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

Before anything, three reminders in plain words:

$h$ = the spinning wheel's angular momentum, a fat arrow pointing along the spin axis. Big because the wheel spins fast.
$\overset{g}{^}$ = the gimbal axis — the hinge line the whole wheel is tilted about.
$Ω = \dot{δ} \overset{g}{^}$ = how fast we tilt, times the hinge direction. $\dot{δ}$ is the gimbal rate.

The output torque is the cross product $Ω \times h$ : perpendicular to the hinge AND perpendicular to the momentum arrow.

The scenario matrix

Cell	What varies / breaks	Example
A. Baseline	$\overset{g}{^} ⊥ h$ , positive rate	Ex 1
B. Sign flip	negative gimbal rate $\dot{δ} < 0$	Ex 2
C. Angle $\neq = 9 0^{\circ}$	$\overset{g}{^}$ not perpendicular → $sin θ$	Ex 3
D. Degenerate zero	$\dot{δ} = 0$ or $\overset{g}{^} ∥ h$ → zero torque	Ex 4
E. Rank-loss singularity	columns become parallel	Ex 5
F. Structural singularity	planar cluster can't torque about $z$	Ex 5
G. Limiting blow-up	eigenvalue $ϵ \to 0$ , rate $\to \infty$	Ex 6
H. SR-inverse rescue	cap the rate with $λ$	Ex 7
I. Real-world word problem	slew a telescope, size the cluster	Ex 8
J. Exam twist	given torque + geometry, back out $\dot{δ}$	Ex 9

Nine examples cover all ten cells (Ex 5 hits both E and F).

Setup figure — the geometry every example uses

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

Look at the three arrows: the fat lavender $h$ (momentum), the mint $\overset{g}{^}$ (hinge), and the coral $τ$ jumping out perpendicular to both. That right-angle corner is the whole story. When $\overset{g}{^}$ tilts away from perpendicular, only the perpendicular component of $h$ contributes — that's where $sin θ$ enters.

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 perpendicular to spin ( $θ = 9 0^{\circ}$ ). Find $∣ τ ∣$ .

Forecast: momentum will be tens of $N\cdotm\cdots$ ; torque will be about half of it. Guess a number before reading.

Convert spin to rad/s. $ω_{w} = 5000 \cdot \frac{2 π}{60} = 523.6 rad/s$ . Why this step? The formula $h = I_{w} ω_{w}$ demands SI angular units; rpm is revolutions per minute, not radians per second.
Momentum. $h = I_{w} ω_{w} = 0.08 \times 523.6 = 41.89 N\cdotm\cdots$ . Why? This is the fat arrow's length — the thing we get to redirect.
Torque. $∣ τ ∣ = h \dot{δ} sin 9 0^{\circ} = 41.89 \times 0.5 \times 1 = 20.94 N\cdotm$ . Why? $Ω \times h$ has magnitude $∣Ω∣∣ h ∣ sin θ$ ; here $θ = 9 0^{\circ}$ so $sin θ = 1$ .

Verify: Units: $(N\cdotm\cdots) \times (rad/s) = N\cdotm$ (radians are dimensionless). ✓ A reaction wheel matching $20.94 N\cdotm$ would need $\overset{ω}{˙} = τ / I_{w} = 262 rad/s^{2}$ — absurd. CMG wins.

Worked example Ex 3 — Cell C: hinge not perpendicular

$h = 41.89 N\cdotm\cdots$ , $\dot{δ} = 0.5 rad/s$ , but the hinge makes $θ = 3 0^{\circ}$ with the spin axis (not the standard $9 0^{\circ}$ ). Find $∣ τ ∣$ .

Forecast: less than the baseline $20.94$ , because we're no longer using the full geometry. Guess the fraction.

Insert $sin θ$ . $∣ τ ∣ = h \dot{δ} sin 3 0^{\circ} = 41.89 \times 0.5 \times 0.5 = 10.47 N\cdotm$ . Why this step? The cross product measures only the part of $h$ perpendicular to $Ω$ . When the hinge lines up more with the spin, less of $h$ is perpendicular, so $sin θ$ shrinks it.
Why $sin$ and not $cos$ ? $θ$ is the angle between $\overset{g}{^}$ and $h$ . At $θ = 9 0^{\circ}$ (perpendicular, best case) we want the maximum, and $sin 9 0^{\circ} = 1$ . At $θ = 0$ (hinge parallel to spin — twisting the arrow about itself does nothing) we want zero, and $sin 0 = 0$ . Only $sin$ has that shape.

Verify: $sin 3 0^{\circ} = 0.5$ exactly, so answer is exactly half of Ex 1's $20.94$ . ✓ And it correctly sits between the $θ = 0$ (zero) and $θ = 9 0^{\circ}$ (max) extremes.

