<p>The algorithms for GMST avoids this ambiguity by using the number of Julian centuries of UT1 since J2000 (or some other epoch) rather than a time and date.  It would only be a problem if someone gave you a date and sidereal time and for some reason wanted you to unambiguously determine the time, which I can't imagine a practical use for.</p>
<p>Here's one algorithm used for computing GMST, it's likely different than what's in Duffett-Smith's book, but reproduced here for others without access to to the book. Example implementation here: <a href="https://www.celestialprogramming.com/snippets/greenwichMeanSiderealTime.html" rel="nofollow noreferrer">Greenwich Mean Sidereal Time</a></p>
<p><span class="math-container">$$
\begin{align*}
\theta &amp;= 0.7790572732640 + 0.00273781191135448D_u + frac(JD(UT1)) \\
T &amp;=\frac{JD_{UT1} - 2451545.0}{36525} \\
GMST &amp;= 2\pi\theta + (0.014506 + 4612.15739966T + 1.39667721T^2 -0.00009344T^3 + 0.00001882T^4)/60/60*\pi/180
\end{align*}
$$</span></p>
<p><span class="math-container">$ D_u $</span>= Number of days of UT1 since J2000 (JD(UT1) - 2451545.0)</p>
<p><span class="math-container">$ \theta $</span> = Earth Rotation Angle (in fraction of full revolutions)</p>
<p>T = Centuries of UT1 since J2000</p>
<p>GMST = Greenwich Mean Sidereal Time (in radians)</p>
<p><span class="math-container">$ frac(JD(UT1)) $</span> = fraction of day</p>

CC BY-SA 4.0
2022-09-01T18:08:39.483Z
Are you sure that formula for Earth Rotation Angle is correct? According to my Astronomical Almanac (my calculation, so it could be wrong) the ERA at J2000 (JD = 2451545.0) is 280.5 degrees. Your formula gives 0.7790572732640/3600 = 0.000216405 degrees.
Peter
CC BY-SA 4.0
2022-09-01T19:31:56.840Z
It is equation 2.11 from USNO Circular 179.  The units are actually full revolutions, so I need to change the description.  So it matches the answer you got.
Greg Miller
CC BY-SA 4.0
2022-09-02T07:51:45.590Z
I don't think $D_{U}$ refers to *whole* days. For JD = 2451545.50, the Earth should have rotated about 180 degrees since JD = 2451545.0. You don't get this result by assuming $D_{U}=0$ in your equation. You do get it by letting $D_{U}=0.5$.
Peter
CC BY-SA 4.0
2022-09-02T14:05:54.317Z
$ D_u $ has to be a whole number, .5 is invalid.  For that case you set $ D_u $ to 0 and add .5.
Greg Miller
CC BY-SA 4.0
2022-09-02T15:09:54.970Z
Grovelling apologies if I've got this wrong, but for JD = 2451545.50, if you set $D_{U}=0$ and $\mathrm{frac\left(JD\left(UT1\right)\right)=0.5,}$ then $\theta=360^{\circ}\left(0.7790572732640+\left(0.00273781191135448\times0\right)+0.5\right)=460.460618^{\circ}.$ Less ERA at epoch ($280.460618^{\circ}$) gives $180.000000^{\circ}$ (wrong). Setting $D_{U}=\mathrm{frac\left(JD\left(UT1\right)\right)=0.5,}$ gives $460.953425^{\circ}$. Less $280.460618^{\circ}$ gives $180.492806^{\circ}$ (correct).
Peter
CC BY-SA 4.0
2022-09-02T20:53:15.800Z
You are correct.  Looks like I implemented it correctly, but said it wrong, I'll update it.  Thanks.
Greg Miller
CC BY-SA 4.0
2022-09-01T16:50:45.330Z
2022-09-02T20:53:34.727Z
2022-09-02T20:53:34.727Z
Greg Miller
<p>I'm reading <em>Practical Astronomy With Your Calculator Or Spreadsheet</em> by Duffett-Smith and Zwart. Discussing the conversion of GST to UT, the authors mention that because the sidereal day is slightly shorter than the solar day, for a given calendar date there is a small range of sidereal times that occur twice. (approximately if <span class="math-container">$0h\leq\mathrm{UT}\leq03m\,56s$</span> and <span class="math-container">$23h\,56m\,04s\leq\mathrm{UT}&lt;0h$</span>).</p>
<p>I can't find much online about this and, as far as I can see, it's not mentioned in the Wikipedia <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="noreferrer">Sidereal Time</a> article.</p>
<p>So my question is, in their day-to-day working lives is this ambiguity a practical problem for astronomers or is it only of theoretical interest?</p>

