Daylight by LatitudeFeb 13, 2025

Every spring, as the seasons change and the sun rises higher each day, I try to figure out how much daylight we are gaining. Turns out there is an equation to calculate it to within a few minutes of accuracy for any latitude!

Astronomy Notes

Static Images:
Total Daylight
Change in Daylight

• The sun rises and sets exactly one day per year at the poles, so they experience either 24 hours of day or night all year round.
• The Arctic and Antarctic Circles are the latitudes where the sun never goes below the horizon for at least one day per year (the Solstice). This is a little different than having exactly one day of complete light or dark, due to refraction, but it is almost the same. You can determine the latitude of the circles in the charts below by the lowest & highest points of the solid-colored sections.
• The latitudes around the Arctic Circle experience the largest changes in daylight per day, and the largest changes occur around the equinoxes.

Hover the charts below for data

Note that the chart may be a little off for high latitudes around the solstices due to approximation in the formula.

The Equation

Source: mathforum.org

Date: 08/17/2000 at 11:13:56
From: David Toomey
Subject: Calculating daylight hours by date and latitude

I did quite a bit of searching, and finally found an article in Ecological Modeling, volume 80 (1995) pp. 87-95, called "A Model Comparison for Daylength as a Function of Latitude and Day of the Year." This article presented a model that apparently does a very good job of estimating the daylight - the error is less than one minute within 40 degrees of the equator, and less than seven minutes within 60 degrees and usually within two minutes for these latitudes.

\(D\) \(L\) \(J\)

Day length Latitude Day of the year

Hours Degrees 1–365

$$ P = \arcsin(0.39795 \cdot \cos( 0.2163108 + 2 \cdot \arctan( 0.9671396 \cdot \tan(0.00860(J - 186)) ) ) $$

$$D = 24 - \frac{{24}}{{\pi}} \cdot \arccos(\frac{{\sin(0.8333 \cdot \frac{{\pi}}{{180}}) + \sin(L \cdot \frac{{\pi}}{{180}} ) \cdot \sin(P)}}{{\cos(L \cdot \frac{{\pi}}{{180}}) \cdot \cos(P)}}) $$

Original ASCII Equations

D = daylength
L = latitude
J = day of the year

P = asin[.39795*cos(.2163108 + 2*atan{.9671396*tan[.00860(J-186)]})]

                          _                                         _
                         / sin(0.8333*pi/180) + sin(L*pi/180)*sin(P) \
D = 24 - (24/pi)*acos   {  -----------------------------------------  }
                         \_          cos(L*pi/180)*cos(P)           _/
                

The Data