Powered descent guidance — G-FOLD algorithm (convex optimization): Question bank

3.5.53 · D5 · Physics › Guidance, Navigation & Control (GNC) › Powered descent guidance — G-FOLD algorithm (convex optimiza

Shuru karne se pehle, do words jo tumhe pin karke rakhne chahiye (neeche sab kuch inhi par tika hai):

Recall Convex set / convex function (ek-line refresher)

Convex set woh hota hai jisme kisi bhi do points ke beech ki seedhi line andar hi rahti hai. Convex function bowl-shaped hoti hai: uska ek lowest point hota hai, koi false valley nahi. Dekho Convex Optimization aur Second-Order Cone Programming. G-FOLD ka poora point yahi hai ki dono constraints (sets) aur cost (function) ko convex banao taaki solver global minimum tak pahunche.

Definition Is page par har symbol, ek baar define kiya gaya

Ye exactly wahi quantities hain parent note se — koi bhi question padhne se pehle inhe pin kar lo.

$r$ — lander ka position vector, ek fixed target-centred frame mein measure kiya gaya jahan $r_{z}$ zameen se upar ki height hai aur $r_{x}, r_{y}$ target se do horizontal offsets hain. $r_{z, land}$ landing surface ki height hai (usually $0$ ).
$v = \dot{r}$ — velocity vector (position ka rate of change).
$m$ — lander ka mass; $z = ln m$ log-mass hai.
$T$ — woh thrust vector jo hum engine se command karte hain; $∥ T ∥$ uski magnitude hai (newtons mein push).
$g$ — planet ki surface ke paas constant gravitational acceleration vector; uski magnitude $g$ neeche point karti hai (jaise Mars par $g \approx 3.71 m/s^{2}$ ).
$T_{m i n}, T_{m a x}$ — engine ki thrust floor aur ceiling (newtons): wo physically $T_{m i n}$ se soft ya $T_{m a x}$ se hard push nahi kar sakti.
$a = T / m$ — acceleration command, thrust per unit mass (units $m/s^{2}$ ); dynamics ko $m$ se divide karne ke baad yahi naya control variable hai.
$Γ$ — slack variable, $∥ T ∥/ m$ (acceleration-magnitude) ka stand-in, $∥ a ∥ \leq Γ$ ke saath. Fuel cost $\int Γ d t$ hai.
$α = 1/ v_{e}$ — mass-consumption coefficient, exhaust speed $v_{e} = I_{s p} g_{0}$ ka reciprocal; ye thrust ko mass-burn rate mein convert karta hai $\overset{m}{˙} = - α ∥ T ∥$ ke zariye. Bada $α$ (slower exhaust) ⇒ zyada fuel burn per newton.
$θ_{g s}$ — glide-slope angle jo safe descent cone define karta hai; $t_{f}$ — total flight time.

Sach ya jhooth — justify karo

Mass equation

\overset{m}{˙} = - α ∥ T ∥

kehti hai ki lander engines off hone par bhi mass khoता hai.

Jhooth. Engines off hone par

∥ T ∥ = 0

, isliye

\overset{m}{˙} = 0

— koi exhaust nahi, koi mass loss nahi. Mass tabhi girti hai jab thrust nonzero ho. (

α = 1/ v_{e} > 0

ek fixed constant hai, isliye ye akele kabhi loss nahi karta.)

Total thrust

\int ∥ T ∥ d t

minimize karna fuel minimize karne ke barabar hai.

Sach true thrust ke liye, lekin trap ye hai ki thrust ko pointwise minimize karna (har jagah low rakhna). Ek lamba, gentle burn jo gravity se zyada der tak ladte hue karta hai, ek short hard burn se zyada total propellant burn karta hai — gravity loss. Fuel integral hai, peak nahi.

Constraint

∥ T ∥ \leq T_{m a x}

(upper bound) hi problem ko non-convex banata hai.

Jhooth. Ek sphere ka andar convex hota hai, isliye upper thrust-ceiling bound theek hai. Non-convex woh lower bound

∥ T ∥ \geq T_{m i n}

(thrust floor) hai — sphere ke bahar ka region — jo non-convex hai.

Lossless convexification ke baad solver aisa solution return kar sakta hai jisme

Γ > ∥ a ∥

ho (slack unused).

Optimum par Jhooth. Cost

\int Γ d t

directly

Γ

ko penalize karti hai, isliye optimizer

Γ

ko tab tak squeeze karta hai jab tak wo

∥ a ∥

ke barabar na ho jaye.

Γ > ∥ a ∥

wala koi bhi interior point strictly worse hai aur reject ho jata hai — yahi Pontryagin Minimum Principle ki guarantee hai.

Change of variable

z = ln m

cost ko nonlinear banata hai.

Jhooth. Ye mass dynamics ko linear banata hai (

\overset{z}{˙} = - α Γ

). Cost

\int Γ d t

Γ

mein linear hai aur log substitution se untouched rehti hai.

Solve karne se pehle flight time

t_{f}

fix karna optimality kho deta hai.

