<p><strong>Note:</strong> I'm accepting this as the answer because the main cause of the problem was that I was counting the days wrong. But there is another more important, more fundamental problem which was correctly pointed out by <code>planetmaker</code> and <code>PM2Ring</code> so make sure you read their answers/comments as well.</p>
<p>I figured what I've been doing wrong.</p>
<p>I was counting the days <strong>until</strong> March 21st, while I should have counted the days <strong>since</strong> March 21st.</p>
<p>So on Feb/10 we are 324 days after March/21. That's 324x4 = 1296min = 21h36m. So now (according to the video) we subtract this from Sirius' right ascension which gives us -14h50m. So Sirius will be at the meridian 14h50m before the sun. Today the sun is at the meridian at 12:18 so last night Sirius would have been there at 21:28. And indeed by checking in Stellarium Sirius was at the meridian at 21:30 last night. Hence tonight it will be there at 21:26.</p>
<p><strong>Edit:</strong> I'm still getting a lot of error with both the 4 minute rule and the supposedly more accurate way of calculating sidereal hours using the formulas used <a href="https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/telling_time_by_the_stars.htm" rel="nofollow noreferrer">here</a>.</p>
<p><strong>Edit2:</strong> I did some reading and some calculations and as correctly pointed out by <code>planetmaker</code> and <code>PM2Ring</code> the reason for the error I'm getting is due to the <a href="https://en.wikipedia.org/wiki/Equation_of_time#Major_components_of_the_equation" rel="nofollow noreferrer">equation of time</a>. In short other than accounting for sidereal time one should also account for the irregularity of Earth's movement around the sun which will cause about +/-16 minutes of error in time keeping.</p>
<p>I wrote a script and the error is indeed at maximum when the equation of time graph is at its max or min values. The error I get is almost exactly the same as the table below:
<a href="https://i.sstatic.net/qsiW8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/qsiW8.png" alt="Equation of Time effects in different dates" /></a></p>
<p>The formula I used is:</p>
<p><code>sidereal days = actual days * 1.0027380</code></p>
<p><code>sidereal hours = (sidereal days - actual days) * 23.9344696</code></p>
<p>Where <code>actual days</code> is number of days since March 21st.</p>
<p>The <code>sidereal hours</code> calculated here will be the Sun's right ascension in that particular day which should be subtracted from the star-of-interest's right ascension as described in the video in my original question.</p>
<p>Note that if you round up the values more you'll get more error.
Also if you use the 4 minute per day rule for calculating the sidereal hours you'll get much more error the closer we get to the end of the year (because the number of days increases and the error accumulates).</p>
<p>This is pretty basic stuff but I'm a noob and just wanted to make sure for myself that the error is indeed due to the equation of time.</p>
<p>As it seems I won't be able to get a <em>very</em> accurate result just by linearly accounting for sidereal time.</p>
<p>And thank you everyone for helping me find the answer.</p>

