# NavList:

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

**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 Ruizwrote: > > 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