London Unified Prayer Timetable
London Unified Prayer Timetable

Technical

The London Unified Prayer Timetable is calculated using standard astronomical formulas (see below) for the base times, and UK observational data for Fajr and Isha.

The calculations performed on this website have been matched to an Excel spreadsheet based on the same formulas, then compared to ensure accuracy and transparency.

Times for sunrise, noon, ‘Asr and sunset

Calculations are based on algorithms and equations from Astronomical Algorithms by Jean Meeus (second edition 1998). The equations are listed below.

Coordinates for calculations

The centre of London has traditionally been considered as Charing Cross. This is still the point from which distances to London are measured.

This is not only a good average point for the mosques in inner London, it also lies exactly between the most eastern and western edges of the M25, which is the boundary of the region covered by this prayer timetable.

Latitude and longitude for Charing Cross: 51.5073°N, 0.12755°W.

Fajr and ‘Isha

The times for Fajr and ‘Isha are based on the UK observations of Hizbul Ulama; they determined the times between Fajr and sunrise, and between sunset and ‘Isha.

Accuracy and geographic coverage

The timetable was calculated initially using Microsoft Excel, then using JavaScript in this website. Times are rounded to the nearest minute.

The National Oceanic & Atmospheric Administration makes the following observation about calculations: “although the sunrise and sunset results are theoretically accurate to within a minute, due to variations in atmospheric composition, temperature, pressure and conditions, observed values may vary from calculations.”

The intention is for these prayer times to be applicable anywhere within the M25 around London. The difference in calculated times between the centre and the farthest points east and west is about 100 seconds. To allow for these differences, also for elevations above sea level and for atmospheric refraction, the following adjustments are made:

  • Sunrise: 3 minutes earlier
  • ‘Asr: 2 minutes later
  • Sunset: 3 minutes later

An allowance is made as follows to avoid Zawal (when the Sun is at its highest point):

  • Zuhr: 5 minutes later

These allowances ensure valid times throughout the London region. There will therefore be a 3-minute difference between the prayer timetable and, for example, the sunrise and sunset times on the BBC Weather website and app.

Formulas used in calculations

There follows the sequence of calculations used. These have been reproduced first using Excel, which then has been used as a basis for JavaScript calculations made on this website. The times in Excel and JavaScript when rounded to the nearest minute match exactly.

The Sun’s mean longitude in degrees (in range 0° to 360°):

L0=280.4664567+360007.6982779 r+0.03032028 r2+r349931−r415300−r52000000L_0 = 280.4664567 + 360007.6982779\,r + 0.03032028\,r^2 + \frac{r^3}{49931} - \frac{r^4}{15300} - \frac{r^5}{2000000}

where r is the time measured in Julian millennia. The last three terms can be omitted without affecting desired accuracy.

The Sun’s mean anomaly in degrees:

M=357.5291092+35999.0502909 T−0.0001536 T2−T324490000M = 357.5291092 + 35999.0502909\,T - 0.0001536\,T^2 - \frac{T^3}{24490000}

where T, the time measured in Julian centuries of 36525 ephemeris days from the epoch J2000.0 (2000 January 1.5 TD), is given by

T=JD−245154536525T = \frac{JD - 2451545}{36525}

where JD is the Julian Day. Note that the Julian Ephemeris Day is not introduced, as the difference in accuracy is only a fraction of a second.

The eccentricity of the Earth's orbit:

e=0.016708634−0.000042037 T−0.0000001267 T2e = 0.016708634 - 0.000042037\,T - 0.0000001267\,T^2

Sun’s equation of centre:

C=(1.914602∘−0.004817∘ T−0.000014∘ T2)sin⁡M+(0.019993∘−0.000101∘ T)sin⁡2M+0.000289sin⁡3MC = (1.914602^{\circ} - 0.004817^{\circ}\,T - 0.000014^{\circ}\,T^2)\sin M + (0.019993^{\circ} - 0.000101^{\circ}\,T)\sin 2M + 0.000289\sin 3M

Sun's true longitude ⊙ and true anomaly v in degrees:

⊙=L0+C\odot = L_0 + C
v=M+Cv = M + C

Sun’s radius vector, or the distance between the centres of the Sun and the Earth, expressed in astronomical units:

