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

3.5.44 · D3 · Physics › Guidance, Navigation & Control (GNC) › Thrust vector control — single-gimbal, dual-gimbal; TVC angl

Yeh page parent topic ka drill floor hai. Hum gimbal maths ke har tarah ke case cover karenge: small angles, bade angles, zero deflection, degenerate "engine dead" case, single- aur dual-gimbal geometry dono, sign flips (left vs right steer karna), ek real-world word problem, aur ek exam twist jo zyaadatar students ko pakad leta hai. Neeche jo bhi numbers hain wo sab machine-checked hain — closing "Machine-check summary" box dekho.

Shuru karne se pehle, yeh do engine formulas hain jinpar poori page chalti hai, saath mein teen auxiliary results jo unse derive hote hain:

Recall Do core formulas (+ teen helpers)

Do engines (neeche sab kuch inhi mein se ek hai, ya inki rearrangement):

• Engine 1 — exact torque magnitude (engine tilting se turning strength): $|M|=\ell T\sin \delta$. Signed torque (kaun si taraf spin hota hai) $M=-\ell T\sin \delta$ hai agle box ke frame mein — signs pe note neeche dekho.
• Engine 2 — small-angle torque magnitude (linearised form jo autopilot use karta hai): $|M|\approx \ell T\delta$, jahan $\delta$ radians mein ho.

Signed vs magnitude — yeh ek baar padho: zyaadatar examples magnitude $|M|=\ell T\sin \delta$ quote karte hain (kitna hard ghoomta hai). Sirf jab direction matter karta hai (Example 4) tab hum cross product ka minus sign carry karte hain, jo signed $M=-\ell T\sin \delta$ deta hai. Physics same hai; sign bas batata hai nose kis taraf jaata hai.

Teen helper results (ye sab engines / basic geometry se aate hain, naya physics nahi):

• Axial (useful, forward) thrust: $T_{\text{axial}}=T\cos \delta$ — usi tilt ka bacha hua hissa.
• Steering mein gayi fractional thrust: $1-\cos \delta \approx {\delta }^2/2$ — upar wali line ka Taylor expansion.
• Dual-gimbal total tilt: ${\delta }_{\text{tot}}=\sqrt{{\delta }_p^2+{\delta }_y^2}$ — do perpendicular tilts par Pythagoras (do angles ${\delta }_p,{\delta }_y$ neeche box mein define hain).

Yahan $T$ thrust magnitude hai, $\ell$ moment arm hai (CoM-to-gimbal distance), aur $\delta$ tilt angle hai. Sab kuch parent note mein scratch se banaya gaya hai.

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Is page par use hone wala Sign convention

Neeche sab kuch ek fixed body frame use karta hai taaki signs kabhi surprise na karein:

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Scenario matrix

Is table ko ek checklist ki tarah socho. Har row ek class ka problem hai jo alag behave karta hai. Last column us example ka naam hai jo ise cover karta hai — jab khatam karo, har cell tick ho jaani chahiye.

#	Case class	Kya khaas hai	Covered by
A	Small angle, forward	$\sin \delta \approx \delta$ valid; linearly solve karo	Example 1
B	Large angle (>10°)	exact $\sin \delta$ use karna padega, $\delta$ nahi	Example 2
C	$\delta =0$ (degenerate)	zero torque, max thrust — baseline	Example 3
D	Sign / direction	left vs right steer, torque ka sign	Example 4
E	Dual-gimbal feasible	vector sum limit ke andar	Example 5
F	Dual-gimbal clipped	vector sum limit se zyaada → scale down	Example 6
G	Limiting value	mechanical hard stop par thrust loss	Example 7
H	Real-world word problem	dead engine (T→0), kya toot ta hai	Example 8
I	Exam twist	degrees-vs-radians trap; roll authority = 0	Example 9

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Example 1 — Case A: small angle, linearly solve karo

Forecast: andaaza lagao — kya $\delta$ degree ka fraction hoga, kuch degrees, ya tens of degrees?

1. Torque needed: $M=I\dot{\omega }=1.5\times 10^5\times 0.4=6.0\times 10^4\text{N⋅m}$. Yeh step kyun? Newton ka rotational law $M=I\dot{\omega }$ chahiye (spin-up rate) ko cause (torque) se jodta hai. Rigid Body Rotational Dynamics dekho.
2. Small-angle law invert karo: $\delta =\frac{M}{\ell T}=\frac{6.0\times 10^4}{5\times 1.0\times 10^6}=0.012\text{rad}$. Yeh step kyun? Hum assume karte hain small angle hai isliye torque linear hai, ek clean division milta hai. Assumption ko Verify mein justify karenge.
3. Degrees mein convert karo: $\delta =0.012\times \frac{180}{\pi }\approx 0.687^{\circ }$. Yeh step kyun? Engineers gimbal limits degrees mein padhte hain; yeh hume typical $\pm 8^{\circ }$ range ke against sanity-check karne deta hai.

