<p>The solution is probably something like this:</p>
<p><span class="math-container">$s = t + a$</span>, where <span class="math-container">$t$</span> is local hour angle and <span class="math-container">$a$</span> is right ascension.</p>
<p><span class="math-container">$t = \arccos(-tg \phi \times tg \delta)$</span></p>
<p><span class="math-container">$t = \arccos(0) = 90 (=6^h), \ \ $</span> since <span class="math-container">$\phi = 0$</span> (equator)</p>
<p>so:</p>
<p><span class="math-container">$s = 6^h45^m + 6^h = 12^h45^m$</span></p>
<p>then:</p>
<p><span class="math-container">$\cos(A) = { \dfrac{\sin(\delta)} {\cos(\phi)} }, \ \ $</span>  where <span class="math-container">$A$</span> is azimuth</p>
<p>I got</p>
<p><span class="math-container">$A = 105$</span></p>

CC BY-SA 4.0
2022-01-22T23:37:13.057Z
Nice answer; spherical trig always makes my head spin. I made some small MathJax formatting changes, hope you don't mind, but for the argument `-tg \phi * tg \delta` ($-tg \phi * tg \delta$) I don't understand what the two terms are exactly or the best way to write them. Is there some subscripting necessary? $t$ is a time, but what is $g$ and how do the terms related to $\delta$ and $\phi$? Also, we usually just put two terms next to each other to indicate multiplication and skip the "*". If absolutely necessary, we use `\times` (which looks like $\times$).
uhoh
CC BY-SA 4.0
2022-01-22T18:20:05.667Z
2022-01-23T06:37:18.200Z
2022-01-23T06:37:18.200Z
ALiCe P.
<p>This is from: <a href="https://www.celestialprogramming.com/risesetalgorithm.html" rel="nofollow noreferrer">https://www.celestialprogramming.com/risesetalgorithm.html</a> .  It has a Javascript implementation and example you can run for the rise, set, and transit times.  Keep in mind that if you use this algorithm for objects that move, like the moon, sun, or planets, you may need to iterate to find an accurate solution.</p>
<p><span class="math-container">$ \cos H_0 = \dfrac{\sin h_0 - \sin \varphi \sin \delta }{\cos \varphi \cos \delta} $</span></p>
<p>If <span class="math-container">$ cos H_0 $</span> &lt; -1 or &gt; 1, the point is either always above or below the horizon.</p>
<p><span class="math-container">$ T=(jd-2451545.0)/36525.0 $</span></p>
<p><span class="math-container">$ \Theta_0 = 280.46061837+360.98564736629*(jd-2451545.0)+0.000387933T^2 - T^3/38710000.0 $</span>
<span class="math-container">$ transit = \dfrac{\delta + L - \Theta_0 }{360^{\circ}} $</span></p>
<p><span class="math-container">$ rise = transit - \dfrac{H_0}{360^{\circ}} $</span></p>
<p><span class="math-container">$ set = transit + \dfrac{H_0}{360^{\circ}} $</span></p>
<p><span class="math-container">$ jd $</span> is the Julian Date for the date in question.</p>
<p><span class="math-container">$ \delta $</span> Declination</p>
<p><span class="math-container">$ L $</span> Longitude</p>
<p><span class="math-container">$ \varphi $</span> Latitude</p>
<p><span class="math-container">$ h_0 $</span> Apparent rise or set angle, -0.8333 for the Sun, +0.125 for the Moon, and -0.5667 for most other objects.</p>
<p><span class="math-container">$ \Theta_0 $</span> Greenwich sidereal time at 0h for the day in question.</p>
<p><span class="math-container">$ A_r $</span> is the azimuth of the object when it rises, and</p>
<p><span class="math-container">$ A_s $</span> is the setting azimuth.</p>
<p><span class="math-container">$ \cos A_r = \dfrac{\sin \delta + \sin h_0 \sin \varphi}{\cos h_0 \cos \varphi} $</span></p>
<p><span class="math-container">$ \cos A_s = 360 - A_r $</span></p>

