Autonomous GNC for reusable rockets — SpaceX approach overview: Exercises

3.5.55 · D4 · Physics › Guidance, Navigation & Control (GNC) › Autonomous GNC for reusable rockets — SpaceX approach overvi

Yeh page ek self-test ladder hai. Har problem apna level batata hai, clean numbers deta hai, aur ek collapsible callout ke andar poora solution chhupata hai. Pehle khud try karo, phir reveal karo. Sab kuch parent overview pe build karta hai.

Shuru karne se pehle, ek picture is page ke har sign convention ko fix kar deti hai. Ise neeche di gayi definition ke saath padho — har arrow ka naam wahan hai.

Figure — Autonomous GNC for reusable rockets — SpaceX approach overview

Definition Sign convention, aur Figure s01 ko kaise padhen

Figure kaise draw ki gayi hai. Picture mein, lamba lal shape booster hai jo black ground line (yaani altitude $h = 0$ ) se kaafi upar hai. Teen black arrows booster se nikalte hain, aur frame ke top-left mein ek chota black arrow axis hai:

sabse left mein upar ki taraf jaata arrow, jo "+ (up positive)" label karta hai, hamari reference direction hai — jo bhi us taraf point karta hai woh positive count hota hai;
booster ke left mein neeche ki taraf jaata arrow, jo "velocity = −v₀" label karta hai, woh velocity hai (yeh girta hai, isliye neeche point karta hai);
booster ke right mein neeche ki taraf jaata arrow, jo "g" label karta hai, woh gravity hai;
booster ke neeche upar ki taraf jaata arrow, jo "Thrust" label karta hai, woh engine push hai.

Toh ek nazar mein: velocity aur gravity neeche point karte hain, thrust upar point karta hai, aur "upar" positive direction hai.

Ab symbols. Upar positive hai. Booster girta hai, isliye uski actual velocity vector neeche point karti hai; hum us velocity ko $- v_{0}$ likhte hain jahan $v_{0} > 0$ neeche ki speed hai (ek simple positive number, "kitni tezi se gir raha hai"). Gravity neeche kheenchti hai: $g = + 9.8 m/s^{2}$ magnitude ke roop mein use kiya jaata hai. Engine thrust upar push karta hai.

Burn ke dauran net upward acceleration hai $a_{n e t} = \frac{T}{m} - g$ jahan $T$ thrust magnitude hai aur $m$ mass hai. Jab $a_{n e t} > 0$ toh rocket apni giraavat decelerate kar raha hai (slow ho raha hai), jo exactly wahi hai jo ek landing burn karta hai.

Kyun hum neeche diye kinematic formulas mein signs drop karte hain. Pure braking burn ke dauran sirf do hi directions occur hoti hain: booster neeche move karta hai, deceleration upar act karta hai. Mid-burn mein kuch bhi direction nahi badalta. Toh minus signs juggle karne ki jagah hum plain positive magnitudes ke saath kaam karte hain: $v$ = current downward speed (m/s, yahan hamesha $\geq 0$ ), $a_{n e t}$ = upward deceleration ki size (m/s², positive), aur $s$ = stopping distance, abhi aur girna baaki kितना (metres, positive). Un magnitudes ke saath standard "brake to a stop" relations kuch yun padhte hain $v = v_{0} - a_{n e t} t, v^{2} = v_{0}^{2} - 2 a_{n e t} s .$ Yahan minus signs simply kehte hain "braking time ke saath aur distance ke saath speed reduce karti hai." Hum $s$ ko burn ke dauran travel ki gayi distance define karte hain; jab burn poora raasta ground tak cover karta hai, $s$ ignition altitude $h_{i g n}$ ke barabar hota hai.

Level 1 — Recognition

Problem L1.1

GNC ke teen letters ka naam batao aur, ek-ek phrase mein, har ek ka woh sawaal jo woh answer karta hai.

Recall Solution

Guidance — Main kahan jaana chahta hoon? (desired trajectory / target state compute karta hai).
Navigation — Main abhi kahan hoon? (sensors se position, velocity, attitude estimate karta hai).
Control — Gap close karne ke liye actuators ko kaise move karun? (commands ko gimbal / fin / thruster deflections mein turn karta hai).

Problem L1.2

Ek booster ka minimum engine thrust, jab almost empty ho, ek thrust-to-weight ratio (TWR) deta hai jo 1 se zyada ho. Ek single word batao jo yeh explain kare ki yeh hover kyun nahi kar sakta?

