ISEGORIA / MATH ENCYCLOPEDIA
Oscillators and escapements: how a watch keeps time
A mechanical watch is a torsional oscillator that has to be pushed without being disturbed. The balance and its hairspring set the frequency, the regulator trims it by a few seconds a day, and the escapement feeds it energy twice per swing. Where and how that push is delivered decides whether the rate changes with amplitude, and an out-of-balance speck of metal makes the watch run differently in every position except at one special amplitude.
Before you begin: Classical mechanics, damped oscillators and Bessel functions
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The balance, the hairspring and the timegrapher
The balance is a wheel of moment of inertia I on a spiral spring of torsional stiffness κ, so it swings with period T = 2π√(I/κ) whatever its amplitude, as long as the spring obeys Hooke’s law. A watch beating 28,800 times an hour has a balance at 4 Hz: each of the 8 beats a second is one tick of the escapement. The regulator index clamps the outer turn of the hairspring and changes its active length L; since κ ∝ 1/L, shortening the spring by one part in ten thousand makes the watch gain 4.3 s a day. Drag the gold regulator along its scale and watch the timegrapher on the right: it plots the time of each tick against a perfect clock, so a gaining watch draws a rising line, and a beat error, when the balance’s rest point is not centred on the escapement, splits the tick and tock into two parallel lines.
Worked example. A balance of \(I=12\ \mathrm{mg\,cm^2}=1.2\times10^{-9}\ \mathrm{kg\,m^2}\) at 4 Hz needs \(\kappa=I\omega_0^2=7.6\times10^{-7}\ \mathrm{N\,m/rad}\). Swinging through 270°, it stores \(E=\tfrac12\kappa A^2=8.4\ \mu\mathrm J\); with a quality factor \(Q=250\) it loses \(2\pi E/Q\) per cycle, so the mainspring must supply \(\omega_0E/Q\approx0.85\ \mu\mathrm W\). Moving the regulator so that \(L\) shrinks by 0.2‰ gives \(86400(\sqrt{1/0.9998}-1)=8.6\ \mathrm{s/day}\).
Watch out. Isochronism, a period independent of amplitude, holds only for an ideal linear spring and a free balance. A real hairspring breathes eccentrically as it winds and unwinds unless its outer end is shaped (a Breguet overcoil) or it is made from a spring whose centre of gravity stays on the axis, and the escapement disturbs the balance as the next experiment shows. The index changes \(L\) but also where the spring is held, so real regulators also shift the beat error slightly. The timegrapher here is simulated from the model, with no measurement noise.
Why does shortening the hairspring make the watch gain, and why is the effect proportional to half the fractional change?
The torque needed to twist a strip through a given angle is inversely proportional to its length, because the twist is shared along it: \(\kappa=EI_{\rm s}/L\) for a flat spiral of Young’s modulus \(E\) and section moment \(I_{\rm s}\). Shorter spring, stiffer spring, higher frequency. Since \(\omega_0\propto\sqrt\kappa\propto L^{-1/2}\), a small change gives \(\Delta\omega/\omega=-\tfrac12\Delta L/L\), and one day is 86,400 s, so the rate is \(43200\,|\Delta L|/L\) seconds a day.
2. Airy’s theorem: where the push goes
The escapement has to give the balance a push twice per cycle to replace what friction takes, and every push shifts the phase of the oscillation a little. Airy showed in 1830 which pushes do no harm: a kick delivered exactly as the balance passes its centre, where the displacement is zero, changes the amplitude but not the timing. A kick before the centre advances the phase and the watch gains; a kick after it retards the phase and the watch loses. On the left the state of the balance is a point going round the phase plane; each kick is a vertical jump in velocity, drawn exaggerated. On the right, the rate error for a kick at a fixed angle from the centre: it grows as the amplitude falls, because the same angle is a larger part of a smaller swing. That is the escapement error watchmakers fight at low mainspring power.
