Autonomous GNC for reusable rockets — SpaceX approach overview: Worked examples

3.5.55 · D3 · Physics › Guidance, Navigation & Control (GNC) › Autonomous GNC for reusable rockets — SpaceX approach overvi

Tools, zero se dobara earn kiye

Kisi bhi example se pehle, yeh pakka karte hain ki har symbol ke saath ek picture ho.

Fig 1 (neeche) kaise padhein: lavender arrow altitude $h$ hai (up-positive); coral arrow ek girte hue booster ki neeche ki velocity hai; slate arrow gravity $- g$ hai; mint arrow net braking acceleration $a_{n e t}$ hai jo sirf tab aata hai jab engine fire kare. Dhyaan do ki coral (fall) aur mint (brake) opposite directions mein point karte hain — yahi opposition poora hoverslam hai.

Figure — Autonomous GNC for reusable rockets — SpaceX approach overview

Definition Sign convention (koi bhi example se pehle yaad karo)

Upar positive hai, neeche negative — hamesha. Is ek rule se sab kuch follow karta hai:

Kisi quantity ka sign uski direction batata hai: $v > 0$ ⇒ chadh raha hai, $v < 0$ ⇒ gir raha hai.
$v^{2} = v_{0}^{2} - 2 a_{n e t} s$ mein, letter $s$ motion ke saath travel ki gayi distance hai (hamesha covered metres ki ek positive number), aur $2 a_{n e t}$ ke aage sign encode karta hai ki acceleration us motion ko help karta hai ya fight karta hai.
- Braking descent: motion neeche hai, $a_{n e t}$ upar point karta hai (opposing) ⇒ minus sign use karo, $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ (speed girti hai).
- Accelerating upward: motion upar hai, $a_{n e t}$ bhi upar (helping) ⇒ plus sign use karo, $v^{2} = v_{0}^{2} + 2 a_{n e t} s$ (speed badhti hai). Jab bhi hum switch karte hain ki hum kaunsa case mein hain (jaise Ex 4 mein), hum dobara bataenge ki humne kaunsa sign chuna aur kyun.

Scenario matrix

Har hoverslam question in cells mein se ek hai (ya blend). Neeche har row ko kam se kam ek worked example cover karta hai.

#	Case class	Kya alag hai	Example
A	Nominal fall	tez gir raha hai, $h_{i g n}$ solve karo	Ex 1
B	Doosra unknown solve karo	$h$ diya, required $a_{n e t}$ / thrust $T$ nikalo	Ex 2
C	Zero / degenerate input	$v_{0} = 0$ (pehle se rest par) — formula kya kahta hai?	Ex 3
D	Bahut HIGH ignite kiya (leftover velocity ka sign $> 0$ , upar ja raha hai)	pad ke upar ruk jaata hai, phir upar jaata hai	Ex 4
E	Bahut LOW ignite kiya (leftover velocity $h = 0$ par bhi neeche)	crash speed	Ex 5
F	Limiting value	throttle floor: minimum $a_{n e t}$ ⇒ maximum survivable $v_{0}$	Ex 6
G	Degenerate: engine bahut weak	$a_{n e t} \leq 0$ (TWR $\leq 1$ ) — formula break ho jaata hai	Ex 7
H	Real-world word problem	crosswind → sideways PID Control correction	Ex 8
I	Exam twist	burn ke dauran mass change hota hai (Rocket Equation (Tsiolkovsky) se link)	Ex 9

Ex 1 — Cell A: nominal hoverslam

Worked example Statement

Ek booster $v_{0} = 250 m/s$ par girta hai. Uska engine $a_{n e t} = 25 m/s^{2}$ produce kar sakta hai. Kaunsi altitude par engine light karni chahiye?

Forecast: andaaza lagao — kya $h_{i g n}$ parent note ke 1000 m se bada hoga ya chhota? (Tez fall lekin strong brake... padhne se pehle andaaza lagao.)