Recall Solution

TWR > 1 ka matlab hai minimum upward thrust force weight force se zyada hai ( $T > m g$ , toh $T / m g > 1$ ). Lowest throttle par bhi net acceleration upward hai — rocket upar accelerate karega, float nahi karega. Hovering ke liye thrust exactly weight ke barabar chahiye (TWR = 1), jo engine itna low nahi ja sakta. Isliye ise ek single timed hoverslam karna padta hai.

Level 2 — Application

Problem L2.1

Ek booster downward speed $v_{0} = 150 m/s$ se gir raha hai. Engine ek net upward acceleration $a_{n e t} = 25 m/s^{2}$ produce kar sakta hai. Konsi altitude $h_{i g n}$ par engine light karna chahiye taaki velocity exactly ground par zero ho jaye?

Recall Solution

$v^{2} = v_{0}^{2} - 2 a_{n e t} s$ kyun? Is kinematic relation mein koi time nahi — yeh speed ko directly travel ki gayi distance $s$ se link karta hai. Hum woh distance chahte hain jis par speed zero ho jaati hai, toh yahi sahi tool hai (time-based $v = v_{0} - a_{n e t} t$ nahi). Yaad raho $s$ = stopping distance (burn kitna vertical distance cover karta hai); yahan burn poora raasta ground tak jaata hai, toh $s = h_{i g n}$ .

Final speed $v = 0$ set karo stopping distance $s = h_{i g n}$ par: $0 = v_{0}^{2} - 2 a_{n e t} h_{i g n} \Rightarrow h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} .$ Plug in karo: $h_{i g n} = \frac{15 0 ^{2}}{2 \cdot 25} = \frac{22500}{50} = 450 m .$ Answer: $h_{i g n} = 450$ m.

Agla figure exactly yahi case dikhata hai ek speed-versus-altitude curve ke roop mein. Horizontal axis altitude $h$ metres mein hai (pad par 0, right ki taraf badh raha hai); vertical axis speed $v$ m/s mein hai. Accent-red line booster ki speed hai jab woh brake karta hai.

Ise right se left padho (girna matlab altitude kam hona): upper right par black dot $h = 450$ m, $v = 150$ m/s par hai — yeh ignition hai. Red curve ko down-and-left follow karo aur yeh exactly origin par land karta hai, dusra black dot $h = 0$ , $v = 0$ par — touchdown. Kyunki curve zero altitude par zero speed reach karta hai aur kahin pehle nahi, woh single meeting point hi ek perfect hoverslam hai.

Problem L2.2

Same booster aur burn ( $v_{0} = 150$ , $a_{n e t} = 25$ ). Burn kitne time tak chalta hai?

Recall Solution

Ab time-form kyun, distance-form nahi? Figure s02 ne speed ko altitude ke against plot kiya — "speed zero kahan hoti hai" ke liye great, lekin uspar koi clock nahi hai, toh tum seconds nahi padh sakte. Sawaal kitna time poochha hai, jo ek clock sawaal hai. Do kinematic tools cleanly split hote hain is basis par ki woh kya chhod dete hain: $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ mein koi time nahi (distances ke liye use karo), jabki $v = v_{0} - a_{n e t} t$ mein koi distance nahi (durations ke liye use karo). Kyunki hum seconds chahte hain, hum woh waala choose karte hain jo actually $t$ contain karta hai.

$v = v_{0} - a_{n e t} t$ use karo final $v = 0$ ke saath: $t = \frac{v _{0}}{a _{n e t}} = \frac{150}{25} = 6 s .$ Answer: $t = 6$ s. (Sanity check: average speed $\frac{150 + 0}{2} = 75$ m/s, 6 s par = 450 m = $h_{i g n}$ . ✓ — yeh distance par wapas bridge exactly wahi hai jahan do forms agree karte hain.)

Problem L2.3

Ignition par booster mass $m = 30, 000$ kg hai, aur $g = 9.8 m/s^{2}$ hai. $a_{n e t} = 25 m/s^{2}$ use karke, required thrust $T$ kya hai?

Recall Solution

$T = m (g + a_{n e t})$ kyun? $a_{n e t} = T / m - g$ se, $T$ ke liye solve karo: engine ko pehle gravity cancel karni hai ( $m g$ ) aur phir extra deceleration supply karni hai ( $m a_{n e t}$ ). $T = m (g + a_{n e t}) = 30000 (9.8 + 25) = 30000 \times 34.8 = 1, 044, 000 N .$ Answer: $T \approx 1.044 \times 1 0^{6}$ N (1044 kN).

