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

Answer: (b). Ek set convex hota hai agar uske kisi bhi do points ko jodne wali seedhi line set ke andar rahe.

• (a) $\Vert \mathbf{T}\Vert \le T_{\max }$ ek solid disk hai (sabhi thrust vectors jo $T_{\max }$ se chhote ya barabar hain) — koi bhi do points uthao, unke beech ka segment andar rahega. Convex hai.
• (c) ek linear equality hai — state-space mein ek flat plane. Convex hai.
• (d) upar ki taraf khuline wala cone hai (glide-slope). Convex hai.
• (b) $\Vert \mathbf{T}\Vert \ge T_{\min }$ circle ke bahar ki sab cheez hai. Ek point left rim par aur ek right rim par lo: unke beech ka segment seedha forbidden hole ke beech se guzarta hai. Set se yeh escape hi non-convex ki definition hai. Upar diye figure mein pink dashed chord dekho — uske do pink endpoints allowed ring par hain, lekin dashed line unke beech se guzar kar yellow inner circle ke forbidden hole mein ghus jaati hai.

• $\mathbf{a}=\mathbf{T}/m$ — acceleration command (thrust per unit mass; rocket actually kya feel karta hai).
• $\Gamma =\Vert \mathbf{T}\Vert /m$ — slack magnitude jo $\Vert \mathbf{a}\Vert$ ko upper-bound karta hai; yeh hai throttle kitna hai, acceleration units mein.
• $z=\ln m$ — mass ka log, is liye choose kiya gaya taaki mass dynamics linear ho jaaye.
• $\alpha =1/v_e=1/(I_{sp}{g}_0)$ — fuel-cost rate: newton-second thrust per kg mass lost, is tarah express kiya gaya taaki $\dot{z}=-\alpha \Gamma$.

Answer line — $\alpha$ ek sentence mein ::: $\alpha$ thrust magnitude ko mass-loss rate mein convert karta hai; yeh exhaust speed $v_e$ ka reciprocal hai.

WHAT karte hain: sabse bada net deceleration dhundho aur constant-acceleration kinematics use karo. Pehle sign handle karo: up-thrust acceleration $+T_{\max }/m$ deta hai; gravity $-g$ deta hai. Toh braking ke dauran net upward acceleration hai: $a=\frac{T_{\max }}{m}-g=15-3.71=11.29{\text{m/s}}^2(>0,\text{ toh fall decelerate hota hai}).$ WHY hum speed $25$ m/s use kar sakte hain: stopping-distance formula $d=v^2/(2a)$ velocity ki magnitude use karta hai kyunki $v^2=(-25{)}^2=25^2$ — square karne par sign disappear ho jaata hai. Velocity signed hai ($v_z=-25$), lekin uska square (kinetic-energy term) direction-free hai, isliye formula sirf speed $|v_z|=25$ maangta hai. $d=\frac{|v_z{|}^2}{2a}=\frac{25^2}{2\times 11.29}=\frac{625}{22.58}\approx 27.68\text{m}.$ Kyunki $27.68\text{m}<150\text{m}$, haan — altitude spare hai, isliye ek feasible (aur fuel-optimal) solution exist karta hai.

Edge case — jab ruk paana impossible ho: net decelerating acceleration hai $a=T_{\max }/m-g$. Agar $T_{\max }/m\le g$, toh $a\le 0$: full throttle par bhi engine gravity ko nahi maar sakta, lander neeche accelerate karta rehta hai, aur $d=v^2/(2a)$ undefined ya negative ho jaata hai — feasibility check fail ho jaata hai aur us state se koi landing exist nahi karta. Toh G-FOLD ki sabse pehli zaroorat yeh hai ki thrust ceiling local gravity se zyada ho, $T_{\max }/m>g$.

WHAT constraint geometrically kehta hai: har altitude $r_z$ par, term $\Vert [r_x,r_y]\Vert$ ek horizontal circle ka radius hai jiske andar lander allowed hai. Kyunki woh allowed radius height ke saath linearly badhta hai, circles ko stack karne par ek upar ki taraf khuline wala cone banta hai jiska tip target par hota hai. Neeche ka figure exactly yahi plot karta hai: vertical axis altitude $r_z$ hai, horizontal axis offset $\Vert [r_x,r_y]\Vert$ hai, aur shaded wedge allowed funnel hai.

