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Thanks, I copied the discussion to Template_Talk:JULIANDAY#Contradiction because the error is in the template. -- Erzbischof ( talk) 09:10, 24 February 2009 (UTC)
Besides UNIX time, there are two time standards defined by Microsoft. The .NET DateTime structure is defined as follows: "Time values are measured in 100-nanosecond units called ticks, and a particular date is the number of ticks since 12:00 midnight, January 1, 0001 A.D. (C.E.) in the GregorianCalendar calendar. For example, a ticks value of 31241376000000000L represents the date, Friday, January 01, 0100 12:00:00 midnight. A DateTime value is always expressed in the context of an explicit or default calendar." This seems to have all the same naive difficulties as POSIX time, it is unclear what they are doing in detail -- that is, I have no idea how to convert .NET time to UTC or TT. DonPMitchell ( talk) 03:26, 24 July 2009 (UTC)
I believe the Julian Day Number of Oct 5, 1582 should be equal to 2299161, but I'm not sure. This is the same as Oct 15, when the switch from Julian to Gregorian calenders occured. My own test code marches a year/month/day count from 4713 BCE and compares a calculated JDN to a running day count. However, some JDN calculators (like isotropic.com) give a JDN of 2299151 for that date, and so does the JULIAN subroutine in the JPL ephemeris software. There is some subtle mistake one way or the other. Strangly, all these programs agree on the JDN of 0.0 and the JDN of modern dates, they just are off by 10 days in 1582. I wonder what the bug is? DonPMitchell ( talk) 15:02, 26 July 2009 (UTC) (See reply below from Mottelg ( talk) 12:16, 15 October 2009 (UTC).)
Reply to "bug" from Mottelg ( talk) 12:16, 15 October 2009 (UTC): It's not a bug. It depends which calendar you are using. Your test code counting forward from 1/1/-4712, counts Julian calendar dates (proleptic until year -45). (Your code will have treated every fourth year as a leap year without exception). October 5, 1582, Julian = JDN 2299161. Oct 5, 1582, Gregorian was JDN 2299151. The 10 day difference was the difference in the 1500's between the Julian calendar and the Gregorian calendar (the latter, proleptic until Oct 5, 1582), i.e. in the year 1582, the Gregorian date, Oct 5, occurred 10 days earlier than Oct 5, Julian. Since then, that difference has grown by 1 day in each of the years 1700, 1800 and 1900 (i.e. in the last year of each century, except for 1600 and 2000, which were leap years in both calendars). Date m/d/y Julian now occurs 13 days after m/d/y Gregorian, so the JDN of m/d/y, Julian minus the JDN of m/d/y, Gregorian = 13 for the 20th and 21st centuries.
The general convention is that, unless otherwise specified, dates > Oct 4, 1582 are assumed to be Gregorian, dates < Oct 5, 1582 are assumed to be Julian. JDN 0 is 1/1/-4712, Julian proleptic = Nov 24 -4713 Gregorian proleptic. Note that at that time, the date m/d/y, Julian proleptic, occurred 38 days before m/d/y Gregorian proleptic. As you go backward from 1582, the difference of 10 days reduces by 3 days every 400 years. In the 200's the difference is zero, and before that, the difference (i.e. the JDN of m/d/y, Julian minus the JDN of m/d/y, Gregorian proleptic) is negative, its absolute value increasing the further back you go. Mottelg ( talk).
Further comment from Gleyshon
Can we get a citation from a published source about that convention dates > Oct 4, 1582 are assumed to be Gregorian, dates < Oct 5, 1582 are assumed to be Julian? It threw me in the table for Alternatives on this page because in the Epoch column, the BC dates are Julian and the other dates are Gregorian. The title 'Julian Day' implied it might be related to Julian Calendar dates... at any rate the definition of Epoch to head the column ought to be made clear.
Gleyshon ( talk) 22:33, 2 July 2011 (UTC)
This edit seems invalid; I can find noting wrong with either the Gregorian date algorithm or the Unix time algorithm. The Unix time algorithm does seem overly complex. -- Jc3s5h ( talk) 15:41, 8 September 2009 (UTC)
Thank you very much to User:Jc3s5h for replacing the formulas with new ones. Unfortunately, my computation shows it is wrong again. The error is only tiny so it's probable there is simply a piece of relevant information missing from the article to explain the difference. I'll freely admit my math is weak but these are simple sums to turn into computer code and I cannot see how I'd have made an error. Either way, please don't remove the dubious tag without explaining why my proof is invalid if you wish to remove it. If you see no problem with my proof then we will have to use the NOAA formula (which I know is inaccurate even in fairly recent history, but it's not inaccurate for today). So, here is my Python code testing the formula against one another:
from math import floor
import time
# Simple, easy conversion from another consecutive source
year, month, day = 2009, 9 , 9
print "Simple conversion from unix-time: %f" % (
2440587.5 + time.mktime((2009,9,9) + (0,)*6) / 86400)
# This is from serveral places, including the American NOAA
y, m, d = (year, month, day)
if m < 2:
y -= 1
month += 12
a = y /100
b = 2 - a + (a / 4)
print "Most common method: %f:" % (floor(365.25 * (y + 4716)) +
floor(30.6001 * (m + 1)) + d + b - 1524.5)
del a, b
# This is your new method, using integer arithmetic (copy/pasted) and with
# 2 regexps applied to change multiplication/subtraction symbols to * and -:
Y, M, D = (year, month, day)
JD = (1461 * (Y + 4800 + (M - 14)/12))/4 +(367 * (M - 2 - 12 * ((M - 14)/12)))/12 - (3 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 32075
print "Your method using integer math: %f" % JD
del Y, M, D, JD
# This is the above method, using floating point math.
# It is just the above formula with a regex applied in vi to make the numbers
# floats: :s/\([0-9]\)\([^0-9]\|$\)/\1.0\2/g
#
# If you don't know python, the map call below will make Y M and D floats too.
Y, M, D = map(float,(year, month, day))
JD = (1461.0 * (Y + 4800.0 + (M - 14.0)/12.0))/4.0 +(367.0 * (M - 2.0 - 12.0 * ((M - 14.0)/12.0)))/12.0 - (3.0 * ((Y + 4900.0 + (M - 14.0)/12.0)/100.0))/4.0 + D - 32075.0
print "Your method using floating point math %f:" % JD
The results of running this program are:
Simple conversion from unix-time: 2455083.500000 Most common method: 2455083.500000: Your method using integer math: 2455086.000000 Your method using floating point math 2455084.248125:
82.132.139.213 ( talk) 09:42, 9 September 2009 (UTC)
JDN = (1461 × (2009 + 4800 + (9 − 14)/12))/4 +(367 × (9 − 2 − 12 × ((9 − 14)/12)))/12 − (3 × ((2009 + 4900 + (9 − 14)/12)/100))/4 + 9 − 32075 JDN = (1461 × (2009 + 4800 + (9 − 14)/12))/4 +(367 × (9 − 2 − 12 × (−5/12)))/12 − (3 × ((2009 + 4900 + −5/12)/100))/4 + 9 − 32075 JDN = (1461 × (2009 + 4800 + −5/12))/4 +(367 × (9 − 2 − 12 × 0))/12 − (3 × (6909/100))/4 + 9 − 32075 JDN = (1461 × 6809)/4 +(367 × 7)/12 − (3 × 69)/4 + 9 − 32075 JDN = 9947949/4 +2569/12 − 207/4 + 9 − 32075 JDN = 2486987 +214 − 51 + 9 − 32075 JDN = 2455084
AD ASTRA SCIENTIA ( talk) 02:12, 29 April 2011 (UTC)
Thank you for the algorithm for converting JDN to Gregorian Date and the accompanying explanation. (The Fliegel & Van Flandern formula on page 604 of the Explanatory Supplement to the Astronomical Almanac (USA) is too cryptic to understand.) However, I think one part of the explanation given here may be wrong (the intention is certainly unclear, IMO.) This is the explanation added in parentheses to dot-point four for the reduction "to a maximum of three".
