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

Hum kya decide karte hain: $25^{\circ }$ pe small-angle shortcut ab bharosemand nahi raha. Kyun: linear law $M\approx \ell T\delta$ sirf $\sin \delta =\delta -{\delta }^3/6+\dots$ ka pehla term hai. Pehle convert karke, $\delta =25^{\circ }\times \frac{\pi }{180}=0.4363$ rad, toh $\sin \delta =0.4226$, aur linear form isse zyada estimate karta hai $\frac{0.4363-0.4226}{0.4226}=3.2\%.$ Answer: exact law $\boxed{M=\ell T\sin \delta }$ use karo. Small-angle law sirf lagbhag $\pm 8^{\circ }$ tak valid hai.

Tilted thrust vector ko vehicle axis ke saath aur across resolve karo (yeh sirf $T$ length ke rotated arrow ki trigonometry hai, exactly figure s01 ke do dotted arrows):

• (a) Axial (useful) thrust: $T_{\text{axial}}=T\cos \delta .$
• (b) Transverse (steering) force: $T_{\perp }=T\sin \delta .$

Steering force $T\sin \delta$ ko arm $\ell$ se multiply karo toh exactly torque $M=\ell T\sin \delta$ milta hai.

Step 1 — angle convert karo. $\delta =4^{\circ }\times \frac{\pi }{180}=0.06981$ rad. Kyun: linear law ko radians chahiye (hamaari units convention). Step 2 — $M\approx \ell T\delta$ apply karo. $M=5\times (500\times 10^3)\times 0.06981=1.745\times 10^5\text{N⋅m}.$ Yeh step kyun: torque woh steering force $T\sin \delta$ hai jo lever $\ell$ ke through act karta hai, yani $M=\ell T\sin \delta$; kyunki $4^{\circ }$ small-angle window ke andar hai hum $\sin \delta$ ko $\delta$ se replace karte hain (error $<0.1\%$), law ko linear product $\ell T\delta$ mein badal dete hain jise directly evaluate kar sakte hain. Hum teen diye gaye quantities multiply karte hain kyunki torque teeno se badhta hai: zyada lever, zyada thrust, ya zyada tilt sab vehicle ko zyada twist karte hain. Answer: $M\approx 1.75\times 10^5$ N·m. Yeh torque directly Rigid Body Rotational Dynamics mein $M=I\dot{\omega }$ ke through jaata hai.

Step 1 — needed torque (Newton's rotational law, Torque and Moment Arm): $M=I\dot{\omega }=8.0\times 10^4\times 0.6=4.8\times 10^4\text{N⋅m}.$ Kyun: rocket ko $\dot{\omega }$ pe spin up karne ke liye exactly torque $M=I\dot{\omega }$ supply karna hoga — yeh $F=ma$ ka rotational twin hai. Step 2 — linear law $M=\ell T\delta$ invert karo: $\delta =\frac{M}{\ell T}=\frac{4.8\times 10^4}{5\times 5\times 10^5}=0.0192\text{rad}.$ Kyun: hume pata hai kaunsa torque chahiye aur fixed hardware ($\ell$, $T$); akela free knob $\delta$ hai, toh hum linear law ko divide karke solve karte hain. Step 3 — degrees mein: $\delta =0.0192\times \frac{180}{\pi }=1.10^{\circ }.$ Answer: lagbhag $1.1^{\circ }$ — itna chhota ki linear law legitimate tha (self-consistent).

Pehle convert karo. $\delta =6^{\circ }\times \frac{\pi }{180}=0.10472$ rad; neeche har $\sin$/$\cos$ is radian value use karta hai (degree label sirf padhne ke liye hai). (a) Axial loss. Fractional loss $=1-\cos \delta$. $\delta =0.10472$ rad ke saath: $\cos \delta =0.99452$, toh loss $=0.00548=0.548\%.$ (b) Side force. $T\sin \delta =5\times 10^5\times \sin (0.10472)=5\times 10^5\times 0.10453=5.226\times 10^4$ N $=52.3$ kN. Compare karo. Hum thrust ka $0.55\%$ kharach karte hain (lagbhag $2.7$ kN forward push, kyunki $0.00548\times 500\text{ kN}$) aur $52.3$ kN steering force gain karte hain — roughly $19:1$ return. Kyun kaam karta hai: loss $1-\cos \delta \approx {\delta }^2/2$ $\delta$ mein second order hai, jabki side force $T\sin \delta \approx T\delta$ first order hai. Chhote $\delta$ pe ek first-order quantity second-order wali ko dwarf kar deti hai — yahi poori wajah hai ki gimbaling efficient hai.

