Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!uunet!mcsun!cernvax!chx400!chx400!ugun2b!ugsc2a!fisher
From: fisher@sc2a.unige.ch
Newsgroups: comp.misc
Subject: Re: Julian day numbers
Message-ID: <1991Apr25.103747.406@sc2a.unige.ch>
Date: 25 Apr 91 08:37:47 GMT
References: <41601@cup.portal.com>
Organization: University of Geneva, Switzerland
Lines: 57

In article <41601@cup.portal.com>, J_Otto_Tennant@cup.portal.com writes:
> In the introductory chapter of _Numerical_Recipies_, programs are 
> presented for the calculation of the Julian day number given a 
> date in the form of month/day/year, and vice-versa.
> 
> The key trick, as described in a recent article in "Dr. Dobb's",
> is to renumber months so that February is last.  Given that wonderful
> hack (which numbers March as "4" and February as "15"), the next step
> is given as:
> 
> 	JULDAY = INT(365.25*JY)+INT(30.6001*JM)+ID+1720995
> 
> [...]
>
> From my point of view, there are two aspects to the incomprehensibility
> of 30.6001.
> 
> [...]

It's easily understood:  as the "year" starts with March you want to add
nothing for March, 31 days for April, 31+30 days for May, etc.  Taking the
INT( (NUM(Month)-NUM(March))*30.5 + 0.5) gives you alternating months of
31 and 30 days.

But every five months, you need to add an extra day, and substract 0.5 from
the original offset of 0.5.  This means you add a half day every five month,
or 0.1 every month.  This is why you take 30.6 instead of 30.5.

Now depending on the hardware (in some cases the software), the result of
30.6 * 5 is 153.0 or 152.99999...  Small difference, but the resulting INT()
changes.  That is the reason for the 30.6001  (INT(JM*30.6+0.001) would work
also, of course).

  Month NUM(Month)  NUM*30.6001      Prev. months   Last month

   Mar.     4        122.4004                0       (28/29)
   Apr.     5        153.0005 *             31         31
   May      6        183.6006               61         30
   Jun.     7        214.2007               92         31
   Jul.     8        244.8008              122         30
   Aug.     9        275.4009              153         31
   Sep.    10        306.0010 *            184         31
   Oct.    11        336.6011              214         30
   Nov.    12        367.2012              245         31
   Dec.    13        397.8013              275         30
   Jan.    14        428.4014              306         31
   Feb.    15        459.0015 *            337         31

The needed column is the total of days in the previous months, given by
INT(NUM(Month)*30.6001) - 122.  Of course, the 122 is included in the
constant 1720995 in the example.

NOTE: the example works only from 1. March 1990 to 28. Feb. 2100...

Regards,

Markus Fischer, Dept of Anthropology, Geneva CH