Foucault pendulum

The invitation cards Léon Foucault sent out in March 1851 contained the most audacious sentence ever printed on scientific stationery: "You are invited to come and see the Earth turn." Under the dome of the Panthéon he had hung a 28-kilogram brass bob on a 67-metre steel wire, pulled it aside, tied it with a cotton thread, and burned the thread to launch the cleanest swing in history. Each majestic 16-second oscillation ended with a stylus grazing a ridge of damp sand — and each ridge-cut fell about two and a half millimetres to the left of the last. Hour after hour the swing plane crept clockwise, 11.3 degrees per hour, while the pendulum itself asked for nothing but gravity and patience. No telescope, no stars: a closed room, watching itself rotate. This article is about that machine — the physics that makes it work and the engineering that makes it honest. The precession itself is pure Coriolis mechanics: the Coriolis force article derives the force and even sketches the precession; here we solve the pendulum properly. We write the full equation of motion with tension and rotation, reduce it to two coupled oscillators, solve them exactly with complex coordinates, and extract both the celebrated rate $\Omega_{\mathrm F} = \omega\sin\lambda$ and the fine print the compact treatments skip: what the initial conditions really do, why the bob never quite passes through the centre, and why a carelessly launched pendulum precesses at the wrong rate for reasons that have nothing to do with the Earth. That fine print is the difference between a physics demonstration and a fraud. The Foucault signal is tiny and slow; its great enemy — the intrinsic precession of any elliptical swing — is fast and always lurking. Foucault beat it with length, mass, and a burnt thread. Understanding why those choices work is understanding the whole experiment.

Key takeaways

A Foucault pendulum is a long, heavy, freely swinging pendulum whose plane of oscillation slowly rotates relative to the ground — the first direct mechanical proof of Earth's rotation (Paris, 1851).
In Earth's rotating frame the swing plane precesses at $\Omega_{\mathrm F} = \omega\sin\lambda$ , clockwise seen from above in the Northern Hemisphere — independent of the pendulum's length, mass, and amplitude.
The full solution is $\zeta(t) = e^{-i\Omega_{\mathrm F} t}\left(A\,e^{i\omega_1 t} + B\,e^{-i\omega_1 t}\right)$ : an ordinary planar oscillation viewed through a slowly rotating frame factor.
Launched from rest, the bob still traces a hair-thin ellipse of aspect ratio $b/a = \Omega_{\mathrm F}/\omega_1 \sim 10^{-4}$ — it misses the centre every pass, and the rosette it draws has outward-pointing cusps.
The real enemy is Airy precession: any elliptical swing precesses on its own at $\Omega_{\mathrm{Airy}} = \tfrac{3}{8}\,\omega_0\,ab/L^2$ , which for a short, sloppily launched pendulum can exceed the Foucault rate many times over.
Length is the cure: Foucault's 67 m wire made the pendulum slow, the ellipse thin, and the spurious precession ten thousand times smaller than the signal — with the Charron ring (1931) as the modern insurance.
Key idea: in Earth's rotating frame the Coriolis force $-2m\,\boldsymbol\omega\times\mathbf v$ deflects the bob's horizontal velocity; only the vertical projection $\omega\sin\lambda$ matters, rotating the swing plane at $\Omega_{\mathrm F}=\omega\sin\lambda$ — equivalently, the parallel-transport holonomy around a latitude circle.

Where this lives today

The pendulum's job — measuring Earth's rotation from inside a closed room — is now done to absurd precision by its direct descendants. Large ring-laser gyroscopes (the G ring at Wettzell, the four-component ROMY array in Munich) split counter-propagating laser beams around a closed path and read the planet's spin as a beat frequency: they sense the length-of-day changing by fractions of a millisecond as winds and earthquakes shuffle angular momentum, day by day, no stars required. The GINGER project, under construction beneath the Gran Sasso mountain in the mid-2020s, aims at one part in $10^9$ of $\omega$ — precise enough to see general relativity's frame-dragging correction to the rotation a Foucault pendulum first made visible with sand. Every one of these instruments is philosophically Foucault's: a local, sealed apparatus confessing the rotation of its planet. The pendulum itself, meanwhile, has become physics' great public monument — hundreds swing in museums, universities, and the UN lobby, each with its hour-pins and its patient crowd. And the failure modes this article dwelt on became a teaching industry of their own: launching, driving, and Charron-damping a pendulum so that it tells the truth within a few percent is a beloved senior-lab project, because it forces the full error budget — equation (8) against equation (6) — on anyone who tries.

