Autonomous GNC for reusable rockets — SpaceX approach overview: Question bank

3.5.55 · D5 · Physics › Guidance, Navigation & Control (GNC) › Autonomous GNC for reusable rockets — SpaceX approach overvi

Neeche har reveal Question ::: Answer use karta hai — left padho, socho, phir right check karo.

Definition Is page par use hone wale symbols (saare vertical, upward positive)

Neeche sab kuch vertical motion ke liye likha gaya hai jo landing pad se measure hota hai, upward ko positive direction maante hue.

$h$ = remaining altitude / stopping distance jo abhi girni baki hai, metres mein. Total flight height nahi — ye $0$ ki taraf shrink hota hai jab booster pad ke paas aata hai.
$v$ = booster ki current downward speed (us speed ke size ke liye ek positive number ke roop mein likha gaya); $v_{0}$ braking burn ke shuru mein woh speed hai.
$g$ = gravitational acceleration ( $\approx 9.8 m/s^{2}$ ), neeche ki taraf pull karta hai.
$T$ = engine thrust (upar push karne wali force); $m$ = booster ka current mass.
$a_{n e t}$ = burn ke dauran net upward acceleration, define kiya gaya hai as $a_{n e t} = \frac{T}{m} - g$ yaani engine se per kilogram upward push, minus jo gravity wapas chura leti hai. Ye actual deceleration hai jo fall ko khatam karne ke liye available hai.

Formula Braking formulae kahan se aate hain (taaki ye page akela khada rahe)

Burn ke dauran hum $a_{n e t}$ ko roughly constant maante hain. Time-free kinematic relation $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ use karte hue (ye speed ko distance se bina clock ke link karta hai), demand karo ki fall ruk jaaye ( $v = 0$ ) remaining distance $h$ girne ke baad: $0 = v_{0}^{2} - 2 a_{n e t} h \Rightarrow a_{n e t} = \frac{v _{0}^{2}}{2 h} .$ Ignition altitude ke liye rearrange kiya gaya — woh height jis par engine light karna zaroori hai: $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} .$ Aur kyunki $a_{n e t} = T / m - g$ , thrust jo controller ko command karni chahiye taaki current speed $v$ current remaining height $h$ par zero ho jaaye: $T_{c m d} = m (g + \frac{v ^{2}}{2 h}) .$ $v^{2} / (2 h)$ piece hi required $a_{n e t}$ hai: ye badhta hai jab tum fast ho ( $v$ bada) ya ground ke close ho ( $h$ chota), kyunki dono harder braking demand karte hain; $+ g$ sirf gravity ko cancel karta hai taaki engine ka net effect us required deceleration ke barabar ho.

True ya false — justify karo

Ek booster touchdown se pehle lineup karne ke liye barge ke bilkul upar hover kar sakta hai.

False — almost-empty booster ka minimum thrust uske weight se zyada hota hai (TWR

> 1

), isliye sabse choti possible push bhi usse upward accelerate karti hai. Ye literally constant altitude par balance nahi kar sakta, isliye use ek timed hoverslam karni padti hai.

Hoverslam kaam karta hai kyunki engine ko bilkul zero thrust tak throttle down kiya ja sakta hai.

False — constraint ye hai ki even lowest usable throttle

T / m > g

deta hai (TWR

> 1

). Agar wo TWR

= 1

tak throttle down kar sakta to hover kar sakta tha; problem poori isiliye exist karti hai kyunki wo nahi kar sakta.

Guidance aur Control ek hi kaam ke do naam hain.

False — Guidance decide karta hai kahan jaana hai (desired trajectory/target state), jabki Control decide karta hai kaise actuators ko move karna hai us trajectory ke gap ko close karne ke liye. Navigation teesra hai: main abhi kahan hoon?

Kalman filter sirf isiliye zaroori hai kyunki GPS noisy hai.

False — ye do complementary flaws ko fuse karta hai: Kalman Filter ek fast-but-drifting IMU ko slow-but-drift-free GPS/radar ke saath blend karta hai, har ek par uski uncertainty ke according trust karta hai. Koi bhi source hatane se estimate degrade ho jaata hai.

Ek baar engine ignition par fuel-optimal descent trajectory compute ho jaaye, use ground tak unchanged follow kiya jaata hai.

False — ye har second mein kai baar recompute hoti hai (closed-loop). Winds, exact mass, aur engine performance kabhi exactly known nahi hote, isliye trajectory fresh state estimates se live re-solve hoti hai.

Ek PID controller ka proportional term akela ek constant crosswind bias eliminate kar sakta hai.

