<p>In other "words", the connection between the time of transit $t_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta_0$ (if you know your local siderial time $\theta$, you don't need $\theta_0$ or $l$):
$$t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - \theta.$$</p>

CC BY-SA 3.0
2014-01-06T16:44:06.370Z
2014-01-06T16:44:06.370Z
AstroFloyd
<p>Let's think it is september 21 midnight AT GREENWHICH (you didn't specify midnight where).</p>

<p>Let's go to <a href="http://www.csgnetwork.com/siderealjuliantimecalc.html" rel="nofollow">http://www.csgnetwork.com/siderealjuliantimecalc.html</a></p>

<p>There input september 21, midnight at Greenwich. It says Sidereal time (ST) is 00:00:6.62 or, in hours, 0.001839h</p>

<p>This number is the difference between any star RA and its Azimuth at Greenwich.</p>

<p>Miaplacidus has a RA of 9h13min (9,216667h) so its Azimuth at Greenwich is 9,218506 at that moment.</p>

<p>Since the star is at Azimuth 0 for you, you are -9,218506h away from greenwich, that is (multiplying by 15) -138,27759º away from Greenwich. This is your Longitude.</p>

CC BY-SA 3.0
2013-12-13T18:22:45.710Z
2013-12-13T18:22:45.710Z
Envite
<p>Considering that a star of certain declination is crossing the local meridian at the observer's zenith at an unknown location on the earth. Here, the declination of the star is equal to the observer's latitude. Knowing the Right Ascension of the star, how could the observer calculate his longitude? </p>

<p>Example: The star Miaplacidus (Dec: -69° 42' and RA: 9h 13m) crossing at the observer's zenith!</p>

<p><strong>Latitude</strong> = 69° 42' S </p>

<p>[ <strong><em>Longitude</strong>= RA(Decimal Conversion) X 15°</em> ... <em>I'm not sure how to proceed from here ...</em> ]</p>

CC BY-SA 3.0
2013-12-12T10:45:15.820Z
You need a clock. Do you have one for this exercise?
Envite
CC BY-SA 3.0
2013-12-12T16:28:45.633Z
Nope :/ Perhaps proceed with a time or consider september 21 midnight... ST=LT
Ken
<star><amateur-observing><observational-astronomy><fundamental-astronomy>
How to calculate Longitude from Right Ascension?
Ken
# How to calculate Longitude from Right Ascension?

(Asked by: username_0 on 2013-12-12)

Considering that a star of certain declination is crossing the local meridian at the observer's zenith at an unknown location on the earth. Here, the declination of the star is equal to the observer's latitude. Knowing the Right Ascension of the star, how could the observer calculate his longitude?

Example: The star Miaplacidus (Dec: -69° 42' and RA: 9h 13m) crossing at the observer's zenith!

**Latitude** = 69° 42' S

[ ***Longitude***= RA(Decimal Conversion) X 15° ... *I'm not sure how to proceed from here ...* ]

**Tags:** star, amateur-observing, observational-astronomy, fundamental-astronomy

> **Comments:**
> **username_1**: You need a clock. Do you have one for this exercise?  
> **username_0**: Nope :/ Perhaps proceed with a time or consider september 21 midnight... ST=LT  

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## 2 answers

### Answer by username_2 on 2014-01-06. Score: 4

In other "words", the connection between the time of transit $t\_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta\_0$ (if you know your local siderial time $\theta$, you don't need $\theta\_0$ or $l$):
$$t\_\mathrm{tr} = \alpha - \theta\_0 - l = \alpha - \theta.$$


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### Answer by username_1 on 2013-12-13. Score: 2

Let's think it is september 21 midnight AT GREENWHICH (you didn't specify midnight where).

Let's go to <http://www.csgnetwork.com/siderealjuliantimecalc.html>

There input september 21, midnight at Greenwich. It says Sidereal time (ST) is 00:00:6.62 or, in hours, 0.001839h

This number is the difference between any star RA and its Azimuth at Greenwich.

Miaplacidus has a RA of 9h13min (9,216667h) so its Azimuth at Greenwich is 9,218506 at that moment.

Since the star is at Azimuth 0 for you, you are -9,218506h away from greenwich, that is (multiplying by 15) -138,27759º away from Greenwich. This is your Longitude.


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