Timeless equation chuno. $0 = v_{0}^{2} - 2 a_{n e t} h$ use karo (minus sign: descent upward $a_{n e t}$ se brake hota hai). Yeh step kyun? Hume woh altitude chahiye jahan speed zero ho — time irrelevant hai, isliye no- $t$ relation sahi tool hai.
$h$ solve karo. $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} = \frac{25 0 ^{2}}{2 \cdot 25} = \frac{62500}{50} = 1250 m$ . Yeh step kyun? Rearrange karke uss ek unknown ko isolate kiya jo poocha gaya tha.
Burn duration (bonus): time relation $v = v_{0} - a_{n e t} t$ se, $v = 0$ set karo: $t = v_{0} / a_{n e t} = 250/25 = 10 s$ . Yeh step kyun? Feasibility sanity-check karta hai — 10 s ka fuel hona chahiye. Hum yahan $v = v_{0} - a_{n e t} t$ pe switch karte hain kyunki ab sawaal kab ka hai, na kahan ka.

Fig 2 kaise padhein: coral curve burn ke dauran speed vs altitude hai. Yeh $h = 1250$ m par 250 m/s se start hoti hai aur exactly pad par zero touch karti hai — perfect hoverslam ka signature. Curve $v$ mein parabola hai kyunki $v^{2}$ (na ki $v$ ) $h$ mein linear hai.

Verify: $h = 1250$ back plug karo: $v^{2} = 25 0^{2} - 2 \cdot 25 \cdot 1250 = 62500 - 62500 = 0$ ✓. Speed sach mein pad par zero hoti hai. Units: $(m/s)^{2} / (m/s^{2}) = m$ ✓.

Ex 2 — Cell B: height diya, thrust nikalo

Worked example Statement

Sensors kehte hain engine ko exactly $h = 800 m$ par light karni hai jabki $v_{0} = 180 m/s$ par gir raha hai. Booster mass $m = 30, 000 kg$ , $g = 9.8 m/s^{2}$ . Kaunsa thrust $T$ chahiye?

Forecast: kya needed thrust booster ke weight $m g$ se zyada hoga ya kam?

Needed $a_{n e t}$ nikalo. $a_{n e t} = \frac{v _{0}^{2}}{2 h} = \frac{18 0 ^{2}}{2 \cdot 800} = \frac{32400}{1600} = 20.25 m/s^{2}$ . Yeh step kyun? Is baar usi equation ko flip kiya acceleration ke liye solve karne ke liye, kyunki $h$ known hai.
Thrust mein convert karo. $T = m (g + a_{n e t}) = 30000 (9.8 + 20.25) = 30000 \cdot 30.05 = 901, 500 N$ . Yeh step kyun? $a_{n e t} = T / m - g$ , isliye $T = m (g + a_{n e t})$ — engine ko gravity beat bhi karni hai aur braking bhi supply karni hai.
Weight se compare karo. $m g = 30000 \cdot 9.8 = 294, 000 N$ ; toh $T / m g \approx 3.07$ — thrust-to-weight $\approx 3$ . Yeh step kyun? Parent-note claim confirm karta hai ki TWR $> 1$ (actually $≫ 1$ ) — tum sach mein hover nahi kar sakte.

Fig 3 kaise padhein: bar chart thrust ke do kaam stack karta hai — weight beat karna ( $m g$ , slate) aur braking supply karna ( $m a_{n e t}$ , mint). Unka sum (lavender total) required thrust hai, weight bar se ~3× oopar. Woh height ratio hi thrust-to-weight ratio hai.

Verify: $a_{n e t} = T / m - g = 901500/30000 - 9.8 = 30.05 - 9.8 = 20.25$ ✓. $T$ ke units: $kg \cdot m/s^{2} = N$ ✓.

Ex 3 — Cell C: degenerate input $v_{0} = 0$

Worked example Statement

Maano booster momentarily at rest ( $v_{0} = 0$ ) hai kisi altitude par. $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ kya kehta hai, aur kya yeh physical hai?

Forecast: kya $h_{i g n}$ zero, infinite, ya undefined hoga?

Substitute karo. $h_{i g n} = \frac{0 ^{2}}{2 a _{n e t}} = \frac{0}{2 a _{n e t}} = 0$ . Yeh step kyun? Degenerate value plug karne se pata chalta hai ki maths kya sochti hai.
Interpret karo. $h_{i g n} = 0$ matlab "tumhe zero braking distance chahiye," yaani tum pehle se ruk gaye ho, toh velocity null karne ke liye koi burn zaroori nahi. Yeh step kyun? Ek formula ko wapas words mein padhna chahiye taaki check ho ki sensible hai ya nahi.
Lekin gravity watch karo. Rest par engine off ke saath, gravity immediately $v < 0$ phir se bana deti hai. Toh "hawa mein rest par" ek stable landing nahi hai — yeh fall aur re-fall ke beech ka instant hai. Hoverslam idea sirf tab apply hota hai jab tumhare paas abhi bhi neeche ki speed ho jo brake karni ho. Yeh step kyun? Yeh trap cover karta hai ki " $v_{0} = 0$ ⇒ done" yeh ignore karta hai ki tum hover nahi kar sakte (TWR $> 1$ ).