False — ek pure proportional term correction produce karne ke liye nonzero error chahta hai, isliye ye ek steady wind ke against residual offset ke saath settle hota hai. Integral term

K_{i} \int e d t

us offset ko accumulate karta hai aur steady-state error ko zero tak drive karta hai.

Convex optimization mainly isliye use hoti hai kyunki ye PID se code karna zyada fast hai.

False — ye isliye use hoti hai kyunki convex problems ka guaranteed global optimum bounded time mein milta hai. Woh determinism essential hai jab ek mission-critical decision har cycle mein produce karni ho; dekho Convex Optimization aur Lossless Convexification.

Landing burn ke dauran steering ke liye grid fins primary hain.

False — kisi bhi powered burn ke dauran engine gimbal steering torque dominate karta hai. Grid fins atmospheric coast/descent ke dauran aerodynamic steering provide karte hain, jab engine kaam nahi kar raha hota tab nahi.

Agar video feed fast enough hoti to tum landing remotely pilot kar sakte.

False — perfect feed ke saath bhi, signal latency plus plasma blackout plus ~250 ms human reaction time, vehicle ke zaroori ~10 ms control loop se zyada ho jaate hain. Decision onboard lena hi padega.

Error dhundho

"Safe rehne ke liye, landing engine ko jitna ho sake utna upar light karo taaki maximum braking distance mile."

Error ye hai: jaldi light karne se velocity zero ground ke upar pohonch jaati hai. Kyunki TWR

> 1

hai to hover nahi kar sakte, isliye tum upar uthte ho ya fuel dump karte ho — ignition altitude

h_{i g n} = v_{0}^{2} / (2 a_{n e t})

(jahan

a_{n e t} = T / m - g

) ek precise target hai, floor nahi.

"IMU acceleration ko do baar integrate karta hai, isliye akela position de sakta hai — GPS sirf ek luxury hai."

Error ye hai: position acceleration ka double integral hai, isliye ek constant acceleration bias

b

, position mein

\frac{1}{2} b t^{2}

contribute karta hai — ek chota offset jo time ke saath balloon ho jaata hai. GPS/radar us drift ko bound karne ke liye zaroori hai, luxury nahi.

"Kinematic relation

v^{2} = v_{0}^{2} - 2 a_{n e t} h

isliye use hoti hai kyunki ye burn time deti hai."

Error ye hai: us relation mein koi time nahi hai — isliye hi use choose kiya gaya hai. Ye speed ko directly distance se link karta hai, jawab deta hai ki speed kahan zero hogi; burn time alag se

t = v_{0} / a_{n e t}

se aata hai. Dekho Kinematics — v²=v0²+2as.

"Boostback burn ke dauran rocket apni vertical velocity cancel karta hai taaki seedha neeche gire."

Error ye hai: boostback burn horizontal velocity cancel ya reverse karta hai taaki landing site ki taraf aim ho sake. Vertical descent speed baad mein entry aur landing burns se bleed off hoti hai.

"Cold-gas thrusters sabse strong actuator hain, isliye ye flip aur landing handle karte hain."

Error ye hai: cold-gas thrusters weak hote hain; inhe sirf vacuum/coast mein use kiya jaata hai jab fins ke liye koi air nahi aur engine off hai. Torque authority environment se choose hoti hai, raw strength se nahi — burns ke liye gimbal, atmosphere ke liye fins.

"Kyunki hum

v = 0

chahte hain

h = 0

par, hum

a_{n e t} = v_{0}^{2} / (2 h)

set karte hain jahan

h

total flight altitude hai."

Error ye hai:

h

us moment par remaining stopping distance hai jab hum evaluate karte hain, total altitude nahi. Controller continuously current

v

aur current

h

ke saath har cycle mein recompute karta hai.

Why questions

Booster simply parachute se neeche kyun nahi aa sakta jaise pehle ke capsules aate the?

Parachute koi lateral steering precision nahi deta aur wind ke through barge-sized target hit karne ka koi tarika nahi hai, aur ek heavy booster ko enormous chutes chahiye. Propulsive landing usi engine se possible hai jo launch kiya, steering aur braking pinpoint tak kar sakta hai.

Landing burn ko "suicide burn" kyun kehte hain?

Kyunki koi margin nahi hoti: TWR

> 1

ke saath tum pause nahi kar sakte correct karne ke liye, isliye ignition timing aisi karni padti hai ki velocity exactly zero altitude par zero ho. Ek pal ki galti matlab mid-air ruk jaana (phir girna) ya ground se takrana.