Jhooth. Har fixed

t_{f}

ke liye hume exact convex optimum milta hai; phir hum

t_{f}

ke saath outer 1-D search karte hain. Fuel-vs-

t_{f}

curve unimodal hai, isliye sweep sach mein best

t_{f}

dhundh leta hai convexity tode bina.

Ek convex problem mein phir bhi kai equally-good landing solutions ho sakte hain.

Sach. Convex ka matlab hai koi worse local minima nahi, lekin optimal points ka set flat ho sakta hai (tie). Jo guarantee hai woh ye hai ki har returned optimum ka globally-best cost same hoga — tum kabhi inferior valley mein nahi phansoge.

Glide-slope constraint

∥ [r_{x}, r_{y}] ∥ \leq tan θ_{g s} (r_{z} - r_{z, land})

tan

ki wajah se non-convex hai.

Jhooth.

tan θ_{g s}

bas ek fixed positive number hai (ek slope), aur

r_{x}, r_{y}, r_{z}

target frame mein position components hain. Ye constraint ek norm

\leq

ek linear function hai — ek convex second-order cone. Ye SOCP mein free mein fit ho jata hai.

Error dhundho

"Humne

∥ a ∥ = Γ

∥ a ∥ \leq Γ

tak relax kiya, isliye model real engine ko ab represent nahi karta."

Error ye hai: relaxation sirf optimum ke bahar loose hai. Optimum par inequality active hoti hai (

∥ a ∥ = Γ

), exactly true engine physics recover karti hai. Kuch bhi nahi khoता — isliye lossless.

Γ

∥ T ∥/ m

define kiya gaya hai, isliye ye bas state ka ek fixed function hai — real variable nahi."

Error ye hai:

∥ T ∥

woh control hai jo hum choose kar rahe hain, isliye

Γ

(acceleration-magnitude slack) ek genuine decision variable hai. Convexification ke baad

Γ

sirf

a

se inequality

∥ a ∥ \leq Γ

ke zariye linked hai, optimizer ko freedom deta hai jise proof phir pin kar deta hai.

m \dot{v} = T + m g

m

se divide karo aur equation mein abhi bhi

m

hai, isliye ye abhi bhi nonlinear hai."

Error ye hai: divide karne se

\dot{v} = a + g

milta hai jahan

a = T / m

naya control hai aur

g

constant gravity vector hai.

m

a

ki definition mein absorb kar liya gaya hai; equation ab chosen variables mein linear hai.

T_{m i n} e^{- z}

ka Taylor expansion ek approximation hai, isliye G-FOLD ka answer sirf approximate hai."

Ye partly trap hai. Floor

T_{m i n} e^{- z}

ko reference log-mass

z_{0}

ke baare mein second order tak expand kiya gaya hai:

T_{m i n} e^{- z_{0}} [1 - (z - z_{0}) + \frac{1}{2} (z - z_{0})^{2}]

. Kyunki

e^{- z}

convex hai, uska second-order Taylor polynomial ek upward parabola hai — khud convex — isliye constraint

Γ \geq (parabola)

convex rehti hai. Ye

z_{0}

ke paas tight hai, exactly wahan jahan true trajectory ka mass rehta hai, aur tighten karne ke liye re-reference kiya ja sakta hai.

"Minimum-fuel descent ko smooth landing ke liye poore raste gently throttle karna chahiye."

Error ye hai: minimum-fuel typically bang-bang hota hai — coast/min-thrust, phir end ke paas ek hard full-thrust brake ("suicide burn"). Smooth gentle throttling extra fuel burn karta hai gravity se zyada der tak ladte hue.

"Kyunki interior-point methods iterative hain, G-FOLD global optimum guarantee nahi kar sakta."

Error ye hai: ek convex program ke liye, interior-point methods guaranteed polynomial time mein global optimum tak converge karte hain. Convexity, iteration nahi, global guarantee deti hai.

Why questions

Mass ka logarithm kyun lete hain, bas

m

use kyun nahi karte?

Kyunki

\overset{m}{˙} / m

log-derivative hai

\frac{d}{d t} ln m

z = ln m

set karne se messy

1/ m

mass dynamics bilkul linear

\overset{z}{˙} = - α Γ

mein badal jaati hai (jahan

α = 1/ v_{e}

) — logs division ko subtraction mein convert karte hain.

Descent ke dauran global optimum (sirf ek achha local nahi) itna important kyun hai?

Trajectory commit karne se pehle roughly ek second compute karne ke liye hota hai. Ek non-convex solver false valley return kar sakta hai ya converge karne mein fail ho sakta hai; ek convex solve reliably har baar single best answer return karta hai, jo safety demand karti hai.

Fuel cost

Γ

ko small kyun "chahti" hai, aur us fact ka kya matlab hai?

Cost

\int Γ d t

hai, isliye smaller

Γ

(acceleration-magnitude slack) matlab kam fuel. Woh downward pressure exactly hi hai jo

Γ

ko uske lower bound

∥ a ∥

par drive karta hai, lossless relaxation ko true equality par wapas collapse karta hai.