CC BY-SA 4.0
2022-09-01T15:11:12.013Z
It's only (potentially) ambiguous if you try to specify an event by giving a sidereal time and a calendar date. So don't do that. ;)
PM 2Ring
CC BY-SA 4.0
2022-09-03T20:11:45.287Z
@PM2Ring invoking [Henny Youngman](https://www.goodreads.com/quotes/7570350-the-patient-says-doctor-it-hurts-when-i-do-this)?
uhoh
<time>
Sidereal times occuring twice a day question
Peter
# Sidereal times occuring twice a day question

(Asked by: username_0 on 2022-09-01)

I'm reading *Practical Astronomy With Your Calculator Or Spreadsheet* by Duffett-Smith and Zwart. Discussing the conversion of GST to UT, the authors mention that because the sidereal day is slightly shorter than the solar day, for a given calendar date there is a small range of sidereal times that occur twice. (approximately if $0h\leq\mathrm{UT}\leq03m\,56s$ and $23h\,56m\,04s\leq\mathrm{UT}<0h$).

I can't find much online about this and, as far as I can see, it's not mentioned in the Wikipedia [Sidereal Time](https://en.wikipedia.org/wiki/Sidereal_time) article.

So my question is, in their day-to-day working lives is this ambiguity a practical problem for astronomers or is it only of theoretical interest?

**Tags:** time
> **Comments:**
> **username_1**: It's only (potentially) ambiguous if you try to specify an event by giving a sidereal time and a calendar date. So don't do that. ;)  
> **username_2**: @PM2Ring invoking [Henny Youngman](https://www.goodreads.com/quotes/7570350-the-patient-says-doctor-it-hurts-when-i-do-this)?  

---

## 1 answers

### Answer by username_3 on 2022-09-01. Score: 4 | Accepted answer

The algorithms for GMST avoids this ambiguity by using the number of Julian centuries of UT1 since J2000 (or some other epoch) rather than a time and date. It would only be a problem if someone gave you a date and sidereal time and for some reason wanted you to unambiguously determine the time, which I can't imagine a practical use for.

Here's one algorithm used for computing GMST, it's likely different than what's in Duffett-Smith's book, but reproduced here for others without access to to the book. Example implementation here: [Greenwich Mean Sidereal Time](https://www.celestialprogramming.com/snippets/greenwichMeanSiderealTime.html)

$$
\begin{align\*}
\theta &= 0.7790572732640 + 0.00273781191135448D\_u + frac(JD(UT1)) \\
T &=\frac{JD\_{UT1} - 2451545.0}{36525} \\
GMST &= 2\pi\theta + (0.014506 + 4612.15739966T + 1.39667721T^2 -0.00009344T^3 + 0.00001882T^4)/60/60\*\pi/180
\end{align\*}
$$

$ D\_u $= Number of days of UT1 since J2000 (JD(UT1) - 2451545.0)

$ \theta $ = Earth Rotation Angle (in fraction of full revolutions)

T = Centuries of UT1 since J2000

GMST = Greenwich Mean Sidereal Time (in radians)

$ frac(JD(UT1)) $ = fraction of day

> **Comments:**
> **username_0**: Are you sure that formula for Earth Rotation Angle is correct? According to my Astronomical Almanac (my calculation, so it could be wrong) the ERA at J2000 (JD = 2451545.0) is 280.5 degrees. Your formula gives 0.7790572732640/3600 = 0.000216405 degrees.  
> **username_3**: It is equation 2.11 from USNO Circular 179.  The units are actually full revolutions, so I need to change the description.  So it matches the answer you got.  
> **username_0**: I don't think $D_{U}$ refers to *whole* days. For JD = 2451545.50, the Earth should have rotated about 180 degrees since JD = 2451545.0. You don't get this result by assuming $D_{U}=0$ in your equation. You do get it by letting $D_{U}=0.5$.  
> **username_3**: $ D_u $ has to be a whole number, .5 is invalid.  For that case you set $ D_u $ to 0 and add .5.  
> **username_0**: Grovelling apologies if I've got this wrong, but for JD = 2451545.50, if you set $D_{U}=0$ and $\mathrm{frac\left(JD\left(UT1\right)\right)=0.5,}$ then $\theta=360^{\circ}\left(0.7790572732640+\left(0.00273781191135448\times0\right)+0.5\right)=460.460618^{\circ}.$ Less ERA at epoch ($280.460618^{\circ}$) gives $180.000000^{\circ}$ (wrong). Setting $D_{U}=\mathrm{frac\left(JD\left(UT1\right)\right)=0.5,}$ gives $460.953425^{\circ}$. Less $280.460618^{\circ}$ gives $180.492806^{\circ}$ (correct).  
> **username_3**: You are correct.  Looks like I implemented it correctly, but said it wrong, I'll update it.  Thanks.  

---

