Package: juliandate
juliandateSimple conversions betweenJulian
Datesand Julian/Gregorian
calendar dates, supporting ancient dates (BCE)Installationpip install juliandateUsageConverting from Julian Date to Gregorian or Julian Calendar DateA Julian Date such as2440423.345486indicates the number of days
(and fractional part) since noon January 1, 4713 BCE (Julian) or
November 24, 4714 BCE (Gregorian), a system proposed byJoseph
Scaligerin 1583.2440423is the number of full days, and the fractional
part,0.345486indicates that a little more than a third of a day
has passed (since noon).To convert this to a Gregorian calendar date:>>> import juliandate as jd
>>> jd.to_gregorian(2440423.345139)
(1969, 7, 20, 20, 17, 0, 9609)The return value is tuple consisting of the year, month, day, hour (in
24-hour format) minutes, seconds, and microseconds All are integers
and the last value is rounded to the nearest microsecond. This value,(1969, 7, 20, 20, 17, 0, 9609), means July 20, 1969 at 20:17 or 8:17
PM (when Apollo 11 touched down on the moon). There is some
imprecision in the seconds due to floating-point division.For ancient dates, conversions to theJulian
calendar(and prior to
8 CE theProleptic Julian
Calendar)
are supported.>>> jd.to_julian(1705426)
(-43, 3, 15, 12, 0, 0, 0)The negative year indicates BCE. 1 BCE is 0, so -43 means 44 BCE, and
this value is March 15, 44 BCE (the Ides of March)Note that since there is no fractional part of a day in1705426,
this comes out to noon, the start of the Julian Day. If we add half a
day or more, we will be into the next calendar day (March 16):>>> jd.to_julian(1705426.5)
(-43, 3, 16, 0, 0, 0.0)Conversions to theProleptic Gregorian
Calendarwork as well. The Julian calendar date March 15, 44 BCE corresponds to
a Gregorian Calendar date of March 13:>>> jd.to_gregorian(1705426)
(-43, 3, 13, 12, 0, 0)Converting from Gregorian or Julian Calendar Date to Julian DateThe reverse functions are:>>> jd.from_gregorian(1969, 7, 20, 20, 17, 0, 0)
2440423.345138889

>>> jd.from_julian(-43, 3, 15, 0, 0, 0, 0)
1705425.5Hours, minutes, seconds, and microseconds are optional so that latter could be:>>> jd.from_julian(-43, 3, 15)
1705425.5The "Astronomical" DayJulian dates begin at noon, the start of theastronomical
day. This can lead to
some confusion since a calendar day, such as March 15 44 BCE, runs
from 1705425.5 (inclusive) to 1705426.5 (exclusive).# Midnight
>>> jd.to_julian(1705425.5) 
(-43, 3, 15, 0, 0, 0, 0)

# 6 AM
>>> jd.to_julian(1705425.75)
(-43, 3, 15, 6, 0, 0, 0)

# Noon
>>> jd.to_julian(1705426)
(-43, 3, 15, 12, 0, 0, 0)

# 9 PM
>>> jd.to_julian(1705426.25)
(-43, 3, 15, 18, 0, 0, 0)

# Midnight, next calendar day
>>> jd.to_julian(1705426.5)
(-43, 3, 16, 0, 0, 0, 0)If all you care about is the calendar day, add 0.5 to the Julian Date
and take the integral part. This will return noon of the calendar
day. For example:# 6 AM
>>> jd.to_julian(int(1705425.75 + 0.5))
(-43, 3, 15, 12, 0, 0)	

# Noon
>>> jd.to_julian(int(1705426 + 0.5))
(-43, 3, 15, 12, 0, 0)

# 9 PM
>>> jd.to_julian(int(1705426.25 + 0.5))
(-43, 3, 15, 12, 0, 0)NodatetimeObjectsjuliandatedoes not use Pythondatetimeobjects because these do
not support dates before 1 CE. Any date thatcanbe represented as a
Pythondatetimecan easily be converted:>>> import juliandate as jd
>>> from datetime import datetime
>>> datetime(*jd.to_gregorian(2440423.345139))
datetime.datetime(1969, 7, 20, 20, 17, 0, 9609)Imprecision and DomainAs noted above, floating-point math causes some imprecision in the
seconds and microseconds. This is unavoidable since 24ths and 60ths
don't divide equally. This round-trip, for instance, ends up being off
by 7 microseconds. Take care if this is important.>>> jd.to_gregorian(jd.from_gregorian(1969, 7, 20, 20, 17))
(1969, 7, 20, 20, 16, 59, 999993)You can checkjuliandate's calculations against the US Naval
Observeratory'sJulian Date
Converter.The conversion formulas used byjuliandateare valid only for JDN ≥ 0. That is,juliandatecalculations cannot be considered correct for dates before Julian January 1, 4713 BCE = Gregorian November 24, 4714 BCE.Command Line Scriptjuliandatecomes with a command-line scriptjdfor converting Juliand Days$ jd 1705426.25
-43, 3, 13, 18, 0, 0, 0Usejd -hfor more usage details.SourcesThe conversion formulas are taken from pages 604 and 606 of:Seidelmann, P.K. 2006Explanatory Supplement to the Astronomical Almanac: A Revision to the Explanatory Supplement to the Astronomical Ephemeris and the American Ephemeris and Nautical Almanac.Sausalito, University Science Books.and pages 617–619 of:Urban, Sean E., and P. Kenneth Seidelmann, eds.Explanatory Supplement to the Astronomical Almanac. 3. ed. Mill Valley, Calif: University Science Books, 2012.ContributingBug reports and pull requests are welcome on GitHub athttps://github.com/seanredmond/juliandate