NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: What is better 5 times 3 or 3 times 5 sights?
From: Bill Lionheart
Date: 2022 Oct 10, 21:58 +0100
From: Bill Lionheart
Date: 2022 Oct 10, 21:58 +0100
Its not an answer to the question but related. If you have three position lines the most probable position, (assuming the errors are identically distributed zero mean normal variables) is the symmedian point which is easy to find by construction...and is surprisingly close to the shortest side of the triangle.
You can also construct a confidence ellipse (but its quite tedious ... I have a you tube video of me doing it.)
So what about more than three lines? What is the equivalent of the symmedian point? Well it is easy enough to solve arithmetically, or with a computer program, but can one draw it?
It turns out you can. The encylopedia of poly figures labels it nL-n-P6 and gives a construction
There is a litttle more in the paper I wrote with Kimberling and Moses in the Jounal of Navigation : preprint here
As to which is best 3 of 5 or 5 of 3 it is worth noting if you took m sets on n parallel lines you get the same answer averaging each of the n lines first as taking the least squares point of mn lines. So going back to the question it is the same as saying "which 15 lines are best?" I would say the one with the most circular ellipse of uncertainty is a good choice and often that would be the one using 5 stars averaging over 3 sights of each.
Bill Lionheart.
On Mon, 10 Oct 2022, 17:02 Herman Dekker, <NoReply_Dekker@fer3.com> wrote:
Franks last comment about the Fairchild Maxson LOP Computer.
Computation is cheap so we can detect errors and anomalies by taking more sights.
Did me think What is better 5 times 3 or 3 times 5 sights?
When I tried to solve a problem from: 100 problems in celestial navigation and had to draw
6 position lines I was overwelmd and had no idea what fix to choose.
Therefore I normally stop when I have 3 (average of 5) sights.
So max. 3 positionlines that is understandable for me.
Or gives 5 sights (Average of 3) a more reliable fix?
For me it gives only more confusion.Is there any logical ot mathematical appoach for this?
regards
hermanD