CC BY-SA 4.0
2021-02-10T12:53:57.577Z
The ~4 minutes per day thing is an average. It's ok for a rough estimate, but not if you want to be more precise. The traditional way to calculate the sidereal time is to interpolate from the Greenwich sidereal time at midnight (GSTM), or at noon; there are some questions about that here, eg https://astronomy.stackexchange.com/q/21002/16685
PM 2Ring
CC BY-SA 4.0
2021-02-10T15:03:20.540Z
There are two (main) reasons why the change in the Sun's RA isn't constant. Firstly, the  Earth's orbit is an ellipse, and it moves fastest when it's closest to the Sun, in early January. So the Sun's motion along the ecliptic is fastest in early January too.
PM 2Ring
CC BY-SA 4.0
2021-02-10T15:03:48.663Z
Secondly, RA is measured along the celestial equator, not the ecliptic. At the equinoxes, the Sun's RA speed equals its ecliptic longitude speed, but as it gets further from the equator each degree of ecliptic motion produces a greater change in RA, with a maximum at the solstices. The explanation here is pretty good: https://en.wikipedia.org/wiki/Equation_of_time#Major_components_of_the_equation
PM 2Ring
CC BY-SA 4.0
2021-02-10T15:17:57.897Z
I see. I actually calculated using 4 minutes for a few different dates and I got horrible results (~30min error). I then tried to calculate the sidereal hours using [this](https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/telling_time_by_the_stars.htm) method but it also gives pretty inaccurate results in latest months of the year. (I assume because of rounding errors).
It's such a letdown because I was hoping to be able to calculate these "in the wild" without using the Internet or anything. But apparently if I want precise locations I'll need to know the sidereal time.
Pouria P
CC BY-SA 4.0
2021-02-10T15:37:23.610Z
Well, you *can* get fairly good values for these things without the internet. The calculations are a bit tedious, especially if you want to do everything from scratch, without any tables. Even in the old days, navigators used almanacs with tables of GSTM, solar RA, etc. And logarithm tables.
PM 2Ring
CC BY-SA 4.0
2021-02-10T15:52:32.510Z
@PM2Ring Thanks that was a lot of useful info. One last quick question: is the GSTM like GMT? As in, can I set the time on a regular clock to the current GSTM time and expect it to always show the correct sidereal time like a normal clock? Or are there tricks to that as well?
Pouria P
CC BY-SA 4.0
2021-02-10T15:55:00.163Z
Sorry, I just realised that I may have said something misleading earlier. The deviation of sidereal time from apparent solar time has those complications due to eccentricity and obliquity of the ecliptic. However, the relationship between sidereal time and *mean* solar time is a lot more regular. So the complication in your method comes from using the time that the Sun transits your meridian.
PM 2Ring
CC BY-SA 4.0
2021-02-10T16:01:43.833Z
No, a normal clock won't show sidereal time because it ticks at the wrong rate. 1 sidereal day (of 24 sidereal hours) equals ~23 hours, 56 minutes, 4.09 seconds (of mean solar time). Take a look at https://en.wikipedia.org/wiki/Sidereal_time So a normal clock will be ~4 minutes per day too fast.
PM 2Ring
CC BY-SA 4.0
2021-02-11T11:05:20.453Z
@PM2Ring Thank you very much for your help. The error turned out to be indeed due to the equation of time. I've updated my answer with the details.
Thanks for the formula and the script too but what I wanted was a straightforward way to calculate the exact position of objects and as it turns out it needs more complicated calculations than I hoped for.
Pouria P
CC BY-SA 4.0
2021-02-11T12:20:59.300Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/119627/discussion-between-pm-2ring-and-pouria-p).
PM 2Ring
CC BY-SA 4.0
2021-02-10T11:56:05.763Z
2021-02-11T11:36:47.850Z
2021-02-11T11:36:47.850Z
Pouria P
<p>The sun's position is irrelevant to when a star will be in its meridian; the Sun's coordinates in the RA/dec grid does change over the year. A star's position is fixed in that system and it will be always at the exact same <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="noreferrer">sidereal time</a> that it culminates (as that's the definition of sidereal time).</p>
<p>The difference of the sidereal day to the <em>mean</em> solar day is ~3:56 minutes (366 earth revolutions in a sidereal year, 365 in a solar year), thus a star culminates (and rises and sets) that much earlier each day.</p>
<p>If you compare it to the true solar time (which determines when the Sun is in the meridian) as you try, you also have to account for the difference of the mean solar time and the true solar time due to the eccentricity of Earth's orbit; this phenomenon visible as the <a href="https://en.wikipedia.org/wiki/Analemma" rel="noreferrer">analemma</a> (which basically visualizes the <a href="https://en.wikipedia.org/wiki/Equation_of_time" rel="noreferrer">equation of time</a>) accounts for a difference of the true noon which varies by about +-20 minutes throughout the year - and can explain the differences you see.
This means that the Sun culminates at different times throughout the year. On March 21st it may be 12:00h. But that is not exactly true 2 month earlier or later, it might be 11:40h or 12:20h.</p>
<p>I think the explanation in the video as linked in your comment as &quot;hours behind the sun&quot; can be a bit mis-leading, and is only exactly true for that one exact date (and defines how the two coordinate systems align with respect to eachother) - but he ignores the equation of time and assumes the mean solar time to be used while his words &quot;behind the sun&quot; imply true solar time. For the task at hand, where to visually find a certain constallation, that is good and accurate enough, though - so not wrong, just mathematically you will have to consider effects he graciously left out.</p>