Verify: $0.012$ rad bahut chhota hai, isliye small-angle khoobsoorti se hold karta hai ($\sin 0.012=0.0119997$, error $<0.003\%$). Back plug karo: $\ell T\sin \delta =5\times 10^6\times \sin (0.012)=5.9998\times 10^4\approx M$. ✓ Units: poore mein $\text{N⋅m}$.

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Example 2 — Case B: large angle, exact law zaroori

Forecast: 20° par, kya small-angle formula <1%, ~2%, ya >5% se galat hai?

1. Radians mein convert karo: $20^{\circ }=20\times \frac{\pi }{180}=0.3491\text{rad}$. Kyun? $\sin$ aur linear form dono ko fair comparison ke liye radians chahiye.
2. Exact torque: $M_{\text{exact}}=\ell T\sin 20^{\circ }=5\times 10^6\times 0.3420=1.710\times 10^6\text{N⋅m}$. Kyun? ~10° se aage approximation $\sin \delta \approx \delta$ toot jaati hai; sach mein transverse thrust $T\sin \delta$ hai.
3. Linear estimate: $M_{\text{lin}}=\ell T\delta =5\times 10^6\times 0.3491=1.745\times 10^6\text{N⋅m}$. Kyun? Yeh woh hai jo ek naive autopilot linear law use karta hua sochega ki usne command kiya.
4. Error: $\frac{M_{\text{lin}}-M_{\text{exact}}}{M_{\text{exact}}}=\frac{1.745-1.710}{1.710}\approx 2.0\%$.

Verify: linear form hamesha over-estimate karta hai kyunki $\delta >\sin \delta$. 20° par 2% error takeaway hai — real-world $8^{\circ }$ par same error $0.4\%$ se kam hogi. Units ratio mein cancel ho jaate hain. ✓

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Example 3 — Case C: degenerate baseline, $\delta =0$

Forecast: kaun max hai aur kaun zero?

1. Torque: $M=\ell T\sin 0=0\text{N⋅m}$. Kyun? $\sin 0=0$: koi tilt nahi matlab koi transverse force nahi, isliye koi moment nahi. Thrust line seedhi... actually gimbal se hoke, phir $-x_b$ along jaati hai — yeh CoM se zero lever ke saath pass hoti hai, isliye koi turning nahi.
2. Axial thrust: $T_{\text{axial}}=T\cos 0=1.0\times 10^6\text{N}$ — poora thrust. Kyun? $\cos 0=1$: kuch bhi sideways nahi gira. Yeh fuel-efficient cruise state hai.

Verify: $\delta =0$ exactly woh state hai jo parent ka Mistake #1 highlight karta hai: forward thrust yahan maximize hota hai, $\delta$ badhane se nahi. Torque $0$, thrust $100\%$. ✓

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Example 4 — Case D: torque ka sign / direction

Forecast: same size, opposite sign — ya kuch aur?

Neeche figure "Sign convention" box ka fixed body frame draw karta hai: white line vehicle axis $x_b$ hai, yellow dot CoM hai, blue dot $(-\ell ,0)$ par gimbal pivot hai. Red thrust arrow $+\delta$ case hai (exhaust upar-aur-peeche pheka, isliye $F_y>0$) aur uska red curved arrow nose neeche pitching dikhata hai. Green thrust arrow $-\delta$ case hai, mirror-imaged, ek green curved arrow ke saath jo nose upar pitching dikhata hai. Do curved arrows padho: same size, opposite spin.

1. Dono cases mein magnitude: $|M|=\ell T\sin 4^{\circ }=6\times 8\times 10^5\times 0.06976=3.348\times 10^5\text{N⋅m}$. Kyun? $|\delta |$ ka $\sin$ size set karta hai; do deflections mirror images hain.
2. Signs (convention box se): $\vec{r}=(-\ell ,0)$ ke saath, planar torque $M=-\ell T\sin \delta$ hai. $+\delta$ se $F_y=T\sin (+\delta )>0$ milta hai, isliye $M<0$ → nose-down (red). $-\delta$ se $F_y$ flip hota hai, isliye $M>0$ → nose-up (green). Kyun? Sign encode karta hai kaun si taraf spin hoga — ise figure mein red vs green curved arrows se match karo.
3. Conclusion: equal-and-opposite torques, $\mp 3.348\times 10^5\text{N⋅m}$.

