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About julian days

  1. Jun 15, 2007 #1
    the julian day formula goes like this:

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

    For a date in the Gregorian calendar (at noon):

    [tex]\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}[/tex]

    For a date in the Julian calendar (at noon):

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

    I do not understand how each step works, so can someone please guide me through?
     
    Last edited by a moderator: Jun 22, 2007
  2. jcsd
  3. Jun 15, 2007 #2

    D H

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    It looks like the latter two are formulae for some kind of modified Julian date rather than a Julian date. They also don't like right. Where did you get these? Much better formulae are at the http://aa.usno.navy.mil/faq/docs/JD_Formula.html" [Broken].
     
    Last edited by a moderator: May 2, 2017
  4. Jun 21, 2007 #3
    See

    http://www.vsg.cape.com/~pbaum/date/date0.htm [Broken]

    for a derivation.
     
    Last edited by a moderator: May 2, 2017
  5. Jun 22, 2007 #4

    Chronos

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    Another source to consider:
    http://aa.usno.navy.mil/faq/docs/JD_Formula.html [Broken]
     
    Last edited by a moderator: May 2, 2017
  6. Jun 24, 2007 #5
    thanks guys.
     
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