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

• Fixed: wheel spin rate ${\omega }_w$ (isliye $h=I_w{\omega }_w$ constant magnitude rakhta hai).
• Changed: gimbal angle $\delta$ — yeh $\vec{h}$ ki direction $\hat{s}(t)$ ko rotate karta hai. Torque ek already-large $\vec{h}$ ko redirect karne se aata hai, wheel ko zyaada fast spin karne se nahi. Yahi ek CMG ka poora point hai reaction wheel ke comparison mein.

Use karo $\vec{\tau }=\dot{\delta }\hat{g}\times \vec{h}=\dot{\delta }h(\hat{z}\times \hat{x})=\dot{\delta }h\hat{y}$. Torque $+\hat{y}$ ke along point karta hai ($\dot{\delta }>0$ ke liye): dono $\hat{g}$ aur $\vec{h}$ ke perpendicular. Gimbal axis ke along nahi — neecha wala mistake dekho.

Step 1 — rpm se rad/s. Kyun? torque formula ko SI angular units chahiye. ${\omega }_w=8000\cdot \frac{2\pi }{60}=837.76\text{rad/s}.$ Step 2 — momentum size. $h=I_w{\omega }_w=0.05\times 837.76=41.89\text{N⋅m⋅s}.$ Step 3 — torque. $\hat{g}\perp \vec{h}$ ke saath, $\sin \theta =1$: $|\tau |=h\dot{\delta }=41.89\times 0.5=20.9\text{N⋅m}.$

General magnitude hai $|\tau |=h\dot{\delta }\sin \theta$. $\sin \theta$ kyun? cross product $|\hat{g}\times \vec{h}|=h\sin \theta$ — sirf $\vec{h}$ ka woh component jo $\hat{g}$ ke perpendicular hai, contribute karta hai. $|\tau |=41.89\times 0.5\times \sin 30^{\circ }=20.9\times 0.5=10.47\text{N⋅m}.$ Torque half ho gaya kyunki $\sin 30^{\circ }=0.5$. Isliye real CMGs $\hat{g}\perp \vec{h}$ ke saath banaye jaate hain — $\sin \theta$ ke maximum pe rehne ke liye.

Reaction-wheel torque hai $\tau =I_w\dot{\omega }$, toh $\dot{\omega }=\frac{\tau }{I_w}=\frac{20.9}{0.05}=418.9{\text{rad/s}}^2.$ CMG ko sirf ek gentle $0.5\text{rad/s}$ tilt chahiya tha; reaction wheel ko ek punishing $419{\text{rad/s}}^2$ spin-up chahiye — motor-power-hungry aur yeh jaldi saturate ho jaata hai. Yahi torque amplification hai.

Step 1 — columns. Kyun? har column $\partial {\vec{h}}_i/\partial {\delta }_i$ woh direction hai jisme CMG $i$ abhi push kar sakta hai. $\frac{\partial {\vec{h}}_i}{\partial {\delta }_i}=h(-\sin {\delta }_i,\cos {\delta }_i,0).$ Step 2 — angles plug in karo (measure ke liye $h=1$ lo): $A=\left[\begin{array}{cc}-\sin 30^{\circ }& -\sin 120^{\circ }\\ \cos 30^{\circ }& \cos 120^{\circ }\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}-0.5& -0.866\\ 0.866& -0.5\\ 0& 0\end{array}\right].$ Step 3 — do columns $90^{\circ }$ apart hain (figure dekho: red arrows perpendicular hain), toh woh poora $xy$-plane span karte hain. $2\times 2$ top block ka determinant hai $\det =(-0.5)(-0.5)-(-0.866)(0.866)=0.25+0.75=1.$ $2\times 2$ block ka $m=\sqrt{\det (A{A}^T)}$ $=|\det |=1\ne 0$. Singular nahi — yeh actually do planar CMGs ke liye best possible spread hai. (Note karo ki poora $z$-row zero hai: ek planar cluster kabhi bhi $\hat{z}$ ke around torque nahi de sakta. Yeh ek structural singularity hai — real clusters 3-D pyramid kyun use karte hain uska reason, rank considerations se.)

Do columns $h(-\sin {\delta }_i,\cos {\delta }_i,0)$ exactly tab parallel hain jab ${\delta }_1={\delta }_2\text{or}{\delta }_1={\delta }_2+\pi .$ Kyun? do unit-length 2-D vectors parallel hain iff unke angles match karein ya $\pi$ se differ karein. Tab rank $A=1$: columns ek line pe collapse ho jaate hain (figure mein red line). Lost torque direction $\hat{u}$ woh in-plane direction hai jo us common column ke perpendicular hai — tum column ke along push kar sakte ho, across kabhi nahi. Numeric check: ${\delta }_1={\delta }_2=40^{\circ }$ pe, $\det \text{(top }2\times 2)=(-\sin 40)(\cos 40)-(-\sin 40)(\cos 40)=0.$ Singular, jaise predict kiya tha.

Pseudo-inverse solution hai $\dot{\vec{\delta }}=-A^T(A{A}^T{)}^{-1}{\vec{\tau }}_{cmd}$. Eigenvalue ${\lambda }_k$ wali ek eigen-direction ke along, $(A{A}^T{)}^{-1}$ $1/{\lambda }_k$ se scale karta hai, aur $A^T$ $\sim \sqrt{{\lambda }_k}$ factor contribute karta hai, toh net gimbal rate scale karta hai $|\dot{\delta }|\sim \frac{\sqrt{{\lambda }_k}}{{\lambda }_k}=\frac{1}{\sqrt{{\lambda }_k}}=\frac{1}{\sqrt{ϵ}}=\frac{1}{\sqrt{0.01}}=10.$ Interpretation: healthy direction mein unit torque $\sim 1$ cost karta hai; usi torque ko weak direction ke along lene mein $\sim 10$ cost aata hai — das guna gimbal rate. Jaise $ϵ\to 0$, yeh diverge karta hai. Wahi divergence hai jo motors ko saturate karta hai.

SR-inverse: $\dot{\vec{\delta }}=-A^T(A{A}^T+\lambda I{)}^{-1}{\vec{\tau }}_{cmd}$. Weak eigenvalue $ϵ$ ban jaata hai $ϵ+\lambda$: $|\dot{\delta }|\sim \frac{\sqrt{ϵ}}{ϵ+\lambda }=\frac{\sqrt{0.01}}{0.01+0.05}=\frac{0.1}{0.06}=\mathrm{1.667.}$ $10$ se $1.67$ tak — gimbal rate mein 6× reduction. Trade-off: hum exactly commanded torque produce nahi karte (chhoti error add hoti hai), lekin motors unsaturated rehte hain. Wall ke paas survivability ke liye accuracy sacrifice ki gayi.

Output torque hai $\vec{\tau }=-A\dot{\vec{\delta }}$. Substitute karo: ${\vec{\tau }}_{null}=-A(I-A^{+}A)\vec{d}=-(A-A{A}^{+}A)\vec{d}.$ Moore–Penrose pseudo-inverse ki defining property hai $A{A}^{+}A=A$. Isliye ${\vec{\tau }}_{null}=-(A-A)\vec{d}=\vec{0}.$ Useful kyun hai: gimbals spacecraft ko torque kiye bina reshuffle ho jaate hain (motion $A$ ke null space mein rehti hai). Iska use karo cluster ko ek approaching singularity se door pre-steer karne ke liye, actually us dead direction mein push karne ki zaroorat se pehle — attitude control clean rehti hai jabki tum wall se bache rehte ho.

(a) Torque momentum change ki rate hai ($\vec{\tau }=-\dot{\vec{H}}$ se), toh constant torque ke over: $\Delta H=|\tau |\cdot t=15\times 8=120\text{N⋅m⋅s}.$ (b) Capacity ka fraction: $\frac{\Delta H}{H_{max}}=\frac{120}{200}=0.60=60\%.$ Cluster ab saturation ke $60\%$ pe hai aur uske gimbals kaafi swing kar chuke hain — likely ek singular region ki taraf. Aage: momentum management ko ek external torque (magnetic torquers ya thrusters) use karke stored momentum dump karna hoga, gimbals ko wapas ek well-conditioned configuration mein reset karne ke liye.

• (i) Hyperbolic / escapable — ordinary gimbal motion se uske around steer karo; null-motion reposition mein help karta hai.
• (ii) Elliptic / internal (trapped) — standard steering stuck ho jaati hai kyunki saari escapes pehle wrong way point karti hain. SR-inverse use karo (thodi torque error accept karo pass through karne ke liye) combined with null-motion cluster ko reshape karne ke liye. Elliptic singularities dangerous wali hoti hain.

$\hat{g}$ aur $\vec{h}$ ke beech angle $\theta$ pe ek CMG ka torque? ::: $|\tau |=h\dot{\delta }\sin \theta$; $\theta =90^{\circ }$ pe maximum. $3\times n$ Jacobian ka sahi rank test? ::: $m=\sqrt{\det (A{A}^T)}=0$ (kabhi $\det A$ nahi — yeh square nahi hai). Weak eigenvalue $ϵ$ ke along gimbal-rate blow-up? ::: $|\dot{\delta }|\sim 1/\sqrt{ϵ}$. Null motion zero torque kyun produce karta hai? ::: Kyunki $A{A}^{+}A=A$, toh $A(I-A^{+}A)=0$. Ek slew ke baad kya hota hai jo $H_{max}$ ka zyaadatar use karta hai? ::: External torque (torquers/thrusters) se momentum desaturation.