Foucault precession rate and latitude dependence

When Léon Foucault unveiled his pendulum at the Panthéon in 1851, the swinging wire proved that the Earth rotates. But the deeper prize was hiding in a single trigonometric factor. Foucault announced — without a full derivation, to the lasting irritation of the mathematicians — that the swing plane turns at a rate proportional to the sine of the latitude: one full turn per day at the pole, 31.8 hours in Paris, and no turn at all on the equator. Replicas sprang up across Europe and America within months, each city clocking a different precession period, and every one of them landed on Foucault's "law of the sine." This article is about that law: $\Omega_F = \Omega\sin\varphi$ . Why does only a fraction of the Earth's rotation show up in the pendulum — and why exactly the sine? The answer is a projection: at latitude $\varphi$ , the planet's angular velocity vector splits into a piece along the local vertical, $\Omega\sin\varphi$ , which steadily steers the bob's horizontal velocity, and a horizontal piece, $\Omega\cos\varphi$ , whose influence on the swing averages to zero. Proving that second half rigorously — not just dropping the term, but showing its effect genuinely cancels over each swing — is the heart of the physics here. The payoff is an instrument. Read the law backwards and a swinging weight becomes a latitude meter: time the precession, divide into a sidereal day, take the arcsine, and you know where on Earth you stand — no stars, no GPS, just a wire, a bob, and a clock.

Key takeaways

A Foucault pendulum's swing plane precesses at $\Omega_F = \Omega\sin\varphi$ , where $\Omega = 7.292\times10^{-5}\ \mathrm{rad/s}$ is Earth's sidereal rotation rate and $\varphi$ the latitude — a full turn takes $23.93\ \mathrm{h}/\sin\varphi$ .
Only the local vertical component of Earth's angular velocity turns the plane; the horizontal component $\Omega\cos\varphi$ produces forces that average to zero over each swing and never accumulate.
The sense is clockwise (seen from above) in the Northern Hemisphere, counterclockwise in the Southern, and the precession vanishes on the equator.
The correct clock is the sidereal day, $86{,}164\ \mathrm{s}$ — Earth's rotation relative to the stars — not the 24-hour solar day; confusing them is a built-in 0.27% error.
Away from the poles the swing plane is not fixed relative to the stars: it returns each day rotated by $2\pi(1-\sin\varphi)$ in inertial space — a geometric deficit developed in the parallel-transport article.
Inverted, the law is a self-contained latitude meter: $\sin\varphi = T_{\mathrm{sid}}/T_F$ .
Key idea: project Earth's angular velocity onto the local vertical — the swing plane precesses at exactly that vertical component, $\Omega_F = \Omega\sin\varphi$ , giving a turn period of one sidereal day divided by $\sin\varphi$ .

The core insight

Project Earth's angular velocity onto the local vertical: the in-plane precession rate of a Foucault pendulum is exactly the vertical component, $\Omega_F=\Omega\sin\varphi$ , giving a turn period of one sidereal day over $\sin\varphi$ .

Where this lives today

The quantity a Foucault pendulum measures — the vertical component of the planet's rotation at your feet — is the beating heart of modern geophysical fluid dynamics, where it appears doubled as the Coriolis parameter $f = 2\Omega\sin\varphi$ . Every weather model integrates $f$ at every grid point; and it is the latitude dependence itself, the derivative $\beta = \mathrm{d}f/\mathrm{d}y$ , that drives Rossby waves — the planetary-scale meanders of the jet stream that set week-ahead weather. Foucault's sine law, differentiated once, is why forecasts beyond a few days are hard. The pendulum's precision descendants are ring-laser gyroscopes: kilometre-scale optical loops (the G ring at Wettzell, GINGERINO under the Gran Sasso mountain) in which two counter-propagating laser beams pick up a Sagnac frequency splitting proportional to the local $\boldsymbol{\Omega}\cdot\hat{\mathbf{n}}$ . They read the same projection Foucault's bob reads, continuously, to parts in $10^{9}$ — sharp enough to watch the day's length wobble by milliseconds and the rotation axis wander by centimetres. Meanwhile the original instrument keeps its public post: hundreds of museum pendulums worldwide, from the restored Panthéon original to the one that turns full circle in 4.4 days in Bengaluru, each displaying its own latitude in the slow arithmetic of $\sin\varphi$ .

