Sunrise, sunset and twilight

Builds on: Sections 1.4, 1.5, 2.2 and 2.3
Reading time: 40 min
Exercises: 4
Olympiad problems: 1

In Ankara the shortest day of the year is 21 December, yet the earliest sunset comes on 8 December, at 17:23, and the latest sunrise not until 4 January, at 08:10. On the shortest day itself the Sun sets three minutes later than on 8 December and rises four minutes earlier than on 4 January. The reason is the equation of time of Section 2.2. In this section we work out the times of sunrise, sunset and twilight by the clock, and see how the air changes them.

2.6.1Sunrise and sunset by the clock

Before you readMake a guess first

In Ankara, when in December does the Sun set earliest by the clock?

(1.21) gives the hour angle $H_{0}$ at which the Sun sets, with $h_{0} = - 0.83 °$ for the top edge of the disc seen through the air. The Sun rises at hour angle $- H_{0}$ , so by (2.3) it rises at an apparent solar time of $12 h - H_{0}$ and sets at $12 h + H_{0}$ , with $H_{0}$ in hours at 15° to the hour. The chain of Section 2.3 then turns apparent solar time into the time on the clock.

ResultSunrise and sunset by the clock(2.13)

\begin{aligned} t_{rise} & = 12 h - H_{0} - E - λ / 15 ° + Z \\ t_{set} & = 12 h + H_{0} - E - λ / 15 ° + Z \end{aligned}

$t_{rise}, t_{set}$	times of sunrise and sunset by the clock
$H_{0}$	hour angle of sunset from (1.21) with $h_{0} = - 0.83 °$ , in hours
$E$	equation of time
$λ$	longitude, positive east
$Z$	offset of the clocks from UT, in hours

Half way between sunrise and sunset is apparent noon, at

12 h - E - λ / 15 ° + Z

by the clock.

The result explains the earliest sunset. Sunrise and sunset sit symmetrically about apparent noon, $H_{0}$ before and after it. In December apparent noon moves later by the clock by about half a minute a day, as the equation of time falls, while $H_{0}$ shrinks until the solstice and then grows. Sunset starts to come later as soon as $H_{0}$ shrinks by less than half a minute a day, which happens about two weeks before the solstice. In the same way, sunrise goes on getting later until about two weeks after it.

Figure 2.7Sunrise, sunset and twilight through the yearinteractive

City

Date in 2026

Sunrise and sunset
Length of the day
Fully dark

The figure shows the times of sunrise and sunset through 2026 by the clocks of the chosen city, drawn heavy, worked out from the Sun's position each day. The thin, dashed and dotted curves show where civil, nautical and astronomical twilight begin in the morning and end in the evening. The clocks are on standard time all year. The slider picks a date. In Ankara the earliest sunset comes on and the latest sunrise on , both away from the solstice. Near the equator, in , sunrise and sunset change little through the year and twilight is short. In , at 64° N, the Sun sets just after midnight and it never gets fully dark.

Example 2.12Sunrise and sunset in Ankara in June

On 21 June the Sun's hour angle at sunset in Ankara (39.93° N, 32.86° E, UT+3) is 112.53°, as found in Example 1.14, and the equation of time is −1.8 min. When do the Sun rise and set by the clock?

Solution

Apparent noon is at $12 : 00 + 1.8 min - 2 : 11.4 + 3 : 00 = 12 : 50.4$ by the clock. The hour angle of sunset is $112.53 / 15 = 7.502$ h, or 7 h 30.1 min, so the Sun rises at $12 : 50.4 - 7 : 30.1 = 5 : 20.3$ and sets at $12 : 50.4 + 7 : 30.1 = 20 : 20.5$ .

AnswerThe Sun rises at about 05:20 and sets at about 20:20, nearly 50 minutes after the times it would have if Ankara's clocks kept its own mean time.

Example 2.13Sunrise in Hanoifinish it

On 28 September 2026 the Sun's declination is −2.02° and the equation of time is +9.3 min. When does the Sun rise in Hanoi (21.03° N, 105.85° E, UT+7)? Give the answer in hours.

The start

By (1.21),

\begin{aligned} \cos H_{0} & = \frac{\sin (- 0.83 °) - \sin 21.03 ° \sin (- 2.02 °)}{\cos 21.03 ° \cos 2.02 °} \\ = - 0.0020, \end{aligned}

so $H_{0} = 90.11 °$ , which is 6.0075 h, or 6 h 0.45 min.

Your stepNow use (2.13) to find the time of sunrise by the clock.

hours

2.6.2Twilight

Before you readMake a guess first

At the equator at an equinox, about how long after sunset does the sky become fully dark?