Worked example Ex 4 — Cell D: degenerate zero inputs

Two dead cases. (a) $\dot{δ} = 0$ (gimbal locked). (b) $\overset{g}{^}$ parallel to $h$ ( $θ = 0^{\circ}$ ), $\dot{δ} = 0.5$ . Find $∣ τ ∣$ for each.

Forecast: both give zero — but for different reasons. Name them.

(a) Locked gimbal. $∣ τ ∣ = h \cdot 0 \cdot sin θ = 0$ . Why? No tilt means $h$ 's direction is frozen, so $d h / d t = 0$ , so no torque. A CMG at rest is just a stored flywheel — no output.
(b) Hinge along spin. $∣ τ ∣ = h \dot{δ} sin 0^{\circ} = 0$ . Why? Spinning a fat arrow about its own axis doesn't move it. $\overset{g}{^} \times h = 0$ when they're parallel. Tilt as fast as you like — nothing comes out.

Verify: Both zero, but case (a) is $\dot{δ} = 0$ and case (b) is $sin θ = 0$ . Reactivating case (a) needs a gimbal command; case (b) needs a re-orientation of the whole gimbal geometry. Different fixes for the same symptom. ✓

Forecast: if the two gimbals point the same way, their reachable pushes are parallel — a wall. Which way does the wall run?

Build $A$ . With $δ_{1} = δ_{2} = 3 0^{\circ}$ , both columns equal $h (- sin 3 0^{\circ}, cos 3 0^{\circ}, 0) = h (- 0.5, 0.866, 0)$ . Why? Each column is that CMG's instantaneous push direction; stacking them is the Jacobian.

Check rank. Both columns identical → rank $= 1 < 2$ . Cell E: rank-loss singularity. Why this step? Parallel columns span only a line, so the reachable torque set collapses from a plane to a line.

Find the singular direction $\overset{u}{^}$ in-plane. Perpendicular to the common column $(- 0.5, 0.866)$ is $(0.866, 0.5)$ . Why? $\overset{u}{^}^{T} A = 0$ means $\overset{u}{^}$ is orthogonal to every column — the direction you cannot push. Check: $(0.866) (- 0.5) + (0.5) (0.866) = 0$ . ✓

The $z$ -wall — Cell F. Every column has a zero $z$ -component, so no gimbal motion ever torques about $z$ . This is structural: it holds for ALL $δ_{i}$ , not just special ones. Why? A planar cluster stores momentum only in the $x y$ -plane; you can't redirect what isn't there. This is exactly why real clusters use a 3-D pyramid.

Verify: $(0.866, 0.5) \cdot (- 0.5, 0.866) = - 0.433 + 0.433 = 0$ ✓ (in-plane wall). And every column's third entry is $0$ ✓ (structural $z$ -wall).

Worked example Ex 6 — Cell G: limiting blow-up near a wall

Suppose $A A^{T}$ (a $3 \times 3$ matrix) has eigenvalues ${1, 1, ϵ}$ with $ϵ = 0.01$ . A commanded torque of unit size lies entirely along the weak eigen-direction. Estimate the required gimbal-rate magnitude $∣ \dot{δ} ∣$ .

Forecast: the tiny eigenvalue will get inverted, producing a big number. How big for $ϵ = 0.01$ ?

Pseudo-inverse scaling. $\dot{δ} = - A^{T} (A A^{T})^{- 1} τ_{c m d}$ . Along the weak direction the factor is $1/ ϵ$ . Why? The pseudo-inverse inverts $A A^{T}$ ; a small eigenvalue becomes a large $1/ ϵ$ .
Plug in. $∣ \dot{δ} ∣ \sim 1/ ϵ = 1/0.01 = 100 rad/s$ . Why this matters: that's ~955 rpm of gimbal rate for a unit torque — motors saturate, control fails.

Verify: As $ϵ \to 0$ , $1/ ϵ \to \infty$ : the exact signature of hitting the singular wall. At $ϵ = 0.01$ the answer is $100$ ; halving $ϵ$ to $0.005$ doubles it to $200$ . ✓ (blows up as $1/ ϵ$ ).

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

Same weak eigenvalue $ϵ = 0.01$ , but now use the singularity-robust inverse with damping $λ = 0.1$ : the scaling becomes $ϵ / (ϵ^{2} + λ)$ per unit torque along the weak direction. Compare the capped rate to Ex 6.

Forecast: the rate should stay finite and modest now. Roughly what?

Damped scaling. Replace the raw $1/ ϵ$ by the damped-least-squares gain $\frac{ϵ}{ϵ ^{2} + λ}$ . Here $\frac{0.01}{0.0001 + 0.1} = \frac{0.01}{0.1001} = 0.0999$ . Why this step? Damped least squares adds $λ I$ inside the inverse so the denominator never hits zero.
Interpret. Gimbal-rate gain $\approx 0.0999$ instead of Ex 6's $100$ — about 1000× smaller. Why the trade? We accept a small torque error (we no longer produce exactly the command) in exchange for finite, safe gimbal rates. Cap $\sim 1/ (2 λ) = 1.58$ is the worst case; we're well under it.