Level 3 — Analysis

Problem L3.1

L2.1 ke numbers use karke ( $v_{0} = 150$ , correct $h_{i g n} = 450$ m), maano computer engine early light karta hai, $h = 600$ m par, same $a_{n e t} = 25$ ke saath. Booster ki velocity kahan zero reach karti hai, aur physically aage kya hota hai?

Recall Solution

Same $a_{n e t}$ aur same starting speed ke saath, stopping distance $s$ (burn speed kill karne mein kitna far jaata hai) fixed hai: $s_{s t o p} = \frac{v _{0}^{2}}{2 a _{n e t}} = \frac{22500}{50} = 450 m .$ 600 m par light karne se, booster braking mein 450 m altitude consume karta hai aur ground se $600 - 450 = 150$ m upar rukta hai. Ab woh momentarily 150 m par rest mein hai engine still on ke saath aur TWR > 1 → net upward acceleration → woh wapas upar jaata hai. Woh hover ke liye throttle nahi kar sakta. Toh early ignition "margin add" nahi karta — yeh rocket ko mid-air mein stranded kar deta hai ya wasteful re-fall force karta hai. Answer: velocity 150 m altitude par zero; phir woh upward accelerate karta hai (hover nahi kar sakta).

Problem L3.2

Ab woh late light karta hai, $h = 300$ m par, same $a_{n e t} = 25$ , same $v_{0} = 150$ . Impact speed kya hai?

Recall Solution

Burn ke paas ab sirf $s = 300$ m stopping distance available hai, lekin usse 450 m chahiye tha. $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ use karo $s = 300$ ke saath: $v^{2} = 15 0^{2} - 2 (25) (300) = 22500 - 15000 = 7500,$ $v = 7500 \approx 86.6 m/s (downward).$ Answer: yeh ground par $\approx 86.6$ m/s se hit karta hai → crash. Yeh razor-thin corridor hai: ignition-altitude error ke $\pm 150$ m ka fark hai wapas uthne aur highway speed par slam in karne mein.

Problem L3.3

Explain karo kyun IMU akele poora descent navigate nahi kar sakta, aur ek constant accelerometer bias $b$ ke position estimate mein growth law do. Phir batao 3D position-error size ka kya hota hai agar har axis ka same bias ho.

Recall Solution

Pehle, notation: ek bold letter jaise $r$ ek vector matlab hai — ek quantity jisme teen components hain (ek $x$ , ek $y$ , ek $z$ ), yahan space mein position; isi tarah $a$ acceleration vector hai. Bold sirf kehta hai "iske teen numbers aur ek direction hai."

IMU acceleration deta hai; position twice integrate karke recover hoti hai ( $r = \iint a d t$ — pehle time ke saath acceleration add karo velocity paane ke liye, phir time ke saath velocity add karo position paane ke liye).

Ek axis explicitly karo. Maano $x$ -accelerometer ek fixed wrong offset $b$ (metres per second², constant) read karta hai. Velocity error ke liye ek baar integrate karo: $Δ v_{x} = \int_{0}^{t} b d t = b t .$ Position error ke liye woh ek aur baar integrate karo: $Δ x = \int_{0}^{t} b t^{'} d t^{'} = \frac{1}{2} b t^{2} .$ Toh har axis par same algebra same shape $\frac{1}{2} b t^{2}$ deta hai — $x$ ke baare mein kuch special nahi tha, toh $y$ aur $z$ bhi $\frac{1}{2} b t^{2}$ produce karte hain. Isliye "argument har axis ke saath identical hai": integration ko pata nahi aur parwah nahi ki woh kis axis par run kar raha hai.

3D error size. Agar teeno axes same bias $b$ carry karte hain, toh error vector hai $Δ r = (\frac{1}{2} b t^{2}, \frac{1}{2} b t^{2}, \frac{1}{2} b t^{2})$ . Uski length (3D mein Pythagoras) hai $∣Δ r ∣ = 3 (\frac{1}{2} b t^{2}) = \frac{3}{2} b t^{2},$ abhi bhi $t^{2}$ ki tarah grow kar raha hai — sirf $3 \approx 1.73$ times akele ek axis se bada.

Kyun yeh matter karta hai: error time ke square ke saath badh ta hai — small bias, long flight → huge error. Toh hum ek drift-free lekin slow reference (GPS/radar) ko Kalman filter ke zariye fuse karte hain drift ko bound karne ke liye. Fast IMU + slow GPS = dono ka best.

