Control moment gyroscopes (CMG) — high torque, singularity: Worked examples

3.5.49 · D3 · Physics › Guidance, Navigation & Control (GNC) › Control moment gyroscopes (CMG) — high torque, singularity

Shuru karne se pehle, teen yaad-dilane plain words mein:

$h$ = spinning wheel ki angular momentum, ek mota arrow jo spin axis ki taraf point karta hai. Bada isliye kyunki wheel tezi se ghoomta hai.
$\overset{g}{^}$ = gimbal axis — wo hinge line jiske baare mein poora wheel tilt hota hai.
$Ω = \dot{δ} \overset{g}{^}$ = kitni tezi se hum tilt karte hain, times hinge direction. $\dot{δ}$ hai gimbal rate.

Output torque cross product hai $Ω \times h$ : hinge ke perpendicular AUR momentum arrow ke perpendicular.

Scenario matrix

Cell	Kya vary/break karta hai	Example
A. Baseline	$\overset{g}{^} ⊥ h$ , positive rate	Ex 1
B. Sign flip	negative gimbal rate $\dot{δ} < 0$	Ex 2
C. Angle $\neq = 9 0^{\circ}$	$\overset{g}{^}$ perpendicular nahi → $sin θ$	Ex 3
D. Degenerate zero	$\dot{δ} = 0$ ya $\overset{g}{^} ∥ h$ → zero torque	Ex 4
E. Rank-loss singularity	columns parallel ho jaate hain	Ex 5
F. Structural singularity	planar cluster $z$ ke baare mein torque nahi kar sakta	Ex 5
G. Limiting blow-up	eigenvalue $ϵ \to 0$ , rate $\to \infty$	Ex 6
H. SR-inverse rescue	rate ko $λ$ se cap karo	Ex 7
I. Real-world word problem	telescope slew karo, cluster size karo	Ex 8
J. Exam twist	given torque + geometry, $\dot{δ}$ back out karo	Ex 9

Nau examples sare das cells cover karte hain (Ex 5 E aur F dono hit karta hai).

Setup figure — har example mein yahi geometry use hoti hai

Figure — Control moment gyroscopes (CMG) — high torque, singularity

Teen arrows dekho: mota lavender $h$ (momentum), mint $\overset{g}{^}$ (hinge), aur coral $τ$ jo dono ke perpendicular bahar kood raha hai. Woh right-angle corner hi poori kahani hai. Jab $\overset{g}{^}$ perpendicular se tilt ho jaata hai, sirf $h$ ka perpendicular component contribute karta hai — yahi jagah hai jahan $sin θ$ aata hai.

Worked examples

Worked example Ex 1 — Cell A: baseline torque

Wheel $I_{w} = 0.08 kg\cdotm^{2}$ , spin $ω_{w} = 5000 rpm$ . Gimbal rate $\dot{δ} = 0.5 rad/s$ , hinge spin ke perpendicular ( $θ = 9 0^{\circ}$ ). $∣ τ ∣$ nikalo.

Forecast: momentum tens of $N\cdotm\cdots$ mein hoga; torque uska roughly aadha. Steps padhne se pehle ek number guess karo.

Spin ko rad/s mein convert karo. $ω_{w} = 5000 \cdot \frac{2 π}{60} = 523.6 rad/s$ . Yeh step kyun? Formula $h = I_{w} ω_{w}$ ko SI angular units chahiye; rpm revolutions per minute hai, radians per second nahi.
Momentum. $h = I_{w} ω_{w} = 0.08 \times 523.6 = 41.89 N\cdotm\cdots$ . Kyun? Yeh mote arrow ki length hai — woh cheez jo hum redirect karte hain.
Torque. $∣ τ ∣ = h \dot{δ} sin 9 0^{\circ} = 41.89 \times 0.5 \times 1 = 20.94 N\cdotm$ . Kyun? $Ω \times h$ ki magnitude $∣Ω∣∣ h ∣ sin θ$ hoti hai; yahaan $θ = 9 0^{\circ}$ hai isliye $sin θ = 1$ .

