Thrust vector control — single-gimbal, dual-gimbal; TVC angles: Visual walkthrough

3.5.44 · D2 · Physics › Guidance, Navigation & Control (GNC) › Thrust vector control — single-gimbal, dual-gimbal; TVC angl

Step 0 — Pehle ek coordinate system aur sign convention fix karo

KYA. Forces ki kisi bhi picture se pehle, hum decide karte hain ki "positive" direction kaun si hai, taaki baad mein koi bhi arrow ya sign kabhi ambiguous na ho.

KYUN. " $+ 5$ " ka torque meaningless hai jab tak hum na batayein ki $+$ ka matlab kya hai. Is page ka har sign (deflection $δ$ , torque $M$ ) us frame ke against measure hota hai jo hum yahan, ek baar aur hamesha ke liye fix karte hain.

Definition Hamara 2D frame aur sign rules (neeche har jagah use honge)

$+ x$ backward point karta hai, tail se bahar (jis direction mein exhaust jaata hai). Toh CoM $x = 0$ par hai aur gimbal $x = + ℓ$ par hai... lekin puraani convention ke hisaab se hum CoM ko origin par rakhte hain aur gimbal position ko $r = (- ℓ, 0)$ likhte hain, jo CoM se gimbal tak nose direction mein measure hoti hai. Ek consistent picture ke liye, hum $+ x$ ko nose ki taraf (forward) point karte hain, toh gimbal, CoM ke peeche hone ki wajah se, $x = - ℓ$ par baithta hai.
$+ y$ upar point karta hai (page ke plane mein).
$+ δ$ = engine is tarah tila hua ki uski sideways thrust $+ y$ direction mein point kare (nozzle ko "page ke neeche" deflect karta hai), jo hum dekhenge ki nose-up rotation produce karta hai. $- δ$ is ka mirror image hai → nose-down.
$+ M$ = page mein counter-clockwise spin (standard math convention, right-hand rule page se bahar).

Yeh chaar rules yaad rakho; baad ka har sign inhi ke against bookkeeping hai.

Step 1 — Rocket draw karo aur har distance ko naam do

KYA. Hum sabse simple possible picture se shuru karte hain: ek rocket, uske balance point ke liye ek dot, aur jahan engine bolt kiya gaya hai wahan ke liye ek dot.

KYUN. Kisi bhi formula se pehle, hum agree karna chahiye ki kya push karta hai aur kahan se push karta hai. Baad mein aane wala har symbol pehle is picture mein koi cheez honi chahiye jis par tum ungli rakh sako.

PICTURE. Figure dekho. Do special points hain:

Rocket ka balance point — woh single jagah jahan agar tum ise ungli par balance karne ki koshish karo toh yeh level baithega. Yeh balance point hi center of mass (CoM) hai; ab se hum sirf "CoM" naam use karenge, aur yahi woh pivot hai jiske around poora rocket rotate karta hai.
Gimbal point — woh mechanical hinge jahan engine attached hai, CoM ke $ℓ$ distance peeche baitha hua. Hamare frame mein (Step 0) yeh use $x = - ℓ$ par rakhta hai.

Figure — Thrust vector control — single-gimbal, dual-gimbal; TVC angles

Is waqt engine seedha peeche point kar raha hai, toh push seedha forward CoM ke through jaati hai. CoM ke seedha aim ki gayi push kuch spin nahi kar sakti — socho ki door ko exactly uski hinge par push karna. Toh abhi: zero steering. Bilkul theek. Yahi hamara baseline hai.

Step 2 — Engine ko tilt karo aur tilt ko ek naam do

KYA. Thrust arrow ko rocket ke axis se thoda sa angle par rotate karo.

KYUN. Pivot ki taraf seedha aimed push kuch nahi karta (Step 1). Muud ne ke liye, hum push ko thodi si side mein aim karna chahiye taaki uska "sideways grab" ho. Tilt ki matra woh ek knob hai jo autopilot ghumaata hai, toh ise ek symbol milna chahiye.

PICTURE. Figure mein thrust arrow (red) dashed axis line se swing off hai. Dashed axis aur red arrow ke beech ka wedge tilt hai.

Step 3 — Tilted push ko do honest pieces mein split karo

KYA. Single slanted red arrow ko right angles par do arrows mein tod do: ek axis ke along, ek across.