The physical picture

Start where the picture is trivial: the North Pole. Hang a pendulum directly over the spin axis and set it swinging. The plane of oscillation holds fixed relative to the stars — nothing twists it — while the planet, and the observer standing on it, rotate underneath once per sidereal day. To the observer the swing plane appears to sweep clockwise, $15.04°$ every hour, completing a lap in $23.93$ hours. At the pole, the pendulum simply is a star-fixed reference watching the Earth go by. Now carry the experiment to the equator. The spin axis lies flat along the ground there — pointing north along the horizon — and however the Earth turns about that horizontal axis, it never pivots the local floor about the vertical. The ground under the pendulum tilts through space but does not twirl. The swing plane has nothing to follow and nothing to lag: it stays put relative to the ground, forever. Between these two extremes the floor twirls at a fraction of the full rate, and the fraction is exactly the projection of the spin vector onto the local vertical: $\sin\varphi$ .

Why the sine and not some other interpolation? Because an angular velocity is a vector, and vectors project. At latitude $\varphi$ the spin vector $\boldsymbol{\Omega}$ makes an angle $90°-\varphi$ with the local vertical, so its component along the vertical is $\Omega\sin\varphi$ and its component along the ground (pointing due north) is $\Omega\cos\varphi$ . The vertical component is a genuine twirl of the local floor — the kind of rotation a swing plane can lag behind. The horizontal component is a tilt — it tips the whole laboratory end over end, the way the equatorial floor tips, and a tilt does not rotate directions drawn on the floor. But there is a real physics question buried in that tidy geometry, and the honest version of this article must face it. The Coriolis force couples the full vector $\boldsymbol{\Omega}$ to the bob's velocity, and at mid-latitudes the discarded horizontal component is not small — at $30°$ it is bigger than the vertical component we keep. The reason it does not matter is not that it is weak but that it is futile: the forces it produces on the bob reverse sign in step with the swing itself, pushing east while the bob falls and west while it rises, so that over each complete oscillation the impulse sums to zero. The vertical component, by contrast, always pushes to the same side of the velocity — a small bias, never undone, compounding swing after swing. Secular beats oscillatory: that is the entire selection principle, and the derivation below makes it exact.

Figure

Why one component wins — steady steering accumulates, oscillating forcing cancels

One more feature deserves early notice because it protects you from a widespread myth. Only at the poles is the swing plane fixed relative to the stars. At any other latitude the plane precesses in inertial space too: after one sidereal day it fails to return to its stellar orientation by the angle $2\pi(1-\sin\varphi)$ . That residue is not a defect of the pendulum — it is pure geometry, the holonomy of dragging a direction around a circle on a curved globe, and it has its own article: parallel transport and the Foucault pendulum.

The sine law made quantitative

The precession rate is the projection of the Earth's angular velocity onto the local vertical. Everything in this article unpacks, defends, or exploits this one line:

\Omega_F = \Omega\sin\varphi \qquad (1)

where: $\Omega_F$ — the angular rate at which the swing plane turns relative to the ground (clockwise from above for $\varphi>0$ ); $\Omega = 7.292\times10^{-5}\ \mathrm{rad/s}$ — Earth's sidereal rotation rate; $\varphi$ — geographic latitude, negative in the south. In everyday units the rate is $15.04°\times\sin\varphi$ per hour: the full $15.04°/\mathrm{h}$ at the pole, $11.3°/\mathrm{h}$ in Paris, nothing on the equator. Invert the rate for the observable a museum visitor actually experiences — the time for the swing plane to come full circle:

