<blockquote>
<pre><code>LST = 100.46 + 0.985647 * d + long + 15*UT
</code></pre>
  
  <p>They don't explain what the two constants are (100.46 and 0.985647), could anyone explain what those constants are and how they were calculated in the first place please?</p>
</blockquote>

<p>There are three constants there, 100.46, 0.985647, and 15.</p>

<p>The value of 100.46 degrees is the value needed to make the expression yield the correct value for GMST at 0<sup>h</sup> UT on 1 January 2000. The value of 0.985647 degrees per day is the number of degrees the Earth rotates in one mean solar day, sans a multiple of 360. The value of 15 degrees per hour is the number of degrees the Earth rotates with respect to the mean fictitious Sun every hour.</p>

<p>Regarding 0.985647: There is one extra sidereal day in a solar year than there are solar days. There are 365.2422 solar days in a year, so the Earth rotates $360*366.2422/365.2422=360.985647332$ degrees per solar day with respect to the stars. That first 360 is irrelevant (the result needs to be taken mod 360 in the end), resulting in the factor of 0.985647 (0.985647332 rounded to six significant digits).</p>

<p>Regarding 15: Note that this is the number of degrees the Earth rotates per hour with respect to the Sun. Multiplying this by $366.2422/365.2422$ yields 15.04106864, the number of degrees the Earth rotates per hour with respect to the stars.</p>

<p>Another way to achieve the same result is to fold that extra 0.04106864 degrees per hour into the number of days since noon on 1 January 2000.  Not surprisingly, 0.04106864*24 = 0.985647. This means that the $d$ in the approximate formula in the question must include the fractional days.</p>

<p>You need to take care with this approximate formula. It is approximately true for the 200 year period centered around midnight on 1 January 2000, and you need to make sure that the $d$ is the number of days from noon on 1 January 2000, including fractional days.</p>

<p><br>
<strong>Addendum</strong>: Showing this is the same as the <em>Astronomical Almanac</em> expression, sans a quadratic term</p>

<p>The Astronomical Almanac gives an expression for approximate mean sidereal time, in hours:
$$\mathit{GMST} = 6.697374558 + 0.06570982441908 D_0 + 1.00273790935 H + 0.000026 T^2$$
Where $\mathit{GMST}$ is the mean sidereal time in hours, $H$ is the universal time at the time in question, $D_0$ is the Julian date on the previous midnight of the time in question less 2451545.0, $D$ is the Julian date at the time in question (including fractional days) less 2451545.0, and $T$ is $D/36525$. The relationship between $D_0$, $D$, and $H$ is quite simple: $D_0 = D - H/24$. Substituting this in the above and omitting the quadratic term yields
$$\begin{aligned}
\mathit{GMST} &amp;= 6.697374558 + 0.06570982441908 (D-H/24) + 1.00273790935 H
\\
&amp;= 6.697374558 + 0.06570982441908D + H
\end{aligned}$$
(Strictly speaking, 1.00273790935-0.06570982441908/24 = 0.9999999999992 rather than 1.0, but that's just because that 1.00273790935 should be 1.0027379093508).</p>

<p>Multiplying by 15 yields the GMST in degrees:
$$\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H$$
This is the expression in the question, sans the longitude and plus some extra digits.</p>

