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    Re: Longitude via lunar altitudes, simplified
    From: Dave Walden
    Date: 2007 Feb 13, 06:51 -0800

    I would suggest Letcher's Feb 1964 article in Yachting is the earliest
    "modern" rediscovery of the longitude by lunar altitude methods.
    As Frank points out, such methods were well known more than a hundred
    of years ago.  The Ashe discussion is from: Monthly Notices of the
    Royal Asronomical Society, Vol XII, nov 1851-june 1852, page 179.
    (Available from Google books.)  Many of the astronomy and navigation
    texts from the 1800's (and there are quite a large number on google
    books) have references to longitude by lunar altitudes.  Since this
    was before LOP, most discuss only longitude assuming latitude by one
    of a number of other methods.  See for example, New Methods of finding
    the Longitude at sea or on shore, by Yarrow and Lynn, 1826, also on
    google books.
    Below is an example using one of these methods with the data from
    Bennett's recent "Longitude from Lunar Altitudes Simplified".
    Find local hour angle of star given altitude, latitude, and
    declination of star.  Find local sidereal time given LHA (sometimes t
    when measured +/-) and RA for star.
    Extend LST to time and place of lunar observation.
    Find LHA of moon given altitude, latitude, and declination of  moon.
    Interpolate into RA of moon as a function of time to find time of RA
    that gives correct LST when combined with LHA.
    HAMEL@ 6 54 23
    h       L       d
    17.440000   41.258333   23.491667
    fm sextant  known lat   alm dec
        cos t=  0.053432    calc fm h L d
        t(deg)= 86.937139   calc fm cos t
        t(hr)=  5.795809    calc fm t(deg)
        RA(hr)= 2.125222    fm almanac@6 54 23
        lst(hr)=7.921031    55.261890   15.713400
                    min     sec
    time dif st 2.246049
    MOON@ 9 7 55
    h       L       d
    37.608333   41.401667   28.538333
    fm sextant  known lat   alm dec
        cos t=  0.446633    calc fm h L d
        t=  63.472146   calc fm cos t
        t(hr)=  4.231476    13.888584   53.315014
    lst(hr)=10.167081   10.024851   1.491079    required
    ra(hr)= 5.935604    56.136268   8.176065    required
    RA@9 0 0    89.008333   5.933889    56.033333   2.000000
    RA@10 0 0   89.598333   5.973222    58.393333   23.600000
    new time=   9.043616    2.616977    37.018614
    delta time= -0.088328   -5.299690   -17.981386
    Final answer, 5 min 18 sec vs published 5 min 47sec.  As close as
    might be expected for the accuracies involved.
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