Here in the northern hemisphere the vernal equinox just passed and the days are
quickly getting longer. One of my colleagues lives in Stavanger, Norway. Our
team’s semi-weekly standup is at 6:30pm his time, so I’ve been accustomed to
seeing the window in his background be pitch black for the past six months.
But from one meeting to the next, his window went from pitch black to bright.
This led me to think about a basic astronomy question I had never given much
thought to before — just how fast do the days get longer? When Spring comes
and the days are getting longer, how many extra minutes of sunlight do we get
from one day to the next? So I built a little interactive graph that shows how
the length of the day changes as a function of latitude, along with how it
changes from one day to the next:

The vertical dashed lines represent the various solstices and equinoxes. As
expected, for northern latitudes the longest day is on the summer solstice and
the shortest on the winter solstice. On the equinoxes the day is exactly 12
hours no matter what latitude you are at, and this is also when the length of
the day is changing the fastest — unless you are very close to the Arctic
circle (latitude 66.55°).

One of the more interesting features I hadn’t appreciated before is that when
you get close to the Arctic circle, the length of the days is essentially a
zigzag, straight up from the winter solstice all the way to the summer solstice
and back down again.

The math behind it

So what calculations are going into producing these curves?

How long is the Sun up for?

The first thing we need to know is how long the Sun is up for on a particular
day of the year. Spherical astronomy has a useful quantity that lets us
determine this, called the hour angle. The hour angle of an object is the
angle it makes with the meridian (the line across the sky going from north to
south). Converting the hour angle into a unit of time (like hours) tells us
how long it will be before the object crosses the meridian (“transits” in the
astronomical parlance). Since we want the time from rising to setting, the
length of daylight will be twice the amount of time it takes between rising and
transit:

[t_{textrm{daylight}} = 2H left( frac{24 , textrm{hr}}{360^{circ}} right)
= frac{2H}{15^{circ}} , textrm{hr}.]

If we can figure out what the Sun’s hour angle is when it rises, we’ll have the
amount of daylight in a day. To get this, we need to know two things: the
observer’s latitude, which we’ll call (lambda); and the declination of the
Sun, (delta), which is the angle of the Sun above the celestial equator.

A diagram (from Prof. Fiona Vincent) helps to illustrate the setup:

The position of the object we’re interested in is labeled with (X) in this
diagram, and it shows the general case where the object is at some arbitrary
altitude (a) above the horizon. (Note that the latitude is labeled (phi)
in this diagram.)

We know the lengths of all three sides of the triangle and want to find the
angle (H). To do this we can turn to the spherical law of cosines:

[cos (90^{circ} – a) = cos (90^{circ} – lambda) cos (90^{circ} – delta)
+ sin(90^{circ} – lambda) sin (90^{circ} – delta) cos H.]

Since we want to know the hour angle of the Sun when it’s rising, we can set
the altitude, (a), to 0. Solving for (H), this becomes

[H = arccos (-tan lambda tan delta).]

This is the so-called “sunrise equation.”

In order to make use of this equation we now need to