Verify karo: Units: $(N\cdotm\cdots) \times (rad/s) = N\cdotm$ (radians dimensionless hote hain). ✓ Ek reaction wheel jo $20.94 N\cdotm$ match kare usse $\overset{ω}{˙} = τ / I_{w} = 262 rad/s^{2}$ chahiye hoga — absurd. CMG jeet jaata hai.

Worked example Ex 2 — Cell B: negative gimbal rate

Wahi wheel Ex 1 jaisa ( $h = 41.89 N\cdotm\cdots$ ), lekin ab $\dot{δ} = - 0.5 rad/s$ (doosri taraf tilt ho raha hai). $∣ τ ∣$ nikalo aur Ex 1 ke relative uski direction batao.

Forecast: magnitude same, direction reverse. Kyun?

Magnitude. $∣ τ ∣ = h ∣ \dot{δ} ∣ sin 9 0^{\circ} = 41.89 \times 0.5 = 20.94 N\cdotm$ . Kyun? Magnitude $∣ \dot{δ} ∣$ use karta hai; sign direction mein rehta hai, size mein nahi.
Direction. $τ = \dot{δ} \overset{g}{^} \times h$ . $\dot{δ}$ ka sign flip karna poore vector ko $18 0^{\circ}$ flip kar deta hai. Yeh kyun matter karta hai: ek spacecraft ek taraf phir wapas slew karta hai gimbal rate ko reverse karke — exactly yahi sign flip hai.

Verify karo: $cos (18 0^{\circ}) = - 1$ , isliye dono torque vectors exact opposites hain; unka dot product $- (20.94)^{2}$ hai. Turning ki same speed, opposite twist. ✓

Worked example Ex 3 — Cell C: hinge perpendicular nahi

$h = 41.89 N\cdotm\cdots$ , $\dot{δ} = 0.5 rad/s$ , lekin hinge spin axis se $θ = 3 0^{\circ}$ angle banata hai (standard $9 0^{\circ}$ nahi). $∣ τ ∣$ nikalo.

Forecast: baseline $20.94$ se kam, kyunki hum poori geometry use nahi kar rahe. Fraction guess karo.

$sin θ$ daalo. $∣ τ ∣ = h \dot{δ} sin 3 0^{\circ} = 41.89 \times 0.5 \times 0.5 = 10.47 N\cdotm$ . Yeh step kyun? Cross product sirf $h$ ka woh part measure karta hai jo $Ω$ ke perpendicular ho. Jab hinge spin ke saath zyada align hoti hai, $h$ ka kam hissa perpendicular hota hai, isliye $sin θ$ use chhhota kar deta hai.
$sin$ kyun aur $cos$ kyun nahi? $θ$ angle hai $\overset{g}{^}$ aur $h$ ke beech. $θ = 9 0^{\circ}$ par (perpendicular, best case) hum maximum chahte hain, aur $sin 9 0^{\circ} = 1$ . $θ = 0$ par (hinge spin ke parallel — arrow ko uski apni axis ke baare mein twisting se kuch nahi milta) hum zero chahte hain, aur $sin 0 = 0$ . Sirf $sin$ ka woh shape hota hai.

Verify karo: $sin 3 0^{\circ} = 0.5$ exactly, isliye answer exactly Ex 1 ke $20.94$ ka aadha hai. ✓ Aur yeh correctly $θ = 0$ (zero) aur $θ = 9 0^{\circ}$ (max) extremes ke beech mein baitha hai.

Worked example Ex 4 — Cell D: degenerate zero inputs

Do dead cases. (a) $\dot{δ} = 0$ (gimbal locked). (b) $\overset{g}{^}$ parallel to $h$ ( $θ = 0^{\circ}$ ), $\dot{δ} = 0.5$ . Dono ke liye $∣ τ ∣$ nikalo.

Forecast: dono zero denge — lekin alag-alag reasons se. Unhe naam do.