Verify: $\sin (-4^{\circ })=-\sin (4^{\circ })$, isliye torques exact negatives hain. Yahi reason hai ki ek single gimbal ek actuator se dono directions mein pitch command kar sakta hai. ✓

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Example 5 — Case E: dual-gimbal, feasible command

Forecast: $6+8=14>12$ — kya isliye infeasible hai? (Dhyaan raho!)

Figure command ko ek deflection plane mein point ki tarah plot karta hai: blue arrow yaw leg ${\delta }_y=8^{\circ }$ hai (horizontal), yellow arrow pitch leg ${\delta }_p=6^{\circ }$ hai (vertical, right angles par), aur red arrow unka vector sum hai — true nozzle tilt. Green dashed quarter-circle mechanical limit hai radius $12^{\circ }$ par; command point iske andar hai, isliye reachable hai.

1. Vector ki tarah combine karo, sum ki tarah nahi: ${\delta }_{\text{tot}}=\sqrt{6^2+8^2}=\sqrt{100}=10^{\circ }$. Kyun? Pitch aur yaw tilts nozzle ke ek physical cone-tilt ke perpendicular components hain (do legs right angle par milte hain, hypotenuse true tilt hai). Ye Pythagoras se jodtey hain, linearly nahi.
2. Limit se compare karo: $10^{\circ }<12^{\circ }$ → feasible. ✓
3. Trap note karo: naive linear sum $14^{\circ }$ ek bilkul legal command ko galat reject kar deta.

Verify: $6$-$8$-$10$ ek scaled 3-4-5 triangle hai, isliye ${\delta }_{\text{tot}}=10^{\circ }$ exactly, $12^{\circ }$ se comfortably kam. ✓

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Example 6 — Case F: dual-gimbal, clipped command

Forecast: legal hai ya scale karna padega?

1. Total tilt: ${\delta }_{\text{tot}}=\sqrt{10^2+10^2}=\sqrt{200}=14.142^{\circ }$. Kyun? Same vector-sum reason jaise Example 5 mein.
2. Check karo: $14.142^{\circ }>12^{\circ }$ → illegal, gimbal itna door nahi pahunch sakta. Kyun? Nozzle physically apne bearing stop se $12^{\circ }$ tilt par takraa jaata hai.
3. Direction rakhte hue clip karo: dono ko $k=\frac{12}{14.142}=0.8485$ se scale karo. ${\delta }_p^{\prime }=10\times 0.8485=8.485^{\circ }$, ${\delta }_y^{\prime }=8.485^{\circ }$. Kyun? Hum tilt vector ko length $12^{\circ }$ tak shrink karte hain par uska angle rakhte hain, isliye commanded torque ka direction preserve hota hai — sirf magnitude cap hoti hai. Yeh standard autopilot saturation strategy hai.

Verify: $\sqrt{8.485^2+8.485^2}=12.000^{\circ }$ ✓ (exactly limit par land karta hai), aur ratio ${\delta }_p^{\prime }/{\delta }_y^{\prime }=1$ preserve hai, isliye torque direction unchanged hai. ✓

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Example 7 — Case G: limiting value, hard stop par thrust loss

Forecast: loss ~0.5%, 2%, ya 10% ke around?

1. Fractional axial loss: $1-\cos 12^{\circ }=1-0.97815=0.02185=2.19\%$. Kyun? Forward thrust $T\cos \delta$ hai; missing piece woh hai jo tilting cost karta hai. Loss $\delta$ mein quadratic hai (parent §3).
2. Side force: $T\sin 12^{\circ }=1.0\times 10^6\times 0.20791=2.079\times 10^5\text{N}=207.9$ kN. Kyun? Magnitude $T$ ka thrust vector, $\delta$ se tilt hoke, ek axial part $T\cos \delta$ (forward) aur ek transverse part $T\sin \delta$ (sideways) mein split hota hai. Woh transverse component hi steering force hai — yeh simply ek thrust vector ko decompose karne ka perpendicular leg hai.
3. Trade read-out: ~2.2% forward loss ~208 kN sideways authority kharidti hai.

Verify: Taylor check ${\delta }^2/2$ ke saath $12^{\circ }=0.2094$ rad deta hai $0.02193$ — $2.19\%$ se do figures tak match karta hai, confirming ki loss genuinely second-order hai. ✓ Units: poore mein N.

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Example 8 — Case H: real-world word problem, dead-engine degeneracy

Forecast: kya torque half ho jaata hai, ya bilkul khatam?

