Exercises — Propulsive forces — thrust misalignment, gimbal angle

Koi torque nahi. CoM ke baare mein torque $\vec{\tau }=\vec{r}\times \vec{F}$ hai, aur ye tabhi nonzero hota hai jab force ka ek component lever arm par act kare (CoM se force ki line tak ki perpendicular doori). Yahan thrust line CoM se guzarti hai, isliye wo perpendicular doori zero hai, aur $\tau =0$. Engine sirf rocket ko aage translate karta hai — koi twist nahi. (Dekho Torque and Moment of Inertia.)

Gimbal angle $\delta$ commanded hota hai — computer jaanbujhkar engine ko tilts karta hai ek steering torque generate karne ke liye. Misalignment ${\delta }_{err}$ ek unintended offset hai jisse control system ko lagaataar fight karna padta hai. Dono ek jaisa torque $LT\sin (\cdot )$ produce karte hain; fark sirf intent mein hai, physics mein nahi.

Step 1 — convert (KYUN: trig ko radians chahiye). $\delta =4\times \frac{\pi }{180}=0.06981\text{rad}.$ Step 2 — torque formula lagao. Torque lateral thrust component $T\sin \delta$ se lever arm $L$ par aata hai: $\tau =LT\sin \delta =15\times 600000\times \sin (0.06981).$ Step 3 — evaluate karo. $\sin (0.06981)=0.06975$, isliye $\tau =15\times 600000\times 0.06975\approx 6.28\times 10^5\text{N⋅m}.$ Ye is stage ko steer karne ke liye available twisting authority hai.

Axial (aage) part $T\cos \delta$ hai; lost part $T(1-\cos \delta )$ hai. $\cos (0.06981)=0.99756\Rightarrow \text{loss fraction}=1-0.99756=0.002436=0.244\%.$ $\text{loss}=600000\times 0.002436\approx 1.46\times 10^3\text{N}=1.46\text{kN}.$ Ek quarter percent — negligible — jabki upar wala torque poora usable hai. Ye asymmetry (loss $\propto {\delta }^2$, torque $\propto \delta$) hi wo poori wajah hai jis se gimballing kaam karta hai.

Step 1 — rotational Newton's law. $\ddot{\theta }=\frac{\tau }{I}=\frac{6.28\times 10^5}{2.5\times 10^6}=0.2513{\text{rad/s}}^2.$ Step 2 — rest se constant acceleration integrate karo (KYUN: constant $\delta$ ⇒ constant $\tau$ ⇒ constant $\ddot{\theta }$): $\dot{\theta }=\ddot{\theta }t=0.2513\times 0.8=0.2010\text{rad/s}.$ Step 3 — degrees/s mein convert karo intuition ke liye: $0.2010\times \frac{180}{\pi }=11.5\text{°/s}.$ Sirf 0.8 s mein rocket pehle se ~11.5°/s se turn kar raha hai — isliye real gimbal commands chote aur brief hote hain, phir rotation rokne ke liye reverse hote hain. (Dekho Attitude Control & Stability.)

Tilt torque (A): ${\tau }_A=LT\sin \delta$. Offset torque (B): yahan thrust axial hai ($+x$) lekin uski line CoM se $d$ miss karti hai; perpendicular lever arm $d$ hai, isliye ${\tau }_B=dT$ (pure moment — axial force khud lever-arm force hai). Barabar set karo: $dT=LT\sin \delta \Rightarrow \boxed{d=L\sin \delta }.$ $T$ cancel ho jaata hai — sirf geometry equivalence set karti hai. Numerically: $d=15\times \sin (2^{\circ })=15\times 0.034899=0.5235\text{m}.$ Toh ~52 cm sideways mounting error ek 2° tilt ke twist ko mimic karta hai. (Note: tilt ke unlike, ek offset koi axial thrust cost nahi karta — force abhi bhi fully forward hai.)

