>From a quick look, it appears the algorithms below don't allow for years
that are multiples of 400 to be leap years, so might not work for 2000 (as
well as 1600, which is not so important, as the Gregorian calendar suffered
a big one-time adjustment jump in 1852, I think. I don't think anyone is
now using the Julian calendar, which preceded the current Gregorian calendar
and had major errors, but are numbering days serially based on the Julian
period, devised by Scaliger and named after his father.
As for modern computer usage of the term Julian to refer to minutes or
seconds since some epoch instant, I regret what I consider a misuse of the
term "Julian", but that's cool!
Russell [EMAIL PROTECTED]
----- Original Message -----
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, November 10, 1999 12:09 PM
Subject: [REBOL] epoch? Re:(10)
> Most of the time related function come from the various
> Asto/Navigation texts. Here is a copy of some of the standard
> functions that I have collected over the years.
>
>
> Andy
>
>
> JD from Calendar Date
>
> Let Y be the year
> M be the month number
> D be the day number (with decimals if any)
> Of the given Calendar date
>
> If M>2 leave Y and M unchanged
>
> If M=1 or 2 replace Y by Y-1 and M by M+12
>
> (ie if the date is in jan or feb it is considered
> to be in the 13th or 14th month of the preceeding year)
>
> In the gregorian calendar calculate:
>
> A = INT (y/100) B= 2 - A + INT (A/4)
>
> In the Julian calendar, take B= 0
>
> The required JD ids then
>
> JD= INT(365.25(Y+4716)) + INT(30.6001(M+1)) + D + B - 1524.5
>
>
> Example: the launch of Sputnik was 1957, Oct 4.81
>
> Y=1957 M = 10 D=4.81
> M>2 so leave Y&M unchanged
>
> As the date is in the Gregorian calendar
>
> A=INT(1957/100) = INT(19.57) = 19
> B = 2 - 19 + INT(19/4) = 2 - 19 + 4 = -13
>
> JD = INT(365.25*6673) + INT(30.6001*11) + 4.81 - 13 - 1524.5
> JD = 2436116.31
>
>
> Example calculate the JD for JAn 27 at 12h of the year 333
>
> M = 1, so Y=333-1 = 332 and M = 1+12 = 13
> Because date is in julian calendar B=0
>
> JD = INT(365.25*5048) + INT(30.6001*14)+27.5+0-1524.5
> JD = 1842713.0
>
> Note if you are only interested in dates between 1900 march 1
> and 2100 feb 28 set B to = -13
>
>
> Calendar date from JD
>
> (not valid for negitive JD's)
>
> Add 0.5 to the JD
> Let z be the integer part and F the fractional part
>
> If z < 2299161 take A = z
> else
>
> a = INT(Z-1867216.25/36524.25)
> A = z + 1 + a - INT(a/4)
>
> Then Calculate
>
> B = A + 1524
> C = INT(B - 122.1/365.25)
> D = INT(365.25 C)
> E = INT(B - D/30.6001)
>
> The day of month with decimals is
>
> B - D - INT(30.6001 E) + F
>
> The month Number is
>
> E - 1 if E < 14
> E - 13 if E = 14 or 15
>
> The year is
> C - 4716 if m>2
> C - 4715 if m = 1 or 2
>
> example
>
> calc the calendar date for JD 2436116.31
>
> 2436116.31 + 0.5 = 2436116.81
> z = 2436116 F = 0.81
>
> z > 2299161 so:
>
> a = INT 2436116 - 1867216.25/36524.25) = 15
>
> A = 2436116+1+15 - INT(15/4) = 2436129
>
> B = 2437653
> C = 6673
> D = 2437313
> E = 11
>
> day of month = 4.81
> Month m = E-1 (because E<14)
> Year = C - 4716 = 1957 (because m > 2)
>
> Day of week
>
> calculate the JD of the date at 0h and add 1.5
>
> Divide the result by 7
>
> the remainder will indicate the day of the week
>
> 0 = sunday
> 1 = mon
> ...
> 6 sat
>
> Day of the year
>
> The nukber N of a day in the year
>
> N = INT(275 M /9) - K * INT(M+9/12) + D - 30
>
> where M is the month number, D the day of the month
> K = 1 for leap year
> K = 2 for common year
>
> (N = 1 for Jan 1st)
>
> Date from Day number
>
> M = INT[ 9(K+N)/275 + 0.98]
> If N<32 then M=1
> D = N - INT(275M/9) + K * INT(M+9/12) + 30
>
> Note that Julian day numbers traditionally begin at
> Greenwich mean NOON that is 12h Universal Time.
>
