Terminal descent — velocity vector alignment, touchdown constraints: Worked examples

3.5.54 · D3 · Physics › Guidance, Navigation & Control (GNC) › Terminal descent — velocity vector alignment, touchdown cons

Yeh page parent topic ka exhaustive drill hai. Hum ek guidance formula aur touchdown envelope lete hain aur unhe har possible case se guzaarte hain: har sign, degenerate inputs (zero altitude gap, zero drift), dangerous limit ( $t_{g o} \to 0$ ), ek real word problem, aur ek exam twist.

Shuru karne se pehle, hum un do objects ko re-anchor karte hain jo hum har jagah reuse karenge, simple words mein.

Definition Symbols, clearly explained

$r$ = current position, $r_{f}$ = target position (pad). $r_{f} - r$ = "gap" arrow jo hamare se pad ki taraf point karta hai.
$v$ = current velocity (ek arrow: kitni fast aur kis direction mein move kar rahe hain), $v_{f}$ = desired touchdown velocity.
$t_{g o} = t_{f} - t$ = time-to-go (touchdown tak kitne seconds bache hain). Dekho Time-to-go Estimation.
$g = (0, 0, - g)$ = gravity, seedha neeche point karta ek arrow. Iske teen components $(g_{x}, g_{y}, g_{z})$ hain; kyunki gravity ka koi sideways part nahi hai, $g_{x} = g_{y} = 0$ aur vertical component $g_{z} = - g$ hai (negative kyunki down $= - z$ ). Mars par $g = 3.71 m/s^{2}$ , toh $g_{z} = - 3.71$ . Jab bhi neeche $g_{z}$ dikhein, iska matlab sirf " $g$ ka $z$ -component" hai, jo $- g$ ke barabar hai.
$T$ = engine se thrust force (ek arrow), $m$ = vehicle ki mass kg mein. Toh $T / m$ woh acceleration hai jo engine produce karta hai (force per kilogram). Yeh law engine ko command karta hai ki exactly $T / m = a_{c m d}$ banao.
$v_{t d}$ = vertical touchdown speed (contact par kitni fast sink karte hain, m/s, jaise 1.5). $v_{h} = v_{x}^{2} + v_{y}^{2}$ = horizontal touchdown speed (contact par sideways drift). $θ$ = body tilt angle vertical se.
Is poori page par sign convention: up $= + z$ , down $= - z$ . Downward speed ek negative number hota hai.

Formula Tilt aur drift kyun add hote hain — the tip-angle rule

Ek lander tab topple hota hai jab uska centre of mass leg footprint ke edge se baar ho jaata hai. Contact par do alag cheezein vehicle ko us taraf lean karaati hain:

the body tilt $θ$ — thrust axis already $θ$ se off-vertical hai;
the velocity lean — velocity arrow vertical sink $v_{t d}$ ko adjacent side aur horizontal drift $v_{h}$ ko opposite side wala right triangle banata hai, toh yeh vertical se $arctan (v_{h} / v_{t d})$ off point karta hai.

Arctan kyun, aur $v_{h} / v_{t d}$ kyun? $tan (angle) = \frac{opposite}{adjacent} = \frac{v _{h}}{v _{t d}}$ measure karta hai ki arrow vertical se kitna lean karta hai; $arctan$ inverse question hai "kis angle ka yeh ratio hai?" — yeh ek slope ko wapas angle mein convert karta hai. Dono leans ek hi taraf push karte hain, toh yeh add hote hain: $θ_{eff} = θ + arctan \frac{v _{h}}{v _{t d}} \leq θ_{t i p}$ Landing tab tip-safe hai jab yeh sum tip threshold $θ_{t i p}$ ke andar rahe. Example 7 ek aisa case work karta hai jahan yeh fail ho jaata hai.

Scenario matrix

Is topic ke baare mein har solvable question in cells mein se kisi ek mein aata hai. Neeche ke examples un cell(s) ke saath tagged hain jo woh cover karte hain, aur saath milke yeh poora table fill karte hain.

