<p>I'm using the <em>Explanatory Supplement to the Astronomical Almanac</em> 3rd ed. as a reference. The glossary tells us that Greenwich Apparent Sidereal Time (GAST) is the GHA of the true equinox of date, that is, "GHA(v. equinox)" in your equation. On page 80 equation 3.14 states</p>

<p>local hour angle = local apparent sidereal time - apparent right ascension (3.14)</p>

<p>If one is located on the Greenwich meridian, then that equation becomes</p>

<p>GHA = GAST - apparent right ascension (3.14a)</p>

<p>solving for apparent RA,</p>

<p>apparent RA = GAST - GHA (3.14b)</p>

<p>Comparing your equation</p>

<p>( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)</p>

<p>So your sign seems to be inverted.</p>

<p>Looking up values using UT1 and the Multiyear Computer Interactive Almanac, rounding to nearest second,</p>

<p>apparent RA 23 h 16 m 38 s
GAST         6 h 40 m 29 s</p>

<p>Substituting into eq. 3.14a, GHA = 7 h 23 m 51 s</p>

CC BY-SA 3.0
2017-02-23T12:47:02.377Z
2017-02-23T12:47:02.377Z
Gerard Ashton
<p>I have zero clue in astronomy and I need astronomical data to write a program. What I learned so far:</p>

<ul>
<li>celestial coordinates are usually measured relatively to vernal equinox aka meridian of First point of Aries (which is not located in Aries anymore)</li>
<li>the GHA of first point of Aries changes constantly</li>
</ul>

<p>I was searching for vernal equinox GHA data and it seems to me that <a href="http://sv-inua.net/the-nautical-almanac" rel="nofollow noreferrer">this personal, generated Almanac version</a> lists it as "Aries" (it was misleading me).</p>

<p>I expected that<br>
( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)<br>
as per <a href="https://en.wikipedia.org/wiki/Right_ascension" rel="nofollow noreferrer">this article</a>. However I have found <a href="http://www.moshier.net/aadoc.html" rel="nofollow noreferrer">a program</a> which calculates declination and right ascension and I have found out that the relation is sort of inverse. For example:</p>

<p><a href="https://sv-inua.net/wp-content/uploads/2016/12/almanac2020.pdf" rel="nofollow noreferrer">https://sv-inua.net/wp-content/uploads/2016/12/almanac2020.pdf</a></p>

<p>And the program output:  </p>

<blockquote>
<pre><code>               The Moon   
</code></pre>
  
  <p>JD 2458849.50,  2019 December 31 Tuesday 23h 58m 49.174s  UT 2020<br>
  <strong>January 1 Wednesday  0h 00m 00.000s  TDT</strong> nutation dRA -1.043s dDec<br>
  -6.12" Geometric lon -13.866 deg, lat -4.894 deg, rad 2.6996e-003 au   Apparent geocentric longitude 346.129 deg   latitude -4.894 deg<br>
  Distance 63.319 Earth-radii Horizontal parallax    0d 54' 17.67"<br>
  Semidiameter    0d 14' 47.60" Elongation from sun 66.20 deg,<br>
  Illuminated fraction 0.30 Phase 2.0 days before First Quarter<br>
      <strong>Apparent:  R.A. 23h 16m 38.283s  Declination -  9d 58' 41.62"</strong><br>
  Local apparent sidereal time   1h 54m 46.005s diurnal aberration dRA<br>
  0.012s dDec -0.03" diurnal parallax dRA -104.977s dDec -2485.74"      atmospheric refraction 0.033 deg  dRA 4.296s dDec 102.06" Topocentric:<br>
  Altitude 25.678 deg, Azimuth 224.447 deg Topocentric: R.A. 23h 14m<br>
  57.615s   Dec. - 10d 38' 25.32" local meridian transit 2019 December 31 Tuesday 21h 16m 02.396s  UT rises 2019 December 31 Tuesday 15h 49m<br>
  26.475s  UT sets 2020 January 1 Wednesday  2h 50m 11.075s  UT Visible hours 11.0124      </p>
</blockquote>

<p>So:
110 deg 56,9 sec - 100 deg 7,1 sec = 10 deg 49,8 sec<br>
which is inverse of R.A. which program lists.</p>

<p>Do I have a misconception?</p>

CC BY-SA 3.0
2017-02-22T12:28:34.003Z
Feel free to fix any misusage of terms.
Euri Pinhollow
CC BY-SA 3.0
2017-02-22T20:08:51.077Z
[Hour angle](https://en.wikipedia.org/wiki/Hour_angle) is measured westward; right ascension is measured eastward.
Mike G
CC BY-SA 3.0
2017-02-23T03:26:39.407Z
@MikeG make it an answer. :)
Euri Pinhollow
<coordinate>
How do GHA of vernal equinox, GHA of astronomical object and RA relate?
Euri Pinhollow
# How do GHA of vernal equinox, GHA of astronomical object and RA relate?

