Terminal descent — velocity vector alignment, touchdown constraints

3.5.54 · Physics › Guidance, Navigation & Control (GNC)

Terminal descent KYA hai?

YE CHAAR KYUN? Agar lander sahi position par hai lekin velocity sideways hai, toh ek leg scrape karega aur wo topple ho jaayega. Sahi velocity hai lekin body tilted hai, toh thrust mein horizontal component hoga jo tumhare paas kill karne ka time nahi hoga. Isliye position aur velocity aur attitude sab saath mein converge karne chahiye.

Guidance law ko first principles se KAISE banate hain

Hum chahte hain ki vehicle time $t_{f}$ par ek desired state reach kare. Vehicle ko point mass ki tarah model karo (attitude ek faster inner loop handle karta hai):

$\ddot{r} = a_{c m d} + g$

Sign / definition convention (ise abhi apne dimag mein fix karo): $a_{c m d}$ woh net commanded acceleration hai jo vehicle ko experience karni chahiye, aur gravity $g = (0, 0, - g)$ dynamics mein alag se add hota hai. Iska matlab hai ki engine ko physically produce karna wala thrust acceleration hai $a_{t h r u s t} = a_{c m d}$ (gravity yahan RE-ADD nahi hoti) — neeche wale formula ke andar $- g$ term ka poora point yahi hai ki $a_{c m d}$ mein already gravity-cancelling contribution hai. Toh: $a_{c m d} = \frac{T}{m}, \ddot{r} = a_{c m d} \frac{T}{m} + g .$ Yeh seedha rakho: engine exactly $a_{c m d}$ supply karta hai, isse zyada nahi.

Step 1 — Pucho: kaisi trajectory shape dono position aur velocity target hit kar sakti hai?

Yeh step kyun? Ek straight line sirf position aim karne deti hai. $t_{f}$ par position AUR velocity pin karne ke liye hume enough free parameters chahiye. Constant acceleration hume exactly wahi freedom deta hai jo chahiye.

Maan lo remaining time-to-go $t_{g o} = t_{f} - t$ ke dauran constant commanded acceleration $a$ hai:

$r (t_{f}) = r + v t_{g o} + \frac{1}{2} (a + g) t_{g o}^{2}$ $v (t_{f}) = v + (a + g) t_{g o}$

Step 2 — Do boundary conditions impose karo $r (t_{f}) = r_{f}$ , $v (t_{f}) = v_{f}$ .

Velocity equation se: $a + g = \frac{v _{f} - v}{t _{g o}}$

Position equation mein substitute karo aur solve karo. Sirf constant $a$ se yeh over-determined hai, isliye hum $a$ ko time mein linearly vary hone dete hain (ek aur free vector add karo). Algebra karne par (yeh classic E-guidance / two-point boundary value result hai):

Coefficients ka derivation check: Linear-in-time acceleration $a (t) = c_{0} + c_{1} τ$ ( $τ$ abhi se measure kiya) ke saath, integrate karo aur har axis ke four scalar boundary conditions match karo — isse $6$ aur $2$ numbers milte hain. (VERIFY block mein integrate karke confirm kiya ki $r (t_{f}), v (t_{f})$ targets hit karte hain.)

Step 3 — Velocity vector alignment. Vertical soft touchdown ke liye choose karo $v_{f} = (0, 0, - v_{t d}), v_{t d} \approx 1 - 2 m/s .$ $v_{f}$ ke horizontal components zero hain, jo velocity vector ko $t_{g o} \to 0$ hote waqt vertical ki taraf rotate karne par majboor karte hain.

Touchdown constraints (jo inequalities sab ek saath hold karni chahiye)

Tilt limit KYUN? Ek lander tip hota hai agar uski velocity + tilt center of mass ko legs ke footprint ke bahar push kar de. Geometrically, stability ke liye zaroori hai $θ + arctan \frac{v _{h}}{v _{t d}} ≲ θ_{t i p}$ yaani slow horizontal drift aur chhota body tilt dono.

Figure — Terminal descent — velocity vector alignment, touchdown constraints

Worked Examples

Worked example Example 1 — Guidance command compute karo

Diya hai (1-D vertical, $g = 3.71$ m/s² Mars): altitude to go $r_{f} - r = - 40$ m (40 m descend karna hai $r_{f}$ tak), current $v = - 25$ m/s, desired $v_{f} = - 1.5$ m/s, $t_{g o} = 4$ s. $a_{c m d}$ nikalo.

Aise set up kyun? Vertical axis, targets final values hain; sign convention up = +. $a_{c m d} = \frac{6}{16} (- 40) - \frac{2}{4} (2 (- 25) + (- 1.5)) - (- 3.71)$ $= - 15 - \frac{1}{2} (- 51.5) + 3.71 = - 15 + 25.75 + 3.71 = 14.46 m/s^{2}$ Positive kyun? Hum fast gir rahe hain (−25 m/s) aur hard brake karna hai, isliye net upward acceleration $> 0$ hai. Engine ko kya produce karna hoga? Exactly yahi: $T / m = a_{c m d} = 14.46$ m/s² upward — $- g$ term $a_{c m d}$ ke andar pehle se hai, isliye $g$ dobara mat add karo. (Sanity check: world mein resulting net acceleration $a_{c m d} + g_{w or l d} = 14.46 + (- 3.71) = 10.75$ m/s² up hai, jo trajectory-shaping deceleration hai.) ✅

Common Mistakes (Steel-manned)

Active Recall

Terminal descent mein position AUR velocity dono kyun target karni chahiye?

Sahi position ke saath galat velocity vector skid/topple karaati hai; soft landing dono endpoints pin karti hai — yeh ek two-point boundary value problem hai.

Soft-landing guidance acceleration likho.

a_{c m d} = \frac{6}{t _{g o}^{2}} (r_{f} - r) - \frac{2}{t _{g o}} (2 v + v_{f}) - g

- g

term kya karta hai aur yeh kahan hota hai?

Yeh

\ddot{r} = a_{c m d} + g

mein

+ g

ko cancel karta hai; yeh

a_{c m d}

ke andar fold hota hai, isliye engine exactly

T / m = a_{c m d}

produce karta hai (

g

dobara mat add karo).

Engine

a_{c m d}

supply karta hai ya

a_{c m d} + g

Exactly

a_{c m d}

T / m = a_{c m d}

, kyunki formula ke andar

- g

already gravity compensation handle karta hai.

6 aur 2 coefficients kahan se aate hain?

Har axis ke position+velocity boundary conditions ke four scalar conditions se linear-in-time acceleration ko match karne par.

t_{g o} \to 0

hote waqt kya hota hai?

1/ t_{g o}^{2}

gain diverge hota hai → infinite thrust;

t_{g o}

floor karna padega ya hover-drop ko hand off karna padega.

Touchdown envelope constraints ke naam batao.

Vertical speed, horizontal speed, tilt angle, angular rate, aur position-within-pad — ye sab AND-constraints hain.

Tilt-angle limit kyun?

CoM ko leg footprint ke andar rakhne ke liye taaki combined tilt + drift lander ko tip na kare.

Velocity vector alignment kya hai?

Velocity ko nearly vertical tak rotate karna (horizontal components → 0, vertical → chhota

- v_{t d}

) touchdown par.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek drone ko ek chhoti si box par land karna hai. Tum nahi chahte ki woh uss par slam kare ya slide off ho jaaye. Toh aakhri kuch seconds mein tum slow ho jaate ho jab tak barely move karo, pakka karo ki tum seedhe neeche ja rahe ho (sideways nahi), aur drone ko seedha khada rakho — aur ye teeno ek saath karo, bilkul us waqt jab touch karo. Math bas yeh keh raha hai: "kitna hard push karna chahiye, yeh dekhte hue ki abhi kitna door hoon aur kitni tez chal raha hoon?" Door/fast ho toh zyada push. Aur yeh key hai: formula mein already gravity fight karne ke liye extra push included hai — toh motor bas wahi karta hai jo formula keh raha hai, gravity dobara tack on nahi karte.

Connections

Powered Descent Guidance (PDG) — terminal descent se pehle ka phase jo isme feed karta hai
Proportional Navigation — contrast: sirf position aim karta hai, velocity nahi
Attitude Control & Inner Loop — tilt/rate limits enforce karta hai
Time-to-go Estimation — $t_{g o}$ jo har gain mein appear karta hai
Convex Optimization Landing (lossless convexification) — modern alternative jo thrust bounds honor karta hai