Level 4 — Synthesis

Problem L4.1

Ek booster altitude $h = 1000$ m par hai aur $v_{0} = 200$ m/s se descend kar raha hai. Engine $a_{n e t} = 30 m/s^{2}$ net upward accel deliver kar sakta hai. (a) Kya yeh already us point se past hai jahan ek $30 m/s^{2}$ burn ise save kar sake? (b) Minimum $a_{n e t}$ kya hoga jo exactly ise yahan se ground par stop kar de?

Recall Solution

(a) $a_{n e t} = 30$ ke liye required ignition altitude (stopping distance $s = h_{i g n}$ ): $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} = \frac{20 0 ^{2}}{2 \cdot 30} = \frac{40000}{60} \approx 666.7 m .$ Kyunki booster 1000 m par hai, jo 666.7 m se upar hai, yeh point se past nahi hai — ise girte rehna chahiye aur baad mein light karna chahiye. Uske paas margin hai.

(b) Minimum $a_{n e t}$ woh hai jo barely ise puri 1000 m stopping distance use karke stop kare: $a_{n e t, m i n} = \frac{v _{0}^{2}}{2 h} = \frac{40000}{2000} = 20 m/s^{2} .$ Answers: (a) point se past nahi (h_ign ≈ 666.7 m < 1000 m); (b) minimum $a_{n e t} = 20 m/s^{2}$ . Interpretation: $\geq 20 m/s^{2}$ capable koi bhi engine ise land kar sakta hai; ek $30 m/s^{2}$ engine bas neechay light karta hai aur chota burn karta hai.

Problem L4.2

Upar wale case ke liye, $a_{n e t} = 30 m/s^{2}$ aur mass $m = 28, 000$ kg ignition par ( $g = 9.8$ ) use karke, (a) burn duration aur (b) required thrust find karo.

Recall Solution

(a) Ignition $h_{i g n} \approx 666.7$ m par hoti hai, speed abhi bhi $v_{0} = 200$ m/s. $t = \frac{v _{0}}{a _{n e t}} = \frac{200}{30} \approx 6.67 s .$ (b) Thrust: $T = m (g + a_{n e t}) = 28000 (9.8 + 30) = 28000 \times 39.8 = 1, 114, 400 N .$ Answers: (a) $t \approx 6.67$ s; (b) $T \approx 1.114 \times 1 0^{6}$ N (1114 kN).

Level 5 — Mastery

Problem L5.1 — Live closed-loop recompute

Real thrust perfectly steady nahi hoti. Maano actual burn under-performs karke sirf $a_{n e t} = 18 m/s^{2}$ deliver karta hai instead of planned 20. Vehicle har cycle mein live state $(v, h)$ se $a_{c m d} = v^{2} / (2 h)$ use karke commanded net accel recompute karta hai. Jis instant usne light kiya (planned), woh $v = 200$ m/s, $h = 1000$ m par tha. (a) Us instant par closed-loop law kya net accel demand karta hai? (b) Agar engine sirf 18 tak pahunch sakta hai, guidance kya karta hai, aur kya landing abhi bhi possible hai? Reason karo.

Recall Solution

(a) $v = 200$ , $h = 1000$ par: $a_{c m d} = \frac{v ^{2}}{2 h} = \frac{20 0 ^{2}}{2 \cdot 1000} = \frac{40000}{2000} = 20 m/s^{2} .$ (b) Engine sirf 18 de sakta hai, yaani woh under-brake karta hai. Toh agly cycle mein vehicle planned se kam speed shed karti hai; naye altitude par ratio $v^{2} / (2 h)$ climb karta hai (speed shrinking altitude ke liye bahut zyada hai). Closed-loop law isliye agly cycle mein zyada accel demand karta hai. Jab tak engine ki ceiling ( $a_{n e t}$ jo woh actually bana sakta hai, throttle aur mass diye hue) rising demand se upar rehti hai, tab tak yeh catch up kar leta hai aur $v = 0$ at $h = 0$ nail karta hai. Agar demand touchdown se pehle engine ceiling se kabhi bhi zyada ho jaaye, toh case L3.2 ke "late ignition" crash ban jaata hai. Yahi exactly kyun guidance closed-loop hai: woh har cycle mein $a_{c m d} = v^{2} /2 h$ re-solve karta hai ek pre-scripted plan par trust karne ki bajay. Fuel-optimal version of this recompute-under-constraints problem ke liye, SpaceX-style guidance ek real-time convex optimization solve karta hai lossless convexification ke zariye, har cycle mein bounded time mein global optimum guarantee karta hai. Answer: (a) 20 m/s² demand karta hai; (b) woh har cycle mein accel up-command karta hai; landing tab survive karti hai jab engine ceiling rising demand se upar rahti hai.

