Autonomous GNC for reusable rockets — SpaceX approach overview: Visual walkthrough

3.5.55 · D2 · Physics › Guidance, Navigation & Control (GNC) › Autonomous GNC for reusable rockets — SpaceX approach overvi

Hum ek boxed result tak build karte hain: $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}}$ Abhi kisi bhi symbol ki chinta mat karo — har ek neeche earn kiya gaya hai.

Step 1 — Ek girne wali cheez mein downward speed hoti hai

KYA. Booster ko apne coast ke baad imagine karo, upar kahin, engine off, seedha neeche gir raha hai. Jis moment hum dekhna shuru karte hain, woh neeche ki taraf kisi speed se move kar raha hai jise hum $v_{0}$ kehte hain.

KYUN. Isse pehle ki hum rokne ki baat karein, hume us cheez ka naam dena hoga jo hum rokna chahte hain: downward speed. Baaki sab kuch is number ko khatam karne ke liye build kiya gaya hai.

PICTURE. Figure mein rocket tall shape hai; mota downward arrow uski velocity hai. Uski length speed hai. Bottom par dashed line ground hai (height $0$ ); jahan rocket baitha hai wahan dashed line current height $h$ hai.

Figure — Autonomous GNC for reusable rockets — SpaceX approach overview

Hum upward ko positive choose karte hain. Toh ek downward fall matlab velocity negative direction mein point karti hai — lekin $v_{0}$ khud ek positive number likha jaata hai (us downward speed ka size). Is convention ko seedha rakhna hi poora game hai.

Step 2 — Engine sirf ek hi taraf push kar sakta hai: upar

KYA. Engine, jab jalta hai, rocket ko upar ek force se push karta hai. Gravity hamesha neeche kheenchti hai. Rocket ka ek mass bhi hai. Yeh teen — up-push, down-pull, mass — decide karte hain ki rocket ki speed kitni jaldi change hoti hai.

KYUN. "Speed kitni jaldi change hoti hai" exactly wahi hai jo humein chahiye, kyunki hum downward speed ko $v_{0}$ se $0$ tak shrink karna chahte hain. Koi cheez jo speed change karti hai woh accelerate kar rahi hoti hai, toh humein acceleration build karni hogi.

PICTURE. Rocket par do arrows: ek lamba burnt-orange arrow upar (thrust $T$ ) aur ek chhota teal arrow neeche (weight). Up arrow jaanbujhkar lamba draw kiya gaya hai — engine gravity se zyada push karta hai. Plum arrow bacha hua upward pull hai, net effect.

Yeh shape kyun aur kuch fancy kyun nahi? Kyunki Newton's law $force = mass \times acceleration$ ko rearrange karne par $acceleration = force / mass$ milta hai. Humne sirf forces add kiye (up minus down) aur mass se divide kar diya. Aur kuch nahi chahiye.

Step 3 — Speed ko marte hue draw karo: speed-vs-height picture

KYA. Jaise rocket apne burn mein upar jaata hai (ya rather, girta rehta hai lekin decelerate karta hai), uski downward speed $v_{0}$ se $0$ tak drop hoti hai. Hum speed ko up-axis par plot karte hain height ke against side-axis par, rocket ko height $h$ se height $0$ tak travel karte hue dekhte hain.

KYUN. Humein parwah hai ki speed zero kahan hogi (kaunsi height), na ki kab (kaun sa second). Toh hum ek picture chahte hain jisme height ek axis par ho aur speed doosri par, time chhupa hua ho. Yahi poora reason hai ki hum Step 4 mein ek time-free formula choose karenge.

PICTURE. Curve point $(h, v_{0})$ se shuru hota hai — upar, fast chalte hue — aur exactly $(0, 0)$ par land karna chahiye — pad height, dead stop. Agar curve zero height se pehle zero speed reach kar le (dotted plum), rocket hawa mein rok jaata hai aur phir gir jaata hai. Agar height zero par bhi speed ho (dotted teal), toh pad se crash. Sirf burnt-orange curve ek acchi landing hai.

Recall Kyun "jaldi rokna" aur "der se rokna" dono fail hote hain

Jaldi rokna ::: velocity 0 hit karta hai airborne rehte hue; TWR>1 matlab wahan ruk nahi sakte, toh upar ya neeche jaoge → miss. Der se rokna ::: height 0 par bhi moving hai → pad se crash.

Step 4 — Woh tool choose karo jisme time nahi hai

KYA. Humein ek equation chahiye jo start speed $v_{0}$ , final speed ( $0$ ), acceleration $a_{n e t}$ , aur travel ki gayi distance $h$ ko link kare — bina time ke.

KYUN. Kyunki Step 3 ki picture mein koi time axis nahi hai. Standard motion equations mein se, exactly ek mein koi $t$ nahi hai: $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ Yeh constant-acceleration relation hai. (Humne minus likha kyunki $a_{n e t}$ upar point karta hai jabki motion neeche hai — acceleration fall se lad raha hai, speed remove kar raha hai.)

PICTURE. Teen motion equations ka ek menu; do ko cross out kiya gaya hai kyunki unme $t$ hai. Survivor circle kiya gaya hai — yeh speed aur distance ko directly connect karta hai.

Step 5 — Perfect landing demand karo aur solve karo

KYA. Step 3 ki landing condition impose karo: speed exactly $v = 0$ honi chahiye jis moment travel ki gayi distance poore drop ke barabar ho, $s = h$ .

KYUN. "Zero height par zero speed" hi soft landing ki definition hai. Ise plug in karne se general relation woh ek number ban jaata hai jis par flight computer dhyan deta hai.

PICTURE. Equation mein $v = 0$ aur $s = h$ substitute kiya gaya, aur arrows $a_{n e t}$ ko isolate karne ke liye use karte hain.

