Why does the terminator pass through both poles only at the equinoxes?

The terminator is the great circle 90° from the subsolar point. It reaches a pole only when the subsolar point lies on the equator, i.e. when δ = 0. At any other date the circle is tilted by δ and misses the poles, leaving a cap of 90° − |δ| around one pole in continuous daylight and the opposite cap in continuous night.

Is the day really 12 hours everywhere at the equinox?

In this geometric model yes: with δ = 0 the formula cos H₀ = −tan φ tan δ gives H₀ = 90° for every latitude, so sunrise is 6 h before solar noon and sunset 6 h after. Real almanacs list a few minutes more because refraction lifts the Sun by about 0.6° at the horizon and sunrise is timed from the upper limb, not the centre.

Why is the subsolar longitude not exactly 0° at 12:00 UTC on the June solstice?

With the equation of time switched off it is exactly 0°. Switching it on applies the real Sun's lead or lag relative to a uniform clock: around 21 June the Sun crosses the Greenwich meridian about 1.5 minutes late, which shifts the subsolar point roughly 0.4° east. In early November the lag reaches −16 minutes, in mid-February +14 minutes.

Why does a city show "polar day" even though the simulator says it is twilight there?

Polar day means the Sun's centre stays at or above the geometric horizon all day, h ≥ 0. Near the limit latitude the midnight Sun just grazes the horizon, so its altitude is close to zero and the shading there looks like twilight. Move the clock to local midnight and compare the altitude readout with 0°.

Day and Night on Earth: The Terminator Line

Sunlit hemisphere from δ(day) and UTC: globe or map, twilight band, polar day/night limits, city sunrise and sunset.

Date & time

Day of year · 21 Jun172

UTC time · 12:0012 h

Clock speed (hours per second)1 h/s

Include equation of time

View

Show subsolar pointShow latitude / longitude gridShow selected city

City

Sunlit side: sin h = sin φ sin δ + cos φ cos δ cos(λ − λ☉). Declination uses the textbook sine of the day of year; sunrise and sunset are geometric (h = 0, no refraction). City clocks use standard time, no daylight saving.

Shortcuts

Space / Enter — run / pause the clock
R — reset date, time and view
Drag the globe to rotate it

Measured values

Solar declination δ+23.44°

Subsolar point23.4°N, 0.0°E

Equation of time+0.0min

polar day north of66.56° N

polar night south of66.56° S

London — Sun altitude now+61.9° · daytime

Day length16 h 24 min

Solar noon (UTC / local)12:01 / 12:01

Sunrise (UTC / local)03:48 / 03:48

Sunset (UTC / local)20:13 / 20:13

Local time now12:00 (UTC+0)

About this model

The simulator shades the Earth by the altitude of the Sun above the local horizon. Two numbers fix the picture: the solar declination δ, taken from the textbook approximation δ ≈ 23.44° · sin[360°(N − 80)/365.25] for day of year N, and the subsolar longitude, which moves west at 15° per hour of UTC (λ☉ = −15° · (t − 12 h), with an optional equation-of-time correction of up to ±16 minutes). For every point the altitude follows from sin h = sin φ sin δ + cos φ cos δ cos(λ − λ☉); h > 0 is day, 0 to −6° is the civil-twilight band, and the h = 0 great circle is the terminator. Dashed circles at |φ| = 90° − |δ| mark the limits of polar day and polar night. For the chosen city the hour angle at sunrise, cos H₀ = −tan φ tan δ, gives the day length 2H₀/15° and the UTC and standard-time clock times of sunrise, solar noon and sunset. The globe is an orthographic projection you can drag; the flat map is equirectangular. Coastlines are a hand-drawn low-resolution polygon set, so shapes are only approximate. Omitted: atmospheric refraction and the Sun's angular size (which add about 7 minutes of daylight at mid-latitudes), orbital eccentricity in the declination formula, daylight-saving time, and the slight year-to-year drift of the calendar. For the orbital geometry behind δ — tilt, orbit position and the seasons — see the companion simulator astronomy/seasons-axial-tilt (Seasons & Axial Tilt).

Who it's for: Grades 7–11 geography, Earth science and astronomy classes learning why day length changes with season and latitude, and how time zones relate to the Sun.

Key terms

Terminator
Solar declination
Subsolar point
Civil twilight
Polar day and polar night
Hour angle
Day length
Equation of time

How it works

Half of the Earth is always lit by the Sun; the boundary between day and night is the terminator. Its position depends on two things: the Sun's declination δ (which latitude has the Sun overhead — it swings between ±23.44° over the year) and the subsolar longitude, which moves west at 15° per hour as the Earth turns. The simulator shades every point of the globe by the Sun's altitude above its horizon, adds a civil-twilight band down to −6°, and marks the latitudes beyond which the Sun does not set or does not rise (90° − |δ|). For the selected city it solves the hour-angle equation for sunrise and sunset.

Key equations

δ ≈ 23.44° · sin[360° · (N − 80) / 365.25]

λ☉ = −15° · (t_UTC − 12 h + EoT)

sin h = sin φ sin δ + cos φ cos δ cos(λ − λ☉)

cos H₀ = −tan φ tan δ, day length = 2H₀ / 15° per hour

polar day / night beyond |φ| = 90° − |δ|

Frequently asked questions

Why does the terminator pass through both poles only at the equinoxes?: The terminator is the great circle 90° from the subsolar point. It reaches a pole only when the subsolar point lies on the equator, i.e. when δ = 0. At any other date the circle is tilted by δ and misses the poles, leaving a cap of 90° − |δ| around one pole in continuous daylight and the opposite cap in continuous night.
Is the day really 12 hours everywhere at the equinox?: In this geometric model yes: with δ = 0 the formula cos H₀ = −tan φ tan δ gives H₀ = 90° for every latitude, so sunrise is 6 h before solar noon and sunset 6 h after. Real almanacs list a few minutes more because refraction lifts the Sun by about 0.6° at the horizon and sunrise is timed from the upper limb, not the centre.
Why is the subsolar longitude not exactly 0° at 12:00 UTC on the June solstice?: With the equation of time switched off it is exactly 0°. Switching it on applies the real Sun's lead or lag relative to a uniform clock: around 21 June the Sun crosses the Greenwich meridian about 1.5 minutes late, which shifts the subsolar point roughly 0.4° east. In early November the lag reaches −16 minutes, in mid-February +14 minutes.
Why does a city show "polar day" even though the simulator says it is twilight there?: Polar day means the Sun's centre stays at or above the geometric horizon all day, h ≥ 0. Near the limit latitude the midnight Sun just grazes the horizon, so its altitude is close to zero and the shading there looks like twilight. Move the clock to local midnight and compare the altitude readout with 0°.

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