Problem L5.2 — Degenerate / edge cases

Ignition formula $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ teen limiting cases mein kya predict karta hai, discuss karo, aur batao ki har ek physical hai ya nahi: (a) $v_{0} = 0$ ; (b) $a_{n e t} \to 0^{+}$ ; (c) $a_{n e t}$ bahut bada.

Recall Solution

(a) $v_{0} = 0$ (already at rest). $v_{0} = 0$ substitute karo: $h_{i g n} = 0^{2} / (2 a_{n e t}) = 0$ . Physical meaning: agar tumhari speed already zero hai toh tumhe koi braking distance nahi chahiye ( $s = 0$ ) — kinematically tum wahan touch down kar sakte ho jahan ho. (Real gravity phir tumhe neeche kheenchna shuru karega, toh tumhe hold karne ke liye abhi bhi thrust chahiye, lekin pure "speed kill karne ki distance" genuinely zero hai.) Formula correct aur sensible hai.

(b) $a_{n e t} \to 0^{+}$ (engine barely gravity se stronger). Jab $a_{n e t}$ positive side se zero ki taraf shrink hota hai, denominator $2 a_{n e t} \to 0^{+}$ jabki numerator fixed rehta hai, toh $h_{i g n} = v_{0}^{2} / (2 a_{n e t}) \to + \infty$ . Physical meaning: ek rocket jo barely decelerate kar sakta hai, stop karne ke liye ek infinite runway chahiye — yaani kisi bhi real corridor ke andar woh fall ko rok hi nahi sakta. Yeh boundary hai jahan engine bahut weak hai (TWR just above 1); landing impossible ho jaati hai, aur formula correctly diverge karke warn karta hai.

(c) $a_{n e t}$ bahut bada (extremely powerful burn). Jab $a_{n e t} \to \infty$ , denominator bina bound ke badh ta hai, toh $h_{i g n} = v_{0}^{2} / (2 a_{n e t}) \to 0^{+}$ . Physical meaning: ek enormously powerful engine ko almost koi stopping distance nahi chahiye — tum almost poora raasta deck tak girte ho, phir ek violent, bahut-short burst tumhe rok deta hai. Mathematically fine, lekin yeh near-perfect timing (ek vanishingly small ignition window) aur punishing g-loads maangta hai, toh real vehicles deliberately ek moderate $a_{n e t}$ choose karte hain taaki timing corridor aur structural loads dono sane rahen.

Takeaway: formula har extreme par sensibly behave karta hai — $v_{0} = 0 \Rightarrow h_{i g n} = 0$ ; weak engine $\Rightarrow h_{i g n} \to \infty$ (impossible); strong engine $\Rightarrow h_{i g n} \to 0$ (twitchy). Koi bhi case surprise nahi karta, jo exactly ise ek trustworthy live decision rule banata hai.

Final figure poori family plot karta hai: $h_{i g n}$ engine ke $a_{n e t}$ ke against, fixed $v_{0} = 200$ m/s ke liye. Horizontal axis net accel $a_{n e t}$ m/s² mein hai, vertical axis ignition altitude $h_{i g n}$ m mein hai, aur accent-red curve hi ignition law hai.

Red curve trace karo: far left par (small $a_{n e t}$ ) yeh infinity ki taraf shoot up karta hai — woh case (b) hai, too-weak engine. Far right par (large $a_{n e t}$ ) yeh zero ki taraf flatten hota hai — woh case (c) hai, twitchy super-strong engine. Do annotated ends exactly wahi limits hain jinke baare mein tumne upar reason kiya.

Active recall

Recall Ladder ka one-line summary

Upar sab kuch ek equation hai jo paanch angles se dekhi gayi hai: $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ . Recognition ise naam deta hai, Application ise plug karta hai, Analysis ise tod ta hai (early/late/IMU drift), Synthesis ise $T = m (g + a_{n e t})$ aur burn time $t = v_{0} / a_{n e t}$ ke saath combine karta hai, aur Mastery ise live (closed-loop recompute) aur robust (edge cases) banata hai. Prerequisite tools Kinematics — v²=v0²+2as, Rocket Equation (Tsiolkovsky), PID Control, Thrust Vectoring (Gimbal), aur Attitude Dynamics & Quaternions mein hain.

Ignition-altitude law ek line mein

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

Required landing-burn thrust

T = m (g + a_{n e t})

Burn duration

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

Closed-loop kyun, open-loop nahi

real mass, wind, aur thrust har flight vary karte hain; error bound karne ke liye har cycle recompute karo