Attitude control modes — spin stabilization, 3-axis active: Exercises

3.5.47 · D4 · Physics › Guidance, Navigation & Control (GNC) › Attitude control modes — spin stabilization, 3-axis active

Intuition Is page ko kaise use karein

Ye problems ek ladder ki tarah upar chadhte hain. L1 sirf check karta hai ki aap words aur pictures pehchante ho. L5 mein aapko ek puri control system scratch se design karni hai. Har problem ka ek collapsible solution hai — pehle khud try karo, phir check karne ke liye click karo. Jo bhi physics chahiye woh parent note par hai; yahan hum ise use karte hain.

Shuru karne se pehle, har symbol ki ek reminder taaki koi pehli line se hi confused na ho:

$L$ = angular momentum (kisi body mein kitni "spinning motion" hai, direction = spin axis).
$I$ (bold) = the moment-of-inertia tensor — "rotational mass" jo spin ko resist karta hai (dekho Moment of inertia tensor & principal axes). Kisi principal axis ke along yeh ek single number $I$ ban jaata hai.
$ω$ = angular velocity (rad/s, kitni tezi se spin kar raha hai).
$τ$ = torque (ek twisting push).
$Ω$ = precession rate (kitni tezi se spin axis khud slowly ghoomta hai — dekho Gyroscopic precession).

Level 1 — Recognition

L1.1 — Mode ka naam batao

Ek sasta cubesat $5 rad/s$ par ek axis ke baare mein spin kiya jaata hai aur phir poori mission ke liye bina kisi active correction ke akela chhod diya jaata hai. Attitude-control mode ka naam batao, aur pointing ki sabse badi ek limitation bhi batao jo tum is mode mein face karte ho.

Recall Solution

Mode: spin stabilization — yeh stored angular momentum $L$ ko passive "gyroscopic stiffness" ke liye use karta hai, koi active actuators fire nahi hote. Limitation: tum sirf spin axis ko freely point kar sakte ho. Baaki do axes ko independently aim nahi kiya ja sakta — ek side face par laga camera sirf sky ko sweep karta hai jab craft rotate karta hai.

L1.2 — Euler equation padhna

Parent note se ek Euler equation dekho: $τ_{1} = I_{1} \overset{ω}{˙}_{1} + (I_{3} - I_{2}) ω_{2} ω_{3}$ Simple words mein batao ki term $(I_{3} - I_{2}) ω_{2} ω_{3}$ kya represent karta hai, aur yeh kab zero ho jaata hai?

Recall Solution

Yeh gyroscopic coupling term hai — woh "free" twisting jo axis 1 par sirf isliye appear hoti hai kyunki body already axes 2 aur 3 ke baare mein spin kar rahi hai. Isme koi actuator effort nahi lagta; rotation khud hi ise generate karta hai. Yeh tab vanish hota hai jab ya to $ω_{2} = 0$ ya $ω_{3} = 0$ (baaki do axes par simultaneously spin nahi), ya jab $I_{2} = I_{3}$ (woh do inertias equal hain, toh unka difference zero hai).

Level 2 — Application

L2.1 — Gyroscopic stiffness

Ek spinning satellite ka $I = 150 kg\cdotm^{2}$ hai aur woh $ω_{s} = 8 rad/s$ par spin karta hai. Usp par ek steady transverse disturbance torque $τ = 0.015 N\cdotm$ act karta hai. Precession rate $Ω$ rad/s mein nikalo.

Recall Solution

Step 1 — spin angular momentum. $L = I ω_{s} = 150 \times 8 = 1200 kg\cdotm^{2} / s$ . Kyun: $L$ "stiffness reservoir" hai; jitna bada hoga, axis ko swing karna utna hi mushkil hoga. Step 2 — precession from $τ = Ω \times L$ . $Ω = \frac{τ}{L} = \frac{0.015}{1200} = 1.25 \times 1 0^{- 5} rad/s .$ Yeh lagbhag $0.0007°/ s$ hai — axis bahut dheere bhatkta hai, yahi pura point hai.

L2.2 — Spin axis choose karna

Ek rocket-body-shaped satellite ka $I_{long} = 40 kg\cdotm^{2}$ hai apni long symmetry axis ke baare mein aur $I_{trans} = 95 kg\cdotm^{2}$ hai ek transverse axis ke baare mein. Isme sloshing fuel (energy dissipation) hai. Ise kis axis ke baare mein spin karna chahiye, aur kyun?

Recall Solution (see figure)

Figure — Attitude control modes — spin stabilization, 3-axis active

Transverse axis ke baare mein spin karo, kyunki

I_{trans} = 95 > I_{long} = 40

, yani yeh maximum-inertia axis hai. Kyun: fixed angular momentum

L

par, kinetic energy

T = L^{2} / (2 I)

hai. Bada

I

⇒ chhoti energy. Koi bhi dissipation (fuel slosh) hamesha energy drain karta hai, body ko uske lowest-energy spin state ki taraf drive karta hai — jo ki max-

I

axis hai. Min-

I

(long) axis par spin karo aur dissipation wobble badhata hai → flat-spin (Explorer 1 flat-spin anomaly).

L2.3 — Reaction-wheel momentum trade

Ek spacecraft body ka $I_{body} = 400 kg\cdotm^{2}$ hai aur initially rest par hai, internal reaction wheel bhi rest par hai. Wheel ( $I_{w} = 0.05 kg\cdotm^{2}$ ) ko $ω_{w} = 300 rad/s$ tak spin up kiya jaata hai. Body kaunsi angular velocity acquire karti hai?

Recall Solution

Conservation of total angular momentum (kuch bhi external nahi, dekho Reaction wheels and Control-Moment Gyros (CMG)): $L_{body} + L_{wheel} = 0 \Rightarrow I_{body} ω_{body} + I_{w} ω_{w} = 0.$ $ω_{body} = - \frac{I _{w} ω _{w}}{I _{body}} = - \frac{0.05 \times 300}{400} = - 0.0375 rad/s .$ Minus sign: body wheel ke opposite counter-rotate karti hai. Momentum trade hua, create nahi — koi fuel use nahi hua.

Level 3 — Analysis

L3.1 — Slew ke liye PD controller design karna

Ek single-axis maneuver $I \ddot{θ} = τ$ obey karta hai jisme $I = 500 kg\cdotm^{2}$ hai. Tum control law $τ = - K_{p} θ - K_{d} \dot{θ}$ use karte ho. Tumhe natural frequency $ω_{n} = 0.2 rad/s$ aur damping ratio $ζ = 0.7$ chahiye. $K_{p}$ aur $K_{d}$ nikalo.

Recall Solution

Closed loop hai $I \ddot{θ} + K_{d} \dot{θ} + K_{p} θ = 0$ , ek second-order system (dekho PID control & second-order systems (damping ratio)). Standard form se match karo $\ddot{θ} + 2 ζ ω_{n} \dot{θ} + ω_{n}^{2} θ = 0$ : $ω_{n}^{2} = \frac{K _{p}}{I} \Rightarrow K_{p} = I ω_{n}^{2} = 500 \times 0. 2^{2} = 20 N\cdotm/rad .$ $2 ζ ω_{n} = \frac{K _{d}}{I} \Rightarrow K_{d} = 2 ζ ω_{n} I = 2 (0.7) (0.2) (500) = 140 N\cdotm\cdots/rad .$ Kyun $ζ = 0.7$ : yeh essentially bina overshoot ke fast settle karta hai — sluggish ( $ζ > 1$ ) aur ringing ( $ζ < 1$ ) ke beech ka sweet spot.

L3.2 — Bang-bang slew time

Ek body $I = 600 kg\cdotm^{2}$ ko $40°$ rotate karna hai. Ek wheel constant torque magnitude $τ = 0.12 N\cdotm$ supply karta hai: pehle aadhe angle ke liye accelerate karo, doosre aadhe ke liye decelerate karo. Total maneuver time aur peak wheel momentum nikalo jo store hoga.

Recall Solution (see figure)

Step 1 — angular acceleration.

α = τ / I = 0.12/600 = 2 \times 1 0^{- 4} rad/s^{2}

. Step 2 — half angle.

40° = 0.6981 rad

, toh half

= 0.3491 rad

. Step 3 — ek half ke liye time

θ_{half} = \frac{1}{2} α t^{2}

se:

t = \frac{2 θ _{half}}{α} = \frac{2 ( 0.3491 )}{2 \times 1 0 ^{- 4}} = 3491 \approx 59.08 s .

Total time

= 2 t \approx 118.2 s

. Step 4 — peak momentum (mid-slew par, top speed

ω_{peak} = α t

h_{peak} = I α t = 600 \times 2 \times 1 0^{- 4} \times 59.08 = 7.09 N\cdotm\cdots .

Kyun accelerate/brake split karo: "bang-bang" ensure karta hai ki craft target par exactly rest mein pahunche, overshoot na kare.

L3.3 — Precession direction

Ek satellite $L$ ke saath $+ \overset{x}{^}$ ke along spin karta hai. Ek disturbance torque $+ \overset{y}{^}$ ke along point karta hai. $τ = \dot{L}$ use karke, $L$ ki tip initially kis direction mein move karti hai, aur kya spin rate change hota hai?

Recall Solution

$\dot{L} = τ$ $+ \overset{y}{^}$ ke along point karta hai, isliye $L$ ki tip $+ \overset{y}{^}$ ki taraf move karti hai — axis $x$ – $y$ plane mein sideways swing karta hai, woh ulta nahi padata. Kyunki change $\dot{L}$ $L$ ke perpendicular hai (ek $x$ ke along, ek $y$ ke along), yeh $L$ ko rotate karta hai bina uski length change kiye. Length $∣ L ∣ = I ω_{s}$ unchanged hai ⇒ spin rate constant rehta hai; sirf direction precess karta hai. Yeh perpendicular-push-turns-not-speeds hi Gyroscopic precession ka essence hai.

Level 4 — Synthesis

L4.1 — Desaturation budget

Ek reaction wheel ne $h_{wheel} = 12 N\cdotm\cdots$ store kiya hai aur saturation ke karib hai. Ek magnetorquer wheel ke momentum ke against maximum external torque $τ_{mag} = 0.004 N\cdotm$ produce karta hai. Ek full "momentum dump" mein kitna time lagta hai, aur wheel apna momentum akele dump kyun nahi kar sakta?

Recall Solution

Stored momentum remove karne ke liye tumhe ek external torque chahiye (sirf external torques total angular momentum change karte hain — wheel aur body sirf internally trade karte hain). Dekho Momentum dumping / desaturation with magnetorquers. $t = \frac{h _{wheel}}{τ _{mag}} = \frac{12}{0.004} = 3000 s \approx 50 min .$ Self-dump kyun nahi: wheel ko slow karna uska momentum body ko transfer karta hai, jo phir rotate karne lagti hai — total conserved hai, kuch remove nahi hota. Sirf ek external agent (magnetorquer ke through Earth ka magnetic field, ya thrusters) poore spacecraft se momentum bahar bhejta hai.

L4.2 — Spin vs 3-axis trade decision

Ek mission ko ek wide-field antenna ko 5 saal tak ek fixed direction mein point karna hai minimal cost aur minimal failure risk ke saath, aur ise kabhi bhi quickly re-point karne ki zaroorat nahi hai. Body inertias hain $I_{1} = 30$ , $I_{2} = 30$ , $I_{3} = 70 kg\cdotm^{2}$ . Ek mode recommend karo, agar koi ho toh spin axis batao, aur woh ek failure mode batao jise tum zaroor guard karo.

Recall Solution

Spin stabilization recommend karo — sasta, passive, robust, aur mission ko sirf ek axis pointed chahiye slow re-pointing ke saath. Yeh spin ki strengths se match karta hai aur 3-axis capability ki zaroorat nahi hai. Spin axis: axis 3, kyunki $I_{3} = 70$ maximum inertia hai (fixed $L$ par min energy) — dissipation ke saath sirf yahi stable choice hai. Sath hi $I_{1} = I_{2}$ hai, isliye transverse axes symmetric hain (ek saaf "top"). Guard karne wala failure mode: flat-spin agar energy dissipation kabhi ise max- $I$ axis se hata de; nutation damper se aur yeh ensure karke guard karo ki $I_{3}$ truly sabse bada rehta hai fuel consume hone ke baad bhi.

L4.3 — Combined slew with gyroscopic disturbance

Ek 3-axis slew ke dauran, body $ω_{2} = 0.05 rad/s$ aur $ω_{3} = 0.04 rad/s$ par do axes ke baare mein spin kar rahi hai jabki ek wheel axis 1 control karta hai. Principal inertias: $I_{1} = 300$ , $I_{2} = 260$ , $I_{3} = 340 kg\cdotm^{2}$ . Axis-1 actuator ko gyroscopic coupling cancel karne ke liye ( $\overset{ω}{˙}_{1} = 0$ taaki) kitna extra torque supply karna hoga?

Recall Solution

Axis-1 Euler equation use karo (dekho Euler's rotational equations of motion): $τ_{1} = I_{1} \overset{ω}{˙}_{1} + (I_{3} - I_{2}) ω_{2} ω_{3} .$ $\overset{ω}{˙}_{1} = 0$ ke liye actuator ko exactly coupling term supply karna hoga: $τ_{1} = (I_{3} - I_{2}) ω_{2} ω_{3} = (340 - 260) (0.05) (0.04) = 80 \times 0.002 = 0.16 N\cdotm .$ Kyun: 3-axis control ko wahi gyroscopic coupling fight karni padti hai jise spin stabilization exploit karta hai — actuator sirf axis ko body ki apni spinning geometry ke against steady rakhne mein effort burn karta hai.