(Asked by: username_0 on 2017-02-22)

I have zero clue in astronomy and I need astronomical data to write a program. What I learned so far:

* celestial coordinates are usually measured relatively to vernal equinox aka meridian of First point of Aries (which is not located in Aries anymore)
* the GHA of first point of Aries changes constantly

I was searching for vernal equinox GHA data and it seems to me that [this personal, generated Almanac version](http://sv-inua.net/the-nautical-almanac) lists it as "Aries" (it was misleading me).

I expected that  
( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)  
as per [this article](https://en.wikipedia.org/wiki/Right_ascension). However I have found [a program](http://www.moshier.net/aadoc.html) which calculates declination and right ascension and I have found out that the relation is sort of inverse. For example:

<https://sv-inua.net/wp-content/uploads/2016/12/almanac2020.pdf>

And the program output:

> ```
>                The Moon   
>
> ```
>
> JD 2458849.50, 2019 December 31 Tuesday 23h 58m 49.174s UT 2020  
> **January 1 Wednesday 0h 00m 00.000s TDT** nutation dRA -1.043s dDec  
> -6.12" Geometric lon -13.866 deg, lat -4.894 deg, rad 2.6996e-003 au Apparent geocentric longitude 346.129 deg latitude -4.894 deg  
> Distance 63.319 Earth-radii Horizontal parallax 0d 54' 17.67"  
> Semidiameter 0d 14' 47.60" Elongation from sun 66.20 deg,  
> Illuminated fraction 0.30 Phase 2.0 days before First Quarter  
> **Apparent: R.A. 23h 16m 38.283s Declination - 9d 58' 41.62"**  
> Local apparent sidereal time 1h 54m 46.005s diurnal aberration dRA  
> 0.012s dDec -0.03" diurnal parallax dRA -104.977s dDec -2485.74" atmospheric refraction 0.033 deg dRA 4.296s dDec 102.06" Topocentric:  
> Altitude 25.678 deg, Azimuth 224.447 deg Topocentric: R.A. 23h 14m  
> 57.615s Dec. - 10d 38' 25.32" local meridian transit 2019 December 31 Tuesday 21h 16m 02.396s UT rises 2019 December 31 Tuesday 15h 49m  
> 26.475s UT sets 2020 January 1 Wednesday 2h 50m 11.075s UT Visible hours 11.0124

So:
110 deg 56,9 sec - 100 deg 7,1 sec = 10 deg 49,8 sec  
which is inverse of R.A. which program lists.

Do I have a misconception?

**Tags:** coordinate

> **Comments:**
> **username_0**: Feel free to fix any misusage of terms.  
> **username_1**: [Hour angle](https://en.wikipedia.org/wiki/Hour_angle) is measured westward; right ascension is measured eastward.  
> **username_0**: @username_1 make it an answer. :)  

---

## 1 answers

### Answer by username_2 on 2017-02-23. Score: 1 | Accepted answer

I'm using the *Explanatory Supplement to the Astronomical Almanac* 3rd ed. as a reference. The glossary tells us that Greenwich Apparent Sidereal Time (GAST) is the GHA of the true equinox of date, that is, "GHA(v. equinox)" in your equation. On page 80 equation 3.14 states

local hour angle = local apparent sidereal time - apparent right ascension (3.14)

If one is located on the Greenwich meridian, then that equation becomes

GHA = GAST - apparent right ascension (3.14a)

solving for apparent RA,

apparent RA = GAST - GHA (3.14b)

Comparing your equation

( GHA(object) - GHA(v. equinox) ) mod 24h = RA (object)

So your sign seems to be inverted.

Looking up values using UT1 and the Multiyear Computer Interactive Almanac, rounding to nearest second,

apparent RA 23 h 16 m 38 s
GAST 6 h 40 m 29 s

Substituting into eq. 3.14a, GHA = 7 h 23 m 51 s


---

