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    Re: navigation problem
    From: Frank Reed CT
    Date: 2006 Feb 25, 02:45 EST

    Dick W wrote:
    "The moon was barely visible  above the fog.  They computed the way back.
    For any meridian the moon  will cross just 49 minutes after the sun crosses
    and the moon falls back 49  minutes every day. By counting the days after new
    moon, multiplying that number  by 49 and dividing by 60, one arrives at the
    time in hours and minutes when the  moon is due south. With south noted and
    knowing the reciprocal bearing back the  ship they could safely return."
    This will get you a rough bearing.  Because, as George H has already pointed
    out, the Moon's motion in hour angle is  not uniform (both because its orbital
    speed is variable and because the path  through the sky is tilted with
    respect to the celestial equator), it will not  cross the meridian with regular
    timing. That said, as long as the Moon is not  too high in the sky, this will
    still yield a decent azimuth, accurate within 10  to 15 degrees or so. Also note
    that this would be a true bearing. If they kept  the bearing outbound with a
    magnetic compass, that can be a significant  correction, but I think we can
    assume that anyone smart to think about the  Moon's bearing is smart enough to
    know the magnetic variation.
    By the  way, I don't think it's true that pocket watches aboard ship were
    uncommon,  especially at the end of the century. But there's another wrinkle
    here. Suppose  I work out the time of the Moon's meridian passage and discover
    that it's due at  02:30. But my watch reads midnight. Do I sit around and wait?
    With yet a little  more knowledge of astronomy and a good estimate of your
    latitude, you could  convert the Moon's hour angle to azimuth even if it's not on
    the  meridian.
    So this method will work --up to a point. But is it plausible?  It depends on
    someone knowing the date of New Moon as well as the 49 minute per  day rule.
    I would suggest that any sailor or young officer  with that  knowledge would
    also be likely to have one other bit of knowledge that's more  immediately
    useful: the Moon's position in the sky from the previous night. If  you spend any
    amount of time on deck at all when the Moon is visible, you'll  probably have
    a pretty good idea of its compass bearing at various times. And  it's easy to
    estimate where it would be one day later.
    By the way, for  the mathematically inclined, there's a nice way to think
    about the delay of the  Moon's arrival at the meridian. The 49 minute rule tracks
    a fictitious "Mean  Moon" in the same way that an ordinary clock tracks the
    "Mean Sun". The delay of  the Sun's arrival at the meridian every day is the
    "Equation of Time" and it  depends on the inclination of the Sun's path to the
    celestial equator and the  eccentricity of the Earth's orbit in about equal
    measures. We can define a  similar "Moon Equation of Time" and calculate it based
    on the inclination of the  Moon's path through the sky and the eccentricity
    of the Moon's orbit. Since the  Moon's orbit is about three times more
    eccentric than the Earth's, this will be  a bigger, dominant contribution for the Moon
    Eq.T (about 1600 seconds in  amplitude vs 350-900 seconds for the inclination
    term). Additionally, there is  no fixed relationship between the phase of the
    inclination contribution and the  eccentricity contribution. But it can all
    be worked out, and if I've done it  right, the result is that a "typical" Moon
    Equation of Time value is about 20  minutes with extreme values around 40
    minutes. In other words, that's how early  or late the Moon will arrive on the
    meridian compared to the fictitious Mean  Moon. Using that typical "20 minute"
    value translates to a typical azimuth error  of about 5 degrees. Add in other
    sources of error, like the uncertainty in the  exact time of New Moon, and the
    average azimuth error will climb to 10 or 15  degrees.
    42.0N 87.7W, or 41.4N  72.1W.

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