CC BY-SA 4.0
2021-02-10T10:55:05.753Z
How is one supposed to use right ascension then? I've been googling since last night and every source just describes the celestial sphere and assumes we are on Marth 21st. How can one use the right ascension on a day other than March 21st?
Pouria P
CC BY-SA 4.0
2021-02-10T11:28:35.723Z
just linearily subtract or add the daily time difference between sidereal and solar day. I made a small edit to the answer which reflects the accuracy level the video targets / assumes
planetmaker
CC BY-SA 4.0
2021-02-10T12:02:52.333Z
Thank you for your answer but I figured what I was doing wrong (I explained it in an answer). The 22 minute was just pure chance otherwise I should have gotten a much bigger error.
I still need to read more about true solar time and different noon times that you mentioned tho. I still haven't fully understood it but it seems to matter if I'm using the sun and noon times for my calculations.
Pouria P
CC BY-SA 4.0
2021-02-11T11:01:51.173Z
Thank you very much, this was very helpful. I read up on the subjects you mentioned and did some calculations and it turns out that the error is indeed due to the equation of time. I reflected my findings in my own answer. I'm choosing my own answer because the main thing I was doing wrong was how I calculated the days but I've also edited in the details about the equation of time as well.
Pouria P
CC BY-SA 4.0
2021-02-11T11:09:42.167Z
You're welcome. Feel free to accept the answer which helped most, that may as well be your own.
planetmaker
CC BY-SA 4.0
2021-02-10T10:33:25.770Z
2021-02-10T11:27:49.627Z
2021-02-10T11:27:49.627Z
planetmaker
<p>The Equation of time has nothing to do with stars. It's just a measure of the irregularity of the Sun's apparent position, which varies over the course of a year. As planetmaker points out, the Sun's position is irrelevant to stars' positions.</p>
<p>You don't need to know sidereal time, as such. You just need a reference time and position of the vernal equinox, which is the zero reference for right ascension. It moves 360 degrees westward every sidereal day (23.9344696 hours). Sirius is 6h 46m behind. The vernal equinox, in a few minutes, at 1900 UTC, will have a Greenwich Hour Angle of 67 deg 3.1 min. That's the west longitude of its meridian at that time.</p>
<p>I assume you know how to crunch the numbers</p>

CC BY-SA 4.0
2021-02-12T11:30:54.390Z
You say the Sun's position is irrelevant to stars' positions but in [this wikipedia article](https://en.wikipedia.org/wiki/Right_ascension) it says clearly that right ascension is the angle from the sun on March equinox. The sun is in the very definition of right ascension and hence position of stars.
Pouria P
CC BY-SA 4.0
2021-02-12T15:06:28.560Z
A lot of online documents, and my answer, refer to the 0 meridian as the vernal equinox. Strictly speaking "equinox" is a time, not a location. It's the time when the sun apparently crosses the celestial equator. The Wikipedia article was using it in that sense. It's at the same place on the celestial sphere every year, but it's not at the same time. If this was a navigation SE I would have called it the First Point of Aries, but I thought that would be confusing. Wikipedia has a good article *First Point of Aries*
stretch
CC BY-SA 4.0
2021-02-11T18:37:29.707Z
2021-02-11T18:37:29.707Z
stretch
<p>I've been watching <a href="https://www.youtube.com/watch?v=kejolGai7X8" rel="noreferrer">an astronomy course</a> on YouTube and I'm struggling to calculate stars' positions based on their right ascension.</p>
<p><strong>What I'm trying to achieve:</strong> Calculate when a star (Sirius in this example) will be exactly on my local meridian.</p>
<p><strong>What I know:</strong> The star's right ascension and the &quot;noon&quot; time (=when the sun will be exactly on my local meridian).</p>
<p><strong>Let's assume we are on March 21st:</strong> According to <a href="https://stellarium-web.org/" rel="noreferrer">https://stellarium-web.org/</a> The sun will be on my meridian at 12:10. Sirius's right ascension is 6h46m, which means Sirius should be at the meridian 6h46m later which is 18:56. According to Stellarium Sirius will indeed be at the meridian at 18:52 (I suppose that 4 minute difference is because the sun is already 4 minute behind due to Earth rotation).</p>
<p>But when I try to do the same calculation for today (Feb 10th) I get a 20 minute error.
The &quot;noon&quot; on Feb 10th where I live is at 12:18. We add Sirius' right ascension to get 19:04. But this time we also have to take into account the fact that we are 41 days away from March 21st. Each day the sun falls behind by ~4 minutes, so that's 2h44m. If we add that to 19:04 we get 21:48. So that's the time Sirius should be at my local meridian on this particular day. But Stellarium shows that Sirius will actually be at the meridian at 21:26. So there's 22 minutes error in my calculations for some reason.</p>
<p>I think the root of the error is the 2h44m that I calculated for the sun's position. According to Stellarium the sun's right ascension on that day is 21h38m. Meaning it is 2h22m behind not 2h44m. And that's why I get that 22 minutes error. But I don't understand how it can be 2h22m because Feb/10 is 41 days away from March/21. That means 41x4 = 164 minutes = 2h44m.</p>
<p>What am I missing here?</p>