KYUN. Ek slanted force ke baare mein seedha reason karna mushkil hai. Physics mein har jagah use hone wala trick: ek awkward arrow ko do easy arrows se replace karo jo milke wapas use banaate hain. Hum woh do directions choose karte hain jo physically kuch matlab rakhti hain — "forward" (useful pushing karta hai) aur "sideways" (steering karta hai).

Trig kyun, aur specifically cosine/sine kyun? Red arrow, forward piece, aur sideways piece ek right triangle banate hain jiska longest side (hypotenuse) length $T$ ka hai. Trigonometry exactly woh tool hai jo angle plus hypotenuse ko do legs ki lengths mein convert karta hai. Yehi woh sawaal hai jo hum pooch rahe hain, toh yehi tool hum use karte hain.

PICTURE. Figure right triangle dikhata hai. Angle $δ$ gimbal par baitha hai.

Angle ke paas wali leg (adjacent) forward piece hai. $cos δ = \frac{adjacent}{hypotenuse} = \frac{forward}{T}$ , toh forward $= T cos δ$ .
Angle ke saamne wali leg (opposite) sideways piece hai. $sin δ = \frac{opposite}{hypotenuse} = \frac{sideways}{T}$ , toh sideways $= T sin δ$ .

Step 4 — Sideways push ko ek twist mein badlo

KYA. Poochhte hain ki woh sideways force poori rocket ke saath kya karta hai, given ki woh CoM se $ℓ$ distance par act karta hai.

KYUN. Pivot se door apply ki gayi sideways force rocket ko sirf slide nahi karti — yeh use rotate karti hai. Us rotation-tendency ki strength ko torque kehte hain, aur torque hi woh cheez hai jo rocket ki heading change karti hai. Hum heading change karna chahte hain, toh torque woh quantity hai jiske peeche hum hain. (Foundations: Torque and Moment Arm.)

"Force × distance" rule kyun? Door ko hinge ke paas push karo aur kuch nahi hota; door ke far edge par push karo aur woh easily swing karta hai — same force, zyada turning. Torque exactly yahi capture karta hai: yeh sideways force ko multiply kiya gaya hai by kitna door yeh act karta hai, moment arm $ℓ$ (wohi black bar jo Step 1 aur Step 4 figures mein label hai).

PICTURE. Figure sirf sideways component $T sin δ$ ko dikhata hai jo gimbal par act kar raha hai, CoM se $ℓ$ distance par (labelled moment arm), rocket ko around curl karte hue.

$∣ M ∣ = ℓ T sin δ$

Yeh parent ka exact result hai (ek magnitude), ab ek ek brick karke assemble kiya gaya.

Step 5 — Sign restore karo: cross product ke zariye wahi twist

KYA. Step 4 ko formal cross-product bookkeeping ke saath redo karo, taaki torque ki direction (sign) honest ho aur Step 0 convention se match kare.

KYUN. Step 4 ne size $∣ M ∣ = ℓ T sin δ$ diya. Lekin ek twist ki direction bhi hoti hai — clockwise ya counter-clockwise. Cross product $r \times F$ woh tool hai jo position aur force se size aur spin direction dono return karta hai. Direction chahiye taaki autopilot jaane ki nose kaun si taraf swing karega. (Machinery: Rigid Body Rotational Dynamics.)

PICTURE. Figure position vector $r = (- ℓ, 0)$ (CoM se gimbal tak, Step 0 se match karta hua: gimbal $x = - ℓ$ par) aur Step 3 se $F$ place karta hai, aur woh curl direction dikhata hai jo cross product predict karta hai.

Step 6 — Small angles: steering almost free kyun hai

KYA. Tiny $δ$ par zoom in karo aur compare karo ki turning kitni fast badhti hai versus forward thrust kitna fast khota hai.

KYUN. Real gimbals sirf kuch degrees move karte hain. Us tiny window mein, curves seedhi lines jaisi lagti hain, aur hum $sin$ aur $cos$ ko simpler expressions se replace kar sakte hain. Yahan trade-off ("cheap steering") reveal hota hai. (Tool: Small-Angle Approximation.)

PICTURE. Figure $δ = 0$ ke paas teen curves overlay karta hai: seedhi line $δ$ , curve $sin δ$ (yeh dono ek doosre se chipke rehte hain), aur $1 - cos δ$ (jo flat rehta hai phir slowly utha hai).

