Gyrocompass

Why a spinning rotor points anywhere at all

A rotor with moment of inertia I spinning at ω carries angular momentum H = Iω (here 0.040 kg m² and 10 000 rpm give about 42 kg m²/s). With no torque on it, H keeps its direction in inertial space. A torque τ does not tilt the axis towards the push: it moves the axis at right angles to the push, dH/dt = τ, so the axis precesses at Ωp = τ/H.

The Earth turns under the gyro

In the local frame (East, North, Up) at latitude φ the ground rotates at Ω = 7.292×10-5 rad/s, with a horizontal north part Ω cos φ and a vertical part Ω sin φ. A ship adds ωE = -vN/R, ωN += vE/R and ωU += vE tan φ / R. The model integrates the spin-axis unit vector s = (E, N, U) exactly (fourth-order Runge-Kutta, no small-angle shortcuts):

ds/dt = -ω × s - a tanθ (s × z) - c θ eθ

The first term is the apparent motion of a gyro fixed in space. The second is the gravity torque of the pendulous bob: its torque is -m g l tanθ (s × z), and dividing by H gives a = m g l / H. The third is the damping torque of the oil dampers acting about the vertical axis, with c = 2 ζ ω0 (referenced to 45° latitude so the slider keeps its meaning).

Small-angle solution (what the plots compare against)

With azimuth ψ (east of north) and tilt θ: ψ' = ωU - aθ and θ' = ωNψ - ωE - cθ. Eliminating θ gives a damped oscillator ψ'' + cψ' + aωNψ = const:

• Natural period T0 = 2π / √(a Ω cos φ). The default rotor and bob give about 85 minutes at 45°: a gyrocompass swings very slowly and needs hours to settle. A heavier rotor (bigger H) lengthens the period, a heavier bob shortens it, and the period grows towards the poles where cos φ shrinks.
• Damped period Td = T0/√(1 - ζ2) and amplitude decay e-ct/2; with ζ ≥ 1 it creeps in without swinging.
• Settling point ψeq = (c ωU/a + ωE) / ωN and θeq = ωU/a. At rest this is the latitude error (c/a) tan φ (zero at the equator, large near the poles, where the compass fails). For a moving ship the second part ωE/ωN is the steaming error, δ ≈ -v cos K / (RΩ cos φ + v sin K) for speed v and course K; navigators correct both from tables.

The three numerical checks

(1) A free gyroscope's integrated drift against Ω sin φ (15.04°/h × sin lat). (2) The measured period of a full nonlinear swing against T0. (3) The centre of an undamped swing of a moving ship against the classical steaming-error formula. All three use the same integrator as the page, but a different closed form as the reference.

Honest simplifications

The rotor is a single rigid body with constant speed; the follow-up servo, ship roll and pitch (gimbals and the ballistic damping cross-coupling), rotor run-up and spin-down, and the speed-and-latitude correction unit of real instruments are not modelled. The float is shown with the pendulosity as a point bob; commercial compasses obtain it from a Schuler-tuned design (period 84.4 min) which this page approximates by the default settings. Geometry, masses and rpm are illustrative values of the right order, not those of a particular manufacturer. The vertical axis damping is an idealised linear torque.