Hum

t_{f}

ke upar bahar se search kyun karte hain instead of andar optimize karne ke?

t_{f}

ko andar vary karne se discretized dynamics bilinear ho jate hain (unknowns ke products), convexity destroy ho jaati hai. Bahar ek unimodal curve par 1-D sweep ke roop mein rakha, har inner solve ki convexity preserve hoti hai.

Fuel-versus-flight-time curve unimodal (ek minimum) kyun hai?

Bahut chhota

t_{f}

samay mein brake karne ke liye violent, near-max thrust demand karta hai — heavy fuel burn (aur eventually infeasibility); bahut lamba

t_{f}

extra seconds tak gravity ke khilaf hover karne mein propellant waste karta hai. Fuel ek sweet spot ke aas-paas girta phir badhta hai, isliye curve mein ek single dip hai — 1-D search se safely bracket kiya ja sakta hai. (Ye closed form mein prove karne ki bajay numerically per problem verify kiya jata hai.)

Hum thrust kyun cap nahi kar sakte aur lower bound

T_{m i n}

bhool kyun nahi jaate?

Real engines physically

T_{m i n}

floor se below throttle nahi kar sakti bina flame out kiye. Ise ignore karna ek aisa solution deta hai jo hardware execute nahi kar sakta; convexification hume floor rakhne deti hai convex rehte hue.

Fuel minimize karna final mass maximize karne ke barabar kyun hai?

Fuel spent equals initial mass minus final mass, aur initial mass fixed hai. Isliye kam fuel burned ⇔ touchdown par zyada mass bacha — dono objectives same direction mein point karte hain.

G-FOLD Apollo/Falcon-style "late burn" behaviour ko bina bataye kyun reproduce karta hai?

Ye dynamics ke under

\int Γ d t

minimize karne se naturally nikalta hai — thrust delay karna gravity se ladte hue time kam karta hai (gravity loss). Dekho Apollo Lunar Descent Guidance aur SpaceX Falcon 9 Landing.

Edge cases

Agar available deceleration gravity se kam ho (

T_{m a x} / m < g

Yahan

T_{m a x}

engine ceiling hai aur

g

gravity magnitude; best upward acceleration

T_{m a x} / m

downward

g

ko overcome nahi kar sakta, isliye lander apna fall nahi rok sakta — problem infeasible hai. G-FOLD ek fake landing ki jagah infeasibility report karta hai, jo sahi safe outcome hai.

Jab thrust exactly

T_{m i n}

ho (engine apne floor par) us instant kya hota hai?

Throttle-box lower constraint

Γ \geq T_{m i n} e^{- z}

active ho jaata hai. Ye convex box ka ek valid, feasible edge hai — solver ise kisi bhi active constraint ki tarah handle karta hai, koi special casing nahi.

Agar target lander ke directly neeche ho (purely vertical descent)?

Glide-slope cone degenerate ho jaata hai:

∥ [r_{x}, r_{y}] ∥ \leq tan θ_{g s} (r_{z} - r_{z, land})

trivially satisfied hai kyunki horizontal offset ~0 hai. Problem 1-D vertical landing mein reduce ho jaati hai — ek valid limiting case.

θ_{g s} \to 9 0^{\circ}

ka glide-slope cone ke liye kya matlab hai?

tan 9 0^{\circ} \to \infty

, isliye horizontal-offset bound unlimited ho jaata hai — cone fully khul jaata hai aur koi restriction impose nahi karta. Physically, "koi bhi approach angle allowed hai."

θ_{g s} \to 0^{\circ}

ka kya matlab hai?

tan 0^{\circ} = 0

force karta hai

∥ [r_{x}, r_{y}] ∥ \leq 0

, yaani target axis ke along purely vertical descent. Cone ek line mein collapse ho jaata hai — extremely restrictive lekin abhi bhi convex.

Agar initial velocity sahi altitude aur position par already zero ho?

Trajectory already ek soft pinpoint state hai; minimum-fuel solution ko bas ek hover-then-touch ya immediate zero divert chahiye, aur cost integral essentially woh propellant hai jo zameen tak gravity ke against hold karne ke liye chahiye.

Agar do alag flight times

t_{f}

near-identical fuel dein?

Unimodal fuel curve apne minimum ke paas flat hai, isliye kai

t_{f}

tie karte hain. Unme se koi bhi acceptable hai; woh choose karo jo extra margin de (jaise actuator response ke liye thoda zyada time).

Recall Poore trap set ka ek-line summary

Non-convexity sirf thrust lower bound mein rehti hai; $z = ln m$ mass linearize karta hai; $Γ$ -relaxation lossless hai kyunki cost $Γ = ∥ a ∥$ push karti hai; $t_{f}$ bahar rehta hai; optimum bang-bang hota hai; aur har degenerate cone/thrust case phir bhi ek convex, globally-solvable program ke andar land karta hai.