Here you will find a method used to convert a certain date given in Julian or in the Gregorian calendar into to the corresponding Julian Day (JD) number, or vice versa.

    • General Remarks:

    The number of the Julian Day or more simply the Julian Day consist in a continuous accounting of days and fractions of days since the start of the year -4712. According to the tradition the Julian Day starts in Greenwich at noon, i.e., at the 12h of Universal Time (UT). If the Julian Day corresponds to an instant measured in Dynamical Time uniform time scale the expression of Ephemeris Day is often used. For instance:

1977 April 26.4 UT = DJ 2443259.9

1977 April 26.4 DT = JDE 2443259.9

    In the following methods, is taken into account the reformulation of the Gregorian calendar, so the day after 4 of October of 1582 (Julian calendar) is the day 15 of October of 1582 (Gregorian calendar).

    The Gregorian calendar was not immediately official adopted by all the countries. It's important to remark that one of the conflicts between astronomers and historians is how to account the years before the year 1. The astronomical method to count negative years only can be used for arithmetical ways.

    Consider the function INT(x), this functional will give the less or equal integer value of the real value x given. For instance:

    Regarding to the negative numbers the correct result of this function should follow the same behaviour. For instance:

INT(-7.83) = -8        (and not equal to –7)

    • Julian Day Calculation:

    The following method it's correct for positive and negative years, but not for negative julian years. Taken Y as the year, M as the number corresponding to the month (1 for January, 2 to February,…), D as the day of the month of a certain date of a calendar, we have:

        If M > 2, we will keep Y and M;

        Else,

          If ((M = 1) or (M = 2)), we change Y to Y – 1, and M to M + 12;

        In other words, if the date is in January or February we consider it as being the thirteenth or fourteenth month from the previous year.

        Calculate in the Gregorian calendar:

        A = INT(Y/100)

        B = 2 - A + INT(A/4)

        In the Julian calendar we make B = 0.

        The corresponding Julian Day is then

        JD= INT(365.25(Y+4716)) + INT(30.6001(M+1)) + D + B - 1524.5

        The number 30.6 (instead of) should generate a correct result, but the value 30.6001 it's used to obtain the integer value.

    In the previous procedure, the constant 4716 was added to the argument of the first INT() function, to avoid the problems regarding to the negative years.

    • When is a year a Leap Year?

    In the Julian calendar a year is called a Leap Year (with 366 days) when it is divisible by 4. All the other years are common years and have only 365 days. For instance the years 900 and 1236 are leap years, while the years 750 and 1429 are common years.

    The same law can be used in the Gregorian calendar with an exception: the centurial years that are not divisible by 400 such as 1700, 1800 and 1900 are common years the other ones are leap years (1600, 2000,…).

    It's often use the term Julian Modified Day (JDM) in the most recent studies, contrarily to the Julian Day, this Day starts at Midnight in Greenwich and it is given by:

JDM = JD - 2400000.5

    Consequently, DJM = 0.0 correspond to the 0h UT of the day 17 of November of 1858.

    • Calculation of the Gregorian Calendar Date from the Julian Day:

    One more time, this method is valid not only for positive years but also for the negative ones, but no for negative julian years.

        Add 0.5 to the JD and let Z be the integer part and F the decimal part of the result.

        If Z < 2299161, we take A = Z;

        Else,

              If (Z ≥ 2291161), we must calculate:

                a = INT((Z-1867216.25)/36524.25)

                    A = Z + 1 + a - INT(a/4)

        Now calculate:

                B = A + 1524

                C = INT((B-122.1)/365.25)

                D = INT(36525×C)

                E = INT((B-D)/30.6001)

        The day of the month (with decimal part in the case of existence) is:

            B – D – INT(30.6001×E) + F

        In this procedure the number 30.6001 cannot be replaced by 30.6 even the computer generate the correct result.