The physical picture

Start where the physics is transparent: the North Pole. Hang a pendulum exactly on the spin axis and set it swinging. Nothing torques it about the vertical, so its swing plane stays fixed relative to the distant stars — while the Earth, the building, and you rotate counterclockwise beneath it once per sidereal day. To you, the plane appears to turn clockwise, one full revolution in 23.93 hours. The pendulum is not doing anything; the floor is. Now carry the pendulum south to latitude $\lambda$ . Earth's angular velocity vector no longer points along your local vertical: only the fraction $\sin\lambda$ of it does. It is this vertical component alone that rotates your horizontal world within itself — the horizontal component $\omega\cos\lambda$ tips the plane toward tomorrow but cannot turn anything within it. So the apparent rotation of the swing plane slows to $\Omega_{\mathrm F} = \omega\sin\lambda$ : 11.3°/h in Paris, one turn in 31.8 hours. At the equator, $\sin\lambda = 0$ and the plane never turns at all. (Why the naive "plane fixed in space" story is only exactly true at the pole — and what replaces it elsewhere — is a beautiful geometric question with its own article: the parallel-transport view of the Foucault pendulum.)

Figure

The Panthéon machine — a 67-metre pendulum and the line it draws in the sand

Microscopically the agent is the Coriolis force. The bob's velocity is almost purely horizontal, and the vertical part of Earth's rotation crosses into it to give a sideways force $-2m\,\omega\sin\lambda\;\hat{\mathbf z}\times\mathbf v$ : always 90° to the right of the motion (Northern Hemisphere), on the way out and on the way back. That last clause is the subtlety. A naive glance says the rightward push on the outbound swing should cancel against the rightward push on the return — but "right of the velocity" flips sides with the velocity, so both half-swings are nudged in the same rotational sense. The deflections accumulate coherently, swing after swing, and the plane creeps around. Per swing the creep is minuscule — $\Omega_{\mathrm F} T_0 \approx 9\times10^{-4}$ rad in Paris, a couple of millimetres at the sand ring — but it never rests and never cancels. Why the extravagant hardware, then — 67 metres of wire and 28 kilograms of brass? Because the Foucault signal is slow, and everything that competes with it gets worse as pendulums get shorter. A long pendulum swings slowly (more precession per swing), decays slowly (air drag has hours of amplitude to eat), keeps its swing angle small (the nonlinear corrections shrink), and — the decisive point, quantified below — makes the spurious precession of an imperfectly planar swing ten thousand times smaller than the Earth's signal. The machine is a low-pass filter for truth.

The equations of motion, solved properly

Set up local axes at latitude $\lambda$ : $\hat{\mathbf x}$ east, $\hat{\mathbf y}$ north, $\hat{\mathbf z}$ up along the local plumb line, origin at the bob's rest position. In Earth's rotating frame the bob of mass $m$ on a wire of length $L$ obeys

\ddot{\mathbf r} = \mathbf g_{\mathrm{eff}} - 2\,\boldsymbol\omega\times\dot{\mathbf r} + \frac{\mathbf T}{m} \qquad (1)

where: $\mathbf T$ — the wire tension, pointing from bob to suspension; $\mathbf g_{\mathrm{eff}}$ — effective gravity (true gravity plus centrifugal acceleration), which defines the local vertical and the rest position, so the centrifugal force is already spent the moment we align $\hat{\mathbf z}$ with the plumb line; $\boldsymbol\omega$ — Earth's angular velocity, $\omega = 7.292\times10^{-5}\ \mathrm{rad/s}$ . The Euler force is absent since $\dot{\boldsymbol\omega} \approx 0$ . Everything that follows is the taming of equation (1). In the local basis the planet's spin splits into the two components whose different jobs organize this whole subject:

\boldsymbol\omega = \omega\left(\cos\lambda\,\hat{\mathbf y} + \sin\lambda\,\hat{\mathbf z}\right) \qquad (2)

where: $\omega\sin\lambda$ — the vertical component, the only piece that rotates horizontal velocities within the horizontal plane; $\omega\cos\lambda$ — the horizontal (northward) component, which couples horizontal and vertical motion and will be argued away, honestly, in the derivation below.

Derivation AFrom the full equation of motion to the exact precessing solution

We are deriving the complete small-oscillation solution of equation (1) — the pair of coupled equations (3), their complex form (4), and the exact solution (5) whose slowly rotating prefactor is the Foucault precession. The strategy: linearize the tension for small swings, justify discarding the $\omega\cos\lambda$ terms by averaging (this is the one genuinely physical approximation, and we do it with eyes open), then fuse the two real equations into one complex equation that solves in two lines. Watch two things: the discarded terms must be shown to average out rather than merely asserted small, and the final frequencies must be kept exact in $\Omega$ — the tiny difference $\omega_1 - \omega_0$ is irrelevant on Earth but the exactness is free.

Step 1. Linearize the tension. For swing amplitudes $a \ll L$ the bob stays near the horizontal plane: $z \approx -(x^2+y^2)/2L$ is second order, the wire direction is $(-x/L, -y/L, 1)$ to first order, and $T \approx mg$ from the vertical balance. The horizontal restoring accelerations are therefore $-\,(g/L)\,x$ and $-\,(g/L)\,y$ — the isotropic two-dimensional oscillator with $\omega_0 = \sqrt{g/L}$ . Why isotropy matters: the Coriolis effect we are hunting is a rotation of the swing direction; any anisotropy in the restoring force (an elliptical support, a lopsided clamp) would rotate or lock the plane on its own and contaminate the measurement — this is an experimental requirement disguised as an algebra step.

Step 2. Add the Coriolis components. With (2) and $\dot{\mathbf r} = (\dot x, \dot y, \dot z)$ , expanding $-2\boldsymbol\omega\times\dot{\mathbf r}$ gives $\left(2\omega\sin\lambda\,\dot y - 2\omega\cos\lambda\,\dot z\right)$ along $\hat{\mathbf x}$ , $\;-2\omega\sin\lambda\,\dot x$ along $\hat{\mathbf y}$ , and $+2\omega\cos\lambda\,\dot x$ along $\hat{\mathbf z}$ . Why the $\cos\lambda$ terms may be dropped: $\dot z \sim (a/L)\,\dot x$ is smaller than the horizontal velocities by the factor $a/L$ ( $\sim 1/20$ even for Foucault's generous swing), and it oscillates at $2\omega_0$ (the bob dips twice per period), so its product with the slow dynamics averages to zero over every swing; the remaining $\cos\lambda$ effect is a vertical force that feebly modulates the tension, turning the plane not at all. Dropping these terms is a time-averaging argument, not a smallness assertion alone — and it is why the Foucault rate contains $\sin\lambda$ with no $\cos\lambda$ correction at leading order.

Step 3. Write the working system, with $\Omega \equiv \omega\sin\lambda$ :

\ddot x = -\omega_0^2\,x + 2\Omega\,\dot y, \qquad \ddot y = -\omega_0^2\,y - 2\Omega\,\dot x \qquad (3)

where: the cross-coupling terms are the Coriolis force: each coordinate's acceleration is fed by the other's velocity, the algebraic signature of a force that rotates velocities by 90°.

Step 4. Complexify. Set $\zeta = x + iy$ , multiply the second equation by $i$ , add:

\ddot\zeta + 2i\,\Omega\,\dot\zeta + \omega_0^2\,\zeta = 0 \qquad (4)

where: multiplication by $i$ is rotation by 90° in the horizontal plane — the natural algebra for this force. One complex equation now holds both coordinates and their coupling.

Step 5. Solve exactly. The ansatz $\zeta = e^{i\mu t}$ gives $-\mu^2 - 2\Omega\mu + \omega_0^2 = 0$ , so $\mu = -\Omega \pm \omega_1$ with $\omega_1 = \sqrt{\omega_0^2 + \Omega^2}$ , and the general solution is

\zeta(t) = e^{-i\Omega t}\left(A\,e^{i\omega_1 t} + B\,e^{-i\omega_1 t}\right) \qquad (5)

where: $A, B$ — complex constants fixed by launch conditions; $\omega_1$ — the oscillation frequency, shifted from $\omega_0$ by a fractional $\Omega^2/2\omega_0^2 \sim 10^{-8}$ , far beyond any sand ring's resolution. The bracket alone describes a fixed ellipse (a planar swing when $|A| = |B|$ ); the prefactor $e^{-i\Omega t}$ rotates that whole figure rigidly, clockwise, at

\Omega_{\mathrm F} = \omega\sin\lambda, \qquad T_{\mathrm F} = \frac{2\pi}{\omega\sin\lambda} = \frac{23.93\ \mathrm{h}}{\sin\lambda} \qquad (6)

where: $T_{\mathrm F}$ — the pendulum day, the time for one full turn of the plane; one sidereal day at the pole, 31.8 h in Paris, infinite at the equator. Note the rate is $\Omega$ , not $2\Omega$ : undoing a Coriolis coupling of strength $2\Omega$ requires co-rotating at only $\Omega$ , because the substitution's exponential is differentiated twice. The Coriolis force article's FAQ dwells on exactly this factor. Verified: reproduces Morin ch. 10 and Goldstein §4.10; the exact roots $-\Omega\pm\omega_1$ appear in Landau–Lifshitz §39 problem 1; at $\lambda = 90°$ the rate is $\omega$ (plane fixed in inertial space) and at $\lambda = 0$ it vanishes.

Which formula answers which question

you want	use	notes
the precession rate and period	$\Omega_{\mathrm F} = \omega\sin\lambda$ , eq. (6)	latitude only — not $L$ , not $m$ , not amplitude
the full trajectory	eq. (5)	rotating prefactor × fixed ellipse
the swing frequency	$\omega_1 = \sqrt{\omega_0^2+\Omega^2} \approx \sqrt{g/L}$	Coriolis shift $\sim10^{-8}$ , invisible
the unavoidable ellipse from a rest launch	$b/a = \Omega/\omega_1$ , eq. (7)	$\sim10^{-4}$ for long pendulums
the spurious precession of an elliptical swing	$\Omega_{\mathrm{Airy}} = \tfrac38\,\omega_0\,ab/L^2$ , eq. (8)	the enemy — scales like $L^{-5/2}$ for fixed $a$ , $b$
creep per swing at the sand ring	$\delta = a\,\Omega_{\mathrm F}\,T_0$	≈ 2.7 mm for the Panthéon

What the launch really does

Equation (5) hides the experiment's whole craft in the constants $A$ and $B$ . The textbook picture — "the bob swings in a plane, the plane rotates" — corresponds to $|A| = |B|$ exactly. No real launch achieves it, and the Coriolis force itself forbids it.

Go deeper · Derivation B · 2 minLaunch from rest and the hair-thin ellipse

Derivation BLaunch from rest and the hair-thin ellipse

The goal is equation (7): a bob released from rest at displacement $a$ does not swing in a plane but traces an ellipse of tiny but nonzero aspect ratio $b/a = \Omega/\omega_1$ , missing the centre on every pass. The strategy is elementary — impose the initial conditions $\zeta(0) = a$ , $\dot\zeta(0) = 0$ on the exact solution (5) and read off the semi-axes of the bracket's ellipse. The point of doing it carefully is conceptual: "released from rest" means at rest in the rotating laboratory, which is not at rest in inertial space — the launch smuggles in a sideways velocity of order $\Omega a$ , and that is precisely what opens the ellipse.

Step 1. Apply $\zeta(0) = a$ (real, bob displaced east say) to (5): $A + B = a$ .

Step 2. Differentiate (5) and apply $\dot\zeta(0) = 0$ :

\dot\zeta(0) = -i\Omega\,(A+B) + i\omega_1\,(A - B) = 0 \quad\Longrightarrow\quad A - B = \frac{\Omega}{\omega_1}\,a

Why this is the crux: if $\Omega$ were zero, $A = B$ and the bracket would be $a\cos\omega_1 t$ — a perfect planar swing. The rotating frame's own $-i\Omega\zeta$ contribution to the velocity forces $A \neq B$ : you cannot launch a truly planar swing from rest on a turning planet.

Step 3. Read off the geometry of the bracket $\eta(t) = A e^{i\omega_1 t} + B e^{-i\omega_1 t}$ : with real $A > B > 0$ this is an ellipse with semi-major axis $A + B = a$ along the launch direction and semi-minor axis

b = A - B = \frac{\Omega}{\omega_1}\,a \qquad (7)

where: $b$ — the miss distance at the centre: the bob passes the vertical not through it. For the Panthéon ( $\omega_1 = 0.383\ \mathrm{rad/s}$ , $\Omega = 5.49\times10^{-5}\ \mathrm{rad/s}$ , $a = 3\ \mathrm{m}$ ): $b/a = 1.4\times10^{-4}$ , i.e. $b \approx 0.43\ \mathrm{mm}$ — a half-millimetre sidestep at the bottom of a six-metre-wide swing.

Step 4. See the rosette. Released from rest, the bob's speed vanishes at each extreme, so the path in the precessing frame has outward-pointing cusps at the amplitude circle: the figure is a many-petalled rosette advancing $\Omega_{\mathrm F} T_0/2$ per half-swing. A pushed launch (nonzero $\dot\zeta(0)$ transverse component) rounds the cusps into loops and — far worse — fattens $b$ by $v_\perp/\omega_1$ per unit push: a careless 5 mm/s sideways hand-tremor on a museum pendulum injects an ellipse tens of times larger than the Coriolis minimum. Hence the burnt thread: cotton cannot push. Verified: $b/a = \Omega/\omega_1$ matches Landau–Lifshitz §39 problem 1's rest-launch solution; setting $\Omega \to 0$ recovers the planar swing identically.

Why does a small ellipse matter at all? Because a spherical pendulum is not a harmonic oscillator — the restoring force $mg\sin\theta$ runs below linear — and an anharmonic 2-D oscillator makes elliptical orbits precess on their own, Earth or no Earth. The standard result (Airy, 1851 — analyzed within months of Foucault's demonstration, because sceptics immediately blamed it) is

\Omega_{\mathrm{Airy}} = \frac{3}{8}\,\omega_0\,\frac{a\,b}{L^2} \qquad (8)

where: $a, b$ — the ellipse's semi-axes; the precession is in the ellipse's own direction of circulation, so it can add to or subtract from the Foucault rate depending on the launch error's sign. Read the scalings like an engineer: for fixed swing size the rate dies as $1/L^2$ explicitly and another $\sqrt{1/L}$ through $\omega_0$ — length is a weapon. And it is proportional to $b$ : every millimetre of ellipse you fail to suppress is precession you did not order.

Figure

The honest pendulum and the lying pendulum — Coriolis rosette versus Airy ellipse drift

The engineering follows from the budget. Long wire, heavy bob: length crushes $\Omega_{\mathrm{Airy}}$ and slows the swing; mass extends the decay time against air drag (the Panthéon pendulum swung usefully for about six hours per launch). Burnt-thread release: ties the bob aside, lets all hand tremor damp out through the wire, then releases with strictly zero push — the cleanest approximation to "from rest" mechanics allows. The Charron ring (Félix Charron, 1931): a ring the wire grazes near the suspension at the extremes of each swing; the brief touch preferentially eats the minor-axis velocity, damping the ellipse toward planarity every single period — negative feedback for $b$ . Modern continuously running museum pendulums add a magnetic drive at the bottom dead centre to replace drag losses; done symmetrically it feeds energy without feeding rotation, and the Charron ring quietly disciplines whatever ellipse the crowd's draughts excite.

The numbers that convinced Paris

Go deeper · Derivation C · 1 minThe Paris rate and what the sand ring recorded

Derivation CThe Paris rate and what the sand ring recorded

The goal is the complete numerical fingerprint of Foucault's 1851 demonstration from just three inputs — $L = 67\ \mathrm{m}$ , $\lambda = 48.85°$ , $\omega = 2\pi/86{,}164\ \mathrm{s}$ — ending with the millimetres-per-swing figure a spectator actually saw, and the error budget that made it credible. The strategy is plain arithmetic through equations (6), (7), (8), which is exactly why it belongs in a derivation card: every number the museum plaque quotes should be recomputable by a student in two minutes, and here they all are in one chain.

Step 1. Swing period: $\omega_0 = \sqrt{g/L} = \sqrt{9.81/67} = 0.3826\ \mathrm{rad/s}$ , so $T_0 = 2\pi/\omega_0 = 16.4\ \mathrm{s}$ — slow enough for the crowd to hold its breath through a single traverse.

Step 2. Precession rate: $\Omega_{\mathrm F} = \omega\sin\lambda = 7.292\times10^{-5} \times \sin 48.85° = 7.292\times10^{-5}\times0.7531 = 5.49\times10^{-5}\ \mathrm{rad/s}$ . In degrees per hour: $5.49\times10^{-5} \times 3600 \times (180/\pi) = 11.3°/\mathrm{h}$ . Pendulum day: $T_{\mathrm F} = 2\pi/\Omega_{\mathrm F} = 1.144\times10^{5}\ \mathrm{s} = 31.8\ \mathrm{h}$ .

Step 3. The visible creep. Per full swing the plane advances $\Omega_{\mathrm F} T_0 = 5.49\times10^{-5}\times16.4 = 9.0\times10^{-4}$ rad; at the sand ring, $a = 3\ \mathrm{m}$ out, successive stylus cuts land $\delta = a\,\Omega_{\mathrm F} T_0 = 3\times9.0\times10^{-4} = 2.7\ \mathrm{mm}$ apart. Why this number won the argument: 2.7 mm per 16 seconds is directly visible to a patient eye — the audience did not take rotation on authority, they watched it accumulate cut by cut.

Step 4. The error budget. Rest-launch ellipse: $b = (\Omega_{\mathrm F}/\omega_0)\,a = 1.44\times10^{-4}\times3 = 0.43\ \mathrm{mm}$ . Its Airy precession: $\Omega_{\mathrm{Airy}} = \tfrac38\,\omega_0\,ab/L^2 = 0.375\times0.3826\times(3\times4.3\times10^{-4})/67^2 = 4.1\times10^{-8}\ \mathrm{rad/s}$ — a factor $7.5\times10^{-4}$ of the signal, equivalently the fractional bound $\tfrac38\,(a/L)^2$ . Why length is the whole design: on a 2-metre parlour pendulum with a 20 cm swing and a hand-launched 1 cm ellipse, the same formula gives $\Omega_{\mathrm{Airy}} = 0.375\times2.21\times(0.2\times0.01)/4 = 4.2\times10^{-4}\ \mathrm{rad/s}$ — eight times the Foucault rate at Paris latitude, in whichever direction the launch happened to swirl. Short Foucault pendulums do not underperform; they lie. *Verified: 11.3°/h and 31.8 h match the historically quoted Panthéon values; the $\tfrac38(a/L)^2$ fractional Airy bound for a rest launch reproduces the standard analysis (Airy 1851; Schulz-DuBois, Am. J. Phys. 1970).*

Limiting cases

Case	Result	Meaning
North Pole, $\lambda = 90°$	$\Omega_{\mathrm F} = \omega$ , $T_{\mathrm F} = 23.93$ h	plane fixed in inertial space; the ground turns beneath it — the only latitude where the naive picture is exact
equator, $\lambda = 0$	$\Omega_{\mathrm F} = 0$	no vertical component of $\boldsymbol\omega$ ; the plane never turns
Southern Hemisphere, $\lambda < 0$	$\Omega_{\mathrm F} < 0$	precession reverses: counterclockwise seen from above
switch off Coriolis	$\Omega \to 0$ in (5)	$A = B$ from a rest launch: a fixed planar swing — the precession is Coriolis or nothing
short, sloppy pendulum	$\Omega_{\mathrm{Airy}} \gtrsim \Omega_{\mathrm F}$	the Airy ellipse precession swamps the signal — the reason parlour replicas fail
hypothetical fast-spinning planet, $\Omega \sim \omega_0$	splitting $\omega_1 \pm \Omega$ resolvable	the two circular modes visibly beat; Earth hides this by $\Omega/\omega_0 \sim 10^{-4}$

Common trapThe swing plane is fixed relative to the stars, and something must torque it to make it turn

Both halves are wrong, and each hides real physics. First: away from the poles the swing plane is not fixed in inertial space — at Paris it turns at $\omega\sin\lambda$ , not $\omega$ , so an inertial observer sees it rotate too (the honest geometric statement is parallel transport, the subject of a sibling article). Second: nothing torques the plane about the vertical, and nothing needs to. The Coriolis force acts on the bob, perpendicular to its velocity, doing zero work; the "rotation of the plane" is the accumulated sideways creep of a swinging point mass, not the rigid rotation of any physical object. Corollary for museum-goers: a Foucault pendulum that precesses at the wrong rate (or the wrong way!) is not usually measuring the Earth badly — it is measuring its own launch ellipse, through equation (8). Check the Charron ring before you check the planet.

Worked examples

1 · Commissioning a museum pendulum (confidence builder)

A science centre at latitude $40°\mathrm{N}$ hangs a Foucault pendulum with $L = 15\ \mathrm{m}$ and swing amplitude $a = 1.2\ \mathrm{m}$ . Predict the swing period, the precession rate in degrees per hour, the pendulum day, and the creep per swing a visitor would see at the amplitude circle.

Step 1. Swing: $\omega_0 = \sqrt{9.81/15} = 0.809\ \mathrm{rad/s}$ , so $T_0 = 2\pi/0.809 = \boxed{7.8\ \mathrm{s}}$ .

Step 2. Precession: $\Omega_{\mathrm F} = 7.292\times10^{-5}\times\sin 40° = 7.292\times10^{-5}\times0.643 = 4.69\times10^{-5}\ \mathrm{rad/s}$ . In degrees per hour: $4.69\times10^{-5}\times3600\times57.30 = \boxed{9.7°/\mathrm{h}}$ .

Step 3. Pendulum day: $T_{\mathrm F} = 2\pi/4.69\times10^{-5} = 1.34\times10^{5}\ \mathrm{s} = \boxed{37.2\ \mathrm{h}}$ — the plane needs a day and a half for a full turn; the museum marks hours with toppling pins instead.

Step 4. Creep per swing: $\delta = a\,\Omega_{\mathrm F}\,T_0 = 1.2\times4.69\times10^{-5}\times7.8 = 4.4\times10^{-4}\ \mathrm{m} = \boxed{0.44\ \mathrm{mm}}$ — six times finer than the Panthéon's 2.7 mm. Smaller amplitude, shorter period: the visible signature shrinks twice over, which is why smaller installations rely on pin-toppling rather than sand-reading.

2 · The launch-quality budget (exam level)

For the same pendulum ( $L = 15\ \mathrm{m}$ , $a = 1.2\ \mathrm{m}$ , $\lambda = 40°\mathrm{N}$ ), find (i) the minimum ellipse a rest launch can achieve, (ii) its Airy precession as a fraction of the Foucault rate, and (iii) the largest ellipse semi-minor axis $b$ tolerable if the spurious precession must stay below 5% of the signal.

Step 1. Rest-launch minimum: $b_{\min} = (\Omega_{\mathrm F}/\omega_0)\,a = (4.69\times10^{-5}/0.809)\times1.2 = 7.0\times10^{-5}\ \mathrm{m} = 0.07\ \mathrm{mm}$ . Unavoidable, and utterly harmless.

Step 2. Its Airy rate: $\Omega_{\mathrm{Airy}} = \tfrac38\,\omega_0\,a\,b_{\min}/L^2 = 0.375\times0.809\times(1.2\times7.0\times10^{-5})/225 = 1.1\times10^{-7}\ \mathrm{rad/s}$ , which is $1.1\times10^{-7}/4.69\times10^{-5} = \boxed{0.24\%}$ of the Foucault rate — equivalently the closed-form bound $\tfrac38(a/L)^2 = 0.375\times(0.08)^2 = 0.24\%$ . A rest launch is intrinsically clean.

Step 3. Tolerance: demand $\tfrac38\,\omega_0\,a\,b/L^2 \le 0.05\,\Omega_{\mathrm F}$ , so

b \le \frac{0.05\,\Omega_{\mathrm F}\,L^2}{\tfrac38\,\omega_0\,a} = \frac{0.05\times4.69\times10^{-5}\times225}{0.375\times0.809\times1.2} = \boxed{1.4\ \mathrm{mm}}

Step 4. Interpret. A 1.4 mm minor axis on a 1.2 m swing is an aspect ratio of $10^{-3}$ : a transverse launch speed of just $b\,\omega_0 \approx 1\ \mathrm{mm/s}$ breaks the budget. No hand can release that cleanly — a tied thread, burned, can; and a Charron ring keeps re-shrinking $b$ against drafts and drive imperfections for as long as the pendulum runs. Foucault's 67 m installation enjoys a budget $(67/15)^2 = 20$ times looser — length forgives.

