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    Re: Sight Reduction Formula Question
    From: Geoffrey Kolbe
    Date: 2005 Jan 8, 09:25 +0000

    George Huxtable wrote:
    >Now consider the other formula, deriving Z from its sine. If, in this case,
    >that formula has given us sin Z = +0.98, what do we do then? Well, one
    >solution is that Z = 88.5 deg, which might well be the right answer. On the
    >other hand, sin 101.5 deg is also +0.98, so that's an equally valid answer.
    >It could be either, then. How do we discover which is the one we want? In
    >this case, there's no simple way, that I know of, to resolve the ambiguity
    >by using inspection and logic to determine whether the wanted result is
    >just North of East or just South of East.
    Geoffrey Kolbe comments:
    The attraction of creating two right angled triangles from the navigational
    triangle and using sines and cosines to solve for Hc and Z is that only one
    table is required. In Ageton's method, for example, a perpendicular is
    dropped from the geographical position of the celestial body to the
    observer's meridian to create the two right angled triangles. Ageton's
    tables are just 36 pages short. For the sight reduction part of my own
    "Long Term Almanac 2000 - 2050" I use what is essentially the Ageton
    method, (but keep the sines and cosines instead of inverting them to create
    secants and cosecants as Ageton did), and the tables occupy just 18 pages.
    George correctly identifies the choice of azimuth quadrant as being a
    potential problem where the celestial body lies near the observer's Prime
    Vertical. The ambiguity should only arise where the declination of the
    celestial body is less than the observer's estimated latitude, both are in
    the same North-South Hemisphere (have the 'same name') and  where the LHA
    is less than 90 degrees or more than 270 degrees. Otherwise, as George
    says, the choice of azimuth quadrant is easily resolved by inspection.
    Where the above conditions apply and there is an ambiguity, it turns out
    that there is a simple test to resolve this problem. One of the angles it
    is necessary to calculate to derive Hc is the co-latitude for the
    intersection of the perpendicular dropped to the observer's meridian. If
    the latitude of this intersection is greater than the observer's estimated
    latitude, the azimuth quadrant will be a Northerly one where the positions
    are in the North, and vice versa.
    Geoffrey Kolbe.

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