A simple sunrise-sunset algorithm taken from http://williams.best.vwh.net/sunrise_sunset_algorithm.htm in JavaScript

computeSunrise.js

function computeSunrise(day, sunrise) {

    /*Sunrise/Sunset Algorithm taken from
        http://williams.best.vwh.net/sunrise_sunset_algorithm.htm
        inputs:
            day = day of the year
            sunrise = true for sunrise, false for sunset
        output:
            time of sunrise/sunset in hours */

    //lat, lon for Berlin, Germany
    var longitude = 13.408056;
    var latitude = 52.518611;
    var zenith = 90.83333333333333;
    var D2R = Math.PI / 180;
    var R2D = 180 / Math.PI;

    // convert the longitude to hour value and calculate an approximate time
    var lnHour = longitude / 15;
    var t;
    if (sunrise) {
        t = day + ((6 - lnHour) / 24);
    } else {
        t = day + ((18 - lnHour) / 24);
    };

    //calculate the Sun's mean anomaly
    M = (0.9856 * t) - 3.289;

    //calculate the Sun's true longitude
    L = M + (1.916 * Math.sin(M * D2R)) + (0.020 * Math.sin(2 * M * D2R)) + 282.634;
    if (L > 360) {
        L = L - 360;
    } else if (L < 0) {
        L = L + 360;
    };

    //calculate the Sun's right ascension
    RA = R2D * Math.atan(0.91764 * Math.tan(L * D2R));
    if (RA > 360) {
        RA = RA - 360;
    } else if (RA < 0) {
        RA = RA + 360;
    };

    //right ascension value needs to be in the same qua
    Lquadrant = (Math.floor(L / (90))) * 90;
    RAquadrant = (Math.floor(RA / 90)) * 90;
    RA = RA + (Lquadrant - RAquadrant);

    //right ascension value needs to be converted into hours
    RA = RA / 15;

    //calculate the Sun's declination
    sinDec = 0.39782 * Math.sin(L * D2R);
    cosDec = Math.cos(Math.asin(sinDec));

    //calculate the Sun's local hour angle
    cosH = (Math.cos(zenith * D2R) - (sinDec * Math.sin(latitude * D2R))) / (cosDec * Math.cos(latitude * D2R));
    var H;
    if (sunrise) {
        H = 360 - R2D * Math.acos(cosH)
    } else {
        H = R2D * Math.acos(cosH)
    };
    H = H / 15;

    //calculate local mean time of rising/setting
    T = H + RA - (0.06571 * t) - 6.622;

    //adjust back to UTC
    UT = T - lnHour;
    if (UT > 24) {
        UT = UT - 24;
    } else if (UT < 0) {
        UT = UT + 24;
    }

    //convert UT value to local time zone of latitude/longitude
    localT = UT + 1;

    //convert to Milliseconds
    return localT * 3600 * 1000;
}

This needs the dayOfYear() function.

function dayOfYear() {
      var yearFirstDay = Math.floor(new Date().setFullYear(new Date().getFullYear(), 0, 1) / 86400000);
      var today = Math.ceil((new Date().getTime()) / 86400000);
      var dayOfYear = today - yearFirstDay;
      return dayOfYear;
  }

I took it from your site… http://thule.mine.nu/html/
Thanks :)

Do you have a unit test? I wrote up the same code but I'm trying to verify that the numbers are correct.

@drewreece nicer implementation:

function dayOfYear() {
                var now = new Date();
		var start = new Date(now.getFullYear(), 0, 0);
		var diff = now - start;
		var oneDay = 1000 * 60 * 60 * 24;
		return Math.floor(diff / oneDay);
	}

from http://stackoverflow.com/a/8619946/861615

If somebody wants to convert return values to date objects, like this: new Date( computeSunrise() ) You need to do following:
Remove these piece of code:

 //convert UT value to local time zone of latitude/longitude
    localT = UT + 1;

And replace return statement by this:

//convert to Milliseconds
	var now = new Date();
	return new Date( now.getFullYear(), now.getMonth(), now.getDate() ).getTime() + ( UT * 3600 * 1000 );

fixed results..
Thanks for sharing the code ;)

Hello,

Does anybody understand this line of the algorithm

//calculate local mean time of rising/setting
T = H + RA - (0.06571 * t) - 6.622;

For me the local mean time should only be a addition of the Sun's local hour and the right ascension. I don't get how the time of the year comes in consideration and where does this constant 6.622 comes from. I found that the 0.06571 is equal to 365.25/24 but it doesn't help me much for my understanding