Propulsive forces — thrust misalignment, gimbal angle: Worked examples

3.4.12 · D3 · Physics › Rocket Flight Mechanics › Propulsive forces — thrust misalignment, gimbal angle

Yeh page parent topic ka drill hall hai. Hum gimbal torque law ko har direction se attack karenge: positive tilt, negative tilt, zero tilt, straight-back offset, woh degenerate case jahan engine bilkul center of mass par baitha ho, ek huge tilt ka limiting case, aur do "real-life" curveballs. Yahan kuch bhi aisa symbol use nahi hoga jo parent note mein earn nahi kiya — lekin shuru karne se pehle un teen symbols ko re-anchor karte hain jo sabse zyada matter karte hain, aur apna coordinate frame pin down karte hain.

Woh ek formula jo hum exercise karte hain, uski full aur small-angle form mein:

The scenario matrix

Har gimbal problem neeche ke cells mein se ek hai. Jo examples follow karte hain woh us cell ke saath labelled hain jo woh cover karte hain, isliye end tak aap ne poora board dekh liya hoga. (Cell D mein use hone wala offset symbol $d$ upar define kiya gaya hai.)

Cell	Situation	Sign / regime	Kya test karta hai
A	Positive tilt $δ > 0$	$τ < 0$ (nose ek taraf)	Basic torque + direction
B	Negative tilt $δ < 0$	$τ > 0$ (nose doosri taraf)	Sign symmetry
C	Zero tilt $δ = 0$	$τ = 0$	Degenerate: koi steering nahi
D	Lateral offset $d$ , koi tilt nahi	$τ = - d T$ (constant)	Alag geometry
E	Engine CoM par, $L = 0$	$τ = 0$ chahe $δ$ kuch bhi ho	Degenerate lever
F	Large tilt (limit)	$sin δ$ vs $δ$ diverge karte hain	Jahan small-angle breaks
G	Real-world word problem	mixed	Time-to-rate, propellant cost
H	Exam twist: burn ke beech CoM shift ho	$L$ barhta hai	Changing authority

Aao har cell clear karte hain.

Cell A — Positive tilt, basic torque aur direction

Worked example A · Ek commanded right-tilt

Ek stage mein $T = 500 kN$ , lever arm $L = 15 m$ hai, aur engine $δ = + 2^{\circ}$ se gimballed hai. Torque magnitude nikalo aur batao nose kis taraf swing karta hai.

Forecast: compute karne se pehle guess karo — kya torque $1 0^{5}$ ke kareeb hoga ya $1 0^{6}$ N·m? Agar exhaust $+ y$ (hamari picture mein upar) ki taraf push ho raha ho toh nose kis taraf jaata hai?

Step 1 — Angle convert karo. $δ = 2 \times \frac{π}{180} = 0.03491 rad .$ Yeh step kyun? $sin$ sirf small-angle rule obey karta hai aur sahi numbers deta hai jab input radians mein ho. Degrees silently wrong honge.

Step 2 — Magnitude form mein plug in karo (minus sign jaanboojh ke drop kiya). $∣ τ ∣ = L T sin δ = 15 \times 500000 \times sin (0.03491) .$ Yeh step kyun? Pehle hum raw turning strength chahte hain, isliye $∣ τ ∣ = L T sin δ$ use karte hain — absolute-value bars remind karte hain ki humne sign abhi side pe rakh diya hai aur Step 4 mein restore karenge. Hum exact $sin$ use karte hain, approximation nahi, taaki hamare paas ek reference value ho.

Step 3 — Evaluate karo. $sin (0.03491) = 0.03490 \Rightarrow ∣ τ ∣ = 15 \times 500000 \times 0.03490 \approx 2.62 \times 1 0^{5} N\cdotm .$

Step 4 — Sign restore karo aur direction padho. Signed law deta hai $τ = - L T sin δ = - 2.62 \times 1 0^{5}$ N·m, yani negative → clockwise → nose neeche. Figure dekho: thrust rocket ke back ko sideways $+ y$ ki taraf push karta hai; tail par $+ y$ ki taraf force tail ko upar swing karti hai aur isliye nose neeche, bilkul wahi clockwise sense jo minus sign predict karta hai.

Figure A: rocket ka side view jisme CoM (yellow dot) origin par hai aur pivot (red square) $x = - L$ par. Yellow double arrow lever $L$ mark karta hai. Green arrow tilted thrust $T$ hai; uska chota red arrow sideways component $T sin δ$ hai jo tail ko $+ y$ ki taraf push karta hai. Pivot par yellow arc gimbal angle $δ$ hai jo dashed axial reference line se measure kiya gaya hai.

