Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!samsung!munnari.oz.au!metro!otc!brendan
From: brendan@otc.otca.oz (Brendan Jones)
Newsgroups: comp.lang.c
Subject: Re: Leap Year Checker.
Message-ID: <1950@otc.otca.oz>
Date: 5 Oct 90 00:26:09 GMT
References: <2188@ukc>
Lines: 34

in article <2188@ukc>, tlg@ukc.ac.uk (T.L.Goodwin) says:
> Aha, but the Russian system (where the year is a leap year iff the
> remainder on division by 7 is 2 or 6) is actually more accurate* than
> the Gregorian system, as well as not having the confusion at the end of
> a century.
> *i.e. better compensates for the fact that 1 year != 365 days

GARBAGE!

The Gregorian system (with 97 leap years over its 400 year repeating cycle) 
makes the length of an average year 365 days 5 hours 49 minutes and 
12.0 seconds.

The length of a siderial year (orbit of Earth around the Sun referenced to
the stars) is 365 days 6 hours 9 minutes 9.55 seconds (increasing by 95
microseconds per year), hence the Gregorian system underestimates this by 
19 minutes 57.5 seconds per year.  I think this standard is called UT0.

The Russian system has 2 leap years every 7 years, or an average year length 
of 365 days 6 hours 51 minutes and 25.7 seconds.  This overestimates the
length of a siderial year by 42 minutes 16.15 seconds per year, much worse
than the Gregorian.

However, the sidereal year is not used any more for accurate scientfic
purposes, UTC is now the time standard.  I believe the Gregorian system 
approaches this year length very closely, hence the need for only about 1 
leap second per year compensation.  Unfortunately, I do not have the reference 
for a year length referred to UT1 or UTC.  Perhaps someone can help.

-- 
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