1. Shutdown se pehle torque: $M=\ell T\sin 5^{\circ }=6\times 8\times 10^5\times 0.08716=4.184\times 10^5\text{N⋅m}$. Kyun? Engine live hone par baseline steering capability.
2. Shutdown ke baad torque: $M=\ell \cdot 0\cdot \sin 5^{\circ }=0\text{N⋅m}$. Kyun? TVC torque thrust ke proportional hota hai — koi thrust nahi matlab gimbal ek bekar lever hai, chahe kitna bhi tilt kiya ho. Yeh degenerate $T=0$ case hai.
3. Kya take over karta hai: koi thrust nahi hone par, sirf Reaction Control System (RCS) (cold-gas/monopropellant thrusters) attitude torque generate kar sakta hai. TVC aur RCS exactly isi reason ke liye complementary hain.

Verify: $M\propto T$, isliye $T=0\Rightarrow M=0$ regardless of $\delta$; $4.184\times 10^5\to 0$. Yeh bhi note karo ki Center of Mass Migration TVC ke liye matter nahi karta jab thrust chali jaaye — arm $\ell$ zero force ko multiply karta hai, isliye irrelevant hai. ✓

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Example 9 — Case I: exam twist (units + roll)

Forecast: kya degree-plugging error ek chhoti si galti hai ya bahut bada factor?

1. Galat (radian formula mein degrees): $M_{\text{wrong}}=6\times 8\times 10^5\times 5=2.4\times 10^7\text{N⋅m}$. Yeh kyun dikhate hain? Trap expose karne ke liye: linear law ko $\delta$ radians mein chahiye.
2. Correct linear law: $5^{\circ }=0.08727$ rad convert karo, phir $M=6\times 8\times 10^5\times 0.08727=4.189\times 10^5\text{N⋅m}$. Kyun? $M\approx \ell T\delta$ $\sin \delta \approx \delta$ se derive hota hai, aur woh approximation sirf tab hold karta hai jab $\delta$ radians mein measure kiya jaaye.
3. Error ka size: galat answer $\frac{180}{\pi }\approx 57.3$ factor se zyaada bada hai. Kyun? Radians→degrees exactly yeh conversion factor hai.
4. Roll authority zero hai: roll ke liye, torque $\vec{r}\times \vec{F}$ roll axis $x_b$ ke baare mein hai. Thrust $(-\ell ,0,0)$ par axis par apply hota hai, aur uska transverse component $y_b$–$z_b$ plane mein hai. Roll ($x_b$) torque ke liye lever arm woh perpendicular distance hai $x_b$ axis se thrust line tak — lekin ek centered engine ki thrust line $x_b$ axis se hoke guzarti hai, isliye woh distance $\approx 0$ hai, jo roll torque $\approx 0$ deta hai. Kyun? Ek axis ke baare mein torque $=$ force $\times$ us axis ki perpendicular distance; yahan distance khatam ho jaati hai. Roll ke liye isliye canted engines, differential gimbaling, ya RCS chahiye.

Verify: $\frac{2.4\times 10^7}{4.189\times 10^5}=57.3=\frac{180}{\pi }$ ✓, confirm karta hai ki galti pure unit error hai. Roll torque $=0$ zero-perpendicular-distance argument se. ✓

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Machine-check summary

Is page par har numeric answer neeche ===VERIFY=== block ke checks se verified hai. Yeh roll-call hai taaki dekh sako kuch skip nahi hua:

Example	Key numbers checked
1	$M=6.0\times 10^4$ N·m, $\delta =0.012$ rad, $0.687^{\circ }$
2	$M_{\text{exact}}=1.710\times 10^6$, $M_{\text{lin}}=1.745\times 10^6$, error $\approx 2\%$
3	$M=0$, $T_{\text{axial}}=T$ at $\delta =0$
4	$
5	${\delta }_{\text{tot}}=10^{\circ }<12^{\circ }$
6	${\delta }_{\text{tot}}=14.142^{\circ }$, $k=0.8485$, clipped legs → $12^{\circ }$
7	loss $=0.0219$, side force $=2.079\times 10^5$ N, Taylor $=0.02193$
8	$M=4.184\times 10^5$ pehle, $0$ baad mein
9	wrong $=2.4\times 10^7$, correct $=4.189\times 10^5$, factor $=57.3$

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Recall Har example ne kaun si cell fill ki? (cover karke recall karo)

Example 1 covers ::: Case A (small angle, linear solve) Example 2 covers ::: Case B (large angle, exact sine) Example 3 covers ::: Case C (δ=0 degenerate baseline) Example 4 covers ::: Case D (sign / steer direction) Example 5 covers ::: Case E (dual-gimbal feasible) Example 6 covers ::: Case F (dual-gimbal clipped) Example 7 covers ::: Case G (thrust loss at the limit) Example 8 covers ::: Case H (dead engine, T→0) Example 9 covers ::: Case I (degrees/radians trap + zero roll)