Welcome to the NavList Message Boards.


A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

Compose Your Message

Add Images & Files
    Re: An analytical solution of the two star sight problem of celestial navigation
    From: Bill Lionheart
    Date: 2019 Mar 9, 13:51 +0000

    I think what you mean is estimating the error in the LOPs using the
    residual square error needs more than two LOPs. Let me elaborate
    (Andres I am sure you now this but maybe not everyone understood, so
    indulge me and see if we agree!) Actually you don’t even need the
    errors to be Gaussian to do this, just mean zero (so of course you
    cant tell if you have index error with with three LOPS). The
    Gauss-Markov theorem is what is needed here.  I attach a snipped from
    the book Åke Björck Numerical Methods in Matrix Computations, Ch 2
    which I recommend. There is a version for "weighted least squares"
    too.  With m>2 LOPS and now assuming the errors are normally
    distributed the residual will be distributed as Chi squared with m-2
    DOF giving more possibilities for analysis.
    However better still is to look at the regression line for a series of
    points for the same body. The residual here gives a better way to
    estimate the variance of the measurements and for n observations with
    Gaussian errors is distributed  as Chi squared with n-2 DOF, in this
    case as the linear fit has two parameters slope and intercept. So even
    for observations of two bodies, but each with multiple sights, we get
    the variance of each LOP. Typically  these will be weighted, so you
    need the weighted least squares approach (weighted symmedian, weighted
    ellipse). Even for two LOPs this is an ellipse (in general) but there
    is an interesting twist. Suppose you are quick at making observations
    at twilight but you only have a few celestial bodies visible and the
    azimuths are not very well distributed. Then you can decide to take
    more sights on some than the others, and so control the variances.
    Then one can choose the ratios of the variances to make the
    probability contours circles.
    Bill Lionheart
    On Thu, 7 Mar 2019 at 20:33, Andrés Ruiz  wrote:
    > Robin,
    > interesting indeed.
    > In a nautical sense the ellipse depends on the standard deviation of the 
    estimated position, and it is proportional to the inverse of the square root 
    of number of sights minus two: sigma = K sqrt(1/(n-2)), so n must be greater 
    than two.
    > Real world, not the imaginary one. Mathematics and the practicality of 
    things. The difference between a theoretical science an engineering...
    > Yes, parallel lines intersect at a point which lies at infinity.
    > I admire your approaches to navigation problems.
    > --
    > Andrés Ruiz
    > Navigational Algorithms
    > http://sites.google.com/site/navigationalalgorithms/
    > View and reply to this message


    Browse Files

    Drop Files


    What is NavList?

    Join NavList

    (please, no nicknames or handles)
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.

    Email Settings

    Posting Code:

    Custom Index

    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site