The Julian Period
2013 is year
in the current Julian Period
The Julian period (and the Julian Day) must not be confused with the Julian calendar.
The French scholar Joseph Justus Scaliger (1540-1609), pictured above, was interested in assigning a positive number to every year without having to worry about BC/AD. He invented what is today known as the Julian Period.
The Julian Period probably takes its name from the Julian calendar, although it has been claimed that it is named after Scaliger’s father, the Italian scholar Julius Caesar Scaliger (1484-1558).
Scaliger’s Julian period starts on 1 January 4713 BC (Julian calendar) and lasts for 7980 years. After 7980 years the number starts from 1 again.
Current Julian Day:
Astronomers have used the Julian period to assign a unique number to every day since 1 January 4713 BC. This is the so-called Julian Day (JD) or Julian Day Number (JDN). JD 0 designates the 24 hours from noon TT on 1 January 4713 BC to noon TT on 2 January 4713 BC. (TT=Terrestrial Time, which is roughly equivalent to UTC. The current difference between the two is about one minute.)
This means that at noon TT on 1 January AD 2000, JD 2,451,545 started.
This can be calculated thus:
From 4713 BC to AD 2000 there are 6712 years. In the Julian calendar, years have 365.25 days, so 6712 years correspond to 6712×365.25=2,451,558 days. Subtract from this the 13 days that the Gregorian calendar is ahead of the Julian calendar, and you get 2,451,545.
Often fractions of Julian Days are used, so that 1 January AD 2000 at 15:00 TT is referred to as JD 2,451,545.125.
Note that some people use the term “Julian day number” to refer to any numbering of days. NASA, for example, uses the term to denote the number of days since 1 January of the current year, counting 1 January as day 1.
The following formula explains how to convert a calendar date to a Julian Day. (Click here for a description of the symbols and and the operator ‘mod’.)
|For a date in the Gregorian calendar:|
|For a date in the Julian calendar:|
JD is the Julian Day that starts at noon TT on the specified date.
The algorithm works fine for AD dates. If you want to use it for BC dates, you must first convert the BC year to a negative year (e.g., 10 BC = -9). The algorithm works correctly for all dates after 4800 BC, i.e. at least for all positive Julian Day.
To convert the other way (i.e., to convert a Julian Day, JD, to a day, month, and year) these formulas can be used:
|For the Gregorian calendar, calculate:|
|For the Julian calendar, calculate:|
|The for both calendars, calculate:|
Sometimes a modified Julian Day (MJD) is used which is 2,400,000.5 less than the Julian Day. This brings the numbers into a more manageable numeric range and makes the day numbers change at midnight TT rather than noon.
MJD 0 thus started on 17 Nov 1858 (Gregorian) at 00:00:00 TT.
The Lilian Day Number is similar to the Julian Day, except that Lilian Day Number 1 started at midnight at the beginning of the first day of the Gregorian calendar, that is, 15 October 1582.
The Lilian Day Number was invented by Bruce G. Ohms of IBM in 1986. It is named after Aloysius Lilius.
In calendrical calculations, we frequently use an operation call integer division.
Ordinarily we would say that, for example, 14 divided by 5 is 2.8. But when we use integer division, we discard the decimal fraction and simply state that 14 divided by 5 is 2.
We indicate that we use integer division by enclosing the division between the symbols for example thus: and ,
14/5 = 2
When we perform integer division, the division leaves a remainder. In the case of 14 divided by 5, the remainder is 4: We can subtract 5 twice from 14, and this leaves us with 4.
We use the mathematical operator ‘mod’ to indicate the remainder. This is known as the modulo operator. We therefore have:
14 mod 5 = 4
If you want to use integer division in computer programs, you must realize that different programs use different notations for the operations. The following table shows how integer division and the modulo operator may be written:
|The Calendar FAQ||14/5||14 mod 5|
|Microsoft Excel (English)||INT(14/5)||MOD(14,5)|
|Visual Basic||14 \ 5||14 Mod 5|
|C, C#, C++||14 / 5||14 % 5|
|^||||http://en.wikipedia.org/wiki/Terrestrial_Time, retrieved 2 September 2011.|