CC BY-SA 4.0
2022-01-28T01:06:24.733Z
2022-01-28T01:06:24.733Z
Greg Miller
<p>I know that right ascension equals to <span class="math-container">$6^h45^m$</span> and declination equals to <span class="math-container">$-16°43'$</span>. I need to calculate local time of sunrise and azimuth of rising point of Sirius on 23 of September for an observator located on the equator. So that means <span class="math-container">$\phi = 0$</span>. Should I use sunrise equation:</p>
<p><span class="math-container">$\cos \omega _{\circ }=-\tan \phi \tan \delta$</span></p>
<p>or is there another way?</p>

<time><celestial-mechanics><declination><right-ascension><azimuth>
How to calculate the local time of sunrise and azimuth of rising point of Sirius?
ALiCe P.
# How to calculate the local time of sunrise and azimuth of rising point of Sirius?

(Asked by: username_0 on 2022-01-22)

I know that right ascension equals to $6^h45^m$ and declination equals to $-16°43'$. I need to calculate local time of sunrise and azimuth of rising point of Sirius on 23 of September for an observator located on the equator. So that means $\phi = 0$. Should I use sunrise equation:

$\cos \omega \_{\circ }=-\tan \phi \tan \delta$

or is there another way?

**Tags:** time, celestial-mechanics, declination, right-ascension, azimuth

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## 2 answers

### Answer by username_0 on 2022-01-22. Score: 1 | Accepted answer

The solution is probably something like this:

$s = t + a$, where $t$ is local hour angle and $a$ is right ascension.

$t = \arccos(-tg \phi \times tg \delta)$

$t = \arccos(0) = 90 (=6^h), \ \ $ since $\phi = 0$ (equator)

so:

$s = 6^h45^m + 6^h = 12^h45^m$

then:

$\cos(A) = { \dfrac{\sin(\delta)} {\cos(\phi)} }, \ \ $ where $A$ is azimuth

I got

$A = 105$

> **Comments:**
> **username_1**: Nice answer; spherical trig always makes my head spin. I made some small MathJax formatting changes, hope you don't mind, but for the argument `-tg \phi * tg \delta` ($-tg \phi * tg \delta$) I don't understand what the two terms are exactly or the best way to write them. Is there some subscripting necessary? $t$ is a time, but what is $g$ and how do the terms related to $\delta$ and $\phi$? Also, we usually just put two terms next to each other to indicate multiplication and skip the "*". If absolutely necessary, we use `\times` (which looks like $\times$).  

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### Answer by username_2 on 2022-01-28. Score: 1

This is from: <https://www.celestialprogramming.com/risesetalgorithm.html> . It has a Javascript implementation and example you can run for the rise, set, and transit times. Keep in mind that if you use this algorithm for objects that move, like the moon, sun, or planets, you may need to iterate to find an accurate solution.

$ \cos H\_0 = \dfrac{\sin h\_0 - \sin \varphi \sin \delta }{\cos \varphi \cos \delta} $

If $ cos H\_0 $ < -1 or > 1, the point is either always above or below the horizon.

$ T=(jd-2451545.0)/36525.0 $

$ \Theta\_0 = 280.46061837+360.98564736629\*(jd-2451545.0)+0.000387933T^2 - T^3/38710000.0 $
$ transit = \dfrac{\delta + L - \Theta\_0 }{360^{\circ}} $

$ rise = transit - \dfrac{H\_0}{360^{\circ}} $

$ set = transit + \dfrac{H\_0}{360^{\circ}} $

$ jd $ is the Julian Date for the date in question.

$ \delta $ Declination

$ L $ Longitude

$ \varphi $ Latitude

$ h\_0 $ Apparent rise or set angle, -0.8333 for the Sun, +0.125 for the Moon, and -0.5667 for most other objects.

$ \Theta\_0 $ Greenwich sidereal time at 0h for the day in question.

$ A\_r $ is the azimuth of the object when it rises, and

$ A\_s $ is the setting azimuth.

$ \cos A\_r = \dfrac{\sin \delta + \sin h\_0 \sin \varphi}{\cos h\_0 \cos \varphi} $

$ \cos A\_s = 360 - A\_r $


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