Verify: small-angle cross-check $L T δ = 15 \times 500000 \times 0.03491 = 2.618 \times 1 0^{5}$ N·m — Step 3 se 0.01% ke andar agree karta hai, confirming karta hai ki angle itna chhota hai. Units: $m \times N = N\cdotm$ . ✓

Cell B — Negative tilt, sign symmetry

Worked example B · Mirror-image tilt

Same stage ( $T = 500$ kN, $L = 15$ m) lekin ab engine doosri taraf tilt hai, $δ = - 2^{\circ}$ . Cell A ke saath signed torque compare karo.

Forecast: kya magnitude bilkul change hogi? Nose same taraf jaayega ya opposite?

Step 1 — Sine ki odd symmetry use karo. $sin (- δ) = - sin (δ) .$ Yeh step kyun? Scratch se recompute karne ki jagah, hum exploit karte hain ki $sin$ ek odd function hai — tilt flip karne se torque ka sign flip hota hai lekin size nahi. Isliye ek actuator left aur right dono taraf steer kar sakta hai.

Step 2 — Signed torque. $τ = - L T sin (- 0.03491) = + L T sin (0.03491) = + 15 \times 500000 \times 0.03490 \approx + 2.62 \times 1 0^{5} N\cdotm .$ Yeh step kyun? Hum master law ka minus sign aur negative input dono rakhte hain, isliye do negatives multiply hokar positive signed torque dete hain — yahi cell ka poora point hai: Cell A ke against ek genuine sign flip, assert kiya nahi balki compute kiya gaya.

Step 3 — Direction. Hamari convention ke mutabik positive $τ$ matlab counter-clockwise = nose upar, bilkul Cell A ke nose-down ka mirror. Yeh step kyun? Number akela meaningless hai jab tak top par fix ki gayi sign convention ke through physical rotation mein map na ho — wahi mapping hai jo " $+ 2.62 \times 1 0^{5}$ " ko ek steering command mein turn karti hai.

Verify: magnitude Cell A ke identical ( $2.62 \times 1 0^{5}$ N·m), sign reversed. ✓ Symmetry matlab controller ko sirf magnitude table aur ek sign bit chahiye.

Cell C — Zero tilt, degenerate "no steering" case

Cell D — Lateral offset, no tilt (different geometry)

Worked example D · Engine jo seedha hai lekin shifted hai

Engine bilkul seedha point kar raha hai ( $δ = 0$ ) lekin uski thrust line CoM se sideways $+ y$ direction mein $d = 0.10 m$ offset hai (ek mounting error). $T = 500$ kN. Signed torque nikalo aur batao nose kis taraf turn karta hai.

Forecast: $δ = 0$ ke saath humne abhi kaha $τ = 0$ . Kya yeh yahan bhi sach hai? Kya change hua?

Step 1 — Recognise karo ki yeh ek different geometry hai. Yahan poora (axial) thrust vector $+ x$ ki taraf point karta hai lekin uski line of action $y = + d$ par shifted hai, isliye woh CoM ko miss karta hai. Lever arm ab perpendicular distance $d$ hai, aur poori force $T$ uske perpendicular act karti hai. Yeh step kyun? Torque hai (perpendicular distance) × (force). Cell A mein perpendicular distance $L$ tha aur sirf lateral component $T sin δ$ matter karta tha. Yahan perpendicular distance $d$ hai aur poora $T$ act karta hai.

Step 2 — 2D cross product kya hai yeh recall karo, phir components name karo. Torque ek force ka turning effect hai, cross product $τ = r \times F$ ke roop mein likha jaata hai, jahan $r$ CoM se force apply hone ki jagah tak ka position vector hai aur $F$ force hai. Hamare flat, two-dimensional picture mein (sab kuch $x$ – $y$ plane mein hai) yeh cross product ek single signed number mein collapse ho jaata hai: $τ = r_{x} F_{y} - r_{y} F_{x},$ jahan $r_{x}, r_{y}$ $r$ ke $x$ - aur $y$ -parts hain aur $F_{x}, F_{y}$ $F$ ke $x$ - aur $y$ -parts hain. Is scenario se padhte hain: force $r = (0, + d)$ par apply hoti hai (CoM ke seedha upar offset $d$ se, koi $x$ -shift nahi kyunki thrust line wahan CoM plane cross karti hai), isliye $r_{x} = 0, r_{y} = + d$ . Force purely axial hai, $F = (T, 0)$ , isliye $F_{x} = T, F_{y} = 0$ . Yeh step kyun? Use karne se pehle formula earn karna padega — $τ = r_{x} F_{y} - r_{y} F_{x}$ sirf full $r \times F$ ka 2D shadow hai, aur charon components ko pin down karna hi allow karta hai ki hum sign guess kiye bina numbers plug in kar sakein.