Step 7 — Ek saath do axes (dual-gimbal) aur circular limit

KYA. Engine ko do perpendicular directions mein tilt karne do — up/down (pitch, angle $δ_{p}$ ) aur left/right (yaw, angle $δ_{y}$ ) — aur poochhte hain ki mechanical hinge kaun sa limit impose karta hai.

KYUN. Ek tilt plane sirf ek plane mein steer karta hai (pitch ya yaw). Ek engine jo do planes mein tilt karta hai dono mein steer kar sakta hai. Lekin hinge sirf itna hi center se kisi bhi direction mein lean ho sakta hai, toh humein combined tilt dhundhni hai.

Square root kyun? Do tilts ek ek physical lean ke perpendicular pieces hain, exactly jaise Step 3 ke triangle ki do legs. Small tilts ke liye true lean $(δ_{p}, δ_{y})$ triangle ki hypotenuse hai — Pythagoras — toh yeh $δ_{p}^{2} + δ_{y}^{2}$ hai, $δ_{p} + δ_{y}$ nahi.

PICTURE. Engine bell ke seedha upar dekho: nozzle ki tip ek $δ_{max}$ radius ke circle ke andar kahin bhi pahunch sakti hai. Command $(δ_{p}, δ_{y})$ ek point hai; center se uski distance total tilt hai.

One-picture summary

KYA. Ek figure jo poori chain carry karta hai: tilt → sideways component → lever → torque, plus linear-torque / quadratic-loss trade.

Ise left se right padhte hain: engine $δ$ se tilt hota hai → push forward $T cos δ$ (CoM ke through, koi turn nahi) aur sideways $T sin δ$ mein split hoti hai → woh sideways push, CoM se $ℓ$ distance (moment arm) par, $ℓ T sin δ$ magnitude ka torque banata hai (hamare frame mein signed $- ℓ T sin δ$ ) → small tilts ke liye radians mein, $∣ M ∣ \approx ℓ T δ$ (bada turn) jabki sirf $\frac{1}{2} δ^{2}$ thrust lost hoti hai (tiny cost).

Recall Feynman retelling — plain words mein bolke dikhao

Socho shopping trolley ko peeche se push kar rahe ho. Agar dead center push karo, toh seedha chalti hai. Agar apni push ko thodi si left mein angle karo, toh trolley almost utni hi fast forward roll karti rehti hai — lekin woh turn bhi karne lagti hai, kyunki teri push ab use ek side par pakad rahi hai. Rocket engine exactly waisa hi karta hai: yeh kuch degrees swing karta hai, toh uski zyaadatar shove forward jaati rehti hai lekin ek slice sideways jaati hai. Woh sideways slice, tail mein kaafi neeche act karti hai (balance point se door), poori rocket ko around twist karti hai — jaise door ko uski hinge ki jagah edge par push karna. Yeh kaun si taraf turn karta hai yeh depend karta hai ki tum kaun si taraf tilt karo: ek taraf tilt karo, nose up; doosri taraf, nose down. Thoda tilt karo, bahut turn; thoda tilt karo, almost koi speed nahi jaati. Ek saath do directions mein karo aur tum nose ko kahin bhi point kar sakte ho — except ise top ki tarah spin karo, kyunki seedhe middle se push karna use roll nahi kara sakta. Yahi hai thrust vector control.

Recall Boxed formula memory se rebuild karo

Tilt $δ$ se shuru karo ::: thrust ko $T cos δ$ (forward) aur $T sin δ$ (sideways) mein split karo Kaun sa piece rocket ko turn karta hai? ::: sideways piece $T sin δ$ Torque get karne ke liye use kisse multiply karo? ::: moment arm $ℓ$ CoM se gimbal tak Exact torque magnitude? ::: $∣ M ∣ = ℓ T sin δ$ Hamare frame mein signed torque (Step 0)? ::: $M = - ℓ T sin δ$ — positive $δ$ clockwise deta hai (nose-down) Small-angle version aur kyun? ::: $∣ M ∣ \approx ℓ T δ$ , kyunki $sin δ \approx δ$ $δ$ ke liye radians mein Two-axis combined limit (small angles)? ::: $δ_{p}^{2} + δ_{y}^{2} \leq δ_{max}$ , circular reach limit