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    Re: Sight Reduction method accuracy
    From: Mike Boersma
    Date: 2004 Apr 5, 00:00 -0400

    Please forgive my ignorance in asking this question.
    What is the difference between using the time of the meridian transit of
    the moon to find longitude and using the lunar distance method of
    determining GMT, thus longitude? The information for determining
    meridian transit is on the daily page of the Nautical almanac as are the
    tables for determining longitude from MT of the moon. Are these methods
    Michael Boersma
    Frank Reed wrote:
    > Hello, everybody. Just a quick note...
    > George H asked:
    > "For example, tell us how you predicted lunar distances, in the days
    > after they
    > disappeared from almanacs, but before calculators existed."
    > Dunraven wrote on this a bit less than a hundred years ago:
    > "Lunar Distances for every third hour of M.T.G. [GMT] used to be
    > published in the Nautical Almanac issued by the Admiralty; for some
    > reason they are now discontinued in that work. They are still
    > tabulated in the official Nautical Almanac of the United States of
    > America; and lunar distances between the Sun and Moon only are printed
    > in Brown's Nautical Almanac, a British publication. It is not,
    > however, a difficult matter to find the true distance between the Moon
    > and any other Heavenly Body, so if you are curious to find your
    > Longitude by this method you can do so without the tabulated
    > Distances."
    > He then goes on to describe the solution of the appropriate spherical
    > triangle.
    > Letcher, writing almost thirty years ago, in his "Self Contained
    > Celestial Navigation with H.O. 208" puts the steps in terms that a
    > modern navigator raised on tabular methods can easily understand. The
    > trick is to imagine yourself to be standing on the Earth directly
    > under the Moon at some instant of time. Then if you want the distance
    > between the Sun and Moon at that moment, you simply calculate the
    > Sun's altitude as seen from that location and time. And since the Moon
    > is at the zenith by setup, the lunar distance you need is just 90 -
    > the calculated altitude. Note that this is the predicted geocentric
    > lunar distance (no adjustments should be made to the Sun's altitude
    > for refraction or any other factors). So let's say that at some
    > specific time the Moon's GHA is 120d 14.5' and its Dec is 5d 31.0'
    > North. And assume that the Sun's GHA is 170d 43.4' and its Dec is 5d
    > 58.7' South. All I have to do to get the predicted lunar distance is
    > enter my sight reduction tables pretending that my position on the
    > Earth is identical to the Moon's position in the sky. That is, I make
    > my DR position 120d 14.5' W and 5d 31.0' N. From that position I
    > calculate the altitude of the Sun (it would be in the west roughly 45
    > degrees high). Finally I subtract that altitude from 90. That's the
    > required lunar distance. Normally you would repeat the calculation for
    > an hour later.
    > Predicting lunars is every bit as easy as calculating altitudes.
    > Having pre-calculated distances in an almanac saves maybe five minutes
    > of table work, but that's all.
    > Frank E. Reed
    > [ ] Mystic, Connecticut
    > [X] Chicago, Illinois

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