(this reduction occurs for the last day of a leap centennial year where c would be 4 if it were not reduced)
Can you provide an algorithm along the same lines for converting JDN to a date in the Julian calendar.
I have code here that works. It is similar to the Fliegel & Van Flandern formula for the Gregorian conversion, but is just as cryptic:
sub JDNtoJulianDate (JDN, Y, M, D)
' ---------------
' Adapted from: http://dev.remotenetworktechnology.com/wsh/jdn.htm
' Similar to the Fliegel & Van Flandern method for Gregorian dates.
' Language: PowerBasic. (The symbol \ denotes an integer division.)
Local J, K, L, N, I
J = JDN + 1402
K = (J - 1) \ 1461
L = J - 1461 * K
N = ((L - 1) \ 365) - (L \ 1461)
I = L - 365 * N + 30
J = (80 * I) \ 2447
D = I - ((2447 * J) \ 80)
I = (J \ 11)
M = J + 2 - 12 * I
Y = 4 * K + N + I - 4716
end sub ' JDNtoJulianDate ---------------------------------------
Thanks. Mottelg ( talk). —Preceding undated comment added 01:52, 16 October 2009 (UTC).
Hi Folks, new to wikipedia here, so bear with me.
I've noticed that in this equation:
JDN = (1461 × (Y + 4800 + (M − 14)/12))/4 +(367 × (M − 2 − 12 × ((M − 14)/12)))/12 − (3 × ((Y + 4900 + (M - 14)/12)/100))/4 + D − 32075
The second term (367 × (M − 2 − 12 × ((M − 14)/12)))/12
reduces to the constant 367 as follows:
(367 × (M − 2 − 12 × ((M − 14)/12)))/12
(367 × (M − 2 − (M − 14)))/12
(367 × (M − 2 − M + 14))/12
(367 × (−2 + 14))/12
(367 × 12)/12
367
Which could then be combined with the final term as such:
JDN = (1461 × (Y + 4800 + (M − 14)/12))/4 − (3 × ((Y + 4900 + (M - 14)/12)/100))/4 + D − 31708
I have verified this result for a handful of dates against the original equation, the Naval Observatory NOVAS software ( http://aa.usno.navy.mil/software/novas/novas_info.php) and their Julian Date page which you already reference ( http://aa.usno.navy.mil/data/docs/JulianDate.php)
I suspect the original equation was transcribed directly from reference cited. Is there any reason not to apply this basic algebraic simplification?
Also note that different characters are used for the minus sign in the page as written, which causes issues when cutting & pasting into some tools. This is what it looks like pasting into nedit - a basic Linux text editor. I got a similar result pasting into excel:
JDN = (1461 × (Y + 4800 + (M \u2212 14)/12))/4 +(367 × (M \u2212 2 \u2212 12 × ((M \u2212 14)/12)))/12 \u2212 (3 × ((Y + 4900 + (M - 14)/12)/100))/4 + D \u2212 32075
If we replace this \u2212 symbol with a minus sign and replace the 'x' symbol with an * then users could simply copy and paste the formula into various tools and have it work, if the variables are defined correctly. The final result would be:
JDN = (1461 * (Y + 4800 + (M - 14)/12))/4 - (3 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 31708
I noticed somebody above posted some code - here's a listing of a program with this equation cut-and-pasted in and the results. I did have to make some of the constants floating point to force it to calculate the floating point answer, or you'd get integer math issues.
#!/usr/bin/python
import sys
import string
Y = string.atoi(sys.argv1])
M = string.atoi(sys.argv2])
D = string.atoi(sys.argv3])
print "Gregorian Date:\t", Y, "-", M, "-", D
#Original Formula
JDN1 = (1461.0 * (Y + 4800 + (M - 14)/12))/4 +(367 * (M - 2 - 12 * ((M - 14)/12)))/12 - (3.0 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 32075
#Simplified Formula
JDN2 = (1461.0 * (Y + 4800 + (M - 14)/12))/4 - (3.0 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 31708
print "Orginal Formula: \t", JDN1
print "Simplified Formula:\t", JDN2
print "Percent Error: \t", 100.0*(JDN1 - JDN2)/JDN1
print "Absolute Error:\t", (JDN1 - JDN2)
My $0.02 Thanks Eric -- Ergeorge ( talk) 21:55, 9 February 2010 (UTC)
Nevermind - I did some more testing and figured it out for myself. There's some integer math voodoo that goes on in that second term - if you do everything in integer math and strict order of operations, it doesn't always drop out. Very elegant - I think I will get that paper and take a look at the derivation. This also explains some of the questions in the section above (which is where this should have been posted - clicked the wrong edit). You can't do this as a floating point calculation. -- Ergeorge ( talk) 22:44, 9 February 2010 (UTC)
As far as I can tell "Chronological Julian Date" was created by Peter Meyer, and all the sources for it seem to either refer to his web page, or are Wikipedia mirrors. I would like to see some independent evidence that this date is widely used by this name.
I wouldn't be surprised if some historians have adapted the concept of a Julian Date to a particular location and considered the day to change at local midnight rather than local noon. What would surprise me is if they consistently referred to this as a Chronological Julian Date. Jc3s5h ( talk) 23:03, 7 February 2010 (UTC)
I recently edited the epoch entry for the Chronological Julian Day (CJD) from "00:00 January 1, 1, Monday" to "00:00 January 1, 4713 BC, Monday". This corrects an error that had slipped in at this edit, and restores the articles agreement with other sources. I am removing the "dubious" template. -- SteveMcCluskey ( talk) 23:11, 7 February 2010 (UTC)
"A Julian date of 2454115.05486 means that the date and Universal Time is Sunday January 14, 2007 at 13:18:59.9."
Shouldn't this be something like 2454115.5486 (note the decimals)? —Preceding unsigned comment added by 77.167.55.199 ( talk) 02:27, 8 February 2010 (UTC)
The history section mentions the link between the JDN count and the Julian Period invented by Joseph Justus Scaliger. As the article states, the first year of the Julian Period (4713 BC) was the last time when all three of the Indiction, Metonic and Solar cycles were in their first year together. This implies that at some stage there was some common convention as to a starting point for those cycles.
My question is where are the references documenting how those cycles were counted? The methods of counting them must have differed between different calendars. (For example, the Metonic cycle was used in the ancient Greek luni-solar calendar, and (much later) in the Jewish luni-solar calendar, and in the Christian ecclesiastical calendar for reckoning the dates of Easter. Yet, nowadays, the "Golden Number" of a year in the Christian calendar (which denotes its position in a Metonic cycle) does not match the position of the corresponding Jewish year in the Metonic cycles of the Jewish calendar.)