T_F = \frac{2\pi}{\Omega_F} = \frac{T_{\mathrm{sid}}}{\sin\varphi} \qquad (2)

where: $T_{\mathrm{sid}} = 86{,}164\ \mathrm{s} = 23.93\ \mathrm{h}$ — the sidereal day. Because $\sin\varphi \le 1$ , the precession period is at least a sidereal day everywhere, stretching without bound as you approach the equator: two days at $30°$ , nearly six days at $10°$ . A Foucault pendulum is a clock that runs at a different speed in every city — that is the law of the sine worn on its face. Both formulas descend from one decomposition. Set up local axes at latitude $\varphi$ : $\hat{\mathbf{x}}$ east, $\hat{\mathbf{y}}$ north, $\hat{\mathbf{z}}$ up. The Earth's angular velocity vector — which points out of the North Pole along the spin axis — resolves in these axes as

\boldsymbol{\Omega} = \Omega\left(\cos\varphi\,\hat{\mathbf{y}} + \sin\varphi\,\hat{\mathbf{z}}\right) \qquad (3)

where: the $\hat{\mathbf{y}}$ piece is the horizontal component, pointing due north toward the pole, and the $\hat{\mathbf{z}}$ piece is the vertical component. At the pole the vector is all $\hat{\mathbf{z}}$ ; on the equator all $\hat{\mathbf{y}}$ . Note what is absent: no east component, ever — the spin axis lies in every local meridian plane.

Which formula answers which question

you want	use	notes
precession rate at your latitude	$\Omega_F = \Omega\sin\varphi$ , eq. (1)	$15.04°/\mathrm{h}\times\sin\varphi$
time for one full turn of the plane	$T_F = 23.93\ \mathrm{h}/\sin\varphi$ , eq. (2)	sidereal hours, not solar
the spin vector in local axes	eq. (3)	vertical part turns the plane
latitude from a timed precession	$\sin\varphi = T_{\mathrm{sid}}/T_F$	the pendulum as instrument
the correct day length	$T_{\mathrm{sid}} = 86{,}164\ \mathrm{s}$ , eq. (7)	derived below

Derivation AOnly the vertical component turns the swing plane

We are deriving the sine law (1) from the rotating-frame equation of motion, and the target is precision about a step most treatments wave through: dropping the horizontal component $\Omega\cos\varphi$ . That component is larger than the one we keep everywhere below $45°$ , so it cannot be discarded for being small — it must be shown to act futilely, its pushes cancelling over each swing. The strategy: write all three Coriolis components in local axes, classify each force term as secular (always the same side of the velocity) or oscillatory (sign-reversing with the swing), average over a period, and keep only what survives. What must be watched: the vertical bob velocity $\dot z$ is second order in the swing amplitude, and its force contribution oscillates at twice the pendulum frequency — both facts are needed to kill it cleanly.

Step 1. Write the rotating-frame equation for the bob — gravity plus tension plus Coriolis (the centrifugal term is already absorbed into effective gravity, and the Euler force vanishes since $\Omega$ is constant):

\ddot{\mathbf{r}} = \mathbf{g}_{\mathrm{eff}} + \frac{\mathbf{T}}{m} - 2\,\boldsymbol{\Omega}\times\dot{\mathbf{r}} \qquad (4)

where: $\mathbf{T}$ — wire tension; $\mathbf{g}_{\mathrm{eff}}$ — local effective gravity defining the vertical. For small swings the first two terms combine into the standard isotropic restoring acceleration $-\omega_0^2(x\hat{\mathbf{x}} + y\hat{\mathbf{y}})$ with $\omega_0 = \sqrt{g/\ell}$ . Why the general Coriolis machinery may be assumed: the transformation producing the $-2\boldsymbol{\Omega}\times\dot{\mathbf{r}}$ term, factor of 2 and all, is the parent article's Derivation A — here we spend it, not re-derive it.

Step 2. Insert the decomposition (3) and expand the cross product component by component. With $\dot{\mathbf{r}} = (\dot x, \dot y, \dot z)$ :

-2\boldsymbol{\Omega}\times\dot{\mathbf{r}} = 2\Omega\begin{pmatrix} \dot y\sin\varphi - \dot z\cos\varphi \\ -\dot x\sin\varphi \\ \dot x\cos\varphi \end{pmatrix}

Why this grouping is the whole story: every appearance of $\sin\varphi$ multiplies a horizontal velocity and produces a horizontal force — the twirl channel; every appearance of $\cos\varphi$ couples the vertical direction to the east–west direction — the tilt channel. The two channels never mix, so they can be judged separately.

Step 3. Judge the $\cos\varphi$ terms. The vertical force $2\Omega\dot x\cos\varphi\,\hat{\mathbf{z}}$ modulates the apparent weight by at most $2\Omega v/g \sim 10^{-5}$ — an imperceptible flutter in the wire tension, absorbed by the constraint. The horizontal force $-2\Omega\dot z\cos\varphi\,\hat{\mathbf{x}}$ needs more care. For a bob swinging with horizontal amplitude $A$ , the height is $z \approx (x^2+y^2)/2\ell$ , so $\dot z$ is second order in $A/\ell$ and oscillates at $2\omega_0$ — up-down-up within each half swing — with zero mean. Why zero mean kills it: a bounded oscillatory acceleration with no secular part produces only a closed micrometre-scale wiggle in the orbit (about $16\ \mu\mathrm{m}$ for a 20 m pendulum swinging 2.5 m); it cannot rotate the swing plane by even one degree over a year.

Step 4. Keep the $\sin\varphi$ terms. They act in the horizontal plane on the horizontal velocity, always at right angles to it, always on the same side — the textbook Coriolis steering, at the reduced strength $\Omega_z = \Omega\sin\varphi$ . The surviving horizontal system is

\ddot x = -\omega_0^2 x + 2\Omega_z\dot y, \qquad \ddot y = -\omega_0^2 y - 2\Omega_z\dot x \qquad (5)

where: $x, y$ — east and north bob displacements. This is exactly the polar Foucault problem, with $\Omega \to \Omega\sin\varphi$ : the latitude enters the dynamics only through this one substitution.

Step 5. Solve by the complex trick — combine $\xi = x + iy$ , so the pair (5) becomes $\ddot\xi + 2i\Omega_z\dot\xi + \omega_0^2\xi = 0$ , whose solution factorizes:

\xi(t) = e^{-i\Omega_z t}\left(A\,e^{i\omega' t} + B\,e^{-i\omega' t}\right), \qquad \omega' = \sqrt{\omega_0^2 + \Omega_z^2} \approx \omega_0 \qquad (6)

where: the bracket is an ordinary planar pendulum oscillation and the prefactor $e^{-i\Omega_z t}$ rigidly rotates that whole pattern clockwise at $\Omega_z$ . Why the substitution rotates at $\Omega_z$ and not $2\Omega_z$ : undoing a Coriolis coupling of strength $2\Omega_z$ requires co-rotating at only $\Omega_z$ — the factor-of-2 bookkeeping is spelled out in the Coriolis-force article's Foucault derivation. The frequency shift is fractionally $\Omega_z^2/2\omega_0^2 \sim 10^{-8}$ — the swinging itself never notices the Earth.

Step 6. Read off the law: the plane precesses at $\Omega_F = \Omega_z = \Omega\sin\varphi$ , clockwise from above for $\varphi > 0$ . For $\varphi < 0$ , $\sin\varphi$ flips sign and the rotation reverses — the Southern Hemisphere pendulum turns counterclockwise. Period: $T_F = 2\pi/\Omega_F = T_{\mathrm{sid}}/\sin\varphi$ . Verified: matches Morin ch. 10, Taylor §9.9, and Landau–Lifshitz §39; limits check — $\varphi = 90°$ gives one turn per sidereal day, $\varphi = 0$ gives no precession, and the discarded terms are $O(\Omega A/\ell)$ oscillatory with zero mean.

The one physical input the sine law still owes you is its clock. Why $23.93$ hours and not $24$ ?

Go deeper · Derivation B · 1 minWhy the pendulum keeps sidereal time

Derivation BWhy the pendulum keeps sidereal time

We are deriving the length of the sidereal day, $T_{\mathrm{sid}} = 86{,}164\ \mathrm{s}$ , and justifying why that — and not the 24-hour solar day — is the period in equation (2). The strategy is bookkeeping: count how many times the Earth actually rotates in a year versus how many sunrises it delivers, and convert. The point to watch is the reference frame: the pendulum's dynamics care about rotation relative to inertial space (that is what the Coriolis force measures), while the 24-hour day is defined by the Sun's position — a moving target, since the Earth orbits it.

Step 1. In one year the Earth makes $365.2422$ solar days — noon-to-noon cycles. But in the same year it also completes one full orbit around the Sun, and that orbital revolution uses up one rotation as seen from the stars: to face the Sun again after a day of orbital motion, the Earth must turn slightly more than $360°$ . Why the correction is one full turn per year: after 365 solar days the extra bits accumulate to exactly one revolution — the orbit itself.

Step 2. So relative to the stars the Earth rotates $366.2422$ times per year. The sidereal day is therefore shorter than the solar day by the ratio of the two counts:

T_{\mathrm{sid}} = 86{,}400\ \mathrm{s}\times\frac{365.2422}{366.2422} = 86{,}164\ \mathrm{s} = 23\ \mathrm{h}\ 56\ \mathrm{min}\ 4\ \mathrm{s} \qquad (7)

where: $86{,}400\ \mathrm{s}$ — the mean solar day. The difference is $3\ \mathrm{min}\ 56\ \mathrm{s}$ per day, or fractionally $0.27\%$ .

Step 3. The Coriolis force in equation (4) is generated by the rotation of the Earth frame relative to inertial space — the frame in which Newton's laws hold without correction. Hence $\Omega = 2\pi/T_{\mathrm{sid}} = 7.292\times10^{-5}\ \mathrm{rad/s}$ , and every period in this article is measured in sidereal time. Why the error matters in practice: a precision measurement of $\varphi$ from a timed precession inherits the full $0.27\%$ if the solar day is used — about $0.13°$ of latitude at mid-latitudes, an error a good pendulum can genuinely resolve (see Worked example 2). *Verified: $2\pi/86{,}164\ \mathrm{s} = 7.2921\times10^{-5}\ \mathrm{rad/s}$ , the standard astronomical value; the year contains exactly one more sidereal than solar day.*

The precession around the globe

Equation (2) turns the world map into a table of clocks. A few stations, all recomputed from $T_F = 23.93\ \mathrm{h}/\sin\varphi$ and rate $= 15.04°/\mathrm{h}\times\sin\varphi$ :

place	latitude	rate	full turn
North or South Pole	$\pm90°$	$15.0°/\mathrm{h}$	$23.9\ \mathrm{h}$
Anchorage	$61.2°\mathrm{N}$	$13.2°/\mathrm{h}$	$27.3\ \mathrm{h}$
Paris (Panthéon)	$48.85°\mathrm{N}$	$11.3°/\mathrm{h}$	$31.8\ \mathrm{h}$
New York	$40.7°\mathrm{N}$	$9.8°/\mathrm{h}$	$36.7\ \mathrm{h}$
Delhi	$28.6°\mathrm{N}$	$7.2°/\mathrm{h}$	$50.0\ \mathrm{h}$
Bengaluru	$13.0°\mathrm{N}$	$3.4°/\mathrm{h}$	$106\ \mathrm{h}\ (4.4\ \mathrm{d})$
Singapore	$1.35°\mathrm{N}$	$0.35°/\mathrm{h}$	$42\ \mathrm{days}$
Equator	$0°$	$0$	never
Sydney	$33.86°\mathrm{S}$	$8.4°/\mathrm{h}$ ccw	$43.0\ \mathrm{h}$

Figure

The law of the sine as a curve — full-turn time against latitude

And what does the pendulum on the equator do? Nothing dramatic — and that nothing is instructive. The swing plane holds its bearing relative to the ground indefinitely. The only Coriolis trace left is the futile channel of the derivation: the horizontal component $\Omega\cos\varphi$ (at full strength $\Omega$ there) couples to the bob's tiny vertical velocity and jiggles it east–west at twice the swing frequency, with an amplitude measured in micrometres. The Earth's rotation has not vanished at the equator — a dropped stone still lands measurably to the east, as the Coriolis force article computes — but the specific observable "swing-plane rotation" reads exactly zero. A Foucault pendulum measures one component of a vector, not the vector itself.

