Control moment gyroscopes (CMG) — high torque, singularity

WHY do we need CMGs at all?

Spacecraft need to reorient (slew) fast — imaging satellites, the ISS, telescopes. Two ways to make internal torque without expelling fuel:

Reaction wheel: change wheel speed $\Rightarrow$ torque $=I\dot\omega$ . To get big torque you need big angular acceleration, which costs a lot of motor power and saturates quickly.
CMG: keep wheel speed fixed, rotate its axis $\Rightarrow$ torque $=\Omega \times h$ , where $h$ is already large. Small gimbal rate $\Omega$ , huge torque. This is torque amplification.

HOW a single CMG makes torque — derive it

First principles. Newton–Euler for rotation says torque = rate of change of angular momentum: $\vec\tau = \frac{d\vec h}{dt}.$

Since $h=|\vec h|$ is constant, only $\hat s$ rotates. If the gimbal rotates $\hat s$ about the gimbal axis $\hat g$ at rate $\Omega = \dot\delta$ (gimbal angle $\delta$ ), then a rotating unit vector obeys $\frac{d\hat s}{dt} = \vec\Omega \times \hat s,\qquad \vec\Omega = \dot\delta\,\hat g.$

Why this step? A vector of fixed length that is being rotated by an angular velocity $\vec\Omega$ always changes as $\vec\Omega\times(\text{vector})$ — that's the definition of rotation. Multiply through by $h$ :

Multiple CMGs: the Jacobian and total momentum

One CMG only torques in a plane. To control all 3 axes you use a cluster (typically 4, in a pyramid). Let each CMG $i$ have gimbal angle $\delta_i$ ; the total stored momentum is $\vec H(\vec\delta)=\sum_i \vec h_i(\delta_i).$

The output torque on the body is $\vec\tau = -\dot{\vec H} = -\sum_i \frac{\partial \vec h_i}{\partial \delta_i}\dot\delta_i = -A(\vec\delta)\,\dot{\vec\delta},$ where the Jacobian $A(\vec\delta) = \left[\frac{\partial \vec h_1}{\partial\delta_1}\ \Big|\ \cdots\ \Big|\ \frac{\partial \vec h_n}{\partial\delta_n}\right]\in\mathbb R^{3\times n}.$

Why a Jacobian? It's just the chain rule: each column tells you which direction that CMG can push right now, given its current gimbal angle. To produce a commanded torque $\vec\tau_{cmd}$ you solve for gimbal rates: $\dot{\vec\delta} = -A^{+}\,\vec\tau_{cmd},\qquad A^{+}=A^{T}(AA^{T})^{-1}\ \text{(pseudo-inverse).}$

WHAT is a singularity?

Definition CMG singularity

A singular configuration is a set of gimbal angles $\vec\delta$ where the Jacobian $A$ loses rank (rank $<3$ ): all the columns $\partial\vec h_i/\partial\delta_i$ become coplanar (or collinear). Then there exists a singular direction $\hat u$ such that $\hat u \cdot \frac{\partial\vec h_i}{\partial\delta_i}=0 \quad\forall i \;\Rightarrow\; \hat u^T A = 0.$ No combination of gimbal motions can produce torque along $\hat u$ . Since each column is $\hat g_i\times\hat h_i$ , singularity occurs when all output-torque directions are perpendicular to $\hat u$ — equivalently, all $\hat h_i$ have the same projection onto $\hat u$ .

A useful singularity measure: $m = \sqrt{\det(A A^{T})}\ \xrightarrow{\text{singular}}\ 0.$

Escapable vs. trapped

Hyperbolic (escapable): nearby gimbal motions exist that reduce momentum away from the wall — steer around it.
Elliptic / internal (trapped): every escape requires momentarily building momentum toward $\hat u$ ; standard steering gets stuck. Handled with null-motion and singularity-robust (SR) inverse: $\dot{\vec\delta} = -A^{T}(AA^{T}+\lambda I)^{-1}\vec\tau_{cmd} + (I-A^{+}A)\,\vec d.$

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

Worked examples

Worked example 1 — Torque from a single CMG

Wheel: $I_w=0.1\ \text{kg·m}^2$ , $\omega_w=6000\ \text{rpm}$ . Gimbal rate $\dot\delta=1\ \text{rad/s}$ , $\hat g\perp\vec h$ . Find output torque.

Step 1 — momentum. $\omega_w = 6000\cdot\frac{2\pi}{60}=628.3\ \text{rad/s}$ . Why? rpm → rad/s. So $h=I_w\omega_w=0.1\times628.3=62.83\ \text{N·m·s}$ . Step 2 — torque. $|\tau|=h\dot\delta\sin90^\circ=62.83\times1=62.8\ \text{N·m}$ . Contrast: a reaction wheel with the same $I_w$ would need $\dot\omega=628\ \text{rad/s}^2$ to match — an enormous, unsustainable acceleration. CMG wins with a gentle 1 rad/s tilt. That's amplification.

Worked example 2 — Detect a singularity (2 CMGs in a plane)

Two CMGs, momenta $\vec h_1 = h(\cos\delta_1,\sin\delta_1,0)$ , $\vec h_2=h(\cos\delta_2,\sin\delta_2,0)$ , gimbal axes both along $\hat z$ . Columns $\partial\vec h_i/\partial\delta_i = h(-\sin\delta_i,\cos\delta_i,0)$ .

