Moon Phase Calculators

A little while ago the task of calculating lunar phases came to mind. After investigating the various algorithms out there, I collected together this little bunch, (mainly wrapped inside complete 'show all the phases for a year' programs), and hacked them about a bit to work nicely with Javascript . They are presented here for your edification, delight and ripping off.

First, a little demonstration. They all return a single value - the phase day (0 to 29, where 0=new moon, 15=full etc.) for the selected date.

12345 678910 1112131415 1617181920 2122232425 262728293031 January February March April May June July August September October November December 20072008200920102008

Simple: Conway: Trig1: Trig2:

The four algorithms are:

Simple

This simply mods the difference between the date and a known new moon date (1970-01-07) by the length of the lunar period.
For this reason, it is only valid from 1970 onwards.

function Simple(year,month,day)
{
	var lp = 2551443; 
	var now = new Date(year,month-1,day,20,35,0);						
	var new_moon = new Date(1970, 0, 7, 20, 35, 0);
	var phase = ((now.getTime() - new_moon.getTime())/1000) % lp;
	return Math.floor(phase /(24*3600)) + 1;
}

Conway

This is based on a 'do it in your head' algorithm by John Conway. In its current form, it's only valid for the 20th and 21st centuries, but I'm sure John's got refinements. :)

function Conway(year, month, day)
{	
	var r = year % 100;
	r %= 19;
	if (r>9){ r -= 19;}
	r = ((r * 11) % 30) + parseInt(month) + parseInt(day);
	if (month<3){r += 2;}
	r -= ((year<2000) ? 4 : 8.3);
	r = Math.floor(r+0.5)%30;
	return (r < 0) ? r+30 : r;
}

Trig1

This is based on some Basic code by Roger W. Sinnot from Sky & Telescope magazine, March 1985. I don't pretend to understand it - something to do with a first-order approximation to the 'real' calculation of the position of the bodies involved - which I'm still working on... :)

function Trig1(year,month,day) {
	var thisJD = julday(year,month,day);
	var degToRad = 3.14159265 / 180;
	var K0, T, T2, T3, J0, F0, M0, M1, B1, oldJ;
	K0 = Math.floor((year-1900)*12.3685);
	T = (year-1899.5) / 100;
	T2 = T*T; T3 = T*T*T;
	J0 = 2415020 + 29*K0;
	F0 = 0.0001178*T2 - 0.000000155*T3 + (0.75933 + 0.53058868*K0) - (0.000837*T + 0.000335*T2);
	M0 = 360*(GetFrac(K0*0.08084821133)) + 359.2242 - 0.0000333*T2 - 0.00000347*T3;
	M1 = 360*(GetFrac(K0*0.07171366128)) + 306.0253 + 0.0107306*T2 + 0.00001236*T3;
	B1 = 360*(GetFrac(K0*0.08519585128)) + 21.2964 - (0.0016528*T2) - (0.00000239*T3);
	var phase = 0;
	var jday = 0;
	while (jday < thisJD) {
		var F = F0 + 1.530588*phase;
		var M5 = (M0 + phase*29.10535608)*degToRad;
		var M6 = (M1 + phase*385.81691806)*degToRad;
		var B6 = (B1 + phase*390.67050646)*degToRad;
		F -= 0.4068*Math.sin(M6) + (0.1734 - 0.000393*T)*Math.sin(M5);
		F += 0.0161*Math.sin(2*M6) + 0.0104*Math.sin(2*B6);
		F -= 0.0074*Math.sin(M5 - M6) - 0.0051*Math.sin(M5 + M6);
		F += 0.0021*Math.sin(2*M5) + 0.0010*Math.sin(2*B6-M6);
		F += 0.5 / 1440; 
		oldJ=jday;
		jday = J0 + 28*phase + Math.floor(F); 
		phase++;
	}
	return (thisJD-oldJ)%30;
}
function GetFrac(fr) {	return (fr - Math.floor(fr));}

Trig2

This is I think another reworking of the maths in Trig1 - done nicer.

function Trig2(year,month,day) {
	n = Math.floor(12.37 * (year -1900 + ((1.0 * month - 0.5)/12.0)));
	RAD = 3.14159265/180.0;
	t = n / 1236.85;
	t2 = t * t;
	as = 359.2242 + 29.105356 * n;
	am = 306.0253 + 385.816918 * n + 0.010730 * t2;
	xtra = 0.75933 + 1.53058868 * n + ((1.178e-4) - (1.55e-7) * t) * t2;
	xtra += (0.1734 - 3.93e-4 * t) * Math.sin(RAD * as) - 0.4068 * Math.sin(RAD * am);
	i = (xtra > 0.0 ? Math.floor(xtra) :  Math.ceil(xtra - 1.0));
	j1 = julday(year,month,day);
	jd = (2415020 + 28 * n) + i;
	return (j1-jd + 30)%30;
}

Both of the Trig functions require the following routine to work out the Julian Day Number.

function julday(year, month, day) {
	if (year < 0) { year ++; }
	var jy = parseInt(year);
	var jm = parseInt(month) +1;
	if (month <= 2) {jy--;	jm += 12;	} 
	var jul = Math.floor(365.25 *jy) + Math.floor(30.6001 * jm) + parseInt(day) + 1720995;
	if (day+31*(month+12*year) >= (15+31*(10+12*1582))) {
		ja = Math.floor(0.01 * jy);
		jul = jul + 2 - ja + Math.floor(0.25 * ja);
	}
	return jul;
}

The two 'Trig' ones are the more accurate, but the 'Simple' and the 'Conway' are quicker. You could use any of them in a loop (say, when printing a monthly calendar), or just to find out the state of the moon today.
All the routines are contained within a single file here. Take which one you like and strip out the rest. Happy coding!