R=1.000001018 (1−e2)1+ecos⁡vR = \frac{1.000001018\,(1 - e^2)}{1 + e\cos v}

Sun’s apparent longitude in degrees:

λ=⊙−0.00569∘−0.00478∘sin⁡(125.04∘−1934.136∘ T)\lambda = \odot - 0.00569^{\circ} - 0.00478^{\circ}\sin(125.04^{\circ} - 1934.136^{\circ}\,T)

Mean obliquity of the ecliptic in degrees:

ε0=23∘26′21.448′′−46.8150′′ T−0.00059′′ T2+0.001813′′ T3\varepsilon_0 = 23^{\circ}26'21.448'' - 46.8150''\,T - 0.00059''\,T^2 + 0.001813''\,T^3

Corrected obliquity of the ecliptic in degrees:

ε=ε0+0.00256∘cos⁡(125.04∘−1934.136∘ T)\varepsilon = \varepsilon_0 + 0.00256^{\circ}\cos(125.04^{\circ} - 1934.136^{\circ}\,T)

Sun’s right ascension α and declination δ in degrees:

tan⁡α=cos⁡ε sin⁡λcos⁡λ\tan \alpha = \frac{\cos \varepsilon \, \sin \lambda}{\cos \lambda}
sin⁡δ=sin⁡ε sin⁡λ\sin \delta = \sin \varepsilon \, \sin \lambda

The equation of time in degrees:

E=ysin⁡2L0−2esin⁡M+4eysin⁡Mcos⁡2L0−12y2sin⁡4L0−54e2sin⁡2ME = y \sin 2L_0 - 2e \sin M + 4ey \sin M \cos 2L_0 - \tfrac{1}{2} y^2 \sin 4L_0 - \tfrac{5}{4} e^2 \sin 2M

where y=tan⁡2 ⁣(ε2)y = \tan^2\!\left(\tfrac{\varepsilon}{2}\right). Multiply E by 4 to convert from degrees to minutes of time.

The hour angle at sunrise in degrees:

H=cos⁡−1 ⁣(sin⁡(−0.833333∘)−sin⁡φ sin⁡δcos⁡φ cos⁡δ)H = \cos^{-1}\!\left( \frac{\sin(-0.833333^{\circ}) - \sin \varphi \, \sin \delta}{\cos \varphi \, \cos \delta} \right)

where φ is the latitude. The value −0.833333° comprises 16′ for the semidiameter of the Sun and 34′ for atmospheric refraction.

Time, in hours, of solar noon:

m0=12−L15−E60m_0 = 12 - \frac{L}{15} - \frac{E}{60}

where L is the geographic longitude in degrees, measured positively east from Greenwich, negatively to the west (note that if comparing with Jean Meeus, he goes against convention by denoting longitude positively to the west).

Time, in hours, when the length of an object’s shadow equals the length of the object itself plus the length of that object’s shadow at noon:

A1=m0+cos⁡−1 ⁣(sin⁡(cot⁡−1(1+tan⁡(φ−δ)))−sin⁡φ sin⁡δcos⁡φ cos⁡δ)⋅115A_1 = m_0 + \cos^{-1}\!\left( \frac{\sin\left( \cot^{-1}(1 + \tan(\varphi - \delta)) \right) - \sin \varphi \, \sin \delta}{\cos \varphi \, \cos \delta} \right) \cdot \frac{1}{15}

where the division by 15 converts from degrees to hours.

In Excel, where the calculation is in radians, and the result is required as a fraction of one day, the division is instead by 2π (first multiply by 360/2π to convert to degrees, then divide by 15 to convert to hours, then divide by 24 to convert to a fraction of one day).

Similarly, when the shadow equals twice the length:

A2=m0+cos⁡−1 ⁣(sin⁡(cot⁡−1(2+tan⁡(φ−δ)))−sin⁡φ sin⁡δcos⁡φ cos⁡δ)⋅115A_2 = m_0 + \cos^{-1}\!\left( \frac{\sin\left( \cot^{-1}(2 + \tan(\varphi - \delta)) \right) - \sin \varphi \, \sin \delta}{\cos \varphi \, \cos \delta} \right) \cdot \frac{1}{15}

Time, in hours, of sunrise Sr and sunset Ss respectively:

Sr=m0−H15S_r = m_0 - \frac{H}{15}
Ss=m0+H15S_s = m_0 + \frac{H}{15}