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

Before you can read the parent page, you must own every letter it writes. We build them in order, each from the one before, each anchored to a picture. Nothing here assumes you have seen a vector, a cross product, or a matrix.

1. Arrows that carry a number and a direction — vectors

Look at the blue arrow in the figure below. To write it down we drop it into a grid and read off how far it reaches along each axis. Those readings — $a_x$ along the horizontal, $a_y$ along the vertical, $a_z$ out of the page — are the components. We stack them: $\vec{a}=(a_x,a_y,a_z).$

A unit vector is an arrow of length exactly 1 — it stores only direction, no size. We mark it with a hat: $\hat{s}$. Read $\hat{s}$ as "the direction s." The parent uses $\hat{s}$ for the spin axis, $\hat{g}$ for the gimbal axis — both are just pure directions.

2. Spinning stores "turning-motion" — angular momentum $\vec{h}$

Imagine the flywheel spinning fast. It resists being twisted — that stubbornness is stored angular momentum.

So $\vec{h}$ is an arrow that points along the spin axis, with length $h=I_w{\omega }_w$ telling you how much stubbornness is stored. In a CMG, $h$ is held constant and only the arrow's direction $\hat{s}$ is tilted.

3. Turning radians per second — angular velocity $\vec{\Omega }$

When you tilt the whole spinning wheel, the tilting itself is a rotation. A rotation also gets an arrow.

Here two new symbols appear:

• $\delta$ (delta) = the gimbal angle — how far the wheel has been tilted from its start.
• $\dot{\delta }$ = the rate of change of that angle. The dot on top means "per second." So $\dot{\delta }$ = how fast you tilt, and the parent also calls this $\Omega =\dot{\delta }$.

The dot is the single most important shorthand on the parent page.

4. Multiplying two arrows to get a third — the cross product $\times$

This is the engine of the whole topic, so we build it slowly.

Why the $\sin \theta$? It measures how spread apart the two arrows are:

• If $\vec{a}$ and $\vec{b}$ point the same way ($\theta =0$), $\sin 0=0$ — the cross product is a zero arrow. Two parallel arrows make no third direction.
• If they are perpendicular ($\theta =90^{\circ }$), $\sin 90^{\circ }=1$ — the cross product is longest. This is exactly the CMG's ideal geometry.

5. The twist that changes rotation — torque $\vec{\tau }$

Read the right side as "how fast the momentum arrow is changing." Since a CMG keeps $|\vec{h}|$ fixed, the only way $\vec{h}$ changes is by swinging its direction. Let us see why that swing is a cross product.

6. Labelling many CMGs at once — the subscript $i$

Before we stack arrows into a grid, we need a way to say "the $i$-th CMG."

So $\vec{H}={\sum }_i{\vec{h}}_i$ on the parent page just reads "total momentum = sum of every CMG's momentum."

7. Bookkeeping many arrows at once — matrices, columns, and the Jacobian $A$

A cluster has several CMGs. We need to line up "which way each one can push" side by side. That table is a matrix.

The parent's Jacobian $A$ has one column per CMG: column $i$ is the arrow $\partial {\vec{h}}_i/\partial {\delta }_i$, meaning "the direction CMG $i$'s momentum moves when you nudge its gimbal." The curly $\partial$ is a derivative just like the dot, but with respect to the angle instead of time. See Jacobian & pseudo-inverse (Moore–Penrose) for the full machinery.

Rank tells you how many genuinely different directions those columns cover. Three independent columns → you can torque any 3-D direction. If the arrows collapse into one plane, they cover only 2 directions — rank drops to 2 — and one direction becomes unreachable. That collapse is precisely the rank loss the parent calls a singularity, with the unreachable line $\hat{u}$.

8. "Un-doing" the matrix to find gimbal rates — the pseudo-inverse $A^{+}$

You command a torque and must solve backward for the gimbal rates $\dot{\vec{\delta }}$. But $A$ is not square (3 rows, often 4 columns), so it has no ordinary inverse — and worse, there are infinitely many gimbal-rate choices that all give the same torque (extra columns = extra freedom).

The danger: $(A{A}^T{)}^{-1}$ requires dividing by how strong the cluster is in each direction. Near a singularity one of those strengths shrinks toward zero, so the division explodes — tiny torque, insane gimbal rates. That is why the parent introduces the singularity-robust inverse, covered next.

9. The identity matrix $I$ and the damping $\lambda$ — taming the blow-up

The parent's rescue formula replaces $(A{A}^T{)}^{-1}$ with $(A{A}^T+\lambda I{)}^{-1}$. Two symbols there are new.

10. How near are we to disaster? — the singularity measure $m$

Prerequisite map

Equipment checklist

Self-test: cover the right side and recall each before revealing.