<p>If you extrapolate, you'll find that those ~4 minutes per day add up to 24 hours per year.
The difference is due to the Earth revolving around the Sun at a rate of almost 1 degree per day.
For every 365$\frac{1}{4}$ solar days, there are 366$\frac{1}{4}$ sidereal days.</p>

<p>Civil time is based on mean
<a href="https://en.wikipedia.org/wiki/Solar_time" rel="nofollow">solar time</a>,
in which the Sun crosses the
<a href="https://en.wikipedia.org/wiki/Meridian_%28astronomy%29" rel="nofollow">celestial meridian</a>
every 24h 0m on average.
In <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="nofollow">sidereal time</a>,
24h 0m is the time between successive meridian transits by any given star, equivalent to 23h 56m 4s of solar time.</p>

<p>Sidereal time also corresponds to the
<a href="https://en.wikipedia.org/wiki/Right_ascension" rel="nofollow">right ascension</a>
of stars crossing the meridian at that time.
If a star in Orion has right ascension 5h 30m, then Orion is on the meridian at 05:30 sidereal time, which occurs 2 hours earlier in the solar day per month.</p>

CC BY-SA 3.0
2016-10-08T17:37:12.580Z
2016-10-08T17:37:12.580Z
Mike G
<p>I am new to this domain and trying to <em>understanding the fundamentals of local sidereal time</em>.
Through an online available calculator <a href="https://www.iiap.res.in/personnel/reks/software/javascript/calclst.php" rel="nofollow">https://www.iiap.res.in/personnel/reks/software/javascript/calclst.php</a> ,I am noting down the <code>local sidereal time</code> for my <code>longitude</code> of interest. I have repeated the exercise for same time for five consecutive days. What I observe is that the <code>local sidereal time</code> is advancing by <code>0.06/07 hours per day</code> for the same time.Like <code>21.88 Hours</code> for current date and time , <code>21.94 Hours</code> for the same time next day ,<code>22.01 Hours</code> for next to next day and the pattern continues. <em>Kindly let me know how this increment is happening</em>?</p>

<fundamental-astronomy><time>
Local sidereal time interpretation
Soumajit
# Local sidereal time interpretation

(Asked by: username_0 on 2016-10-08)

I am new to this domain and trying to *understanding the fundamentals of local sidereal time*.
Through an online available calculator <https://www.iiap.res.in/personnel/reks/software/javascript/calclst.php> ,I am noting down the `local sidereal time` for my `longitude` of interest. I have repeated the exercise for same time for five consecutive days. What I observe is that the `local sidereal time` is advancing by `0.06/07 hours per day` for the same time.Like `21.88 Hours` for current date and time , `21.94 Hours` for the same time next day ,`22.01 Hours` for next to next day and the pattern continues. *Kindly let me know how this increment is happening*?

**Tags:** fundamental-astronomy, time


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## 1 answers

### Answer by username_1 on 2016-10-08. Score: 2

If you extrapolate, you'll find that those ~4 minutes per day add up to 24 hours per year.
The difference is due to the Earth revolving around the Sun at a rate of almost 1 degree per day.
For every 365$\frac{1}{4}$ solar days, there are 366$\frac{1}{4}$ sidereal days.

Civil time is based on mean
[solar time](https://en.wikipedia.org/wiki/Solar_time),
in which the Sun crosses the
[celestial meridian](https://en.wikipedia.org/wiki/Meridian_%28astronomy%29)
every 24h 0m on average.
In [sidereal time](https://en.wikipedia.org/wiki/Sidereal_time),
24h 0m is the time between successive meridian transits by any given star, equivalent to 23h 56m 4s of solar time.

Sidereal time also corresponds to the
[right ascension](https://en.wikipedia.org/wiki/Right_ascension)
of stars crossing the meridian at that time.
If a star in Orion has right ascension 5h 30m, then Orion is on the meridian at 05:30 sidereal time, which occurs 2 hours earlier in the solar day per month.


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