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B Day-related calculations for BC dates

  1. Apr 1, 2016 #1
    Dear All,

    Apologies if this question isn’t really in the right place. I wanted to check I’ve understood some data correctly. The link below shows lunar cycles all the way back to 2000 BC: http://astropixels.com/ephemeris/phasescat/phasescat.html. Suppose I click on a table and get a date of 1st June 750 BC for some event or other. How could I work out the number of days between that date and, say, 1st June 751 AD? Is it simply 1500 * 365.25? Or do I need to use the length of a tropical year in such calculations.

    James.
     
  2. jcsd
  3. Apr 1, 2016 #2
    No, you have to understand the intricacies of the changes to the calendar over the years. It changed radically several time. Personally, I'd use the Jewish calendar instead, AD/BC makes date calculations annoying.
     
  4. Apr 1, 2016 #3
    Thanks for the reply. Which calendar to use, though, isn't really my choice to make, as I want to work with the dates as stated here: http://astropixels.com/ephemeris/phasescat/phasescat.html. Are you saying those dates already factor in all the intricate changes you're referring to, and I now need to unwind them?
     
  5. Apr 1, 2016 #4

    jim mcnamara

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    Staff: Mentor

    These kinds of calculations use Julian days. Otherwise astronomers would go insane.

    In the past I've written conversion software between various calendric systems. You have to realize that there are all kinds of gotchas with calendars.

    Calendars are political/religious "footballs" kicked around mercilessly for reasons both good and bad. Plus the terminology used by people in that field are arcane since it dates from Roman times and the Middle Ages (IMO). From 1582 until 1923 there were different calendars (Julian vs. Gregorian) in Christian Europe - same for Westernized North America which all got on the same "page" in 1752. Example of the nuttiness: UNIX designers decided that September 1752 was "the" month when the changeover occurred.
    Code (Text):

    $ cal 9 1752
      September 1752
    Su Mo Tu We Th Fr Sa
      1  2 14 15 16
    17 18 19 20 21 22 23
    24 25 26 27 28 29 30
     
    Fun huh?

    Christian Western Europe and the Colonial Americas a crazy quilt of countries, colonies, cities, and principalities all using one of two calendars:
    Julian Calendar (from Roman time), or the Gregorian Calendar (Pope Gregory 1582). Plus in your case the Gregorian calendar did not exit 2000 years ago. You have to use a fabricated Proleptic Gregorian calendar -
    https://en.wikipedia.org/wiki/Proleptic_Gregorian_calendar
    Here is someone with your problem:
    http://raywoodcockslatest.blogspot.com/2012/05/million-day-calendar-with-explicit.html

    Your problem is a potential headache. Assuming you know the actual date in a known system try:

    http://www.hermetic.ch/jgdc/jgdc.htm

    The UNIX "cal" program will convert 'old dates' to the Proleptic Gregorian. You can install Cygwin to run Unix under Windows
    https://www.cygwin.com/
     
  6. Apr 1, 2016 #5
    Hi Jim. Wow. Thanks a lot for the info. Unless I’ve misunderstood you, though, my problem’s more straightforward than you suggest, isn’t it? As long as I stick with dates provided from the astropixels.com site, I can work out the number of days between them just by equating a year with 365.25 days. It’s only if I have a date provided derived by some other means that life becomes difficult, right?
     
  7. Apr 1, 2016 #6
    That is...for dates pre 16th cent. AD at least
     
  8. Apr 1, 2016 #7

    jim mcnamara

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    Staff: Mentor

    No! If that is the case just use Julian days like astronomers do. No years, no months, nothing, just days.
    From wikipedia - ignore proleptic for now.

    Use this instead: http://numerical.recipes/julian.html
    Code (Text):

    Get date1 -> JD number1
    Get date2 -> JD number2
    Get the difference as number of days between by subtraction.
     
     
  9. Apr 1, 2016 #8
    Gotcha. Thanks a lot. The link's helpful too.
     
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