Latitude and Longitude and Daylight HoursDate: 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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