Step 1 — ek physical tilt mein combine karo (assumption state karo). Dono deflections ek single tilt ke perpendicular components hain. Assumption used: small angles ke liye, ek axis ke baare mein ${\delta }_p$ tilt aur ek perpendicular axis ke baare mein ${\delta }_y$ tilt first order mein magnitude $\sqrt{{\delta }_p^2+{\delta }_y^2}$ ke ek tilt mein compose hote hain. Second-order coupling pe warning: agar ${\delta }_p$ ya ${\delta }_y$ bade ho jaate (tens of degrees), toh do rotations commute nahi karti aur true combined tilt mein order ${\delta }_p{\delta }_y$ ke second-order cross-terms aa jaate; tab clean Pythagorean sum sirf approximate hota. Yahan dono angles $\le 5^{\circ }$ hain, toh woh cross-terms $\lesssim 0.4\%$ hain aur hum safely first-order rule use kar sakte hain. Kyunki yeh Pythagorean sum ek linear (degree-homogeneous) operation hai — koi $\sin$/$\cos$ nahi aata — ise degrees ya radians mein karna sirf common factor $\frac{\pi }{180}$ se differ karta hai, jo ${\delta }_{\text{tot}}$ vs ${\delta }_{\max }$ comparison mein cancel ho jaata hai. Toh hum degrees mein reh sakte hain: ${\delta }_{\text{tot}}=\sqrt{{\delta }_p^2+{\delta }_y^2}=\sqrt{5^2+5^2}=7.071^{\circ }.$ Step 2 — stop se compare karo. $7.071^{\circ }>7^{\circ }$ → reachable nahi, clip karna hoga. Step 3 — same direction mein clip karo. Dono components ko same factor $k=\frac{{\delta }_{\max }}{{\delta }_{\text{tot}}}=\frac{7}{7.071}=0.9899$ se scale karo taaki tilt direction preserve ho jabki uski length exactly stop pe land kare: ${\delta }_p^{\prime }={\delta }_y^{\prime }=5\times 0.9899=4.950^{\circ },\sqrt{4.950^2+4.950^2}=7.000^{\circ }.✓$ Answer: $(4.95^{\circ },4.95^{\circ })$ pe clip karo.

(a) Early burn. Required torque $M=I\dot{\omega }=1.5\times 10^5\times 1.0=1.5\times 10^5$ N·m. Angle $\delta =\frac{M}{\ell T}=\frac{1.5\times 10^5}{7\times 9\times 10^5}=0.02381$ rad $=1.364^{\circ }$. Comfortable hai. (b) Late burn. $M=6.0\times 10^4\times 1.6=9.6\times 10^4$ N·m. $\delta =\frac{9.6\times 10^4}{4\times 7\times 10^5}=0.03429$ rad $=1.964^{\circ }$. (c) Margin. Dono $1.364^{\circ }$ aur $1.964^{\circ }$ $8^{\circ }$ stop ke neeche comfortably baithe hain, toh haan — margin bana rehta hai. Lekin trend notice karo: chhota arm aur kam thrust har degree gimbal ko kam effective bana dete hain, isliye required angle badhta hai chahе vehicle lighter kyun na ho jaaye. Tightest gimbal demand aksar burn ke ant mein hoti hai, exactly tab jab $\ell$ aur $T$ shrink ho chuke hote hain. Yahi synthesis lesson hai: control authority $=\ell T\delta$ teeno factors pe depend karta hai, aur unmen se do flight ke dauran decay karte hain.

Relative error set up karo. Linear model $\ell T\delta$ deta hai; truth $\ell T\sin \delta$ hai. Overprediction fraction: $\epsilon (\delta )=\frac{\ell T\delta -\ell T\sin \delta }{\ell T\sin \delta }=\frac{\delta -\sin \delta }{\sin \delta }=\frac{\delta }{\sin \delta }-1.$ Pehle, ek rough locator (honest error warning ke saath). Series $\frac{\delta }{\sin \delta }=1+\frac{{\delta }^2}{6}+\frac{7{\delta }^4}{360}+\dots$ suggest karta hai $\frac{{\delta }^2}{6}\approx 0.05\Rightarrow \delta \approx \sqrt{0.3}=0.5477$ rad. Lekin is $\delta$ pe next term $\frac{7{\delta }^4}{360}=\frac{7(0.5477{)}^4}{360}=0.00175$ pehle se $\sim 3.5\%$ hai us $0.05$ ka jise hum match kar rahe hain — toh two-term series yahan sirf kuch percent tak trustworthy hai aur hume ise final answer ke taur pe nahi report karna chahiye. Yeh sirf starting guess hai. Exactly solve karo (numerically). Hum $\frac{\delta }{\sin \delta }=1.05$ solve karte hain. Guess $0.5477$ rad se starting Newton iteration converge karta hai $\boxed{\delta =0.5401\text{rad}=30.9^{\circ }}$ — crude series guess se lagbhag $1.4\%$ neeche, upar di gayi truncation warning ke consistent. (Check: $\sin (0.5401)=0.5144$, aur $0.5401/0.5144=1.050$ ✓.) Answer: linear law lagbhag $\boxed{31^{\circ }}$ pe $5\%$ overprediction reach karta hai. Insight: yeh $\sim 31^{\circ }$ tak $5\%$ ke andar rehta hai — kisi bhi real gimbal stop ($\lesssim 8$–$10^{\circ }$) se bahut aage. Yahi wajah hai ki flight software safely linear form use kar sakta hai: real deflections kabhi wahan tak nahi pahunchti jahan yeh toota ho.

(a) Simplify karo. Half-angle identities $\sin \delta =2\sin \frac{\delta }{2}\cos \frac{\delta }{2}$ aur $1-\cos \delta =2{\sin }^2\frac{\delta }{2}$ use karo: $\eta =\frac{2\sin \frac{\delta }{2}\cos \frac{\delta }{2}}{2{\sin }^2\frac{\delta }{2}}=\frac{\cos \frac{\delta }{2}}{\sin \frac{\delta }{2}}=\cot \frac{\delta }{2}.$ Kyunki $\cot$ $(0,\pi )$ pe decrease karta hai, $\eta$ monotonically girता hai jaisa $\delta$ badhta hai — chhoti deflections hamesha zyada efficient hoti hain per unit thrust sacrificed ke. (b) Values (pehle convert karo: $2^{\circ }=0.03491$ rad, $8^{\circ }=0.13963$ rad, toh half-angles $1^{\circ }$ aur $4^{\circ }$ hain). $\eta (2^{\circ })=\cot (1^{\circ })=\cot (0.017453\text{ rad})=\mathrm{57.29.}$ Har unit thrust kho ke $\approx 57$ units steering force milti hai. $\eta (8^{\circ })=\cot (4^{\circ })=\cot (0.069813\text{ rad})=\mathrm{14.30.}$ Ab sirf $\approx 14$. Interpret karo: steering zyada deflect karne pe kam thrifty hoti jaati hai — marginal return $2^{\circ }$ aur $8^{\circ }$ ke beech chaar guna gir jaata hai. Isliye autopilots kai chhoti corrections prefer karte hain kuch badi ki jagah, aur large-authority needs Reaction Control System (RCS) ko de dete hain.

Exact vs linear torque law ::: $M=\ell T\sin \delta$ exact; $M\approx \ell T\delta$ $\delta \lesssim 8^{\circ }$ ke liye. Steering cheap kyun hai ::: side force $O(\delta )$ hai, thrust loss $O({\delta }^2)$ hai. Torque ka sign ::: $\delta$ mein odd hai — $\delta$ flip karo aur $M$ sense reverse karta hai, same magnitude; $\delta =0$ zero torque deta hai. Dual-gimbal limit constraint ::: $\sqrt{{\delta }_p^2+{\delta }_y^2}\le {\delta }_{\max }$ (vector magnitude, sum nahi). Over-command clip kaise karein ::: direction rakhne ke liye dono components ko ${\delta }_{\max }/{\delta }_{\text{tot}}$ se scale karo. Linear law finally 5% kahan jhooth bolta hai ::: $\delta \approx 31^{\circ }$ ke paas, kisi bhi real gimbal stop se bahut aage. Steering efficiency form ::: $\eta =\cot (\delta /2)$, decreasing — chhoti deflections thriftiest hain.