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

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

```Andres

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.
>
> --
> Andrés Ruiz
>
> View and reply to this message

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