(a) If 4713 BC was year 1 of a 28-year Solar cycle of the Julian calendar and of a Metonic cycle, it follows that (counting from that year), the first year of Solar cycle 170 is year 20 AD, and the first year of Metonic cycle 250 is year 19 AD. However:
(b) in the Jewish calendar, the Jewish year corresponding to AD 21 was year 1 of a 28-year Solar Cycle, all of which begin in Julian years of the form AD 28n+21. And the Jewish year corresponding to AD 22 is the first year of a Metonic Cycle, all of which begin with an AD year of the form 19n+3.
So, since there are clearly different conventions regarding this, according to which convention(s) used during or before Scaliger's time is (a) correct, and did any of them survive into the present era, i.e. are there any systems still in use today in which Solar Cycles of the Julian Calendar begin with AD years of the form 28n+20 and where Metonic Cycles begin with AD years of the form 19n? Mottelg ( talk) 11:56, 2 March 2010 (UTC)
I think there is an error in the Gregorian - JDN calculation. The final constant should be 32077 to remove an error of 2 days. There is an even stranger error with the Julian - JDN calculation. According to http://aa.usno.navy.mil/data/docs/JulianDate.php :
JD | Date/Time from USNO | JD from Gregorian formula | JD from Julian formula |
---|---|---|---|
0.0 | BCE 4713 January 01 12:00:00 | ||
1721423.00000 | BCE 1 December 31 12:00:00 | 1721423 | |
1721424.00000 | CE 1 January 01 12:00:00 | ||
2299160.00000 | CE 1582 October 04 12:00:00 | 2299160 | |
2299161.00000 | CE 1582 October 15 12:00:00 | ||
2455315.00000 | CE 2010 April 28 12:00:00 (today) |
Unfortunately, I don't have the reference used for these calculations (L. E. Doggett, Ch. 12, "Calendars", p. 604/6, in Seidelmann 1992) to check if the article has been vandalised sometime in the past. Astronaut ( talk) 18:25, 28 April 2010 (UTC)
The program listed at http://www.astro.uu.nl/~strous/AA/en/reken/juliaansedag.html does not work correctly. It passes the few test vectors given, but if increment a JD value for a few years, you will eventually hit dates that generate nonsense like February 31. DonPMitchell ( talk) 05:32, 14 May 2010 (UTC)
In the formula for converting julian days to Gregorian dates, there appears to be a bug on this line:
When this formula is used, it sometimes gives dates like April 0 instead of March 31. The cause is the imprecise nature of floating-point computations. It is effectively dividing 5 by 153, or dividing the unmultiplied number by 30.6. On a modern computer, 153/5 does not equal 30.6 precisely. In his book Astronomical Algorithms, Meeus discusses this problem and avoids it by using the constant 30.6001 instead of 30.6 (which is equal to 153/5).
Meeus' intermediate formula that extracts the month is E = INT(some value/30.6001). It does not perform integer division, but accomplishes the same effect by using the INT function, which is the same as the FLOOR function for positive numbers.-- B.D.Mills ( T, C) 23:31, 24 May 2010 (UTC)
int j, g, dg, c, dc, b, db, a, da, y, m; double J, d, T; J = jDNum + 0.5; j = J + 32044; g = j / 146097; dg = j % 146097; c = (dg / 36524 + 1) * 3 / 4; dc = dg - c * 36524; b = dc /1461; db = dc % 1461; a = (db / 365 + 1) * 3 / 4; da = db - a * 365; y = g * 400 + c * 100 + b * 4 + a m = (da * 5 + 308) / 153 - 2; d = da - (m + 4) * 153 / 5 + 122; //Won't correctly calculate the fraction of a day T = jDNum - (int)jDNum; //fractional part of the Julian day number T -= 0.5; //because the Julian day starts at noon if(x < 0.0) T += 1; d += T; //add fraction of day *year = y - 4800 + (m + 2) / 12; *month = (m + 2) % 12 + 1; *day = d + 1; //Will be the correct day number if d is calculated as an int
I wrote a C function to convert dates in the proleptic Gregorian calendar to Julian day numbers: double ProlepticGregorianDateToJDN(int year, int month, double day) {
int a, y, m; a = (14 - month) / 12; y = year + 4800 - a; m = month + 12 * a - 3; return((day + ((153 * m + 2) / 5) + (365 * y) + (y / 4) - (y / 100) + (y / 400) - 32045) - 0.5);
} Notice that one has to subtract 0.5 from the result of the last calculation in the formula given in the article to make the fraction of a day come out correctly. This is because the algorithm fails to account for the fact that the Julian day starts at noon. Senor Cuete ( talk) 14:18, 30 November 2011 (UTC)Senor Cuete
Apparently there is another, less involved use of the term "Julian Date" in industry which simply seems to be the counting of days from January 1 of each calendar year, ending with 365 (366 in leap years) on December 31. I was made aware of this because of articles in the media on the recent salmonella contamination of eggs from Iowa. This seems to be a legitimate inclusion on this page, or there needs to be at least a further disambiguation article about this usage at the top of this article. Robert Johnson ( talk) 10:14, 20 August 2010 (UTC
In this edit User:RadioFan has changed from a clearly reliable source, The Explanatory Supplement to the Astronomical Almanac, to a self-published website by Claus Tøndering. Granted, this website is often cited by serious publications, but still, the WP:IRS guideline would clearly rank it below a book by a recognized scientific publisher.
In addition, the new edit does not follow the established citation style of this article.
So my question is, should the edit stand? Jc3s5h ( talk) 16:11, 21 September 2010 (UTC)
In the food and grocery industries the term "Julian date" is used as follows: 001 is January 1 and 365 is December 31 (unless it is a leap year and then it is 366). An example of the term being used in such a manner by a respectable source is [1]. Typically a Julian date is used to code a pack date. (On the rare occasions which dates are not coded in this manner, it is my experience that consumers mistake it for the expiration date.) Maybe the article should mention this usage of the term as well. 68.97.10.193 ( talk) 23:22, 3 February 2011 (UTC)
This algorithm pertains to the Gregorian Proleptic Calendar. The months (M) January to December are 1 to 12. For the year (Y) astronomical year numbering is used, thus 1 BC is 0, 2 BC is −1, and 4713 BC is −4712. D is the day of the month from 1 to 31 as applicable. JDN is the Julian Day Number, which pertains to the noon occurring in the corresponding calendar date.
It is necessary to use the function for Long Integers in the first line of code. This enables integers in the range of -2,147,483,648 to 2,147,483,647
[2] to be processed.
Integer division is commanded by using the back slash "\" character instead of the forward slash "/".