Step 3 — Signed torque. Charon components substitute karte hain: $τ = r_{x} F_{y} - r_{y} F_{x} = (0) (0) - (d) (T) = - d T = - 0.10 \times 500000 = - 5.0 \times 1 0^{4} N\cdotm .$ Yeh step kyun? Hum sign rakhte hain taaki rotation direction state kar sakein, sirf size nahi.

Step 4 — Direction aur contrast. Negative $τ$ → clockwise → nose neeche: CoM ke upar apply hua ek forward push nose ko neeche tip karta hai, bilkul jaise khade pencil ke top ko push karne se woh aapke se door gir jaata hai. Koi axial loss nahi ( $cos 0 = 1$ ), phir bhi ek real, constant disturbance torque exist karta hai. Isliye offset ( $d$ ) aur tilt ( $δ$ ) do alag failure modes hain. Yeh step kyun? Nonzero torque tabhi useful (ya dangerous) hai jab hum jaanein ki woh vehicle ko kis taraf rotate karta hai — warna controller correcting sign choose nahi kar sakta.

Verify: units $m \times N = N\cdotm$ ; magnitude $5.0 \times 1 0^{4}$ N·m. ✓ Note karo ki isme $sin$ use nahi hota — ek classic trap hai.

Cell E — Engine exactly CoM par (degenerate lever)

Cell F — Large tilt: jahan small-angle rule breaks (aur $δ$ ka full domain)

Worked example F · Linear regime se aage jaana

Teen tilts par exact torque magnitude aur small-angle estimate compare karo: $δ = 5^{\circ}, 2 0^{\circ}, 4 5^{\circ}$ . Use karo $L = 15$ m, $T = 500$ kN. Axial thrust loss bhi report karo. Aakhir mein note karo ki signed law poore domain $0 \leq δ \leq 2 π$ mein kya karta hai.

Forecast: kis angle par cheap approximation $sin δ \approx δ$ 1% se zyada jhooth bolne lagti hai?

Step 1 — Exact vs approximate tabulate karo. Har $δ$ ke liye (radians mein) $sin δ$ aur $δ$ compute karo, aur fractional error $(δ - sin δ) / sin δ$ .

$δ$	$δ$ (rad)	$sin δ$	error
$5^{\circ}$	$0.08727$	$0.08716$	$0.13%$
$2 0^{\circ}$	$0.34907$	$0.34202$	$2.06%$
$4 5^{\circ}$	$0.78540$	$0.70711$	$11.1%$

Yeh step kyun? Yeh dikhata hai kyun real gimbals ~10° se neeche rehte hain: linear control math (torque $\propto δ$ ) wahan trustworthy hai aur aage dangerously optimistic ho jaata hai.

Step 2 — Axial loss $4 5^{\circ}$ par: $loss = T (1 - cos 4 5^{\circ}) = 500000 \times (1 - 0.70711) = 1.46 \times 1 0^{5}$ N ≈ 29% thrust gone. $5^{\circ}$ par yeh sirf $T (1 - cos 5^{\circ}) = 1.90 \times 1 0^{3}$ N ≈ 0.38% hai. Yeh step kyun? Steering torque $sin δ$ ki tarah grow karta hai (first order) lekin forward-thrust jo aap sacrifice karte ho $1 - cos δ$ ki tarah grow karta hai (second order); same angle par dono numbers compare karna hi ek tarika hai yeh dekhne ka ki bada tilt extra turning buy karta hai lekin suddenly-steep fuel-efficiency price par — woh trade-off jo hardware limit set karta hai.

Step 3 — $δ$ ka full domain. Signed law $τ = - L T sin δ$ kisi bhi angle ke liye hold karta hai, sirf chhote walo ke liye nahi. Kyunki $sin$ periodic hai: torque $δ = π$ par phir zero ho jaata hai (engine seedha forward point kar raha hai — physically real nozzle ke liye impossible lekin ek valid limit), yeh $π < δ < 2 π$ ke liye sign reverse karta hai (lateral component ab $- y$ ki taraf point karta hai), aur har $2 π$ pe repeat hota hai. Peak torque magnitude $δ = π /2$ ( $9 0^{\circ}$ ) par hoti hai, jahan thrust entirely lateral hai aur axial thrust zero hai. Real hardware sirf $δ = 0$ ke pass chhote sliver mein rehta hai, lekin formula khud global hai. Yeh step kyun? Ek "drill hall" koi scenario uncovered nahi chhod sakta; ek reader jo problem mein $δ > 9 0^{\circ}$ se milta hai use pata hona chahiye ki same formula apply hota hai, sirf abhi describe kiye sign flips ke saath.

