儒略日

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儒略日（Julian day）是指由公元前4713年 1月1日，協調世界時中午12時開始所經過的天數，多為天文學家採用，用以作為天文學的單一曆法，把不同曆法的年表統一起來。

儒略日是一种不用年月的长期纪日法，简写为JD。是由法国学者Joseph Justus Scliger（1540年-1609年）在1583年所创，这名称是为了纪念他的父亲——意大利学者Julius Caesar Scaliger（1484年-1558年）。

儒略日的起点订在公元前4713年（天文学上记为 -4712年）1月1日格林威治时间平午（世界时12:00），即JD 0 指定为UT时间B.C.4713年1月1日12:00到UC时间B.C.4713年1月2日12:00的24小时。每一天赋予了一个唯一的数字，顺数而下，如：1996年1月1日12:00:00的儒略日是2450084。这个日期是考虑了太阳、月亮的轨道运行周期，以及当时收税的间隔而订出来的。Joseph Scliger定义儒略周期为7980年，是因28、19、15的最小公倍数为28×19×15=7980。其中：

28年为一太阳周期(solar cycle)，经过一太阳周期，则星期的日序与月的日序会重复。

19年为一太阴周期，或称默冬章(Metonic cycle)，因235朔望月=19回归年，经过一太阴周期则阴历月年的日序重复。

15年为一小纪(indiction cycle)，此为罗马皇帝君士坦丁一世（Constantine I）所颁，每15年评定财产价值以供课税，成为古罗马用的一个纪元单位，

故以7980年为一儒略周期，而所选的起点公元前4713年，则是这三个循环周期同时开始的最近年份。

以儒略日计日是为方便计算年代相隔久远或不同历法的两事件所间隔的日数。

由于儒略日数字位数太多，国际天文学联合会于1973年采用简化儒略日（MJD），其定义为 MJD = JD - 2400000.5。MJD相应的起点是1858年 11月17日世界时0时。

[编辑] 換算

[编辑] 公曆轉儒略日

$\begin{matrix}a & = & \left\lfloor\frac{14 - month}{12}\right\rfloor \\ \\y & = & year + 4800 - a \\ \\m & = & month + 12a - 3 \\\end{matrix}$

適用於格里曆日期（中午）：

$\begin{matrix}JDN & = & day + \left\lfloor\frac{153m + 2}{5}\right\rfloor + 365y + \left\lfloor\frac{y}{4}\right\rfloor - \left\lfloor\frac{y}{100}\right\rfloor + \left\lfloor\frac{y}{400}\right\rfloor - 32045\end{matrix}$

適用於儒略曆日期（中午）：

$\begin{matrix}JDN & = & day + \left\lfloor\frac{153m + 2}{5}\right\rfloor + 365y + \left\lfloor\frac{y}{4}\right\rfloor - 32083\end{matrix}$

一些非中文的文字因为尚未翻譯而被隐藏，歡迎參與翻譯。

The constants used at the end of the 格里和儒略曆轉換公式必須求得相同的JDN for the same date in both calendars 200年 3月1日與300年 2月28日之間。 The constants are the JDNs of 2月29日, −4800 in each calendar. In the proleptic Gregorian calendar the Julian day zero is 公元前4714年 11月24日, which is 32045 days apart from the start of the Gregorian quadricentennial cycle (即400年 cycle starting and ending in a year divisible by 400) containing the Julian day zero, which begins on 公元前4801年 3月1日 in the proleptic Gregorian calendar.

For the full Julian date, not counting 閏秒 (divisions are real numbers):

$\begin{matrix}JD & = & JDN + \frac{hour - 12}{24} + \frac{minute}{1440} + \frac{second}{86400}\end{matrix}$

So, for example, 2000年 1月1日中午的JD = 2451545.0。

The day of the week can be determined from the 儒略日 by calculating it modulo 7, where 0代表星期一。

JDN mod 7	0	1	2	3	4	5	6
星期	一	二	三	四	五	六	天

[编辑] 儒略日轉公曆

一些非中文的文字因为尚未翻譯而被隐藏，歡迎參與翻譯。

Let J be the Julian day number from which we want to compute the date components.
With J, compute a relative Julian day number j from a Gregorian epoch starting on March 1 −4800 (i.e. March 1 4801 BC in the proleptic Gregorian Calendar), the beginning of the Gregorian quadricentennial 32,044 days before the epoch of the Julian Period.
With j, compute the number g of Gregorian quadricentennial cycles elapsed (there are exactly 146,097 days per cycle) since the epoch; subtract the days for this number of cycles, it leaves dg days since the beginning of the current cycle.
With dg, compute the number c (from 0 to 4) of Gregorian centennial cycles (there are exactly 36,524 days per Gregorian centennial cycle) elapsed since the beginning of the current Gregorian quadricentennial cycle, number reduced to a maximum of 3 (this reduction occurs for the last day of a leap centennial year where c would be 4 if it were not reduced); subtract the number of days for this number of Gregorian centennial cycles, it leaves dc days since the beginning of a Gregorian century.
With dc, compute the number b (from 0 to 24) of Julian quadrennial cycles (there are exactly 1,461 days in 4 years, except for the last cycle which may be incomplete by 1 day) since the beginning of the Gregorian century; subtract the number of days for this number of Julian cycles, it leaves db days in the Gregorian century.
With db, compute the number a (from 0 to 4) of Roman annual cycles (there are exactly 365 days per Roman annual cycle) since the beginning of the Julian quadrennial cycle, number reduced to a maximum of 3 (this reduction occurs for the leap day, if any, where a would be 4 if it were not reduced); subtract the number of days for this number of annual cycles, it leaves da days in the Julian year (that begins on March 1).
Convert the four components g, c, b, a into the number y of years since the epoch, by summing their values weighted by the number of years that each component represents (respectively 400 years, 100 years, 4 years, and 1 year).
With da, compute the number m (from 0 to 11) of months since March (there are exactly 153 days per 5-month cycle; however, these 5-month cycles are offset by 2 months within the year, i.e. the cycles start in May, and so the year starts with an initial fixed number of days on March 1, the month can be computed from this cycle by a Euclidian division by 5); subtract the number of days for this number of months (using the formula above), it leaves d days past since the beginning of the month.
The Gregorian date (Y, M, D) can then be deduced by simple shifts from (y, m, d).

We can then develop these formulas into a single inlined formula per component, computed as above. All this computing requires only integers and so is not sensitive to rounding errors caused by floating point approximations (most decimal fractions have an inexact representation within the binary format used by floating point arithmetic used by most computer software, so using them would produce false results on some dates because of roundoff errors).

The formulae below (which use Euclidian division — integer division (div) and modulo (mod) — without any negative numbers) are valid for the whole range of dates since −4800. For dates before 1582, the resulting date components are valid only in the Gregorian proleptic calendar. This is based on the Gregorian calendar but extended to cover dates before its introduction, including the pre-Christian era. For dates in that era (before year 1 CE), astronomical year numbering is used. This includes a year zero, which immediately precedes 1 CE. Astronomical year zero is 1 BCE in the proleptic Gregorian calendar and, in general, year n BCE = astronomical year 1 − n, and for astronomical year A (A < 1), the BCE year is 1 + abs(A).

J = Julian day number

j = J + 32044

g = j div 146097

dg = j mod 146097

c = (dg div 36524 + 1) × 3 div 4

dc = dg − c × 36524

b = dc div 1461

db = dc mod 1461

a = (db div 365 + 1) × 3 div 4

da = db − a × 365

y = g × 400 + c × 100 + b × 4 + a

m = (da × 5 + 308) div 153 − 2