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
DEFLNG A-Z
INPUT "Day";d
INPUT "Month";m
INPUT "Year";y
a=(14-m)\12
y=y+4800-a
m=m+12*a-3
j=d+(153*m+2)\5+365*y+y\4-y\100+y\400-32045
PRINT "Julian Day Number =";j
END
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The Algorithm should return the following values as a cross check:-
JDN of 24 NOV, -4713 is 0 (4714 BC Gregorian Proleptic Calendar)
JDN of 15 OCT, 1582 is 2,299,161 (beginning of the Gregorian Calendar)
JDN of 16 NOV, 1858 is 2,400,000
JDN of 01 JAN, 2000 is 2,451,545
JDN of 31 AUG, 2132 is 2,500,000
Summerdrought (
talk)
21:20, 6 March 2011 (UTC)
I've checked the above algorithm but used Excel Visual Basic instead, so had to do some slight syntax adaptations, finding it works fine for all dates since oct. 15 1582. When tested with oct. 4 1582 it returns 2299150 when 2299160 is expected. Jeep IKA ( talk) 13:25, 26 May 2011 (UTC)
The article defines JDN as "The Julian day number (JDN) is the integer part of the Julian date (JD). The day commencing at the above-mentioned epoch is JDN 0. Negative values can be used for preceding dates..." According to this, the day preceding Julian day 0 is also Julian day 0. For example, while the integer part of Julian date 0.3 is 0, so is the integer part of -0.3. Clearly, what is intended is not the integer part but the "floor", that is, the greatest integer not greater than the Julian date. Unfortunately, this is not nearly as easy to express as "the integer part" is. I shall try a wording that gets around this. —Preceding unsigned comment added by 208.53.195.38 ( talk) 17:40, 25 March 2011 (UTC)
I wonder if this article should use the more international date format of day month year. Julian days are used in astronomy, which is an international science. After all, the United States is nearly alone in its use of month day, year. —Preceding unsigned comment added by Kaptain Cirk ( talk • contribs) 00:08, 19 May 2011 (UTC)
The Chronological Julian Day as suggested by Peter Meyer is a useful concept but I believe that its use is beyond that of the day in a particular time zone. It can be based on any form of local time, Local Mean Time, for example. Local time is an integral part of any calendar. The nychthemeron may begin at any point in the day/night cycle. The most common times to begin a cycle are midnight, noon, sunrise and sunset. Any of these points may be used in a Chronological Julian Day numbering system. Alexselkirk1704 ( talk) 14:43, 27 July 2011 (UTC)
Do we really need the BASIC-ish pseudocode conversion methods in this article? It's not something that is terribly useful, from either a practical or programming standpoint, and in any case this seems to be the main point of contention in the article. Surely the formula given in a standard mathematical format is both more useful for non-programmers and programmers alike.
In addition, I don't see much in the article that clearly points out that the Julian day is not really a practical date system for general use, but is almost exclusively used for a reference point in orbital elements and ephimerides. The Julian day is not "heliocentric", does not rely on or make accomodations to the vagaries of Earth's orbit (referenced to TT and not UTC), and does not easily map to a calendar (It's "year" has fractional days, not leap days). Instead there seems to be much about the Julian calendar, which is certainly related, but not the topic of this article.
What do you think? Msaunier ( talk) 11:19, 13 September 2011 (UTC)
The article currently says
and
These two seem to contradict each other; shouldn't the latter be Terrestrial Time? This amounts to a difference of over a minute. AxelBoldt ( talk) 21:20, 19 November 2011 (UTC)
"The use of Julian date to refer to the day-of-year (ordinal date) is usually considered to be incorrect although it is widely used that way in the earth sciences, computer programming, military and the food industry.[1]"
This reference is misleading. While reading one would think that the contentious statement "is usually considered to be incorrect" would be the reason for the reference. But all the article does is reference the use of Julian day in best-used-by dating for eggs, which, really, is totally irrelevant to the use Julian day and whether it is an acceptable form ordinal days.
Davisrunning ( talk) 18:19, 9 July 2012 (UTC)david july 9, 2012
The table on "Alternatives" gives two definitions for TJD, one as JD - 2440000.5, "as introduced by NASA", and the other as JD mod 10000, "NIST definition". I have been unable to find the second definition anywhere online, at NIST or elsewhere. I recall when TJD rolled over in 1995, both definitions were used by different sites, even at NASA ( A four-decimal day count originating at midnight 1968-05-23,24) but the four-decimal part may have referred only to TJD before 1995 since it became self-contradictory afterwards. Earlier versions of this article stated, "NIST treats it as cyclical". IIRC some did suppose that TJD was cyclical based upon the statement that it had only four digits, but I don't know any authoritative source, and it seems that Wikipedia and its copies are now the only reference to this definition. I can find references to five-digit TJDs, but no shorter ones after 1995. Therefore, unless anyone can find a reference I will delete the second definition. -- Nike ( talk) 05:54, 21 July 2012 (UTC) TJD 16129.245
This article as a whole seems confusing and disorganized. I will attempt to clean it up a bit. I don't intend on making major changes, but you'll let me know if I get it wrong. -- Nike ( talk) 23:17, 21 July 2012 (UTC) JD 2456130.467
Strange, I don't think I changed much, aside from removing a few redundancies and moving things around. Please point out what specifically I changed that you don't like. If anything, I think I made it less exclusive. I changed "for scientific use by the astronomy community...recommended for astronomical use by the International Astronomical Union" (which suggests only astronomers) to "used primarily by astronomers".
In fact, Julian Dates are mostly used by astronomers. Julian day numbering was introduced by an astronomer, for astronomers, to date astronomical observations, and that is the way it has been used for over a century. The definition is taken directly from the International Astronomical Union. There is nothing in the definition which precludes non-astronomers from using Julian Dates, and I have not "dismissed" anybody. In fact, there is a mention of Chronological Julian Dates being used by calendar students, despite previous controversy about their notability, and I happened to expand on that just a bit. BTW, the article is Julian day, not Julian period. The Julian Period existed nearly three centuries before Julian days.
I have not come across any historians using Julian Dates, and just a few calendar students as are already mentioned. When you look at the literature (books and journals) practically all the notable references are astronomical. But we cannot say what any other groups are doing or how Julian Dates are used or defined in any other field without knowing what that is. If you wish to research how historians, etc., have used the Julian Period or Julian Dates, you are free to add to the article. Perhaps there could be a separate article for Julian Period, although I don't know what else there is to say about it. -- Nike ( talk) 13:51, 22 July 2012 (UTC) JD 2456131.077
Since "Julian Period" redirects here, perhaps we should include more about it. I looked in Britannica and it says, "Julian period, chronological system now used chiefly by astronomers and based on the consecutive numbering of days..." I am now reading Scaliger's book. -- Nike ( talk) 20:09, 22 July 2012 (UTC) JD 2456131.339
I was having a hard enough time with Latin... -- Nike ( talk) 06:58, 23 July 2012 (UTC) JD 2,456,131.79
Google Translate works much better with German than Latin. But from what I can tell, it's not much of an issue. He does, at one point, talk about Julian days starting at midnight, so .25 is 6 am, .5 is noon, etc. -- Nike ( talk) 01:13, 24 July 2012 (UTC) JD 2456132.55
Why don't you run it through Google Translate, as I suggested? In any event, I don't see how it's relevant to the article. -- Nike ( talk) 21:04, 24 July 2012 (UTC) JD 2456133.377
I don't see how, and one author a century ago hardly constitutes a community in any event. But you might like this page which states, "The Julian Period Year is rarely used (except in the Old Farmer’s Almanac), but the strict count of days is widely used among astronomers, calendar freaks and Mayanists (not that that really encompasses that wide a constituency, by most definitions)." -- Nike ( talk) 21:47, 24 July 2012 (UTC) JD 2456133.407
I have reverted a new algorithm introduced by User:Marlin Woks for several reasons:
I changed the algorithm for the following reasons:
Sorry, I didn't realise that Wikipedia articles couldn't be referenced. Marlin Woks ( talk) 15:20, 21 August 2012 (UTC)
"If you are writing about your algorithm, I see no need to put it in Wikipedia. We have an algorithm for converting Julian dates to Julian day number. We have one for converting Gregorian dates to Julian day numbers. It is trivial for a programmer to write a program that determines whether the date is Gregorian or Julian in the situations the programmer is interested, and then invoke the appropriate algorithm. Wikipedia is an encyclopedia, not a collection of computer algorithms.
