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    Re: using old nautical almanacs
    From: George Huxtable
    Date: 2009 Mar 10, 10:55 -0000

    Bill asked a question, copied below, about a discrepancy. I had quoted the
    length of a year as 365.2422 days (to 4 places- it can go much further!).
    Bill pointed out that the instructions that come with the Nautical Almanac,
    for extrapolating Sun predictions to the following year (page 261), allow
    for an offset in GHA of 87�, or in time of 5 hours 48 minutes, which would
    correspond to 365.2417 days, not 365.2422 , and asked the reason for the
    discrepancy.
    
    The Almanac is approximating here, providing a figure that's "good enough 
    for
    government work". They could well have stated a more precise time difference
    to be, not 5 hours 48 minutes, but 5 hours, 48 minutes, 0.76 seconds, if
    they had thought fit. That's a quantity that has been well known for
    millennia. Indeed, Julius Caesar, when establishing his calendar, could have
    got something like that from Hipparchus, if the Romans had chosen to ask the
    Greeks.
    
    But what would be the point? As the almanac states, the resulting difference
    in the Sun's position would be negligible compared with the pulling-about of
    the Sun's position by other planets, which differs from year to year so it
    can't be predicted in a simple way, and can amount to 0.4'.
    
    ===================
    
    Bill may find it of interest to see how our Gregorian Calendar adjusts to
    cater for that figure of 365.7422, by throwing in an occasional leap year
    day, or sometimes omitting it..
    
    If we never had a leap year, all years would be 365 days long.
    
    Throwing in a leap year of 366 days, one year in four, would mean that each
    4-year period would have 1461 days, an average of 365.25.
    
    However, omitting that leap year in each century-year would mean that each
    century would contain 36424 days, an average of 365.24.
    
    But then reinstating that leap year when the century year divides by 400 (as
    happened on Feb 29 2000) means that in 400 years there are 146097 days,
    averaging 365.2425. That, I think, is where we have got to as of now, and
    you can see we are getting pretty close to the intended value of 365.2422.
    
    The next step would be to choose, every 4th millennium, to omit the leap
    year that would otherwise occur every 400 years, which would bring the
    four-millennium total to 1460969, and the average to 365.24225. I don't know
    whether the Almanac compilers, calendar makers, and computer system builders
    have been alerted to that one yet, but they had better be careful not to be
    taken by surprise in AD 4000!
    
    George.
    
    contact George Huxtable, at  george@hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    
    
    ----- Original Message ----- 
    From: "Bill" 
    To: 
    Sent: Tuesday, March 10, 2009 8:01 AM
    Subject: [NavList 7618] Re: FW: [San Francisco Sailing] reusing old nautical
    almanacs
    
    
    |
    | George wrote
    |
    | > Motion of Sun and stars is very regular and predictable. The only real
    | > problem is the awkwardness of our calendar, because a year doesn't
    | > correspond to a whole number of days. The Sun doesn't return to the same
    | > point in the sky until about 365 and-a-quarter days have passed. So
    after
    | > most calendar years of 365 days, it has lagged by a quarter-day, or so.
    At
    | > the fourth year, we put in an extra day, to put it right again.
    | >
    | > If it was exactly a quarter-day out, then choosing a four-year gap would
    do
    | > the trick rather exactly, but actually a year is more like 365.2422
    days,
    | > not 365.25, which gives rise to the residual error.
    |
    | I mostly understand the argument regarding 365.25 days per year given some
    | years are missing leap years to adjust. I'll accept, given the model used
    in
    | cel nav, "The Sun doesn't return to the same point in the sky until about
    | 365 and-a-quarter days have passed" as being equivalent to the earth not
    | reaching the same position in its orbit around the sun until *about* 365
    | and-a-quarter days have passed.
    |
    | Yet I am missing something subtle. To use the almanac for the next (non
    | leap) year the NA uses an 87 degree offset, nominally 5 hours and 48
    minutes
    | (5.8 hours) using 15d per hour as opposed to 6 hours (if it did indeed
    take
    | the sun and extra quarter day to return to the same orbital position).
    | 5.8/24 = an extra .2417 days (total 365.2417), while George stated
    365.2422.
    | In either case, baring surprises, the current leap years system seems to
    be
    | good for thousands of years before it runs into trouble.
    |
    | Given George's batting average, I will accept his figure and ask what
    | factor(s) are responsible for the difference between my calculation based
    on
    | the NA and his figure.
    |
    | Bill B.
    |
    |
    |
    |
    | |
    |
    |
    
    
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