<p>A highly authoritative source is <em>Explanatory Supplement to the Astronomical Almanac</em> 3rd ed., Urban &amp; Seidelmann, 2013.</p>
<p>Chapter 1 gives the introductory explanations, without all the refinements if you want millisecond accuracy. Page 12 explains that</p>
<p>local sidreal time = Greenwich sidereal time + east longitude</p>
<p>(This book assumes the reader is able to convert between angle notation and time notation whenever necessary.)</p>
<p>If accuracy of 0.1 second is sufficient, the formulas given by <a href="https://aa.usno.navy.mil/faq/GAST" rel="nofollow noreferrer" title="the US Naval Observatory">the US Naval Observatory</a> should do the job. Note that for subsecond accuracy you will have to determine UT1, which can be up to 0.9 second different from the commonly available UTC. You will have to look up the difference each time you do the calculation because the difference cannot be predicted more than about 6 months into the future.</p>

CC BY-SA 4.0
2015-10-04T16:23:13.603Z
2023-07-10T13:35:39.860Z
2023-07-10T13:35:39.860Z
Gerard Ashton
<p>Sidereal time is the time reckoned with respect to the stars rather than the sun. </p>

<p>We normally take the length of the day to be the 24 hours it takes the sun to go from mid-day to mid-day. But as the Earth is spinning on its axis and moving around the sun, the time it takes for a star to appear to move in a full circle once is slightly less: 23.9344 hours. This is called the sidereal day. Sidereal time is time reckoned by this "day" length. Consult <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="nofollow">Wikipedia</a>, for more details</p>

<p>The three constants A, B and C in the linked document are chosen to convert a day, and and local time to sidereal time. They have no special meaning, but note that $C=1+\frac{1}{\rm number\ of\ days\ in\ year}$.</p>

<p>Implementation is straightforward. You will however need to know, or calcualate</p>

<ul>
<li>the day number and the decimal hour. But these can be easily found from the calendar date and time in hours minutes and seconds</li>
<li>The difference between the the time zone and local mean time. This depends on location, and whether daylight saving time is in operation.</li>
<li>that the algorithm returns a decimal hour. You may want to convert to hours minutes and seconds.  </li>
</ul>

CC BY-SA 3.0
2015-10-03T20:55:02.000Z
2015-10-03T20:55:02.000Z
James K
<p>I am trying to follow this guide and move to a programming language
but I have doubts.</p>

<p>What means the strings A, B and C in this manual? 
<a href="http://aastro.org/wp-content/uploads/2010/11/Calculation-of-Local-Sidereal-Time.pdf" rel="nofollow">http://aastro.org/wp-content/uploads/2010/11/Calculation-of-Local-Sidereal-Time.pdf</a></p>

CC BY-SA 3.0
2015-10-03T20:56:17.393Z
do you have doubts about the correctness of this procedure? Is the reason for your doubts that the document doesn't justify the values of A, B and C?
James K
CC BY-SA 3.0
2015-10-03T22:27:59.103Z
Yes the document doesn't justify the values of A, B and C.
Rui
<amateur-observing>
Calculate Local Sidereal Time
Rui
# Calculate Local Sidereal Time

(Asked by: username_0 on 2015-10-03)

I am trying to follow this guide and move to a programming language
but I have doubts.

What means the strings A, B and C in this manual?
<http://aastro.org/wp-content/uploads/2010/11/Calculation-of-Local-Sidereal-Time.pdf>

**Tags:** amateur-observing

> **Comments:**
> **username_1**: do you have doubts about the correctness of this procedure? Is the reason for your doubts that the document doesn't justify the values of A, B and C?  
> **username_0**: Yes the document doesn't justify the values of A, B and C.  

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## 2 answers

### Answer by username_2 on 2015-10-04. Score: 1

A highly authoritative source is *Explanatory Supplement to the Astronomical Almanac* 3rd ed., Urban & Seidelmann, 2013.

Chapter 1 gives the introductory explanations, without all the refinements if you want millisecond accuracy. Page 12 explains that

local sidreal time = Greenwich sidereal time + east longitude

(This book assumes the reader is able to convert between angle notation and time notation whenever necessary.)

If accuracy of 0.1 second is sufficient, the formulas given by [the US Naval Observatory](https://aa.usno.navy.mil/faq/GAST "the US Naval Observatory") should do the job. Note that for subsecond accuracy you will have to determine UT1, which can be up to 0.9 second different from the commonly available UTC. You will have to look up the difference each time you do the calculation because the difference cannot be predicted more than about 6 months into the future.


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### Answer by username_1 on 2015-10-03. Score: 0

Sidereal time is the time reckoned with respect to the stars rather than the sun.

We normally take the length of the day to be the 24 hours it takes the sun to go from mid-day to mid-day. But as the Earth is spinning on its axis and moving around the sun, the time it takes for a star to appear to move in a full circle once is slightly less: 23.9344 hours. This is called the sidereal day. Sidereal time is time reckoned by this "day" length. Consult [Wikipedia](https://en.wikipedia.org/wiki/Sidereal_time), for more details

The three constants A, B and C in the linked document are chosen to convert a day, and and local time to sidereal time. They have no special meaning, but note that $C=1+\frac{1}{\rm number\ of\ days\ in\ year}$.

Implementation is straightforward. You will however need to know, or calcualate

* the day number and the decimal hour. But these can be easily found from the calendar date and time in hours minutes and seconds
* The difference between the the time zone and local mean time. This depends on location, and whether daylight saving time is in operation.
* that the algorithm returns a decimal hour. You may want to convert to hours minutes and seconds.


---

