   Associated Topics || Dr. Math Home || Search Dr. Math ### Latitude and Longitude and Daylight Hours

```
Date: 09/21/98 at 16:51:06
From: Hilde Fiva
Subject: Geometry, hours of daylight

Dear Dr. Math,

I am working on a math-project and I was hoping you could help me with
a problem I have.

Do you have a formula for calculating the number of hours of daylight,
the time of the sunrise, and the time of the sunset, given the
latitude and longitude of a certain point on the globe?

Thank you very much.

Sincerely,
Hilde Fiva
```

```
Date: 09/22/98 at 12:40:09
From: Doctor Rick
Subject: Re: Geometry, hours of daylight

Hi, Hilde. What you're asking for isn't simple. I can show you the
geometry of the hours of daylight, but it won't be complete, and
sunrise and sunset times involve additional complications.

The Analemma Web site gives explanations and math for the declination
of the sun (which you will need for my calculation below) and for the
"equation of time," which tells when noon really occurs. Sunrise and
sunset will be equal time intervals before and after noon, but noon is
not at 12:00:

http://www.analemma.com/Pages/framesPage.html

Here is a brief explanation of the length of the day.

On any given day, the sun is at a particular declination (which is the
latitude of the point on earth where the sun is directly overhead).
The terminator (the line connecting all places where the sun is
setting or rising) is a circle that is tilted from a north-south plane
by an angle equal to the declination. Here is a side view:

****|****   Terminator      SUN
*\*    |    ***
**  \ dec|       **
**     \ B |    C    **
*--------\--+-----------o You
*          \ |A      /    *
*            \|   /   lat   *
*             +--------------
*             |\            *
*            | \          *
*           |  \        *
**         |   \     **
**       |    \  **
***    |    *\*
****|****

The terminator is shown edge-on, at the angle of declination from the
vertical. Your latitude is shown as a horizontal line. We are
interested in the point of intersection of the terminator and your
latitude, because that is where you experience sunrise or sunset.

If we call the radius of the earth 1 unit, then distance A is sin(lat).
Therefore distance B is sin(lat)*tan(dec). Distance C, the radius of
the latitude circle, is cos(lat).

Now look down from the north on your latitude circle:

oooo|oooo
**o    |    ooo
**| \    |       oo
**  |  \C  |         oo
*    |   \  |           o
*     |    \ |            o
*      |  B  \|             o
* night+------+      day    o
*      |     /|             o
*     |    / |            o
*    |   /  |           o
**  |  /   |         oo
**| /    |       oo
**o    |    ooo
oooo|oooo

The angle between sunrise and sunset (on the day side of the angle) is

d = 2 cos^-1(-B/C)
= 2 cos^-1(-tan(dec)tan(lat))

The length of the day is 24 hours * d/360. But remember to consult the
Web site above to see why this isn't quite right.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

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

Hello!

This is more of an *answer* than a question. I'm doing some research
related to seasonal affective disorder, and I needed information about
how many hours of sunlight different latitudes get at different times
of the year. I searched your web site, and it looked like a couple
people had asked this question, but it was hard to answer. Apparently
it is very difficult to answer because the earth bulges out in the
middle and hence isn't a sphere and because the earth's orbit is an
ellipse and not a circle and because the earth wobbles a little, etc.

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.

I figured that if other people were having trouble finding this
information, too, maybe it would be worth saving them some time by
letting you know what I found.  So, here's the model:

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)           _/

Use a radian mode here, but latitude should be entered in degrees.

I hope this is helpful!

-David
```

```
Date: 08/17/2000 at 15:03:44
From: Doctor Rick
Subject: Re: Calculating daylight hours by date and latitude

Hi, David. Thanks for the information. I plotted the results of the
formula and, as a sanity check, used the day length to back-calculate
the declination of the sun using the formula in my response above.

I got a curve that looks reasonable, varying between -22.249 degrees on
Dec. 22 and 24.628 on Jun 22 (working with a latitude of 40 degrees).
There is an asymmetry apparent: the earth's tilt is 23.5 degrees, so
the declination should vary between -23.5 and 23.5 degrees. The sun is
apparently shifted about 1.2 degrees to the north. Most likely the
formula takes into account the refraction of the earth's atmosphere,
which makes the sun appear above the horizon when a straight-line
calculation would place it below the horizon.

Yes, there are lots of factors that must have been taken into account
in developing this formula.

I will suggest to our Dr. Math Archivists that they add your information
to one of the items on this subject in our Archives. Thanks again.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/17/2000 at 15:50:26
From: David Toomey
Subject: Re: Calculating daylight hours by date and latitude

Hi, I'm glad the formula seems to work out for you, too. Yes, it does
take into account refraction. To get when the top of the sun is at the
earth's surface using a straight-line calculation, the article suggests
using 0.26667 in place of the constant of 0.8333, or 0.0 for when the
center of the sun is even with the horizon.

Take care!
-David

```
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