Attitude control modes — spin stabilization, 3-axis active: Worked examples

Before anything else, the symbols must be earned so every line below reads cleanly.

Definition The core quantities we keep using

$I$ (moment of inertia, units $kg\cdotm^{2}$ ) — "rotational heaviness": how hard it is to change a body's spin. Big $I$ = sluggish to turn. See Moment of inertia tensor & principal axes.
$I_{1}, I_{2}, I_{3}$ (principal moments of inertia) — when we line the measuring axes up with the body's natural symmetry axes (the principal axes), the inertia "tensor" collapses to just three numbers: $I_{1}$ about the first body axis, $I_{2}$ about the second, $I_{3}$ about the third. Think of them as the three separate "rotational heavinesses" of the body — one per axis. See Moment of inertia tensor & principal axes.
$ω$ (angular velocity, units $rad/s$ ) — how fast something is turning. One full turn is $2 π \approx 6.283$ radians.
$L = I ω$ (angular momentum, units $kg\cdotm^{2} / s$ ) — "spin momentum" of the whole body. It plays the role that ordinary momentum $m v$ plays for straight-line motion.
$h$ (wheel angular momentum, units $kg\cdotm^{2} / s = N\cdotm\cdots$ ) — the spin momentum stored inside a reaction wheel, a separate spinning flywheel bolted to the body. Keep $h$ (wheel) and $L$ (whole body) distinct: a wheel spins up ( $h$ rises) and the body counter-rotates so the total is conserved. See Reaction wheels and Control-Moment Gyros (CMG).

Definition Torque, the precession

vector, and the cross product

$τ$ (torque, units $N\cdotm$ ) — a twist: the rotational version of a push.
Newton's rotational law: $τ = \frac{d L}{d t}$ — a twist changes the spin momentum.
$Ω$ (precession rate, a vector, units $rad/s$ ) — points along the axis the spin momentum $L$ slowly cones around. The little arrow on top () means "this has a direction in space," not just a size.
$Ω \times L$ is the cross product: it produces a new vector perpendicular to both $Ω$ and $L$ , with size $Ω L sin (angle between them)$ . Its direction follows the right-hand rule (fingers from $Ω$ toward $L$ , thumb gives the result).
The precession law is $τ = Ω \times L$ : a sideways torque does not spin $L$ up, it cones it. Taking sizes (when $Ω ⊥ L$ ) gives the scalar shortcut == $Ω = τ / L$ ==. Picture a leaning spinning top tracing a slow circle instead of falling: Gyroscopic precession.

Definition Control-law symbols (used in the PD examples)

When a spacecraft actively corrects an attitude error $θ$ , its onboard computer computes a corrective torque from a control law. The simplest is proportional–derivative (PD): $τ = - K_{p} θ - K_{d} \dot{θ}$ .

$K_{p}$ (proportional gain, units $N\cdotm/rad$ ) — a spring-like pull back toward the target: the bigger the error angle $θ$ , the harder it pushes back.
$K_{d}$ (derivative gain, units $N\cdotm\cdots/rad$ ) — a shock-absorber-like resistance to the speed $\dot{θ}$ : it damps out wobble so the craft doesn't overshoot forever.
$ω_{n} = K_{p} / I$ (natural frequency, $rad/s$ ) — how fast the corrected axis wants to oscillate.
$ζ = \frac{K _{d}}{2 K _{p} I}$ (damping ratio, dimensionless) — how "cushioned" the response is. $ζ < 1$ rings (under-damped), $ζ = 1$ is critically damped, $ζ > 1$ crawls in slowly (over-damped). The sweet spot is $ζ \approx 0.7$ . See PID control & second-order systems (damping ratio).

The scenario matrix

Every problem in this chapter is one (or a blend) of these cells:

#	Cell class	What changes / the "sign or edge"	Example
A	Gyroscopic stiffness, normal case	positive spin, sideways torque	Ex 1
B	Zero-spin limit	$ω_{s} \to 0$ : stiffness vanishes	Ex 2
C	Spin-axis choice (max vs min $I$ )	which principal axis is stable	Ex 3
D	Degenerate inertia ( $I_{1} = I_{2}$ )	symmetric body, coupling term dies	Ex 4
E	Reaction-wheel slew (bang-bang)	accelerate then brake, sign flip	Ex 5
F	Wheel saturation / desaturation	wheel hits max RPM, external dump	Ex 6
G	PD control tuning (damping ratio)	choose $ζ$ : under/critical/over	Ex 7
H	Real-world word problem	disturbance budget over an orbit	Ex 8
I	Exam twist (sign / quadrant trap)	precession direction by right-hand rule	Ex 9

We work all nine so every cell is filled.

Cell A — Gyroscopic stiffness (normal case)

Worked example Ex 1 — How fast does a spinner drift?

A drum satellite has $I = 200 kg\cdotm^{2}$ and spins at $ω_{s} = 10 rad/s$ . A steady sideways disturbance torque $τ = 0.02 N\cdotm$ acts. Find the precession (drift) rate and comment.

Forecast: guess — will it tumble in seconds, or drift for hours? Write down your gut number.

Spin momentum $L = I ω_{s} = 200 \times 10 = 2000 kg\cdotm^{2} / s$ . Why this step? $L$ is the "stiffness reservoir"; the bigger it is, the more a sideways twist just deflects it slightly.
Precession rate $Ω = \frac{τ}{L} = \frac{0.02}{2000} = 1 \times 1 0^{- 5} rad/s$ . Why this step? From the precession law $τ = Ω \times L$ (defined above), a sideways $τ$ is perpendicular to $L$ , so $τ = Ω L$ and $Ω = τ / L$ — it cones $L$ rather than spinning the body up.
Convert: $1 \times 1 0^{- 5} rad/s \times \frac{180}{π} \times 3600 \approx 2.06 deg/hour$ .

Verify: units $\frac{N\cdotm}{kg\cdotm ^{2} / s} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = 1/ s$ ✓ (a rate). About $2 deg/hr$ — extremely stable, exactly why spinners are used for cheap, robust pointing.

Cell B — The zero-spin limit

Worked example Ex 2 — What if the spin dies to zero?

Same drum, same torque $τ = 0.02 N\cdotm$ , but now $ω_{s} = 0$ (a de-spun, non-rotating body). What happens under the same disturbance for $t = 60 s$ ?

Forecast: stiffness $Ω = τ / L$ has $L = 0$ in the denominator — precession blows up. So the "cone slowly" picture must break. Guess: does it cone infinitely fast, or do something completely different?

With $ω_{s} = 0$ , $L = 0$ , so $Ω = τ / L \to \infty$ . The precession formula is undefined — it no longer describes the motion. Why this step? Precession assumes a large existing $L$ to cone. With none, there is nothing to cone.
Instead use the raw law $τ = I \ddot{θ}$ (no gyroscopic term because $ω = 0$ ), where $θ$ is the turn angle defined above. Angular acceleration $\ddot{θ} = τ / I = 0.02/200 = 1 \times 1 0^{- 4} rad/s^{2}$ . Why this step? At zero spin the Euler equation cross-terms $(\dots) ω_{j} ω_{k}$ vanish; only $I \overset{ω}{˙} = I \ddot{θ}$ survives.
It simply accelerates and tumbles: $θ = \frac{1}{2} \ddot{θ} t^{2} = \frac{1}{2} (1 \times 1 0^{- 4}) (60)^{2} = 0.18 rad \approx 10.3°$ after one minute, and growing.

Verify: the two regimes agree in spirit — huge $L$ ⇒ tiny drift (Ex 1); zero $L$ ⇒ runaway tumble (here). The "stiffness" really is proportional to $L$ . This is why a non-spinning craft must run 3-axis active control.

