Worked examples — Powered descent guidance — G-FOLD algorithm (convex optimization)

3.5.53 · D3 · Physics › Guidance, Navigation & Control (GNC) › Powered descent guidance — G-FOLD algorithm (convex optimiza

Yeh page ek drill room hai. Parent note ne theory build ki; yahan hum problem ke har shape ko exhaust karte hain: signs, degenerate inputs, limiting cases, ek word problem, aur ek exam twist. Har ek ko steps padhne se pehle khud try karo.

Shuru karne se pehle, ek one-line recap un symbols ki jo hum har jagah reuse karte hain (kabhi bhi naam liye bina use nahi kiye):

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The scenario matrix

Har powered-descent question cells ki ek choti family mein se ek (ya blend) hoti hai. Koi table yaad karne ki jagah, neeche diya hua map imagine karo. Iske do bade branches hain — left par vertical (1-D) feasibility aur right par glide-slope geometry — dono sign cases, degenerate cases, aur limits mein split hote hain. Map ka sketch, taaki tum image file khole bina follow kar sako:

Figure (matrix): wahi map scale par draw kiya gaya — top par root, blue vertical-feasibility branch (leaves C0–C5) left par, green glide-slope branch (leaves C6–7) right par, aur orange/red fuel-and-exam leaves (C8–C10) bottom se hanging. Har leaf ek worked example hai neeche.

• C0 — degenerate quadrant $v_z>0$ (abhi bhi climb kar raha hai): kya hota hai agar hum upar move karte shuru karein? → Example 0
• C1 — net acceleration ka sign (thrust vs gravity): kya hum gravity se lad bhi sakte hain? → Example 1
• C2 — stopping distance vs altitude: kya burn fit hota hai? → Example 2
• C3 — degenerate $v_z=0$ (hover): kaun sa thrust bas rokke rakhta hai? → Example 3
• C4 — limiting $T_{\min }\to 0$ (deep throttle): coast-then-burn kab aata hai? → Example 4
• C5 — bang-bang timing: suicide burn kab shuru karein? → Example 5
• C6 — glide-slope offset (horizontal position): cone constraint → Example 6
• C7 — degenerate cone ${\theta }_{gs}\to 90°$ aur $\to 0°$: funnel ki limits → Example 7
• C8 — fuel via $z=\ln m$: log-mass bookkeeping → Example 8
• C9 — word problem (real Mars/F9 numbers): sab kuch saath lagao → Example 9
• C10 — exam twist (infeasible input): "koi landing exist nahi karta" pehchano → Example 10

Numbers jo cells mein reuse hote hain (Mars-ish): $g=-3.71{\text{m/s}}^2$, $T_{\max }/m=12{\text{m/s}}^2$, $T_{\min }/m=3{\text{m/s}}^2$ jab tak kuch aur na bola jaye.

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Example 0 — Initial upward velocity, $v_z>0$ edge case (Cell C0)

Forecast: Guess karo ki extra climb se eventual landing aasaan hogi ya mushkil.

1. $v_z=0$ tak upar coast karo. Sign convention use karte hue, gravity climb ko decelerate karti hai: extra height $\Delta h=\frac{v_z^2}{2|g|}=\frac{15^2}{2(3.71)}=30.3$ m. Yeh step kyun? Positive $v_z$ bilkul naya quadrant hai — lander pehle pad se door jaata hai. Kinematics turning point deta hai jahan upward speed zero ho jaati hai.
2. Peak height. $r_{z,\text{peak}}=100+30.3=130.3$ m. Yeh step kyun? Peak woh hai jahan saari upward speed height mein convert ho gayi ho; yahan se problem ordinary descent ban jaati hai.
3. 100 m par wapas speed. Free fall ki symmetry se, yeh $r_z=100$ m par $v_z=-15$ m/s ke saath wapas aata hai. Yeh step kyun? Energy conservation (koi thrust nahi, koi drag nahi) down-trip ko up-trip ka mirror banata hai — isliye initial $+15$ ek ordinary $-15$ descent ban jaata hai, jise hum pehle se land karna jaante hain (Examples 2, 5).