Step 4 — Graph padho. Figure dikhata hai $sin δ$ (curve) origin ke paas straight line $δ$ se chipki hui aur phir door ho jaati hai — visual reason ki approximation "chhote tilt par cheap, bade tilt par wrong" hai. Yeh step kyun? Teen numbers ki ek table bina samjhe memorize ki ja sakti hai; yeh dekhna ki do curves origin par touch karte hain aur phir visibly alag ho jaate hain — yeh "error angle ke saath barhta hai" ko ek aisi intuition mein turn kar deta hai jo aap reconstruct kar sako, taaki aap linear rule ko kabhi bhi at a glance tab tak trust na karo jab tak lines ek doosre ke upar na hon.

Figure F: gimbal angle ke against degrees mein do curves — blue $sin δ$ (exact) aur dashed yellow straight line $δ$ (small-angle estimate). $5^{\circ}$ , $2 0^{\circ}$ , $4 5^{\circ}$ par coloured stems unke beech vertical gap mark karte hain; gap $5^{\circ}$ (green) par invisible hai, $2 0^{\circ}$ par chhota hai, aur $4 5^{\circ}$ (red) par ek bada 11% hai, exactly dikhata hai kahan linear rule trustworthy rehna band karta hai.

Verify: $sin 2 0^{\circ} = 0.34202$ , error $2.06%$ ; $cos 4 5^{\circ} = 0.70711$ , loss fraction $0.29289$ ; $sin π = 0$ second zero-torque point confirm karta hai. ✓

Cell G — Real-world word problem

Worked example G · Gimbal kitni der tak hold kar sakte hain?

Ek stage apna attitude correct karne ke liye $δ = 3^{\circ}$ hold karta hai. Data: $T = 800$ kN, $L = 20$ m, pitch inertia $I = 3 \times 1 0^{6} kg\cdotm^{2}$ . Guidance limit ek pitch rate of $1 0^{\circ} / s$ hai. Rest se shuru karke, gimbal kitni der tak held reh sakta hai is rate tak pahunchne se pehle?

Forecast: seconds, ya tens of seconds?

Step 1 — Torque magnitude. $δ = 3^{\circ} = 0.05236$ rad. $∣ τ ∣ = L T sin δ = 20 \times 800000 \times sin (0.05236) = 20 \times 800000 \times 0.05234 = 8.37 \times 1 0^{5} N\cdotm .$ Yeh step kyun? Hume sirf turning effect ka size chahiye yeh jaanne ke liye ki rate kitni tezi se build hoti hai, isliye hum magnitude form $∣ τ ∣ = L T sin δ$ use karte hain (minus sign jaanboojh ke drop kiya); direction time ko affect nahi karta kisi given rate tak pahunchne mein.

Step 2 — Angular acceleration. $\ddot{θ} = \frac{∣ τ ∣}{I} = \frac{8.37 \times 1 0 ^{5}}{3 \times 1 0 ^{6}} = 0.2791 rad/s^{2} .$ Yeh step kyun? Attitude $τ = I \ddot{θ}$ ke zariye evolve hota hai; torque ko inertia se divide karna force-world se rotation-world ka bridge hai, aur yeh ek hi tarika hai newton-metres ko turning ki rate mein convert karne ka.

Step 3 — Rate limit ko radians mein convert karo. $1 0^{\circ} / s = 0.1745 rad/s$ . Yeh step kyun? $\ddot{θ}$ rad/s² mein nikla; target rate same radian units mein hona chahiye warna Step 4 mein division incompatible quantities mix kar degi.

Step 4 — Rest se time ( $\dot{θ} = \ddot{θ} t$ ): $t = \frac{θ ˙ _{m a x}}{θ ¨} = \frac{0.1745}{0.2791} = 0.625 s .$ Yeh step kyun? Rest se constant acceleration ke saath, rate linearly time ke saath barhti hai, isliye rate limit hit karne ka time sirf limit divided by acceleration hai — fastest possible check ki ek held gimbal safe hai.