• ${\theta }_{\text{gs}}=45^{\circ }$: $\tan 45^{\circ }=1$, toh radius $=1\times 320=320$ m. Figure mein, dotted $r_z=320$ line par yellow cone edge $320$ m par hai.
• ${\theta }_{\text{gs}}=30^{\circ }$: $\tan 30^{\circ }\approx 0.5774$, toh radius $=0.5774\times 320\approx 184.75$ m — blue cone narrower hai.

WHY tan aur sin nahi: right triangle par (altitude $r_z$ = adjacent vertical side, horizontal offset = opposite side, slant descent line = hypotenuse), $\tan \theta =\text{opposite}/\text{adjacent}=\text{horizontal}/\text{altitude}$. Woh ratio hi cone ka slope hai, jise hum cap karna chahte hain. Chhota ${\theta }_{\text{gs}}$ ek narrower, steeper funnel banata hai — ridges ke against safer hai lekin maneuver karne ki kam jagah.

Step — log differentiate karo (WHAT): chain rule se, $\dot{z}=\frac{d}{dt}\ln m=\frac{\dot{m}}{m}.$ Step — mass law substitute karo (WHAT): $\dot{z}=\frac{-\alpha \Vert \mathbf{T}\Vert }{m}=-\alpha \frac{\Vert \mathbf{T}\Vert }{m}=-\alpha \Gamma .$ WHY log kaam karta hai: acceleration law $\dot{\mathbf{v}}=\mathbf{T}/m$ mein $m$ denominator mein hai — woh $1/m$ hi nonlinear coupling hai jise hamare maarna hai. Mass law ka ratio $\dot{m}/m$ exactly $\ln m$ ka chain-rule derivative hai. $\dot{m}=-\alpha \Vert \mathbf{T}\Vert$ daalne par, $m$'s milkar clean control $\Gamma =\Vert \mathbf{T}\Vert /m$ ban jaate hain, aur hume $\dot{z}=-\alpha \Gamma$ milta hai: constant times control, denominator mein koi state nahi. WHY $z=1/m$ fail karta hai (concrete alternative): try karo. $\dot{z}=\frac{d}{dt}{m}^{-1}=-m^{-2}\dot{m}=-m^{-2}(-\alpha \Vert \mathbf{T}\Vert )=\alpha \Vert \mathbf{T}\Vert /m^2$. Isme abhi bhi $m^{-2}$ aur $\Vert \mathbf{T}\Vert$ alag-alag hain — inhe single control $\Gamma =\Vert \mathbf{T}\Vert /m$ mein fold karna possible nahi bina leftover $1/m$ ke. Toh $z=1/m$ ek state-dependent factor chhodta hai aur dynamics nonlinear rehti hai. Sirf log hi ratio $\dot{m}/m$ produce karta hai jo control $\Gamma$ already match karta hai — yahi sense mein log uniquely clean hai.