(a) Locked gimbal. $∣ τ ∣ = h \cdot 0 \cdot sin θ = 0$ . Kyun? Koi tilt nahi matlab $h$ ki direction frozen hai, isliye $d h / d t = 0$ , isliye koi torque nahi. Ek ruka hua CMG sirf ek stored flywheel hai — koi output nahi.
(b) Hinge along spin. $∣ τ ∣ = h \dot{δ} sin 0^{\circ} = 0$ . Kyun? Ek mote arrow ko uski apni axis ke baare mein spinning karne se woh hilta nahi. $\overset{g}{^} \times h = 0$ jab woh parallel hon. Jitna chaaho tilt karo — kuch bahar nahi aata.

Verify karo: Dono zero hain, lekin case (a) mein $\dot{δ} = 0$ hai aur case (b) mein $sin θ = 0$ hai. Case (a) ko reactivate karne ke liye gimbal command chahiye; case (b) ke liye poori gimbal geometry ki re-orientation chahiye. Same symptom ke liye alag fixes. ✓

Forecast: agar dono gimbals ek hi taraf point karein, toh unke reachable pushes parallel hain — ek wall. Wall kis taraf hai?

$A$ banao. $δ_{1} = δ_{2} = 3 0^{\circ}$ ke saath, dono columns equal hain $h (- sin 3 0^{\circ}, cos 3 0^{\circ}, 0) = h (- 0.5, 0.866, 0)$ . Kyun? Har column us CMG ki instantaneous push direction hai; unhe stack karna Jacobian hai.

Rank check karo. Dono columns identical → rank $= 1 < 2$ . Cell E: rank-loss singularity. Yeh step kyun? Parallel columns sirf ek line span karte hain, isliye reachable torque set ek plane se simat kar ek line ban jaata hai.

In-plane singular direction $\overset{u}{^}$ nikalo. Common column $(- 0.5, 0.866)$ ke perpendicular hai $(0.866, 0.5)$ . Kyun? $\overset{u}{^}^{T} A = 0$ matlab $\overset{u}{^}$ har column ke orthogonal hai — woh direction jisme tum push nahi kar sakte. Check karo: $(0.866) (- 0.5) + (0.5) (0.866) = 0$ . ✓

$z$ -wall — Cell F. Har column ka $z$ -component zero hai, isliye koi bhi gimbal motion kabhi $z$ ke baare mein torque nahi karta. Yeh structural hai: yeh SAARE $δ_{i}$ ke liye hold karta hai, sirf special ones ke liye nahi. Kyun? Ek planar cluster momentum sirf $x y$ -plane mein store karta hai; jo wahan hai hi nahi usse redirect nahi kar sakte. Yahi exact reason hai kyun real clusters 3-D pyramid use karte hain.

Verify karo: $(0.866, 0.5) \cdot (- 0.5, 0.866) = - 0.433 + 0.433 = 0$ ✓ (in-plane wall). Aur har column ki teesri entry $0$ hai ✓ (structural $z$ -wall).

Worked example Ex 6 — Cell G: wall ke paas limiting blow-up

Maano $A A^{T}$ (ek $3 \times 3$ matrix) ke eigenvalues hain ${1, 1, ϵ}$ jahan $ϵ = 0.01$ . Ek commanded torque of unit size poori tarah weak eigen-direction mein hai. Required gimbal-rate magnitude $∣ \dot{δ} ∣$ estimate karo.

Forecast: chota eigenvalue invert hoga, ek bada number produce karega. $ϵ = 0.01$ ke liye kitna bada?

Pseudo-inverse scaling. $\dot{δ} = - A^{T} (A A^{T})^{- 1} τ_{c m d}$ . Weak direction mein factor $1/ ϵ$ hai. Kyun? Pseudo-inverse $A A^{T}$ ko invert karta hai; ek chota eigenvalue bada $1/ ϵ$ ban jaata hai.
Plug in karo. $∣ \dot{δ} ∣ \sim 1/ ϵ = 1/0.01 = 100 rad/s$ . Yeh kyun matter karta hai: woh unit torque ke liye ~955 rpm gimbal rate hai — motors saturate ho jaate hain, control fail ho jaata hai.