Level 5 — Mastery

L5.1 — Full single-axis control design aur settling check

Tumhe $I = 800 kg\cdotm^{2}$ wale axis par $60°$ slew ke liye ek reaction-wheel controller design karna hai. (a) $ω_{n} = 0.15 rad/s$ , $ζ = 0.7$ ke liye gains choose karo. (b) Linear model hai $I \ddot{θ} + K_{d} \dot{θ} + K_{p} θ = 0$ jo target se measure hota hai. Settling time do (2% criterion, $t_{s} \approx 4/ (ζ ω_{n})$ ). (c) Wheel $h_{m a x} = 15 N\cdotm\cdots$ par saturate hota hai. Response ke dauran wheel ko jo sabse zyada momentum absorb karna padta hai woh roughly $h \approx I ω_{n} θ_{0}$ hai jahan $θ_{0} = 60° = 1.047 rad$ initial error hai. Kya wheel limits ke andar hai?

Recall Solution (see figure)

(a) Gains.

K_{p} = I ω_{n}^{2} = 800 \times 0.1 5^{2} = 800 \times 0.0225 = 18 N\cdotm/rad .

K_{d} = 2 ζ ω_{n} I = 2 (0.7) (0.15) (800) = 168 N\cdotm\cdots/rad .

(b) Settling time.

t_{s} \approx \frac{4}{ζ ω _{n}} = \frac{4}{0.7 \times 0.15} = \frac{4}{0.105} \approx 38.1 s .

(c) Peak wheel momentum.

h \approx I ω_{n} θ_{0} = 800 \times 0.15 \times 1.047 = 125.7 N\cdotm\cdots .

Yeh

h_{m a x} = 15 N\cdotm\cdots

se bahut zyada upar hai — wheel massively saturate ho jaata hai. Interpretation: linear design paper par theek hai, lekin physical wheel yeh momentum store nahi kar sakta. Tumhe ya toh maneuver slow karna hoga (lower

ω_{n}

), bada wheel use karna hoga, ya slew split karke beech mein desaturate karna hoga. Yahi real engineering reality hai: actuator limits, control law nahi, dominate karte hain.

L5.2 — Fuel burn ke baad stability margin

Ek dual-spin satellite $I_{spin} = 120$ aur sabse bade transverse $I_{trans} = 100 kg\cdotm^{2}$ ke saath launch hota hai, spin axis ke baare mein spinning (abhi max hai). Spin axis par tanks mein fuel consume hota hai, mission ke dauran $I_{spin}$ ko $25 kg\cdotm^{2}$ kam karta hai jabki $I_{trans}$ $100$ rehta hai. Kya spin stability end-of-life tak preserve hoti hai? Start aur end par margin $I_{spin} - I_{trans}$ nikalo.

Recall Solution

Dissipation ke saath spin stability require karti hai $I_{spin} = I_{m a x}$ , yani margin $M = I_{spin} - I_{trans} > 0$ . Start: $M_{0} = 120 - 100 = + 20 kg\cdotm^{2}$ → stable. End: $I_{spin} = 120 - 25 = 95 kg\cdotm^{2}$ , isliye $M_{1} = 95 - 100 = - 5 kg\cdotm^{2}$ → unstable! End-of-life tak spin axis zyada maximum-inertia axis nahi raha; energy dissipation nutation ko flat-spin mein grow kar dega. Design fix: ya toh poori fuel budget mein $M$ positive rakho (mass distribution/ballast), ya active nutation damping ki taraf switch karo. Yahi exactly Explorer 1 flat-spin anomaly mechanism hai, jo fuel consumption ke through sneakily aata hai.

Recall Self-test summary (cloze)

Spin stabilization sirf ::: maximum ::: -inertia axis ke baare mein stable hai jab dissipation present ho. Transverse torque ke under spinning body ke liye precession rate hai $Ω =$ ::: $τ / (I ω_{s})$ :::. Ek reaction wheel body torque produce karta hai ::: total angular momentum exchange (conserve) karke — koi external torque nahi :::. Wheels apni authority kho dete hain jab woh ::: saturate ::: ho jaate hain, fixed karte hain ::: momentum dumping / desaturation with magnetorquers or thrusters :::. PD-controlled axis ke liye, $K_{p} = I ω_{n}^{2}$ aur $K_{d} =$ ::: $2 ζ ω_{n} I$ :::.