Straightforward derivation of the sunrise equation

Question

I'm looking for a reasonably straightforward derivation of the sunrise equation$$\cos\omega=-\tan\phi\tan\delta$$as referenced in the Wikipedia article here. I've dipped into Astronomical Algorithms by Jean Meeus, where the formula is given on page 101. However, as far as I can see, the only explanation Meeus gives is by setting $h=0$ in his earlier equation 13.6, which he simply states as if obvious. I've also found an online paper called Derivation of Solar Position Formulae by Ross Ure Anderson, where there is a fuller derivation (starts on page 11), but which I'm having trouble following.

Answer 1

We can start by converting between equatorial coordinates (right ascension $\alpha$ and declination $\delta$) to horizontal coordinates (azimuth $\text{Az}$ and elevation/altitude $a$). If you want to go past this, feel free to skip to the end. We draw a spherical triangle on the sky, with the three points being the object of interest at a given point $X$, the north celestial pole $P$, and the zenith angle of the observer $Z$. Just by the definitions of elevation and declination, you should be able to see that the distance between $X$ and $Z$ is $90^{\circ}-a$ and the distance between $X$ and $P$ is $90^{\circ}-\delta$. The distance between $P$ and $Z$ is then $90^{\circ}-\phi$, with $\phi$ the latitude of the observer. Here's a nice diagram showing the geometry:

^{Image credit: Fiona Vincent, University of St. Andrews.}

The angle between $PZ$ and $PX$ is the hour angle, $H$. We can plug all of the above into the spherical law of cosines to get $$\cos(90^{\circ}-a)=\cos(90^{\circ}-\delta)\cos(90^{\circ}-\phi)+\sin(90^{\circ}-\delta)\sin(90^{\circ}-\phi)\cos(H)$$ or $$\sin a=\sin\delta\sin\phi+\cos\delta\cos\phi\cos H$$ which is the formula you say Meeus gave. Now, at sunrise or sunset, the Sun appears at the horizon, meaning it has an elevation of $a=0$; that's just the definition of the horizon and any object on it. This in turn means that $\sin a=0$, so our equation becomes $$0=\sin\delta\sin\phi+\cos\delta\cos\phi\cos H\implies-\sin\delta\sin\phi=\cos\delta\cos\phi\cos H$$ Rearranging this gives $$\cos H=-\tan\delta\tan\phi$$ as requested.