# Lesson 32 Flashcards

Click a question to reveal the answer.

<div data-flashcards data-deck="b4l32"></div>

<details>
<summary><strong>1. Why is a threat ring on a planning chart not really a circle?</strong></summary>
<div class="card-answer"><p>Because the circle is one special case: one radar's detection range at one aspect (usually nose-on), at an altitude where the Earth does not block the line of sight, against a missile shot at a target flying straight at the site. Your heading changes the aspect the radar sees at every bearing, your altitude brings in the horizon, and what you do after launch changes the missile's reach.</p></div>
</details>

<details>
<summary><strong>2. An aircraft on heading <span class="math notranslate nohighlight">\(\psi\)</span> sits at bearing <span class="math notranslate nohighlight">\(\varphi\)</span> from a radar. What aspect does the radar see?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\theta = \mathrm{wrap}(\varphi + 180^\circ - \psi)\)</span>, measured clockwise from the nose. From the aircraft the radar lies at bearing <span class="math notranslate nohighlight">\(\varphi + 180^\circ\)</span>; subtracting the heading measures that from the nose. <span class="math notranslate nohighlight">\(\theta = 0^\circ\)</span> is nose-on, <span class="math notranslate nohighlight">\(180^\circ\)</span> tail-on.</p></div>
</details>

<details>
<summary><strong>3. Write the aspect-dependent detection contour.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R(\varphi) = R_\text{ref}\,[\sigma_b(\theta(\varphi))/\sigma_\text{ref}]^{1/4}\)</span>: the R⁴ law evaluated, bearing by bearing, at the aspect that bearing produces, with <span class="math notranslate nohighlight">\(\sigma_b\)</span> read from the signature pattern in the radar's band.</p></div>
</details>

<details>
<summary><strong>4. On one contour, which quantity is the same everywhere and which one changes?</strong></summary>
<div class="card-answer"><p>The heading is the same everywhere; only the bearing from the radar changes, and the bearing alone decides which face the radar sees — nose-on upstream of the site on your track, the beam abeam of it, the tail past it.</p></div>
</details>

<details>
<summary><strong>5. What shape is the B-21's X-band contour against Project 3's TTR, and why?</strong></summary>
<div class="card-answer"><p>A cross: 45.6 km nose-on, needles to 91.0 km at the two beams, 60.8 km at the tail, and pinched to 21.6 km in the quarters between them. At X-band shaping works, and the notional pattern gathers its energy into a small nose spike (−15 dBsm) and large beam spikes (−3 dBsm) over a −28 dBsm floor.</p></div>
</details>

<details>
<summary><strong>6. Why is the VHF contour nearly a circle?</strong></summary>
<div class="card-answer"><p>At VHF the airframe's features are comparable to the wavelength (the resonance region), so shaping cannot steer the echo: the pattern moves only about 2 dB (+3 to +5 dBsm) around the compass, and the fourth root turns 2 dB into about a 12% change in range (89.9 to 100.9 km).</p></div>
</details>

<details>
<summary><strong>7. Why can routing on the nose-on ring alone mislead you?</strong></summary>
<div class="card-answer"><p>It is honest only along the line straight at the site. Passing abeam, you present your beam and the contour spikes far outside the ring; in the quarters the real contour sits well inside it. A route planned to skirt the nose-on ring can be detected much earlier than planned.</p></div>
</details>

<details>
<summary><strong>8. State the two-ended radar horizon and its units.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(d = 4.12\left(\sqrt{h_r} + \sqrt{h_t}\right)\)</span> km with <span class="math notranslate nohighlight">\(h_r\)</span> (radar mast) and <span class="math notranslate nohighlight">\(h_t\)</span> (aircraft altitude) in meters, using the 4/3-Earth refraction allowance. L11's one-ended <span class="math notranslate nohighlight">\(4.12\sqrt{h}\)</span> is one of the two terms.</p></div>
</details>

<details>
<summary><strong>9. How do the horizon and the R⁴ law combine?</strong></summary>
<div class="card-answer"><p>Detection needs both an echo strong enough and a line of sight, so the effective range is the smaller: <span class="math notranslate nohighlight">\(R_\text{eff}(\varphi) = \min(R(\varphi), d)\)</span>. The two bounds never add.</p></div>
</details>

