Terminal descent — velocity vector alignment, touchdown constraints: Visual walkthrough

3.5.54 · D2 · Physics › Guidance, Navigation & Control (GNC) › Terminal descent — velocity vector alignment, touchdown cons

Yeh parent note ka picture-first companion hai. Agar tum "kyun chaar constraints" wali kahani chahte ho, pehle woh padho; yahan hum sirf acceleration command kis tarah derive hoti hai — isi pe focus karenge.

Step 1 — Problem draw karo: do dots aur do arrows

KYA. Lander ko ek single dot (uska center of mass) imagine karo. Abhi woh ek jagah par hai jise hum $r$ kehte hain ("woh kahan hai") aur ek velocity $v$ ke saath move kar raha hai ("kitni tezi se, aur kis direction mein"). Landing instant $t_{f}$ par hum chahte hain ki woh $r_{f}$ jagah par ho (pad) aur uski velocity $v_{f}$ ho (seedha neeche, bahut slow).

KYUN. Har guidance problem yahi hai: "is state se us state tak jao." Yahan state ka matlab hai ek dot AUR uska arrow saath mein. Humne dono draw kiye kyunki poori mushkil yahi hai — hume ek saath do cheezein match karni hain, ek nahi.

PICTURE. Red arrow current velocity hai — lamba aur tilted (khatarnak). Pad pe chhota kala arrow target velocity hai — chhota aur vertical (safe).

Figure — Terminal descent — velocity vector alignment, touchdown constraints

$dot now r, arrow now v t_{g o} dot at pad r_{f}, arrow at pad v_{f}$

Step 2 — Clock ka naam rakho: time-to-go

KYA. $t$ ko current time aur $t_{f}$ ko landing time maano. Time-to-go define karo: $t_{g o} = t_{f} - t .$ Yeh simply "kitne seconds bacha hai" hai. Shuru mein yeh kuch seconds hota hai; touchdown par $0$ ho jaata hai.

$t$ ki jagah yeh kyun? "Main apni velocity aur kitna badal sakta hoon" ki physics sirf remaining time pe depend karti hai, is waqt clock kya show kar raha hai — uss pe nahi. $t_{g o}$ use karne se equations clean rehti hain: finish line hamesha $t_{g o} = 0$ par hoti hai.

PICTURE. Ek shrinking horizontal bar: shuru mein pura, touchdown par khali.

Step 3 — Trajectory shape guess karo. "Acceleration" kyun?

KYA. Hum vehicle ko uski acceleration $a$ choose karke control karte hain — yani woh rate jis par velocity arrow badalta hai. Physics Newton's law hai, apne point mass ke liye rewrite kiya: $\ddot{r} = a_{c m d} + g,$ jahan $\ddot{r}$ ka matlab "acceleration" hai (position ka second time-derivative), $a_{c m d}$ woh hai jo engine command karta hai, aur $g = (0, 0, - g)$ gravity hai jo neeche kheenchti hai.

"Bas ek path choose karo" ki jagah acceleration kyun? Kyunki engine thrust produce karta hai, aur thrust ÷ mass acceleration hi hai. Hum dot ko teleport nahi kar sakte; hum bas use push kar sakte hain. Isliye natural unknown $a_{c m d}$ hai.

Ek constant-ish acceleration kyun? Socho: woh sabse simple push kaun si hai jo dono — ek target position aur ek target velocity — hit kar sake? Ek single fixed direction (seedha ek dhakka) sirf unme se ek aim kar sakta hai. Agle step mein degrees of freedom count karte hain — woh count hi sab kuch decide karta hai.

PICTURE. Velocity arrow ko ek chhota acceleration arrow second by second ek naye direction mein mod raha hai.

Step 4 — Freedom count karo: constant acceleration kyun kaafi nahi hai

KYA. Maano acceleration poore $t_{g o}$ ke liye ek constant $a$ hai. Tab standard "constant-acceleration" motion equations (gravity ke neeche phenki gayi ball wali) finish par deti hain: $v (t_{f}) = v + (a + g) t_{g o}$ $r (t_{f}) = r + v t_{g o} + \frac{1}{2} (a + g) t_{g o}^{2}$

Velocity line annotate karo: $end mein arrow v (t_{f}) = abhi arrow v + total push (a + g) waqt jitna lagta hai t_{g o}$

Count kyun? Hare paas har axis ke liye do targets hain: final position aur final velocity. Yeh do equations hain. Ek constant $a$ humein ek free vector deta hai (teen numbers, ek per axis) — lekin per axis yeh do equations ke liye ek unknown hai. Ek knob, do locks. Yeh generally dono satisfy nahi kar sakta.

PICTURE. Ek single fixed acceleration path ko mod deta hai — lekin endpoint arrow galat nikalta hai (target velocity overshoot ho jaati hai) chahe endpoint dot sahi jagah land kare.

