3.5.47 · D3Attitude control modes — spin stabilization, 3-axis active
Intuition Yeh page kya haiParent note Attitude control modes — spin stabilization, 3-axis active ne tumhe machinery di thi: Euler's equations , gyroscopic precession, reaction wheels, aur PID control. Yahan hum ise drill karte hain. Hum ek scenario matrix banate hain — har tarah ke problem ki ek checklist — phir har cell ka ek example solve karte hain taaki koi bhi exam question surprise na kar sake.
Shuru karne se pehle, symbols ko samajhna zaroori hai taaki neeche ki har line clean lage.
Definition Core quantities jo hum baar baar use karte hain
I kg⋅m 2 I Moment of inertia tensor & principal axes .I 1 , I 2 , I 3 principal moments of inertia ) — jab hum measuring axes ko body ke natural symmetry axes (principal axes ) ke saath align karte hain, to inertia "tensor" sirf teen numbers mein collapse ho jata hai: I 1 I 2 I 3 Moment of inertia tensor & principal axes .ω rad/s 2 π ≈ 6.283 L = I ω kg⋅m 2 / s m v L h kg⋅m 2 / s = N⋅m⋅s reaction wheel ke andar stored hota hai — ek alag spinning flywheel jo body se judi hoti hai. h L h total conserved rahe. Dekho Reaction wheels and Control-Moment Gyros (CMG) .
Definition Rotation angle, uski rates, aurα
θ rad body apne target orientation se kitna door ghoom gayi hai kisi axis ke baare mein. Compass ki needle ka sochte hain: θ θ ˙ rad/s θ ¨ angular acceleration hai, rad/s 2 α α ≡ θ ¨ . α θ ¨ ek hi cheez hain; α
Definition Torque, precessionvector , aur cross product
τ N⋅m twist : push ka rotational version.Newton ka rotational law: τ = d t d L change karta hai.
Ω vector , units rad/s L Ω × L cross product hai: yeh Ω L dono ke perpendicular ek naya vector produce karta hai, jiska size Ω L sin ( angle between them ) Ω L Precession law hai τ = Ω × L L Ω ⊥ L Ω = τ / L Gyroscopic precession .
Definition Control-law symbols (PD examples mein use hote hain)Jab ek spacecraft attitude error θ control law se compute karta hai. Sabse simple hai proportional–derivative (PD): τ = − K p θ − K d θ ˙
K p proportional gain , units N⋅m/rad θ K d derivative gain , units N⋅m⋅s/rad speed θ ˙ ω n = K p / I natural frequency , rad/s ζ = 2 K p I K d damping ratio , dimensionless) — response kitna "cushioned" hai. ζ < 1 ζ = 1 ζ > 1 ζ ≈ 0.7 PID control & second-order systems (damping ratio) .
Is chapter ka har problem in cells mein se ek (ya blend) hai:
#
Cell class
Kya change hota hai / "sign or edge"
Example
A
Gyroscopic stiffness, normal case
positive spin, sideways torque
Ex 1
B
Zero-spin limit ω s → 0 Ex 2
C
Spin-axis choice (max vs min I
kaunsa principal axis stable hai
Ex 3
D
Degenerate inertia (I 1 = I 2 symmetric body, coupling term die ho jata hai
Ex 4
E
Reaction-wheel slew (bang-bang)
accelerate phir brake, sign flip
Ex 5
F
Wheel saturation / desaturation
wheel max RPM hit karta hai, external dump
Ex 6
G
PD control tuning (damping ratio)
choose ζ
Ex 7
H
Real-world word problem
disturbance budget over an orbit
Ex 8
I
Exam twist (sign / quadrant trap)
precession direction by right-hand rule
Ex 9
Hum saare nine solve karte hain taaki har cell filled ho.
Worked example Ex 1 — Ek spinner kitni tezi se drift karta hai?Ek drum satellite ka I = 200 kg⋅m 2 ω s = 10 rad/s τ = 0.02 N⋅m
Forecast: guess karo — kya wo seconds mein tumble karega, ya ghanton mein drift karega? Apna gut number likh lo.
Spin momentum L = I ω s = 200 × 10 = 2000 kg⋅m 2 / s Yeh step kyun? L
Precession rate Ω = L τ = 2000 0.02 = 1 × 1 0 − 5 rad/s Yeh step kyun? Precession law τ = Ω × L τ L τ = Ω L Ω = τ / L L
Convert karo: 1 × 1 0 − 5 rad/s × π 180 × 3600 ≈ 2.06 deg/hour
Verify karo: units kg⋅m 2 / s N⋅m = kg⋅m 2 / s kg⋅m 2 / s 2 = 1/ s 2 deg/hr
Worked example Ex 2 — Agar spin zero ho jaye to kya hoga?Same drum, same torque τ = 0.02 N⋅m ω s = 0 t = 60 s
Forecast: stiffness Ω = τ / L L = 0 toot jaani chahiye. Guess karo: kya wo infinitely tezi se cone karta hai, ya kuch bilkul alag hota hai?
ω s = 0 L = 0 Ω = τ / L → ∞ undefined hai — yeh ab motion describe nahi karta.
Yeh step kyun? Precession assume karta hai ek bada existing L Iske bajay raw law use karo τ = I θ ¨ ω = 0 θ θ ¨ = τ / I = 0.02/200 = 1 × 1 0 − 4 rad/s 2 Yeh step kyun? Zero spin par Euler equation cross-terms ( … ) ω j ω k I ω ˙ = I θ ¨
Yeh simply accelerate aur tumble karta hai: θ = 2 1 θ ¨ t 2 = 2 1 ( 1 × 1 0 − 4 ) ( 60 ) 2 = 0.18 rad ≈ 10.3°
Verify karo: dono regimes spirit mein agree karte hain — bada L L L zaroor 3-axis active control chalana padhta hai.
Worked example Ex 3 — Kaunsa axis dissipation survive karta hai?Ek cylinder ka I a = 50 kg⋅m 2 I t = 120 kg⋅m 2
Figure caption: Cylinder ka ek rectangular side-view. Body ki length ke along padi black dashed double-arrow long (symmetry) axis hai, jiska moment of inertia chhota value I a = 50 kg⋅m 2 red vertical arrow transverse axis hai, jiska moment of inertia bada value I t = 120 kg⋅m 2
Forecast: intuition kehta hai "lamba pencil-jaisa axis sabse zyada stable lagta hai." Likh lo — phir dekho wo galat nikalta hai.
Figure par do candidate axes identify karo: black dashed arrow long (symmetry) axis I a = 50 red vertical arrow transverse axis I t = 120 L T = 2 I L 2 Same L L = 100 Yeh step kyun? Dissipation (slosh, flexing) energy remove karta hai lekin L minimum T
Long axis ke baare mein: T a = 2 ( 50 ) 10 0 2 = 100 J T t = 2 ( 120 ) 10 0 2 = 41.7 J Yeh step kyun? Denominator mein bada I I minimum-energy state hai.
Minimum energy = T t transverse (max-I — red arrow. Wahi ek dissipation-stable choice hai; black dashed long axis unstable hai.
Verify karo: T t = 41.7 < T a = 100 Explorer 1 flat-spin anomaly ek line mein hai: Explorer 1 apne minimum -I Fix rule: I m a x
Worked example Ex 4 — Symmetric body, torque-freeEk perfectly symmetric spinner ke principal moments I 1 = I 2 = 100 kg⋅m 2 I 3 = 60 I 1 , I 2 , I 3 Euler equations use karke, spin-axis rate ω 3
Forecast: kya ω 3
Teesra Euler equation (torque-free): I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2 Yeh step kyun? Axis 3 par gyroscopic coupling difference I 1 − I 2
Degeneracy: I 1 = I 2 ⇒ I 1 − I 2 = 0 ω ˙ 3 = 0 Yeh step kyun? Jab do moments of inertia equal hain, to wo coupling term exactly zero ho jata hai — case class "collapse" ho jaata hai.
Isliye ω 3 = const exactly conserved hai, chahe ω 1 , ω 2
Verify karo: ( I 1 − I 2 ) ω 1 ω 2 = 0 ⋅ ω 1 ω 2 = 0 precession behaviour.
Worked example Ex 5 — 30° turn karne mein kitna time?I = 500 kg⋅m 2 30° τ = 0.1 N⋅m brake (sign reverse) doosre half ke liye taaki wo rest par pahunche. Total slew time aur peak wheel momentum nikalo. Dekho Reaction wheels and Control-Moment Gyros (CMG) .
Forecast: total time nearest 30 s tak guess karo.
Angular acceleration α = θ ¨ = τ / I = 0.1/500 = 2 × 1 0 − 4 rad/s 2 α ≡ θ ¨ Yeh step kyun? Wheel torque body par momentum exchange ke zariye act karta hai: τ body = − h ˙ h
Half-angle = 15° = 0.262 rad θ = 2 1 α t 2 t = 2 θ / α = 2 ( 0.262 ) / ( 2 × 1 0 − 4 ) ≈ 51.2 s Yeh step kyun? Constant-torque kinematics turn angle θ
Total time = 2 × 51.2 ≈ 102.4 s sign flip "bang-bang" hai: maximum push, phir maximum brake.
Peak wheel momentum h = I α t = 500 ( 2 × 1 0 − 4 ) ( 51.2 ) ≈ 5.12 N⋅m⋅s Yeh step kyun? Yeh mid-slew angular momentum hai: body kuch rate tak spin up ho gayi hai, aur conservation se wheel equal-and-opposite h
Verify karo: midpoint par body rate θ ˙ = α t = 2 × 1 0 − 4 × 51.2 = 0.01024 rad/s I θ ˙ = 500 × 0.01024 = 5.12 N⋅m⋅s
Worked example Ex 6 — Momentum dump kab karna hoga?Ex 5 ka wheel maximum storable momentum h m a x = 4 N⋅m⋅s τ d = 5 × 1 0 − 4 N⋅m Momentum dumping / desaturation with magnetorquers .
Forecast: ghante? din? ek orbit?
Wheel momentum rate h ˙ = τ d = 5 × 1 0 − 4 N⋅m⋅s per s h Yeh step kyun? Constant external twist ke against attitude hold karne ke liye, wheel ko wo momentum lete rehna hai (total L = L body + h
Bharne ka time: t sat = h m a x / h ˙ = 4/ ( 5 × 1 0 − 4 ) = 8000 s ≈ 2.22 hours Yeh step kyun? Constant absorption ⇒ h h m a x
Dhyan do Ex 5 ka peak slew momentum 5.12 N⋅m⋅s > 4 N⋅m⋅s already saturate kar deta. Wheel ek one-shot 30° Yeh step kyun? Hamesha slew demand ko rating ke against cross-check karo.
Desaturation: magnetorquers/thrusters fire karo taaki ek external torque apply ho jo wheel ko wapas spin karae (h Yeh step kyun? Sirf ek external torque total L
Verify karo: units N⋅m N⋅m⋅s = s 8000 s /3600 = 2.22 hr
Worked example Ex 7 — Achhi settling ke liye derivative gain choose karoEk single axis I θ ¨ = τ I = 500 kg⋅m 2 θ τ = − K p θ − K d θ ˙ K p = 20 N⋅m/rad ζ = 0.7 K d K p K d ω n ζ PID control & second-order systems (damping ratio) .
Forecast: K d
PD law ko I θ ¨ = τ I θ ¨ = − K p θ − K d θ ˙ I θ ¨ + K d θ ˙ + K p θ = 0 θ damped oscillator , jahan ω n = K p / I ζ = 2 K p I K d Yeh step kyun? Small-angle linearization gyroscopic terms drop kar deta hai; jo bachta hai wo classic second-order system hai jiske do knobs K p , K d ω n ζ
Natural frequency ω n = K p / I = 20/500 = 0.04 = 0.2 rad/s Yeh step kyun? ω n ζ K p , I
ζ = 2 K p I K d K d K d = 2 ζ K p I = 2 ( 0.7 ) 20 × 500 = 1.4 10000 = 1.4 × 100 = 140 N⋅m⋅s/rad Yeh step kyun? ζ = 0.7 Recommendation: K d = 140 N⋅m⋅s/rad K p = 20 ω n = 0.2 rad/s ζ = 0.7
Verify karo: back-substitute: ζ = 2 20 × 500 140 = 2 × 100 140 = 0.7 K d ζ < 1 ζ > 1 0.7
Worked example Ex 8 — Ek orbit mein solar-pressure driftEk 3-axis craft ko ek full orbit T orb = 5400 s τ SRP = 3 × 1 0 − 5 N⋅m h m a x = 0.2 N⋅m⋅s Gravity-gradient & solar-radiation disturbance torques .
Forecast: kya ek orbit ka push wheel ke andar fit hota hai?
Orbit ke dauran accumulated momentum: Δ h = τ SRP T orb = 3 × 1 0 − 5 × 5400 = 0.162 N⋅m⋅s h Yeh step kyun? Constant torque ⇒ momentum linearly badhta hai; body ko still rakhne ke liye wheel sab kuch absorb karta hai.
Capacity se compare karo: 0.162 < 0.2 hold karta hai — bas barely (81% full).
Yeh step kyun? Mission planners is margin ko dhyan mein rakhte hue har orbit mein ek desaturation schedule karte hain.
Margin: 0.2 − 0.162 = 0.038 N⋅m⋅s 0.038/ ( 3 × 1 0 − 5 ) ≈ 1267 s ≈ 21 min
Verify karo: 0.162/0.2 = 0.81 N⋅m × s = N⋅m⋅s ∼ 1 0 − 5 N⋅m
L L + z ^ + x ^ τ = Ω × L Ω direction nikalo (trap: students expect karte hain ki L
Figure caption: Do black arrows givens dikhate hain — L + z ^ τ + x ^ red arrow answer hai: precession vector Ω + y ^ 90° nahi hai.
Forecast: guess karo: kya Ω + y ^ − y ^ + x ^
Figure se known vectors likho: L = L z ^ τ = τ x ^ Ω τ = Ω × L Yeh step kyun? Precession vector is cross-product relation se define hota hai, isliye direction — sirf size nahi — solve karna hoga.
Ek direction guess karo aur right-hand rule se test karo. Try Ω = Ω y ^ Ω × L = Ω L ( y ^ × z ^ ) = Ω L x ^ Yeh step kyun? Right-hand rule deta hai y ^ × z ^ = x ^ + x ^
Yeh exactly + x ^ Ω > 0 Ω + y ^ red arrow . Spin axis + y ^ 90° + x ^ nahi .
Yeh step kyun? Yeh perpendicular response poora gyroscopic surprise hai: spinning top ko sideways push karo aur wo apni push ke right angles par move karta hai.
Magnitude, as always: Ω = τ / L Ω ⊥ L + y ^ τ / L Yeh step kyun? Exam dono chahta hai — number aur correct axis; sirf Ω = ( τ / L ) y ^
Verify karo: cross-product check y ^ × z ^ = x ^ Ω = τ / L 90° Gyroscopic precession ka counter-intuitive heart.
Recall Self-test — answers cover karoZero-spin limit: kya precession formula Ω = τ / L L = 0 τ = I θ ¨ I I I L I 1 = I 2 ω 3 I 1 − I 2 = 0 L h L h α α ≡ θ ¨ rad/s 2 external torque (magnetorquer/thruster) ise dump karta hai.
ζ = K d / ( 2 K p I ) ζ 0.7 90° τ = Ω × L
Mnemonic Matrix yaad rakhne ki ek line"Spin stiff, zero tumbles, max-I ζ = 0.7
Back to parent: Attitude control modes — spin stabilization, 3-axis active .