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    Re: Ellipse of Confidence in position finding
    From: Bill Lionheart
    Date: 2019 Mar 13, 07:51 +0000
    Andres

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    On Tue, 12 Mar 2019 at 02:33, Andrés Ruiz <NoReply_AndresRuiz@fer3.com> wrote:
    Bill you wrote http://fer3.com/arc/m2.aspx/An-analytical-solution-two-star-sight-problem-celestial-Lionheart-mar-2019-g44520

    Yes, I got "Numerical Methods in Matrix Computations", is a very good reference.
    There, Theorem 2.1.1
    m = number of sights and n = 2 

    I recommend Chapter 2 of this book to anyone who wants to understand a bit more about least squares solutions and residuals. It is important to understand that there is some merit in least squares even for non Gaussian errors!

     

    I try to explain it.
    To draw a circle in a paper, you need 3 restrictions. Like in a 2D CAD program. A circle is a special case of an ellipse.
    It is a 2D problen with 3 DOF, degrees of freedom
    1- pin the fist point of the circle -> the circle can rotate around this pin (now we have 2 DOF)
    2- pin a different second point of the circle -> it can rotate through the center through the two fixed points (now we have 1 DOF)
    3 - pin the 3rd point ->  No DOF, fixed circle.


    Andres I expect you know this well but let me try to re-explain it. One thing to think about here is for any number m of LoPs,  the matrix A^TA is 2 by 2  symmetric so has three degrees of freedom,  as it has three distinct elements. One can also use two eigenvalues and the direction of one eigenvector as the 3 DOF. These also define the lengths of the semi major and minor axes and the direction of one axis, hence an ellipse with a given centre. Except in the case where all the LOPS are parallel (including m=1) when the matrix has rank 1, one of the eigenvalues is zero and the ellipse is degenerate. 


     
    Another important thing about ellipses. Remember my post with an example where the EoC, the symmedian point, and others give us the wrong position.
    Fix by bisectors


    I had a bit of a look at that thread and saw a lot of confusion.  You get a probability density function and that is correct given the assumptions. As navigators we have sometimes to choose a point estimate from the PDF, for example as departure point for our next dead reckoning. One should always understand a "fix" as just a point estimate of a probability distribution and so consider the risks to navigation of being at the "most dangerous" location within a probability contour.

    I think practically the common reason the symmedian is not a good point estimate is that the variances of the LoPs are different, eg they were obtained from linear regression fit for a different number of altitude observation of the same body. Then we need the weighted symmedian (weighted least squares solution) as our maximum likelihood estimate. It is still inside the cocked hat for three LoPs but it could be anywhere in the interior depending on the weights.

    I hope this is adding to clarity rather than causing more confusion!

    Bill Lionheart


     

       
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