Jc3s5h (
talk)
17:23, 21 August 2012 (UTC)
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![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 | Archive 5 |
Thanks, I copied the discussion to Template_Talk:JULIANDAY#Contradiction because the error is in the template. -- Erzbischof ( talk) 09:10, 24 February 2009 (UTC)
Besides UNIX time, there are two time standards defined by Microsoft. The .NET DateTime structure is defined as follows: "Time values are measured in 100-nanosecond units called ticks, and a particular date is the number of ticks since 12:00 midnight, January 1, 0001 A.D. (C.E.) in the GregorianCalendar calendar. For example, a ticks value of 31241376000000000L represents the date, Friday, January 01, 0100 12:00:00 midnight. A DateTime value is always expressed in the context of an explicit or default calendar." This seems to have all the same naive difficulties as POSIX time, it is unclear what they are doing in detail -- that is, I have no idea how to convert .NET time to UTC or TT. DonPMitchell ( talk) 03:26, 24 July 2009 (UTC)
I believe the Julian Day Number of Oct 5, 1582 should be equal to 2299161, but I'm not sure. This is the same as Oct 15, when the switch from Julian to Gregorian calenders occured. My own test code marches a year/month/day count from 4713 BCE and compares a calculated JDN to a running day count. However, some JDN calculators (like isotropic.com) give a JDN of 2299151 for that date, and so does the JULIAN subroutine in the JPL ephemeris software. There is some subtle mistake one way or the other. Strangly, all these programs agree on the JDN of 0.0 and the JDN of modern dates, they just are off by 10 days in 1582. I wonder what the bug is? DonPMitchell ( talk) 15:02, 26 July 2009 (UTC) (See reply below from Mottelg ( talk) 12:16, 15 October 2009 (UTC).)
Reply to "bug" from Mottelg ( talk) 12:16, 15 October 2009 (UTC): It's not a bug. It depends which calendar you are using. Your test code counting forward from 1/1/-4712, counts Julian calendar dates (proleptic until year -45). (Your code will have treated every fourth year as a leap year without exception). October 5, 1582, Julian = JDN 2299161. Oct 5, 1582, Gregorian was JDN 2299151. The 10 day difference was the difference in the 1500's between the Julian calendar and the Gregorian calendar (the latter, proleptic until Oct 5, 1582), i.e. in the year 1582, the Gregorian date, Oct 5, occurred 10 days earlier than Oct 5, Julian. Since then, that difference has grown by 1 day in each of the years 1700, 1800 and 1900 (i.e. in the last year of each century, except for 1600 and 2000, which were leap years in both calendars). Date m/d/y Julian now occurs 13 days after m/d/y Gregorian, so the JDN of m/d/y, Julian minus the JDN of m/d/y, Gregorian = 13 for the 20th and 21st centuries.
The general convention is that, unless otherwise specified, dates > Oct 4, 1582 are assumed to be Gregorian, dates < Oct 5, 1582 are assumed to be Julian. JDN 0 is 1/1/-4712, Julian proleptic = Nov 24 -4713 Gregorian proleptic. Note that at that time, the date m/d/y, Julian proleptic, occurred 38 days before m/d/y Gregorian proleptic. As you go backward from 1582, the difference of 10 days reduces by 3 days every 400 years. In the 200's the difference is zero, and before that, the difference (i.e. the JDN of m/d/y, Julian minus the JDN of m/d/y, Gregorian proleptic) is negative, its absolute value increasing the further back you go. Mottelg ( talk).
Further comment from Gleyshon
Can we get a citation from a published source about that convention dates > Oct 4, 1582 are assumed to be Gregorian, dates < Oct 5, 1582 are assumed to be Julian? It threw me in the table for Alternatives on this page because in the Epoch column, the BC dates are Julian and the other dates are Gregorian. The title 'Julian Day' implied it might be related to Julian Calendar dates... at any rate the definition of Epoch to head the column ought to be made clear.
Gleyshon ( talk) 22:33, 2 July 2011 (UTC)
This edit seems invalid; I can find noting wrong with either the Gregorian date algorithm or the Unix time algorithm. The Unix time algorithm does seem overly complex. -- Jc3s5h ( talk) 15:41, 8 September 2009 (UTC)
Thank you very much to User:Jc3s5h for replacing the formulas with new ones. Unfortunately, my computation shows it is wrong again. The error is only tiny so it's probable there is simply a piece of relevant information missing from the article to explain the difference. I'll freely admit my math is weak but these are simple sums to turn into computer code and I cannot see how I'd have made an error. Either way, please don't remove the dubious tag without explaining why my proof is invalid if you wish to remove it. If you see no problem with my proof then we will have to use the NOAA formula (which I know is inaccurate even in fairly recent history, but it's not inaccurate for today). So, here is my Python code testing the formula against one another:
from math import floor
import time
# Simple, easy conversion from another consecutive source
year, month, day = 2009, 9 , 9
print "Simple conversion from unix-time: %f" % (
2440587.5 + time.mktime((2009,9,9) + (0,)*6) / 86400)
# This is from serveral places, including the American NOAA
y, m, d = (year, month, day)
if m < 2:
y -= 1
month += 12
a = y /100
b = 2 - a + (a / 4)
print "Most common method: %f:" % (floor(365.25 * (y + 4716)) +
floor(30.6001 * (m + 1)) + d + b - 1524.5)
del a, b
# This is your new method, using integer arithmetic (copy/pasted) and with
# 2 regexps applied to change multiplication/subtraction symbols to * and -:
Y, M, D = (year, month, day)
JD = (1461 * (Y + 4800 + (M - 14)/12))/4 +(367 * (M - 2 - 12 * ((M - 14)/12)))/12 - (3 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 32075
print "Your method using integer math: %f" % JD
del Y, M, D, JD
# This is the above method, using floating point math.
# It is just the above formula with a regex applied in vi to make the numbers
# floats: :s/\([0-9]\)\([^0-9]\|$\)/\1.0\2/g
#
# If you don't know python, the map call below will make Y M and D floats too.
Y, M, D = map(float,(year, month, day))
JD = (1461.0 * (Y + 4800.0 + (M - 14.0)/12.0))/4.0 +(367.0 * (M - 2.0 - 12.0 * ((M - 14.0)/12.0)))/12.0 - (3.0 * ((Y + 4900.0 + (M - 14.0)/12.0)/100.0))/4.0 + D - 32075.0
print "Your method using floating point math %f:" % JD
The results of running this program are:
Simple conversion from unix-time: 2455083.500000 Most common method: 2455083.500000: Your method using integer math: 2455086.000000 Your method using floating point math 2455084.248125:
82.132.139.213 ( talk) 09:42, 9 September 2009 (UTC)
JDN = (1461 × (2009 + 4800 + (9 − 14)/12))/4 +(367 × (9 − 2 − 12 × ((9 − 14)/12)))/12 − (3 × ((2009 + 4900 + (9 − 14)/12)/100))/4 + 9 − 32075 JDN = (1461 × (2009 + 4800 + (9 − 14)/12))/4 +(367 × (9 − 2 − 12 × (−5/12)))/12 − (3 × ((2009 + 4900 + −5/12)/100))/4 + 9 − 32075 JDN = (1461 × (2009 + 4800 + −5/12))/4 +(367 × (9 − 2 − 12 × 0))/12 − (3 × (6909/100))/4 + 9 − 32075 JDN = (1461 × 6809)/4 +(367 × 7)/12 − (3 × 69)/4 + 9 − 32075 JDN = 9947949/4 +2569/12 − 207/4 + 9 − 32075 JDN = 2486987 +214 − 51 + 9 − 32075 JDN = 2455084
AD ASTRA SCIENTIA ( talk) 02:12, 29 April 2011 (UTC)
Thank you for the algorithm for converting JDN to Gregorian Date and the accompanying explanation. (The Fliegel & Van Flandern formula on page 604 of the Explanatory Supplement to the Astronomical Almanac (USA) is too cryptic to understand.) However, I think one part of the explanation given here may be wrong (the intention is certainly unclear, IMO.) This is the explanation added in parentheses to dot-point four for the reduction "to a maximum of three".
