Reaction wheels — momentum management, zero-crossing: Worked examples

This page is a drill deck. The parent note tells the full story; this page is fully self-contained, so you can drill here alone. We hammer every case a wheel problem can throw at you — both signs, zero and saturated wheels, degenerate geometry, limiting behaviour, a real-world word problem, and an exam twist.

Before we start, five symbols we will reuse — each defined in plain words, no prior page assumed:

Definition The five quantities we juggle

Picture the spacecraft as a big box (the body) with a small flywheel (the wheel) spinning inside it.

$Ω_{w}$ = wheel spin rate in radians per second (rad/s) — how fast the little flywheel turns. One radian ≈ 57°; "rad/s" = radians turned each second.
$ω_{b}$ = body angular rate (rad/s) — how fast the whole box slowly rotates. This is the thing we ultimately want to control (where the telescope points).
$I_{w}$ = wheel's spin inertia ( $kg\cdotm^{2}$ ) — the flywheel's resistance to being spun up.
$I_{b}$ = body's inertia ( $kg\cdotm^{2}$ ) about the same axis — the big box's resistance to being turned. It is much larger than $I_{w}$ (a box is far heavier than its little wheel).
$τ$ = torque (units $N\cdotm$ = newton-metre) = a twist. Torque is the rate of change of angular momentum: $τ = \dot{H} = \frac{d H}{d t}$ . The little dot means "per second."

Combined: $H_{w} = I_{w} Ω_{w}$ = wheel angular momentum (units $N\cdotm\cdots$ ) — how much spinning-oomph is stored in the wheel. Likewise $H_{b} = I_{b} ω_{b}$ for the body.

The single pair of equations everything below descends from:

The scenario matrix

Every reaction-wheel numerical problem is one of these cells. The examples below are labelled with the cell(s) they hit.

Cell	What makes it special	Example
A. Positive slew	Spin wheel one way, body turns the other (sign of $ω_{b}$ )	Ex 1
B. Negative / reverse slew	Opposite sign — check the minus really flips	Ex 2
C. Constant torque → saturation	$τ$ = const, linear growth, finite time	Ex 3
D. Sign of external torque	Torque helps the wheel vs fights it (bias direction)	Ex 4
E. Zero-crossing / degenerate $Ω_{w} = 0$	Wheel passes through zero; stiction; why bias	Ex 5
F. Geometry limit: $m \times B$	Angle $θ$ ; $sin θ$ ; the $θ \to 0$ degenerate case	Ex 6
G. Real-world word problem	Combine two torques over an orbit	Ex 7
H. Exam twist: multi-wheel array	Momentum shared so no wheel crosses zero	Ex 8

Worked examples

Ex 1 — Positive slew (cell A)

Worked example Spin the wheel, turn the body

$I_{b} = 500 kg\cdotm^{2}$ , $I_{w} = 0.05 kg\cdotm^{2}$ . Wheel goes $0 \to + 200 rad/s$ . No external torque, starts at rest. Find the body rate $ω_{b}$ .

Forecast: guess the sign and rough size of $ω_{b}$ before reading on. (Hint: the wheel's inertia is 10 000× smaller than the body's.)

Step 1 — Total momentum is fixed at 0. Why this step? Conservation of Angular Momentum says with no external torque the total $H$ never leaves its starting value, which is $0$ (everything at rest). $I_{b} ω_{b} + I_{w} Ω_{w} = 0$

Step 2 — Solve for the body rate. Why this step? We want $ω_{b}$ alone; algebraically isolate it. $ω_{b} = - \frac{I _{w} Ω _{w}}{I _{b}} = - \frac{0.05 \times 200}{500} = - 0.02 rad/s$

Verify: The wheel spun positive, body rate is negative → they oppose, exactly as the minus sign in the master relation demands. Size check: body is $I_{b} / I_{w} = 1 0^{4}$ times heavier in spin, so $∣ ω_{b} ∣ = 200/1 0^{4} = 0.02$ . ✓ Units: $\frac{kg\cdotm ^{2} \cdot rad/s}{kg\cdotm ^{2}} = rad/s$ . ✓

Ex 2 — Reverse slew (cell B)

Ex 3 — Constant torque to saturation (cell C)

Worked example How long until the wheel is "full"?