Step 1. Form $A=h\begin{bmatrix}-\sin\delta_1 & -\sin\delta_2\\ \cos\delta_1 & \cos\delta_2\\ 0&0\end{bmatrix}$ . Why? stack the reachable directions. Step 2. These two 2-D columns are parallel when $\delta_1=\delta_2$ or $\delta_1=\delta_2+\pi$ . Then rank $=1$ : no torque in the direction perpendicular to the common column. (Also note the whole $z$ -row is zero — a planar cluster can never torque about $z$ ; that's a structural singularity, why you need a 3-D pyramid.)

Active recall

Recall Cover the answers first — predict, then verify (Forecast-then-Verify)

What quantity is held constant in a CMG, and what is varied? ⇒ spin rate constant; gimbal angle varied.
Give the torque formula and its direction. ⇒ $\vec\tau=\vec\Omega\times\vec h$ , perpendicular to both $\hat g$ and $\vec h$ .
Define a singularity in one line. ⇒ Jacobian $A$ loses rank; a direction $\hat u$ exists with no achievable torque.
Name two escape strategies. ⇒ SR-inverse ( $\lambda I$ ) and null motion $(I-A^{+}A)\vec d$ .

Recall Feynman: explain to a 12-year-old

Imagine spinning a bicycle wheel and holding it by its axle. It fights you when you try to turn the axle — and when you do turn it, you feel it push sideways in a weird direction. A CMG uses that weird sideways push to steer a spacecraft. It's efficient because the wheel is already spinning hard, so a gentle tilt gives a big push. But if you have four such wheels and they all end up able to push only left–right, then "up" becomes impossible — that stuck situation is the "singularity," and clever software wiggles the wheels around to avoid it.

Connections

Reaction wheels — momentum storage & saturation
Gyroscopic precession & rigid-body Euler equations
Jacobian & pseudo-inverse (Moore–Penrose)
Spacecraft attitude control & slew maneuvers
Null space & rank of a matrix
Singularity-robust inverse & damped least squares
Momentum management & desaturation (magnetic torquers, thrusters)

CMG core idea: what's constant, what's actuated?

Wheel spin rate is constant; the gimbal angle (spin-axis orientation) is actuated.

CMG output torque formula

\vec\tau = \vec\Omega\times\vec h = \dot\delta\,\hat g\times\vec h

, magnitude

h\dot\delta\sin\theta

Direction of CMG output torque

Perpendicular to both the gimbal axis

\hat g

and the wheel momentum

\vec h

Why CMG gives "high torque" vs a reaction wheel

It redirects an already-large stored momentum

h

with a small gimbal rate, so torque

=h\dot\delta

amplifies a small input; reaction wheels need large

\dot\omega

Definition of a CMG singularity

Gimbal configuration where the Jacobian

A

loses rank; a direction

\hat u

exists with

\hat u^T A=0

so no torque is achievable there.

Singularity measure

m=\sqrt{\det(AA^{T})}

; it goes to 0 at a singularity.

Why gimbal rates blow up near singularity

A^{+}

inverts

AA^{T}

; a near-zero eigenvalue makes the inverse huge, demanding infinite gimbal rate.

Two singularity-avoidance methods

Singularity-robust (damped) inverse

A^T(AA^T+\lambda I)^{-1}

and null motion

(I-A^{+}A)\vec d

What is null motion?

Gimbal motion in the null space of

A

that reconfigures the cluster while producing zero net output torque.

Escapable vs trapped singularity

Hyperbolic = escapable via nearby gimbal motion; elliptic/internal = trapped, needs SR-inverse/null motion.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, CMG ka funda simple hai: ek flywheel constant speed pe ghoom raha hai, uska angular momentum $\vec h$ bahut bada hai. Hum wheel ki speed nahi badalte — bas uske spin axis ko thoda tilt karte hain (gimbal). Jab aap ek badi momentum vector ko ghumate ho to $\vec\tau=\vec\Omega\times\vec h$ ke according ek bada torque milta hai, wo bhi choti si gimbal rate se. Isi liye CMG "high torque" deta hai — reaction wheel ki tarah pura wheel spin-up nahi karna padta, sirf tilt karna hai. Yeh lever jaisa amplification hai.

Ek important baat: torque hamesha gimbal axis aur $\vec h$ dono ke perpendicular nikalta hai. Log yahan galti karte hain — sochte hain motor jis axis pe twist karta hai, torque bhi wahin milega. Lekin cross product ki wajah se output 90 degree side me chala jaata hai. Yeh gyroscopic precession hai, aur yahi CMG ka jaadu bhi hai aur problem bhi.

Ab singularity: teen axes control karne ke liye 4 CMG ka cluster (pyramid) lagate hain. Har CMG sirf apne plane me push kar sakta hai. Jab saare CMG aise arrange ho jayein ki sabka push ek hi plane me aa jaaye, to ek direction $\hat u$ aisi ho jaati hai jahan koi bhi combination torque nahi de sakta — yeh singularity hai. Maths me matrix $A$ ka rank gir jaata hai, aur pseudo-inverse me chota eigenvalue hone se gimbal rate infinity tak chala jaata hai (motor saturate ho jaata hai).

Isse bachne ke do trick: singularity-robust inverse ( $\lambda I$ add karke rate ko finite rakhna, thodi accuracy sacrifice karke), aur null motion — matlab gimbals ko aise hilaana ki net torque zero rahe par cluster acchi position me aa jaaye. Yeh sab isliye important hai kyunki satellites, telescopes aur ISS bina fuel jalaye fast slew karne ke liye CMG use karte hain.

Control moment gyroscopes (CMG) — high torque, singularity

WHY do we need CMGs at all?

HOW a single CMG makes torque — derive it

Multiple CMGs: the Jacobian and total momentum

WHAT is a singularity?

Escapable vs. trapped

Worked examples

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections