Day Length & the Equation of Time

Where the Sun rises and sets on any date, and why a sundial and a clock drift a quarter of an hour apart.

Worked example — Day Length & the Equation of Time

Sunrise	06:02
Sunset	18:13
Day length	12h 10m
Solar noon	12:07
Solar declination	−0.04
Equation of time	−7.44
Tilt term	−0.04
Orbit term	−7.42

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Nobody on Earth gets a twelve-hour equinox 🖖

Set the date to 20 March 2026 and the day length card reads 12h 07m at the equator, 12h 10m in London, 12h 16m at 66° N. The extra time is in the definition. An almanac calls it sunrise when the Sun's upper limb touches the horizon, and refraction near the horizon lifts the whole disc by roughly 34 arcminutes, so the centre of the Sun is really 0.833° below the horizon at both ends of the day. Switch the control to Centre on the horizon and the same date drops to 12h 00m at the equator and 11h 59m at 66° N; that last minute is the calendar day not landing on the exact instant of the equinox. Between the polar circles the excess never falls below 6 minutes 39 seconds, and at 66° S it reaches 17 minutes 10 seconds. The same 0.833° allowance is why Tromsø's midnight sun runs 69 days in 2026 while its polar night runs only 48.

The earliest sunset comes days before the shortest day 🖖

Click New York, earliest sunset. The tool puts sunset at 16:28 on 8 December 2026, and on 21 December — the shortest day of that year — sunset has already crept back to 16:32. Sunset is solar noon plus half the day, and both parts move. Through the first three weeks of December the day is still shortening, but solar noon is sliding later by about half a minute a day, because the equation of time is falling from +10.92 min on 1 December to +1.84 min on 21 December. Approaching the solstice the day length stops changing while that slide does not, so sunset turns around first. Drag the latitude and watch the gap close: thirteen days at 40° N, five days at 60° N. At the equator, where day length barely moves all year, the earliest sunset falls on 2 November, seven weeks early. Sweep the whole span from 66° S to 66° N and sunset is still getting later on 21 December at every one of them, so nowhere in that span does the earliest sunset land on the solstice.

The tilt moves the Sun's clock further than the ellipse does 🖖

Ask why a sundial runs ahead of a clock in November and the usual answer is Earth's elliptical orbit. The ellipse is the smaller of the two causes. Writing the equation of time as E = L₀ − α splits it almost exactly in two: the reduction to the ecliptic, which the 23.44° axial tilt produces and the tool prints as the tilt term, and the equation of the centre, which is the ellipse. Over 2026 the tilt term swings ±9.86 minutes and the orbit term ±7.65, a ratio of 1.29. They also run at different rates — the tilt term completes two cycles a year, the orbit term one — and in 2026 their peaks miss each other: their amplitudes sum to 17.51 minutes, and the year's real maximum is +16.46, on 3 November. Add the two cards on any date and you recover the equation-of-time card to within 1.1 seconds. The leftover is the aberration and nutation corrections folded into the Sun's apparent longitude.

Problems solved in full

Thirteen days between New York's earliest sunset and shortest day in 2026 6 steps
New York's shortest day in 2026 is 21 December, and its earliest sunset falls on 8 December. Where do those thirteen days come from, and does every latitude have them?
1. Sunset is solar noon plus half the day. Solar noon in clock time is 720 − 4λ − E + 60Z minutes after midnight, with λ the longitude east, Z the standard-time offset and E the equation of time. Neither λ nor Z changes through the year, so only two quantities can move sunset: E, and the day length D.
  
  $t_{set} = (720 - 4 λ + 60 Z) - E + \frac{1}{2} D$
2. Take the three weeks before the solstice. The tool puts solar noon at 11:45 on 1 December and 11:54 on 21 December, which is 705.1 and 714.2 minutes after midnight. Noon has slipped 9.1 minutes later, and the only term in it that moves is the equation of time, falling from +10.92 min to +1.84 min.
  
  $Δ (- E) = 714.2 - 705.1 = + 9.1 min$
3. Across those same weeks the day shortens from 568.4 to 555.0 minutes, which the tool shows as 9h 28m and 9h 15m. Sunset carries half of that change, so it loses 6.7 minutes.
  
  $\frac{1}{2} Δ D = \frac{1}{2} (555.0 - 568.4) = - 6.7 min$
4. Add the two. Sunset moves 2.4 minutes later, from 989.3 to 991.7 minutes after midnight, over three weeks in which the day got 13.3 minutes shorter.
  
  $Δ t_{set} = + 9.1 - 6.7 = + 2.4 989.3 \to 991.7 min$
5. The turning point is the date where the two rates cancel. On 8 December dE/dn is −0.441 minutes per day and half of dD/dn is −0.433, near enough equal, and the tool prints sunset at 16:28 with a day of 9h 21m. That is the earliest sunset of the year.
  
  $\frac{d E}{d n} = \frac{1}{2} \frac{d D}{d n} ⟹ - 0.441 \approx - 0.433$
6. Now every latitude at once. D depends on latitude, but its change through the year is driven by the declination alone, and the declination is stationary at a solstice, so dD/dn all but vanishes there whatever your latitude. What is left is dt/dn = −dE/dn, which is +0.497 minutes per day on 21 December and carries no latitude in it. Swept from 66° S to 66° N in hundredths of a degree the real rate runs from +0.497 at the equator to +0.411 at the ends, never turning negative, so the earliest sunset lands on the December solstice at none of those latitudes. The argument frays exactly where "all but vanishes" stops being true. At 65.72° N on the June solstice the day is 23h 47m long, and two hundredths of a degree further north it is a full 24 h; there dD/dn is not negligible at all, and sunset really is getting earlier, by 0.11 minutes a day. That band is a few hundredths of a degree wide.
  
  $\frac{d D}{d n}_{n_{sol}} \to 0 \Rightarrow \frac{d t _{set}}{d n}_{n_{sol}} = - \frac{d E}{d n} = + 0.497 > 0$
Answer

16:28 on 8 December, thirteen days ahead of the solstice, because the equation of time outruns the change in day length. The tool shows one place at a time; the last step covers all of them.
A sundial 16 minutes fast, measured in kilometres 6 steps
On 3 November, with the clock kept on GMT, London’s solar noon falls at 11:44. Work out how far east you would have to move for a mean-time clock to agree with the sundial, then decide whether a sundial could ever have found longitude at sea.
1. Solar noon is mean noon shifted by two things, and only one of them matters here. London sits within a thousandth of a degree of the meridian, so the whole displacement is the equation of time, which the panel prints as +16.46 minutes.
  
  $t_{noon} = 720 - 4 λ - E + 60 Z = 720 - 4 (- 0.0015) - 16.46 + 0 = 703.5 min$
2. Time and longitude are one quantity in two units, at a rate the Earth’s rotation fixes: a full turn in 1,440 minutes.
  
  $\frac{1440 min}{36 0 ^{\circ}} = 4 min per degree$
3. The sundial’s lead is therefore worth 4.11° of longitude. A clock kept on mean time at that longitude strikes noon exactly when a London sundial does.
  
  $\frac{16.46}{4} = 4.1 1^{\circ}$
4. In ground distance, at London’s latitude, that is 285 km east.
  
  $111.32 cos 51.4 8^{\circ} = 69.3 km per degree, 4.11 \times 69.3 = 285 km$
5. And it reverses. The equation of time bottoms out at −14.18 minutes on 11 February, so across a year the sundial swings 30.64 minutes, which is 7.66° of longitude.
  
  $E_{m a x} - E_{m i n} = 16.46 - (- 14.18) = 30.64 min = 7.6 6^{\circ}$
6. At the equator that swing is 853 km, and a single minute of clock error is 27.8 km. Those two numbers are why the equation of time is tabulated in every almanac rather than waved away.
  
  $7.6 6^{\circ} \times 111.32 = 853 km, 1 min \to 0.2 5^{\circ} \to 27.8 km$
Answer

4.11° east, about 285 km. The equation of time is a correction rather than an obstacle: it is tabulated, and a dial read to the minute and corrected gives local apparent time to a quarter of a degree, 28 km at the equator. What no dial supplies is the other half of the sum. Longitude is the difference between local time and the time at an agreed place, and a sundial only ever knows that it is noon here. That is why the answer to the longitude problem was a clock that carries the reference, and not a better dial.

$Δ λ = \frac{E}{4 min / ^{\circ}}, Δ x = Δ λ \times 111.32 cos φ$

By the 42Math team · Updated 2026-09-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Equator, March equinox - The March equinox on the equator runs 12h 07m, not twelve hours, because sunrise is counted from the Sun's upper limb through 34 arcminutes of refraction.
London, June solstice - London on the June solstice with the clocks on BST: sunrise 04:43, sunset 21:21, 16h 38m of daylight and solar noon at 13:02.
New York, earliest sunset - New York's earliest sunset of 2026 is 16:28 on 8 December, thirteen days before the shortest day, on which sunset is already back to 16:32.
Greenwich, the sundial's biggest lead - 3 November 2026, the day the equation of time peaks: at Greenwich a sundial reads 16.46 minutes ahead of the clock and solar noon falls at 11:44.
Tromsø, midnight sun - Tromsø on the June solstice: the Sun never sets, so there is no sunrise and no sunset to print, a day length of 24h 00m, and solar noon still at 11:46.