# Light day

The clear day , or just day, describes the period from sunrise to sunset . The clear day corresponds to the apparent day orbit of the sun on which it is above the horizon . This path of the sun is also known as the day arc .

Duration of the light day in hours (English: Hours of daylight) depending on the calendar day (English: Day of year) and north latitude (English: Latitude).

The length of the day depends primarily on the geographical latitude of the observation site. Away from the equator , the bright day, which is illuminated by daylight , lasts for different lengths of time in relation to night . In general, the further the latitude of the observation site is from the equator, the greater the seasonal differences between day and night (see figure).

## Duration

Duration of the clear day in hours for 50 ° north latitude
Daily changes in the duration of the clear day in minutes for 50 °  N

In Germany the length of the day is

• at the northernmost point (about 55.05 ° N) near List on Sylt :
• at the southernmost point (about 47.27 ° N) at Haldenwanger Eck :
• on the days of the summer solstice 15 h 43 min
• on the days of the winter solstice 8 h 17 min.

Due to the north-south extension of around 7.8 ° N, the duration of a clear day differs by a maximum of around one and a half hours within Germany.

The seasonal fluctuations in the length of the day between the days of the solstices also show clear differences depending on the width with over 10 hours (List) compared to less than 7½ hours (Haldenwanger Eck); at the 50th degree of latitude (50 ° north, e.g. in Mainz ) the daylight fluctuates by around 8¼ hours over the course of a six-month period (see first illustration in this chapter).

The duration of the clear day (day length) does not change equally quickly over the course of a year. The changes in the length of the day are most pronounced around the date of the equinox: the following day is then almost four and a half minutes longer or shorter at a location with a geographical latitude of 50 ° (see second illustration in this chapter).

### calculation

The time span from sunrise (center of the sun on the horizon) to the highest position of the sun or from the highest position of the sun to sunset (center of the sun on the horizon) - i.e. half the length of a day - can be calculated using the following formula (slight extension due to the refraction of light near the horizon not taken into account):

${\ displaystyle {\ text {Daylight half (in hours [h])}} = {\ frac {\ arccos (- \ tan \ varphi \ cdot \ tan \ delta)} {15 ^ {\ circ}}} \, \ mathrm {h}}$

With

• the latitude of the observation site${\ displaystyle \ varphi}$
• the declination of the sun between 23.44 ° north (northern beginning of summer ) and 23.44 ° south (northern beginning of winter or southern beginning of summer); a certain declination corresponds to one or two certain dates in the year (see first figure of this article)${\ displaystyle \ delta}$
• with a uniform rotation of 360 ° in 24 hours, 15 ° corresponds to one hour (1 h).

If one of the two angles is equal to or equal to zero, the product of the tangent values ​​becomes zero and thus arccos (0) = 90 °, i.e. H. half clear day length = 90 ° h / 15 ° = 6 h and whole clear day length = 12 h. This is the case all year round at the equator ( ) and twice a year at all other locations: on the equinox days or equinoxes (then it is always ). ${\ displaystyle \ varphi}$${\ displaystyle \ delta}$${\ displaystyle \ varphi = 0}$${\ displaystyle \ delta = 0}$

The product of the tangent values ​​becomes 1 when the two angles and add up to ± 90 °. The whole clear day then arithmetically lasts 0 or 24 hours. Such days do not exist in reality because they are constantly changing. The statement applies to places between an arctic circle and the neighboring pole in the sense that the polar day or the polar night begins or ends there. Both apply to both, depending on whether it rises or falls. ${\ displaystyle \ varphi}$${\ displaystyle \ delta}$${\ displaystyle \ delta}$${\ displaystyle \ delta}$

### Deriving the equation

Sky tent at sunrise

The duration of the clear day can, similar to the calculation of the azimuth of the rising sun , be calculated with the side cosine law of spherical trigonometry . To do this, consider the triangle that is spanned in the sky by the north pole , the zenith of the observer and the position of the sun at sunrise. is the geographical latitude of the observer and the current declination of the sun. is the zenith distance of the sun as seen from the observer. The three sides of the triangle are and . At sunrise is the zenith distance of the sun . ${\ displaystyle (NP)}$ ${\ displaystyle (Z)}$${\ displaystyle \ varphi}$${\ displaystyle \ delta}$${\ displaystyle c}$${\ displaystyle a, b, c}$${\ displaystyle a = (90 ^ {\ circ} - \ delta)}$${\ displaystyle b = (90 ^ {\ circ} - \ varphi)}$${\ displaystyle c = 90 ^ {\ circ}}$

${\ displaystyle \ gamma}$is the angle at the North Pole between the observer's meridian and the current sun meridian at sunrise. From the moment of sunrise to noon, i.e. the moment when the sun is on the observer's meridian, the earth rotates around the angle . The earth rotates at an angular speed of . The half of the day therefore lasts and the duration of the clear day is twice that${\ displaystyle \ gamma}$${\ displaystyle 15 ^ {\ circ} / h}$${\ displaystyle {\ frac {\ gamma} {15 ^ {\ circ} / h}}}$${\ displaystyle {\ frac {2 \ gamma} {15 ^ {\ circ} / h}}}$

The angle is calculated as follows: ${\ displaystyle \ gamma}$

 The spherical side cosine law is: ${\ displaystyle \ cos c}$ = ${\ displaystyle \ cos a \ cdot \ cos b + \ sin a \ cdot \ sin b \ cdot \ cos \ gamma}$ For follows: ${\ displaystyle \ gamma}$ ${\ displaystyle \ gamma}$ = ${\ displaystyle \ arccos \ left [{\ frac {\ cos c- \ cos a \ cdot \ cos b} {\ sin a \ cdot \ sin b}} \ right]}$ With : ${\ displaystyle a = (90 ^ {\ circ} - \ delta), b = (90 ^ {\ circ} - \ varphi), c = 90 ^ {\ circ}}$ ${\ displaystyle \ gamma}$ = ${\ displaystyle \ arccos \ left [- \ tan \ delta \ cdot \ tan \ varphi \ right]}$
Dauer des lichten Tages[h] ${\displaystyle ={\frac {2\cdot \arccos \left[-\tan \delta \cdot \tan \varphi \right]}{15^{\circ }/h}}}$

With this formula the duration of the clear day is calculated as a function of latitude and declination . ${\ displaystyle \ varphi}$${\ displaystyle \ delta}$

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## annotation

1. day, m. . In: Jacob Grimm , Wilhelm Grimm : German Dictionary . Hirzel, Leipzig 1854–1961 ( woerterbuchnetz.de , University of Trier).