Cell	Kya alag banaata hai	Example
A — Vertical, braking	Fast gir raha hai, slow karna hai: $a_{c m d}$ positive (up) aata hai	Ex 1
B — Vertical, mild	Slow descent, gap chota: $a_{c m d}$ hover ke paas	Ex 2
C — Full 3-D, mixed signs	Horizontal + vertical, negative gaps dono taraf	Ex 3 (figure)
D — Degenerate: zero gap	$r_{f} = r$ : position term vanish ho jaata hai	Ex 4
E — Degenerate: zero drift	$v_{x} = v_{y} = 0$ already: alignment term horizontally kuch nahi karta	Ex 4
F — Limiting: $t_{g o} \to 0$	The $1/ t_{g o}^{2}$ blow-up; the floor/handoff fix	Ex 5 (figure)
G — Envelope AND-check	Saari 5 touchdown inequalities, ek fail	Ex 6
H — Tilt + drift geometry	Combined tip angle $θ + arctan (v_{h} / v_{t d})$	Ex 7 (figure)
I — Word problem (real)	English → numbers → command translate karo	Ex 8
J — Exam twist	Ulta solve karo required $t_{g o}$ ke liye	Ex 9

Worked Examples

Worked example Example 1 — Cell A: vertical braking (positive command)

1-D vertical, Mars $g = 3.71 m/s^{2}$ . Gap $r_{f} - r = - 40$ m (40 m descend karna hai), current $v = - 25$ m/s, target $v_{f} = - 1.5$ m/s, $t_{g o} = 4$ s. $a_{c m d}$ find karo.

Forecast: hum 25 m/s par plunge kar rahe hain — kya engine command up (positive) hogi ya down? Padhne se pehle guess karo.

Position term $\frac{6}{t _{g o}^{2}} (r_{f} - r) = \frac{6}{16} (- 40) = - 15 m/s^{2}$ . Yeh step kyun? Yeh measure karta hai ki hum kitna mispositioned hain; yahaan pad hamare neeche hai toh yeh command ko down pull karta hai.
Velocity term $- \frac{2}{t _{g o}} (2 v + v_{f}) = - \frac{2}{4} (2 (- 25) + (- 1.5)) = - \frac{1}{2} (- 51.5) = + 25.75 m/s^{2}$ . Yeh step kyun? Hum desired se bahut fast move kar rahe hain; yeh term speed bleed karne ke liye up push karta hai.
Gravity cancel $- g_{z} = - (- 3.71) = + 3.71 m/s^{2}$ (yaad raho $g_{z} = - g$ $g$ ka vertical component hai). Yeh step kyun? Woh $+ g$ remove karta hai jo duniya add karti hai, taaki trajectory intended braking shape follow kare.
Sum: $a_{c m d} = - 15 + 25.75 + 3.71 = 14.46 m/s^{2}$ (up). Inhe add kyun karo? Teen terms independent jobs hain (position, velocity, gravity) jo same axis par act karti hain, toh engine ko produce karna hua total acceleration sirf unka sum hai — ek line of motion par superposition.

Verify: braking term (+25.75) position term (−15) ko dominate karta hai kyunki hum hard overspeeding kar rahe hain — physically sensible. Net world acceleration $= a_{c m d} + g_{w or l d} = 14.46 - 3.71 = 10.75 m/s^{2}$ up, yani real deceleration. Units: saare terms m/s². ✅

Worked example Example 2 — Cell B: mild descent near hover

Vertical, Moon $g = 1.62 m/s^{2}$ . Gap $r_{f} - r = - 2$ m, $v = - 1.0$ m/s, $v_{f} = - 1.0$ m/s, $t_{g o} = 2$ s.

Forecast: hum already target speed par hain aur almost wahan pahunch gaye — kya command sirf gravity cancel karne ke paas honi chahiye? Guess karo.

