NavList:

Antoine,
Thank you for posting your calculation results. Looks good! Very close to the center of the short side of the triangle.

You noted:
"Results from own Software used since the early 1980's"

Wow. Early 1980s? That's impressive. I hope you are running it on newer hardware?? :) Oh, but I can't get the image out of my head: I'm now imagining you with one of those Compaq clones of the original IBM PC, carefully preserved in a closet somewhere, and a 5.25-inch floppy disk chattering away in a disk drive as it loads DOS 3.3 and runs your old code... Ha ha.

You had it even earlier, but the mathematical procedure has been widely, almost universally available since the late 1980s in the back of every copy of the Nautical Almanac. I think that was when I first encountered it. But amazingly enough (and really it's rather astonishing) most navigators and celestial navigation enthusiasts today in the year 2025 have no idea how or why that math works, and they simply ignore it. The argument is a bit hypocritical: they may claim that they're "not interested in fancy math" or maybe "nobody trusts statistics"... but then they'll actually brag about the fact that they can "work with spherical trigonometry". Showing off "fancy math" in one breath and denouncing it in the next breath!

So I have been on a quest ...an extremely slow-moving quest. :) For some years now, when I find the time, I have been coming up with techniques that could give us these fixes in the cases where "dropping a dot" is insufficient. For example: consider a simple right triangle, like the case I have posted here. Suppose its long side height is h and its shorter side, its base, has length b. If we make some simple assumption that produces a "thin" triangle, h ≫ b, then we have a relatively simple math problem. If I drop a point inside the triangle, what location minimizes the sum of the squared distances from the sides of that triangle? It's a little geometry to set up the triangle and distances and then a little calculus to minimize the sum of the squares. The result is clean and simple, and it covers a great many of the cases of interest in real navigation fixes.

Details later today.

You concluded:
"Not claiming that it is the actual position though because the Error Ellipse is important, but it should be reasonable for most practical purposes."

Well, this point is the center of the error ellipse, so it's certainly important, right? Our goal from any set of sights is to get an error ellipse (or family of ellipses), which is defined by its semimajor and semiminor axes and its center. We need that center! After we have that ellipse, the lines of position that entered into it... evaporate.

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA