Exercises — Attitude control modes — spin stabilization, 3-axis active

Mode: spin stabilization — it uses stored angular momentum $\vec{L}$ for passive "gyroscopic stiffness", no active actuators firing. Limitation: you can only freely point the spin axis. The other two axes cannot be independently aimed — a camera on a side face just sweeps the sky as the craft rotates.

It is the gyroscopic coupling term — the "free" twisting that appears on axis 1 simply because the body is already spinning about axes 2 and 3. It costs no actuator effort; the rotation itself generates it. It vanishes when either ${\omega }_2=0$ or ${\omega }_3=0$ (no simultaneous spin on the other two axes), or when $I_2=I_3$ (those two inertias equal, so their difference is zero).

Step 1 — spin angular momentum. $L=I{\omega }_s=150\times 8=1200{\text{kg⋅m}}^2/\text{s}$. Why: $L$ is the "stiffness reservoir"; the bigger it is, the harder the axis is to swing. Step 2 — precession from $\vec{\tau }=\vec{\Omega }\times \vec{L}$. $\Omega =\frac{\tau }{L}=\frac{0.015}{1200}=1.25\times 10^{-5}\text{rad/s}.$ That is about $0.0007°/\text{s}$ — the axis wanders extremely slowly, which is the whole point.

Conservation of total angular momentum (nothing external, see Reaction wheels and Control-Moment Gyros (CMG)): $L_{\text{body}}+L_{\text{wheel}}=0\Rightarrow I_{\text{body}}{\omega }_{\text{body}}+I_w{\omega }_w=0.$ ${\omega }_{\text{body}}=-\frac{I_w{\omega }_w}{I_{\text{body}}}=-\frac{0.05\times 300}{400}=-0.0375\text{rad/s}.$ The minus sign: the body counter-rotates against the wheel. Momentum was traded, not created — no fuel used.

The closed loop is $I\ddot{\theta }+K_d\dot{\theta }+K_p\theta =0$, a second-order system (see PID control & second-order systems (damping ratio)). Match to standard form $\ddot{\theta }+2\zeta {\omega }_n\dot{\theta }+{\omega }_n^2\theta =0$: ${\omega }_n^2=\frac{K_p}{I}\Rightarrow K_p=I{\omega }_n^2=500\times 0.2^2=20\text{N⋅m/rad}.$ $2\zeta {\omega }_n=\frac{K_d}{I}\Rightarrow K_d=2\zeta {\omega }_{n}I=2(0.7)(0.2)(500)=140\text{N⋅m⋅s/rad}.$ Why $\zeta =0.7$: it settles fast with essentially no overshoot — the sweet spot between sluggish ($\zeta >1$) and ringing ($\zeta <1$).

$\dot{\vec{L}}=\vec{\tau }$ points along $+\hat{y}$, so the tip of $\vec{L}$ moves toward $+\hat{y}$ — the axis swings sideways in the $x$–$y$ plane, it does not flip over. Because the change $\dot{\vec{L}}$ is perpendicular to $\vec{L}$ (one along $x$, one along $y$), it rotates $\vec{L}$ without changing its length. Length $|\vec{L}|=I{\omega }_s$ is unchanged ⇒ the spin rate stays constant; only the direction precesses. This perpendicular-push-turns-not-speeds is the essence of Gyroscopic precession.

To remove stored momentum you need an external torque (only external torques change total angular momentum — the wheel and body just trade internally). See Momentum dumping / desaturation with magnetorquers. $t=\frac{h_{\text{wheel}}}{{\tau }_{\text{mag}}}=\frac{12}{0.004}=3000\text{s}\approx 50\text{min}.$ Why not self-dump: slowing the wheel down transfers its momentum to the body, which then starts rotating — total is conserved, nothing is removed. Only an external agent (Earth's magnetic field via the magnetorquer, or thrusters) sends momentum out of the whole spacecraft.

Recommend spin stabilization — cheap, passive, robust, and the mission only needs one axis pointed with slow re-pointing. That matches spin's strengths and doesn't need its 3-axis capability. Spin axis: axis 3, because $I_3=70$ is the maximum inertia (min energy at fixed $L$) — the only dissipation-stable choice. Handily $I_1=I_2$, so the transverse axes are symmetric (a clean "top"). Failure mode to guard against: flat-spin if energy dissipation ever pushes it off the max-$I$ axis; guard with a nutation damper and by ensuring $I_3$ truly stays the largest even after fuel is consumed.

Use the axis-1 Euler equation (see Euler's rotational equations of motion): ${\tau }_1=I_1{\dot{\omega }}_1+(I_3-I_2){\omega }_2{\omega }_3.$ For ${\dot{\omega }}_1=0$ the actuator must exactly supply the coupling term: ${\tau }_1=(I_3-I_2){\omega }_2{\omega }_3=(340-260)(0.05)(0.04)=80\times 0.002=0.16\text{N⋅m}.$ Why: 3-axis control must fight the very gyroscopic coupling that spin stabilization exploits — the actuator burns effort just to hold the axis steady against the body's own spinning geometry.

Spin stability with dissipation requires $I_{\text{spin}}=I_{\max }$, i.e. margin $M=I_{\text{spin}}-I_{\text{trans}}>0$. Start: $M_0=120-100=+20{\text{kg⋅m}}^2$ → stable. End: $I_{\text{spin}}=120-25=95{\text{kg⋅m}}^2$, so $M_1=95-100=-5{\text{kg⋅m}}^2$ → unstable! By end-of-life the spin axis is no longer the maximum-inertia axis; energy dissipation will grow nutation into a flat-spin. Design fix: either keep enough spin-axis inertia (mass distribution/ballast) so $M$ stays positive across the whole fuel budget, or switch to active nutation damping. This is precisely the Explorer 1 flat-spin anomaly mechanism, sneaking in through fuel consumption.

Spin stabilization is stable only about the ::: maximum ::: -inertia axis when dissipation is present. Precession rate for a spinning body under transverse torque is $\Omega =$ ::: $\tau /(I{\omega }_s)$ :::. A reaction wheel produces body torque by ::: exchanging (conserving) total angular momentum — no external torque :::. Wheels lose authority when they ::: saturate :::, fixed by ::: momentum dumping / desaturation with magnetorquers or thrusters :::. For a PD-controlled axis, $K_p=I{\omega }_n^2$ and $K_d=$ ::: $2\zeta {\omega }_{n}I$ :::.