(a) Disturbance torque. ${\delta }_{err}=0.6\times \frac{\pi }{180}=0.010472\text{rad},\sin {\delta }_{err}=\mathrm{0.010472.}$ ${\tau }_{err}=LT\sin {\delta }_{err}=15\times 600000\times 0.010472\approx 9.42\times 10^4\text{N⋅m}.$ (b) Cancelling command. Disturbance null karne ke liye, commanded gimbal ko equal-and-opposite torque produce karna hoga. Kyunki dono same $L$ aur $T$ use karte hain: $LT\sin {\delta }_c=LT\sin {\delta }_{err}\Rightarrow \sin {\delta }_c=\sin {\delta }_{err}\Rightarrow {\delta }_c={\delta }_{err}=0.6^{\circ }.$ Authority consumed: $\frac{0.6^{\circ }}{5^{\circ }}=0.12=12\%.$ Toh steering budget ka 12% permanently sirf still khade rehne mein kharach ho jaata hai — misalignment ki ek real cost jo control margin kha jaati hai. (Dekho Attitude Control & Stability.)

Torque linearly $L$ ke saath scale karta hai ($T$ aur $\delta$ unchanged hain): $\frac{{\tau }_2}{{\tau }_1}=\frac{L_2}{L_1}=\frac{22}{15}=\mathrm{1.467.}$ Wahi gimbal command burn ke end mein 47% zyada torque produce karta hai. Matlab: controller "commanded angle" aur "resulting turn" ke beech fixed relationship use nahi kar sakta — jaise $L$ badhta hai, gimbal ka har degree zyada kaatnay lagta hai, isliye guidance law ko flight ke dauran apne gains adapt karne honge. (Ye exactly Center of Mass shift during burn problem hai jiske around TVC systems tune hote hain.)

(a) Required angular acceleration. Constant $\ddot{\theta }$ rest se $t$ tak rate $\dot{\theta }=\ddot{\theta }t$ tak pahunchta hai. Pehle target rate rad/s mein convert karo: ${\dot{\theta }}_{\text{target}}=10\times \frac{\pi }{180}=0.17453\text{rad/s}.$ $\ddot{\theta }=\frac{{\dot{\theta }}_{\text{target}}}{t}=\frac{0.17453}{1.2}=0.14544{\text{rad/s}}^2.$ (b) Zaroorat ka torque (rotational Newton's law): $\tau =I\ddot{\theta }=3.2\times 10^6\times 0.14544=4.654\times 10^5\text{N⋅m}.$ (c) Commanded gimbal. $\tau =LT\sin \delta$ ko invert karo: $\sin \delta =\frac{\tau }{LT}=\frac{4.654\times 10^5}{18\times 750000}=\frac{4.654\times 10^5}{1.35\times 10^7}=\mathrm{0.034473.}$ $\delta =\arcsin (0.034473)=0.034480\text{rad}=1.975^{\circ }.$ $\pm 7^{\circ }$ limit ke andar aaram se — bahut margin hai. (Yahan $\arcsin$ sawaal ka jawaab deta hai "kaun sa angle ka ye sine hai?" — ye woh $\sin$ undo karta hai jo lateral thrust component ko torque mein daalta hai.) (d) Axial loss $\delta =0.034480\text{rad}$ par: $1-\cos \delta =1-0.999406=0.000594=0.0594\%.$ $\sim 2°$ par steer karne ke liye ek tenth percent se kam lost — gimballing ka payoff ek number mein.

Gimbal ko $-\delta$ par reverse karne se torque $-\tau$ ho jaata hai, jo $\ddot{\theta }=-0.14544{\text{rad/s}}^2$ deta hai (same magnitude, opposite sign). $\dot{\theta }=0.17453\text{rad/s}$ se shuru karke $0$ tak decelerate karte hue: $t_{\text{stop}}=\frac{\dot{\theta }}{|\ddot{\theta }|}=\frac{0.17453}{0.14544}=1.2\text{s}.$ Symmetric isliye hai kyunki deceleration magnitude acceleration magnitude ke barabar hai (same $L,T,\delta ,I$), isliye rate build karne aur khatam karne mein equal time lagta hai — classic bang–bang steering pulse: ek taraf tilt karo turn shuru karne ke liye, mirror-image taraf tilt karo use rokne ke liye. (Dekho Attitude Control & Stability.)