Verify: $\Delta h=15^2/(2\cdot 3.71)=30.32$ m ✔; peak $=130.3$ m; return speed $=15$ m/s (magnitude) ✔.

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Example 1 — Net acceleration ka sign (Cell C1)

Forecast: Guess karo ki lander dono throttle settings par decelerate (apni fall slow) kar sakta hai ya nahi.

1. Full throttle par net. $a_{\text{net}}=a_{\text{cmd}}+g=12+(-3.71)=8.29{\text{m/s}}^2$ (upar). Yeh step kyun? Newton accelerations ko vectors ki tarah add karta hai; upar $+$ hai, gravity $-$ hai, isliye hum literally signed numbers add karte hain.
2. Min throttle par net. $a_{\text{net}}=3+(-3.71)=-0.71{\text{m/s}}^2$ (neeche). Yeh step kyun? Throttle floor par engine gravity se zyada push nahi kar sakta — sign negative ho jaata hai, matlab yeh abhi bhi neeche accelerate kar raha hai.

Verify: $12-3.71=8.29$ ✔; $3-3.71=-0.71$ ✔. Units: sab ${\text{m/s}}^2$. Min-throttle net ka negative hona confirm karta hai ki engine ka floor hover nahi kar sakta — C3 dekho.

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Example 2 — Kya burn fit hota hai? (Cell C2)

Forecast: Stopping distance guess karo — 100 m se zyada ya kam?

Yahan $d$ vertical distance hai jo lander brake karte waqt stop tak travel karta hai — jahan burn shuru hoti hai wahan se neeche measure kiya gaya. Kyunki lander pad ke seedha upar hai, yeh $d$ height mein change hai, yani $r_z$ mein change: agar burn $r_z=100$ par shuri hui aur poore $100$ m use kar diye, toh yeh $r_z=0$ par exactly touch down karega. Isliye $d$ ko starting height $r_z$ se compare karna poora feasibility test hai.

1. Magnitudes par switch karo. Upar diye gaye sign-convention box ke according, signed state $v_z=-20$ entry speed $v_0=|v_z|=20$ m/s ban jaata hai, aur signed net $a_{\text{net}}=+8.29$ (upar, fall ka virodh karte hue) braking deceleration $a=8.29{\text{m/s}}^2$ ban jaata hai. Yeh step kyun? Braking ek seedha downward segment hai jahan velocity aur motion direction share karte hain; direction pehle se "up is $+$" se fix hai, isliye sirf sizes matter karte hain aur hum signs drop kar sakte hain.
2. Tool chunno: kinematics $v^2=v_0^2-2ad$. Yahan $v$ braking ke end mein speed hai (hum chahte hain $v=0$), $v_0=20$ m/s, $a=8.29{\text{m/s}}^2$, aur $d$ upar defined braking distance hai. Yeh tool kyun? Hum start speed jaante hain, constant braking deceleration jaante hain, aur $v=0$ reach karne ki distance chahiye. Yeh equation exactly un teeno ko bina time ke link karta hai — yahan sabse saaf question-answerer.
3. Final speed ko zero set karo aur $d$ solve karo. $0=20^2-2(8.29)d\Rightarrow d=\frac{400}{16.58}=24.1$ m. Yeh step kyun? $v=0$ "just stopped" condition hai; solve karne se max brake par sabse choti possible stopping distance milti hai.
4. $d$ ko starting height $r_z$ se compare karo. $24.1\text{m}<100\text{m}$. ✔ Feasible. Yeh step kyun? $d$ woh height hai jo stop karte waqt khat ho jaati hai; agar $d\le r_z$ hai toh lander pad par ya upar zero speed tak pahunch jaata hai, isliye soft landing physically possible hai. Agar instead $d>r_z$ hota toh yeh ground level par abhi bhi move kar raha hota — crash. Inequality $d\le r_z$ hi feasibility criterion hai.

