Block 2 · Lesson 14 · Where We Left Off
L13 turned a measured phase into a bearing. But one bearing is a line — the emitter is somewhere along it, near or far, we cannot tell.
Cross a second bearing from a second receiver and the lines meet. That intersection is a fix.
Block 2 · Lesson 14 · Objectives
By the end of this lesson, you will be able to:
Block 2 · Lesson 14 · Part One
Two receivers, two bearings, one point — if the geometry cooperates. That last clause is the whole lesson.
Block 2 · Lesson 14 · Triangulation
Each receiver gives a line of position (LOP) — the emitter lies somewhere along its bearing. Two LOPs cross at the fix.
Two receivers, two bearings, one point — if the geometry cooperates.
Block 2 · Lesson 14 · The Cut
Each bearing carries a noise band. The fix lives where the two bands overlap.
A wide cut overlaps in a small, near-circular patch. What happens to that patch as the two bearings swing toward parallel?
Block 2 · Lesson 14 · The Cut
Same receivers, same bearing noise — a shorter baseline, and the bands now slide along each other.
The patch is 5× longer than it is wide, and the long axis points down-range — toward and away from the receivers. Push it all the way to collinear and you get a seam: no cut at all, and range is unobservable.
Block 2 · Lesson 14 · Fusion
With three or more bearings, the lines almost never meet at a single point — noise pulls them apart.
Block 2 · Lesson 14 · Part Two
Crossing bearings across platforms is how standoff geolocation works — and the bookkeeping that feeds it is where the whole thing can quietly go wrong.
Block 2 · Lesson 14 · Networked ESM
Crossing bearings across platforms is how standoff geolocation works:
Block 2 · Lesson 14 · The Chain
Block 2 · Lesson 14 · Association
Two receivers each report a bearing. Crossing them assumes they heard the same emitter. In a dense scene, that is a real question.
L14_TwoReceiverCrossFix.m
Block 2 · Lesson 14 · Quick Exercise
Type this with me — a B-21 standoff leg, RWR bearings good to 0.5°.
sig = deg2rad(0.5); % RWR bearing 1-sigma
wp = [-60 -25; -60 25]; % WP1, WP2 as [N E], km
d = [0 0] - wp; % [dN dE] to the emitter at (0,0)
R = hypot(d(:,1), d(:,2)); % range from each waypoint, km
th = atan2d(d(:,2), d(:,1)); % bearing, North toward East
cut = abs(th(1) - th(2)); % crossing ("cut") angle, deg
ecr = R(1)*sig*1e3; % one bearing, across the LOS, m
a = ecr/(sqrt(2)*sind(cut/2)); % down-range semi-axis, m
b = ecr/(sqrt(2)*cosd(cut/2)); % cross-range semi-axis, m
fprintf('R = %.1f km bearings %+.1f / %+.1f deg\n', R(1), th);
fprintf('cut = %.1f deg e_cr = %.0f m\n', cut, ecr);
fprintf('ellipse a = %.0f m b = %.0f m CEP = %.0f m\n', ...
a, b, 0.59*(a+b));
The ellipse comes out 2.4× longer one way. Which way, and why?
Then: double the leg, or halve the standoff — which buys more?
Block 2 · Lesson 14 · Wrap-Up