DataMuseum.dk

top - metrics - download

    Length: 4018 (0xfb2)
    Types: TextFile
    Names: »moontx.c«

└─⟦b20c6495f⟧ Bits:30007238 EUUGD18: Wien-båndet, efterår 1987
    └─⟦this⟧ »EUUGD18/General/Date/moontx.c« 

TextFile

#
/*
 *              Phase of the Moon Conversion.
 *
 * Synopsis
 *
 *      char    *moontxt(buffer, year, month, day)
 *      char    *buffer;        -- Output goes here.
 *      int     year;           -- Year         1979 = 1979
 *      int     month;          -- Month        Jan. = 1
 *      int     day;            -- Day          1    = 1
 *
 * The phase of the moon (in readable ascii) is written into the buffer,
 * followed by a null.  The routine returns a pointer to the trailing null.
 *
 * External routines called:
 *
 *      copyst		(String copy routine)
 */

extern char		*copyst();	/* String output routine        */

static char *phasetxt[] = {
	"new",
	"waxing crescent",
	"in its first quarter",
	"waxing gibbous",
	"full",
	"waning gibbous",
	"in its last quarter",
	"waning crescent"
};

static int day_year[] = {		/* Days in year for each month	*/
	-1, -1, 30, 58, 89, 119, 150, 180, 211, 241, 272, 303, 333
};					/* Note: Jan. 1 will equal zero	*/

char *moontxt(buffer, year, month, day)
char    *buffer;                        /* Where to put the text        */
int     year;                           /* Year, 1978 = 1978            */
int     month;                          /* Month, Jan = 1               */
int     day;                            /* Day, 1 = 1                   */
/*
 * Output the phase of the moon for the given year, month, day.
 * The routine calculates the year's epact (the age of the moon on Jan 1.),
 * adds this to the number of days in the year, and calculates the phase
 * of the moon for this date.
 *
 * In the algorithm:
 *
 *	diy	Is the day of the year - 1 (i.e., Jan 1 is day 0).
 *
 *	golden	Is the number of the year in the Mentonic cycle, used to
 *		determine the position of the calender moon.
 *
 *	epact	Is the age of the calender moon (in days) at the beginning
 *		of the year.  To calculate epact, two century-based
 *		corrections are applied:
 *		Gregorian:	(3 * cent)/4 - 12
 *			is the number of years such as 1700, 1800 when
 *			leap year was not held.
 *		Clavian:	(((8 * cent) + 5) / 25) - 5
 *			is a correction to the Mentonic cycle of about
 *			8 days evry 2500 years.  Note that this will
 *			overflow 16 bits in the year 409600.  Beware.
 *
 * The algorithm is accurate for the Gregorian calender only.
 *	
 * The magic numbers used in the phase calculation are as follows:
 *	 29.5		The moon's period in days.
 *	177		29.5 scaled by 6
 *	 22		(29.5 / 8) scaled by 6 (this gets the phase)
 *	 11		((29.5 / 8) / 2) scaled by 6
 *
 * Theoretically, this should yield a number in the range 0 .. 7.  However,
 * two days per year, things don't work out too well.
 *
 * Epact is calculated by the algorithm given in Knuth vol. 1 (calculation
 * of Easter).  See also the article on Calenders in the Encyclopaedia
 * Britannica and Knuth's algorithm in CACM April 1962, page 209.
 */
{
	int		phase;		/* Moon phase			*/
	register int	cent;		/* Century number (1979 = 20)	*/
	register int	epact;		/* Age of the moon on Jan. 1	*/
	register int	diy;		/* Day in the year		*/
	int		golden;		/* Moon's golden number		*/

	if (month < 0 || month > 12) month = 0;	/* Just in case		*/
	diy = day + day_year[month];		/* Day in the year	*/
	if ((month > 2) && ((year % 4 == 0) && 
			((year % 400 == 0) || (year % 100 != 0))))
		diy++;				/* Leapyear fixup	*/
	cent = (year / 100) + 1;		/* Century number	*/
	golden = (year % 19) + 1;		/* Golden number	*/
	epact = ((11 * golden) + 20		/* Golden number	*/
		+ (((8 * cent) + 5) / 25) - 5	/* 400 year cycle	*/
		- (((3 * cent) / 4) - 12)) % 30;/* Leap year correction	*/
	if (epact <= 0)
		epact += 30;			/* Age range is 1 .. 30	*/
	if ((epact == 25 && golden > 11) || epact == 24)
		epact++;
	/*
	 * Calculate the phase, using the magic numbers defined above.
	 * Note that (phase and 7) is equivalent to (phase mod 8) and
	 * is needed on two days per year (when the algorithm yields 8).
	 */
	phase = (((((diy + epact) * 6) + 11) % 177) / 22) & 7;
	return(copyst(buffer, phasetxt[phase]));
}