NavList:

A few days ago, I wrote:
"Here's a key piece of it... For "thin" triangles with short side or "base" of length b, and height equal to h, when the ratio h/b is considerably greater than one... effectively defining a "thin" triangle, the true least squares fix is located just above the mid-point of the base at a height, y, relative to the base length, b, equal to half the base divided by the height:
    y / b = (1/2) b / h
This is an exact result for the limiting case where b/h is small and a surprisingly accurate result for much more moderate values of b/h. In words: "the true least-squares fix in a thin triangle of three lines of position is located just inside the triangle above mid-point of the short side with height above the short side relative to the length of the short side equal to one-half the short side length divided by the long side length"

And I added: "It's not hard to prove this rule for "thin" isosceles triangles by a little brute force --yet relatively simple-- calculus work. Similarly it is not hard to prove this for "thin" right triangles by a bit more calculus. But the real fun comes next... There is a really slick way of demonstrating this rule and a bunch of other fascinating transformations, in the specific case of "right" triangles for fixes."

The slick way of demonstrating this rule that I promised is an application of a simple rotation.

Suppose I have two lines of position crossing at 90°. For the initial setup, let's suppose that one LOP runs north-south and the other runs east-west. The fix that those two LOPs generate is simply the crossing point. And the error ellipse around that fix is a perfect circle. Now suppose Instead that one LOP runs NW-SE and the other runs NE-SW. The two LOPs are still separated by 90°. The fix is in the same location, and the error ellipse is the same perfect circle. Rotating this pair or LOPs has yielded exactly the same navigational outcome. Got it??

This equivalence lets us replace any pair of lines of position with a 90° separation by any other pair of LOPs centered at the same location with the same 90° separation, but rotated to any convenient angle. The navigational information produced by those rotated LOPs, namely the fix and the error ellipse, are identical. Also notice that the 90° angular separation does not have to be perfect. A few degrees difference witll negligibly change the shape of the error ellipse. You can think of any pair of LOPs in a plot with a separation of 90° like a rotor on the page. Pick it up and spin it. Turn the pair to any angle that makes the geometry simpler... and drop it. You can convert a problem with no obvious solution into a beautifully simple case with a trivial solution. It's a miracle.

And now we can solve the "thin triangle" that started all of this. I pick up the pair at the right angle corner... and I turn it until it's aligned parallel with the remaining side of the triangle. Drop it there. The equivalent lines of position have the simplest possible three-body crossing arrangement: two parallel LOPs with a third crossing both at a right angle. The fix falls halfway between the two parallel LOPs. It's that easy. Try other long thin triangles, and you'll discover that the numbers also match the simple analytical rule that I described previously: y / b = (1/2) b / h.

Ideally, I would make a neat animation of this, but for the time being a few clean, unanimated plot diagrams should be enough. See below...

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA

File:

File:

File: