Calculate daylight hours Based on gegraphical coordinates - algorithm

Question

I want to calculate Daylight hours based on given Latitude and Longitude and DateTime
I mean calculate the time of sunrise and the time of sunset in a specefic Date and based on gegraphic coordinate.

Answer 1

Check this Latitude and Longitude and Daylight Hours
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) _/

Answer 2

Here is a python function that returns the number of hours of daylight with arguments of latitude and day of the year(number between 1-356):
import math
def Daylight(latitude,day):
P = math.asin(0.39795 * math.cos(0.2163108 + 2 * math.atan(0.9671396 * math.tan(.00860 * (day - 186)))))
pi = math.pi
daylightamount = 24 - (24 / pi) * math.acos(
(math.sin((0.8333 * pi / 180) + math.sin(latitude * pi / 180) * math.sin(P)) / (math.cos(latitude * pi / 180) * math.cos(P))))
return daylightamount

Answer 3

I just answered another question, and think that my solution is fitting here aswell. It's a Javascript solution, so you should be able to convert easily to other languages if you need.
I've created a repository under GitHub Sundial it is licenced under the permissive modified BSD license, so you can use it freely in your own projects.
It should be accurate to 0.0001 minutes and takes into account the axial tilt of the earth, and the equation of time.
Sundial AMD Loadable Sun Day Light Calculator
/* Credit and References */
// http://lexikon.astronomie.info/zeitgleichung/ EOT
// http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1989MNRAS.238.1529H&db_key=AST&page_ind=2&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES
// http://code.google.com/p/eesim/source/browse/trunk/EnergySim/src/sim/_environment.py?spec=svn6&r=6
// http://mathforum.org/library/drmath/view/56478.html
// http://www.jgiesen.de/elevaz/basics/meeus.htm
// http://www.ehow.com/how_8495097_calculate-sunrise-latitude.html
// http://www.jgiesen.de/javascript/Beispiele/TN_Applet/DayNight125d.java
// http://astro.unl.edu/classaction/animations/coordsmotion/daylighthoursexplorer.html
// http://www.neoprogrammics.com/nutations/Nutation_In_Longitude_And_RA.php
(function (factory) {
if (typeof define === 'function' && define.amd ) {
// AMD. Register as module
if(typeof dojo === 'object') {
define(["dojo/_base/declare"], function(declare){
return declare( "my.calc.Sun", null, factory());
});
} else {
define( 'Sundial', null, factory());
}
} else {
Sun = new factory();
}
}(function () {
return {
date : new Date(),
getDate : function(){
return this.date;
},
setDate : function(d){
this.date = d;
return this;
},
getJulianDays: function(){
this._julianDays = Math.floor(( this.date / 86400000) - ( this.date.getTimezoneOffset() / 1440) + 2440587.5);
return this._julianDays;
},
// Calculate the Equation of Time
// The equation of time is the difference between apparent solar time and mean solar time.
// At any given instant, this difference will be the same for every observer on Earth.
getEquationOfTime : function (){
var K = Math.PI/180.0;
var T = (this.getJulianDays() - 2451545.0) / 36525.0;
var eps = this._getObliquity(T); // Calculate the Obliquity (axial tilt of earth)
var RA = this._getRightAscension(T);
var LS = this._getSunsMeanLongitude(T);
var deltaPsi = this._getDeltaPSI(T);
var E = LS - 0.0057183 - RA + deltaPsi*Math.cos(K*eps);
if (E>5) {
E = E - 360.0;
}
E = E*4; // deg. to min
E = Math.round(1000*E)/1000;
return E;
},
getTotalDaylightHoursInYear : function(lat){
// We can just use the current Date Object, and incrementally
// Add 1 Day 365 times...
var totalDaylightHours = 0 ;
for (var d = new Date(this.date.getFullYear(), 0, 1); d <= new Date(this.date.getFullYear(), 11, 30); d.setDate(d.getDate() + 1)) {
this.date = d;
// console.log( this.getDaylightHours(lat) );
totalDaylightHours += this.getDaylightHours(lat);
}
return totalDaylightHours;
},
getDaylightHours : function (lat) {
var K = Math.PI/180.0;
var C, Nenner, C2, dlh;
var T = (this.getJulianDays() - 2451545.0) / 36525.0;
this._getRightAscension(T); // Need to get the Suns Declination
Nenner = Math.cos(K*lat)*Math.cos(K*this._sunDeclination);
C = -Math.sin(K*this._sunDeclination)*Math.sin(K*lat)/Nenner;
C2=C*C;
// console.log( T, C2, C, Nenner, lat, K, Math.cos(K*lat) );
if ((C>-1) && (C<1)) {
dlh=90.0 - Math.atan(C / Math.sqrt(1 - C2)) / K;
dlh=2.0*dlh/15.0;
dlh=Math.round(dlh*100)/100;
}
if (C>1) {
dlh=0.0;
}
if (C<-1) {
dlh=24.0;
}
return dlh;
},
_getRightAscension : function(T) {
var K = Math.PI/180.0;
var L, M, C, lambda, RA, eps, delta, theta;
L = this._getSunsMeanLongitude(T); // Calculate the mean longitude of the Sun
M = 357.52910 + 35999.05030*T - 0.0001559*T*T - 0.00000048*T*T*T; // Mean anomoly of the Sun
M = M % 360;
if (M<0) {
M = M + 360;
}
C = (1.914600 - 0.004817*T - 0.000014*T*T)*Math.sin(K*M);
C = C + (0.019993 - 0.000101*T)*Math.sin(K*2*M);
C = C + 0.000290*Math.sin(K*3*M);
theta = L + C; // get true longitude of the Sun
eps = this._getObliquity(T);
eps = eps + 0.00256*Math.cos(K*(125.04 - 1934.136*T));
lambda = theta - 0.00569 - 0.00478*Math.sin(K*(125.04 - 1934.136*T)); // get apparent longitude of the Sun
RA = Math.atan2(Math.cos(K*eps)*Math.sin(K*lambda), Math.cos(K*lambda));
RA = RA/K;
if (RA<0) {
RA = RA + 360.0;
}
delta = Math.asin(Math.sin(K*eps)*Math.sin(K*lambda));
delta = delta/K;
this._sunDeclination = delta;
return RA;
},
// Calculate the Mean Longitude of the Sun
_getSunsMeanLongitude : function(T){
var L = 280.46645 + 36000.76983*T + 0.0003032*T*T;
L = L % 360;
if (L<0) {
L = L + 360;
}
return L;
},
// Nutation in ecliptical longitude expressed in degrees.
_getDeltaPSI : function(T){
var K = Math.PI/180.0;
var deltaPsi, omega, LS, LM;
LS = this._getSunsMeanLongitude(T);
LM = 218.3165 + 481267.8813*T;
LM = LM % 360;
if (LM<0) {
LM = LM + 360;
}
// Longitude of ascending node of lunar orbit on the ecliptic as measured from the mean equinox of date.
omega = 125.04452 - 1934.136261*T + 0.0020708*T*T + T*T*T/450000;
deltaPsi = -17.2*Math.sin(K*omega) - 1.32*Math.sin(K*2*LS) - 0.23*Math.sin(K*2*LM) + 0.21*Math.sin(K*2*omega);
deltaPsi = deltaPsi/3600.0;
return deltaPsi;
},
// T = Time Factor Time factor in Julian centuries reckoned from J2000.0, corresponding to JD
// Calculate Earths Obliquity Nutation
_getObliquity : function (T) {
var K = Math.PI/180.0;
var LS = this._getSunsMeanLongitude(T);
var LM = 218.3165 + 481267.8813*T;
var eps0 = 23.0 + 26.0/60.0 + 21.448/3600.0 - (46.8150*T + 0.00059*T*T - 0.001813*T*T*T)/3600;
var omega = 125.04452 - 1934.136261*T + 0.0020708*T*T + T*T*T/450000;
var deltaEps = (9.20*Math.cos(K*omega) + 0.57*Math.cos(K*2*LS) + 0.10*Math.cos(K*2*LM) - 0.09*Math.cos(K*2*omega))/3600;
return eps0 + deltaEps;
}
};
}));
Demo jsFiddle
You can check out a demo of how you might use it on jsfiddle.
http://jsfiddle.net/wjKRw/
And then when I get around to it, check out the sample use cases at the repository.
GitHub Sundial

Answer 4

sin24+(24cos-18^12)^(day number of the year)+(latitude)^24= #of daylight hours