<p>Short answers: It's not just angular velocity of Earth's rotation times elapsed time. You do need to factor in declination and latitude.</p>
<p>Using your representative date and observer location:
The apparent Sun, on the day of the summer solstice, crossed the Greenwich
meridian at 12:02 GMT/UT in 2021, then proceeded westward directly above the
Tropic of Cancer at the rate of 15 degrees per hour.</p>
<p>A spherical triangle with
vertices at the observer's position, the Sun's geographical position (GP), and
the North Pole has two sides known and you're looking for the relationship
between the third side and the angle at the pole. Using the law of cosines for
sides and plugging in the observer latitude <span class="math-container">$L$</span> and The Sun's declination <span class="math-container">$D$</span> gives:</p>
<p><span class="math-container">$$\cos(a) = \cos(90-L)\cos(90-D) + \sin(90-L)\sin(90-D)\cos(A)$$</span></p>
<p>Here, <span class="math-container">$a$</span> is the arc distance from observer to Sun's GP (AKA altitude) and <span class="math-container">$A$</span> is the angle at the pole.</p>
<p>For an observer at 80 degrees N latitude:</p>
<p><span class="math-container">$$\cos(a) = 0.06645228 + 0.90984373\cos(A)$$</span></p>
<p>When <span class="math-container">$a$</span> is 90 degrees, the observer would see half the Sun above the horizon,
were it not for the pesky atmosphere. Going just beyond, to 90 degrees plus 50
minutes for refraction and semidiameter puts the Sun out of sight.</p>
<p><span class="math-container">$$\cos(A) = \frac{-0.01744867-0.06645228}{0.90984373}; A = 84.7089^\circ$$</span></p>
<p>If the observer is at, say, 75 degrees W longitude, the Sun's Greenwich hour
angle would be 159.7089 degrees, and 10 hours and 39 minutes would have elapsed
since meridian passage of the Sun at Greenwich. GMT/UT is 22:41. Observer's
local civil time (EDT) is 6:41 PM.</p>

CC BY-SA 4.0
2021-06-23T18:29:12.890Z
To add maths you can use Mathjax, which is $\TeX$ like markup.  so write \$\cos(A)\$  for $\cos(A)$
James K
CC BY-SA 4.0
2021-06-24T08:33:06.417Z
I just did that...
B--rian
CC BY-SA 4.0
2021-06-23T16:46:29.763Z
2021-06-24T08:32:54.987Z
2021-06-24T08:32:54.987Z
stretch
<p>The <a href="https://en.wikipedia.org/wiki/Celestial_coordinate_system#Equatorial_%E2%86%94_horizontal" rel="noreferrer">transformation</a> from the equatorial coordinates to <a href="https://en.wikipedia.org/wiki/Horizontal_coordinate_system" rel="noreferrer">altitude</a> (azimuth is not important) is given by
<span class="math-container">$$\sin{a}=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$</span></p>
<p>Let's say that <span class="math-container">$t=0\rm\, s$</span> at <a href="https://en.wikipedia.org/wiki/Sun_transit_time" rel="noreferrer">local solar noon</a>. Then, <a href="https://en.wikipedia.org/wiki/Hour_angle" rel="noreferrer">the hour angle</a> is given by
<span class="math-container">$$h=\frac{360°\cdot t}{86400\rm\,s}$$</span>
and we can reformulate the first equation using above expression to
<span class="math-container">$$a=\arcsin{(\cos{\frac{360°\cdot t}{86400\rm\,s}}\cos{23.5°}\cos{10°}+\sin{23.5°}\sin{10°})}$$</span>
<span class="math-container">$$a=\arcsin{(0.90313\cos{\frac{t}{240\rm\,s}}+0.069242)}$$</span>
Your calculator must be set in degrees and <span class="math-container">$t$</span> is in seconds.</p>
<hr />
<p>You can also determine the time in seconds since local solar noon of sunset. We just write the above formula
<span class="math-container">$$\sin{a}=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$</span>
and the altitude <span class="math-container">$a$</span> must be 0° since the Sun is on the horizon on the sunset (note that we ignore refraction and the angular size of the solar disc, if you want to include that as well, set <span class="math-container">$a=-50'=-0.833°$</span>).
<span class="math-container">$$\sin{0°}=0=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$</span>
Then we just express the hour angle <span class="math-container">$h$</span>:
<span class="math-container">$$h=\arccos{(-\tan\delta\tan\phi)}$$</span>
and the time since the local noon:
<span class="math-container">$$t_{sunset}=\frac{86400\rm\,s\cdot h}{360°}$$</span></p>

CC BY-SA 4.0
2021-06-22T16:26:07.637Z
2021-06-22T16:49:04.600Z
2021-06-22T16:49:04.600Z
User123
<p>Let's say it is the day of the summer solstice and the Sun is setting at 23.5 degrees declination. I am at let's say, 10 degrees latitude. How would I go about calculating the angle of the Sun below the horizon as a function of time? Would it simply be the <code>angular velocity of Earth's rotation * elapsed time</code>? Or would I need to factor in the declination of the Sun or the latitude that I am currently at?</p>

