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    Re: Baffled by Baffin
    From: George Huxtable
    Date: 2005 Nov 24, 00:47 -0000

    Tony Crowley asks a really interesting question about Baffin's lunar
    observations, which I have not read (but perhaps should).
    
    It might help our reconstruction if he gave us more details: the date and
    the estimasted latitude. Then we can work out the phase of the Moon, etc..
    
    My comments are in square brackets.
    
    I have been reading the OVoyages of William Baffin? (Hakluyt Society), but
    am having a problem with the description of his initial attempt to find his
    longitude on the west coast of Greenland (pp 20 & 21). Putting aside the
    shortcomings of the lunar culmination method,  his data and description
    suggest that he saw the moon as outstripping the sun in its daily passage
    around the earth.
    
    ?Thursday, the ninth day, very early in the morning, I went on shore the
    island, being a faire morning, and observed till the moone came just upon
    the meridian.
    
    [does he give any clue as to how he could tell tnat the Moon was on the
    meridian? Had he previously marked a North-South line? That's the tricky
    part of Baffin's method.]
    
    At which instant I observed the sunne?s height, and found it 8
    51.? He then goes on to describe that by the use of spherical trignometry
    involving his latitude, the sun?s declination and the sun?s altitude, he
    found the sun?s local hour angle. Converting this to local time, he found
    the time of the observation to be 4h 17m 24s.
    
    [Presumably, times in that era were measured from noon. In that case his
    deduced time of 4h 17m 24s was after noon though he says it was morning.. If
    you use a trig formula to get LHA  Sun from alt, lat, and dec, it gives a
    value for LHA that is the angular difference from noon, without
    distinguishing between before noon and after noon. So he seems to be telling
    us that the time of the Moon being on the meridian was actually 4h 17m 24s
    before noon-by-the-Sun, or 7h 42m 36s am by modern reckoning? Would that fit
    his altitude of 8 deg 51'? Can we work it backwards, given the missing
    data?]
    
    My problem is what follows next. ?Which when I had done, I found by mine
    ephemerides, that the moone came to the meridian at London that morning at
    4h 25m 34s which 17 minutes 24 seconds subtracted from 25minutes 34 seconds
    leaveth 8min 10secs of time for the difference of longitude betwixt the
    meridian of London (for which the ephemerides was made) and the meridian
    passing by this place in Groenland.? He then describes the moon?s motion
    that day as 48mins and 29 secs and describes it as Othe time that the moone
    commeth to the meridian sooner that day then she did the day before, give
    360, the whole circumference of the earth; what shall 8 minutes 10 seconds
    give, to wit 60 degrees, 30 minutes.?
    
    [ Again, a time at London of 4h 25m 34s would not be in the morning, unless
    he is assessing it as LHA=4h 25m 34s before noon, i.e. 7h 34m 26s am, by our
    modern way of reckoning time.]
    
    Has anyone else seen this and can they offer any explanation. Baffin writes
    with much clarity and the problem does not arise elesewhere in his account.
    As his pioneering attempt to measure longitude was so very far out, I have
    wondered if he looked up the wrong data in Searle?s Ephemerides.
    
    ====================
    Comment from George.
    
    How I see it is this; The Moon was presumably at the time West of the Sun,
    in Last Quarter. The Sun was lagging the Moon across the sky,  by an amount
    somewhat greater than 4 hours. That lag was reducing by about an hour each
    day, because the Moon, like the tides it causes, is about an hour later each
    day (actually in this case later each day by 48m 29s, not an hour, according
    to Baffin)  So if the lag of the Sun following the moon has decreased by 8m
    10s, in the time it's taken to get round to Greenland, then that puts
    Greenland at a longitude of 360 x 8m 10s / 48m 29s. It all seems perfectly
    logical, and would indeed be a good way to determine lunars without a
    timekeeper, if the difficulty of observing that Moon culmination could be
    overcome.
    
    However, I can't reconcile that with the quoted statement that "the Moone
    cometh to the meridian sooner that day than she did the day before". It
    seems to me that Baffin has made a mistake about that, and got the sign
    wrong.
    
    George
    
    contact George Huxtable at george---.u-net.com
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    
    
    

       
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