Step 5 — Ek knob add karo: acceleration ko linearly ramp karne do

KYA. Acceleration ko linearly time ke saath vary karne do. Abhi se ( $τ = 0$ ) touchdown tak ( $τ = t_{g o}$ ) ek fresh clock $τ$ measure karte hue: $a (τ) = c_{0} + c_{1} τ .$

Linear kyun, curved kyun nahi? Yeh waoh simplest shape hai jiske paas enough freedom hai. Ek extra parameter ( $c_{1}$ ) exactly woh ek extra equation khareed ta hai jo Step 4 mein miss thi — na zyada, na kam. Curvature add karna unused freedom de deta.

PICTURE. Acceleration ko time ke against plot karo — ab yeh ek sloped line hai (start value $c_{0}$ , slope $c_{1}$ ), flat nahi.

Step 6 — Integrate karo aur chaar boundary conditions match karo

KYA. "Integrate" ka matlab hai: chote chote pushes add karo dekho ki velocity aur position time ke saath kaise build hoti hai. $g$ ko ek moment ke liye ignore karte hain (baad mein fold back karenge), $a (τ) = c_{0} + c_{1} τ$ integrate karo:

$gained velocity = \int_{0}^{t_{g o}} a d τ = c_{0} t_{g o} + \frac{1}{2} c_{1} t_{g o}^{2}$ $gained position = \int_{0}^{t_{g o}} (velocity) d τ = v t_{g o} + \frac{1}{2} c_{0} t_{g o}^{2} + \frac{1}{6} c_{1} t_{g o}^{3}$

Ab demand karo ki dono targets hit hon:

Velocity target: $v + c_{0} t_{g o} + \frac{1}{2} c_{1} t_{g o}^{2} = v_{f}$
Position target: $r + v t_{g o} + \frac{1}{2} c_{0} t_{g o}^{2} + \frac{1}{6} c_{1} t_{g o}^{3} = r_{f}$

Do vector equations, do unknowns $(c_{0}, c_{1})$ . Solve karne pe (VERIFY block mein kiya gaya hai) woh acceleration milti hai jo hume abhi command karni chahiye, jo $c_{0}$ hai:

$c_{0} = \frac{6}{t _{g o}^{2}} (r_{f} - r) - \frac{2}{t _{g o}} (2 v + v_{f}) .$

Formula Magic numbers kahan se aate hain

$a_{c m d} = position gap khatam karo \frac{6}{t _{g o}^{2}} (r_{f} - r) - velocity shape karo \frac{2}{t _{g o}} (2 v + v_{f}) - gravity cancel karo g$

6 aur 2 guess nahi kiye gaye — yeh seedha integrals mein $\frac{1}{6}$ aur $\frac{1}{2}$ se nikalta hai jab tum do equations solve karte ho. Alag integrals ⇒ alag numbers; yeh forced hain.
Position term par $\frac{1}{t _{g o}^{2}}$ : ek position error ko acceleration se fix karna padta hai, aur acceleration → position (time)² ki tarah scale karta hai, isliye undo karne ke liye $1/ t_{g o}^{2}$ chahiye.
Velocity term par $\frac{1}{t _{g o}}$ : velocity errors acceleration se ek integration door hain, isliye $1/ t_{g o}$ scale karte hain.
$2 v + v_{f}$ : current velocity ko target se do guna zyada weight diya gaya hai kyunki yahi woh cheez hai jo humein sabse zyada bleed off karni hai. (Yahi structure Powered Descent Guidance (PDG) aur E-guidance family ke neeche hai.)

$- g$ wapas kyun lagate hain? Dynamics kehte hain $\ddot{r} = a_{c m d} + g$ . Upar ke hamare integrals ne pure $a$ assume kiya tha bina gravity ke. Real world ko (jisme $+ g$ hai) woh gravity-free plan follow karana ke liye, command mein $- g$ hona chahiye taaki woh cancel ho. Engine tab exactly $T / m = a_{c m d}$ produce karta hai — gravity andar hai, use do baar add mat karo.

PICTURE. Do curves converge ho rahi hain: position curve $r_{f}$ par land karti hai aur uska slope (velocity) usi instant mein $v_{f}$ par land karta hai.

Step 7 — Velocity vector alignment: arrow rotate hote dekho

KYA. Target velocity vertical choose karo: $v_{f} = (0, 0, - v_{t d})$ jahan $v_{t d} \approx 1$ – $2$ m/s. Command ka horizontal part dekho. $v_{f, x} = 0$ ke saath, horizontal shaping term hai: $a_{c m d, x} (shaping part) = - \frac{2}{t _{g o}} (2 v_{x} + 0) = - \frac{4 v _{x}}{t _{g o}} .$

$badhta hai jab t_{g o} \to 0 - \frac{4 v _{x}}{t _{g o}}$

Velocity arrow vertical kyun ho jaata hai? Jab $t_{g o} \to 0$ , factor $1/ t_{g o}$ badh jaata hai, isliye koi bhi bacha hua sideways speed $v_{x}$ ek baar se zyada strong opposing push paata hai. Yeh touchdown par exactly zero tak squeeze ho jaata hai — isliye velocity arrow tilted se seedha-neeche rotate ho jaata hai. Yahi woh "velocity vector alignment" hai jo parent note demand karta hai, aur yeh closely related hai ki kaise Proportional Navigation line-of-sight rotation null karta hai.