Verify: $400/(2\cdot 8.29)=24.126\text{m}$ ✔, aur $24.1<100$. Units: $(\text{m/s}{)}^2/({\text{m/s}}^2)=\text{m}$ ✔.

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Example 3 — Hover, degenerate $v_z=0$ case (Cell C3)

Forecast: Compute karne se pehle number guess karo.

1. Hover ka matlab zero net acceleration. Require karo $a_{\text{cmd}}+g=0\Rightarrow a_{\text{cmd}}=-g=3.71{\text{m/s}}^2$. Yeh step kyun? "Jagah par rehna" degenerate case hai jahan velocity kabhi nahi badalti, isliye net exactly zero hona chahiye — thrust gravity ko cancel kare.
2. Box check karo. Kya $3\le 3.71\le 12$ hai? Haan. Yeh step kyun? $[T_{\min }/m,T_{\max }/m]$ se bahar command physically impossible hai; yahan yeh floor ke theek upar baith jaata hai, isliye hover achievable hai.

Verify: $-g=3.71$ ✔ aur $3\le 3.71\le 12$ ✔.

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Example 4 — Limiting case $T_{\min }\to 0$ (Cell C4)

Forecast: Kya zero-floor throttle coasting ke liye zyada ya kam altitude available karta hai?

1. Max brake unchanged hai. Full-throttle net abhi bhi $12-3.71=8.29$ hai; stopping distance abhi bhi $24.1$ m. Yeh step kyun? Upper limit braking power set karta hai; floor ko lower karna max deceleration nahi badalta.
2. Floor kya control karta hai. $T_{\min }/m=0$ ke saath engine fully off switch ho sakta hai, net $=g=-3.71$ par pure coast deta hai. Yeh step kyun? Parent note ki obstacle yaad karo: constraint $\Vert \mathbf{T}\Vert \ge T_{\min }$ thrust space mein $T_{\min }$ radius ke ball ke andar ke saare thrust vectors forbid karta hai, yaani woh region us sphere ke bahar hi chhodta hai — ek non-convex hole. $T_{\min }/m\to 0$ bhejne se woh forbidden ball ek point tak shrink ho jaata hai, isliye hole gayab ho jaata hai aur $\Gamma$ $0$ tak free hai (engine fully off).
3. Available coast distance. $r_z=100$ se, ship freely coast karti hai, phir uske paas $\ge 24.1$ m brake karne ke liye hona chahiye. Isliye yeh mandatory burn se pehle $100-24.1=75.9$ m tak coast kar sakta hai. Yeh step kyun? Yahi bang-bang ka geometric meaning hai: jitna ho sake coast karo, phir last legal instant par brake lagao.

Verify: $100-24.1=75.9$ m coast ✔; $75.9+24.1=100$ ✔.

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Example 5 — Kab fire karein: bang-bang timing (Cell C5)

Figure (Example 5): vertical axis par height $r_z$ (m) vs horizontal axis par speed $|v_z|$ (m/s). Blue curve free-fall coast hai (speed badhti hai jaise height girti hai); red curve full-throttle brake hai (speed zero tak shrink hoti hai). Dono orange dot par milte hain — burn-start handoff $v\approx 28.1$ m/s, height $\approx 47.6$ m par. Blue dot top-right start hai (v=20, r=100); green dot origin par soft touchdown hai (v=0, r=0). Words mein: ise time mein right-to-left padho — ship blue coast curve par speed gain karte hue slide karti hai, orange corner par hit karti hai, phir green origin tak red brake curve par ride karti hai.

Forecast: Guess karo ki coast 2 seconds se zyada chalti hai ya kam.