Setup: kisi bhi instant par, $\Gamma$ ko touch karne wale constraints hain $\Gamma \ge \Vert \mathbf{a}\Vert$ aur $\Gamma \ge {\Gamma }_{\min }(z)$ (throttle floor). Cost $+\Gamma$ per unit time pay karti hai. Argument (WHY $\Gamma$ $\Vert \mathbf{a}\Vert$ se chipak jaata hai): maan lo kisi instant par $\Gamma >\max (\Vert \mathbf{a}\Vert ,{\Gamma }_{\min })$ strictly hai. Tab hum $\Gamma$ ko ek tiny $\epsilon$ se ghata sakte hain bina kisi lower bound ko violate kiye. Isse cost integral $\epsilon dt$ se reduce hota hai aur — kyunki $\Gamma$ dynamics mein sirf $\dot{z}=-\alpha \Gamma$ ke through aata hai — mass bhi bachti hai (final $z$ badhta hai), jo sirf behtar hai. Toh strictly-slack $\Gamma$ hamesha suboptimal hai: optimizer $\Gamma$ ko uski do floors mein se tighter par dabata hai. Jahan floor inactive ho (${\Gamma }_{\min }<\Vert \mathbf{a}\Vert$), binding floor $\Vert \mathbf{a}\Vert$ hai, isliye $\Gamma =\Vert \mathbf{a}\Vert$ — exact engine physics recover ho jaati hai. Isliye relaxing "costs nothing" = lossless hai. Pontryagin last gap kaise close karta hai (sketch, sirf citation nahi): upar ki hand-wave assume karti hai ki hum $\Gamma$ ko freely ek single instant par nudge kar sakte hain. Lekin $\Gamma$ dynamics ke through state se coupled hai, toh purely local nudge in principle blocked ho sakti hai. Pontryagin Minimum Principle ise properly handle karta hai: woh Hamiltonian $H=\Gamma +\mathbf{\lambda }\cdot (\text{dynamics})$ banata hai aur poochtha hai ki optimal path ke along $(\mathbf{a},\Gamma )$ ka kaun sa choice $H$ ko pointwise minimize karta hai. Woh minimization work out karne par pata chalta hai ki $\Gamma$ multiply karne wala coefficient strictly positive hai (yeh $1$ plus ek mass-costate term hai jo positive rehta hai), toh $H$ minimize hota hai $\Gamma$ ko uske lower bounds ki allow karne wali minimum tak le jaane se — yaani $\Gamma =\max (\Vert \mathbf{a}\Vert ,{\Gamma }_{\min })$ har jagah, yeh rule out karte hue ki koi aisa interval ho jahan $\Gamma$ strictly upar float kare. Woh positivity hi "no pathological case" guarantee hai.

Forward Euler = "new value = old value + step $\times$ rate." Har continuous law par apply karte hue: ${\mathbf{r}}_{k+1}={\mathbf{r}}_k+h{\mathbf{v}}_k$ ${\mathbf{v}}_{k+1}={\mathbf{v}}_k+h({\mathbf{a}}_k+\mathbf{g})$ $z_{k+1}=z_k-h\alpha {\Gamma }_k$ Per-node constraints (poori list): $\Vert {\mathbf{a}}_k\Vert \le {\Gamma }_k\text{(second-order cone — yeh relaxed thrust magnitude hai)}$ ${\Gamma }_k\ge {\Gamma }_{\min }(z_k)\text{(throttle floor — }{T}_{\min }\text{ inner circle recover karta hai)}$ ${\Gamma }_k\le {\Gamma }_{\max }(z_k)\text{(throttle ceiling — }{T}_{\max }\text{ outer circle)}$ $\Vert [r_{x,k},r_{y,k}]\Vert \le \tan {\theta }_{\text{gs}}(r_{z,k}-r_{z,\text{land}})\text{(glide-slope cone)}$ Floor aur ceiling lines ke bina, discrete model sirf thrust ko upar se cone ke through cap karega aur annulus bilkul kho dega — toh dono mandatory hain. Convexity check: teeno Euler updates linear equalities hain (flat sets → convex). ${\Gamma }_{\min },{\Gamma }_{\max }$ parent note ke convexified (Taylor) bounds hain, toh woh dono convex hain. Do norm constraints second-order cones hain — convex. Isliye poora discretized problem ek SOCP hai jo Interior-Point Methods se solve hota hai. Yeh continuous physics se us cheez ka bridge hai jo Convex Optimization / Second-Order Cone Programming solvers khaate hain.

Step — log undo karo (WHY): $z=\ln m$, toh $m=e^z$. Final mass: $m(t_f)=e^{6.7}\approx 812.41\text{kg}.$ Initial mass $m(0)=e^{\ln 1000}=1000$ kg. Fuel burned: $\Delta m=1000-812.41\approx 187.59\text{kg}.$ Step — objective recover karo: $\dot{z}=-\alpha \Gamma$ ko $[0,t_f]$ par integrate karte hue: $z(t_f)-z(0)=-\alpha {\int }_0^{t_f}\Gamma dt\Rightarrow {\int }_0^{t_f}\Gamma dt=\frac{z(0)-z(t_f)}{\alpha }.$ ${\int }_0^{t_f}\Gamma dt=\frac{\ln 1000-6.7}{5\times 10^{-4}}=\frac{6.9078-6.7}{5\times 10^{-4}}\approx 415.5\text{(m/s)}.$ Cross-check: $\int \Gamma dt=(1/\alpha )\ln (m_0/m_f)$ exactly Tsiolkovsky $\Delta v$ ideal budget hai — yeh reassuring hai ki "minimize $\int \Gamma dt$" literally velocity-change cost minimize karta hai.