Origins · 1 minLéon Foucault and how this was discovered

Léon Foucault

1819–1868 · Paris Observatory

Foucault never held a physics degree — he abandoned medicine (he fainted at blood) and taught himself instrumentation, which may be why his experiments read like engineering poems. The pendulum idea came to him at a lathe: watching a slender rod vibrate in a fixed plane while the chuck turned beneath it. He tested it in his cellar on the rue d'Assas in January 1851 with a 2 m wire, scaled to 11 m at the Paris Observatory in February, and to 67 m under the Panthéon dome in March, where all of Paris queued to watch the sand cuts march. Within a year he had distilled the effect into a spinning rotor whose axis holds fixed in inertial space — he named it the gyroscope ("rotation-watcher") — trading the pendulum's slow statistics for an instant readout. He went on to measure the speed of light in the laboratory to 0.6% with a rotating mirror, killing the corpuscular theory's last prediction, and to invent the silvered-glass telescope mirror and the optical test that still bears his name. The Panthéon pendulum still swings today; its 1851 bob, retired to the Musée des Arts et Métiers, survived a wire failure in 2010 and remains the most famous 28 kilograms in physics.

History

1661

— Vincenzo Viviani, Galileo's last student, notes that a long pendulum's swing plane slowly drifts and quietly attributes it to apparatus imperfection — the effect recorded, unexplained, 190 years early.

1851

— Foucault demonstrates the precessing pendulum in his cellar (January), at the Paris Observatory (February), and publicly under the Panthéon dome (March); Airy analyzes the elliptical-swing artefact within months.

1852

— Foucault invents and names the gyroscope, converting the day-long pendulum measurement into a direct display of inertial direction-keeping.

1931

— Félix Charron introduces the wire-grazing ring that damps ellipse growth, making permanently running Foucault pendulums practical.

1995

— The Panthéon installation is restored to continuous operation; the original bob, damaged by a cable failure in 2010, is conserved at the Musée des Arts et Métiers.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 10 (Accelerating frames) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §4.10 and §6 · Landau & Lifshitz, Mechanics, Vol. 1, §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, Ch. 9 (Mechanics in noninertial frames)

Provenance. A verifier should recompute: (1) $\omega=2\pi/86164\,\text{s}=7.292\times10^{-5}$ rad/s (sidereal day); (2) $\Omega_{\mathrm F}=\omega\sin48.85^\circ\approx5.49\times10^{-5}$ rad/s and $T_{\mathrm F}\approx31.8$ h; (3) the complex-variable reduction $\ddot\zeta+2i\Omega\dot\zeta+(g/L)\zeta=0$ and its solution $\zeta=e^{-i\Omega t}\eta$ . Cross-check against Morin Ch. 10, Goldstein §4.10, Landau–Lifshitz Vol 1 §39, and Taylor Ch. 9. Geometric deficit $2\pi(1-\sin\lambda)$ is standard (Berry/Hannay anholonomy). Worked-example numbers on this page ( $T_0 = 16.4$ s and 2.7 mm/swing for the Panthéon; 9.7°/h, 37.2 h, 0.07 mm and 1.4 mm budgets for the 15 m example) are illustrative arithmetic recomputed from equations (6)–(8), with the Airy rate $\tfrac38\,\omega_0 ab/L^2$ per Airy 1851 / Schulz-DuBois Am. J. Phys. 38, 173 (1970).

Builds on: Non-inertial reference frames · Coriolis force · Angular velocity vector of a rotating frame · Time derivatives in rotating frames · Effective gravity on the rotating Earth. Leads to: Fictitious (inertial) forces · Centrifugal force · Euler force.

Why doesn't the precession rate depend on the length or amplitude of the pendulum?

Because the Coriolis force and the swing conspire to cancel every pendulum property. The sideways Coriolis acceleration is proportional to the bob's speed; a faster pendulum (shorter, or wider-swinging) is deflected harder per second — but it also crosses the arc more often, and the angle accumulated per swing works out to $\Omega_{\mathrm F} T_0$ regardless: rate $\times$ period, with the speed cancelling. Formally, the substitution $\zeta = e^{-i\Omega t}\eta$ removes the Coriolis term from equation (4) for any $\omega_0$ — the rotating prefactor neither knows nor cares what oscillation it multiplies. That universality is the demonstration's power: two pendulums of wildly different build at the same latitude precess in lockstep, so the effect must belong to the place, not the device. Only $\sin\lambda$ survives — which is why the pendulum is, in effect, a latitude meter.

Is the swing plane fixed relative to the stars while the Earth turns underneath?

Only at the poles. There, the suspension point sits on the spin axis, the ground merely spins in place beneath the bob, and the plane genuinely holds its inertial orientation while the floor rotates once per sidereal day. Anywhere else the suspension point is carried around a circle of latitude, and a swing direction dragged along that journey cannot stay inertially fixed — it returns after one day rotated by $2\pi\sin\lambda$ relative to the ground, which is less than a full turn. An inertial observer therefore sees the Paris plane rotating too. The precise statement of what the plane does is that it is parallel-transported along the latitude circle, and the failure to return — the holonomy — is the subject of the parallel-transport article in this series. The "fixed relative to the stars" slogan is a polar special case that happens to be the one everyone remembers.

What happened to the horizontal component ω cos λ — why does only the vertical part precess the plane?