<details>
<summary><strong>10. At what altitude does the horizon stop binding?</strong></summary>
<div class="card-answer"><p>Where <span class="math notranslate nohighlight">\(d = R\)</span>: <span class="math notranslate nohighlight">\(h^\ast = (R/4.12 - \sqrt{h_r})^2\)</span>. Below it the Earth sets detection; above it the signature does. For the TTR on a 10 m mast seeing the B-21 nose-on at 45.6 km, <span class="math notranslate nohighlight">\(h^\ast \approx 62.5\)</span> m; against the VHF EWR's 89.9 km it is about 350 m.</p></div>
</details>

<details>
<summary><strong>11. Why do low observability and low altitude interact?</strong></summary>
<div class="card-answer"><p>A stealthier signature shortens the R⁴ ranges, which lowers the crossing altitude <span class="math notranslate nohighlight">\(h^\ast\)</span>: the quieter you are, the lower you must fly before the horizon adds anything. Terrain-following buys most against the long-range, low-band layers and little against a TTR that already sees you late.</p></div>
</details>

<details>
<summary><strong>12. Define the MEZ and the LEZ in the constant-speed model.</strong></summary>
<div class="card-answer"><p>MEZ: the reach against a target flying straight at the site, <span class="math notranslate nohighlight">\(R_\text{MEZ} = (v_m + v_t)\,t_\text{max}\)</span>. LEZ (the lethal or no-escape zone): the reach against a target that turns directly away, <span class="math notranslate nohighlight">\(R_\text{LEZ} = (v_m - v_t)\,t_\text{max}\)</span>. It is a teaching model — real envelopes come from fly-out models.</p></div>
</details>

<details>
<summary><strong>13. What does turning directly away do between the LEZ and the MEZ, and inside the LEZ?</strong></summary>
<div class="card-answer"><p>Between the rings the turn escapes: the missile must chase at <span class="math notranslate nohighlight">\(v_m - v_t\)</span> and runs out of flight time. Inside the LEZ even a fleeing target is caught — no escape. The turn costs progress toward the target and shows the radar your tail.</p></div>
</details>

<details>
<summary><strong>14. What is the heading-dependent envelope, and what angle does it use?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R(\alpha) \approx (v_m + v_t\cos\alpha)\,t_\text{max}\)</span>, the radial approximation, with <span class="math notranslate nohighlight">\(\alpha\)</span> the angle between your velocity and the direction to the site — the same angle as the aspect <span class="math notranslate nohighlight">\(\theta\)</span>. It gives the MEZ at <span class="math notranslate nohighlight">\(0^\circ\)</span>, the LEZ at <span class="math notranslate nohighlight">\(180^\circ\)</span>, and <span class="math notranslate nohighlight">\(v_m t_\text{max}\)</span> abeam.</p></div>
</details>

<details>
<summary><strong>15. What is a site's lethal zone, and what sets the first shot for the B-21 at altitude?</strong></summary>
<div class="card-answer"><p>Inside the tracking radar's effective contour, inside the missile envelope, and outside the minimum range <span class="math notranslate nohighlight">\(R_\text{min}\)</span>. Nose-on against the TTR the track starts at 45.6 km inside a 50 km MEZ, so the track — not the missile — sets the first shot.</p></div>
</details>

<details>
<summary><strong>16. Define an overlap, a gap and a seam in a layered laydown.</strong></summary>
<div class="card-answer"><p>Overlap: a point inside two or more lethal zones. Gap: inside early-warning coverage but outside every lethal zone — seen, not shot. Seam: the boundary where one site's coverage hands off to the next, where tracks drop and cues arrive late. All three are heading-dependent.</p></div>
</details>

<details>
<summary><strong>17. Why does a gap between two SAMs change when you change heading?</strong></summary>
<div class="card-answer"><p>Both halves of each lethal zone depend on heading: the TTR contour through the aspect's <span class="math notranslate nohighlight">\(\sigma\)</span>, and the missile envelope through <span class="math notranslate nohighlight">\(\cos\alpha\)</span>. In the lesson's laydown the gap is 20 km at 090°, 24 km at 000°, and 57 km at 045°, where both sites see a quiet quarter of the X-band pattern.</p></div>
</details>

<details>
<summary><strong>18. In Joint doctrine, what is a "missile engagement zone," and how does this course use the term?</strong></summary>
<div class="card-answer"><p>In Joint doctrine (JP 3-01; the DoD Dictionary) it is airspace in which engagement responsibility rests with surface-to-air missiles — an airspace-control measure. Threat analysis and mission planning borrow the letters for a site's kinematic reach, which is the sense used here; air-to-air literature calls the inner zone the no-escape zone.</p></div>
</details>