CC BY-SA 4.0
2020-04-20T02:09:56.563Z
I've [written this](https://astronomy.stackexchange.com/a/35871/7982) based in part on this
uhoh
CC BY-SA 4.0
2022-06-02T10:19:14.643Z
Hi, you mentioned that "There are 365.2422 solar days in a year". However, should one use the tropical or sidereal year? why? Thanks!
Cheng
CC BY-SA 4.0
2022-06-03T10:31:23.260Z
@Cheng It's the tropical (or solar) year. The reason why is that the 86400 second long mean solar day (now 86400 seconds plus a millisecond or two)  is defined as the average length of a solar day over the course of a tropical year. The length of an apparent solar day (the timespan from one solar noon to the next) varies over the course of a year (tropical year) primarily due to the equation of time.
David Hammen
CC BY-SA 4.0
2022-12-02T23:07:43.263Z
@David Hammen Can you explain what the fractional day is - is it not the time dealt with later in the OPs equation? or is it taking account of the half day start point of J2000, what form should the hours take and how are the seconds dealt with? wikipedia has decimal time as a 10hr100m100s format used in the French revolution so totally confused.
user36093
CC BY-SA 4.0
2022-12-04T12:32:48.577Z
@user36093 Forget about the French revolution's concept of decimal time. It never took hold. The relationship here is simple: $D_0=D-H/24$ (or alternatively, $D=D_0+H/24$). $D_0$ is the Julian date at midnight, less 2451545.0, so it will always be some integer plus a half. For example, 7:30 PM on some day is 19.5 hours after midnight and 7.5 hours after noon. Thus $H$ will be 19.5, $D$ will be some integer (the Julian day number at noon on that day) plus 0.3125 (7.5/24), and $D_0$ will be that same integer minus 0.5.
David Hammen
CC BY-SA 3.0
2018-02-02T19:14:20.153Z
2018-02-02T20:11:05.543Z
2018-02-02T20:11:05.543Z
David Hammen
<p>I am trying to understand how to calculate local sidereal time and have found the following formula:</p>

<p>$$\text{LST} = 100.46 + 0.985647 \cdot d + \text{long} + 15 \cdot \text{UT}$$</p>

<p>Here, $d$ is the number of days from J2000, including the fraction of a day<br>
UT is the universal time in decimal hours<br>
long is your longitude in decimal degrees, East positive.</p>

<p>They don't explain what the two constants are (<code>100.46</code> and <code>0.985647</code>), could anyone please explain what those constants are and how they were calculated in the first place?</p>

CC BY-SA 3.0
2018-02-02T16:48:43.957Z
Just a guess off the top of my head, with currently no research to back this up, but the `0.985647` value might be the conversion from solar to sidereal day.
zephyr
CC BY-SA 3.0
2018-02-02T16:51:27.763Z
It's really close, but I think `0.9972695663290843` is the ratio between the solar and sidereal day. I had thought that the `100.46` was decimal days from the start of a year until the vernal equinox, but that's about 20 days out too :/
user3574623
<coordinate><time>
Local Sidereal Time
user3574623
# Local Sidereal Time

(Asked by: username_0 on 2018-02-02)

I am trying to understand how to calculate local sidereal time and have found the following formula:

$$\text{LST} = 100.46 + 0.985647 \cdot d + \text{long} + 15 \cdot \text{UT}$$

Here, $d$ is the number of days from J2000, including the fraction of a day  
UT is the universal time in decimal hours  
long is your longitude in decimal degrees, East positive.

They don't explain what the two constants are (`100.46` and `0.985647`), could anyone please explain what those constants are and how they were calculated in the first place?

**Tags:** coordinate, time

> **Comments:**
> **username_1**: Just a guess off the top of my head, with currently no research to back this up, but the `0.985647` value might be the conversion from solar to sidereal day.  
> **username_0**: It's really close, but I think `0.9972695663290843` is the ratio between the solar and sidereal day. I had thought that the `100.46` was decimal days from the start of a year until the vernal equinox, but that's about 20 days out too :/  

---

## 1 answers

### Answer by username_2 on 2018-02-02. Score: 10 | Accepted answer

LST = 100.46 + 0.985647 * d + long + 15*UT

They don't explain what the two constants are (100.46 and 0.985647), could anyone explain what those constants are and how they were calculated in the first place please?

