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

• Fixed: the wheel spin rate ${\omega }_w$ (so $h=I_w{\omega }_w$ keeps constant magnitude).
• Changed: the gimbal angle $\delta$ — this rotates the direction $\hat{s}(t)$ of $\vec{h}$. Torque comes from redirecting an already-large $\vec{h}$, not from spinning the wheel faster. That is the whole point of a CMG versus a reaction wheel.

Use $\vec{\tau }=\dot{\delta }\hat{g}\times \vec{h}=\dot{\delta }h(\hat{z}\times \hat{x})=\dot{\delta }h\hat{y}$. The torque points along $+\hat{y}$ (for $\dot{\delta }>0$): perpendicular to both $\hat{g}$ and $\vec{h}$. Not along the gimbal axis — see the mistake below.

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

The general magnitude is $|\tau |=h\dot{\delta }\sin \theta$. Why $\sin \theta$? the cross product $|\hat{g}\times \vec{h}|=h\sin \theta$ — only the component of $\vec{h}$ perpendicular to $\hat{g}$ contributes. $|\tau |=41.89\times 0.5\times \sin 30^{\circ }=20.9\times 0.5=10.47\text{N⋅m}.$ The torque halved because $\sin 30^{\circ }=0.5$. This is why real CMGs are built with $\hat{g}\perp \vec{h}$ — to sit at the maximum of $\sin \theta$.

Reaction-wheel torque is $\tau =I_w\dot{\omega }$, so $\dot{\omega }=\frac{\tau }{I_w}=\frac{20.9}{0.05}=418.9{\text{rad/s}}^2.$ The CMG needed only a gentle $0.5\text{rad/s}$ tilt; the reaction wheel needs a punishing $419{\text{rad/s}}^2$ spin-up — motor-power-hungry and it saturates fast. This is torque amplification.

Step 1 — columns. Why? each column $\partial {\vec{h}}_i/\partial {\delta }_i$ is the direction CMG $i$ can currently push. $\frac{\partial {\vec{h}}_i}{\partial {\delta }_i}=h(-\sin {\delta }_i,\cos {\delta }_i,0).$ Step 2 — plug in angles (take $h=1$ for the measure): $A=\left[\begin{array}{cc}-\sin 30^{\circ }& -\sin 120^{\circ }\\ \cos 30^{\circ }& \cos 120^{\circ }\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}-0.5& -0.866\\ 0.866& -0.5\\ 0& 0\end{array}\right].$ Step 3 — the two columns are $90^{\circ }$ apart (look at the figure: the red arrows are perpendicular), so they span the full $xy$-plane. The $2\times 2$ top block has determinant $\det =(-0.5)(-0.5)-(-0.866)(0.866)=0.25+0.75=1.$ $m=\sqrt{\det (A{A}^T)}$ of the $2\times 2$ block $=|\det |=1\ne 0$. Not singular — this is actually the best possible spread for two planar CMGs. (Note the entire $z$-row is zero: a planar cluster can never torque about $\hat{z}$. That is a structural singularity — the reason real clusters use a 3-D pyramid, from rank considerations.)

The two columns $h(-\sin {\delta }_i,\cos {\delta }_i,0)$ are parallel exactly when ${\delta }_1={\delta }_2\text{or}{\delta }_1={\delta }_2+\pi .$ Why? two unit-length 2-D vectors are parallel iff their angles match or differ by $\pi$. Then rank $A=1$: the columns collapse onto one line (red line in the figure). The lost torque direction $\hat{u}$ is the in-plane direction perpendicular to that common column — you can push along the column, never across it. Numeric check: at ${\delta }_1={\delta }_2=40^{\circ }$, $\det \text{(top }2\times 2)=(-\sin 40)(\cos 40)-(-\sin 40)(\cos 40)=0.$ Singular, as predicted.

The pseudo-inverse solution is $\dot{\vec{\delta }}=-A^T(A{A}^T{)}^{-1}{\vec{\tau }}_{cmd}$. Along an eigen-direction with eigenvalue ${\lambda }_k$, $(A{A}^T{)}^{-1}$ scales by $1/{\lambda }_k$, and $A^T$ contributes a factor $\sim \sqrt{{\lambda }_k}$, so the net gimbal rate scales as $|\dot{\delta }|\sim \frac{\sqrt{{\lambda }_k}}{{\lambda }_k}=\frac{1}{\sqrt{{\lambda }_k}}=\frac{1}{\sqrt{ϵ}}=\frac{1}{\sqrt{0.01}}=10.$ Interpretation: a unit torque in a healthy direction costs $\sim 1$; the same torque along the weak direction costs $\sim 10$ — ten times the gimbal rate. As $ϵ\to 0$, this diverges. That divergence is what saturates the motors.

SR-inverse: $\dot{\vec{\delta }}=-A^T(A{A}^T+\lambda I{)}^{-1}{\vec{\tau }}_{cmd}$. The weak eigenvalue $ϵ$ becomes $ϵ+\lambda$: $|\dot{\delta }|\sim \frac{\sqrt{ϵ}}{ϵ+\lambda }=\frac{\sqrt{0.01}}{0.01+0.05}=\frac{0.1}{0.06}=\mathrm{1.667.}$ Down from $10$ to $1.67$ — a 6× reduction in gimbal rate. The trade: we no longer produce exactly the commanded torque (small error added), but the motors stay unsaturated. Accuracy sacrificed for survivability near the wall.

Output torque is $\vec{\tau }=-A\dot{\vec{\delta }}$. Substitute: ${\vec{\tau }}_{null}=-A(I-A^{+}A)\vec{d}=-(A-A{A}^{+}A)\vec{d}.$ A defining property of the Moore–Penrose pseudo-inverse is $A{A}^{+}A=A$. Therefore ${\vec{\tau }}_{null}=-(A-A)\vec{d}=\vec{0}.$ Why useful: the gimbals reshuffle without torquing the spacecraft (the motion lives in the null space of $A$). You use this to pre-steer the cluster away from an approaching singularity before you actually need to push in the dead direction — attitude control stays clean while you dodge the wall.

(a) Torque is the rate of momentum change (from $\vec{\tau }=-\dot{\vec{H}}$), so over constant torque: $\Delta H=|\tau |\cdot t=15\times 8=120\text{N⋅m⋅s}.$ (b) Fraction of capacity: $\frac{\Delta H}{H_{max}}=\frac{120}{200}=0.60=60\%.$ The cluster is now at $60\%$ of saturation and its gimbals have swung far — likely toward a singular region. Next: momentum management must dump the stored momentum using an external torque (magnetic torquers or thrusters), resetting the gimbals back to a well-conditioned configuration.

• (i) Hyperbolic / escapable — steer around it with ordinary gimbal motion; null-motion helps reposition.
• (ii) Elliptic / internal (trapped) — standard steering gets stuck because all escapes point the wrong way first. Use the SR-inverse (accept a little torque error to pass through) combined with null-motion to reshape the cluster. Elliptic singularities are the dangerous ones.

Torque of one CMG at angle $\theta$ between $\hat{g}$ and $\vec{h}$? ::: $|\tau |=h\dot{\delta }\sin \theta$; maximal at $\theta =90^{\circ }$. Correct rank test for a $3\times n$ Jacobian? ::: $m=\sqrt{\det (A{A}^T)}=0$ (never $\det A$ — it isn't square). Gimbal-rate blow-up along a weak eigenvalue $ϵ$? ::: $|\dot{\delta }|\sim 1/\sqrt{ϵ}$. Why does null motion produce zero torque? ::: Because $A{A}^{+}A=A$, so $A(I-A^{+}A)=0$. What follows a slew that uses most of $H_{max}$? ::: Momentum desaturation via external torque (torquers/thrusters).