Verify: $v^{2} = 0^{2} - 2 a_{n e t} \cdot 0 = 0$ ✓ — consistent hai lekin trivially. Degenerate case well-behaved hai (0 deta hai, divide-by-zero nahi).

Ex 4 — Cell D: BAHUT ZYADA HIGH ignite karna (leftover upward velocity)

Worked example Statement

$v_{0} = 200, a_{n e t} = 20$ ke liye nominal ignition $h_{i g n} = 1000 m$ hai (parent note). Maano computer early light karta hai, $h = 1200 m$ par. Jab yeh $h = 0$ reach karta hai toh velocity kya hai, aur uska sign kya hai?

Forecast: pad par, kya yeh abhi bhi gir raha hoga, exactly ruka hoga, ya upar ja raha hoga?

Total braking distance ab 1200 m hai. Speed sirf $Δ h = v_{0}^{2} /2 a_{n e t} = 1000$ m braking ke baad zero hoti hai — woh altitude $1200 - 1000 = 200 m$ par hoga. Yahan hum minus sign use karte hain (descent braked): $0 = v_{0}^{2} - 2 a_{n e t} Δ h$ . Yeh step kyun? Brake ko speed null karne ke liye exactly 1000 m chahiye; 200 m zyada oopar se start karne ka matlab hai zero-speed 200 m upar hogi.
Sign flip — ab yeh chadh raha hai. Us point ke upar engine abhi bhi fire kar raha hai (TWR $> 1$ ), isliye $h = 200 m$ se rocket upar accelerate karta hai. Motion aur $a_{n e t}$ ab same direction mein hain, isliye hum plus sign pe switch karte hain, $v^{2} = v_{0}^{2} + 2 a_{n e t} s$ (speed badhti hai). Rest se start karke aakhiri $s = 200 m$ pad ki taraf wapas girne par: $v^{2} = 0^{2} + 2 a_{n e t} (200) = 2 \cdot 20 \cdot 200 = 8000 \Rightarrow v = 8000 \approx 89.4 m/s (upward) .$ Yeh step kyun? Hum sign rule explicitly restate karte hain: same direction ⇒ plus; $v > 0$ ⇒ chadh raha hai.
Consequence. Tum upar jaate hue pad reach nahi kar sakte — tum wapas sky mein chale jaoge. Real GNC yahan throttle down karta hai, lekin agar $m g$ se neeche throttle nahi kar sakta, toh early ignition fail hoga, exactly jaise parent note warn karta hai. Yeh step kyun? $v$ ke sign ko physical failure mode se connect karta hai.

Fig 4 kaise padhein: coral curve ko 1200 m se neeche follow karo — yeh 200 m par zero speed hit karta hai (lavender dot, "too high!"), phir mint dashed branch dikhata hai booster upar speeding up, ~89 m/s par galat direction mein pad arrive karta hai. Dono branches stop point par milti hain jahan speed momentarily zero hoti hai.

Verify: stop altitude $= 1200 - 1000 = 200$ ✓; leftover upward speed $2 \cdot 20 \cdot 200 = 8000 \approx 89.44$ m/s, sign $+$ (up) ✓.

Ex 5 — Cell E: BAHUT NEECHE ignite karna (leftover downward velocity = crash)

Worked example Statement

Wohi booster ( $v_{0} = 200, a_{n e t} = 20$ , 1000 m chahiye). Woh late light karta hai, $h = 900 m$ par. $h = 0$ par impact speed kya hai?

Forecast: kya rocket pad se pehle rukega, ya abhi bhi neeche move karte hue hit karega?

Available braking distance = sirf 900 m. 900 m braking ke baad speed compute karo (descent, minus sign): $v^{2} = v_{0}^{2} - 2 a_{n e t} h = 20 0^{2} - 2 \cdot 20 \cdot 900 = 40000 - 36000 = 4000.$ Yeh step kyun? Hum poochh rahe hain: saari 900 m use karne ke baad kitni speed bachi hai? Positive $v^{2}$ matlab woh kabhi zero nahi hua.
Root lo. $v = 4000 \approx 63.2 m/s$ , abhi bhi neeche (yeh kabhi zero cross nahi hua, isliye sign negative rehta hai). Yeh step kyun? Pad par $v^{2} > 0$ ⇒ nonzero impact speed ⇒ crash. Sign hai "abhi bhi gir raha hai."
Ruk jaane ke liye actually kitna altitude chahiye tha? $v_{0}^{2} /2 a_{n e t} = 1000 m > 900$ ⇒ 100 m shortfall — isliye zor se hit karta hai. Yeh step kyun? Required vs available distance compare karna "too low" ka clean diagnostic hai.

Fig 5 kaise padhein: coral curve 200 m/s aur $h = 900$ m se start hoti hai lekin pad se pehle axis kabhi touch nahi karti — yeh $h = 0$ cross karti hai abhi bhi ~63 m/s par. Curve ke foot aur ground ke beech red gap impact speed hai: ek crash.

Verify: $v = 40000 - 36000 = 4000 \approx 63.25$ m/s, downward ✓. Sanity: $v_{0} = 200$ se kam, isliye yeh did brake kiya, bas kaafi nahi ✓.

Ex 6 — Cell F: limiting value (throttle floor max fall speed set karta hai)

Worked example Statement

Ek engine infinitely throttle nahi kar sakta. Maano landing burn ke dauran uski minimum deliverable net acceleration $a_{n e t}^{m i n} = 18 m/s^{2}$ hai, aur pad ground se pehle maximum braking distance $h_{m a x} = 1500 m$ allow karta hai. Sabse zyada fall speed $v_{0}$ kya hai jis par booster abhi bhi time par ruk sakta hai?

Forecast: jaise allowed distance badhti hai, kya survivable speed linearly badhti hai, ya slower/faster than linear?

Boundary case set karo. Sabse bura survivable case saari height sabse weak brake ke saath use karta hai: $0 = v_{0}^{2} - 2 a_{n e t}^{m i n} h_{m a x}$ . Yeh step kyun? Limit wahan hai jahan equality hold kare — thoda tez aur crash, thoda slow aur margin hai.
$v_{0}$ solve karo. $v_{0} = 2 a_{n e t}^{m i n} h_{m a x} = 2 \cdot 18 \cdot 1500 = 54000 \approx 232.4 m/s$ . Yeh step kyun? Timeless equation rearrange karke max entry speed dono limits ka function milta hai.
Scaling insight. Kyunki $v_{0} \propto h$ , runway double karne se survivable speed sirf $2 \approx 1.41$ se multiply hoti hai — sub-linear. Speed kill karna expensive hai. Yeh step kyun? Reveal karta hai kyun entry burns exist karte hain: altitude se zyada braking nahi khareed sakte, isliye tum pehle se slow arrive karo.

Fig 6 kaise padhein: lavender curve survivable $v_{0} = 2 a_{n e t}^{m i n} h$ ko available height $h$ ke against plot karta hai. Uska flatten hona sub-linear $h$ law hai: double speed survive karne ke liye tumhe char guna height chahiye hogi (mint guide lines).

Verify: $v_{0} = 2 \cdot 18 \cdot 1500 = 54000 \approx 232.38$ m/s ✓. Check back: $232.3 8^{2} - 2 \cdot 18 \cdot 1500 \approx 0$ ✓.

Ex 7 — Cell G: engine bahut weak hai ( $a_{n e t} \leq 0$ , TWR ≤ 1)

Worked example Statement

Ek hypothetical booster (ya badly-throttled) ka thrust $T = 280, 000 N$ , mass $m = 30, 000 kg$ , $g = 9.8$ hai. Woh $v_{0} = 200 m/s$ par gir raha hai. $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ kaunsa ignition altitude deta hai, aur iska matlab kya hai?

Forecast: kya formula ek sensible positive height dega, ya kuch impossible?

$a_{n e t}$ compute karo. $a_{n e t} = \frac{T}{m} - g = \frac{280000}{30000} - 9.8 = 9.333 - 9.8 = - 0.467 m/s^{2}$ . Yeh step kyun? Hoverslam formula trust karne se pehle net acceleration ka sign check karo — sab kuch isi par depend karta hai.
$a_{n e t} < 0$ matlab bilkul braking nahi. Thrust ( $9.33 m/s^{2}$ ) gravity ( $9.8$ ) se kam hai, isliye engine booster ko hold bhi nahi kar sakta — woh neeche accelerate karta rehta hai. $h_{i g n} = v_{0}^{2} /2 a_{n e t} = 20 0^{2} / (2 \cdot - 0.467) \approx - 42, 900 m$ mein plug karne par negative altitude milti hai — physically meaningless. Yeh step kyun? Negative $h_{i g n}$ maths ki cheekh hai ki "impossible": koi altitude nahi hai jahan yeh engine fall rok sake.
Boundary $a_{n e t} = 0$ (TWR exactly 1). Yahan $h_{i g n} = v_{0}^{2} /0 = + \infty$ : infinite height bhi speed null nahi kar sakti, kyunki zero net deceleration kabhi slow nahi karta. Toh hoverslam formula sirf $a_{n e t} > 0$ (TWR $> 1$ ) ke liye valid hai — exactly isliye real boosters ka thrust weight se zyada hona chahiye, aur isliye woh hover nahi kar sakte. Yeh step kyun? Degenerate limit cleanly cover karta hai: $a_{n e t} > 0$ ⇒ finite positive $h_{i g n}$ ; $a_{n e t} = 0$ ⇒ $\infty$ ; $a_{n e t} < 0$ ⇒ negative/nonsense.

Fig 7 kaise padhein: curve $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ ko $a_{n e t}$ ke against plot kiya gaya hai. Right side ( $a_{n e t} > 0$ ) sensible positive heights deta hai jo infinity blow up hoti hain as $a_{n e t} \to 0$ (mint asymptote). Left side ( $a_{n e t} < 0$ , coral shaded "no-landing zone") negative dive karta hai — impossible. Sirf mint region real hoverslam hai.

Verify: $a_{n e t} = 280000/30000 - 9.8 \approx - 0.4667$ ✓ (negative); $h_{i g n} = 20 0^{2} / (2 \cdot - 0.4667) \approx - 42857$ m ✓ (negative ⇒ non-physical). $a_{n e t} = 0$ par, $h_{i g n} \to \infty$ ✓.

Ex 8 — Cell H: real-world crosswind (sideways PID correction)

Worked example Statement

Landing burn ke dauran ek steady crosswind booster ko sideways push karta hai. Lateral position error $e$ metres hai (target se right). PID Control gimbal law $δ = K_{p} e + K_{d} \overset{e}{˙} + K_{i} \int e d t$ use karke, $K_{p} = 0.5, K_{d} = 1.2, K_{i} = 0.05$ ke saath, ek instant par $e = 4 m$ , $\overset{e}{˙} = - 1.5 m/s$ (pehle se target ki taraf wapas ja raha hai), aur running integral $\int e d t = 8 m\cdots$ hai. Kaunsa gimbal deflection $δ$ command kiya jaata hai?

Forecast: right side error ke saath lekin closing, kya teen terms add up honge ya ek doosre se fight karenge?

Present term. $K_{p} e = 0.5 \cdot 4 = 2.0$ (deg). React karta hai abhi error kahan hai. Yeh step kyun? Proportional term current 4 m offset ke against push karta hai.
Damping term. $K_{d} \overset{e}{˙} = 1.2 \cdot (- 1.5) = - 1.8$ . Kyunki $\overset{e}{˙} < 0$ hum pehle se correct kar rahe hain; yeh term overshoot avoid karne ke liye pull back karta hai. Yeh step kyun? Derivative term isliye hai ki rocket oscillate nahi karta — yeh motion anticipate karta hai.
Bias term. $K_{i} \int e d t = 0.05 \cdot 8 = 0.4$ . Constant wind se steady offset remove karta hai. Yeh step kyun? Ek persistent crosswind ek steady error chhodta hai jise sirf integral term erase kar sakta hai.
Sum. $δ = 2.0 - 1.8 + 0.4 = 0.6$ deg. Yeh step kyun? PID output teen ka sum hai; yahan woh partly cancel karte hain, ek gentle net command dete hain — exactly $K_{d}$ ka point.

