Control moment gyroscopes (CMG) — high torque, singularity: Question bank

Before we start, a quick shared vocabulary so no symbol appears unexplained. The picture below anchors the three key arrows — keep it in mind for every question.

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

$I$ = the wheel's moment of inertia — a number saying how hard the wheel is to spin up (big, heavy rim = big $I$ ).
$ω$ (or $ω_{w}$ ) = the flywheel's spin rate in rad/s — how fast the wheel itself is turning.
$h = I ω \overset{s}{^}$ = a single wheel's stored angular momentum — an arrow of fixed length $h$ (spin speed constant) whose direction is what the gimbal swings.
$\overset{g}{^}$ = the gimbal axis, the hinge line the whole spinning wheel is tilted about.
$Ω = \dot{δ} \overset{g}{^}$ = the gimbal rate — how fast the hinge turns ( $δ$ = gimbal angle).
$τ = Ω \times h$ = the output torque on the spacecraft; the " $\times$ " is the cross product, always perpendicular to both arrows.
$n$ = the number of CMGs in the cluster (typically 4). Each CMG contributes one gimbal angle $δ_{i}$ , so there are $n$ of them, $i = 1, \dots, n$ .
$A$ = the Jacobian, a table with $3$ rows (one per spatial axis $x, y, z$ ) and $n$ columns (one per CMG). Column $i$ is the arrow $\partial h_{i} / \partial δ_{i} = \overset{g}{^}_{i} \times h_{i}$ — "which way CMG $i$ can push right now". So $A$ is a $3 \times n$ matrix.

Definition The pseudo-inverse, the null space, and the singularity measure

To command a torque we must solve $τ = - A \dot{δ}$ for the gimbal rates $\dot{δ}$ . Since $A$ is $3 \times n$ with $n > 3$ (more unknowns than equations), we use the pseudo-inverse — see Jacobian & pseudo-inverse (Moore–Penrose): $A^{+} = A^{T} (A A^{T})^{- 1} .$ Here $A^{T}$ is $A$ with rows and columns swapped ( $n \times 3$ ), and the product $A A^{T}$ is a small $3 \times 3$ matrix — that is why we can take its determinant. $A^{+}$ is the "best-fit undo" of $A$ : it returns the smallest gimbal rates that produce the wanted torque.

The null space of $A$ — see Null space & rank of a matrix — is the set of gimbal-rate combinations $\dot{δ}$ with $A \dot{δ} = 0$ : motions that produce zero torque. With $n = 4 > 3$ it is non-empty, giving null motion room to reshuffle gimbals silently.

The singular direction $\overset{u}{^}$ is the unit vector with $\overset{u}{^}^{T} A = 0$ : the output axis no gimbal motion can reach. It is found as the eigenvector of the $3 \times 3$ matrix $A A^{T}$ whose eigenvalue is zero (or smallest).

The singularity measure $m = det (A A^{T})$ is, geometrically, the volume of the parallelepiped (box) spanned by the columns of $A$ . When the columns are spread out in 3-D the box is fat and $m$ is large; when they flatten into a plane the box collapses and $m \to 0$ — exactly the loss of rank. So $det (A A^{T})$ literally measures how much 3-D volume the reachable directions still fill.

True or false — justify

True or false: A CMG creates torque by speeding its flywheel up and down.

False — that is a reaction wheel, see Reaction wheels — momentum storage & saturation. A CMG holds spin speed

ω

constant and tilts the spin axis; the torque is

Ω \times h

, not

I \overset{ω}{˙}

True or false: Because you only tilt an already-huge momentum, a small gimbal motor delivers a large torque.

True — this is torque amplification:

∣ τ ∣ = h \dot{δ}

, and

h

is large, so even a gentle

\dot{δ}

gives a big

τ

. Storing momentum is cheap; redirecting it gives leverage.

True or false: The output torque of a single CMG points along the gimbal axis

\overset{g}{^}

False — the torque is

Ω \times h

, which is perpendicular to

\overset{g}{^}

(and to

h

). The motor twists about

\overset{g}{^}

; precession delivers the torque sideways.

True or false: A single CMG can produce torque in any of the three spatial directions.

False — one CMG can only push in the plane perpendicular to its gimbal axis. You need a cluster (usually 4) to cover all three axes.

True or false: At a singularity the CMGs stop working entirely and can produce no torque at all.

False — they lose torque only along one singular direction

\overset{u}{^}

; every other direction is still reachable. It is a missing direction, not a dead cluster.

True or false: The singularity measure

m = det (A A^{T})

equals zero exactly when

A

loses rank.

True —

det (A A^{T})

is the volume of the box spanned by

A

's columns (a product of squared singular values); if any singular value hits zero, that box flattens, rank drops below 3, and

m \to 0

True or false: The CMG spins the flywheel to store energy, and that stored energy is what torques the ship.

False — it stores angular momentum, not energy for torque. The torque comes from changing the direction of that momentum (

\dot{h}

), not from the kinetic energy of the spin.

True or false: Null motion changes the total stored momentum

H

of the cluster.

False — null motion lives in the null space of

A

(the zero-torque motions), so by definition it produces zero net torque and leaves

H

unchanged; it only reshuffles the internal gimbal angles.

True or false: A flat (planar) cluster of CMGs, all with vertical gimbal axes, can torque about the vertical axis.

False — if every

\overset{g}{^}_{i}

is vertical, every torque

\overset{g}{^}_{i} \times h_{i}

is horizontal. The vertical axis is a structural singularity; that's why real clusters are 3-D pyramids.

True or false: Near a singularity the required gimbal rates

\dot{δ}

get smaller.

False — they blow up. The pseudo-inverse divides by a nearly-zero eigenvalue of the

3 \times 3

matrix

A A^{T}

, so a tiny commanded torque demands enormous gimbal rates, saturating the motors.

Spot the error

"Since

∣ h ∣

is constant,

\dot{h} = 0

, so a CMG produces no torque." — where's the flaw?

∣ h ∣

(the length) is constant, but

h

is a vector whose direction rotates. A fixed-length rotating vector still has

\dot{h} = Ω \times h \neq = 0

"To escape a singularity, just take the pseudo-inverse

A^{+} = A^{T} (A A^{T})^{- 1}

as usual." — what breaks?

At a singularity the

3 \times 3

matrix

A A^{T}

is not invertible (determinant zero, its box has collapsed), so

A^{+}

doesn't exist. You need the singularity-robust form

A^{T} (A A^{T} + λ I)^{- 1}

instead — see Singularity-robust inverse & damped least squares.

"An elliptic (internal) singularity is easy to leave — just steer around it." — correct this.

That describes a hyperbolic singularity, where nearby gimbal motions reduce momentum away from the singular direction

\overset{u}{^}

, so you can steer around it. An elliptic one is trapped: every escape route requires momentarily building momentum toward

\overset{u}{^}

, so ordinary steering gets stuck; you need null motion or the SR-inverse.

"The Jacobian column for CMG

i

is just its momentum

h_{i}

." — fix it.

The column is

\partial h_{i} / \partial δ_{i} = \overset{g}{^}_{i} \times h_{i}

, the derivative — the direction the momentum tip moves as the gimbal turns — not the momentum itself.

"The SR-inverse's damping

λ I

makes the torque perfectly accurate near the singularity." — what's wrong?

It's the opposite trade:

λ

deliberately lets a small torque error creep in so gimbal rates stay finite. You buy safety by sacrificing a little accuracy.

"Because torque is

Ω \times h

, doubling the gimbal rate always doubles the torque." — when does this fail?

Only when

\overset{g}{^} ⊥ h

. In general

∣ τ ∣ = h \dot{δ} sin θ

; if

\overset{g}{^}

swings toward being parallel to

h

sin θ

shrinks and the same

\dot{δ}

gives less torque.

"Singularity means all the momentum vectors

h_{i}

point the same way." — sharpen this.

Not the momenta — the torque directions

\overset{g}{^}_{i} \times h_{i}

become coplanar, sharing a common perpendicular

\overset{u}{^}

. Equivalently, all

h_{i}

have the same projection onto $\overset{u}{^}$ , which is a subtler condition than "all parallel".

Why questions

Why choose the cross product

Ω \times h

to describe the torque rather than a simple scalar multiply?

Because a rotating vector of fixed length always changes along

Ω \times (vector)

— the cross product is the exact mathematical fingerprint of "rotating without stretching", which is what a gimbal does to

h

Why does the CMG give torque amplification while a reaction wheel does not?

A reaction wheel's torque is capped by how fast the motor can spin the wheel up (

I \overset{ω}{˙}

, with

I

the inertia and

ω

the wheel speed). A CMG's torque acts on the whole stored

h

by merely tilting it, so a small gimbal motor commands the leverage of a large momentum.

Why do we build the Jacobian

A

from partial derivatives

\partial h_{i} / \partial δ_{i}

It's the chain rule: total torque is

- \dot{H} = - \sum (\partial h_{i} / \partial δ_{i}) \dot{δ}_{i}

. Each partial is the instantaneous push-direction of one CMG, so

A

collects "who can push where" right now.

Why does a tiny eigenvalue of

A A^{T}

cause huge gimbal rates?

The pseudo-inverse inverts the

3 \times 3

matrix

A A^{T}

, and inverting a near-zero eigenvalue gives a near-infinite one. The command along that weak direction is scaled by

1/ ϵ

, so

\dot{δ} \to \infty

ϵ \to 0

Why is null motion useful even though it produces zero torque?

Precisely because it's torque-free (it lives in the null space of

A

), it lets you pre-steer the cluster away from an approaching singularity without disturbing the attitude — reshaping the gimbals before you need that dead direction. See Spacecraft attitude control & slew maneuvers.

Why is a CMG's torque always perpendicular to both

\overset{g}{^}

and

h

Because it equals a cross product

Ω \times h

with

Ω ∥ \overset{g}{^}

, and a cross product is by definition perpendicular to both of its factors.

Why do clusters use 4 CMGs to control only 3 axes?

With

n = 4 > 3

the extra CMG gives a non-trivial null space, providing null-motion freedom to steer around singularities that a minimal 3-CMG set could not escape.

How is the damping

λ

actually chosen, and what does it trade off?

You scale

λ

with the closeness to singularity — e.g.

λ = λ_{0} (1 - m / m_{0})

so it is

\approx 0

when well-conditioned and grows only near the wall. Bigger

λ

caps gimbal rates at

\sim 1/ λ

but leaves a larger torque error

\sim λ

; smaller

λ

is accurate but risks huge rates. It is the classic accuracy-vs-safety knob.

Edge cases

What happens to the output torque when

\overset{g}{^}

is exactly parallel to

h

(

θ = 0

∣ τ ∣ = h \dot{δ} sin 0 = 0

— tilting a spin axis about itself just re-spins the wheel and produces no output torque.

At the instant

\overset{g}{^} ⊥ h

(

θ = 9 0^{\circ}

), what is special?

This is the maximal-torque geometry,

∣ τ ∣ = h \dot{δ}

— the standard CMG design point, giving the biggest torque per unit gimbal rate.

If the wheel spin rate

ω

drifts to zero, what does the CMG become?

With

h = I ω = 0

there is no stored momentum to redirect, so

τ = Ω \times 0 = 0

— the "gyroscope" is just a dead hinge with no gyroscopic torque.

For the singularity measure

m = det (A A^{T})

, what value signals a perfectly conditioned cluster vs. a singular one?

A large

m

(fat parallelepiped, all singular values healthy) means torque is easy in every direction;

m \to 0

(flattened box) means one direction has collapsed and you're at the wall.

Two coplanar CMGs share a

\overset{z}{^}

gimbal axis with

δ_{1} = δ_{2}

— reachable rank?

Their two push-columns are parallel, so rank

= 1

: they can only torque along one line in the plane, none in the perpendicular in-plane direction (and never about

z

What is the limiting behaviour of the SR-inverse

A^{T} (A A^{T} + λ I)^{- 1}

λ \to 0

It smoothly reduces to the ordinary pseudo-inverse

A^{+}

wherever

A

is full-rank — so the SR-inverse only differs meaningfully near singularities, exactly where you want it to.

What if a commanded torque

τ_{c m d}

lies entirely along the singular direction

\overset{u}{^}

No gimbal combination can produce it:

\overset{u}{^}^{T} A = 0

means the cluster is blind to that direction, and the SR-inverse will return a small, damped, inevitably inaccurate response.

Recall One-line self-test

Which single fact seeds both the amplification and the singularity of a CMG? ::: The torque is $Ω \times h$ — a cross product: its large length (leverage on big $h$ ) amplifies, and its forced perpendicularity to $\overset{g}{^}$ is exactly what lets output directions collapse into a plane and create a dead axis.