NavList:

Hello Noah.

I think you went down the wrong road by trying to turn this back into latitude and longitude on the sphere There's no trig required in this!

At the scale of the "direct LOP solution" --which Frank Reed also calls the "ABC" solution in those workshops-- you simple have two crossing lines in the plane. IT IS JUST TWO CROSSING LINES. I worked out the algebra on this myself after one of Frank's workshops a few years ago because I thought it might be even easier than drawing two lines (tldr: -MAYBE- but lines really are easy). The algebra NOT TRIG goes like this:

You have two LOPs. Call them P and Q. Each line is entirely defined by two points: P1, P2, and Q1, Q2. The coordinates of those two points on P are LATp1, LONp1, LATp2, LONp2, and similar for Q. Then calculate:

Line P coord diffs:
dLATp = LATp2 - LATp1
dLONp = LONp2 - LONp1

Line Q coord diffs:
dLATq = LATq2 - LATq1
dLONq = LONq2 - LONq1

PQ initial point diffs:
dLATx = LATq1 - LATp1
dLONx = LONq1 - LONp1

numerator, denominator of frac:
num = dLONx * dLATq - dLATx * dLONq
den = dLONp * dLATq - dLATp * dLONq

if den is zero, no intersection! (lines are parallel)

frac = num / den

LAT = LATp1 + frac * dLATP
LON = LONp1 + frac * dLONp

And that's it. That's the lat, lon of the crossing point. THAT is the TWO-BODY FIX.

Is that easir than drawing two lines on a piece of graph paper? Depends... For some navigators with some computing available, it would be. But working with a simple calculator, it could get messsy. It's nice to have this option available. I had to dig up my old spreadsheet on this to write up just now, and I'm sure I would need notes to do it in real navigation. But drawing lines? No notes needed!

Josh