ECE 495 — Electromagnetic Warfare and Survivability

L31 wrote survivability as a chain of links, $P_K = P_D \cdot P_{T|D} \cdot P_{I|T} \cdot P_{H|I} \cdot P_{K|H}$, and multiplied the encounters into $P_S$.

Today we draw that map — the ring each site throws — and find out why it is never the circle on the planning chart.

By the end of this lesson, you will be able to:

• Draw an aspect-dependent detection contour from the R⁴ law and the B-21 pattern
• Bound it with the radar horizon, and say which bound binds at an altitude
• Distinguish the MEZ from the LEZ, and what turning away does in each
• Read a layered laydown for overlaps, gaps, and seams

The X-band TTR on a 10 m mast, the B-21 nose-on: R⁴ gives 45.6 km. At 30 m the Earth cuts that to 35.6 km.

Now climb to 150 m. Which bound takes over — and where do the two cross?

At 150 m the horizon is out at 63.5 km; the signature, not the Earth, now sets 45.6 km.

The crossing sits where $d = R$: $h^\ast = \left(R/4.12 - \sqrt{h_r}\right)^2$. The quieter you are, the lower it falls — the same mast crosses the VHF EWR near 350 m.

A missile flies at an average speed $v_m$ for at most $t_{\max}$; the target flies at $v_t$.

• The MEZ is the outer boundary; the LEZ is the no-escape zone inside it
• Inside $R_{\min}$ the missile cannot guide or arm in time — a hole at the center

A constant-speed teaching model. "MEZ" is also Joint airspace-control vocabulary (JP 3-01); air-to-air literature says "no-escape zone."

Type this with me — one radar, one aspect, three altitudes.

% L32 -- which bound binds: the horizon or the R^4 law?
Rref = 80;  sref = 1;          % radar: 80 km on a 1 m^2 target
sig  = 0.5;                    % RCS it sees at this aspect [m^2]
hr   = 12;                     % radar mast height [m]
ht   = [30 150 600];           % three target altitudes [m]
Rdet = Rref*(sig/sref)^(1/4);  % R^4 law: one range [km]
dhor = 4.12*(sqrt(hr) + sqrt(ht));   % 4/3-Earth horizon [km]
for k = 1:3
  if dhor(k) < Rdet, lim = 'horizon'; else, lim = 'R^4'; end
  fprintf('%3d m: horizon %5.1f  R^4 %4.1f -> %4.1f km (%s)\n', ...
          ht(k), dhor(k), Rdet, min(dhor(k), Rdet), lim);
end

At 150 m the horizon still wins, by 2.6 km. What altitude flips it?

Then: raise the mast to 30 m. Which row changes its answer?

A low-level ingress against one SAM site: its TTR, and the missile behind it.