Cell C — Choosing the stable spin axis

Worked example Ex 3 — Which axis survives dissipation?

A cylinder has $I_{a} = 50 kg\cdotm^{2}$ about its long (symmetry) axis and $I_{t} = 120 kg\cdotm^{2}$ about a transverse axis. It has fuel slosh (energy dissipation). Which axis is a stable spin axis?

Figure caption: A rectangular side-view of the cylinder. The black dashed double-arrow lying along the body length is the long (symmetry) axis, whose moment of inertia is the small value $I_{a} = 50 kg\cdotm^{2}$ — labelled "MIN, unstable." The single red vertical arrow through the middle is the transverse axis, whose moment of inertia is the large value $I_{t} = 120 kg\cdotm^{2}$ — labelled "MAX, STABLE." The red arrow marks the one axis the body will settle onto once energy is dissipated; the black dashed axis is the tempting-but-wrong choice.

Forecast: intuition says "the long pencil-like axis looks most stable." Write that down — then watch it be wrong.

Identify the two candidate axes on the figure: the black dashed arrow is the long (symmetry) axis $I_{a} = 50$ , and the red vertical arrow is the transverse axis $I_{t} = 120$ . At fixed $L$ , kinetic energy $T = \frac{L ^{2}}{2 I}$ . Compare the two axes at the same $L$ (say $L = 100$ ). Why this step? Dissipation (slosh, flexing) removes energy but cannot change $L$ (no external torque). The body drifts to minimum $T$ , so we rank the axes by their energy.
About long axis: $T_{a} = \frac{10 0 ^{2}}{2 ( 50 )} = 100 J$ . About transverse: $T_{t} = \frac{10 0 ^{2}}{2 ( 120 )} = 41.7 J$ . Why this step? Bigger $I$ in the denominator ⇒ smaller energy. The max- $I$ axis (the red arrow in the figure) is the minimum-energy state.
Minimum energy = $T_{t}$ , the transverse (max- $I$ ) axis — the red arrow. That is the only dissipation-stable choice; the black dashed long axis is unstable.

Verify: $T_{t} = 41.7 < T_{a} = 100$ ✓. This is the Explorer 1 flat-spin anomaly in one line: Explorer 1 was spun about its minimum- $I$ (pencil) axis, dissipated energy, and flat-spun. Fix rule: spin about $I_{m a x}$ .

Cell D — Degenerate (symmetric) inertia

Worked example Ex 4 — Symmetric body, torque-free

A perfectly symmetric spinner has principal moments $I_{1} = I_{2} = 100 kg\cdotm^{2}$ (equal transverse) and $I_{3} = 60$ (spin axis), using the $I_{1}, I_{2}, I_{3}$ principal-axis notation defined above. No external torque. Using the Euler equations, what happens to the spin-axis rate $ω_{3}$ ?

Forecast: does $ω_{3}$ stay constant, drift, or oscillate?

Third Euler equation (torque-free): $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2}$ . Why this step? The gyroscopic coupling on axis 3 is driven by the difference $I_{1} - I_{2}$ .
Degeneracy: $I_{1} = I_{2} \Rightarrow I_{1} - I_{2} = 0$ , so $\overset{ω}{˙}_{3} = 0$ . Why this step? When two moments of inertia are equal, that coupling term is exactly zero — the case class "collapses."
Therefore $ω_{3} = const$ : the spin rate about the symmetry axis is exactly conserved, no matter what $ω_{1}, ω_{2}$ do.

Verify: $(I_{1} - I_{2}) ω_{1} ω_{2} = 0 \cdot ω_{1} ω_{2} = 0$ ✓. This is why real spinners are built axially symmetric — you get one guaranteed-constant spin rate and clean precession behaviour.

Cell E — Reaction-wheel slew (bang-bang, sign flip)

Worked example Ex 5 — Time to turn 30°

A body with $I = 500 kg\cdotm^{2}$ must rotate $30°$ . A reaction wheel gives constant torque magnitude $τ = 0.1 N\cdotm$ : accelerate for the first half, then brake (reverse sign) for the second half so it arrives at rest. Find total slew time and peak wheel momentum. See Reaction wheels and Control-Moment Gyros (CMG).

Forecast: guess the total time to the nearest 30 s.

Angular acceleration $α = \ddot{θ} = τ / I = 0.1/500 = 2 \times 1 0^{- 4} rad/s^{2}$ (recall $α \equiv \ddot{θ}$ , defined above). Why this step? Wheel torque acts on the body by momentum exchange: $τ_{body} = - \dot{h}$ , where $h$ is the wheel momentum defined above.
Half-angle $= 15° = 0.262 rad$ . Using $θ = \frac{1}{2} α t^{2}$ : $t = 2 θ / α = 2 (0.262) / (2 \times 1 0^{- 4}) \approx 51.2 s$ . Why this step? Constant-torque kinematics with the turn angle $θ$ ; the accelerate phase covers exactly half the angle.
Total time $= 2 \times 51.2 \approx 102.4 s$ . The sign flip at the midpoint is the "bang-bang": maximum push, then maximum brake.
Peak wheel momentum $h = I α t = 500 (2 \times 1 0^{- 4}) (51.2) \approx 5.12 N\cdotm\cdots$ . Why this step? This is the mid-slew angular momentum: the body has spun up to some rate, and by conservation the wheel holds an equal-and-opposite $h$ — which must be below the wheel's rating.

Verify: at the midpoint body rate $\dot{θ} = α t = 2 \times 1 0^{- 4} \times 51.2 = 0.01024 rad/s$ ; body momentum $I \dot{θ} = 500 \times 0.01024 = 5.12 N\cdotm\cdots$ ✓ matches step 4.

Cell F — Saturation & desaturation

Worked example Ex 6 — When must we dump momentum?

The wheel in Ex 5 has a maximum storable momentum $h_{m a x} = 4 N\cdotm\cdots$ . A steady disturbance torque $τ_{d} = 5 \times 1 0^{- 4} N\cdotm$ is being absorbed by the wheel (spinning it up to hold attitude). How long until saturation, and what dumps it? See Momentum dumping / desaturation with magnetorquers.

Forecast: hours? days? one orbit?

The wheel absorbs momentum at rate $\dot{h} = τ_{d} = 5 \times 1 0^{- 4} N\cdotm\cdots per s$ , where $h$ is the wheel momentum. Why this step? To hold attitude against a constant external twist, the wheel must keep taking up that momentum (total $L = L_{body} + h$ is conserved internally).
Time to fill: $t_{sat} = h_{m a x} / \dot{h} = 4/ (5 \times 1 0^{- 4}) = 8000 s \approx 2.22 hours$ . Why this step? Constant absorption ⇒ linear fill of $h$ up to its ceiling $h_{m a x}$ .
Note Ex 5's peak slew momentum was $5.12 N\cdotm\cdots > 4 N\cdotm\cdots$ — that single slew would already saturate this wheel. The wheel is undersized for a one-shot $30°$ slew here. Why this step? Always cross-check slew demand against the rating.
Desaturation: fire magnetorquers/thrusters to apply an external torque that spins the wheel back down ( $h$ falls) while an equal external torque holds the body — dumping the stored momentum to Earth's field / propellant. Why this step? Only an external torque can change total $L$ ; wheels alone cannot un-saturate themselves.

Verify: units $\frac{N\cdotm\cdots}{N\cdotm} = s$ ✓. $8000 s /3600 = 2.22 hr$ ✓.

Cell G — PD tuning (damping ratio)

Worked example Ex 7 — Choose the derivative gain for good settling

A single axis obeys $I \ddot{θ} = τ$ with $I = 500 kg\cdotm^{2}$ , where $θ$ is the attitude error angle defined above. A PD law $τ = - K_{p} θ - K_{d} \dot{θ}$ is used with $K_{p} = 20 N\cdotm/rad$ . Choose $K_{d}$ for damping ratio $ζ = 0.7$ (fast, no ringing). All of $K_{p}$ , $K_{d}$ , $ω_{n}$ , $ζ$ are defined in the control-law definition above. See PID control & second-order systems (damping ratio).

Forecast: guess $K_{d}$ to the nearest 50.

Substitute the PD law into $I \ddot{θ} = τ$ : $I \ddot{θ} = - K_{p} θ - K_{d} \dot{θ}$ , i.e. $I \ddot{θ} + K_{d} \dot{θ} + K_{p} θ = 0$ — a damped oscillator in the angle $θ$ , with $ω_{n} = K_{p} / I$ and $ζ = \frac{K _{d}}{2 K _{p} I}$ . Why this step? Small-angle linearization drops gyroscopic terms; what remains is the classic second-order system whose two knobs $K_{p}, K_{d}$ set exactly $ω_{n}$ and $ζ$ .
Natural frequency $ω_{n} = K_{p} / I = 20/500 = 0.04 = 0.2 rad/s$ . Why this step? $ω_{n}$ sets how fast the axis wants to respond; we compute it first because $ζ$ depends on the same $K_{p}, I$ .
Solve $ζ = \frac{K _{d}}{2 K _{p} I}$ for $K_{d}$ : $K_{d} = 2 ζ K_{p} I = 2 (0.7) 20 \times 500 = 1.4 10000 = 1.4 \times 100 = 140 N\cdotm\cdots/rad$ . Why this step? $ζ = 0.7$ is the sweet spot: near-fastest settling with only a small, brief overshoot — the standard aerospace pick.
Recommendation: set $K_{d} = 140 N\cdotm\cdots/rad$ (with $K_{p} = 20$ ). This gives $ω_{n} = 0.2 rad/s$ and $ζ = 0.7$ : quick, well-damped pointing.

Verify: back-substitute: $ζ = \frac{140}{2 20 \times 500} = \frac{140}{2 \times 100} = 0.7$ ✓. If $K_{d}$ were smaller ( $ζ < 1$ ) it rings; larger ( $ζ > 1$ ) it crawls in slowly — $0.7$ balances both.

Cell H — Real-world disturbance budget

Worked example Ex 8 — Solar-pressure drift over one orbit

A 3-axis craft suffers a steady solar-radiation-pressure torque $τ_{SRP} = 3 \times 1 0^{- 5} N\cdotm$ about one axis for a full orbit $T_{orb} = 5400 s$ (90 min). Its wheel can store $h_{m a x} = 0.2 N\cdotm\cdots$ . Does it survive one orbit before needing a dump? See Gravity-gradient & solar-radiation disturbance torques.

Forecast: does one orbit's worth of push fit inside the wheel?

Momentum accumulated over the orbit: $Δ h = τ_{SRP} T_{orb} = 3 \times 1 0^{- 5} \times 5400 = 0.162 N\cdotm\cdots$ , added to the wheel momentum $h$ . Why this step? Constant torque ⇒ momentum builds linearly; the wheel soaks it all up to keep the body still.
Compare to capacity: $0.162 < 0.2$ , so the wheel holds — just barely (81% full). Why this step? Mission planners schedule a desaturation once per orbit with this margin in mind.
Margin: $0.2 - 0.162 = 0.038 N\cdotm\cdots$ spare, i.e. $0.038/ (3 \times 1 0^{- 5}) \approx 1267 s$ ( $\approx 21 min$ ) of extra runway before it would saturate.

Verify: $0.162/0.2 = 0.81$ ✓. Units $N\cdotm \times s = N\cdotm\cdots$ ✓. Realistic: SRP torques of $\sim 1 0^{- 5} N\cdotm$ force a dump roughly once per orbit on small craft.

Cell I — Exam twist: precession direction

Worked example Ex 9 — Which way does

$L$ cone? A spinner has spin momentum $L$ pointing "up" along $+ \overset{z}{^}$ (spin axis vertical). A disturbance torque points sideways along $+ \overset{x}{^}$ . Using $τ = Ω \times L$ , find the direction of the precession vector $Ω$ (the trap: students expect $L$ to move toward the torque — it does not). See figure.

Figure caption: Two black arrows show the givens — $L$ pointing up along $+ \overset{z}{^}$ (the spin axis) and $τ$ pointing right along $+ \overset{x}{^}$ (the disturbance torque). The single red arrow is the answer: the precession vector $Ω$ pointing along $+ \overset{y}{^}$ (out to the upper-left), exactly $90°$ from the torque. The whole point of the picture is that the red response arrow is not aligned with the black torque arrow.

Forecast: guess: does $Ω$ point $+ \overset{y}{^}$ , $- \overset{y}{^}$ , or along the torque $+ \overset{x}{^}$ ? Write it down.

Write the known vectors from the figure: $L = L \overset{z}{^}$ (black up-arrow) and $τ = τ \overset{x}{^}$ (black right-arrow). We must find $Ω$ satisfying $τ = Ω \times L$ . Why this step? The precession vector is defined by this cross-product relation, so direction — not just size — must be solved for.
Guess a direction and test it with the right-hand rule. Try $Ω = Ω \overset{y}{^}$ : then $Ω \times L = Ω L (\overset{y}{^} \times \overset{z}{^}) = Ω L \overset{x}{^}$ . Why this step? The right-hand rule gives $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ ; we check whether this reproduces the given torque direction $+ \overset{x}{^}$ .
It matches $+ \overset{x}{^}$ exactly when $Ω > 0$ . Therefore $Ω$ points along $+ \overset{y}{^}$ — the red arrow in the figure. The spin axis cones toward $+ \overset{y}{^}$ , at $90°$ to the torque, not in the torque's direction $+ \overset{x}{^}$ . Why this step? This perpendicular response is the entire gyroscopic surprise: push a spinning top sideways and it moves at right angles to your push.
Magnitude, as always: $Ω = τ / L$ (from taking sizes since $Ω ⊥ L$ ). Direction $+ \overset{y}{^}$ , size $τ / L$ — the full vector answer. Why this step? An exam wants both the number and the correct axis; only $Ω = (τ / L) \overset{y}{^}$ is complete.

Verify: cross-product check $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ ✓ reproduces the torque axis. Magnitude $Ω = τ / L$ agrees with Ex 1. Gyroscopes always respond $90°$ away from the applied twist — the counter-intuitive heart of Gyroscopic precession.

Recall Self-test — cover the answers

Zero-spin limit: does the precession formula $Ω = τ / L$ still apply? ::: No — $L = 0$ makes it undefined; use $τ = I \ddot{θ}$ and the body tumbles. Which axis is dissipation-stable for spin, min- $I$ or max- $I$ ? ::: Max- $I$ (minimum energy at fixed $L$ ). For a symmetric body $I_{1} = I_{2}$ , what is conserved without torque? ::: The spin-axis rate $ω_{3}$ (coupling term $I_{1} - I_{2} = 0$ ). What is the difference between $L$ and $h$ ? ::: $L$ is the whole body's angular momentum; $h$ is the momentum stored inside a reaction wheel — they trade so the total is conserved. What does the symbol $α$ mean here? ::: Angular acceleration, $α \equiv \ddot{θ}$ , in $rad/s^{2}$ . Why does a wheel saturate, and how is it fixed? ::: It absorbs disturbance momentum until max RPM; an external torque (magnetorquer/thruster) dumps it. In $ζ = K_{d} / (2 K_{p} I)$ , what $ζ$ gives fast settling without ringing? ::: About $0.7$ . Precession direction relative to the applied torque? ::: Perpendicular ( $90°$ away), by $τ = Ω \times L$ .

Back to parent: Attitude control modes — spin stabilization, 3-axis active.