NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Frank Reed
Date: 2010 Dec 7, 18:20 -0800
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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