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    Re: Latitude by Lunar Distance
    From: George Huxtable
    Date: 2006 Oct 7, 11:23 +0100

    Peter Fogg has put his finger on the principle involved, without being
    aware of it.
    Yes, if parallax and refraction are ignored, the separation from any
    point on Earth will be exactly the same, and should correspond to the
    predicted separation, calculated from the positions of the two bodies
    taken from the almanac, at a known GMT. It's on that
    parallax-and-refraction difference that the proposed method depends.
    That discrepancy, between that measured separation and the predicted
    separation, will depend on the parallax and the refraction (and the
    azimuth angle between the bodies). If a best-case situation is chosen
    (and there are plenty of stars to choose from) the azimuths of the two
    bodies will be closely the same, or nearly opposite. In that case, any
    arc-minute change in the Moon's parallax will be reflected in a
    discrepancy of a similar amount, between the observed distance and the
    almanac distance at that same moment.
    The Moon's parallax can vary, according to Cos (Moon altitude) x Moon
    HP. The Moon's HP (Horizontal Parallax, the parallax when the Moon is
    on the horizon, at zero altitude) is always somewhere near 1 degree.
    Let's take it to be exactly 1 degree.
    For example, take the situation when the Moon is high, say at 80
    degrees, for which you would need to be near the tropics. That's when
    the proposed method would work best, because parallax is then changing
    most quickly with altitude when the Moon is highest.
    Then the parallax for 80 degrees alt is cos 80 x 1 degree =0 .1736
    deg. or 10.41 minutes
    For 81 degrees it would be 0.1564, or 9.36 minutes, so just over 1
    minute difference, for an altitude change of 1 degree.
    That means, if you can deduce the parallax to 1 arc-minute, then at
    those high Moon altitudes you can deduce the Moon altitude to the
    nearest degree, If you can do better with the parallax, by measuring
    lunar distance more precisely than 1 minute, you can determine Moon
    altitude correspondingly better. And this without being able to see a
    horizon. At lower altitudes, the method would become more insensitive.
    Having done that, starting from an estimated position, you can draw a
    position line on the chart, knowing the Moon's predicted GHA and dec,
    from the almanac, for that known GMT, and the deduced Moon altitude,
    just as if you had observed that Moon altitude, though a lot less
    precisely. As Alex has pointed out, even at its best, it would be
    twice as inaccurate as determining a lunar longitude by lunar distance
    without GMT, and that's difficult enough.
    I can see, then, how Frank's proposal could give one position line,
    placed at right angles to the azimuth of the Moon, on the chart,
    because the Moon's altitude has been deduced from its parallax. But
    it's only the Moon's altitude that can be deduced by this method,
    because it has such a large parallax. No information is provided on
    the altitudes of stars.
    I haven't yet understood Frank's explanation of how a second position
    line can be deduced by measuring the distance between the Moon and a
    second star, and, I too, ask for a more detailed explanation with a
    numerical example.
    contact George Huxtable at george@huxtable.u-net.com
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ----- Original Message -----
    From: "Peter Fogg" 
    Sent: Saturday, October 07, 2006 1:39 AM
    Subject: [NavList 1373] Re: Latitude by Lunar Distance
    | On 10/5/06, Frank Reed wrote:
    | >
    | >
    | > Here's an example of fixing your position using angles measured
    | > the Moon to stars only, no altitudes whatsoever. This approach
    | > that GMT is a known quantity ...
    | If the effects of parallax and refraction are ignored, then the
    | distance between two celestial bodies is going to be identical from
    | point on earth from which it can be observed at that moment? In
    other words;
    | this angular distance is invariant with respect to position?
    | For this method to work the angular distance would need to change
    | to the observer's position?
    | Could Frank perhaps explain this process in more detail, including a
    | worked example?
    | |
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