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?

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    $\begingroup$ 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. $\endgroup$ Jun 22 at 16:18

The transformation from the equatorial coordinates to altitude (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. Then, the 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: $$t_{sunset}=\frac{86400\rm\,s\cdot h}{360°}$$


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.

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    $\begingroup$ To add maths you can use Mathjax, which is $\TeX$ like markup. so write \$\cos(A)\$ for $\cos(A)$ $\endgroup$
    – James K
    Jun 23 at 18:29
  • $\begingroup$ I just did that... $\endgroup$
    – B--rian
    Jun 24 at 8:33

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