, including day length charts and tables with latitudes for world cities and US towns.http://dc.metblogs.com/2006/12/22/the-longest-night-is-over/ The longest night is over

The actual day of year and the latitude (0

The perceived way of the sun around the planet can be viewed at as the boundary circle of the planet's disc. However, this constellation (in which the sun apparently circles along the disc's boundary) applies only at equinoxes and only at the North pole. The further away one is from the North pole (towards the equator), the more the surrounding circle is tilted along the West-East axis, until it is completely upright (perpendicular to the planet's disc) at the equator.

Furthermore, there is also a shift of the circle away from the disc, along the obliquity of the ecliptic
(connecting the centers of the two circles at an angle of 23.439

The following image shows the tilted and shifted solar circle for the Winter Solstice at 45

The following table calculates the exposed part b in relation to the whole circle. The formulas mention 3 parameters, which signify:

Thanks to David X. Callaway to point this out early in the text to avoid confusion.

Note: The expression "observer" in the remarks refers to a hypothetical observer located on the center of the planet's "disc".

Angle between observer and sun's zenith:

Thanks to Andrew Green for spotting an error which was introduced
while translating from HTML to XLM in formulas

Latitude of observer:

$c\; =\; -Lat$Angle between solar disc and sun's zenith:

$a\; =\; z\; -\; c$Distance from observer to sun's zenith:

$d\; =nbsp1sin(a)$Distance from observer to the center of the sun's circle:

$t\; =\; cos(a)d$Exposed radius part between sun's zenith and sun's circle:

$m\; =\; 1\; +\; tan(c)t$Adjust range:

Angle between center of sun's disc and sunrise or sunset point on the solar circle (not the planet's disc), resp.:

$f\; =\; arccos(1\; -\; m)$Exposed fraction of the sun's circle (0=never...1 = whole day):

$b\; =nbspf180$To get the number of hours the sun shines at the given

The calculation of

and

$m\; =\; 1\; +nbsptan(-Lat)cos(a)sin(a)$Because

Since

The expression

This reduces the calculation of

(Note, that the argument of the *radians*, whereas the arguments
of the *degrees*.

Thanks to Kim Mackay for pointing this out.

)Adjust the limits of

completes the calculation.

Notice, that depending on what your plotting software accepts (deg/rad), you might need to modify the
*Microsoft Excel*, which simplifies to

Thanks to reader justanote for this observation.

Above formulas for the Length of Day

Thanks to Martin Bonda for reminding me to make this clear.

The sun does not appear or disappear just so, a shorter or longer twilight period begins
*before* the start of the day and ends *after* the end of the day, i.e. the twilight affects
the duration of the "dark" night, never the duration of the "bright" day.

For most purposes, it is sufficient to take into consideration the *Civil Twilight* plus the
*Nautical Twilight*, but not the *Astronomical Twilight* (which latter would be interpreted
as fully dark anyway for casual observers).

Civil Twilight is defined as the sun being 6

To some extent the angle also depends from the day of year: It is at the equinoxes that the angle is steepest for any latitude, and on the Northern hemisphere the summer solstice is flattest (also the winter solstice is flatter than at the equinoxes, but not so flat as at the summer solstice). However, the differences along a year are short and extend over some minutes only.

When the planet's so far flat disc is given some height

The twilight angle (sun below horizon), as per above definition:

$t\; =\; 12$Thickness of the planet's disc:

$h\; =\; tan(t)$The angle

Knowing the angle

The whole radius fraction

*Adjust range:* 0...2 is the valid range (see comments
in the formula table of the preceding section).
Note, that the

Angle between center of sun's disc and lower twilight point on the *solar* circle
(not the planet's disc):

Exposed fraction of the sun's circle (0=never...1=whole day).
The arc describes the daytime plus both twilight zones (

with the constants

$h\; =\; 12deg$ $j\; =nbsppi182.625$ $Axis\; =\; 23.439deg$This is the calculation of the twilight arc (comprising both twilight durations *and*
the daylength).

Then

$b\; +\; 2e\; =nbsparccos(1\; -\; n)180$completes the calculation.

Some individual twilight durations

Also note, that at and near the Pole there are phases with no twilight, because the sun is present all the day,
or circles too far below the horizon. The values are given as 0, because half the difference
between the daylength b and the arc comprising the day length and the 2 twilights

The following table shows the twilight duration for the civil twilight (6