Fig 8 kaise padhein: teen stacked bars — lavender (present, $+ 2.0$ ), coral (damping, $- 1.8$ , neeche point karta hai kyunki oppose karta hai), butter (bias, $+ 0.4$ ) — aur unka net (mint, $+ 0.6$ ). Near-cancellation dikhata hai ki controller ease off kar raha hai kyunki booster pehle se target par close aa raha hai.

Verify: $0.5 \cdot 4 + 1.2 \cdot (- 1.5) + 0.05 \cdot 8 = 2.0 - 1.8 + 0.4 = 0.6$ ✓. Sign $> 0$ ⇒ target ki taraf correct; magnitude chhoti kyunki vehicle pehle se close aa raha hai ✓.

Ex 9 — Cell I: exam twist — burn ke dauran mass drop hota hai

Worked example Statement

10 s burn ke dauran booster ka mass $m_{0} = 30, 000$ kg se $m_{1} = 27, 000$ kg ho jaata hai jab fuel jalta hai. Agar thrust $T$ constant $T = 900, 000 N$ par rakha jaaye aur $g = 9.8$ ho, toh burn ke start vs end mein $a_{n e t}$ compare karo. Kya deceleration strong hoti hai ya weak?

Forecast: jaise fixed thrust ke saath mass drop hota hai, kya $a_{n e t} = T / m - g$ rise hogi ya fall?

Start. $a_{n e t, 0} = \frac{900000}{30000} - 9.8 = 30 - 9.8 = 20.2 m/s^{2}$ . Yeh step kyun? Ignition par baseline braking, full mass ke saath.
End. $a_{n e t, 1} = \frac{900000}{27000} - 9.8 = 33.33 - 9.8 = 23.53 m/s^{2}$ . Yeh step kyun? Wohi thrust chhote mass se divide hota hai ⇒ zyada acceleration; change quantify karte hain.
Interpret karo. Braking fuel jalne ke saath $\approx 3.3 m/s^{2}$ se strong hoti hai. Toh Ex 1–6 mein constant- $a_{n e t}$ assumption burn ke baad mein braking ki mild under-estimate hai; real GNC har cycle mein $T_{c m d} = m (g + v^{2} /2 h)$ recompute karta hai ise absorb karne ke liye — exact reason parent note live recompute karta hai. Mass depletion khud Rocket Equation (Tsiolkovsky) se governed hai. Yeh step kyun? Numeric drift ko wapas kyun ek live closed loop fixed formula se better hai, se connect karta hai.

Fig 9 kaise padhein: $a_{n e t}$ ke liye do mint bars — taller wala (burn ka end) shorter (start) ke upar baitha hai, coral arrow $\approx 3.3 m/s^{2}$ gain mark karta hai. Wohi thrust, lighter rocket ⇒ strong push, isliye ek fixed formula slowly drift karta hai aur loop ko recompute karna padta hai.

Verify: $a_{n e t, 0} = 900000/30000 - 9.8 = 20.2$ ✓; $a_{n e t, 1} = 900000/27000 - 9.8 \approx 23.533$ ✓; difference $\approx 3.33 m/s^{2}$ , increasing ✓.

Active recall

Recall Kaunsi equation, aur usme time kyun nahi?

Hoverslam ko kaunsa kinematic relation drive karta hai, aur $v = v_{0} - a t$ ke upar ise kyun prefer kiya jaata hai? ::: $v^{2} = v_{0}^{2} - 2 a_{n e t} h$ — yeh speed ko directly altitude se link karta hai bina kisi time term ke, isliye hum solve karte hain kahan speed zero hoti hai.

Recall Pad par leftover velocity ka sign

Agar ek booster bahut high ignite kare, toh pad par (agar) pahunche toh uski velocity ka sign kya hoga? ::: Positive (upward) — woh ground ke upar ruk jaata hai aur, TWR $> 1$ ke saath, wapas chadh jaata hai.

Recall Hoverslam formula kab break karta hai?

$a_{n e t}$ ki kaunsi values ke liye $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ physical answer deta hai? ::: Sirf $a_{n e t} > 0$ (TWR $> 1$ ). $a_{n e t} = 0$ par yeh $+ \infty$ hai; $a_{n e t} < 0$ par yeh negative jaata hai — koi landing possible nahi.

Recall Scaling law

Agar allowed braking distance double ho jaaye, toh maximum survivable entry speed kitne factor se badhti hai? ::: $2 \approx 1.41$ se, kyunki $v_{0} \propto h$ — sub-linear.