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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

    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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