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    Re: Can I Navigate Without an Assumed Position?
    From: Bill Lionheart
    Date: 2017 Sep 24, 18:34 +0100

    For the case of three circles of position (generally not great
    circles) I wonder if there is an equivalent geometric construction to
    the symmedian point giving the least squares solution. I would expect
    it to have the same tendency to be closer to the shorter side of the
    "triangle".
    
    Bill
    
    On 24 September 2017 at 17:38, Frank Reed  wrote:
    > John Howard, you wrote:
    > "Just as an exercise, I used a globe and string to get an AP at 60° N and
    > 115° W."
    >
    > But that's way off! May I suggest that you try this again? With a common
    > globe, you should expect to get a position this way within one or two
    > degrees. Even if you mark up a common ball (like a 10-inch diameter ball
    > that you might find in the toy department of Walmart) with a bit of care,
    > you can get a position this way within 3-5°.
    >
    > And by the way, for others following along who may not realize this, this is
    > not technically an "AP". It's a fix. The globe is serving as an analog
    > computer, and the result, when done right, is your position. Given that the
    > accuracy is rather low, you could then feed that into a standard celestial
    > calculation and use this as a starting point or "assumed position".
    >
    > To reiterate, in the real world, you're never this lost. These are game
    > problems, not navigation. Also in the real world, a sight consists of the
    > altitude as well as an approximate azimuth, which really ought to be given
    > in the statement of the problem. You know what direction you were facing
    > within 10 or 20 degrees, and this significantly constrains the calculation,
    > whether you're working by direct computation or by analog on a globe. If you
    > know that Spica, e.g. was in the SSE at some altitude, h, then you measure
    > off a distance of 90°-h away from Spica's GHA and dec (its "GP") and draw an
    > arc of a circle. That much is obvious. But you don't need to draw the whole
    > circle; you just draw that portion that "sees" Spica in the south-southeast,
    > or equivalently, that portion of the circle which is perpendicular to SSE.
    >
    > Frank Reed
    >
    > View and reply to this message
    
    
    
    -- 
    Professor of Applied Mathematics
    University of Manchester
    http://www.maths.manchester.ac.uk/bl
    

       
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