Position term $\frac{6}{4} (- 2) = - 3.0 m/s^{2}$ . Kyun? Chota gap, chota correction — lekin note karo yeh still non-zero hai, hume pad par neeche tug karta hai.
Velocity term $- \frac{2}{2} (2 (- 1.0) + (- 1.0)) = - (1) (- 3.0) = + 3.0 m/s^{2}$ . Kyun? $2 v + v_{f} = - 3$ zero nahi hai even though $v = v_{f}$ : " $2 v$ " current speed ko twice weight karta hai, toh yeh abhi bhi braking command karta hai.
Gravity cancel $- g_{z} = - (- 1.62) = + 1.62$ .
Sum: $a_{c m d} = - 3.0 + 3.0 + 1.62 = 1.62 m/s^{2}$ (up). Inhe add kyun karo? Same superposition: position pull (−3) aur velocity push (+3) exactly cancel ho jaate hain, toh jo bacha woh pure gravity cancellation hai — hover.

Verify: yeh exactly $g$ aaya — position pull aur velocity push cancel ho gaye, sirf hover bacha. Yahi "mild" signature hai. ✅

Worked example Example 3 — Cell C: full 3-D, mixed signs (figure)

Earth-like $g = 9.81$ , $g = (0, 0, - 9.81)$ . Current $r = (30, - 10, 100)$ m, target $r_{f} = (0, 0, 0)$ (pad at origin, ground level). $v = (- 4, 2, - 20)$ m/s, $v_{f} = (0, 0, - 1.5)$ m/s, $t_{g o} = 5$ s. $a_{c m d}$ find karo.

Forecast: $x$ mein hum $+ 30$ par hain aur $- 4$ se move kar rahe hain (already wapas ja rahe hain). Kya $x$ -command positive hogi ya negative? Guess karo.

Neeche ka figure isko $x$ – $z$ plane par project karta hai: yellow dot pad hai, blue dot vehicle hai, green arrow position gap $r_{f} - r$ hai, blue arrow current velocity hai, aur red arrow resulting command $a_{c m d}$ hai — dekho kaise yeh vertical pad axis ki taraf wapas lean karta hai aur upar point karta hai.

Figure s01 — Ex 3: command arrow (red) pad ke vertical axis ki taraf wapas aur upar lean karta hai, sideways offset aur fall dono kill karta hai.

$x$ -axis: gap $= 0 - 30 = - 30$ . $\frac{6}{25} (- 30) = - 7.2$ . Velocity: $- \frac{2}{5} (2 (- 4) + 0) = - \frac{2}{5} (- 8) = + 3.2$ . $x$ mein gravity nahi. $a_{x} = - 7.2 + 3.2 = - 4.0$ m/s². Kyun? Hum pad ke right hain, toh position left pull karta hai (−); hum already left drift karte hain, toh velocity term right nudge karta hai (+). Position jeetta hai → net left.
$y$ -axis: gap $= 0 - (- 10) = + 10$ . $\frac{6}{25} (10) = + 2.4$ . Velocity: $- \frac{2}{5} (2 (2) + 0) = - 1.6$ . $a_{y} = + 2.4 - 1.6 = + 0.8$ m/s². Kyun? Symmetric logic, opposite sign — yeh axis $+ y$ push karta hai.
$z$ -axis: gap $= - 100$ . $\frac{6}{25} (- 100) = - 24$ . Velocity: $- \frac{2}{5} (2 (- 20) + (- 1.5)) = - \frac{2}{5} (- 41.5) = + 16.6$ . Gravity: $- g_{z} = - (- 9.81) = + 9.81$ . $a_{z} = - 24 + 16.6 + 9.81 = + 2.41$ m/s².
Result: $a_{c m d} = (- 4.0, + 0.8, + 2.41) m/s^{2}$ . Axes alag kyun rakhein? Formula componentwise act karta hai — har axis ka apna gap, velocity aur (sirf $z$ ke liye) gravity hai — toh hum teen independent 1-D problems solve karte hain aur answers ko ek vector mein stack karte hain.

Verify: har axis independent hai (formula componentwise act karta hai), toh hum unhe Ex 1 ki tarah alag check kar sakte hain. Figure dekho: red command arrow pad axis ki taraf wapas aur upar lean karta hai — exactly jo ek lander ko chahiye. ✅

Worked example Example 4 — Cells D & E: zero gap aur zero drift

Vertical, $g = 3.71$ . Hum exactly pad ke upar hain zero altitude gap par: $r_{f} - r = 0$ . Aur zero horizontal drift ( $v_{x} = v_{y} = 0$ ). Vertical $v = - 2$ m/s, $v_{f} = - 1.5$ m/s, $t_{g o} = 2$ s.

Forecast: koi position gap nahi hai, kya position term poori tarah disappear ho jaati hai, sirf braking + gravity bacha rehta hai? Guess karo.

Position term $\frac{6}{4} (0) = 0$ . Kyun? Yeh degenerate cell D hai: kuch bhi zero gap se multiply karo aur woh vanish ho jaata hai. Formula gracefully woh job drop kar deta hai.
Horizontal axes (cell E): $v_{x} = v_{y} = 0$ aur $v_{f} = 0$ , toh velocity term $= - \frac{2}{2} (0) = 0$ , aur horizontal mein gravity nahi → $a_{x} = a_{y} = 0$ . Kyun? Align karne ko kuch nahi; jab drift nahi hai alignment term silent rehta hai. Accha hai — yeh koi sideways kick invent nahi karega.
Vertical velocity term $- \frac{2}{2} (2 (- 2) + (- 1.5)) = - (- 5.5) = + 5.5$ .
Gravity $- g_{z} = + 3.71$ . $a_{z} = 0 + 5.5 + 3.71 = 9.21$ m/s² up. Inhe add kyun karo? Same superposition: position term zero ho gaya, toh sirf braking push (+5.5) aur gravity cancellation (+3.71) bachi, aur yeh vertical axis par sum hoti hain.

Verify: command purely vertical hai, koi phantom horizontal thrust nahi — confirm karta hai ki dono degenerate cells sahi behave karte hain. Units m/s². ✅

Worked example Example 5 — Cell F:

$t_{g o} \to 0$ blow-up (figure) Ex 1 ke numbers waisa hi ( $r_{f} - r = - 40$ ) lekin ab imagine karo ki hum $t_{g o}$ ko 0 ki taraf shrink karne dete hain. Dikhao kyun command explode karta hai, aur ek floor $t_{g o}^{m i n} = 0.5$ s par apply karo aur thrust ceiling $150 m/s^{2}$ ke saath.

Forecast: position term $\frac{6}{t _{g o}^{2}} (- 40)$ hai. Jab $t_{g o} \to 0$ , kya yeh $1/ t_{g o}$ ya $1/ t_{g o}^{2}$ ki tarah grow karta hai? Guess karo — jawab batata hai blow-up kitna violent hai.

Neeche ka figure us position term ki magnitude ko $t_{g o}$ ke against plot karta hai: blue curve $t_{g o} \to 0$ ke saath upar rocket karti hai, yellow dashed line thrust saturation ceiling $150 m/s^{2}$ mark karti hai jo engine physically exceed nahi kar sakta, aur red dashed line $t_{g o} = 0.5$ s floor mark karti hai jahan hum law par trust karna band kar dete hain aur hand off karte hain.

Figure s02 — Ex 5: position term $1/ t_{g o}^{2}$ ki tarah grow karta hai; $t_{g o}$ ko 1 s se 0.5 s par half karna ise quadruple kar deta hai (240 → 960), 150 m/s² saturation ceiling se bahar punch karke.

$t_{g o} = 1$ s par: position term $= 6 (- 40) = - 240$ m/s². $t_{g o} = 0.5$ par: $= \frac{6}{0.25} (- 40) = - 960$ m/s². Yeh step kyun? $t_{g o}$ half karne se term quadruple ho gayi — yahi $1/ t_{g o}^{2}$ signature hai. Yeh hazaaron $g$ hai: physically impossible, engine (ceiling $150 m/s^{2}$ ) saturate ho jaata hai.
Fix — $t_{g o}$ floor karo aur altitude se hand off karo. $t_{g o} \leftarrow max (t_{g o}, 0.5 s)$ clamp karo taaki denominator kabhi collapse na ho, aur guidance mode switch karo jab altitude $h_{h o} = 5$ m se neeche jaye (equivalently jab E-guidance command $150 m/s^{2}$ ceiling exceed kar le). Uske neeche, ek constant-velocity gravity-turn drop chalao: vertical speed $v_{t d} = 1.5$ m/s par hold karo aur gravity cancel karne plus ek chota settling term ke liye bas itna thrust command karo. Yeh thresholds kyun? $5$ m altitude / $150 m/s^{2}$ ceiling / $0.5$ s floor teen equivalent triggers hain; jo bhi pehle fire kare woh $t_{g o}$ -driven law ko impossible demand karne se pehle khatam kar deta hai.
Floor par command Ex 1 ke dusre terms $t_{g o} = 0.5$ tak scale karke: velocity term $- \frac{2}{0.5} (2 (- 25) + (- 1.5)) = - 4 (- 51.5) = + 206$ . Position $- 960$ . Gravity $+ 3.71$ . Sum $= - 750.3$ m/s² — abhi bhi saturating ( $150 m/s^{2}$ ceiling se bahut zyada), yahi wajah hai ki hum formula par trust karne ki jagah hand off karte hain. Phir bhi sum kyun report karo? Yeh dikhane ke liye ki floor par bhi formula impossible demand karta hai, proving karta hai ki floor akela kaafi nahi hai — tumhe sirf number clamp nahi karna, guidance laws switch karne hote hain.

Verify: figure ka curve $t_{g o} \to 0$ ke saath unboundedly rise karta hai; flat dashed line $150 m/s^{2}$ thrust saturation ceiling hai; $0.5$ s par vertical marker hai jahan hum law par trust karna band karte hain. $4 \times$ jump ( $240 \to 960$ ) numerically quadratic growth confirm karta hai. ✅

Worked example Example 6 — Cell G: touchdown envelope, ek constraint fail

Contact par: $v_{t d} = 2.7$ m/s, $v_{h} = 0.4$ m/s, tilt $θ = 3^{\circ}$ , angular rate $∣ ω ∣ = 1^{\circ} / s$ , position pad centre se $0.5$ m. Limits (upar touchdown-envelope definition se): $v_{t d}^{m a x} = 2.5$ , $v_{h}^{m a x} = 1.0$ , $θ_{m a x} = 8^{\circ}$ , $ω_{m a x} = 5^{\circ} / s$ , pad radius $R = 2$ m.

Forecast: chaar numbers theek lagte hain; ek over hai. Kaun sa? Check karne se pehle guess karo.

Vertical: $2.7 \leq 2.5$ ? NO — leg-crush limit violated. Yeh step kyun? Yeh upar define kiya AND envelope hai: har inequality hold karni chahiye; ek failure = unsafe.
Horizontal: $0.4 \leq 1.0$ ✅. Tilt: $3 \leq 8$ ✅. Rate: $1 \leq 5$ ✅. Position: $0.5 \leq 2$ ✅. Baaki chaar kyun check karo? Exactly locate karne ke liye ki kaun sa clause fail karta hai — engineers ko guidance fix karne ke liye specific violated constraint chahiye, sirf pass/fail flag nahi.

Verify: exactly ek failure ( $v_{t d}$ ), toh verdict hai unsafe touchdown — ek broken AND-clause poori landing condemn kar deta hai. Yahi wajah hai ki guidance $v_{t d} \approx 1.5$ target karta hai aur $2.5$ se neeche margin rakhta hai. ✅

Worked example Example 7 — Cell H: combined tip geometry (figure)

Body tilt $θ = 4^{\circ}$ vertical se, horizontal speed $v_{h} = 0.9$ m/s, vertical $v_{t d} = 1.8$ m/s. Tip threshold $θ_{t i p} = 2 0^{\circ}$ . Kya combined effect margin ke andar hai?

Forecast: velocity ek effective extra lean contribute karti hai. Kya $θ$ + woh lean $20°$ se kam rahega? Guess karo.

Neeche ka figure velocity ko right triangle ki tarah draw karta hai: vertical sink $v_{t d}$ adjacent side hai, drift $v_{h}$ opposite side hai, aur blue velocity arrow hypotenuse hai. Red arc lean angle $arctan (v_{h} / v_{t d})$ hai; green dashed line $2 0^{\circ}$ tip limit hai — dekho velocity arrow already iske past lean kar raha hai.

Figure s03 — Ex 7: slow descent (1.8 m/s) ke saath bhi 0.9 m/s drift velocity ko 26.6° lean kara deta hai, aur 4° body tilt add karne se combined angle $20°$ tip line se bahut past ho jaata hai.

Velocity lean $= arctan \frac{v _{h}}{v _{t d}} = arctan \frac{0.9}{1.8} = arctan (0.5) = 26.56 5^{\circ}$ . Arctan kyun, aur $v_{h} / v_{t d}$ kyun? Velocity arrow right triangle banata hai: horizontal $v_{h}$ opposite side hai, vertical $v_{t d}$ adjacent side hai. $tan$ = opposite/adjacent encode karta hai ki arrow vertical se kitna lean karta hai; $arctan$ answer karta hai "kis angle ka yeh ratio hai?" — figure ka triangle dekho.
Combined $θ_{eff} = θ + arctan (v_{h} / v_{t d}) = 4^{\circ} + 26.56 5^{\circ} = 30.56 5^{\circ}$ . Add kyun karo? Body tilt aur velocity lean dono centre of mass ko same taraf push karte hain, leg footprint edge ki taraf — exactly woh tip-angle rule jo envelope section mein derive ki gayi.
Compare $30.56 5^{\circ}$ vs $θ_{t i p} = 2 0^{\circ}$ : exceeds → tips over. Kyun compare, sirf compute nahi? Tip test khud ek inequality hai (envelope ki tarah): number sirf threshold ke relative matter karta hai, aur yahaan yeh over hai.

Verify: $4^{\circ}$ body tilt ke saath bhi, slow $1.8$ m/s descent par $0.9$ m/s drift velocity ko huge $26. 6^{\circ}$ lean kara deta hai (kyunki ratio $0.5$ bada hai). Figure red combined angle ko green tip line se past punch karta dikhata hai. Lesson: slow descent mein, chota horizontal drift ek bada angle ban jaata hai. ✅

Worked example Example 8 — Cell I: real-world word problem

"Ek Mars lander pad se $60$ m upar hai, $18$ m/s seedha neeche descend kar raha hai aur $2$ m/s eastward drift hai. Guidance chahta hai $1.5$ m/s vertical touchdown, koi drift nahi, $t_{g o} = 6$ s mein. East axis ke along, yeh kya command karta hai?" ( $g = 3.71$ , east $= + x$ .)

Forecast: eastward drift $+ 2$ m/s aur target $0$ — east command negative (westward) honi chahiye. Kitni?

Numbers mein translate karo. East axis: $r_{f} - r = 0 - 0 = 0$ (hum assume karte hain east–west already lined up hain; sirf drift matter karta hai). $v_{x} = + 2$ , $v_{f, x} = 0$ . Kyun? Word problems: gap, current velocity, target velocity, $t_{g o}$ har axis ke liye extract karo. Koi east position error stated nahi → gap 0.
Position term $\frac{6}{36} (0) = 0$ .
Velocity term $- \frac{2}{6} (2 (2) + 0) = - \frac{1}{3} (4) = - 1.333$ m/s². Kyun? Is axis par sirf kaam hai $+ 2$ drift kill karna → command west (−) point karta hai, jaise forecast kiya.
$x$ mein koi gravity nahi. $a_{x} = - 1.333$ m/s² (westward). Add kyun karo (yahaan sirf ek non-zero term)? Position aur gravity is axis par dono zero hain, toh sum sirf velocity term tak reduce ho jaata hai — drift-killer.

