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    Lunar longitudes, not by lunar distance. Was- Re: Working a lunar
    From: George Huxtable
    Date: 2009 Aug 7, 14:59 +0100

    Brad Morris has tried generalising this discussion to simply undefined
    "objects", in the sky.
    
    But that general case is of little interest; one of those objects simply
    MUST be the Moon, because of its fast motion with respect to everything
    else.! What Brad says may be correct; but if two stars are chosen, the
    difference in time of their meridiam transits is the same, not only along
    your meridian, but the whole World over, for ever (neglecting the minute
    affects of precession and nutation). So there's no position information to
    be gained from it.
    
    George.
    
    contact George Huxtable, at  george{at}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: "Brad Morris" 
    To: 
    Sent: Thursday, August 06, 2009 9:34 PM
    Subject: [NavList 9392] Re: Working a lunar - a PS
    
    
    Hi Hanno
    
    I have been considering your statement "If I see things right, there must be
    a LOP which connects all those locations on Earth that have a given, fixed
    difference DT between the meridian passages of sun and moon"
    
    Just to be sure I understand your statement, I will re-write it.  The first
    object crosses your meridian. Let us assume that it is the sun.  While this
    is LAN, we don't care about that, you merely start your timepiece stopwatch.
    Next we wait for the second object to cross your meridian.  When it does,
    you stop your timepiece.  We observe the delta time.  From this one data
    point, we are expecting a LOP.
    
    This has nothing to do with the altitudes of the objects, just the Delta
    Time of the meridian crossing.
    
    One problem (not insurmountable) is that the two celestial objects have
    apparent diameters.  As such, we must perform a few more measurements.  That
    is, assuming you are using an older theodolite with 5 wires, you would
    record the 5 times that the leading limb crosses each wire and the 5 times
    that the trailing limb crosses each wire, and then mathematically determine
    when the object was on your meridian.
    
    The next problem (not insurmountable) is to align the theodolite to your
    meridian. Even Bowditch in the 1800's knew how to do this.  However, the
    requirement to align the theodolite to the meridian precludes any use of
    this method whilst at sea.
    
    Now which LOP corresponds to the Delta Time (DT)?  I suggest to you that it
    is your MERIDIAN.  Anyone, at any other latitude, that is on your longitude
    will measure the same precise value that you do.  Now there's an interesting
    outcome!  Sure, the altitudes will be different, but the DT will be the
    same.
    
    Can we tell which meridian?  Considering that the moon essentially travels
    it's diameter in an hour, we run right into the resolution problem, very
    similar to the Lunar lack of resolution.   Since we will be measuring with a
    theodolite, we will have a very good measurement for the time of meridian
    crossings, assuming that the theodolite is aligned to the meridian to
    perfection.   Assuming you measure precisely and accurately, then the answer
    is yes.
    
    Best Regards
    Brad
    
    
    
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