Julian Date

There are several methods for finding the Julian date. The following one, developed by Fliegel and Van Flandern in 1968, is well adapted for computer programs. Let y be the year (with all four digits), m the month and d the day. The Julian date J at noon is then

- (3{[y + (m - 9)/7]/100 +1})/4 + 275m/9 + d +1721029 .

The division here means an integer division, the decimal part being truncated: e. g. 7/3 = 2 and -7/3 = -2.

Example. Find the Julian date on January 1, 1990. Now y = 1990, m = 1 and d = 1.

- 3 x {[1990 + (1 - 9)/7]/100 +1}/4 + 275 x 1/9 +1 +1,721,029

= 730,330 - 3482 - 15 + 30 +1 +1,721,029 = 2,447,893 .

Astronomical tables usually give the Julian date at 0 UT. In this case that would be 2,447,892.5.

The inverse procedure is a little more complicated. In the following J is the Julian date at noon (so that it will be an integer):

a = J + 68,569 , b = (4a)/146,097 , c = a - (146,097b + 3)/4 , d = [4000(c +1)]/1,461,001, e = c - (1461d)/4 + 31, f = (80e)/2447, day = e - (2447f)/80 , g = f/11, month = f + 2 - 12g , year = 100(b - 49) + d + g .

Example. In the previous example we got J = 2,447,893. Let's check this by calculating the corresponding calendar date:

a = 2,447,893 + 68,569 = 2,516,462 , b = (4 x 2,516,462)/146,097 = 68 , c = 2,516,462 - (146,097 x 68 + 3)/4 = 32,813 , d = [4000(32,813 +1)]/1,461,001 = 89 , e = 32,813 - (1461 x 89)/4 + 31 = 337 , f = (80 x 337)/2447 = 11, day = 337 - (2447 x 11)/80 = 1, g = 11/11 = 1, month = 11 + 2 - 12 x 1 = 1, year = 100(68 - 49) + 89 +1 = 1990 .

Thus we arrived back to the original date.

Since the days of the week repeat in seven day cycles, the remainder of the division J/7 unambiguously determines the day of the week. If J is the Julian date at noon, the remainder of J/7 tells the day of the week in the following way:

Example. The Julian date corresponding to January 1, 1990 was 2,447,893. Since 2,447,893 = 7 x 349,699, the remainder is zero, and the day was Monday. Telescopes Mastery

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