Verify: $0.01/ (0.0 1^{2} + 0.1) = 0.0999$ ✓, and $0.0999 ≪ 100$ ✓ — the rescue works. Sanity: as $λ \to 0$ the gain returns toward $1/ ϵ = 100$ (no rescue), as it must. ✓

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

A survey telescope must slew $9 0^{\circ} = 1.571 rad$ in $20 s$ . Spacecraft moment of inertia about the slew axis $J = 800 kg\cdotm^{2}$ . Model the slew as constant-magnitude acceleration for the first half, deceleration for the second (bang-bang). What peak torque must the CMG cluster deliver? If each CMG has $h = 50 N\cdotm\cdots$ , what peak gimbal rate does one CMG need to supply it alone?

Forecast: the required torque will be modest (a few N·m); the gimbal rate a fraction of a rad/s. Guess both.

Half-time angle relation. Accelerate for $t_{1} = 10 s$ through $θ /2 = 0.785 rad$ . For constant $α$ : $θ /2 = \frac{1}{2} α t_{1}^{2}$ . Why bang-bang? It's the fastest slew for a torque limit — accelerate then brake.
Solve $α$ . $α = \frac{θ}{t _{1}^{2}} = \frac{1.571}{100} = 0.01571 rad/s^{2}$ . Why? $θ /2 = \frac{1}{2} α t_{1}^{2} \Rightarrow α = θ / t_{1}^{2}$ .
Peak torque. $τ = J α = 800 \times 0.01571 = 12.57 N\cdotm$ . Why? Newton–Euler for the body: torque = inertia × angular acceleration.
Gimbal rate for one CMG. $τ = h \dot{δ} \Rightarrow \dot{δ} = τ / h = 12.57/50 = 0.2513 rad/s$ . Why? At $θ = 9 0^{\circ}$ geometry, torque magnitude is $h \dot{δ}$ .

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

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

Exam question: "A single CMG delivers $∣ τ ∣ = 15 N\cdotm$ with hinge at $θ = 4 5^{\circ}$ to the spin axis and $h = 30 N\cdotm\cdots$ . What gimbal rate $\dot{δ}$ was commanded?" Then: "If the operator thought $θ$ was $9 0^{\circ}$ , what rate would they wrongly compute?"

Forecast: solving for $\dot{δ}$ needs dividing by $sin θ$ ; the $9 0^{\circ}$ mistake underestimates the needed rate. Why underestimate?

Rearrange the torque law. $∣ τ ∣ = h \dot{δ} sin θ \Rightarrow \dot{δ} = \frac{∣ τ ∣}{h sin θ}$ . Why this step? We're given the output and geometry; invert to find the input.
Correct value. $\dot{δ} = \frac{15}{30 \times sin 4 5 ^{\circ}} = \frac{15}{30 \times 0.7071} = 0.7071 rad/s$ . Why? $sin 4 5^{\circ} = 1/ 2 = 0.7071$ .
The wrong ( $9 0^{\circ}$ ) computation. $\dot{δ}_{wrong} = \frac{15}{30 \times 1} = 0.5 rad/s$ . Why the error? Assuming $sin θ = 1$ pretends the geometry is ideal, so it under-commands the rate — the real torque would fall short.

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

Active recall

Recall Predict, then reveal — one 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): what changes when $\dot{δ} \to - \dot{δ}$ ? ::: magnitude same, direction reverses $18 0^{\circ}$ Off-perpendicular (Ex 3): why $sin θ$ not $cos θ$ ? ::: $θ$ is the angle between $\overset{g}{^}$ and $h$ ; max at $9 0^{\circ}$ where $sin = 1$ , zero at $0^{\circ}$ Two zero cases (Ex 4): the two reasons torque vanishes? ::: $\dot{δ} = 0$ (locked) or $θ = 0$ ( $\overset{g}{^} ∥ h$ ) Singular direction (Ex 5): $\overset{u}{^}$ for two columns along $(- 0.5, 0.866)$ ? ::: $(0.866, 0.5)$ , the perpendicular Blow-up (Ex 6): rate for $ϵ = 0.01$ , unit torque? ::: $\sim 100 rad/s$ Rescue (Ex 7): SR gain for $ϵ = 0.01, λ = 0.1$ ? ::: $\approx 0.0999$ Slew (Ex 8): peak torque for $9 0^{\circ}$ in $20 s$ , $J = 800$ ? ::: $12.57 N\cdotm$ Exam twist (Ex 9): correct $\dot{δ}$ at $4 5^{\circ}$ vs wrong at $9 0^{\circ}$ ? ::: $0.7071$ vs $0.5 rad/s$

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