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    Re: Lunar longitudes, not by lunar distance.
    From: George Huxtable
    Date: 2009 Aug 9, 20:58 +0100

    Sorry, all.
    
    A few minutes ago, I posted the following, in which there was a silly error. 
    It doesn't (I think, at first sight) affect the conclusions, but that will 
    call for a bit more thought.
    
    When I wrote "That shift (in subtended angle in the sky, not in azimuth) 
    isn't far from 15� per hour, mainly due to the Earth's rotation.". Clearly, 
    that was quite wrong, as I realised as soon as I re-read it when posted back 
    to me!
    
    The shift of 15� per hour is shift about the Earth's polar axis, not shift 
    in subtended angle in the sky (nor shift in azimuth). The hourly movement in 
    subtended angle is roughly similar in amount, for bodies at lowish 
    declinations, such as Sun, Moon, planets, but clearly much less for 
    higher-declination stars.
    
    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.
    
    ==================================
    
    I wrote
    
    | >Indeed, that would be the case, if we were measuring azimuths, but we're
    | >simply timing the Moon as it passes a vertical line, so the correction is
    | >constant in time, independent of altitude.
    
    and Geoffrey replied-
    
    | Sorry George, am I missing something here? The point I was alluding
    | to is that when you time the edge of the moon as it touches the
    | vertical hairline, the centre of the moon is then not just a
    | 'semi-diameter' away, but a 'semi-diameter divided by the cosine of
    | the altitude' away in azimuth. Since the tables for the moon's
    | position are for the centre of the moon, that is the correction you
    | must make to determine the LHA between the moon and the star at that
    | moment in time. Is that not so?
    
    Well, if you wanted to correct for semidiameter by timing the Moon limb as
    it passed an azimuth that had been deliberately offset slightly from the
    meridian, then you would indeed have to correct that azimuth for meridian
    altitude, as Geoffrey explains. But why would you do it that way? It's the
    difference in hour-angles (or else celestial longitudes) between Moon and
    star, that's needed in the end to deduce the time; not the difference
    between azimuths, which has only local significance.
    
    Around transit, the motion of all sky objects is close to being exactly
    horizontal, altitude changing little. That shift (in subtended angle in the
    sky, not in azimuth) isn't far from 15� per hour, mainly due to the Earth's
    rotation. A limb of the Sun  is replaced by the other limb, 0.5� or so away,
    about 2 minutes later in time. It doesn't matter how high in the sky the Sun
    is when observed (which depends on observer's latitude) that fact remains.
    Similar considerations apply to the Moon. So the correction IN TIME between
    the Moon's limb crossing the true meridian, and the moment the centre would
    cross it, will always amount to about 1 minute of time. This will be quite
    independent of the observer's latitude. Of course, to predict it exactly
    right all sorts of effects have to be allowed for, but these present no
    serious problems.
    
    With Sun transits you can simply split the difference, always about 2
    minutes, between the times of the two limbs crossing the meridian, to get
    the moment the centre would cross it. However, that isn't possible with the
    Moon, which is why a numerical allowance must be predicted, instead.
    
    
    
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