Verify karo: Jaise $ϵ \to 0$ , $1/ ϵ \to \infty$ : singular wall hit karne ka exact signature. $ϵ = 0.01$ par answer $100$ hai; $ϵ$ ko $0.005$ par halving karna ise $200$ par double kar deta hai. ✓ ( $1/ ϵ$ ki tarah blow up hota hai).

Worked example Ex 7 — Cell H: SR-inverse rescue

Wahi weak eigenvalue $ϵ = 0.01$ , lekin ab singularity-robust inverse use karo damping $λ = 0.1$ ke saath: scaling $ϵ / (ϵ^{2} + λ)$ ho jaati hai weak direction mein unit torque ke liye. Ex 6 se capped rate compare karo.

Forecast: rate ab finite aur modest rehni chahiye. Roughly kitni?

Damped scaling. Raw $1/ ϵ$ ko damped-least-squares gain $\frac{ϵ}{ϵ ^{2} + λ}$ se replace karo. Yahaan $\frac{0.01}{0.0001 + 0.1} = \frac{0.01}{0.1001} = 0.0999$ . Yeh step kyun? Damped least squares inverse ke andar $λ I$ add karta hai taaki denominator kabhi zero na ho.
Interpret karo. Gimbal-rate gain $\approx 0.0999$ Ex 6 ke $100$ ki jagah — roughly 1000× chhota. Yeh trade kyun? Hum ek choti torque error accept karte hain (hum exactly command produce nahi karte) finite, safe gimbal rates ke badle mein. Cap $\sim 1/ (2 λ) = 1.58$ worst case hai; hum us se neeche hain.

Verify karo: $0.01/ (0.0 1^{2} + 0.1) = 0.0999$ ✓, aur $0.0999 ≪ 100$ ✓ — rescue kaam karta hai. Sanity: jaise $λ \to 0$ gain $1/ ϵ = 100$ ki taraf wapas jaata hai (koi rescue nahi), jaisa hona chahiye. ✓

Worked example Ex 8 — Cell I: real-world word problem

Ek survey telescope ko $9 0^{\circ} = 1.571 rad$ $20 s$ mein slew karna hai. Slew axis ke baare mein spacecraft moment of inertia $J = 800 kg\cdotm^{2}$ . Slew ko pehle aadhe mein constant-magnitude acceleration, doosre mein deceleration (bang-bang) ke roop mein model karo. CMG cluster ko peak torque kitna deliver karna hoga? Agar har CMG ka $h = 50 N\cdotm\cdots$ ho, toh ek CMG akele supply karne ke liye peak gimbal rate kitna chahiye?

Forecast: required torque modest hoga (kuch N·m); gimbal rate ek rad/s ka fraction. Dono guess karo.

Half-time angle relation. $t_{1} = 10 s$ tak $θ /2 = 0.785 rad$ se accelerate karo. Constant $α$ ke liye: $θ /2 = \frac{1}{2} α t_{1}^{2}$ . Bang-bang kyun? Yeh torque limit ke liye fastest slew hai — accelerate phir brake.
$α$ solve karo. $α = \frac{θ}{t _{1}^{2}} = \frac{1.571}{100} = 0.01571 rad/s^{2}$ . Kyun? $θ /2 = \frac{1}{2} α t_{1}^{2} \Rightarrow α = θ / t_{1}^{2}$ .
Peak torque. $τ = J α = 800 \times 0.01571 = 12.57 N\cdotm$ . Kyun? Body ke liye Newton–Euler: torque = inertia × angular acceleration.
Ek CMG ke liye gimbal rate. $τ = h \dot{δ} \Rightarrow \dot{δ} = τ / h = 12.57/50 = 0.2513 rad/s$ . Kyun? $θ = 9 0^{\circ}$ geometry par, torque magnitude $h \dot{δ}$ hai.

