp-content/uploads/2016/09/planisphere-set-20-nov.jpg         Are all Planespheres created equal???? The hour angles might not be marked on the map but what is there on the outside of the ring is enough for the general purpose of the Planesphere.  

---

