Sun angle and daylight·50°S · October 15
50°S · October 15
13 h 33 min
Noon altitude 49.6°
Declination -9.6° · 05:14 – 18:46

50°S · October 15 — noon altitude 49.6°, daylight 13 h 33 min

At 50°S · October 15 the noon sun stands 49.6° above the horizon and the day runs 13 h 33 min. That altitude is 90° − |latitude − declination| with a declination of -9.6° for the date, and the times are apparent solar time — no refraction, no equation of time.

Latitude
50°S
Date
October 15
Day of the year
288
Declination
-9.6°
Noon altitude
49.6°
Sun at noon
To the north
Length of day
13 h 33 min
Sunrise and sunset
05:14 – 18:46
Half-day angle
101.62°
Shadow of a 1 m pole
0.851 m

Times are apparent solar time, with noon set to 12:00 when the sun crosses the meridian.

A shadow is the tangent of the altitude, read backwards

If the noon altitude is h, a pole 1 m tall casts 1 ÷ tan h metres of shadow. Measure the shadow instead and the altitude falls out, and from the altitude you can work back to a date or a latitude — that is how Eratosthenes measured the earth in antiquity. Shadows stretch fast as the sun drops: at 45° the shadow equals the height, at 30° it is 1.7 times as long, and at 5° more than eleven times.

  • 1 m ÷ tan 49.6°0.851 m

What this leaves out

Sunrise here means the centre of the sun crossing the geometric horizon, an altitude of 0°. The usual convention adds refraction and the sun’s own radius and uses −0.833°, which makes the day 8 to 10 minutes longer at mid-latitudes. The clock is apparent solar time: to reach civil time you still have to add the equation of time (±16 minutes), the gap between your longitude and your time zone, and any summer time. The declination formula carries an error of up to 1.5°, leap years are not counted, so dates after February sit one day out, and neither the observer’s elevation nor hills and buildings on the horizon are in here.

A high sun is a strong sun

The higher the sun, the less air its light passes through, and the stronger the ultraviolet. As a rule of thumb, when your shadow is shorter than you are, the ultraviolet is near its peak for the day — how long skin lasts before it reddens depends on the index and the skin type.

See burn times by UV index

Nearby cells

Same latitude, other dates

Same date, other latitudes

Worth knowing

  • Declination δ = 23.44° × sin(360° × (284 + N) ÷ 365), where N counts the days from 1 January.
  • Noon altitude = 90° − |latitude − declination|; at the June solstice that is 90° − φ + 23.44° in the north.
  • Length of day = 2H ÷ 15 hours with cos H = −tan(latitude) × tan(declination); beyond |cos H| = 1 lies polar day or night.
  • Shadow = height ÷ tan(altitude). Times are apparent solar time, with refraction and the equation of time left out.

Common questions

Q. How high is the noon sun at 50°S · October 15?

49.6° above the horizon. That is 90° − |latitude − declination|, with a declination of -9.6° that day, and the day itself runs 13 h 33 min.

Q. How long is the day at 50°S · October 15?

13 h 33 min. In solar time the sun rises at 05:14 and sets at 18:46, and the half-day angle is 101.62°.

Q. How long is the noon shadow?

A 1 m pole casts 0.851 m at noon. It is 1 ÷ tan 49.6°, so multiplying the shadow by the tangent of the altitude gives the 1 m back.