1. Speed jab braking shuru ho. $v_b$ woh speed ho jis waqt burn shuru ho. Brake segment ko baaki height par saara $v_b$ bleed off karna hoga: $\frac{v_b^2}{2(8.29)}=$ (height remaining). Yeh step kyun? Total altitude coast aur brake segments ke beech share hota hai; hum speed aur height ko handoff point par match karte hain (figure mein orange dot).
2. Do segments set up karo. Coast height $h_c$ drop karta hai speed gain karte hue $v_b^2=20^2+2(3.71)h_c$. Brake ko chahiye $\frac{v_b^2}{2(8.29)}=100-h_c$. Yeh step kyun? Do equations, do unknowns ($v_b,h_c$) — figure mein red brake curve ka blue coast curve se milne ki geometry.
3. Solve karo. Substitute karo: $\frac{400+7.42h_c}{16.58}=100-h_c\Rightarrow 400+7.42h_c=1658-16.58h_c\Rightarrow 24h_c=1258\Rightarrow h_c=52.42$ m. Tab $v_b=\sqrt{400+7.42(52.42)}=\sqrt{788.9}=28.09$ m/s. Yeh step kyun? Algebra exact handoff altitude find karta hai — figure mein corner.
4. Coast time. $v_b=20+3.71t_c\Rightarrow t_c=(28.09-20)/3.71=2.18$ s. Yeh step kyun? Constant coast accel ke under, speed linearly badhti hai, isliye time = speed gain over accel.

Verify: $h_c=52.42$ m, $v_b=28.09$ m/s, aur $52.42+\frac{28.09^2}{2\cdot 8.29}=52.42+47.58=100$ ✔. Coast time $2.18$ s ✔.

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Example 6 — Glide-slope cone, horizontal offset (Cell C6)

Figure (Example 6): glide-slope funnel ka side view. Horizontal axis = horizontal radius $\sqrt{r_x^2+r_y^2}$ (m); vertical axis = height $r_z$ (m). Do blue lines $45°$ cone walls hain; unke beech shaded band allowed funnel hai. Red dot lander ko radius $200$ m, height $200$ m par mark karta hai — exactly wall par baitha hua. Green dot origin par target hai. Words mein: $45°$ cone ek perfect "V" hai jisiki walls har ek metre out ke liye ek metre upar jaati hain, isliye height 200 m par sabse wide legal position exactly 200 m out hai.

Forecast: Guess karo ki $(120,160)$ 200 m par 45° funnel clear karta hai ya nahi.

1. Cone constraint likho. $\sqrt{r_x^2+r_y^2}\le \tan {\theta }_{gs}(r_z)$. Yeh tool kyun? $\tan \theta =\frac{\text{horizontal}}{\text{vertical}}$ funnel ki steepness measure karta hai; cone kehta hai "tumhari horizontal distance tumhari height par allowed se zyada nahi ho sakti." (Parent ki arctan story jaise hi "opposite over adjacent".)
2. Horizontal radius compute karo. $\sqrt{120^2+160^2}=\sqrt{14400+25600}=\sqrt{40000}=200$ m. Yeh step kyun? Pythagoras do horizontal offsets ko ek radius mein turn karta hai — woh quantity jo cone limit karta hai.
3. Allowance compute karo. $\tan 45°\cdot 200=1\cdot 200=200$ m. Yeh step kyun? $45°$ par allowed radius exactly height ke barabar hoti hai.
4. Compare karo. $200\le 200$ — cone surface par, yani bilkul barely legal. Yeh step kyun? Equality matlab lander funnel wall par baitha hai; kuch aur bahar infeasible hai.

Verify: $\sqrt{120^2+160^2}=200$ ✔; $\tan 45°\cdot 200=200$ ✔; $200\le 200$ ✔.