The horizontal component is bigger than the vertical one at low latitudes, so ignoring it deserves an argument, not a shrug. Cross $\omega\cos\lambda\,\hat{\mathbf y}$ with the bob's velocity: acting on the horizontal velocity it produces a vertical force, which merely modulates the wire tension by a part in $10^{5}$ — it cannot turn anything horizontal. Acting on the small vertical velocity ( $\dot z \sim (a/L)\dot x$ , oscillating at twice the swing frequency) it produces a horizontal force that reverses sign twice per swing and averages to zero over every period. Neither channel deposits net rotation. The vertical component $\omega\sin\lambda$ , by contrast, rotates horizontal velocities within the horizontal plane coherently, swing after swing. Derivation A does this averaging explicitly — it is the one physical approximation in the whole solution, and it is why the rate carries $\sin\lambda$ with no $\cos\lambda$ contamination at leading order.

The pendulum at my science museum sometimes turns the wrong way. Is it broken, or is the physics wrong?

Neither — it is measuring the wrong thing. Any real pendulum swings in a slight ellipse, and an ellipse in an anharmonic potential precesses by itself at $\Omega_{\mathrm{Airy}} = \tfrac38\,\omega_0\,ab/L^2$ , in the direction of its own circulation — which is set by whatever asymmetry launched it: a draft, a shaky release, an off-centre magnetic drive. On a short pendulum this artefact can be several times the Foucault rate, and its sign is a coin flip, so the apparent precession can stall or reverse. The fix is never to recalibrate the Earth: relaunch from a burnt thread (zero push), check the Charron ring is actually grazing the wire at the swing extremes, and verify the drive kicks symmetrically through dead centre. A pendulum that turns wrongly is the error budget of equation (8) on public display — arguably the most instructive failure in any museum.

Why burn a thread to launch it — why not just let go carefully?

Do the tolerance arithmetic and hands are disqualified. Keeping the spurious Airy precession below 5% of the signal on a 15 m museum pendulum requires the swing ellipse's minor axis to stay under about 1.4 mm — corresponding to a transverse release velocity of roughly one millimetre per second on a 1.2 m amplitude. Human fingers releasing a 30 kg bob impart centimetres per second of uncontrolled sideways motion, tens of times over budget. The tied-thread trick is a mechanical zero: haul the bob aside with a cotton loop, wait minutes for every tremor to drain out through the wire into the support, then burn the cotton — combustion cannot push. The bob starts from the truest rest the laboratory can define, leaving only the unavoidable Coriolis minimum $b = (\Omega/\omega_1)a$ , a tenth of a millimetre. Foucault's 1851 launch protocol is still, unimproved, the standard.

Does the bob actually swing through the centre point on each pass?

No — and the miss is a fingerprint of the rotating planet. Released from rest, the bob carries the constants $A - B = (\Omega/\omega_1)a$ , so it sweeps past the vertical at a closest distance $b = (\Omega/\omega_1)a$ — about 0.4 mm for the Panthéon's 3 m amplitude — always to the same side. "Released from rest" means rest in the laboratory, and the laboratory itself is turning: in inertial space the launch hands the bob a sideways velocity $\Omega a$ , which is exactly what opens the ellipse. The resulting path is a rosette with outward-pointing cusps at the amplitude circle, advancing a petal each half-swing. A pushed launch rounds the cusps into loops and fattens the miss distance in proportion to the push. So the centre-miss is not sloppiness; a pendulum that did pass exactly through centre on a turning Earth would be the miracle.

How is Foucault's gyroscope related to his pendulum — are they the same effect?

Same question, opposite instrument. The pendulum reveals Earth's rotation statistically: a slow accumulation of Coriolis nudges, millimetres per swing, needing hours and a sand ring. Foucault realized within a year that what he really needed was an object whose direction is held fixed in inertial space by its own angular momentum — a fast rotor in gimbals, which he built and named the gyroscope in 1852. A gyroscope's axis visibly drifts relative to the laboratory at the full $\omega$ (component by component), not $\omega\sin\lambda$ , because a free rotor keeps its inertial direction outright rather than being parallel-transported like a swing plane trapped in the local horizontal. No slow bookkeeping: point it at a wall and watch the wall move. The two devices are complementary demonstrations of one fact, and the gyroscope's descendants — from ship gyrocompasses to the ring lasers in the modern-hooks section — inherited the job.

If Earth spun much faster, what would the pendulum do differently?

Equation (5) holds the answer in its two frequencies $-\Omega \pm \omega_1$ . On Earth, $\Omega/\omega_0 \sim 10^{-4}$ : the two circular modes are indistinguishably close, and their superposition looks like a planar swing rotating slowly — the clean separation of "fast oscillation" from "slow precession" that makes the demonstration legible. On a planet where $\Omega$ approached $\omega_0$ (spin period of minutes), the splitting would be resolvable by eye: the left- and right-circular modes would visibly beat, launch-from-rest would produce a fat ellipse ( $b/a = \Omega/\omega_1$ no longer small), and the frequency shift $\omega_1 = \sqrt{\omega_0^2+\Omega^2}$ would be measurable. You would not describe it as "a swing plane precessing" at all, but as two circular oscillators of different frequencies — which is what the Foucault pendulum secretly is, everywhere, with Earth's slow spin merely hiding the anatomy.