10
$\begingroup$

I'm looking for a formula that will return the number of hours per day given a specific location. I was thinking that can be calculated as a difference of sunrise and sunset, but I see that there are some other ways, like in this topic.

What is the best, fast and correct way to calculate this?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Sunrise_equation $\endgroup$
    – John Rennie
    Commented May 18, 2012 at 13:50
  • $\begingroup$ en.wikipedia.org/wiki/Day_length - Do you need analytical solution? I think such function would be very complicated... $\endgroup$
    – Pygmalion
    Commented May 18, 2012 at 16:38
  • $\begingroup$ @Pygmalion I need something that I can further program using simple math functions available in PHP. I have the lat/lng and the date as starting points and I need to calculate some monthly averages, while keeping the individual values also on daily basis. $\endgroup$
    – Elzo Valugi
    Commented May 18, 2012 at 17:26
  • $\begingroup$ You can find declination formula needed in John's reference here en.wikipedia.org/wiki/Declination. Therefore the solution between Artic and Antartic circle is finalized. I am puzzled about the rest. $\endgroup$
    – Pygmalion
    Commented May 18, 2012 at 17:50
  • $\begingroup$ jgiesen.de/daylight $\endgroup$
    – anna v
    Commented May 18, 2012 at 18:41

5 Answers 5

Reset to default
9
$\begingroup$

I think that

provide enough information. You put the equation from the second link into the equation from the first link. You get hours by multiplying the positive solution $\omega_0$ by $2 \cdot \frac{24\text{h}}{2\pi}$. If the equation from the first link has no solution ($\tan\phi \cdot \tan\delta>1$ ), this means day is either $24\text{h}$ or $0\text{h}$ long.

As far as I checked equations' output, they seem to be consistent.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ The interpretation that $\tan\phi \cdot \tan\delta>1$ means continual day was very helpful. I would add that $\tan\phi \cdot \tan\delta<-1$ means continual night. $\endgroup$
    – Green
    Commented Jun 15, 2017 at 23:09
3
$\begingroup$

The approximations work very poorly for anything beyond mid latitudes. Why not use the exact solution which is actually not complicated? For the exact solution, calculate the hour angle between sunrise and sunset. The Day Length in hours is then equal to (https://www.quora.com/How-can-you-calculate-the-length-of-the-day-on-Earth-at-a-given-latitude-on-a-given-date-of-the-year):

Day Length (hours) = 2 * ha / 15, (equation 1)

where ha is the hour angle of sunrise (sunset) from noon and is equal to (http://www.jgiesen.de/astro/suncalc/calculations.htm):

ha = arccos (cos(90.833) /(cos(L)*cos(δ)) - tan(L)*tan(δ)). (equation 2)

This uses 90.833 as the zenith angle of the sun at sunrise/sunset (i.e., it takes into account a standard value of the refraction at sunrise/sunset for the entire world, which is of course only approximate), L is the latitude, δ is the solar declination. The fraction of daylight in the day at latitude L is then,

2 * ha / (15 * 24). (equation 3)

Equation 3 can be used to plot the day-night terminator.

Note, that when the term in the arccos in equation 2 is greater or lesser than 1.0, then there is either total day or total night depending on the value of the declination). As a guide, δ is zero at equinoxes, positive in the northern summer, and negative in the northern winter (see reference 1). Equation 3 applies for all latitudes and for all times!

There are many javascript librarires that calculate δ, and then use it to plot the day-night terminator on google, bing, openstreet, etc. maps. I frankly don't understand why they don't use the exact expression above, which is not any more complicated than the one they typically use that ignores atmospheric refraction altogether (which is not physical at all).

To prove the point I modified the terminator code for Leaflet written by Jorg Dietrich. An expert Javascript coder should optimize my code. Here is the link.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Note that we have MathJax here, so you can use equations rather than bolded text. See here for more details. $\endgroup$
    – Kyle Kanos
    Commented Mar 20, 2017 at 9:56
1
$\begingroup$

Number of hours of sunlight on nth day of the year =

12+(Max hrs of sunlight -min hrs of sunlight in the year)/2 * sin[(2π/365)*(n-t) ] where t is that day that has 12 hours of sunlight.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Of course this formula is valid between Arctic and Antarctic circle $\endgroup$
    – Pygmalion
    Commented May 18, 2012 at 16:55
  • $\begingroup$ I am actually interested in something that will work and above Arctic circle as I am in North Norway and the data is from here. $\endgroup$
    – Elzo Valugi
    Commented May 18, 2012 at 17:24
  • 1
    $\begingroup$ @ElzoValugi Above the Arctic circle, the sun stays up 24 hours a day for several days. Do you want the answer "24" or do you want the number of hours the sun stays up total, something like "48" to mean "up for 48 hours in a row"? $\endgroup$
    – user854
    Commented May 21, 2017 at 15:11
  • $\begingroup$ Based on the way he phrased the question, I suspect he just wanted the hours of daylight for a single day. $\endgroup$
    – Josh C
    Commented Jul 29, 2020 at 21:13
1
$\begingroup$

I have done some number crunching and found an equation for this exact question. For the town I live in the equation is 730-198sin ((2Pit/365)+(Pi/2) For any given location the 730 is the average between the longest and shortest day in minutes. This number may vary just slightly. For the 198sin, the coefficient 198 is the difference between the longest and shortest day in minutes. The variable is t for days from the winter solstice (winter solstice is day 0). The remainder of the equation is just a constant.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ You should include your reasonings and calculations here to improve your question. $\endgroup$
    – rmhleo
    Commented Jan 21, 2016 at 14:51
  • 1
    $\begingroup$ wordpress.barrycarter.org/index.php/2012/02/13/… has a similar but not identical calculation for Albuquerque. The further north you are, the less accurate a simple sinusoidal approximation will be. $\endgroup$
    – user854
    Commented May 21, 2017 at 15:13
1
$\begingroup$

\begin{align}T_{0} &=\frac{T}{\pi}\arccos\left(\frac{\tan\varphi\sin\theta}{\sqrt{\tan^{2}\alpha+\cos^{2}\theta}}\right) \\ \textrm{daylength} &=\begin{cases} T-T_{0} & \frac{\pi}{2}\le\alpha\le3\frac{\pi}{2}\\ T_{0} & \textrm{otherwise} \end{cases}\end{align} where $T=24\ \textrm{hours}$, $\varphi$ is latitude, $θ = 23.4°$ is Earth's axial tilt, and $α$ is the angle from Sun to Earth, calibrated to zero when (Sun)-(Earth's South Pole)-(Earth's North Pole) are on the same plane. Conversion from $α$ to day in year can be done directly.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.