Controller

T_{c m d} = m (g + v^{2} /2 h)

command kyun karta hai, fixed thrust kyun nahi?

Kyunki required deceleration

v^{2} / (2 h)

hai: harder braking tab zaroori hai jab speed

v

high ho ya remaining height

h

choti ho, aur dono har instant change hote hain.

+ g

gravity cancel karta hai taaki engine ka net effect us required value ke barabar ho; ek fixed thrust

v

aur

h

ke evolve hone par over- ya under-brake karta.

Attitude ke liye teen angles ki jagah quaternion kyun use karte hain?

Quaternions un gimbal-lock singularities aur discontinuities se bachate hain jo Euler angles vertical orientations ke paas suffer karte hain, jo matter karta hai jab booster large attitude changes ke through flip karta hai; dekho Attitude Dynamics & Quaternions.

Teen alag actuators blend karne ki jagah ek best wala hamesha kyun use nahi karte?

Kyunki har ek ki authority sirf uske regime mein hoti hai: gimbal ko engine lit chahiye, grid fins ko air chahiye, cold-gas ko kuch nahi chahiye lekin weak hai. GNC inhe flight phase ke according schedule karta hai taaki kuch na kuch control hamesha available rahe.

Derivative term

K_{d} \overset{e}{˙}

ek gimbal controller ke liye kyun matter karta hai?

Ye respond karta hai ki error kitni fast change ho rahi hai, motion ko damp karta hai taaki rocket target attitude overshoot karke oscillate na kare. Iske bina, proportional term akela ring karne lagta hai.

Edge cases

h_{i g n} = v_{0}^{2} / (2 a_{n e t})

kya predict karta hai agar descent speed

v_{0}

zero ho?

h_{i g n} = 0

deta hai: koi speed kill karne ke liye nahi hai to kabhi braking ke liye ignite karne ki zaroorat nahi. Formula sensibly degenerate hota hai "no burn required" par.

Required thrust

T = m (g + v_{0}^{2} /2 h)

ka kya hota hai jab remaining stopping distance

h \to 0

hoti hai aur

v_{0} > 0

hai?

Demanded thrust infinity tak diverge hota hai — physically impossible. Isliye ignition itna pehle hona chahiye ki

h

kabhi zero ki taraf squeeze na ho jabki still fast move kar rahe ho.

Ek booster jo exactly

= 1

TWR rakhta, woh kya kar sakta jo real wala nahi kar sakta?

T / m = g

ke saath uska net acceleration

a_{n e t} = 0

hoga, isliye wo truly hover kar sakta tha, jitna slowly chahiye utna dhire neeche aa sakta tha aur pad par ease kar sakta tha. Real boosters is hovering threshold se upar hain (TWR

> 1

), aur yahi wajah hai ki precisely-timed hoverslam forced hai.

Agar burn ke mid mein engine ka achievable

a_{n e t}

suddenly drop ho jaaye (underperformance), to closed loop kya karta hai?

Chota

a_{n e t}

matlab zaroori ignition altitude assumed se zyada thi, isliye vehicle ab behind hai — wo maximum available thrust command karega, aur agar woh bhi remaining

h

mein rok nahi sakta, to koi control law recover nahi kar sakta. Isliye margin trajectory mein design kiya jaata hai, improvise nahi.

Perfect vacuum mein engine off ke saath, kis actuator ke paas koi authority hai?

Sirf cold-gas thrusters — grid fins ke liye koi air nahi aur gimbal karne ke liye exhaust nahi. Exactly yahi regime hai jiske liye ye exist karte hain, jaise post-separation flip.

Agar target directly neeche hai aur zero horizontal error hai — kya gimbal still active hai?

Haan: zero lateral error ke saath bhi ye disturbances (wind gusts, thrust misalignment) ke against attitude hold karta hai.

e = 0

ke saath bhi ek PID controller koi bhi

\overset{e}{˙}

jo appear ho usse counter karta hai, correction ready rakhta hai.

Jis exact instant par velocity zero altitude par zero hoti hai, kya net acceleration zero hota hai?

Nahi — net upward acceleration still

a_{n e t} = T / m - g > 0

hai kyunki TWR

> 1

hai. Isliye engine exactly touchdown par cut off karna padta hai; use lit chhod dene par booster wapas upar push ho jaata.

Recall Har trap ka ek-line summary

Recurring theme: TWR $> 1$ hovering forbid karta hai (isliye timing exact honi chahiye), loop closed hai (live recomputed, replay nahi), sensors fused hone chahiye (IMU drifts), actuators environment-scheduled hain, aur convex optimization bounded time mein ek guaranteed answer deti hai.