Upar positive lo. Commanded up-thrust $\Gamma$ ke under net vertical acceleration hai $a=\Gamma -g$.

Strategy A — constant thrust $\gamma$: humein $200$ m fall mein $30$ m/s descent kill karni hai. Net decelerating acceleration hai $\gamma -g$, aur $|v_z{|}^2=2(\gamma -g)d$: $\gamma -g=\frac{|v_z{|}^2}{2d}=\frac{30^2}{2\times 200}=2.25\Rightarrow \gamma =2.25+3.71=5.96{\text{m/s}}^2.$ Descent time $t_A=|v_z|/(\gamma -g)=30/2.25=13.33$ s. Fuel proxy $=\int \Gamma dt=\gamma t_A=5.96\times 13.33\approx 79.47\text{(m/s)}$.

Strategy B — bang-bang. Coast ke dauran $\Gamma =2$: net accel $a_1=2-3.71=-1.71$ (abhi bhi neeche speed badh rahi hai). Brake ke dauran $\Gamma =14$: net accel $a_2=14-3.71=10.29$ (decelerate ho raha hai). Maan lo coast distance $d_1$ cover karta hai (speed $30$ se $v_1$ ho jaati hai), brake $d_2=200-d_1$ cover karta hai (speed $v_1\to 0$). Coast: $v_1^2=30^2+2(1.71)d_1$ (speed magnitude badhti hai). Brake: $0=v_1^2-2(10.29)d_2\Rightarrow v_1^2=2(10.29)(200-d_1)$. Equal set karo: $900+3.42d_1=20.58(200-d_1)=4116-20.58d_1$. $24d_1=3216\Rightarrow d_1=134.0\text{m},d_2=66.0\text{m}.$ Phir $v_1^2=900+3.42(134.0)=1358.3\Rightarrow v_1=36.86$ m/s. Times: coast $t_1=(v_1-30)/1.71=6.86/1.71=4.01$ s. Brake $t_2=v_1/10.29=36.86/10.29=3.58$ s. Fuel proxy $={\Gamma }_1{t}_1+{\Gamma }_2{t}_2=2(4.01)+14(3.58)=8.02+50.17\approx 58.19\text{(m/s)}$.

Verdict: $58.19<79.47$ — bang-bang jeet jaata hai, $\approx 21$ units fuel proxy bachata hai ($\approx 27\%$). Yahi woh "coast-then-suicide-burn" hai jo G-FOLD rediscover karta hai, aur isliye SpaceX Falcon 9 Landing aur Mars EDL ignition delay karte hain. Pehle jalana propellant gravity loss par waste karta hai (gravity ke against altitude hold karna zyada der tak).

Idea: Model Predictive Control baar baar re-plan karta hai. Har control tick par tum actual state $(\mathbf{r},\mathbf{v},m)$ measure karte ho, phir us state se target tak G-FOLD SOCP re-solve karte ho, sirf pehla thrust command apply karte ho, baaki discard kar dete ho, aur agla tick aane par repeat karte ho. Kyun convexity enabler hai: har re-solve ek SOCP hai jiska unique global optimum hota hai jo Interior-Point Methods se guaranteed bounded time mein reach hota hai — toh ~1 s ke budget ke andar tum hamesha ek certifiably optimal, feasible plan paate ho, kabhi koi stuck local minimum nahi. Ek non-convex re-solve mid-descent mein garbage return kar sakta hai ya time out ho sakta hai; convexity hi closed loop ko fly karne ke liye safe banati hai. Yeh design lineage hai Apollo Lunar Descent Guidance (open-loop polynomial) se modern re-solving guidance tak.