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Example 7 — Degenerate cones: ${\theta }_{gs}\to 90°$ aur $\to 0°$ (Cell C7)

Figure (Example 7): same side-view axes (horizontal radius m mein vs height $r_z$ m mein). Teen cone walls origin se draw ki gayi hain: red = $10°$ (ek narrow near-vertical chute), blue = $45°$, green = $80°$ (ek wide, near-flat funnel). Orange dot test point hai radius $200$ m, height $200$ m par; dashed gray line radius $=200$ mark karta hai. Point green cone ke andar hai, red ke bahar. Words mein: ${\theta }_{gs}$ ko shrink karna wall ko vertical ki taraf tip karta hai (ek straw jisme tumhe seedhe neeche utarna ho); ise badhana wall ko flat ki taraf tip karta hai (almost kuch bhi allowed).

Forecast: Kaun sa lander ko near-vertical chute mein trap karta hai?

1. Case (a) ${\theta }_{gs}=80°$ (wide funnel). Allowed radius $=\tan 80°\cdot 200=5.671\cdot 200=1134.3$ m. Kyunki $200<1134.3$, deeply feasible. Yeh step kyun? Jaise ${\theta }_{gs}\to 90°$, $\tan \to \infty$: cone flat khulta hai, almost koi bhi horizontal position allow karta hai — hills se koi protection nahi.
2. Case (b) ${\theta }_{gs}=10°$ (narrow chute). Allowed radius $=\tan 10°\cdot 200=0.1763\cdot 200=35.27$ m. Kyunki $200>35.27$, infeasible — lander ek steep vertical funnel ke bahar hai. Yeh step kyun? Jaise ${\theta }_{gs}\to 0°$, $\tan \to 0$: cone ek vertical line mein collapse hota hai, almost purely vertical descent force karta hai.

Verify: $\tan 80°\cdot 200\approx 1134.3$ ✔; $\tan 10°\cdot 200\approx 35.27$ ✔; $200<1134.3$, $200>35.27$ ✔.

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Example 8 — Fuel via log-mass (Cell C8)

Forecast: 40 kg propellant se zyada ya kam?

1. Log-mass law yaad karo. Parent note se, $\dot{z}=-\alpha \Gamma$. $\Gamma$ constant hone par yeh integrate hota hai $z(t_f)=z_0-\alpha \Gamma t_f$, jahan $z_0=\ln m_0$. Yeh tool kyun? Substitution $z=\ln m$ messy $\dot{m}/m$ ko ek straight line mein turn karta hai — integration trivial hai jab $\Gamma$ constant hoti hai.
2. Numbers assemble karo. $\alpha =1/v_e=1/2000=5\times 10^{-4}\text{s/m}$, aur $z_0=\ln 1000=6.9078$. Toh $z(t_f)=6.9078-(5\times 10^{-4})(8)(10)=6.9078-0.04=6.8678$. Yeh step kyun? Teen given quantities ($\alpha$ from $v_e$, $\Gamma$, $t_f$) ka linear law mein direct substitution.
3. Log undo karo. $m(t_f)=e^{z(t_f)}=e^{6.8678}=960.79$ kg, kyunki $z=\ln m$ matlab $m=e^z$. Propellant used $=m_0-m(t_f)=1000-960.79=39.21$ kg. Yeh step kyun? Exponential log-mass se real mass recover karta hai; subtract karne se fuel spent milta hai.

Verify: $\ln 1000=6.9078$; $6.9078-0.04=6.8678$; $e^{6.8678}=960.79$ kg; burned $39.21$ kg ✔ (< 40, as forecast). Units: $\alpha \Gamma t_f=\frac{\text{s}}{\text{m}}\cdot \frac{\text{m}}{{\text{s}}^2}\cdot \text{s}=$ dimensionless ✔ (jaisa $z$ hona chahiye).

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Example 9 — Word problem, Falcon-9 style (Cell C9)

Forecast: Real boosters tiny margins ke saath land karte hain — guess karo ki 500 m enough hai ya nahi.