(this reduction occurs for the last day of a leap centennial year where c would be 4 if it were not reduced)
Can you provide an algorithm along the same lines for converting JDN to a date in the Julian calendar.
I have code here that works. It is similar to the Fliegel & Van Flandern formula for the Gregorian conversion, but is just as cryptic:
sub JDNtoJulianDate (JDN, Y, M, D)
' ---------------
' Adapted from: http://dev.remotenetworktechnology.com/wsh/jdn.htm
' Similar to the Fliegel & Van Flandern method for Gregorian dates.
' Language: PowerBasic. (The symbol \ denotes an integer division.)
Local J, K, L, N, I
J = JDN + 1402
K = (J - 1) \ 1461
L = J - 1461 * K
N = ((L - 1) \ 365) - (L \ 1461)
I = L - 365 * N + 30
J = (80 * I) \ 2447
D = I - ((2447 * J) \ 80)
I = (J \ 11)
M = J + 2 - 12 * I
Y = 4 * K + N + I - 4716
end sub ' JDNtoJulianDate ---------------------------------------
Thanks. Mottelg ( talk). —Preceding undated comment added 01:52, 16 October 2009 (UTC).
Hi Folks, new to wikipedia here, so bear with me.
I've noticed that in this equation:
JDN = (1461 × (Y + 4800 + (M − 14)/12))/4 +(367 × (M − 2 − 12 × ((M − 14)/12)))/12 − (3 × ((Y + 4900 + (M - 14)/12)/100))/4 + D − 32075
The second term (367 × (M − 2 − 12 × ((M − 14)/12)))/12
reduces to the constant 367 as follows:
(367 × (M − 2 − 12 × ((M − 14)/12)))/12
(367 × (M − 2 − (M − 14)))/12
(367 × (M − 2 − M + 14))/12
(367 × (−2 + 14))/12
(367 × 12)/12
367
Which could then be combined with the final term as such:
JDN = (1461 × (Y + 4800 + (M − 14)/12))/4 − (3 × ((Y + 4900 + (M - 14)/12)/100))/4 + D − 31708
I have verified this result for a handful of dates against the original equation, the Naval Observatory NOVAS software ( http://aa.usno.navy.mil/software/novas/novas_info.php) and their Julian Date page which you already reference ( http://aa.usno.navy.mil/data/docs/JulianDate.php)
I suspect the original equation was transcribed directly from reference cited. Is there any reason not to apply this basic algebraic simplification?
Also note that different characters are used for the minus sign in the page as written, which causes issues when cutting & pasting into some tools. This is what it looks like pasting into nedit - a basic Linux text editor. I got a similar result pasting into excel:
JDN = (1461 × (Y + 4800 + (M \u2212 14)/12))/4 +(367 × (M \u2212 2 \u2212 12 × ((M \u2212 14)/12)))/12 \u2212 (3 × ((Y + 4900 + (M - 14)/12)/100))/4 + D \u2212 32075
If we replace this \u2212 symbol with a minus sign and replace the 'x' symbol with an * then users could simply copy and paste the formula into various tools and have it work, if the variables are defined correctly. The final result would be:
JDN = (1461 * (Y + 4800 + (M - 14)/12))/4 - (3 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 31708
I noticed somebody above posted some code - here's a listing of a program with this equation cut-and-pasted in and the results. I did have to make some of the constants floating point to force it to calculate the floating point answer, or you'd get integer math issues.
#!/usr/bin/python
import sys
import string
Y = string.atoi(sys.argv1])
M = string.atoi(sys.argv2])
D = string.atoi(sys.argv3])
print "Gregorian Date:\t", Y, "-", M, "-", D
#Original Formula
JDN1 = (1461.0 * (Y + 4800 + (M - 14)/12))/4 +(367 * (M - 2 - 12 * ((M - 14)/12)))/12 - (3.0 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 32075
#Simplified Formula
JDN2 = (1461.0 * (Y + 4800 + (M - 14)/12))/4 - (3.0 * ((Y + 4900 + (M - 14)/12)/100))/4 + D - 31708
print "Orginal Formula: \t", JDN1
print "Simplified Formula:\t", JDN2
print "Percent Error: \t", 100.0*(JDN1 - JDN2)/JDN1
print "Absolute Error:\t", (JDN1 - JDN2)
My $0.02 Thanks Eric -- Ergeorge ( talk) 21:55, 9 February 2010 (UTC)
Nevermind - I did some more testing and figured it out for myself. There's some integer math voodoo that goes on in that second term - if you do everything in integer math and strict order of operations, it doesn't always drop out. Very elegant - I think I will get that paper and take a look at the derivation. This also explains some of the questions in the section above (which is where this should have been posted - clicked the wrong edit). You can't do this as a floating point calculation. -- Ergeorge ( talk) 22:44, 9 February 2010 (UTC)
As far as I can tell "Chronological Julian Date" was created by Peter Meyer, and all the sources for it seem to either refer to his web page, or are Wikipedia mirrors. I would like to see some independent evidence that this date is widely used by this name.
I wouldn't be surprised if some historians have adapted the concept of a Julian Date to a particular location and considered the day to change at local midnight rather than local noon. What would surprise me is if they consistently referred to this as a Chronological Julian Date. Jc3s5h ( talk) 23:03, 7 February 2010 (UTC)
I recently edited the epoch entry for the Chronological Julian Day (CJD) from "00:00 January 1, 1, Monday" to "00:00 January 1, 4713 BC, Monday". This corrects an error that had slipped in at this edit, and restores the articles agreement with other sources. I am removing the "dubious" template. -- SteveMcCluskey ( talk) 23:11, 7 February 2010 (UTC)
"A Julian date of 2454115.05486 means that the date and Universal Time is Sunday January 14, 2007 at 13:18:59.9."
Shouldn't this be something like 2454115.5486 (note the decimals)? —Preceding unsigned comment added by 77.167.55.199 ( talk) 02:27, 8 February 2010 (UTC)
The history section mentions the link between the JDN count and the Julian Period invented by Joseph Justus Scaliger. As the article states, the first year of the Julian Period (4713 BC) was the last time when all three of the Indiction, Metonic and Solar cycles were in their first year together. This implies that at some stage there was some common convention as to a starting point for those cycles.
My question is where are the references documenting how those cycles were counted? The methods of counting them must have differed between different calendars. (For example, the Metonic cycle was used in the ancient Greek luni-solar calendar, and (much later) in the Jewish luni-solar calendar, and in the Christian ecclesiastical calendar for reckoning the dates of Easter. Yet, nowadays, the "Golden Number" of a year in the Christian calendar (which denotes its position in a Metonic cycle) does not match the position of the corresponding Jewish year in the Metonic cycles of the Jewish calendar.)
