Solar time and the equation of time

Builds on: Sections 1.6 and 2.1
Reading time: 40 min
Exercises: 5
Olympiad problems: 0

A sundial and a clock agree on only four days of the year. In early November a sundial runs about 16 minutes ahead of a clock that keeps the mean time of its own place, and in mid-February it falls about 14 minutes behind. The difference comes from the Sun itself. A sundial follows the real Sun, whose day from one noon to the next changes length through the year, while a clock divides every day into the same 24 hours. In this section we see why the Sun's day varies, how far the two kinds of time drift apart, and how to convert between them.

2.2.1Apparent and mean solar time

Before you readMake a guess first

When in the year is a sundial furthest ahead of a clock?

The time a sundial shows is called apparent solar time. It is 12:00 when the Sun crosses the meridian, and in general it is 12 hours more than the Sun's hour angle, so when the Sun is 3h past the meridian it is 15:00. In Section 2.1 the solar day was the time from one noon to the next. Timed with an accurate clock it is not always the same. Around 22 December it lasts about 30 seconds more than 24 hours, and around 16 September about 21 seconds less. These are small differences, but they add up day after day to many minutes.

To keep time with equal hours, astronomers use the mean Sun, an imaginary Sun that moves round the celestial equator at a steady rate and goes round once in the same year as the real Sun. Mean solar time is 12 hours more than the hour angle of the mean Sun, so its days all last exactly 24 hours. Clocks keep mean solar time, or a time based on it (Section 2.3). The difference between apparent and mean solar time is called the equation of time, from an old use of the word equation for a correction.

ResultApparent and mean solar time(2.3)

\begin{aligned} T_{app} & = H_{⊙} + 12 h \\ E & = T_{app} - T_{mean} \end{aligned}

$T_{app}$	apparent solar time, as a sundial shows it
$H_{⊙}$	hour angle of the Sun
$T_{mean}$	mean solar time, as a clock keeping the mean time of the place shows it
$E$	equation of time

With this sign, E is positive when a sundial is ahead of the clock. Some books and some problems use the opposite sign, so a solution should say which sign it uses.

Figure 2.2The equation of time through the yearinteractive

Date in 2026

Equation of time $E$
Part from the shape of the orbit
Part from the tilt of the axis

The figure shows the equation of time

E

through 2026, worked out from the Sun's position. The heavy curve is

E

itself, the thin curve is the part that comes from the shape of Earth's orbit, and the dashed curve is the part that comes from the tilt of Earth's axis. The heavy curve is very nearly the sum of the other two. The slider picks a date, and the table gives the three values on that date.

E

is largest on , at +16.4 min, smallest on , at −14.2 min, and zero four times a year, near , 13 June, 2 September and 25 December.

Quick checkUse the result above

On 26 July the equation of time is −6.6 min. How many minutes after 12:00 by mean solar time does the Sun cross the meridian?

minutes

2.2.2Why the Sun's day changes length

Before you readMake a guess first

Earth is closest to the Sun in early January. How fast does the Sun then move along the ecliptic, compared with early July?

In Section 2.1 the solar day was longer than the sidereal day because the Sun moves eastward among the stars, so Earth has to turn a little more than once before the Sun is back on the meridian. What matters is how far the Sun moves eastward in right ascension each day, since right ascension measures position round the celestial equator, the direction in which the sky turns. If that motion were steady, every solar day would have the same length. It is not steady, for two reasons.

The first is the shape of Earth's orbit. The orbit is an ellipse, and Earth moves fastest when it is closest to the Sun, at perihelion around 3 January, and slowest at aphelion in early July (Chapter 3). Seen from Earth, the Sun moves along the ecliptic by about 1.019° a day in early January and only 0.953° a day in early July. In January Earth has to turn further each day to catch up with the Sun, so the solar days are longer than average. From January to July the real Sun is east of where a steadily moving Sun would be, and it crosses the meridian late. This part of $E$ , the thin curve in Figure 2.2, swings once a year by about 7.7 minutes either way.

The second is the tilt of Earth's axis. Even if the Sun moved along the ecliptic at a steady rate, its motion in right ascension would not be steady. Near the equinoxes the ecliptic crosses the celestial equator at 23.44°, so part of the Sun's motion carries it north or south instead of east, and its right ascension grows by only $\cos 23.44 ° = 0.917$ of its motion along the ecliptic. Near the solstices the Sun moves parallel to the equator, but 23.44° away from it, where the hour circles are closer together, and the same motion along the ecliptic covers $1 / \cos 23.44 ° = 1.090$ times as much right ascension. This part of $E$ , the dashed curve, swings twice a year by about 9.9 minutes either way.

The two parts simply add. Both make the solar days long around the December solstice, when Earth is also near perihelion, so the longest solar days of the year come in late December.