CC BY-SA 4.0
2021-06-22T16:18:49.240Z
You'll certainly need the latitude - you can watch the sun set from 10 degrees latitude, but if you're inside the Arctic Circle on the summer solstice, you won't see the sun dip below the horizon *at all*.
Nuclear Hoagie
<observational-astronomy><the-sun><sunrise><sunset>
Angle the Sun makes with the horizon during sunset
I Love Math
# Angle the Sun makes with the horizon during sunset

(Asked by: username_0 on 2021-06-22)

Let's say it is the day of the summer solstice and the Sun is setting at 23.5 degrees declination. I am at let's say, 10 degrees latitude. How would I go about calculating the angle of the Sun below the horizon as a function of time? Would it simply be the `angular velocity of Earth's rotation * elapsed time`? Or would I need to factor in the declination of the Sun or the latitude that I am currently at?

**Tags:** observational-astronomy, the-sun, sunrise, sunset

> **Comments:**
> **username_1**: You'll certainly need the latitude - you can watch the sun set from 10 degrees latitude, but if you're inside the Arctic Circle on the summer solstice, you won't see the sun dip below the horizon *at all*.  

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## 2 answers

### Answer by username_2 on 2021-06-23. Score: 5

Short answers: It's not just angular velocity of Earth's rotation times elapsed time. You do need to factor in declination and latitude.

Using your representative date and observer location:
The apparent Sun, on the day of the summer solstice, crossed the Greenwich
meridian at 12:02 GMT/UT in 2021, then proceeded westward directly above the
Tropic of Cancer at the rate of 15 degrees per hour.

A spherical triangle with
vertices at the observer's position, the Sun's geographical position (GP), and
the North Pole has two sides known and you're looking for the relationship
between the third side and the angle at the pole. Using the law of cosines for
sides and plugging in the observer latitude $L$ and The Sun's declination $D$ gives:

$$\cos(a) = \cos(90-L)\cos(90-D) + \sin(90-L)\sin(90-D)\cos(A)$$

Here, $a$ is the arc distance from observer to Sun's GP (AKA altitude) and $A$ is the angle at the pole.

For an observer at 80 degrees N latitude:

$$\cos(a) = 0.06645228 + 0.90984373\cos(A)$$

When $a$ is 90 degrees, the observer would see half the Sun above the horizon,
were it not for the pesky atmosphere. Going just beyond, to 90 degrees plus 50
minutes for refraction and semidiameter puts the Sun out of sight.

$$\cos(A) = \frac{-0.01744867-0.06645228}{0.90984373}; A = 84.7089^\circ$$

If the observer is at, say, 75 degrees W longitude, the Sun's Greenwich hour
angle would be 159.7089 degrees, and 10 hours and 39 minutes would have elapsed
since meridian passage of the Sun at Greenwich. GMT/UT is 22:41. Observer's
local civil time (EDT) is 6:41 PM.

> **Comments:**
> **username_3**: To add maths you can use Mathjax, which is $\TeX$ like markup.  so write \$\cos(A)\$  for $\cos(A)$  
> **username_4**: I just did that...  

---

### Answer by username_5 on 2021-06-22. Score: 5

The [transformation](https://en.wikipedia.org/wiki/Celestial_coordinate_system#Equatorial_%E2%86%94_horizontal) from the equatorial coordinates to [altitude](https://en.wikipedia.org/wiki/Horizontal_coordinate_system) (azimuth is not important) is given by
$$\sin{a}=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$

Let's say that $t=0\rm\, s$ at [local solar noon](https://en.wikipedia.org/wiki/Sun_transit_time). Then, [the hour angle](https://en.wikipedia.org/wiki/Hour_angle) is given by
$$h=\frac{360°\cdot t}{86400\rm\,s}$$
and we can reformulate the first equation using above expression to
$$a=\arcsin{(\cos{\frac{360°\cdot t}{86400\rm\,s}}\cos{23.5°}\cos{10°}+\sin{23.5°}\sin{10°})}$$
$$a=\arcsin{(0.90313\cos{\frac{t}{240\rm\,s}}+0.069242)}$$
Your calculator must be set in degrees and $t$ is in seconds.

---

You can also determine the time in seconds since local solar noon of sunset. We just write the above formula
$$\sin{a}=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$
and the altitude $a$ must be 0° since the Sun is on the horizon on the sunset (note that we ignore refraction and the angular size of the solar disc, if you want to include that as well, set $a=-50'=-0.833°$).
$$\sin{0°}=0=\cos{h}\cos{\delta}\cos{\phi}+\sin{\delta}\sin{\phi}$$
Then we just express the hour angle $h$:
$$h=\arccos{(-\tan\delta\tan\phi)}$$
and the time since the local noon:
$$t\_{sunset}=\frac{86400\rm\,s\cdot h}{360°}$$


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