NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Peter Hakel
Date: 2010 Dec 7, 22:42 -0800
Since the location of these "Poles" is really arbitrary, we can choose a customized coordinate system such that one of its Poles coincides with the first fix given by the two mutually perpendicular LOPs. The azimuth of the star is now equivalent to the choice of a "meridian" radiating from this "Pole." The rotational symmetry of this arrangement is now clear. On such a "polar chart" the 3-body fix is always halfway between the 2-body fix point and the star LOP, regardless of its azimuth ("meridian"), i.e. Antoine's answer. This geometric situation is similar to John Karl's estimated position construction with the DR position being the "Pole."
I cannot recall where I read it but some text described the use of a (real) Pole as the assumed position for the purposes of St. Hilaire.
The plotting sheets we use for St. Hilaire are based on the Mercator projection. The cocked hat (and hence the location of the fix thus derived) will be distorted on such a map. If the star LOP is "almost parallel" to one of the Sun LOPs, then their intersection on the chart will be far away in a region where the map's distortion is too big to ignore, at which point the cocked hat becomes unusable.
Peter Hakel
From: Frank Reed <FrankReed@HistoricalAtlas.com>
To: NavList@fer3.com
Sent: Tue, December 7, 2010 6:20:47 PM
Subject: [NavList] Re: A simple three-body fix puzzle
Antoine, you wrote:
"At a distance d/2 from point P ?" and "... and azimut equal to the star azimut +/- 180 degrees."
Yep! That's it.
Gary and Greg both gave methods that are very close, certainly good enough for normal chart plotting, but there's an interesting feature that some of the common graphical methods miss out on: the correct answer should be rotationally symmetrical. In other words, the azimuth of the third LOP doesn't affect the distance of the fix from the crossing point of the two perpendicular LOPs. It's d/2 in any direction. The distance obviously has that value in the special case where the third LOP is parallel to either of the others. But I don't think it's obvious, on the face of it, that it should be the same distance when the third LOP is at some other angle. If you want to work out the exact least squares solution in the general case, turn to the back of your Nautical Almanac and have a look at those equations for generating a fix from a set of intercepts p_i and azimuths Z_i. Set the first two intercepts to zero, the azimuths to 0 and 90 degrees, and leave the third as p and z. There's a little algebra, but it all drops out nicely in the end. Oh, and just ignore those longitude factors to avoid any extra complication; do the problem on the equator.
So WHY is this solution rotationally symmetrical? Why is the three-LOP fix located at a distance d/2 from the two-LOP fix (when the first two LOPs are perpendicular --not otherwise) for any azimuth despite the fact that the "cocked hat" diagrams are so very different for different azimuths? I have one answer that leads to some interesting tricks (and has some bearing on John Karl's "running fix" technique), but before I describe that, does anyone else have an easy way of explaining why this should be rotationally symmetrical?
-FER
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