Verify: sign negative hai (west), physical need se match karta hai eastward drift erase karne ki. Magnitude modest kyunki $t_{g o} = 6$ s generous hai. Units m/s². ✅

Worked example Example 9 — Cell J: exam twist, ulta solve karo

"Ek vertical overspeed problem diya gaya hai jahan velocity term akele $+ 206$ m/s² braking supply karni hai, $v = - 25$ , $v_{f} = - 1.5$ ke saath, required $t_{g o}$ find karo." (Yeh usual computation ka inverse hai.)

Forecast: chota $t_{g o}$ → bada braking. Toh large $206$ ek chota $t_{g o}$ demand karta hai. Roughly kitna chota?

Velocity term ko equation ki tarah likho: $- \frac{2}{t _{g o}} (2 v + v_{f}) = 206$ . Kyun? Exam twists formula invert karte hain — $t_{g o}$ plug in karne ki jagah, hum uske liye solve karte hain.
Knowns plug karo: $2 v + v_{f} = 2 (- 25) + (- 1.5) = - 51.5$ . Toh $- \frac{2}{t _{g o}} (- 51.5) = \frac{103}{t _{g o}} = 206$ .
Solve karo: $t_{g o} = \frac{103}{206} = 0.5$ s. Dividing kyun kaam karta hai yahaan? Rearranging $\frac{103}{t _{g o}} = 206$ matlab hai $t_{g o}$ woh hai jo 103 ko 206 mein ek baar fit karata hai — single-unknown linear solve.

Verify: back plug karo: $- \frac{2}{0.5} (- 51.5) = - 4 (- 51.5) = + 206$ ✅. Note karo yeh Ex 5 ke floor value se match karta hai — cells ke across ek accha consistency check. ✅

Active Recall

Recall Kaun sa matrix cell hai "command

$g$ ke equal aati hai, pure hover"? Cell B (mild descent) — jab position pull aur velocity push cancel ho jaate hain. ::: Cell B, Example 2.

Recall

$t_{g o} \to 0$ term $1/ t_{g o}$ nahi balki $1/ t_{g o}^{2}$ ki tarah kyun blow up karta hai? Kyunki position term $6/ t_{g o}^{2}$ factor carry karta hai, toh $t_{g o}$ denominator mein squared appear karta hai: $t_{g o}$ half karna $(1/2)^{2} = 1/4$ se divide karta hai, yani term ko 4 se multiply karta hai. Velocity term mein sirf $1/ t_{g o}$ hai, toh woh slower grow karta hai aur position term blow-up mein jeett jaata hai. ::: The $6/ t_{g o}^{2}$ position term dominate karta hai; $t_{g o}$ half karna ise quadruple karta hai (240 → 960 in Ex 5).

Recall Touchdown envelope mein kaun si paanch inequalities hain?

Vertical speed $v_{t d} \leq v_{t d}^{m a x}$ , horizontal speed $v_{h} \leq v_{h}^{m a x}$ , tilt $θ \leq θ_{m a x}$ , angular rate $∣ ω ∣ \leq ω_{m a x}$ , aur position pad radius $R$ ke andar. ::: Saari paanch ek saath hold karni chahiye (AND) — ek failure landing condemn kar deta hai.

Recall Tip geometry mein slow descent small drift ko dangerous kyun banata hai?

Lean $arctan (v_{h} / v_{t d})$ hai; denominator mein chota $v_{t d}$ ratio — aur angle — bada bana deta hai. ::: Slow $v_{t d}$ ⇒ same drift ke liye bada lean angle.

Recall Guidance

$t_{g o}$ -driven law se hand off kis conditions mein karta hai? Jab altitude lagbhag $5$ m se neeche jaaye, ya equivalently jab $t_{g o}$ $0.5$ s floor hit kare ya command $150 m/s^{2}$ thrust ceiling exceed kare — tab $v_{t d} = 1.5$ m/s par constant-velocity gravity-turn drop par switch karo. ::: Below ~ $5$ m / $t_{g o} = 0.5$ s floor / $150$ m/s² ceiling → constant-velocity drop.

Cross-links: Powered Descent Guidance (PDG), Proportional Navigation, Attitude Control & Inner Loop, Time-to-go Estimation, Convex Optimization Landing (lossless convexification).