Terminal descent — velocity vector alignment, touchdown constraints: Question bank

3.5.54 · D5 · Physics › Guidance, Navigation & Control (GNC) › Terminal descent — velocity vector alignment, touchdown cons

Quick symbol reminder taaki kuch bhi unclear na rahe:

$r$ = position, $v = \dot{r}$ = velocity (motion ka arrow), $a$ = acceleration.
$t_{g o} = t_{f} - t$ = time-to-go, time $t_{f}$ par touchdown hone tak baaki seconds.
$a_{c m d}$ = woh net acceleration jo guidance law command karta hai; engine physically exactly $T / m = a_{c m d}$ supply karta hai.
$v_{t d}$ = downward touchdown speed, $v_{h} = v_{x}^{2} + v_{y}^{2}$ = horizontal drift speed, $θ$ = thrust axis ka vertical se tilt.

True or false — justify karo

Trajectory ko pad position ki taraf aim karna akela soft landing guarantee karta hai.

False — position par pad hit karna velocity ke baare mein kuch nahi kehta; tum pad par 20 m/s sideways aake land kar sakte ho. Soft landing ek two-point boundary value problem hai jo

r_{f}

aur

v_{f}

dono ko pin karta hai.

Kyunki guidance formula mein already ek

- g

term hai, engine ko gravity se ladne ke liye

a_{c m d} + g

produce karna padta hai.

False —

a_{c m d}

ke andar ka

- g

wahi hai jo

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

mein

+ g

ko cancel karta hai. Engine exactly

T / m = a_{c m d}

supply karta hai;

g

dobara add karna gravity ko double-count kar deta hai.

Touchdown par perfectly vertical velocity vector guarantee karta hai ki lander tip nahi karega.

False — body phir bhi tilted ho sakti hai (

θ

bada) even with zero horizontal drift; tipping

θ

aur drift dono par depend karta hai,

θ + arctan (v_{h} / v_{t d}) ≲ θ_{t i p}

ke through.

v_{f} = (0, 0, - v_{t d})

set karna zero horizontal components ke saath wahi hai jo velocity vector ko straight-down ki taraf rotate karne par majboor karta hai.

True — target mein koi sideways component nahi hai, isliye jab

t_{g o} \to 0

hota hai toh velocity-shaping term

v_{x}, v_{y} \to 0

drive karta hai, sirf downward part bacha rehta hai.

Touchdown envelope constraints alternatives hain — unme se koi bhi ek satisfy karna kaafi hai.

False — ye AND-constraints hain: vertical speed, horizontal speed, tilt, angular rate aur position sab simultaneously contact par limits ke andar hone chahiye.

Jaise lander approach karta hai,

t_{g o}

ko poora zero tak shrink hone dena harmless hai.

False — position term

1/ t_{g o}^{2}

ki tarah scale karta hai, isliye command infinite thrust tak blow up ho jaati hai, jo engine supply nahi kar sakta. Tum

t_{g o}

ko floor karo ya hover-then-drop phase par hand off karo.

Vehicle ko point mass model karna attitude requirement ko throw away karta hai.

False — attitude ignore nahi hota, ise ek faster inner loop handle karta hai; point-mass guidance

a_{c m d}

produce karta hai, jis direction ko inner loop thrust axis point karke realise karta hai.

v_{f}

ke liye exactly

0

m/s ka desired vertical touchdown speed ideal target hai.

False — thoda sa positive

v_{t d} \approx 1 - 2

m/s deliberate hota hai; exactly zero command karna engine se indefinitely hover karne aur kabhi contact na karne ki request hogi, aur koi bhi error tumhe upar drift ya stall kar dega.

Spot the error

"Humne pad hit kiya, toh terminal guidance successful rahi — velocity irrelevant hai."

Galti: velocity ek alag boundary condition hai. Is named constraint ka naam velocity vector alignment hai; galat velocity ke saath position hit karna ek leg scrape ya crash karta hai.

"Thrust command

= a_{c m d} + g

kyunki rocket ko weight overcome karna hai."

Galti: gravity already

a_{c m d}

mein

- g

term ke through fold ho chuki hai. Correct command

T / m = a_{c m d}

hai;

g

dobara add karna over-thrust karta hai.

"Ek single constant commanded acceleration dono final position aur final velocity pin karne ke liye kaafi hai."

Galti: yeh per axis over-determined hai. Chaar scalar conditions match karne ke liye ek acceleration chahiye jo linearly time mein vary kare, yahi woh jagah hai jahan se coefficients

6

aur

2

aate hain.

"Example 1 mein answer negative hai kyunki vehicle gir raha hai."

Galti:

a_{c m d}

ka sign braking direction reflect karta hai, fall nahi.

- 25

m/s par girna matlab tum upward brake karo, isliye

a_{c m d} = + 14.46

m/s² positive hai.

"Formula ka

1/ t_{g o}

velocity term ek fixed gain hai, isliye yeh early aur late descent ko same treat karta hai."

Galti: gain

t_{g o} \to 0

hone par badhta hai, isliye

v_{x}, v_{y}

par correction touchdown ke paas strong hoti jaati hai — woh badhti authority exactly wahi hai jo drift ko time par zero drive karti hai.

"Kyunki gravity hamesha neeche pull karti hai, hum

- g

term drop kar sakte hain aur phir bhi

T / m = a_{c m d}

command kar sakte hain."

Galti: ek convention chuno. Agar tum formula se

- g

drop karte ho toh thrust size karte waqt gravity dobara add karni padegi; dono conventions mix karna har cycle mein gravity error deta hai.

Ek tilted thrust axis ka horizontal component hota hai jise tum contact se pehle cancel nahi kar sakte, aur drift ke saath combine hoke yeh center of mass ko leg footprint ke bahar push karta hai, tip-over cause karta hai.

Position-error coefficient

1/ t_{g o}^{2}

kyun carry karta hai jabki velocity term

1/ t_{g o}

carry karta hai?

Position acceleration ka double integral hai (do time integrations), isliye shrinking window par position gap correct karne ke liye time-squared inverse scale karne wali term chahiye; velocity ek single integral hai, jo

1/ t_{g o}

ki ek power deta hai.

Constant acceleration ek position-and-velocity target hit kyun kar sakta hai lekin ek straight line nahi?

Ek straight line mein sirf position aim karne ki enough freedom hoti hai. Constant (phir linear) acceleration woh free parameters add karta hai jo final velocity fix karne ke liye bhi chahiye — zyada conditions ke liye zyada knobs.

v_{t d}

small but nonzero (1–2 m/s) kyun choose hota hai rather than large?

Small legs ko crush se bachata hai aur

v_{t d}^{m a x}

ke andar rakhta hai; nonzero ensure karta hai ki vehicle actually descend kare aur firm contact kare instead of hover karne ya command error par upar float karne ke.

Drift

v_{h}

ke saath approach karna effectively tilt budget mein kyun add hota hai?

Incoming velocity vertical se

arctan (v_{h} / v_{t d})

angle banati hai; yeh angle body tilt

θ

ke saath tip threshold ke against stack hota hai, isliye drift wahi margin khaata hai jo tilt khaata hai.

Edge cases

Exactly jis instant

t_{g o} = 0

hai, kya guidance formula valid hai?

Nahi — wahan yeh singular hai (

1/ t_{g o}^{2} \to \infty

). Tumhe pehle hi transition karna hoga ek floored-

t_{g o}

ya hover-then-drop phase mein, literally zero par kabhi evaluate mat karo.

Kya hoga agar

v_{h} = 0

ho lekin

v_{t d}

bhi contact par near zero ho?

Incoming-angle term

arctan (v_{h} / v_{t d})

becomes

arctan (0/0)

, undefined; physically iska matlab negligible motion ke saath near-hover hai, isliye tip risk drift ki bajaye akele body tilt

θ

se dominate hota hai.

Agar horizontal drift

v_{x}

terminal descent ke start mein already zero hai, toh kya velocity-shaping term uske saath kuch karti hai?

Nahi —

v_{x} = 0

aur target

v_{f, x} = 0

ke saath term us axis par kuch contribute nahi karti; yeh sirf existing drift hatane ke liye act karti hai, already-vertical velocity ko vertical rakhti hai.

Kya ho agar vehicle pad par aise velocity aur tilt ke saath pahunche jo individually limits ke andar hain lekin unka combined incoming angle

θ_{t i p}

exceed karta ho?

Yeh phir bhi tip kar sakta hai. Stability condition sum

θ + arctan (v_{h} / v_{t d})

par hai; har single-limit check pass karna necessary hai lekin sufficient nahi.

Agar commanded acceleration engine ko touchdown ke paas saturate kare, toh physically trajectory ka kya hoga?

Vehicle guidance shape follow nahi kar sakta, isliye boundary conditions miss ho jaate hain — typically ek harder ya off-target landing; yahi reason hai kyun

t_{g o}

floor hota hai aur kyun convex landing jaise methods explicitly thrust bounds enforce karte hain.

Recall Ek-line self-summary

Is page ki har cheez teen facts se aati hai: soft landing ek two-point (position aur velocity) problem hai, $- g$ $a_{c m d}$ ke andar rehta hai isliye engine exactly $a_{c m d}$ supply karta hai, aur $1/ t_{g o}$ terms touchdown ke paas badhte hain — shaping ke liye powerful, lekin dangerous agar $t_{g o} \to 0$ unbounded ho jaaye.

Question bank — Terminal descent — velocity vector alignment, touchdown constraints

True or false — justify karo

Spot the error

Why questions

Edge cases