1. Net brake. $30-9.81=20.19{\text{m/s}}^2$. Yeh step kyun? Same signed-sum jaisi C1 mein par Earth gravity ke saath.
2. Stopping distance. Magnitudes par switch karte hue ($v_0=|-60|=60$, $a=20.19$), $d=\frac{60^2}{2(20.19)}=\frac{3600}{40.38}=89.15$ m, vertical braking drop. Yeh tool kyun? Kinematics phir se — known speed, known brake, want distance.
3. Feasibility. $89.15<500$ — aaram se fit hota hai. Yeh step kyun? $d$ woh height hai jo stop karte waqt use hoti hai; $d\le r_z$ matlab pad ke upar ruk jaata hai.
4. Hover check. Need $a_{\text{cmd}}=9.81$; box hai $[12,30]$. $9.81<12$, isliye hover impossible hai — engine ka floor gravity se zyada hai. Yeh step kyun? Real high-thrust boosters hover nahi kar sakte; unhe ek continuous fast burn mein land karna hota hai (yahi wajah hai ki Falcon 9 hoverslam karta hai, hover nahi).

Recall Falcon 9 hover kyun nahi kar sakta?

Uska minimum throttle acceleration ($\ge 12$) Earth gravity ($9.81$) se zyada hai, isliye koi bhi thrust ise climb karata hai — ise ek single decelerating burn time karni hoti hai taaki exactly pad par zero speed par pahunche. ::: Engine floor $|g|$ se upar hai, isliye hover thrust $9.81$ throttle box se neeche hai.

Verify: $30-9.81=20.19$; $3600/(2\cdot 20.19)=89.15$ m; $89.15<500$; hover $9.81\notin [12,30]$ ✔.

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Example 10 — Exam twist: infeasibility pehchano (Cell C10)

Forecast: Trap alert — kya burn sirf 20 m mein fit hoti hai?

1. Current max brake par stopping distance. Net $=12-3.71=8.29$; $v_0=|-25|=25$ ke saath, $d=\frac{25^2}{2(8.29)}=\frac{625}{16.58}=37.7$ m. Yeh step kyun? C2 se feasibility test — pehle braking drop $d$ compute karo.
2. Compare karo. $37.7>20$ — infeasible. Rocket ko $37.7$ m chahiye rukne ke liye par sirf $20$ m height hai, isliye jab yeh pad reach karta hai tab bhi fast move kar raha hota hai: crash. Yeh step kyun? Jab braking drop $d$ altitude $r_z$ se zyada ho, toh limits ke andar koi bhi thrust profile soft land nahi kar sakta — SOCP infeasible return karta hai.
3. Required max thrust. Demand karo ki burn just fit ho, $d\le 20$: $\frac{625}{2a_{\text{net}}}\le 20\Rightarrow a_{\text{net}}\ge \frac{625}{40}=15.625{\text{m/s}}^2$. Tab $T_{\max }/m\ge a_{\text{net}}+|g|=15.625+3.71=19.335{\text{m/s}}^2$. Yeh step kyun? Stopping-distance formula invert karo minimum net brake find karne ke liye jo fit hoti hai, phir gravity wapas add karo net acceleration ko required thrust mein convert karne ke liye.

Verify: $625/(2\cdot 8.29)=37.7$ m $>20$ (infeasible) ✔; required $a_{\text{net}}=625/40=15.625$; $T_{\max }/m\ge 19.335$ ✔.

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Recall

Recall Har example kaun sa cell cover karta hai?

C0→Ex0, C1→Ex1, C2→Ex2, C3→Ex3, C4→Ex4, C5→Ex5, C6→Ex6, C7→Ex7, C8→Ex8, C9→Ex9, C10→Ex10. ::: Ek example per scenario cell — matrix fully covered hai.

Related: Tsiolkovsky Rocket Equation ($z=\ln m$ mass law), Second-Order Cone Programming (glide-slope cone), SpaceX Falcon 9 Landing aur Mars EDL (Examples 9–10), Apollo Lunar Descent Guidance (bang-bang heritage).