There are three constants there, 100.46, 0.985647, and 15.
The value of 100.46 degrees is the value needed to make the expression yield the correct value for GMST at 0h UT on 1 January 2000. The value of 0.985647 degrees per day is the number of degrees the Earth rotates in one mean solar day, sans a multiple of 360. The value of 15 degrees per hour is the number of degrees the Earth rotates with respect to the mean fictitious Sun every hour.
Regarding 0.985647: There is one extra sidereal day in a solar year than there are solar days. There are 365.2422 solar days in a year, so the Earth rotates $360*366.2422/365.2422=360.985647332$ degrees per solar day with respect to the stars. That first 360 is irrelevant (the result needs to be taken mod 360 in the end), resulting in the factor of 0.985647 (0.985647332 rounded to six significant digits).
Regarding 15: Note that this is the number of degrees the Earth rotates per hour with respect to the Sun. Multiplying this by $366.2422/365.2422$ yields 15.04106864, the number of degrees the Earth rotates per hour with respect to the stars.
Another way to achieve the same result is to fold that extra 0.04106864 degrees per hour into the number of days since noon on 1 January 2000.  Not surprisingly, 0.04106864*24 = 0.985647. This means that the $d$ in the approximate formula in the question must include the fractional days.
You need to take care with this approximate formula. It is approximately true for the 200 year period centered around midnight on 1 January 2000, and you need to make sure that the $d$ is the number of days from noon on 1 January 2000, including fractional days.

Addendum: Showing this is the same as the Astronomical Almanac expression, sans a quadratic term
The Astronomical Almanac gives an expression for approximate mean sidereal time, in hours:
$$\mathit{GMST} = 6.697374558 + 0.06570982441908 D_0 + 1.00273790935 H + 0.000026 T^2$$
Where $\mathit{GMST}$ is the mean sidereal time in hours, $H$ is the universal time at the time in question, $D_0$ is the Julian date on the previous midnight of the time in question less 2451545.0, $D$ is the Julian date at the time in question (including fractional days) less 2451545.0, and $T$ is $D/36525$. The relationship between $D_0$, $D$, and $H$ is quite simple: $D_0 = D - H/24$. Substituting this in the above and omitting the quadratic term yields
$$\begin{aligned}
\mathit{GMST} &= 6.697374558 + 0.06570982441908 (D-H/24) + 1.00273790935 H
\\
&= 6.697374558 + 0.06570982441908D + H
\end{aligned}$$
(Strictly speaking, 1.00273790935-0.06570982441908/24 = 0.9999999999992 rather than 1.0, but that's just because that 1.00273790935 should be 1.0027379093508).
Multiplying by 15 yields the GMST in degrees:
$$\mathit{GMST}_{\text{deg}} = 100.4606184 + 0.9856473662862 D + 15 H$$
This is the expression in the question, sans the longitude and plus some extra digits.

> **Comments:**
> **username_3**: I've [written this](https://astronomy.stackexchange.com/a/35871/7982) based in part on this  
> **username_4**: Hi, you mentioned that "There are 365.2422 solar days in a year". However, should one use the tropical or sidereal year? why? Thanks!  
> **username_2**: @username_4 It's the tropical (or solar) year. The reason why is that the 86400 second long mean solar day (now 86400 seconds plus a millisecond or two)  is defined as the average length of a solar day over the course of a tropical year. The length of an apparent solar day (the timespan from one solar noon to the next) varies over the course of a year (tropical year) primarily due to the equation of time.  
> **username_5**: @username_2 Can you explain what the fractional day is - is it not the time dealt with later in the OPs equation? or is it taking account of the half day start point of J2000, what form should the hours take and how are the seconds dealt with? wikipedia has decimal time as a 10hr100m100s format used in the French revolution so totally confused.  
> **username_2**: @username_5 Forget about the French revolution's concept of decimal time. It never took hold. The relationship here is simple: $D_0=D-H/24$ (or alternatively, $D=D_0+H/24$). $D_0$ is the Julian date at midnight, less 2451545.0, so it will always be some integer plus a half. For example, 7:30 PM on some day is 19.5 hours after midnight and 7.5 hours after noon. Thus $H$ will be 19.5, $D$ will be some integer (the Julian day number at noon on that day) plus 0.3125 (7.5/24), and $D_0$ will be that same integer minus 0.5.  

---