ResultEquation of time from its two causes(2.4)

\begin{aligned} E & \approx - 7.66 min \times \sin M \\ + 9.86 min \times \sin 2 λ_{⊙} \end{aligned}

$M$	mean anomaly, the angle that Earth would have moved round the Sun since perihelion if it moved at a steady rate, $360 °$ times the days since about 3 January divided by 365.26
$λ_{⊙}$	the Sun's ecliptic longitude (Section 1.6), which is 0° at the March equinox

The first term comes from the shape of the orbit and the second from the tilt of the axis. Together they stay within about 20 seconds of the accurate value all year.

Show the derivation

The mean Sun moves at a steady rate, so its right ascension equals the longitude $L$ of a Sun moving steadily along the ecliptic. A body further east crosses the meridian later by 4 minutes of time for each degree, so the equation of time is $E = 4 min \times (L - α)$ , with angles in degrees. We split $L - α$ into two parts,

L - α = (L - λ) + (λ - α) .

The first part comes from the shape of the orbit. For an orbit of small eccentricity $e$ , the true longitude runs ahead of the steady one by $2 e \sin M$ radians (Chapter 3), so $L - λ = - 2 e \sin M$ . With $e = 0.01671$ this is $0.03342$ rad, or 1.915°, times $\sin M$ , and multiplying by 4 min gives $- 7.66 min \times \sin M$ .

The second part comes from the tilt. A Sun at longitude $λ$ on the ecliptic has $\tan α = \cos ε \tan λ$ by (1.24), and to a good approximation

λ - α = \tan^{2} (ε / 2) \sin 2 λ .

For $ε = 23.44 °$ , $\tan^{2} (ε / 2) = 0.0430$ rad, or 2.47°, and multiplying by 4 min gives $9.86 min \times \sin 2 λ_{⊙}$ .

Example 2.4The equation of time on 1 May

On 1 May 2026 the Sun's mean anomaly is 116.0° and its ecliptic longitude is 41.1°. Use (2.4) to estimate the equation of time.

Solution

The two terms are

\begin{aligned} E & \approx - 7.66 \times \sin 116.0 ° + 9.86 \times \sin 82.2 ° \\ = - 6.885 + 9.769 = 2.884 min . \end{aligned}

AnswerA sundial is about 2.9 min ahead of mean time on 1 May. The orbit makes it 6.9 min behind and the tilt 9.8 min ahead, and the accurate value is also +2.9 min.

Example 2.5The largest equation of timefinish it

On 3 November 2026 the Sun's mean anomaly is 299.4° and its ecliptic longitude is 221.1°. Estimate the equation of time.

The start

Here $2 λ_{⊙} = 442.2 °$ , which is the same direction as 82.2°, so by (2.4)

E \approx - 7.66 \times \sin 299.4 ° + 9.86 \times \sin 82.2 ° .

Your stepNow work out the two terms and add them.

minutes

2.2.3The analemma

If we photograph the Sun at the same mean time every few days through a year, from the same spot, its positions trace a tall figure of eight called the analemma. The height of each point is set by the Sun's declination, which carries it up and down between the solstices. Its position to either side of the meridian is set by the equation of time. At 12:00 mean time the apparent solar time is $12 h + E$ , so by (2.3) the Sun's hour angle is $E$ , and the Sun is west of the meridian when $E$ is positive and east of it when $E$ is negative. Sundials built to show mean time often carry an analemma on their noon line for this reason.

Figure 2.3The Sun at noon by the clock, through a year

The figure shows where the Sun stands in the sky over Ankara (39.9° N) at 12:00 local mean time on every day of 2026, looking south, with altitude up and the angle from the meridian across, on the same scale, as a camera pointed south would record it. The dotted line is the meridian. The Sun is highest at the June solstice, at an altitude of 73.5°, and lowest at the December solstice, at 26.6°. It is furthest east of the meridian on 12 February and furthest west on 1 November, within two days of the smallest and largest equation of time. The dates differ slightly because the sideways angle is the hour angle shrunk by the cosine of the declination.

Quick checkUse the result above

At 12:00 local mean time on 11 February, where is the Sun?

Common errorsSection 2.2

Books do not agree on the sign of the equation of time. A solution should say whether $E$ is apparent minus mean time, as here, or the reverse, before using it.
A sundial shows apparent solar time for its own meridian, while a watch keeps the mean time of a zone meridian. Comparing the two also needs the longitude correction of Section 2.3.
The equation of time is not zero at the equinoxes or the solstices. It is zero near 16 April, 13 June, 2 September and 25 December.
The mean anomaly is counted from perihelion in early January, and the ecliptic longitude from the March equinox, 76 days later. Counting both from the same date puts the two parts of $E$ in the wrong place.

ExercisesSection 2.25 questions

2.2.1
On 3 November the equation of time is +16.4 min. When a sundial shows 12:00, how many minutes before 12:00 is it by mean solar time?
minutes
2.2.2
Which time does an ordinary sundial show?
2.2.3
At the March equinox the Sun's ecliptic longitude is 0°, and in 2026 its mean anomaly is then 74.8°. Use (2.4) to estimate the equation of time.
minutes
2.2.4
Why are the solar days longest in late December?
2.2.5
On 11 February the equation of time is −14.2 min. How many degrees east of the meridian is the Sun at 12:00 local mean time?
degrees