<p>I think that it depends on the Aries point you are considering to measure RA and sidereal time. If you use the &quot;mean&quot; Aries point (relative to the mean equator), in the equation you will have the LST. If you use the true Aries point, you will have the LAST.</p>

CC BY-SA 4.0
2024-12-24T10:27:25.657Z
Sorry I had to read a bit more since I'm pretty new to astronomy. I'm working on a day/night fullscreen clock which is on realtime. The idea is to track the position of the sun and moon according to the user location. The book only mentions "measured from the vernal equinox of the date" on most calculations. Is this what you refer to?. My intention is to obtain the most precise calculation I can get without sacrificing resources, so, the data should be precise, but not to JPL Horizons level though.
RicardoTM
CC BY-SA 4.0
2024-12-23T19:30:05.320Z
2024-12-23T19:30:05.320Z
Al01
<p>So, the book from where I'm basing my calculations (Astronomical Algorithms) takes Local Hour Angle as:</p>
<p>H = LST - RA</p>
<p>However my question is, shouldn't it use LAST instead? I mean, for context, I'm calculating the Sun's Azimuth and Altitude. For that I have calculated Right Ascension and Declination which are affected by nutation.</p>
<p>Shouldn't the Hour Angle be affected by it as well?</p>
<p>Or, since RA is already taking them into account there's no necessity of using LAST?</p>

CC BY-SA 4.0
2024-12-24T00:25:49.980Z
There is "apparent" and "mean" sidereal time.  Which to use is usually a matter of how much computation you want to do.  Just specifying "sidereal time" means you can use which ever you prefer, the algorithm works with both.
Greg Miller
CC BY-SA 4.0
2024-12-24T10:33:45.737Z
Hey @GregMiller , thanks as always for your answers. You've been really helpful to me. I was wondering if there's a way to contact you privately? I'd would be honored if you could test my project and let me know your opinions on it. It's still a work in progress so I can't make it public yet.
RicardoTM
CC BY-SA 4.0
2024-12-24T16:02:36.233Z
My email is on celestialprogramming.com. I wouldn't mind having a look at your project.  But for astronomy questions, it's better to keep them public. With all the junk email I get, I'm probably more likely to see it here.
Greg Miller
CC BY-SA 4.0
2024-12-24T16:34:04.437Z
@GregMiller Yes, for questions I also prefer doing it publicly so others can benefit as well. I tried sending you an email, however it gets rejected and I get a mail back with subject "Mail Delivery System" saying "This is the mail system at host dev. I'm sorry to have to inform you that your message could not
be delivered to one or more recipients. It's attached below. [..]"
RicardoTM
CC BY-SA 4.0
2024-12-25T05:02:38.060Z
Hmm, e-mail seems to be working fine, I'll look more into it.  You can PM me as gmiller123456 on cloudynights.com for now.
Greg Miller
CC BY-SA 4.0
2024-12-25T16:10:36.190Z
I will, thank you Greg. Merry Christmas.
RicardoTM
<algorithm>
Should Hour Angle be calculated from Local Apparent Sidereal Time?
RicardoTM
# Should Hour Angle be calculated from Local Apparent Sidereal Time?

(Asked by: username_0 on 2024-12-23)

So, the book from where I'm basing my calculations (Astronomical Algorithms) takes Local Hour Angle as:

H = LST - RA

However my question is, shouldn't it use LAST instead? I mean, for context, I'm calculating the Sun's Azimuth and Altitude. For that I have calculated Right Ascension and Declination which are affected by nutation.

Shouldn't the Hour Angle be affected by it as well?

Or, since RA is already taking them into account there's no necessity of using LAST?

**Tags:** algorithm
> **Comments:**
> **username_1**: There is "apparent" and "mean" sidereal time.  Which to use is usually a matter of how much computation you want to do.  Just specifying "sidereal time" means you can use which ever you prefer, the algorithm works with both.  
> **username_0**: Hey @GregMiller , thanks as always for your answers. You've been really helpful to me. I was wondering if there's a way to contact you privately? I'd would be honored if you could test my project and let me know your opinions on it. It's still a work in progress so I can't make it public yet.  
> **username_1**: My email is on celestialprogramming.com. I wouldn't mind having a look at your project.  But for astronomy questions, it's better to keep them public. With all the junk email I get, I'm probably more likely to see it here.  
> **username_0**: @GregMiller Yes, for questions I also prefer doing it publicly so others can benefit as well. I tried sending you an email, however it gets rejected and I get a mail back with subject "Mail Delivery System" saying "This is the mail system at host dev. I'm sorry to have to inform you that your message could not be delivered to one or more recipients. It's attached below. [..]"  
> **username_1**: Hmm, e-mail seems to be working fine, I'll look more into it.  You can PM me as gmiller123456 on cloudynights.com for now.  
> **username_0**: I will, thank you Greg. Merry Christmas.  

---

## 1 answers

### Answer by username_2 on 2024-12-23. Score: 0

I think that it depends on the Aries point you are considering to measure RA and sidereal time. If you use the "mean" Aries point (relative to the mean equator), in the equation you will have the LST. If you use the true Aries point, you will have the LAST.

> **Comments:**
> **username_0**: Sorry I had to read a bit more since I'm pretty new to astronomy. I'm working on a day/night fullscreen clock which is on realtime. The idea is to track the position of the sun and moon according to the user location. The book only mentions "measured from the vernal equinox of the date" on most calculations. Is this what you refer to?. My intention is to obtain the most precise calculation I can get without sacrificing resources, so, the data should be precise, but not to JPL Horizons level though.  

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