After sunset the sky stays lit by sunlight scattered from the upper air, and it darkens gradually as the Sun sinks. Twilight is divided into three stages by the Sun's depth below the horizon. Civil twilight lasts until the centre of the Sun is 6° below the horizon. Until then there is enough light to work outdoors, and only the brightest stars and planets show. Nautical twilight lasts until the Sun is 12° below, the depth at which the horizon at sea becomes too dark for sailors to measure the altitudes of stars against it. Astronomical twilight lasts until the Sun is 18° below, when the sky is as dark as it will get and the faintest stars can be seen. The same stages happen in reverse before sunrise. The times follow from (1.21) with $h_{0}$ equal to −6°, −12° or −18°.

How long twilight lasts depends on how steeply the Sun sinks. At the equator at an equinox the Sun goes straight down, and it falls the 17.17° from sunset to the end of astronomical twilight in 69 minutes. Further from the equator the Sun sinks at a slant, so twilight lasts longer. In Ankara it takes 91 minutes at the equinoxes and just over two hours at the June solstice. North of latitude 48.6° the Sun stays less than 18° below the horizon all night around the June solstice, and the white nights of Section 1.4 follow.

ResultRate of change of altitude(2.14)

\frac{d h}{d t} = 15 ° per hour \times \cos φ \sin A

$h$	altitude of a body, here the Sun
$A$	its azimuth, measured from north through east
$φ$	latitude of the observer

The rate is positive in the east, where the body rises, and negative in the west. For a star the rate is 15.04° per hour instead of 15°. A body rises or sets fastest at the equator and when it is due east or due west, and near the poles it skims along the horizon.

Show the derivation

Differentiating (1.19), $\sin h = \sin φ \sin δ + \cos φ \cos δ \cos H$ , with respect to time, with only $h$ and $H$ changing, gives

\cos h \frac{d h}{d t} = - \cos φ \cos δ \sin H \frac{d H}{d t} .

The sine rule in the astronomical triangle gives $\cos δ \sin H = - \cos h \sin A$ , with $A$ measured from north through east. Substituting and dividing by $\cos h$ leaves $d h / d t = \cos φ \sin A d H / d t$ , and $d H / d t$ is 15° per hour for the Sun.

Quick checkUse the result above

The Sun's disc is 0.53° across. At an equinox the Sun sets due west. How many minutes does the disc take to set in Ankara (39.93° N)?

minutes

2.6.3Refraction at the horizon

The air bends light from a body near the horizon so that the body looks higher than it is, by about 34′ at the horizon itself, which is why sunrise is counted at $h_{0} = - 0.83 °$ . The bending is smaller a little higher up, about 29′ half a degree above the horizon, so the lower edge of the rising Sun is lifted more than the upper edge. The Sun therefore looks squashed, about a fifth shorter than it is wide. The amount of bending also changes with the temperature and pressure of the air, so a predicted time of sunrise cannot be trusted to better than a minute or so.

Common errorsSection 2.6

Sunrise and sunset are counted when the top edge of the Sun touches the horizon seen through the air, with $h_{0} = - 0.83 °$ . Taking $h_{0} = 0$ makes the day about ten minutes too short in mid-latitudes.
The Sun's hour angle grows at 15° per hour of ordinary time. The sidereal rate of 15.04° per hour is for stars.
The time on a clock follows from apparent solar time only once the equation of time, the longitude and the zone offset have all been applied, each with its sign.
The three kinds of twilight are set by the depth of the Sun below the horizon, 6°, 12° and 18°, not by fixed lengths of time.

ExercisesSection 2.64 questions

2.6.1
How far below the horizon is the centre of the Sun when astronomical twilight ends?
2.6.2
At latitude 60° at an equinox, the Sun sets due west. Use (2.14) to estimate how long civil twilight lasts after sunset, from $h_{0} = - 0.83 °$ to $- 6 °$ .
minutes
2.6.3
How long is the Sun above the horizon at the equator at an equinox, counting sunrise and sunset at $h_{0} = - 0.83 °$ ?
hours
2.6.4
Why does the earliest sunset of the year come about two weeks before the December solstice?

Olympiad problemsPast IOAA problems, restated1 problem

2018The rising Sun through the air10 points

Adapted from IOAA 2018 (Beijing), theory problem 3. We have restated it in our own words and kept the original data. The original statement of (a) is unclear about where the distances are measured from, and we measure them from the centre of the disc as seen, the reading for which the official answer holds. The solution was last checked on 17 October 2026.

Consider sunrise at Beijing ( $φ = 40 °$ N) on the day of the March equinox.

(a) Just after sunrise, let $r_{l}$ , $r_{d}$ , $r_{r}$ and $r_{u}$ be the distances from the centre of the Sun's disc as seen through the air, where refraction places it, to the edge of the disc towards the left, down, right and up. Put the four in order, using less than, equal to and more than. [4 points]

(b) By how much does the air make the top edge of the Sun appear earlier than it would without an atmosphere? Take the refraction near the horizon as 35′, and consider only the daily motion of the Sun. [6 points]

I solved it without the solution

Answer(a) $r_{d} < r_{u} < r_{l} = r_{r}$ . (b) The top edge appears about 3.05 minutes earlier. The official solution uses the stars' rate of 15.04′ per minute and gets 3.04 minutes, which makes no real difference.