Simple conversions between Julian Dates and Julian/Gregorian calendar dates, supporting ancient dates (BCE)
pip install juliandate
A Julian Date such as 2440423.345486 indicates the number of days
(and fractional part) since noon January 1, 4713 BCE (Julian) or
November 24, 4714 BCE (Gregorian), a system proposed by Joseph
Scaliger
in 1583. 2440423 is the number of full days, and the fractional
part, 0.345486 indicates 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 the Julian calendar (and prior to 8 CE the Proleptic 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 in 1705426,
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 the Proleptic Gregorian Calendar work 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)
Both to_julian() and to_gregorian() raise a ValueError passed a negative number. The published algorithm is not defined for numbers less than 0.
The 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)
1705425.5
Hours, minutes, seconds, and microseconds are optional so that latter could be:
>>> jd.from_julian(-43, 3, 15)
1705425.5
Julian dates begin at noon, the start of the astronomical 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)
juliandate does not use Python datetime objects because these do
not support dates before 1 CE. Any date that can be represented as a
Python datetime can 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)
As 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 check juliandate's calculations against the US Naval
Observeratory's Julian Date
Converter.
The conversion formulas used by juliandate are valid only for JDN ≥ 0. That is, juliandate calculations cannot be considered correct for dates before Julian January 1, 4713 BCE = Gregorian November 24, 4714 BCE.
juliandate comes with a command-line script jd for converting Juliand Days
$ jd 1705426.25
-43, 3, 13, 18, 0, 0, 0
Use jd -h for more usage details.
The conversion formulas are taken from pages 604 and 606 of:
Seidelmann, P.K. 2006 Explanatory 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.
Bug reports and pull requests are welcome on GitHub at https://github.com/seanredmond/juliandate