In FORTRAN integer arithmetic, multiplication and division are performed left to right in the order of occurrence, and the absolute value of each result is truncated to the next lower integer value after each operation, so that both 2/12 and -2/12 become 0.
The algorithm[61] is valid for all (possibly proleptic) Gregorian calendar dates after November 23, −4713. Divisions are integer divisions, fractional parts are ignored.
\begin{aligned}
A & = \text{INT} \left( \frac{Y}{100} \right) \\
B & = 2 - A + \text{INT} \left( \frac{A}{4} \right) \\
JD & = \text{INT} (365.25 (Y + 4716)) + \text{INT} (30.6001(M + 1)) + D + B - 1524.5
\end{aligned}
Let Y be the year, M the month number (1 for January, 2 for February, etc., to 12 for December), and D the day of month (with decimals, if any) of the given calendar date.
INT(x) the greatest integer less than or equal to x.
CelesTrak / Revisiting Spacetrack Report #3 - AIAA 2006-6753 - Source code (C++)
Vallado, David A., Paul Crawford, Richard Hujsak, and T.S. Kelso, "Revisiting Spacetrack Report #3," presented at the AIAA/AAS Astrodynamics Specialist Conference, Keystone, CO, 2006 August 21–24.
\begin{aligned}
J_0 = & 367 y - \text{INT} \frac{7 y + \text{INT} \frac{m + 9}{12}}{4} + \text{INT} \frac{275 m}{9} + d + 1,721,013.5 \\
& 1901 \le y \le 2099 \\
& 1 \le m \le 12 \\
& 1 \le d \le 31
\end{aligned}
INT(x) means retaining only the integer portion of x, without rounding (or, in other words, round toward zero). For example, INT(− 3.9) = −3 and INT(3.9) = 3.
\begin{aligned}
J_0 = & 367 y - \text{INT} \left\{ \frac{7 \left[ y + \text{INT} \left( \frac{m + 9}{12} \right) \right] }{4} \right\} + \text{INT} \left( \frac{275 m}{9} \right) + d + 1,721,013.5 \\
& 1901 \le y \le 2099 \\
& 1 \le m \le 12 \\
& 1 \le d \le 31
\end{aligned}
INT (x) means to retain only the integer portion of x, without rounding (or, in other words, round toward zero), that is, INT (−3.9) = −3 and INT (3.9) = 3
//! Convert a ymd_type into a day number/*! The day number is an absolute number of days since the start of count
*/template<typenameymd_type_,typenamedate_int_type_>BOOST_DATE_TIME_INLINEdate_int_type_gregorian_calendar_base<ymd_type_,date_int_type_>::day_number(constymd_type&ymd){unsignedshorta=static_cast<unsignedshort>((14-ymd.month)/12);unsignedshorty=static_cast<unsignedshort>(ymd.year+4800-a);unsignedshortm=static_cast<unsignedshort>(ymd.month+12*a-3);unsignedlongd=ymd.day+((153*m+2)/5)+365*y+(y/4)-(y/100)+(y/400)-32045;returnstatic_cast<date_int_type>(d);}//! Convert a year-month-day into the julian day number/*! Since this implementation uses julian day internally, this is the same as the day_number.
*/template<typenameymd_type_,typenamedate_int_type_>BOOST_DATE_TIME_INLINEdate_int_type_gregorian_calendar_base<ymd_type_,date_int_type_>::julian_day_number(constymd_type&ymd){returnday_number(ymd);}
他と比較しやすいよう表現を変えると、
\begin{aligned}
a & = \text{INT} \left( \frac{14 - M}{12} \right) \\
y & = Y + 4800 - a \\
m & = M + 12 a - 3 \\
JD & = \text{INT} \left( D + \frac{153m + 2}{5} + 365 y + \frac{y}{4} - \frac{y}{100} + \frac{y}{400} - 32045 \right)
\end{aligned}
/* {{{ proto int gregoriantojd(int month, int day, int year)
Converts a gregorian calendar date to julian day count */PHP_FUNCTION(gregoriantojd){zend_longyear,month,day;if(zend_parse_parameters(ZEND_NUM_ARGS(),"lll",&month,&day,&year)==FAILURE){RETURN_THROWS();}RETURN_LONG(GregorianToSdn(year,month,day));}/* }}} */
zend_longGregorianToSdn(intinputYear,intinputMonth,intinputDay){zend_longyear;intmonth;/* check for invalid dates */if(inputYear==0||inputYear<-4714||inputMonth<=0||inputMonth>12||inputDay<=0||inputDay>31){return(0);}/* check for dates before SDN 1 (Nov 25, 4714 B.C.) */if(inputYear==-4714){if(inputMonth<11){return(0);}if(inputMonth==11&&inputDay<25){return(0);}}/* Make year always a positive number. */if(inputYear<0){year=inputYear+4801;}else{year=inputYear+4800;}/* Adjust the start of the year. */if(inputMonth>2){month=inputMonth-3;}else{month=inputMonth+9;year--;}return(((year/100)*DAYS_PER_400_YEARS)/4+((year%100)*DAYS_PER_4_YEARS)/4+(month*DAYS_PER_5_MONTHS+2)/5+inputDay-GREGOR_SDN_OFFSET);}
defjdays2000(utc_time):"""Get the days since year 2000.
"""return_days(dt2np(utc_time)-np.datetime64('2000-01-01T12:00'))defjdays(utc_time):"""Get the julian day of *utc_time*.
"""returnjdays2000(utc_time)+2451545def_days(dt):"""Get the days (floating point) from *d_t*.
"""returndt/np.timedelta64(1,'D')
date that is an interval of x Julian years of 365.25 days away from the epoch J2000 = JD 2451545.0 (TT), still corresponding (in spite of the use of the prefix "J" or word "Julian") to the Gregorian calendar date of January 1, 2000, at 12h TT (about 64 seconds before noon UTC on the same calendar day).10 Like the Besselian epoch, an arbitrary Julian epoch is therefore related to the Julian date by
