How to calculate the number of hours of sunlight on a certain day of the year in our partnerschools ?
by Jolien De Boodt and Kato De Malsche

1) What suitable mathematical algorithm should be used  ? 

From geography we learnt that locations with different latitude have different hours of sunlight on each day of the year.

Starting from these data we tried to find a suitable mathematical algorithm in order to be able to calculate the number of hours of sunlight  in each location at any day of the year. On the calendar in each place we found the max and min number of hours of sunlight  around resp June 22 and December 22. We took half of the difference as amplitude and half of the sum as the number of hours of sunlight at the equinox moments and the number of the day throughout the year (1 for Jan 1) as x.     

Partnerschool Vittorio Veneto 
(N 45,983°)
(N 51,168°)
(N 59,275°)
(N 61,683°)


max of sunlight

min of sunlight

03:20  until  19:05

06:50  until  15:29

15,75 h

8,65 h

5:29  until  22:00

08:43  until  16:39  

16,52 h

7,94 h

3:54  until  22:44

9:19  until  15:13

18,60 h

6,11 h

3:28  until  22:53

9:29  until  14:50

19,57 h

5,28 h


                                                               22/06                                                        22/12  

Starting from the max of sunlight and min of sunlight for each of the locations, we could use the general sine algorithm to make a graphical representation of the hours of sunlight from Jan 1 until Dec 31

y = 12.20 + 3.55*sin(2pi/365.25*(x-81))

y = 12.23 + 4.29*sin(2pi/365.25*(x-81))

 y = 12.35 + 6.25*sin(2pi/365.25*(x-81))

y = 12.43 + 6.99*sin(2pi/365.25*(x-81))

Taking a sine curve for the representation of the hours of sunlight happens to be far from reality with greater error the higher the latitude. 

Unfortunately there is no single formula that can be used to accurately predict times of this phenomenon over an acceptably wide range of dates and places. However there exists a fine book about Astronomical Algorithms for performing a wide variety of celestial calculations. It is written by Jean Meeus, a Belgian astronomer from Bruges.

We were given the same better formula by the Astronomy Applications  Department of the university of Gent, Leuven and Brussels. 

Y = 12 + (24/pi)*asin(tan(a°)* tan(d°) )    where    sin(d°) = sin(e°)* sin((x-81)*(360/365.25))  and  a° =  latitude and  e° = 23.44°.

In the following graphes you can see the difference between the sine curve and the graph corresponding to the
better formula.

y(VI) = 12.20 + 24/pi*asin(tan(45.983*pi/180)*sin(23.44*pi/180)*sin((x-81)*2pi/365.25)*((1-(sin(23.44*pi/180)*sin((x-81)*2pi/365.25)^2)^(-1/2)))

y(SN) = 12.23 + 24/pi*asin(tan(51.168*pi/180)*sin(23.44*pi/180)*sin((x-81)*2pi/365.25)*((1-(sin(23.44*pi/180)*sin((x-81)*2pi/365.25)^2)^(-1/2)))

y(GR) = 12.35 + 24/pi*asin(tan(59.275*pi/180)*sin(23.44*pi/180)*sin((x-81)*2pi/365.25)*((1-(sin(23.44*pi/180)*sin((x-81)*2pi/365.25)^2)^(-1/2)))

y(MI)  = 12.43 + 24/pi*asin(tan(61.683*pi/180)*sin(23.44*pi/180)*sin((x-81)*2pi/365.25)*((1-(sin(23.44*pi/180)*sin((x-81)*2pi/365.25)^2)^(-1/2)))


                                             22/06                                        22/12  
Vittorio Veneto (N 45,983°)

                                             22/06                                       22/12    Sint-Niklaas(N 51,168°)


                                              22/06                                       22/12
                                Greaker (N 59,275°)

                                              22/06                                        22/12                                   Mikkeli (N 61,683°)

From the University Almanac Office in Helsinki we got more information about the hours of sunlight  throughout the year for places situated on the polar circle (66.56°NL) and Utsjoki located at 69.87°NL.
Below you'll find the associated graphs (better formula). 

y(UT)  = 13 + 24/pi*asin(tan(69.87*pi/180)*sin(23.44*pi/180)*sin((x-81)*2pi/365.25)*((1-(sin(23.44*pi/180)*sin((x-81)*2pi/365.25)^2)^(-1/2)))

y(POI) = 12.85 + 24/pi*asin(tan(66.56*pi/180)*sin(23.44*pi/180)*sin((x-81)*2pi/365.25)*((1-(sin(23.44*pi/180)*sin((x-81)*2pi/365.25)^2)^(-1/2)))

                                                   22/06                                              22/12

In Utsjoki (69.87° NL) the last half of May, the whole of June and the greatest part of July  it will be fully sunlighted and a small part of November,  the whole of December and half of January it will be completely dark 


For locations at the polar circle (66.56 °NL) the graph shows that around June 22 a few days will be  totally sunlighted 

2) How can it be that the number of hours of sunlight on a certain day of the year is so different from one place to another ?

1) The apparent position of the sun is determined not just by the rotation of the earth about its axis, but also by the revolution of the earth around the sun. One of the reasons is that the plane of the equator is not the same as the plane of the earth's orbit around the sun, but is offset from it by the angle of obliquity.

2) A second reason is that the orbit of the earth around the sun is an ellipse and not a circle, and the apparent motion of the sun is thus not exactly equal throughout the year. The sun appears to be moving fastest when the earth is closest to the sun. The sum of effect 1) + 2)  is called the Equation of Time.

3) But that is not all. Also the longest day is longer than the longest night, and the shortest day is longer than the shortest night, for the reason that sunrise occurs when the upper edge of the disk of the Sun appears on the horizon, and sunset is at the moment when the upper edge disappears below the horizon. These are the instants of first and last direct sunlight; but at these times the center of the Sun's disk is already some minutes of arc vertically below the horizon. In addition the Sun is seen some extra minutes of arc above its actual geometric position on account of atmospheric refraction. Consequently, the length of every day exceeds the time that the center of the Sun is geometrically above the horizon by the intervals of time required for the Sun to move through these extra  minutes of arc in altitude at both rising and setting and this effect shortens the night by the same amount.

This excess is greater the higher the latitude because the path of the Sun at rising and setting is at a smaller angle with the horizon, and more time is required in this slanting motion to cover a given vertical distance.
For observers within a couple of degrees of the equator, the period from sunrise to sunset is always several minutes longer than the night. At higher latitudes in the northern hemisphere, the date of equal day and night occurs before the March equinox. Daytime continues to be longer than nighttime until after the September equinox. In the southern hemisphere, the dates of equal day and night occur before the September equinox and after the March equinox.


Astronomical Algorithms, Jean Meeus, ed.Willmann-Bell, Inc.

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