<p><a href="https://i.sstatic.net/gLmO3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gLmO3.png" alt="enter image description here" /></a></p>
<p>I've drawn in the position of the zenith, a line from the pole star to the zenith and the position of the autumnal equinox (RA=12)</p>
<p>I notice that the &quot;pointers&quot; are on the meridian, and the pointers have a right ascension of 11hr.  (I don't know the RA of all the stars, but it is easy to check with a star chart)</p>
<p>It is therefore 11 hours after the culmination of the vernal equinox.</p>

CC BY-SA 4.0
2022-02-04T21:05:41.360Z
2022-02-04T21:05:41.360Z
James K
<p>I have this picture:</p>
<p><a href="https://i.sstatic.net/1kFQZ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1kFQZ.jpg" alt="Fish-eye photo of the night sky" /></a></p>
<p>I need to determine what sidereal time it is (in hours), or how long has it been since the upper culmination of the point of the vernal equinox. How can I do that?</p>

CC BY-SA 4.0
2022-02-04T09:32:17.157Z
Why do you need this and what have you tried so far?
Dr Chuck
CC BY-SA 4.0
2022-02-04T10:13:18.627Z
It a task in astronomy class, I see Ursa Major constellation so I found Polar star and calculated the latitude of the place. It's approximately 41 degrees. Declination of point of vernal equinox is 0, so I can determine the height of culmination, it's 90 - $\phi$ = 49. But I'm not sure what to do next.
ALiCe P.
CC BY-SA 4.0
2022-02-04T10:19:28.100Z
For time calculations, you need to think about right ascension, not declination.
Dr Chuck
CC BY-SA 4.0
2022-02-04T11:51:11.330Z
I found that local sidereal time = right rscension of whichever stars are on the meridian.  I see that Polar star is on a meridian, so it's RA is the local time and equals to $2^h$ and $41^m$.
ALiCe P.
CC BY-SA 4.0
2022-02-04T11:58:47.340Z
That should be correct.
Tosic
CC BY-SA 4.0
2022-02-04T12:53:31.570Z
I'd probably use a different star than Polaris; It's less than 0.75° from the North Celestial Pole, making determining whether it's _exactly_ on the meridian require a _very_ accurate measurement on the image.
notovny
CC BY-SA 4.0
2022-02-04T13:08:39.577Z
In other words, Polaris is nice for letting you know where the meridian _is_, it's almost worthless for finding out local sidereal time, because it will be in almost exactly the same place on the sky from that location regardless of local sidereal time.
notovny
CC BY-SA 4.0
2022-02-04T13:17:02.847Z
@ Alice P  You are along the right lines.  Just look south, using the end 2 stars in UMa.  A star chart chaart will help unless you are familiar with the brighter constellations.
Dr Chuck
CC BY-SA 4.0
2022-02-04T14:26:40.597Z
You need to establish a North-South line, the RA of any star on that line is the sidereal time.  The further the star you pick is from Polaris, the more accurate you'll be.
Greg Miller
CC BY-SA 4.0
2022-02-04T15:30:10.137Z
Correct answer is 11 p.m. It's almost exactly the RA of $\alpha$ UMa.
ALiCe P.
CC BY-SA 4.0
2022-02-04T17:03:57.307Z
11 p.m. is the wrong answer. (1) Sidereal time is not divided into a.m. and p.m. (2) Even if 11 hours sidereal time is the correct answer, that would be "11 a.m." 23 hours sidereal time would be "11 p.m.", if a.m. and p.m. were used.
JohnHoltz
<time><positional-astronomy><equinox>
How to determine sidereal time just looking at the picture?
ALiCe P.
# How to determine sidereal time just looking at the picture?

(Asked by: username_0 on 2022-02-04)

I have this picture:

[![Fish-eye photo of the night sky](https://i.sstatic.net/1kFQZ.jpg)](https://i.sstatic.net/1kFQZ.jpg)

I need to determine what sidereal time it is (in hours), or how long has it been since the upper culmination of the point of the vernal equinox. How can I do that?

**Tags:** time, positional-astronomy, equinox
> **Comments:**
> **username_1**: Why do you need this and what have you tried so far?  
> **username_0**: It a task in astronomy class, I see Ursa Major constellation so I found Polar star and calculated the latitude of the place. It's approximately 41 degrees. Declination of point of vernal equinox is 0, so I can determine the height of culmination, it's 90 - $\phi$ = 49. But I'm not sure what to do next.  
> **username_1**: For time calculations, you need to think about right ascension, not declination.  
> **username_0**: I found that local sidereal time = right rscension of whichever stars are on the meridian.  I see that Polar star is on a meridian, so it's RA is the local time and equals to $2^h$ and $41^m$.  
> **username_2**: That should be correct.  
> **username_3**: I'd probably use a different star than Polaris; It's less than 0.75° from the North Celestial Pole, making determining whether it's _exactly_ on the meridian require a _very_ accurate measurement on the image.  
> **username_3**: In other words, Polaris is nice for letting you know where the meridian _is_, it's almost worthless for finding out local sidereal time, because it will be in almost exactly the same place on the sky from that location regardless of local sidereal time.  
> **username_1**: @ Alice P  You are along the right lines.  Just look south, using the end 2 stars in UMa.  A star chart chaart will help unless you are familiar with the brighter constellations.  
> **username_4**: You need to establish a North-South line, the RA of any star on that line is the sidereal time.  The further the star you pick is from Polaris, the more accurate you'll be.  
> **username_0**: Correct answer is 11 p.m. It's almost exactly the RA of $\alpha$ UMa.  
> **username_5**: 11 p.m. is the wrong answer. (1) Sidereal time is not divided into a.m. and p.m. (2) Even if 11 hours sidereal time is the correct answer, that would be "11 a.m." 23 hours sidereal time would be "11 p.m.", if a.m. and p.m. were used.  

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## 1 answers

### Answer by username_6 on 2022-02-04. Score: 2 | Accepted answer

[![enter image description here](https://i.sstatic.net/gLmO3.png)](https://i.sstatic.net/gLmO3.png)

I've drawn in the position of the zenith, a line from the pole star to the zenith and the position of the autumnal equinox (RA=12)

I notice that the "pointers" are on the meridian, and the pointers have a right ascension of 11hr. (I don't know the RA of all the stars, but it is easy to check with a star chart)

It is therefore 11 hours after the culmination of the vernal equinox.


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