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    Re: Graphical Solution to Double Altitude Sight using Stereographic Projection
    From: Robin Stuart
    Date: 2010 Jun 14, 16:34 -0700

    George,
    Thank you for your input and for bringing my post and query again to the fore. It risked being completely swept aside by the active exchanges on equal vs. double altitude methods!

    I have had a limited number of conversations at Mystic and offlist but so far no one recalls having seen this method.

    You ask:
    "I wonder if there's an analogy, here, with the use of the astrolabe (astronomers' not mariners'), which maps the stars of the Northern sky with exactly such a projection"

    I would expect that these sorts of ruler and compass methods occur fairly quickly to anyone who gets "hands on" with a stereographic grid and indeed occurred to me while I was preparing the figure (attached to previous post) for the Navigation weekend at Mystic. I would also expect that these general sorts of methods might have been employed in the practical construction of astrolabes. The principal difference that I see between mariners and astronomers is that small circles are the domain of the former while, with the possible exception of parallels of latitude/declination, astronomers tend to deal with great circles.

    It should be noted that although, as you correctly say, the purely graphical treatment is "applicable only to a restricted set of practical cases" the basic approach can be extended to situations were one of the vertices of the circle of position (CoP) lies at a high latitude (and is consequently off the chart) if one is willing to engage in a very limited amount of straightforward calculation (far less than that needed for a full blown double altitude sight reduction). The latitude of the center of the CoP can be found from

    tan( 45 + Lc / 2 ) = [ tan( 45 + ( dec + ZD ) / 2 ) + tan( 45 + ( dec - ZD ) / 2 ) ] / 2

    where Lc is the required latitude of the centre of the CoP on the stereographic chart, dec is the declination of the Sun or other celestial body and ZD is its measured zenithal distance. With the exception of 3 additions/subtractions and a division by 2 the rest of the calculation could be relegated to a table in tan( 45 + x/2 ). The drawing compass could then be set to the circle's radius using the calculated position of the centre and the one "on chart" vertex and the CoP drawn as before,

    Regards,

    Robin Stuart
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