Exercise:
The Sun is at declination -14°.
What will be its hour angle at sunrise
(the moment the top
edge of the Sun first appears over the horizon),
at a latitude of
+56°20'?
At sunrise, the true altitude
of the Sun is a = -0°50'
(allowing for semi-diameter and
horizontal refraction).
Use the formula
cos(H) = {
sin(a) - sin(φ) sin(δ)
} / cos(φ) cos(δ)
where
φ = +56°20' and δ
= -14°.
This gives cos(H) = 0.35,
so H = 69.7° or
290.3°
= 4h39m or 19h21m.
To decide which,
note that
the Sun is to the east of the meridian at sunrise,
so H =
19h21m.
If the Sun is on the local
meridian at 12:03,
what time is sunrise?
and what time is
sunset?
The semi-diurnal arc is
4h39m.
Sunrise is at (12:03 - 4h39m) = 07:24.
Sunset is at
(12:03 + 4h39m) = 16:42.
And when will astronomical twilight start and finish?
It is astronomical twilight
if
the Suns altitude is -18° or higher.
Set a = -18° and use the
same formula again, to obtain
H = 101.55° = 6h46m.
So twilight starts at (12:03 -
6h46m) = 05:17,
and ends at (12:03 + 6h46m) = 18:49.
Back to "sunrise, sunset and twilight".