SunPosition.org

How to calculate the sun’s position: a worked example

Every number on this site comes from a short chain of formulas from Jean Meeus’s Astronomical Algorithms. Here is the whole chain, step by step, for London at 13:00 UTC on 15 October 2026, with every intermediate value, so you can follow it with a calculator or check your own code against it.

Read the method and accuracy notes Start reading

The question

Where is the sun over London at 14:00 on 15 October 2026?

London: latitude 51.5074° north, longitude 0.1278° west. The time: 13:00 UTC, 14:00 British Summer Time. The answer at the end of the chain: azimuth 200.69°, elevation 27.79°, or 27.82° as seen through the atmosphere.

Angles are in degrees; convert them to radians before taking a sine or a cosine. Longitudes east of Greenwich are positive here, so London’s is −0.1278°. Meeus counts them the other way round, which is the most common source of sign errors.

Step 1

Time as a Julian Day

Step 1: time
QuantityFormulaValue
Julian Day (UT)days since noon on 1 January 4713 BC; from Unix time: ms / 86,400,000 + 2,440,587.52461329.04167
Delta Tuniform time minus Earth-rotation time; about 69 s in 2015–203569 s
Julian Ephemeris DayJD + ΔT / 86,4002461329.04247
TJulian centuries from J2000: (JDE − 2,451,545) / 36,5250.267872484

The sun’s motion is computed in uniform time (JDE); the Earth’s rotation, in step 5, uses ordinary UT.

Step 2

Where the sun is on the ecliptic

Step 2: the sun’s true longitude (Meeus, chapter 25)
QuantityFormulaValue
Mean longitude L₀280.46646 + 36,000.76983 T + 0.0003032 T²204.08211°
Mean anomaly M357.52911 + 35,999.05029 T − 0.0001537 T²280.68411°
Eccentricity e0.016708634 − 0.000042037 T − 0.0000001267 T²0.0166974
Equation of the centre C(1.914602 − 0.004817 T − 0.000014 T²) sin M + (0.019993 − 0.000101 T) sin 2M + 0.000289 sin 3M−1.88717°
True longitudeL₀ + C202.19494°
Distance R1.000001018 (1 − e²) / (1 + e cos(M + C))0.997176 AU

Reduce every angle to 0–360° before you go on; T is small, but 36,000 T is not.

Step 3

Nutation, obliquity and aberration

Step 3: from true to apparent longitude (Meeus, chapters 22 and 25)
QuantityFormulaValue
Longitude of the moon’s node Ω125.04452 − 1,934.136261 T326.94264°
Nutation in longitude Δψ−17.20″ sin Ω − 1.32″ sin 2L☉ − 0.23″ sin 2L☾ + 0.21″ sin 2Ω+8.10″
Nutation in obliquity Δε9.20″ cos Ω + 0.57″ cos 2L☉ + 0.10″ cos 2L☾ − 0.09″ cos 2Ω+7.97″
Mean obliquity ε₀23°26′21.448″ − 46.8150″ T − 0.00059″ T² + 0.001813″ T³23.435808°
True obliquity εε₀ + Δε23.438020°
Aberration−20.4898″ / R−20.55″
Apparent longitude λtrue longitude + Δψ + aberration202.19148°

L☉ and L☾ are the mean longitudes of the sun and the moon, 280.4665 + 36,000.7698 T and 218.3165 + 481,267.8813 T. Nutation is the nodding of the Earth’s axis; aberration is the shift caused by the Earth’s own motion. Each is worth only seconds of arc, but together with the small terms of the orbit they decide whether a calculation is good to a thousandth of a degree or a hundredth.

Step 4

Right ascension and declination

Step 4: to the celestial equator (Meeus, chapter 13)
QuantityFormulaValue
Right ascension αatan2(cos ε sin λ, cos λ)200.51896°
Declination δasin(sin ε sin λ)−8.64049°

Use the two-argument arctangent, atan2, so the right ascension lands in the right quadrant. A declination of −8.6° means the sun is overhead at 8.6° south that day, so in London it stays fairly low.

Step 5

Sidereal time and the hour angle

Step 5: the Earth’s rotation (Meeus, chapter 12)
QuantityFormulaValue
Days from J2000 (UT) dJD − 2,451,5459784.04167
Mean sidereal time at Greenwich280.46061837 + 360.98564736629 d + 0.000387933 T² − T³ / 38,710,000219.07555°
Apparent sidereal timemean + Δψ cos ε219.07761°
Local hour angle Hsidereal time + longitude − α, reduced to −180° … +180°18.43085°

The hour angle says how far the sun has moved west of the meridian: 18.43° is about 1 hour 14 minutes after solar noon, which in London that day came at 12:46 BST.

Step 6

Elevation and azimuth

Step 6: to the horizon (Meeus, chapters 13, 16 and 40)
QuantityFormulaValue
Elevation from the Earth’s centreasin(sin φ sin δ + cos φ cos δ cos H)27.78786°
Azimuth from northatan2(sin H, cos H sin φ − tan δ cos φ) + 180°200.69028°
Parallax for the ground8.794″ / R × cos(elevation)7.80″
Elevation from the groundelevation − parallax27.78570°
Refraction (Saemundsson)1.02′ / tan(h + 10.3 / (h + 5.11))0.03184°
Apparent elevationelevation + refraction27.81754°

The azimuth formula measures from the south; adding 180° turns it into the usual compass bearing from north. The parallax correction is tiny for the sun, but it is the same correction that moves the moon by up to a degree.

Checking your work

Common mistakes, and how good the result is

  • Degrees and radians mixed up inside a sine or cosine: the most common bug of all.
  • The sign of the longitude: east positive here, west positive in Meeus.
  • atan instead of atan2, which puts the right ascension or the azimuth in the wrong quadrant half the time.
  • Unreduced angles: reduce to 0–360° after every step that can run large.
  • UT and uniform time: the sun’s position uses uniform time, sidereal time uses UT.

The same chain gives the equation of time, the difference between sundial and clock: L₀ − 0.0057183° − α + Δψ cos ε, here 3.56°, or 14.24 minutes. Against the reference values in Meeus’s worked examples and in NREL’s Solar Position Algorithm, this method lands within a few thousandths of a degree. The sun position map runs exactly these steps for any place and moment, and the article on elevation and azimuth explains what the two final numbers mean.

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