(a) If 4713 BC was year 1 of a 28-year Solar cycle of the Julian calendar and of a Metonic cycle, it follows that (counting from that year), the first year of Solar cycle 170 is year 20 AD, and the first year of Metonic cycle 250 is year 19 AD. However:
(b) in the Jewish calendar, the Jewish year corresponding to AD 21 was year 1 of a 28-year Solar Cycle, all of which begin in Julian years of the form AD 28n+21. And the Jewish year corresponding to AD 22 is the first year of a Metonic Cycle, all of which begin with an AD year of the form 19n+3.
So, since there are clearly different conventions regarding this, according to which convention(s) used during or before Scaliger's time is (a) correct, and did any of them survive into the present era, i.e. are there any systems still in use today in which Solar Cycles of the Julian Calendar begin with AD years of the form 28n+20 and where Metonic Cycles begin with AD years of the form 19n? Mottelg ( talk) 11:56, 2 March 2010 (UTC)
I think there is an error in the Gregorian - JDN calculation. The final constant should be 32077 to remove an error of 2 days. There is an even stranger error with the Julian - JDN calculation. According to http://aa.usno.navy.mil/data/docs/JulianDate.php :
JD | Date/Time from USNO | JD from Gregorian formula | JD from Julian formula |
---|---|---|---|
0.0 | BCE 4713 January 01 12:00:00 | ||
1721423.00000 | BCE 1 December 31 12:00:00 | 1721423 | |
1721424.00000 | CE 1 January 01 12:00:00 | ||
2299160.00000 | CE 1582 October 04 12:00:00 | 2299160 | |
2299161.00000 | CE 1582 October 15 12:00:00 | ||
2455315.00000 | CE 2010 April 28 12:00:00 (today) |
Unfortunately, I don't have the reference used for these calculations (L. E. Doggett, Ch. 12, "Calendars", p. 604/6, in Seidelmann 1992) to check if the article has been vandalised sometime in the past. Astronaut ( talk) 18:25, 28 April 2010 (UTC)
The program listed at http://www.astro.uu.nl/~strous/AA/en/reken/juliaansedag.html does not work correctly. It passes the few test vectors given, but if increment a JD value for a few years, you will eventually hit dates that generate nonsense like February 31. DonPMitchell ( talk) 05:32, 14 May 2010 (UTC)
In the formula for converting julian days to Gregorian dates, there appears to be a bug on this line:
When this formula is used, it sometimes gives dates like April 0 instead of March 31. The cause is the imprecise nature of floating-point computations. It is effectively dividing 5 by 153, or dividing the unmultiplied number by 30.6. On a modern computer, 153/5 does not equal 30.6 precisely. In his book Astronomical Algorithms, Meeus discusses this problem and avoids it by using the constant 30.6001 instead of 30.6 (which is equal to 153/5).
Meeus' intermediate formula that extracts the month is E = INT(some value/30.6001). It does not perform integer division, but accomplishes the same effect by using the INT function, which is the same as the FLOOR function for positive numbers.-- B.D.Mills ( T, C) 23:31, 24 May 2010 (UTC)
int j, g, dg, c, dc, b, db, a, da, y, m; double J, d, T; J = jDNum + 0.5; j = J + 32044; g = j / 146097; dg = j % 146097; c = (dg / 36524 + 1) * 3 / 4; dc = dg - c * 36524; b = dc /1461; db = dc % 1461; a = (db / 365 + 1) * 3 / 4; da = db - a * 365; y = g * 400 + c * 100 + b * 4 + a m = (da * 5 + 308) / 153 - 2; d = da - (m + 4) * 153 / 5 + 122; //Won't correctly calculate the fraction of a day T = jDNum - (int)jDNum; //fractional part of the Julian day number T -= 0.5; //because the Julian day starts at noon if(x < 0.0) T += 1; d += T; //add fraction of day *year = y - 4800 + (m + 2) / 12; *month = (m + 2) % 12 + 1; *day = d + 1; //Will be the correct day number if d is calculated as an int
I wrote a C function to convert dates in the proleptic Gregorian calendar to Julian day numbers: double ProlepticGregorianDateToJDN(int year, int month, double day) {
int a, y, m; a = (14 - month) / 12; y = year + 4800 - a; m = month + 12 * a - 3; return((day + ((153 * m + 2) / 5) + (365 * y) + (y / 4) - (y / 100) + (y / 400) - 32045) - 0.5);
} Notice that one has to subtract 0.5 from the result of the last calculation in the formula given in the article to make the fraction of a day come out correctly. This is because the algorithm fails to account for the fact that the Julian day starts at noon. Senor Cuete ( talk) 14:18, 30 November 2011 (UTC)Senor Cuete
Apparently there is another, less involved use of the term "Julian Date" in industry which simply seems to be the counting of days from January 1 of each calendar year, ending with 365 (366 in leap years) on December 31. I was made aware of this because of articles in the media on the recent salmonella contamination of eggs from Iowa. This seems to be a legitimate inclusion on this page, or there needs to be at least a further disambiguation article about this usage at the top of this article. Robert Johnson ( talk) 10:14, 20 August 2010 (UTC
In this edit User:RadioFan has changed from a clearly reliable source, The Explanatory Supplement to the Astronomical Almanac, to a self-published website by Claus Tøndering. Granted, this website is often cited by serious publications, but still, the WP:IRS guideline would clearly rank it below a book by a recognized scientific publisher.
In addition, the new edit does not follow the established citation style of this article.
So my question is, should the edit stand? Jc3s5h ( talk) 16:11, 21 September 2010 (UTC)
In the food and grocery industries the term "Julian date" is used as follows: 001 is January 1 and 365 is December 31 (unless it is a leap year and then it is 366). An example of the term being used in such a manner by a respectable source is [1]. Typically a Julian date is used to code a pack date. (On the rare occasions which dates are not coded in this manner, it is my experience that consumers mistake it for the expiration date.) Maybe the article should mention this usage of the term as well. 68.97.10.193 ( talk) 23:22, 3 February 2011 (UTC)
This algorithm pertains to the Gregorian Proleptic Calendar. The months (M) January to December are 1 to 12. For the year (Y) astronomical year numbering is used, thus 1 BC is 0, 2 BC is −1, and 4713 BC is −4712. D is the day of the month from 1 to 31 as applicable. JDN is the Julian Day Number, which pertains to the noon occurring in the corresponding calendar date.
It is necessary to use the function for Long Integers in the first line of code. This enables integers in the range of -2,147,483,648 to 2,147,483,647
[2] to be processed.
Integer division is commanded by using the back slash "\" character instead of the forward slash "/".