$v = 0$ , $s = h$ set karo: $0 = v_{0}^{2} - 2 a_{n e t} h$ Acceleration term ko across move karo (dono sides mein $2 a_{n e t} h$ add karo): $2 a_{n e t} h = v_{0}^{2}$ Dono sides ko $2 h$ se divide karo: $a_{n e t} = \frac{v _{0}^{2}}{2 h}$

Ise padhna: speed $v_{0}$ se height $h$ ke andar rokne ke liye, tumhe exactly itni deceleration chahiye. Do guna fast giro → chaar guna deceleration chahiye (square ki wajah se). Do guna height do apne aap ko → aadha chahiye.

Aur kyunki $a_{n e t} = T / m - g$ , engine ko jo thrust produce karni hogi woh hai $T = m (g + \frac{v _{0}^{2}}{2 h}) .$ Yahan $m (g)$ sirf rocket ko gravity ke against hold karta hai; extra $m \cdot \frac{v _{0}^{2}}{2 h}$ braking part hai.

Step 6 — Question flip karo: main engine kab jalaaun?

KYA. Flight mein engine ka achievable $a_{n e t}$ roughly fixed hai. Jo computer nahi jaanta jab tak woh moment na aaye woh hai height jis par use ignite karna hai. Toh boxed result ko $a_{n e t}$ ki jagah $h$ ke liye solve karo.

KYUN. Rocket gir raha hai aur har kuch milliseconds mein apni speed aur height dekh raha hai. Use ek trigger chahiye: "jis instant meri height is value tak gire, jala do." Woh trigger ignition altitude hai.

PICTURE. Wohi rocket girta hua; ek horizontal plum trip-line height $h_{i g n}$ par baitha hai. Line ke upar, engine off (grey). Jis moment rocket ise cross karta hai, engine flash on hota hai (orange).

Yeh ek live, closed-loop decision hai — har cycle mein recompute hota hai jaise $v_{0}$ aur mass change hota hai. Ek baar jalne ke baad, thrust har cycle mein present $v$ aur $h$ use karke $T_{c m d} = m (g + \frac{v ^{2}}{2 h})$ se nudge kiya jaata hai (PID Control aur Convex Optimization fine corrections aur fuel-optimality handle karte hain; Thrust Vectoring (Gimbal) drift cancel karne ke liye push sideways aim karta hai).

Step 7 — Numbers, aur do failure edges

KYA. Real values daalo: $v_{0} = 200$ m/s se girte hue, engine $a_{n e t} = 20$ m/s² deta hai.

KYUN. Ek formula jisme tum numbers nahi daal sakte woh abhi tumhara nahi hai. Aur do "thoda sa off" cases dikhate hain ki timing exact kyun honi chahiye.

PICTURE. Altitude ki ek number line. $1000$ m par correct ignition (orange) cleanly $0$ par land karti hai. "100 m late" case (teal, 900 m par light) pad par abhi bhi speed hai. "100 m early" case (plum, 1100 m par light) airborne rok jaata hai.

$h_{i g n} = \frac{20 0 ^{2}}{2 \cdot 20} = \frac{40000}{40} = 1000 m$

Burn time (time-wali equation $v = v_{0} - a_{n e t} t$ use karke, $v = 0$ set karo): $t = \frac{v _{0}}{a _{n e t}} = \frac{200}{20} = 10 s .$

Exactly 1000 m par light karo → 10 s baad, speed $0$ height $0$ par. Perfect.
Der se light karo (900 m) → enough braking distance nahi → touchdown par abhi bhi moving → crash.
Jaldi light karo (1100 m) → speed $0$ hit karta hai ~100 m upar → lekin TWR>1 matlab hover nahi kar sakta → phir se utha / fuel waste → miss.

Degenerate check — agar $v_{0} = 0$ ho? Tab $h_{i g n} = 0$ : agar tumhare paas pehle se koi downward speed nahi hai toh tumhe koi braking distance nahi chahiye. Formula common sense se agree karta hai. Aur agar $a_{n e t} \to 0$ (engine barely gravity se jeette), $h_{i g n} \to \infty$ — tumhe impossibly high se braking shuru karni padti; ek weak engine land nahi kar sakta, exactly isliye boosters TWR comfortably 1 se upar rakhte hain.

Ek-picture summary

Upar sab kuch, compress kiya gaya: girta rocket, $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ par trip-line, speed curve origin mein glide karti hui, aur ek single boxed formula.

Recall Feynman: poora walkthrough plain words mein

Tum ek ball gira rahe ho aur chahte ho ki woh exactly ek coin par ruke, lekin tumhara ek hi tool hai — uske neeche ek spring jo itni strong hai ki ball float nahi kar sakti. Toh tum use dhire se nahi utar sakte. Iske bajaye tum ise girne dete ho, aur ek perfect height par spring fire karte ho taaki ball ki speed exactly tab khatam ho jab woh coin ko touch kare. Woh height dhundne ke liye: girne ki speed ka naam lo ( $v_{0}$ ). Pata karo ki tum kitna brake kar sakte ho — engine push minus gravity, mass se divide karke ( $a_{n e t}$ ). Phir woh ek motion rule use karo jo time ignore karta hai aur sirf speed ko distance se relate karta hai. Demand karo "speed zero at height zero," aur nikal aata hai: brake distance $= v_{0}^{2} /2 a_{n e t}$ . Woh distance hi engine light karne ki height hai. Faster giro aur tumhe chaar guna zyada room chahiye (yeh squared hai). Rocket khud apni speed aur height hazaaron baar ek second mein measure karta hai aur jaise hi woh us line ko cross karta hai trigger pull karta hai.