A constant disturbance $τ_{ext} = + 5 \times 1 0^{- 5} N\cdotm$ (say a fixed Solar Radiation Pressure offset). Wheel saturates at $H_{m a x} = I_{w} Ω_{m a x} = 0.05 \times 600 = 30 N\cdotm\cdots$ . Wheel starts at bias $H_{w} (0) = 0$ . Time to saturation?

Forecast: days or hours? Guess before computing.

Step 1 — Integrate a constant torque. Why this step? Master relation: $H_{w} (t) = H_{w} (0) + \int_{0}^{t} τ d t^{'}$ . A constant under an integral gives $τ \cdot t$ — a straight line. $H_{w} (t) = τ_{ext} t$

The picture below plots this straight line: the horizontal axis is time in days, the vertical axis is stored wheel momentum $H_{w}$ in $N\cdotm\cdots$ . The blue line climbs steadily; the dashed red line is the redline $H_{m a x} = 30$ ; the yellow dot marks the instant they meet — saturation.

Step 2 — Set equal to the redline and solve for $t$ . Why this step? Saturation is the instant $H_{w}$ reaches $H_{m a x}$ . $t_{sat} = \frac{H _{m a x}}{τ _{ext}} = \frac{30}{5 \times 1 0 ^{- 5}} = 6 \times 1 0^{5} s \approx 6.94 days$

Verify: Units: $\frac{N\cdotm\cdots}{N\cdotm} = s$ . ✓ Cross-check by scaling: a smaller torque $2 \times 1 0^{- 5}$ would give $30/ (2 \times 1 0^{- 5}) = 1.5 \times 1 0^{6} s \approx 17.4$ d; here the torque is $2.5 \times$ larger, so the time should be $17.4/2.5 = 6.94$ d. ✓ Consistent.

Ex 4 — Sign of the external torque & bias direction (cell D)

Worked example Does the disturbance help or fight the wheel?

The wheel runs at bias $Ω_{bias} = + 400 rad/s$ , so $H_{w} (0) = 0.05 \times 400 = + 20 N\cdotm\cdots$ . Redline $H_{m a x} = + 30$ , and negative redline $- 30 N\cdotm\cdots$ . A Gravity Gradient Torque gives $τ_{ext} = - 1 \times 1 0^{- 4} N\cdotm$ (opposite sign to the bias). How long until saturation, and which redline does it hit?

Forecast: does a negative torque on a positive wheel drive it toward $+ 30$ or $- 30$ ?

Step 1 — Write the accumulation with the correct signs. Why this step? Signs are the whole point of this cell; keep them. $H_{w} (t) = + 20 + (- 1 \times 1 0^{- 4}) t = 20 - 1 0^{- 4} t$

Step 2 — Find where it hits a redline. Why this step? $H_{w}$ is decreasing, so it heads toward the negative redline $- 30$ . $- 30 = 20 - 1 0^{- 4} t \Rightarrow t = \frac{20 - ( - 30 )}{1 0 ^{- 4}} = \frac{50}{1 0 ^{- 4}} = 5 \times 1 0^{5} s \approx 5.79 days$

Verify: It travels a total span of $50 N\cdotm\cdots$ (from $+ 20$ down to $- 30$ ). A negative torque fights the positive bias — it first unwinds the wheel toward zero, crosses zero (⚠️ see Ex 5), then piles up negatively. This is why bias direction is chosen against the known secular disturbance, so the disturbance walks the wheel away from zero, not through it. ✓ Units check as before. ✓

Ex 5 — Zero-crossing & stiction (cell E, degenerate $Ω_{w} = 0$ )

Worked example The wheel gets stuck at zero

Continue Ex 4: the wheel is walking down through $Ω_{w} = 0$ . Fine-pointing needs a control torque of only $τ_{cmd} = 3 \times 1 0^{- 4} N\cdotm$ . The bearing's static (stick) friction can hold up to $τ_{stick} = 8 \times 1 0^{- 4} N\cdotm$ near zero speed (Bearing Friction and Stiction). Can the motor keep the wheel moving smoothly through zero?

Forecast: will the wheel glide through zero or stick?

Step 1 — Compare command torque to stiction torque. Why this step? At $Ω_{w} = 0$ the wheel only moves if the applied torque exceeds static friction; below that it locks. $τ_{cmd} = 3 \times 1 0^{- 4} N\cdotm < τ_{stick} = 8 \times 1 0^{- 4} N\cdotm$

Step 2 — Interpret the degenerate case. Why this step? Since command < stiction, the wheel hangs at $Ω_{w} = 0$ . Pointing error grows until the accumulated demand exceeds $8 \times 1 0^{- 4}$ , then the wheel snaps free — an impulsive jitter.

The picture below shows this. Horizontal axis = time (s), vertical axis = wheel speed $Ω_{w}$ (rad/s). The dashed blue line is the ideal smooth glide straight through zero; the solid red line is reality — it flattens onto the yellow zero-line (the wheel is stuck), then suddenly snaps downward (the jerk that shakes the telescope).

Step 3 — The fix: bias away from zero. Why this step? If we hold the wheel at $Ω_{bias}$ where friction is small and unidirectional, the wheel never sees the sign-flip. At $Ω_{w} > 0$ only steady kinetic friction acts, which the motor easily overcomes.

Verify: The ratio $τ_{stick} / τ_{cmd} = 8/3 \approx 2.67 > 1$ confirms the stick. If instead $τ_{cmd}$ were $10 \times 1 0^{- 4} N\cdotm$ , the ratio $8/10 = 0.8 < 1$ and it would glide through — showing the threshold is exactly $τ_{cmd} = τ_{stick}$ . ✓

Ex 6 — Magnetorquer geometry & the $θ \to 0$ limit (cell F)

Worked example Torque vs. angle between dipole and field

A magnetorquer produces a magnetic dipole $m = 40 A\cdotm^{2}$ . Earth's field Magnetorquers and Earth's Magnetic Field is $B = 3 \times 1 0^{- 5} T$ . The torque is $τ = m B sin θ$ , where $θ$ is the angle between $m$ and $B$ . Find $τ$ for $θ = 9 0^{\circ}$ , $3 0^{\circ}$ , and $0^{\circ}$ .

Forecast: which angle gives zero torque, and why?

Step 1 — Recall why $sin θ$ appears. Why this step? Torque is the cross product $τ = m \times B$ ; a cross product's magnitude is $∣ m ∣∣ B ∣ sin θ$ . Only the part of $m$ perpendicular to $B$ twists — the parallel part does nothing.

The picture below makes this concrete. The blue arrow is $B$ (the field, pointing along the horizontal axis "along B"); the red arrow is $m$ (the dipole) tilted by the yellow angle $θ$ ; the green dashed arrow is the perpendicular part $m sin θ$ — this piece is what does the twisting. As $θ \to 0$ the red arrow lies flat on the blue one and the green piece shrinks to nothing.

Step 2 — Plug the three angles. $θ = 9 0^{\circ} : τ = 40 \times 3 \times 1 0^{- 5} \times sin 9 0^{\circ} = 1.2 \times 1 0^{- 3} N\cdotm$ $θ = 3 0^{\circ} : τ = 1.2 \times 1 0^{- 3} \times sin 3 0^{\circ} = 1.2 \times 1 0^{- 3} \times 0.5 = 6 \times 1 0^{- 4} N\cdotm$ $θ = 0^{\circ} : τ = 1.2 \times 1 0^{- 3} \times sin 0^{\circ} = 0 N\cdotm$

Verify: At $θ = 0$ the dipole is along $B$ — nothing perpendicular, zero torque. This is the degenerate case that forces you to wait for the orbit to rotate $B$ before you can dump along that axis. Max at $9 0^{\circ}$ , exactly the cross-product peak. ✓ Units: $A\cdotm^{2} \cdot T = A\cdotm^{2} \cdot \frac{kg}{A\cdots ^{2}} = N\cdotm$ . ✓

Ex 7 — Real-world word problem (cell G)

Worked example A telescope over one orbit

An Earth-imaging satellite in LEO feels Aerodynamic Drag Torque $τ_{1} = + 8 \times 1 0^{- 6} N\cdotm$ (always same sign, ram-facing) and an averaged Solar Radiation Pressure torque $τ_{2} = + 4 \times 1 0^{- 6} N\cdotm$ (same axis, same sign). One orbit lasts $T = 5400 s$ (90 min). How much wheel momentum accumulates per orbit, and per day?

Forecast: roughly how many $N\cdotm\cdots$ per orbit — closer to $0.01$ , $0.1$ , or $1$ ?

Step 1 — Add the two constant torques (same axis, same sign). Why this step? Torques on the same axis add as scalars; both push the wheel the same way, so they reinforce — worst case. $τ_{tot} = 8 \times 1 0^{- 6} + 4 \times 1 0^{- 6} = 1.2 \times 1 0^{- 5} N\cdotm$

Step 2 — Integrate over one orbit. Why this step? Constant torque over time $T$ gives $Δ H = τ_{tot} T$ . $Δ H_{orbit} = 1.2 \times 1 0^{- 5} \times 5400 = 6.48 \times 1 0^{- 2} N\cdotm\cdots$

Step 3 — Scale to a day. Why this step? $86400/5400 = 16$ orbits per day. $Δ H_{day} = 6.48 \times 1 0^{- 2} \times 16 = 1.0368 N\cdotm\cdots/day$

Verify: With $H_{m a x} = 30 N\cdotm\cdots$ , saturation time $\approx 30/1.0368 \approx 28.9$ days — so a daily dump keeps a huge margin. ✓ Magnitude $\sim 0.065 N\cdotm\cdots/orbit$ matches the forecast bracket ("closer to $0.1$ "). ✓

Ex 8 — Exam twist: 4-wheel array, no zero-crossing (cell H)

Worked example Share momentum so nobody crosses zero

A maneuver demands a body momentum change of $Δ H_{body} = 12 N\cdotm\cdots$ along one axis. Two wheels are aligned on that axis, each biased at $H_{i} (0) = + 5 N\cdotm\cdots$ , redline $\pm 30$ . The demand is negative on the wheels (the wheels must give up $12 N\cdotm\cdots$ total to turn the body). Can we split it so neither wheel crosses $Ω_{w} = 0$ ? Compare: (a) one wheel does all of it, (b) split equally.

Forecast: which strategy keeps both wheels safely positive?

Step 1 — Option (a): one wheel absorbs all $- 12$ . Why this step? Test the naive single-wheel plan against the zero line. $H_{1} = 5 - 12 = - 7 N\cdotm\cdots (< 0!)$ This wheel crosses zero — stiction jitter (Ex 5). ✗

Step 2 — Option (b): split $- 12$ as $- 6$ each. Why this step? Control Moment Gyroscopes (CMG)-style redundancy: distribute the demand so each wheel stays on one side of zero. $H_{1} = 5 - 6 = - 1, H_{2} = 5 - 6 = - 1 N\cdotm\cdots$ Both cross zero too! ✗ Equal split isn't enough because the demand ( $12$ ) exceeds the total positive bias headroom before zero ( $5 + 5 = 10$ ).

Step 3 — Correct fix: raise the bias first. Why this step? To keep both wheels $\geq 0$ after removing $12$ total, each needs bias $\geq 6$ . Set $H_{i} (0) = + 7$ ; then $H_{i} = 7 - 6 = + 1 > 0$ . ✓

Verify: Required minimum bias per wheel = $Δ H_{body} / (n wheels) = 12/2 = 6$ , so bias $> 6$ needed; $+ 7$ works, and both stay below redline $+ 30$ . This is exactly the design rule: set biases so no wheel crosses zero during a maneuver. ✓

Recall Self-test (click a line to reveal its answer)

Wheel spins $+ 150 rad/s$ , $I_{w} = 0.05$ , $I_{b} = 500$ — body rate? ::: $ω_{b} = - 0.05 \times 150/500 = - 0.015 rad/s$ Constant $τ = 1 0^{- 4} N\cdotm$ , $H_{m a x} = 30$ — time to saturate? ::: $t = 30/1 0^{- 4} = 3 \times 1 0^{5} s \approx 3.47$ days Magnetorquer $m = 40 A\cdotm^{2}$ , $B = 3 \times 1 0^{- 5} T$ , angle $θ = 0^{\circ}$ — torque? ::: $τ = m B sin 0^{\circ} = 0 N\cdotm$ (dipole parallel to field, nothing perpendicular to twist) Why bias a wheel away from zero? ::: to avoid the sign-flip of bearing stiction that causes jitter at $Ω_{w} = 0$

Related: Attitude Determination and Control System (ADCS) · PID and Feedback Control · Conservation of Angular Momentum