PICTURE. Successive instants par velocity arrows ka ek fan, baar baar aur zyada upright hota hua, aakhri wala pad pe seedha neeche pointing karta.

Step 8 — Degenerate case: $t_{g o} \to 0$ blow up karta hai

KYA. Fir se do coefficients dekho: $\frac{6}{t _{g o}^{2}}$ aur $\frac{2}{t _{g o}}$ . Jab $t_{g o} \to 0$ toh yeh infinity ki taraf explode karte hain. Commanded acceleration (aur isliye thrust) engine ki capacity se zyada demand kar legi.

Floor kyun? Physically tum arbitrarily little time mein arbitrarily large error correct nahi kar sakte — uske liye infinite force chahiye. Floor demand ko finite rakhta hai aur Attitude Control & Inner Loop ko saath rehne deta hai.

PICTURE. Coefficient $6/ t_{g o}^{2}$ ko $t_{g o}$ ke against plot karo: zero ke paas sky ki taraf rocket karti ek curve, ek red "floor" line ke saath jahan hum clamp karte hain.

Ek-picture summary

Upar sab kuch, compressed: current state (dot + tilted red arrow) ko ek linearly ramping acceleration se ek curved trajectory mein mod diya jaata hai jo pad par ek chhote vertical arrow ke saath land karti hai, exactly jab $t_{g o} \to 0$ — jabki coefficients $6/ t_{g o}^{2}$ aur $2/ t_{g o}$ position-fix aur velocity-shape ka kaam karte hain, aur $- g$ gravity cancel karta hai taaki engine sirf $a_{c m d}$ output kare.

Recall Feynman retelling — plain words mein zor se bolo

Hamare paas ek dot hai jo ek arrow ke saath move kar rahi hai, aur hum chahte hain ki woh ek pad par pahunche ek tiny downward arrow ke saath. Do cheezein ek saath hit karne ke liye — jagah aur arrow — ek steady push kaafi nahi (ek knob, do locks). Isliye hum push ko remaining time ke saath smoothly change hone dete hain: ek start value aur ek ramp — do knobs do locks ke liye. Jab hum un pushes ko add (integrate) karte hain aur finish ko dono targets se match karne ke liye force karte hain, toh numbers 6 aur 2 nikal aate hain: $6/ t_{g o}^{2}$ distance gap mitaata hai, $2/ t_{g o}$ speed ko target ki taraf bleed karta hai, aur kyunki duniya mein gravity hai jabki haare plan mein nahi thi, hum $g$ subtract karte hain taaki engine ka real thrust exactly command ke barabar ho — gravity do baar add mat karo. Jab clock khatam hoti hai, $1/ t_{g o}$ factor badh jaata hai aur kisi bhi sideways speed ko zero karta hai, isliye arrow bilkul touch karte waqt seedha ho jaata hai. Lekin zero ke paas woh factors explode karte hain, isliye hum clock clamp karte hain aur last moment ke liye ek gentle drop par switch karte hain.

Recall Quick self-test

Ek constant acceleration position aur velocity dono kyun hit nahi kar sakta? ::: Yeh per axis ek knob deta hai lekin per axis do targets hain (jagah aur speed); humhe ek second free parameter chahiye, isliye hum acceleration ko linearly ramp karte hain. Numbers 6 aur 2 kahan se aate hain? ::: Position/velocity integrals mein $\frac{1}{6}$ aur $\frac{1}{2}$ se, jab do boundary conditions ko current acceleration $c_{0}$ ke liye solve kiya jaata hai. End ke paas velocity arrow vertical kyun ho jaata hai? ::: Horizontal shaping term $- 4 v_{x} / t_{g o}$ $t_{g o} \to 0$ ke saath badhta hai, touchdown par kisi bhi sideways speed ko zero kar deta hai. $t_{g o} \to 0$ ke saath kya blow up hota hai aur isse kaise handle karte hain? ::: Coefficients $6/ t_{g o}^{2}$ aur $2/ t_{g o}$ blow up karte hain; hum $t_{g o}$ ko floor karte hain ya constant-velocity drop par hand off karte hain, aur thrust cap karte hain. Kya thrust size karte waqt $a_{c m d}$ ke upar gravity add karte hain? ::: Nahi — $- g$ term pehle se $a_{c m d}$ ke andar hai; engine exactly $T / m = a_{c m d}$ produce karta hai.