Verify: $0.625$ s mein rate $0.2791 \times 0.625 = 0.1745$ rad/s $= 1 0^{\circ} / s$ tak pahunchti hai. ✓ Isliye real gimbal pulses ek second ke fraction tak rehte hain — authority enormous hai.

Cell H — Exam twist: burn ke dauran CoM shift ho jaata hai

Worked example H · Same command, barhta hua lever

Burn ke shuru mein CoM-to-pivot lever $L_{1} = 18 m$ hai; burn ke ant mein, propellant forward drain hone ke baad, yeh $L_{2} = 24 m$ hai. Gimbal command aur thrust unchanged hain: $δ = 2^{\circ}$ , $T = 600$ kN. Control torque kitne factor se barhta hai, aur har value kya hai?

Forecast: compute karne se pehle numbers se ratio $τ_{2} / τ_{1}$ guess karo.

Step 1 — Note karo kya change hota hai. Sirf $L$ change hota hai; $T$ aur $δ$ fixed hain, isliye torque linearly $L$ ke saath scale karta hai: $∣ τ ∣ \propto L$ . Yeh step kyun? $∣ τ ∣ = L T sin δ$ mein, teen factors mein se do frozen hain toh ratio $L_{2} / L_{1}$ tak collapse ho jaata hai — ise recognize karna trig redo karne se bachata hai aur underlying proportionality expose karta hai jo exam actually test kar raha hai.

Step 2 — Early torque magnitude. $δ = 2^{\circ} = 0.03491$ rad, $sin δ = 0.03490$ . $∣ τ_{1} ∣ = 18 \times 600000 \times 0.03490 = 3.769 \times 1 0^{5} N\cdotm .$ Yeh step kyun? Pehle ek concrete anchor value compute karte hain taaki Step 4 ka ratio ke peeche physical numbers hon, sirf ek abstract fraction nahi — aur taaki units sanity-check kar sakein (N·m).

Step 3 — Late torque magnitude. $∣ τ_{2} ∣ = 24 \times 600000 \times 0.03490 = 5.026 \times 1 0^{5} N\cdotm .$ Yeh step kyun? Late-burn case explicitly evaluate karna, same $T$ aur $δ$ ke saath lekin bade $L$ ke saath, CoM shift ko change ka sole cause isolate karta hai — twist ka pedagogical heart.

Step 4 — Ratio. $\frac{∣ τ _{2} ∣}{∣ τ _{1} ∣} = \frac{24}{18} = 1.333.$ Iska importance: same stick-deflection burn ke ant mein 33% zyada turning produce karta hai. Controller ko time ke saath gimbal command reduce karna padega warna woh over-rotate karega — yeh CoM-shift problem ek line mein.

Verify: $∣ τ_{2} ∣/∣ τ_{1} ∣ = 24/18 = 4/3 = 1.3333$ ; $∣ τ_{1} ∣ = 3.769 \times 1 0^{5}$ , $∣ τ_{2} ∣ = 5.026 \times 1 0^{5}$ N·m. ✓

Recall check

Recall Main kaunse cell mein hoon?

Engine seedha point kar raha hai lekin CoM ke side mein 8 cm pe bolted hai — kaunsa formula? ::: Cell D, lateral offset: $τ = - d T$ (koi $sin$ nahi, koi $L$ nahi); sign rotation direction deta hai. Same gimbal command flight ke ant mein zyada torque deta hai — kyun? ::: Cell H: CoM aage move karta hai, lever $L$ barhta hai, aur $∣ τ ∣ \propto L$ . Tilt $+ 4^{\circ}$ ki jagah $- 4^{\circ}$ hai — kya change hota hai? ::: Cell B: same magnitude, torque sign flip hoti hai (sine odd hai) → nose opposite taraf turn karta hai. CoM par exactly mounted engine steer kyun nahi kar sakta? ::: Cell E: $L = 0$ , isliye $τ = L T sin δ = 0$ kisi bhi tilt ke liye — pure translation. $4 5^{\circ}$ tilt par roughly kitna thrust lost hota hai? ::: Cell F: $T (1 - cos 4 5^{\circ}) \approx 29%$ — isliye bade tilts avoid kiye jaate hain. $δ = 0$ ke alawa torque aur kahan zero hota hai? ::: Cell F: $δ = π$ par (engine seedha aage); phir sign $π < δ < 2 π$ ke liye reverse ho jaata hai.

Yeh bhi dekho: Newton's Third Law · Thrust Vector Control (TVC) · Rocket Equation