Exercises — Reaction wheels — momentum management, zero-crossing

Total angular momentum is fixed. It started at zero, so it must stay zero: $I_b{\omega }_b+I_w{\Omega }_w=0.$ If the wheel spins in the negative sense (${\Omega }_w<0$), then ${\omega }_b$ must carry the opposite sign, so ${\omega }_b>0$ — the body turns in the positive (right-hand) sense about $+z$. Opposite senses about the same fixed axis: that is the whole mechanism. What it looks like: see figure s01 — the orange wheel-arrow and the teal body-arrow always point opposite ways about the plum $+z$ axis, like two gears that must cancel.

The wheel speed grows without bound because $H_w(t)=H_w(0)+{\int }_0^t{\tau }_{\text{ext}}d{t}^{\prime }$ integrates a constant-sign torque forever, so it hits maximum speed (saturation) and can absorb no more. You must perform momentum dumping with an external actuator (magnetorquer or thruster).

Conservation about the shared axis, total starts at zero: ${\omega }_b=-\frac{I_w{\Omega }_w}{I_b}=-\frac{0.04\times 500}{400}=-0.05\text{rad/s}.$ The body rotates at $0.05\text{rad/s}$ in the negative sense (opposite the wheel). Why the sign: to point toward a target, spin the wheel away from it.

Torque = inertia times angular acceleration. The wheel's acceleration is ${\dot{\Omega }}_w=500/20=25{\text{rad/s}}^2$: ${\tau }_m=I_w{\dot{\Omega }}_w=0.04\times 25=1.0\text{N⋅m}.$ What we did & why: $\Delta H=\tau \Delta t$ rearranged gives $\tau =I_w\Delta \Omega /\Delta t$ for a constant push.

Constant torque means linear accumulation, $H_w(t)={\tau }_{\text{ext}}t$: $t_{\text{sat}}=\frac{H_{\max }}{{\tau }_{\text{ext}}}=\frac{20}{4\times 10^{-5}}=5\times 10^5\text{s}\approx 5.79\text{days}.$

Torque of a magnetic dipole in a field is $\vec{\tau }=\vec{m}\times \vec{B}$, magnitude $\tau =mB\sin \theta$, maximal at $\theta =90^{\circ }$: ${\tau }_{\max }=40\times 3\times 10^{-5}=1.2\times 10^{-3}\text{N⋅m}.$ Time to remove the momentum at this torque: $t=\frac{H}{{\tau }_{\max }}=\frac{24}{1.2\times 10^{-3}}=2.0\times 10^4\text{s}\approx 5.56\text{h}.$ Why it matters: real dumps take hours because $\vec{B}$ is weak and only the perpendicular component acts (see Magnetorquers and Earth's Magnetic Field).

Why a dot product is the right test: the dot product $\vec{v}\cdot \hat{B}$ measures how much of $\vec{v}$ lies along the direction $\hat{B}$ — it is literally the length of the shadow $\vec{v}$ casts on the $\hat{B}$ axis. So if we want to ask "does the torque have any piece pointing along $\vec{B}$?", the honest way to check is to project it onto $\hat{B}$ with a dot product. If that projection is zero, there is no component in that direction at all. Now compute it. By definition of the cross product, $\vec{m}\times \vec{B}$ is perpendicular to $\vec{B}$: ${\vec{\tau }}_{\text{mag}}\cdot \hat{B}=(\vec{m}\times \vec{B})\cdot \hat{B}=0,$ because the cross product is orthogonal to both of its factors. The projection onto $\hat{B}$ vanishes, so the component of desired momentum change parallel to $\vec{B}$ cannot be touched at this instant. Fix in practice: you rely on the orbital rotation of $\vec{B}$ — as the spacecraft moves, $\vec{B}$ points different ways, and over an orbit the previously-blocked direction becomes accessible. See figure s02.

A stuck wheel breaks free only if the commanded torque exceeds static friction: is ${\tau }_{\text{cmd}}>{\tau }_s$? Here $5\times 10^{-4}<8\times 10^{-4}$, so no — the wheel sticks. The pointing error grows until the controller ramps the command above $8\times 10^{-4}\text{N⋅m}$; then the wheel snaps free. Why the jerk is even bigger than "just breaking stiction": the moment it slips, friction drops from static ${\tau }_s$ to the lower kinetic value ${\tau }_k$. That sudden shortfall of resisting torque is the impulse thrown into the spacecraft: $\Delta \tau ={\tau }_s-{\tau }_k=8\times 10^{-4}-5\times 10^{-4}=3\times 10^{-4}\text{N⋅m},$ released suddenly as jitter. Because kinetic friction is usually less than static, the wheel over-accelerates the instant it lets go — the classic stick-slip snap of Bearing Friction and Stiction behind zero-crossing.

(a) Bias momentum $H_{\text{bias}}=I_w{\Omega }_{\text{bias}}=0.05\times 200=10\text{N⋅m⋅s}$. (b) New momentum $H^{\prime }=10-6=4\text{N⋅m⋅s}$, so ${\Omega }^{\prime }=\frac{H^{\prime }}{I_w}=\frac{4}{0.05}=80\text{rad/s}.$ (c) $80\text{rad/s}>0$ — the wheel stayed on the same side of zero (still positive sense about $+z$), no sign change, so friction never reverses, no jitter. That is exactly the point of running a bias. See figure s03.

Currently the pair holds $9+(-3)=+6\text{N⋅m⋅s}$. We must reach $+2\text{N⋅m⋅s}$, a change of $-4\text{N⋅m⋅s}$. Route it so neither wheel changes sign. One safe split: take it all from A.

• $H_A:9\to 5\text{N⋅m⋅s}$, giving ${\Omega }_A=5/0.05=100\text{rad/s}$.
• $H_B$ unchanged at $-3\text{N⋅m⋅s}$, giving ${\Omega }_B=-3/0.05=-60\text{rad/s}$. Check total: $5+(-3)=2\text{N⋅m⋅s}$. ✓ What "$-60\text{rad/s}$" really means: the minus is only the right-hand-rule direction — wheel B is spinning perfectly happily in the negative sense about $+z$. A negative speed is not dangerous; only passing through zero (a sign reversal) triggers stiction, because that is when the friction torque itself flips direction. Wheel B never approaches zero, so it stays smooth even though its speed is negative. Redundant arrays exploit exactly this — the array trick from Attitude Determination and Control System (ADCS).

The number $25\text{N⋅m⋅s}$ is the headroom above the bias: the disturbance adds momentum starting from $H_{\text{bias}}$, and saturation is reached after an extra $25\text{N⋅m⋅s}$ has accumulated. We never need the bias value itself — only that margin. (a) Saturation time: usable margin over disturbance, $t_{\text{sat}}=\frac{25}{3\times 10^{-5}}=8.333\times 10^5\text{s}\approx 9.65\text{days}.$ (b) Dump time: momentum removed over dump torque, $t_{\text{dump}}=\frac{25}{6\times 10^{-4}}=4.167\times 10^4\text{s}\approx 11.6\text{h}.$ (c) Keep-up test compares the two torques: $\frac{{\tau }_{\text{dump}}}{{\tau }_{\text{ext}}}=\frac{6\times 10^{-4}}{3\times 10^{-5}}=20.$ The dump capability is 20× the disturbance, so momentum is removed 20 times faster than it accumulates. The system is comfortably stable: schedule a dump roughly every few days and it clears in half a day.

Momentum is not destroyed — it is transferred out of the closed spacecraft system into Earth, through the magnetic interaction. The magnetorquer's dipole pushes against Earth's field, and by Newton's third law an equal-and-opposite reaction acts on the field's source — the enormous Earth. Earth's rotation changes by an utterly negligible amount because its moment of inertia is astronomically larger. So the "sink" for wheel momentum is the planet itself, exactly as Conservation of Angular Momentum demands: the total (spacecraft + Earth) is unchanged; only the spacecraft's share drops. A thruster-based dump instead throws the momentum away on ejected propellant mass.

The syntax below is Question ::: Answer — a reveal line: read the question, then uncover the part after the three colons. Why deliberately spin wheels at a non-zero bias? ::: To keep them away from ${\Omega }_w=0$, where a sign reversal makes bearing stiction reverse and inject impulsive jitter — the zero-crossing disturbance.