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A gyrocompass uses a spinning gyroscope and Earth's rotation to identify geographical north. In an ideal model whose axle is constrained to the local horizontal plane but free to turn about the vertical, the northward direction of the spin angular momentum is stable away from the poles; the opposite spin direction is unstable. A real settling instrument also requires damping. This is different from merely preserving an initial inertial orientation with a freely mounted gyroscope.

For a rapidly spinning gyrocompass with axle constrained horizontally, take azimuth $\theta$ positive from north towards west and positive spin angular momentum $J_s=I_1\omega$. In the terrestrial rotating reference frame, the vertical component of the torque equation, with zero external vertical torque, is $I_2\ddot{\theta }+J_s{\Omega }_E\cos \lambda \sin \theta \simeq 0$. Here $I_2$ is the vertical moment of inertia and terms of order ${\Omega }_E^2$ have been neglected. The small-angle approximation about $\theta =0$ gives simple harmonic motion with the displayed squared angular frequency. Linearizing about $\theta =\pi$ reverses its sign and gives an unstable equilibrium. At the poles this restoring effect vanishes. For a thin rotationally symmetric flywheel, the perpendicular axis theorem gives $I_1=2I_2$.

For this gyrocompass, let $\mathbf{n},\mathbf{w},\mathbf{z}$ be local north, west and upward unit vectors, forming a right-handed basis. Earth's rotation has local componentsLet the spin axis point along $\mathbf{u}=\cos \theta \mathbf{n}+\sin \theta \mathbf{w}$, and write its dominant angular momentum as $J_s\mathbf{u}$, where $J_s=I_1\omega >0$. In mechanical equilibrium relative to Earth, this leading angular momentum is fixed in the terrestrial rotating reference frame. The bearings can exert horizontal torques to keep the axle horizontal, but free rotation about the vertical implies $G_z=0$. The vertical component of the rotating-frame derivative formula therefore requiresAway from either geographical pole, $\cos \lambda \ne 0$, henceThis is a necessary equilibrium condition; the next part distinguishes its stable and unstable spin orientations. The rapid-spin approximation neglects angular momentum terms of order $I_2{\Omega }_E$ compared with $I_1\omega$. At either pole the horizontal component of Earth's rotation vanishes, and this north-seeking mechanism gives no preferred horizontal direction. The paper explicitly excludes $90^{\circ }$; if latitude is signed, $-90^{\circ }$ must likewise be excluded.

For the gyrocompass oscillation, use the same local basis and define $\theta$ about the stable orientation, with spin angular momentum directed north at $\theta =0$. Because the vertical is perpendicular to the horizontal axle, the extra vertical angular momentum from the slow azimuthal motion is $I_2\dot{\theta }$. To leading order in Earth's small angular velocity,A constant vertical contribution $I_2{\Omega }_E\sin \lambda$ from Earth's rotation has zero terrestrial time derivative; the transverse terrestrial contributions affect the following equation only at order ${\Omega }_E^2$. Taking the vertical component of the torque equation givesThe terrestrial derivative of $J_s\mathbf{u}$ is horizontal, as is ${\mathbf{\Omega }}_E\times (I_2\dot{\theta }\mathbf{z})$, so neither contributes to this component. Thus the small angle satisfies, using the small-angle approximation $\sin \theta \simeq \theta$,This is simple harmonic motion, $\theta =A\cos ({\omega }_{\mathrm{shm}}t+\varphi )$, with positive squared angular frequency away from the poles. The approximation requires rapid spin, small angular displacement and negligible bearing friction. Near the opposite orientation, putting $\theta =\pi +\delta$ instead yields $\ddot{\delta }-{\omega }_{\mathrm{shm}}^2\delta \simeq 0$, which is unstable. Consequently, the paper's oscillation claim applies to the stable equilibrium, not both north--south spin directions. Without damping the ideal gyrocompass continues to oscillate rather than settling.

For a thin gyrocompass flywheel with rotational symmetry, the perpendicular axis theorem gives $I_1=I_x+I_y=2I_2$. This does not require the mass to be concentrated at the rim. Convert the specified spin to angular velocity and use one sidereal day for Earth's rotation:At the given latitude,The associated oscillation period is $T=2\pi /{\omega }_{\mathrm{shm}}\simeq 29.0\mathrm{s}$. This uses the stated neglect of the support assembly's contribution to the moment of inertia; a substantial additional vertical moment of inertia would lengthen the oscillation period.