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
DEFLNG A-Z
INPUT "Day";d
INPUT "Month";m
INPUT "Year";y
a=(14-m)\12
y=y+4800-a
m=m+12*a-3
j=d+(153*m+2)\5+365*y+y\4-y\100+y\400-32045
PRINT "Julian Day Number =";j
END
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The Algorithm should return the following values as a cross check:-
JDN of 24 NOV, -4713 is 0 (4714 BC Gregorian Proleptic Calendar)
JDN of 15 OCT, 1582 is 2,299,161 (beginning of the Gregorian Calendar)
JDN of 16 NOV, 1858 is 2,400,000
JDN of 01 JAN, 2000 is 2,451,545
JDN of 31 AUG, 2132 is 2,500,000
Summerdrought (
talk)
21:20, 6 March 2011 (UTC)
I've checked the above algorithm but used Excel Visual Basic instead, so had to do some slight syntax adaptations, finding it works fine for all dates since oct. 15 1582. When tested with oct. 4 1582 it returns 2299150 when 2299160 is expected. Jeep IKA ( talk) 13:25, 26 May 2011 (UTC)
The article defines JDN as "The Julian day number (JDN) is the integer part of the Julian date (JD). The day commencing at the above-mentioned epoch is JDN 0. Negative values can be used for preceding dates..." According to this, the day preceding Julian day 0 is also Julian day 0. For example, while the integer part of Julian date 0.3 is 0, so is the integer part of -0.3. Clearly, what is intended is not the integer part but the "floor", that is, the greatest integer not greater than the Julian date. Unfortunately, this is not nearly as easy to express as "the integer part" is. I shall try a wording that gets around this. —Preceding unsigned comment added by 208.53.195.38 ( talk) 17:40, 25 March 2011 (UTC)
I wonder if this article should use the more international date format of day month year. Julian days are used in astronomy, which is an international science. After all, the United States is nearly alone in its use of month day, year. —Preceding unsigned comment added by Kaptain Cirk ( talk • contribs) 00:08, 19 May 2011 (UTC)
The Chronological Julian Day as suggested by Peter Meyer is a useful concept but I believe that its use is beyond that of the day in a particular time zone. It can be based on any form of local time, Local Mean Time, for example. Local time is an integral part of any calendar. The nychthemeron may begin at any point in the day/night cycle. The most common times to begin a cycle are midnight, noon, sunrise and sunset. Any of these points may be used in a Chronological Julian Day numbering system. Alexselkirk1704 ( talk) 14:43, 27 July 2011 (UTC)
Do we really need the BASIC-ish pseudocode conversion methods in this article? It's not something that is terribly useful, from either a practical or programming standpoint, and in any case this seems to be the main point of contention in the article. Surely the formula given in a standard mathematical format is both more useful for non-programmers and programmers alike.
In addition, I don't see much in the article that clearly points out that the Julian day is not really a practical date system for general use, but is almost exclusively used for a reference point in orbital elements and ephimerides. The Julian day is not "heliocentric", does not rely on or make accomodations to the vagaries of Earth's orbit (referenced to TT and not UTC), and does not easily map to a calendar (It's "year" has fractional days, not leap days). Instead there seems to be much about the Julian calendar, which is certainly related, but not the topic of this article.
What do you think? Msaunier ( talk) 11:19, 13 September 2011 (UTC)
The article currently says
and
These two seem to contradict each other; shouldn't the latter be Terrestrial Time? This amounts to a difference of over a minute. AxelBoldt ( talk) 21:20, 19 November 2011 (UTC)
"The use of Julian date to refer to the day-of-year (ordinal date) is usually considered to be incorrect although it is widely used that way in the earth sciences, computer programming, military and the food industry.[1]"
This reference is misleading. While reading one would think that the contentious statement "is usually considered to be incorrect" would be the reason for the reference. But all the article does is reference the use of Julian day in best-used-by dating for eggs, which, really, is totally irrelevant to the use Julian day and whether it is an acceptable form ordinal days.
Davisrunning ( talk) 18:19, 9 July 2012 (UTC)david july 9, 2012
The table on "Alternatives" gives two definitions for TJD, one as JD - 2440000.5, "as introduced by NASA", and the other as JD mod 10000, "NIST definition". I have been unable to find the second definition anywhere online, at NIST or elsewhere. I recall when TJD rolled over in 1995, both definitions were used by different sites, even at NASA ( A four-decimal day count originating at midnight 1968-05-23,24) but the four-decimal part may have referred only to TJD before 1995 since it became self-contradictory afterwards. Earlier versions of this article stated, "NIST treats it as cyclical". IIRC some did suppose that TJD was cyclical based upon the statement that it had only four digits, but I don't know any authoritative source, and it seems that Wikipedia and its copies are now the only reference to this definition. I can find references to five-digit TJDs, but no shorter ones after 1995. Therefore, unless anyone can find a reference I will delete the second definition. -- Nike ( talk) 05:54, 21 July 2012 (UTC) TJD 16129.245
This article as a whole seems confusing and disorganized. I will attempt to clean it up a bit. I don't intend on making major changes, but you'll let me know if I get it wrong. -- Nike ( talk) 23:17, 21 July 2012 (UTC) JD 2456130.467
Strange, I don't think I changed much, aside from removing a few redundancies and moving things around. Please point out what specifically I changed that you don't like. If anything, I think I made it less exclusive. I changed "for scientific use by the astronomy community...recommended for astronomical use by the International Astronomical Union" (which suggests only astronomers) to "used primarily by astronomers".
In fact, Julian Dates are mostly used by astronomers. Julian day numbering was introduced by an astronomer, for astronomers, to date astronomical observations, and that is the way it has been used for over a century. The definition is taken directly from the International Astronomical Union. There is nothing in the definition which precludes non-astronomers from using Julian Dates, and I have not "dismissed" anybody. In fact, there is a mention of Chronological Julian Dates being used by calendar students, despite previous controversy about their notability, and I happened to expand on that just a bit. BTW, the article is Julian day, not Julian period. The Julian Period existed nearly three centuries before Julian days.
I have not come across any historians using Julian Dates, and just a few calendar students as are already mentioned. When you look at the literature (books and journals) practically all the notable references are astronomical. But we cannot say what any other groups are doing or how Julian Dates are used or defined in any other field without knowing what that is. If you wish to research how historians, etc., have used the Julian Period or Julian Dates, you are free to add to the article. Perhaps there could be a separate article for Julian Period, although I don't know what else there is to say about it. -- Nike ( talk) 13:51, 22 July 2012 (UTC) JD 2456131.077
Since "Julian Period" redirects here, perhaps we should include more about it. I looked in Britannica and it says, "Julian period, chronological system now used chiefly by astronomers and based on the consecutive numbering of days..." I am now reading Scaliger's book. -- Nike ( talk) 20:09, 22 July 2012 (UTC) JD 2456131.339
I was having a hard enough time with Latin... -- Nike ( talk) 06:58, 23 July 2012 (UTC) JD 2,456,131.79
Google Translate works much better with German than Latin. But from what I can tell, it's not much of an issue. He does, at one point, talk about Julian days starting at midnight, so .25 is 6 am, .5 is noon, etc. -- Nike ( talk) 01:13, 24 July 2012 (UTC) JD 2456132.55
Why don't you run it through Google Translate, as I suggested? In any event, I don't see how it's relevant to the article. -- Nike ( talk) 21:04, 24 July 2012 (UTC) JD 2456133.377
I don't see how, and one author a century ago hardly constitutes a community in any event. But you might like this page which states, "The Julian Period Year is rarely used (except in the Old Farmer’s Almanac), but the strict count of days is widely used among astronomers, calendar freaks and Mayanists (not that that really encompasses that wide a constituency, by most definitions)." -- Nike ( talk) 21:47, 24 July 2012 (UTC) JD 2456133.407
I have reverted a new algorithm introduced by User:Marlin Woks for several reasons:
I changed the algorithm for the following reasons:
Sorry, I didn't realise that Wikipedia articles couldn't be referenced. Marlin Woks ( talk) 15:20, 21 August 2012 (UTC)
"If you are writing about your algorithm, I see no need to put it in Wikipedia. We have an algorithm for converting Julian dates to Julian day number. We have one for converting Gregorian dates to Julian day numbers. It is trivial for a programmer to write a program that determines whether the date is Gregorian or Julian in the situations the programmer is interested, and then invoke the appropriate algorithm. Wikipedia is an encyclopedia, not a collection of computer algorithms.
Jc3s5h (
talk)
17:23, 21 August 2012 (UTC)
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