Worked example. With \(Q=250\), amplitude 270° and the impulse centred \(2^\circ\) before the line of centres, \(\sin\Phi_i=-2/270\) and the rate is \(+43200\times0.0074/250=+1.28\ \mathrm{s/day}\). If the mainspring runs down and the amplitude falls to 180°, the same impulse gives \(+1.92\ \mathrm{s/day}\): the watch gains more as it runs down. With the impulse exactly at the centre the error is zero at every amplitude.
Watch out. The formula treats each impulse as instantaneous and small, and the losses as viscous. A real lever escapement spreads its impulse over about 10° to 15° of balance rotation, roughly symmetric about the centre, so most of its effect cancels; what remains comes mainly from the unlocking, a braking resistance just before the centre, which by the same rule makes the watch lose as amplitude falls. The picture draws the loss and the kick schematically large so they can be seen; the readout uses the true sizes.
Why does a kick at the centre change the amplitude but not the phase?
In the phase plane the state is \((\theta,\dot\theta/\omega)=A(\sin\Phi,\cos\Phi)\). A kick changes only the velocity coordinate, moving the point vertically. At the centre the point is on the vertical axis, \(\Phi=0\), so a vertical move keeps it on the same ray from the origin: the radius \(A\) changes, the angle \(\Phi\) does not. Anywhere else the vertical move also rotates the ray, by \(-\varepsilon\sin\Phi\) to first order, and that rotation is a time shift that accumulates every cycle.
3. Positional error and the 220° amplitude
If the balance’s centre of mass is off its axis by a speck, gravity pulls on it whenever the watch is vertical and adds a pendulum-like torque to the hairspring’s. How much this changes the period depends on where the heavy spot sits relative to gravity and on the amplitude, through a Bessel function: the rate error is proportional to J₁(A)/A, with A in radians. That function changes sign at A = 3.832 rad, 219.5°, so at that amplitude the unbalance has no effect on the rate in any position, and above it the error reverses. Drag the heavy spot around the rim, pick a position of the watch, and read the four vertical-position curves on the right: the spread between them is what a timegrapher reports as positional variation.
Worked example. For \(I=12\ \mathrm{mg\,cm^2}\) at 4 Hz, an unbalance \(mr=20\ \mu\mathrm g\,\mathrm{mm}\) gives \(mgr/(I\omega_0^2)=2.6\times10^{-4}\). With the heavy spot straight down (\(\alpha=0\)) and amplitude 150°, \(J_1(2.618)/2.618=0.177\), so the watch gains \(86400\times2.6\times10^{-4}\times0.177=4.0\ \mathrm{s/day}\). Turned so the spot is straight up it loses the same amount; at 270° amplitude the two errors are \(\mp1.3\ \mathrm{s/day}\), reversed in sign.
Watch out. The formula is first order in the unbalance, from averaging the gravity torque over one swing; it is excellent for real unbalances, which are tiny. With the watch flat (dial up or down) gravity acts along the axis and this error vanishes; the dial-up and dial-down rates differ for another reason, friction at the pivot ends, which is not modelled here. Watchmakers exploit the 219.5° zero by adjusting the mainspring and escapement so the amplitude in the vertical positions sits near it, and poise the balance by removing metal where the heavy spot is.
Where does the Bessel function come from?
Averaging the gravity torque against the motion \(\theta=A\cos\phi\) over one cycle needs \(\int_0^{2\pi}\sin(A\cos\phi)\cos\phi\,d\phi\), the component of the torque in phase with the displacement, which is what changes the effective stiffness. That integral is the definition of \(2\pi J_1(A)\). Dividing by \(A\) turns a torque into a stiffness. Small amplitudes give \(J_1(A)/A\to\tfrac12\), an ordinary pendulum stiffness; large ones sample the torque \(\sin(\theta+\alpha)\) all the way round, and the average can change sign.
In practice
Where this mathematics and physics is at work, in explainers that take the real thing apart.