CC BY-SA 4.0
2021-02-10T10:59:26.700Z
Welcome to astronomy SE! Would you mind sharing which exact course did you watch?
B--rian
CC BY-SA 4.0
2021-02-10T11:10:01.773Z
@B--rian This is the part of the course about right ascension: https://www.youtube.com/watch?v=kejolGai7X8
As far as I can tell I am calculating things the way he teaches but I'm getting 20 minutes error. And now the user _planetmaker_ says the sun's position is irrelevant which made me even more confused.
Pouria P
<right-ascension>
Calculating the time a star is at my local meridian based on its right ascension
Pouria P
# Calculating the time a star is at my local meridian based on its right ascension

(Asked by: username_0 on 2021-02-10)

I've been watching [an astronomy course](https://www.youtube.com/watch?v=kejolGai7X8) on YouTube and I'm struggling to calculate stars' positions based on their right ascension.

**What I'm trying to achieve:** Calculate when a star (Sirius in this example) will be exactly on my local meridian.

**What I know:** The star's right ascension and the "noon" time (=when the sun will be exactly on my local meridian).

**Let's assume we are on March 21st:** According to <https://stellarium-web.org/> The sun will be on my meridian at 12:10. Sirius's right ascension is 6h46m, which means Sirius should be at the meridian 6h46m later which is 18:56. According to Stellarium Sirius will indeed be at the meridian at 18:52 (I suppose that 4 minute difference is because the sun is already 4 minute behind due to Earth rotation).

But when I try to do the same calculation for today (Feb 10th) I get a 20 minute error.
The "noon" on Feb 10th where I live is at 12:18. We add Sirius' right ascension to get 19:04. But this time we also have to take into account the fact that we are 41 days away from March 21st. Each day the sun falls behind by ~4 minutes, so that's 2h44m. If we add that to 19:04 we get 21:48. So that's the time Sirius should be at my local meridian on this particular day. But Stellarium shows that Sirius will actually be at the meridian at 21:26. So there's 22 minutes error in my calculations for some reason.

I think the root of the error is the 2h44m that I calculated for the sun's position. According to Stellarium the sun's right ascension on that day is 21h38m. Meaning it is 2h22m behind not 2h44m. And that's why I get that 22 minutes error. But I don't understand how it can be 2h22m because Feb/10 is 41 days away from March/21. That means 41x4 = 164 minutes = 2h44m.

What am I missing here?

**Tags:** right-ascension

> **Comments:**
> **username_1**: Welcome to astronomy SE! Would you mind sharing which exact course did you watch?  
> **username_0**: @username_1 This is the part of the course about right ascension: https://www.youtube.com/watch?v=kejolGai7X8 As far as I can tell I am calculating things the way he teaches but I'm getting 20 minutes error. And now the user _planetmaker_ says the sun's position is irrelevant which made me even more confused.  

---

## 3 answers

### Answer by username_0 on 2021-02-10. Score: 3 | Accepted answer

**Note:** I'm accepting this as the answer because the main cause of the problem was that I was counting the days wrong. But there is another more important, more fundamental problem which was correctly pointed out by `username_3` and `username_2` so make sure you read their answers/comments as well.

I figured what I've been doing wrong.

I was counting the days **until** March 21st, while I should have counted the days **since** March 21st.

So on Feb/10 we are 324 days after March/21. That's 324x4 = 1296min = 21h36m. So now (according to the video) we subtract this from Sirius' right ascension which gives us -14h50m. So Sirius will be at the meridian 14h50m before the sun. Today the sun is at the meridian at 12:18 so last night Sirius would have been there at 21:28. And indeed by checking in Stellarium Sirius was at the meridian at 21:30 last night. Hence tonight it will be there at 21:26.

**Edit:** I'm still getting a lot of error with both the 4 minute rule and the supposedly more accurate way of calculating sidereal hours using the formulas used [here](https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/telling_time_by_the_stars.htm).

**Edit2:** I did some reading and some calculations and as correctly pointed out by `username_3` and `username_2` the reason for the error I'm getting is due to the [equation of time](https://en.wikipedia.org/wiki/Equation_of_time#Major_components_of_the_equation). In short other than accounting for sidereal time one should also account for the irregularity of Earth's movement around the sun which will cause about +/-16 minutes of error in time keeping.

I wrote a script and the error is indeed at maximum when the equation of time graph is at its max or min values. The error I get is almost exactly the same as the table below:
[![Equation of Time effects in different dates](https://i.sstatic.net/qsiW8.png)](https://i.sstatic.net/qsiW8.png)

The formula I used is:

`sidereal days = actual days * 1.0027380`

`sidereal hours = (sidereal days - actual days) * 23.9344696`

Where `actual days` is number of days since March 21st.

The `sidereal hours` calculated here will be the Sun's right ascension in that particular day which should be subtracted from the star-of-interest's right ascension as described in the video in my original question.

Note that if you round up the values more you'll get more error.
Also if you use the 4 minute per day rule for calculating the sidereal hours you'll get much more error the closer we get to the end of the year (because the number of days increases and the error accumulates).

This is pretty basic stuff but I'm a noob and just wanted to make sure for myself that the error is indeed due to the equation of time.

As it seems I won't be able to get a *very* accurate result just by linearly accounting for sidereal time.

And thank you everyone for helping me find the answer.

> **Comments:**
> **username_2**: The ~4 minutes per day thing is an average. It's ok for a rough estimate, but not if you want to be more precise. The traditional way to calculate the sidereal time is to interpolate from the Greenwich sidereal time at midnight (GSTM), or at noon; there are some questions about that here, eg https://astronomy.stackexchange.com/q/21002/16685  
> **username_2**: There are two (main) reasons why the change in the Sun's RA isn't constant. Firstly, the  Earth's orbit is an ellipse, and it moves fastest when it's closest to the Sun, in early January. So the Sun's motion along the ecliptic is fastest in early January too.  
> **username_2**: Secondly, RA is measured along the celestial equator, not the ecliptic. At the equinoxes, the Sun's RA speed equals its ecliptic longitude speed, but as it gets further from the equator each degree of ecliptic motion produces a greater change in RA, with a maximum at the solstices. The explanation here is pretty good: https://en.wikipedia.org/wiki/Equation_of_time#Major_components_of_the_equation  
> **username_0**: I see. I actually calculated using 4 minutes for a few different dates and I got horrible results (~30min error). I then tried to calculate the sidereal hours using [this](https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/telling_time_by_the_stars.htm) method but it also gives pretty inaccurate results in latest months of the year. (I assume because of rounding errors). It's such a letdown because I was hoping to be able to calculate these "in the wild" without using the Internet or anything. But apparently if I want precise locations I'll need to know the sidereal time.  
> **username_2**: Well, you *can* get fairly good values for these things without the internet. The calculations are a bit tedious, especially if you want to do everything from scratch, without any tables. Even in the old days, navigators used almanacs with tables of GSTM, solar RA, etc. And logarithm tables.  
> **username_0**: @username_2 Thanks that was a lot of useful info. One last quick question: is the GSTM like GMT? As in, can I set the time on a regular clock to the current GSTM time and expect it to always show the correct sidereal time like a normal clock? Or are there tricks to that as well?  
> **username_2**: Sorry, I just realised that I may have said something misleading earlier. The deviation of sidereal time from apparent solar time has those complications due to eccentricity and obliquity of the ecliptic. However, the relationship between sidereal time and *mean* solar time is a lot more regular. So the complication in your method comes from using the time that the Sun transits your meridian.  
> **username_2**: No, a normal clock won't show sidereal time because it ticks at the wrong rate. 1 sidereal day (of 24 sidereal hours) equals ~23 hours, 56 minutes, 4.09 seconds (of mean solar time). Take a look at https://en.wikipedia.org/wiki/Sidereal_time So a normal clock will be ~4 minutes per day too fast.  
> **username_0**: @username_2 Thank you very much for your help. The error turned out to be indeed due to the equation of time. I've updated my answer with the details. Thanks for the formula and the script too but what I wanted was a straightforward way to calculate the exact position of objects and as it turns out it needs more complicated calculations than I hoped for.  
> **username_2**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/119627/discussion-between-username_2-and-username_0-p).  

---

### Answer by username_3 on 2021-02-10. Score: 5

The sun's position is irrelevant to when a star will be in its meridian; the Sun's coordinates in the RA/dec grid does change over the year. A star's position is fixed in that system and it will be always at the exact same [sidereal time](https://en.wikipedia.org/wiki/Sidereal_time) that it culminates (as that's the definition of sidereal time).

The difference of the sidereal day to the *mean* solar day is ~3:56 minutes (366 earth revolutions in a sidereal year, 365 in a solar year), thus a star culminates (and rises and sets) that much earlier each day.

If you compare it to the true solar time (which determines when the Sun is in the meridian) as you try, you also have to account for the difference of the mean solar time and the true solar time due to the eccentricity of Earth's orbit; this phenomenon visible as the [analemma](https://en.wikipedia.org/wiki/Analemma) (which basically visualizes the [equation of time](https://en.wikipedia.org/wiki/Equation_of_time)) accounts for a difference of the true noon which varies by about +-20 minutes throughout the year - and can explain the differences you see.
This means that the Sun culminates at different times throughout the year. On March 21st it may be 12:00h. But that is not exactly true 2 month earlier or later, it might be 11:40h or 12:20h.

I think the explanation in the video as linked in your comment as "hours behind the sun" can be a bit mis-leading, and is only exactly true for that one exact date (and defines how the two coordinate systems align with respect to eachother) - but he ignores the equation of time and assumes the mean solar time to be used while his words "behind the sun" imply true solar time. For the task at hand, where to visually find a certain constallation, that is good and accurate enough, though - so not wrong, just mathematically you will have to consider effects he graciously left out.

> **Comments:**
> **username_0**: How is one supposed to use right ascension then? I've been googling since last night and every source just describes the celestial sphere and assumes we are on Marth 21st. How can one use the right ascension on a day other than March 21st?  
> **username_3**: just linearily subtract or add the daily time difference between sidereal and solar day. I made a small edit to the answer which reflects the accuracy level the video targets / assumes  
> **username_0**: Thank you for your answer but I figured what I was doing wrong (I explained it in an answer). The 22 minute was just pure chance otherwise I should have gotten a much bigger error. I still need to read more about true solar time and different noon times that you mentioned tho. I still haven't fully understood it but it seems to matter if I'm using the sun and noon times for my calculations.  
> **username_0**: Thank you very much, this was very helpful. I read up on the subjects you mentioned and did some calculations and it turns out that the error is indeed due to the equation of time. I reflected my findings in my own answer. I'm choosing my own answer because the main thing I was doing wrong was how I calculated the days but I've also edited in the details about the equation of time as well.  
> **username_3**: You're welcome. Feel free to accept the answer which helped most, that may as well be your own.  

---

### Answer by username_4 on 2021-02-11. Score: 1

The Equation of time has nothing to do with stars. It's just a measure of the irregularity of the Sun's apparent position, which varies over the course of a year. As username_3 points out, the Sun's position is irrelevant to stars' positions.

You don't need to know sidereal time, as such. You just need a reference time and position of the vernal equinox, which is the zero reference for right ascension. It moves 360 degrees westward every sidereal day (23.9344696 hours). Sirius is 6h 46m behind. The vernal equinox, in a few minutes, at 1900 UTC, will have a Greenwich Hour Angle of 67 deg 3.1 min. That's the west longitude of its meridian at that time.

I assume you know how to crunch the numbers

> **Comments:**
> **username_0**: You say the Sun's position is irrelevant to stars' positions but in [this wikipedia article](https://en.wikipedia.org/wiki/Right_ascension) it says clearly that right ascension is the angle from the sun on March equinox. The sun is in the very definition of right ascension and hence position of stars.  
> **username_4**: A lot of online documents, and my answer, refer to the 0 meridian as the vernal equinox. Strictly speaking "equinox" is a time, not a location. It's the time when the sun apparently crosses the celestial equator. The Wikipedia article was using it in that sense. It's at the same place on the celestial sphere every year, but it's not at the same time. If this was a navigation SE I would have called it the First Point of Aries, but I thought that would be confusing. Wikipedia has a good article *First Point of Aries*  

---