Verify karo: Units: $J α = kg\cdotm^{2} \cdot rad/s^{2} = N\cdotm$ ✓. Reasonable: $0.25 rad/s \approx 1 4^{\circ} / s$ gimbal — gentle, motor limits ke andar easily. ✓

Worked example Ex 9 — Cell J: exam twist (input back out karo)

Exam question: "Ek single CMG $∣ τ ∣ = 15 N\cdotm$ deliver karta hai jahan hinge spin axis se $θ = 4 5^{\circ}$ par hai aur $h = 30 N\cdotm\cdots$ hai. Kaunsa gimbal rate $\dot{δ}$ command kiya gaya tha?" Phir: "Agar operator sochta ki $θ$ $9 0^{\circ}$ tha, toh woh galat rate kya compute karta?"

Forecast: $\dot{δ}$ solve karne ke liye $sin θ$ se divide karna hoga; $9 0^{\circ}$ mistake needed rate ko underestimate karti hai. Underestimate kyun?

Torque law rearrange karo. $∣ τ ∣ = h \dot{δ} sin θ \Rightarrow \dot{δ} = \frac{∣ τ ∣}{h sin θ}$ . Yeh step kyun? Hume output aur geometry di gayi hai; input nikalne ke liye invert karo.
Correct value. $\dot{δ} = \frac{15}{30 \times sin 4 5 ^{\circ}} = \frac{15}{30 \times 0.7071} = 0.7071 rad/s$ . Kyun? $sin 4 5^{\circ} = 1/ 2 = 0.7071$ .
Galat ( $9 0^{\circ}$ ) computation. $\dot{δ}_{wrong} = \frac{15}{30 \times 1} = 0.5 rad/s$ . Error kyun? $sin θ = 1$ assume karna geometry ko ideal maanta hai, isliye rate ko under-command karta hai — real torque short fall karega.

Verify karo: Ratio $\dot{δ}_{correct} / \dot{δ}_{wrong} = 0.7071/0.5 = 2 = 1.414$ ✓ — exactly $1/ sin 4 5^{\circ}$ penalty. $0.7071$ wapas plug karo: $30 \times 0.7071 \times 0.7071 = 15$ ✓.

Active recall

Recall Pehle predict karo, phir reveal karo — ek line each

Baseline (Ex 1): $I_{w} = 0.08, ω_{w} = 5000 rpm, \dot{δ} = 0.5$ , torque? ::: $20.94 N\cdotm$ Sign flip (Ex 2): $\dot{δ} \to - \dot{δ}$ hone par kya badalta hai? ::: magnitude same, direction $18 0^{\circ}$ reverse ho jaati hai Off-perpendicular (Ex 3): $sin θ$ kyun, $cos θ$ kyun nahi? ::: $θ$ $\overset{g}{^}$ aur $h$ ke beech ka angle hai; $9 0^{\circ}$ par max jahan $sin = 1$ , $0^{\circ}$ par zero Do zero cases (Ex 4): torque vanish hone ke do reasons? ::: $\dot{δ} = 0$ (locked) ya $θ = 0$ ( $\overset{g}{^} ∥ h$ ) Singular direction (Ex 5): $(- 0.5, 0.866)$ ke saath do columns ke liye $\overset{u}{^}$ ? ::: $(0.866, 0.5)$ , the perpendicular Blow-up (Ex 6): $ϵ = 0.01$ , unit torque ke liye rate? ::: $\sim 100 rad/s$ Rescue (Ex 7): $ϵ = 0.01, λ = 0.1$ ke liye SR gain? ::: $\approx 0.0999$ Slew (Ex 8): $9 0^{\circ}$ $20 s$ mein, $J = 800$ ke liye peak torque? ::: $12.57 N\cdotm$ Exam twist (Ex 9): $4 5^{\circ}$ par correct $\dot{δ}$ vs $9 0^{\circ}$ par wrong? ::: $0.7071$ vs $0.5 rad/s$

Yeh bhi dekho: Reaction wheels — momentum storage & saturation, Gyroscopic precession & rigid-body Euler equations, Spacecraft attitude control & slew maneuvers, Momentum management & desaturation (magnetic torquers, thrusters), aur parent Control moment gyroscopes (CMG) — high torque, singularity.