A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Andrés Ruiz
Date: 2006 Oct 23, 11:09 +0200
I don’t know, but I do not receive some e-mails.
I see the 1447 and others in http://www.fer3.com/arc/
[NavList 1447] Re:
Navigational Algorithms - required assumed position for SR ?
Date: 19 Oct 2006 11:49
Andres Ruiz wrote-
Indeed using the "DeWit/USNO/Compac Data", least squares algorithm for
reduction of n sights, you can suppose any assumed position far away
from the real position. Because it do an iterative find of the fix, if
a solution exist, after few iterations it is found.
If the assumed position is near the estimated one, DR, or the
celestial position, only one or two iterations are required.
Try it with the software available at my web page: n LOPs Fix
The algorithm is described in the Nautical Almanac and at [n LOPs Fix]:
Comment from George-
In the case when there are only two bodies observed, then there must
always be two solutions, at the two crossing-points of the circles,
which are usually easy to distinguish, but under unusual circumstances
may be close together.
Depending on the assumed position you choose to start with, a
least-squares algorithm homes in on one or the other, which may
possibly not be the one that's relevant.
How do you ensure that both possible solutions have been flushed out,
I have discussed this some time ago with Herbert Prinz, who, as best I
remember, suggested that having found one solution, starting from your
best guess at assumed position, you then started it off again from the
antipode of that assumed position, which should be sure to end up at
the other solution. It made sense, to me.
Mathematically an iterative process finds the roots o an equation by randomly supposing the starting point.
The bisection and the Newton’s method work like this.
In the world of celestial navigation, if two bodies are shot there are two or one solutions,
if errors are acceptable there must be solution.
*Two solutions: The two COPs have two points of intersection. Our DR position determines what the real one is.
*One solution: if the two COPs are tangents. This theorist case is very improbable in navigation.
For the 2 solutions case I have tested what George proposed for several examples. For the 1st solution the starting point is (B,L),
and for 2nd one, the antipodal point Ba = -B, La = L+180º
This assumption work well for the line connecting the two solution points oriented N-S, W-E, NE-SW, and NW-SE,
but I don’t know if there are some cases that fail.
Visually You can check the solutions, (Mike Dorl has done a good program, Sights.exe, that plots the COPs in a world chart),
and decide the initial point in the iterative procedure for each solution.
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