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    Re: Lunars using Bennett
    From: Alexandre Eremenko
    Date: 2008 Apr 3, 22:11 -0400

    Dear D. Walden,
    Thanks for your interesting message.
    You cite my old message saying that Bennet
    is apparently not good for the Lunars.
    Let me make more precise what I meant.
    The tables in Bennett do not have sufficient
    accuracy for the Lunars, because they are
    rounded to 1'. This applies both to the reduction
    tables and to the almanach part.
    Thus using these tables
    you cannot
    do better than 1' of distance in principle
    (and perhaps 2' in practice, because
    you need to use the tables and the almanach
    several times in one
    and the errors will accumulate).
    Such errors in the distance in general will give you
    an error of 30'-60'  in the longitude which is not
    very good.
    Still, if I remember correctly, even for such accuracy
    a (partial) Longitude Prize was offered:-)
    As I understand, in your example, the error in the true
    distance was 0.6 (this is because of the rounding in
    the Bennet almanac) and the error in reduction was
    1'.4 (which is better than I expected).
    I have one question on your calculation: when using
    Frank'a calculator, why did you convert your center-to-center
    distance to  the apparent
    distance by subtracting 2SD? You had to subtract
    (Sun SD + Moon SD) instead.
    P.S. I am also a big fan of Bennet,
    especially when I have to
    travel by an airplane. Then I carry a pocket sextant
    (scale divided to 1') and Bennett, and nothing else.
    This saves a lot of weight in my luggage, and the pocket
    sextant is an ideal match for the Bennet tables.
    I also use them with my bubble horizon aircraft sextant.
    But when using a full size real sextant for the Lunars
    I usually struggle for the tenths of a minute in the
    distance, and
    here Bennett is not appropriate.
    Navigation List archive: www.fer3.com/arc
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