Limiting cases

Case	Result	Meaning
pole, $\varphi = \pm90°$	$\Omega_F = \Omega$ , $T_F = 23.93\ \mathrm{h}$	maximal precession; plane star-fixed, Earth turns beneath — the clean inertial picture
equator, $\varphi = 0$	$\Omega_F = 0$ , $T_F \to \infty$	vertical spin component vanishes; plane never turns, only micrometre east–west jiggle remains
Paris, $\varphi = 48.85°$	$11.3°/\mathrm{h}$ , $T_F = 31.8\ \mathrm{h}$	Foucault's 1851 demonstration — $0.19°$ per minute against the floor scale
Southern Hemisphere, $\varphi < 0$	$\Omega_F < 0$	sense reverses: counterclockwise from above, magnitude $\Omega\sin\lvert\varphi\rvert$
small latitude, $\varphi \ll 1$	$T_F \approx T_{\mathrm{sid}}/\varphi_{\mathrm{rad}}$	period diverges as $1/\varphi$ ; at Singapore a full turn takes six weeks
solar-day error	$\Omega$ overestimated by $0.27\%$	systematic latitude error $\approx 0.13°$ at mid-latitudes — resolvable by a good pendulum

Common trapThe swing plane stays fixed relative to the stars

True at the poles — false everywhere else, and believing it at mid-latitude is the most common way to get this subject wrong. If the plane were star-fixed, every pendulum would show a 23.93-hour period regardless of location, and the sine law would be inexplicable. What is actually fixed is nothing: at latitude $\varphi$ the plane turns at $\Omega\sin\varphi$ relative to the ground and fails to close on itself in inertial space, returning each sidereal day short by the angle $2\pi(1-\sin\varphi)$ . Two related traps: the reference day is sidereal ( $86{,}164\ \mathrm{s}$ ), not solar — the pendulum answers to the stars, not the Sun; and the precession rate is $\Omega\sin\varphi$ , not $2\Omega\sin\varphi$ — the factor 2 in the Coriolis force does not survive into the precession rate, for the reason traced in the Coriolis-force article's Foucault derivation.

Worked examples

1 · Reading a museum pendulum (confidence builder)

A science museum at latitude $\varphi = 40°\mathrm{N}$ installs a Foucault pendulum of length $\ell = 20\ \mathrm{m}$ swinging with horizontal amplitude $A = 2.5\ \mathrm{m}$ . Find the precession rate, the full-turn time, and how far the swing tip migrates during a 10-minute school visit. Would the class see it move?

Step 1. Rate from (1): $\Omega_F = \Omega\sin 40° = 7.292\times10^{-5}\times0.643 = 4.69\times10^{-5}\ \mathrm{rad/s}$ , which is $15.04°/\mathrm{h}\times0.643 = \boxed{9.7°\ \text{per hour}}$ , clockwise seen from above.

Step 2. Full turn from (2): $T_F = 23.93\ \mathrm{h}/0.643 = \boxed{37.2\ \mathrm{h}}$ — a day and a half. Note the swinging itself has period $2\pi\sqrt{\ell/g} = 2\pi\sqrt{20/9.81} = 9.0\ \mathrm{s}$ : the plane turns a mere $0.024°$ per swing. The precession is invisible per oscillation and unmissable per hour.

Step 3. Tip migration: the end of the swing arc sits $A = 2.5\ \mathrm{m}$ from center, so it sweeps sideways at $A\,\Omega_F = 2.5\times4.69\times10^{-5} = 1.17\times10^{-4}\ \mathrm{m/s} \approx 7\ \mathrm{mm}$ per minute. In 10 minutes: $\boxed{7\ \mathrm{cm}}$ — enough to topple the next domino in the classic ring of pegs. The class sees it.

Step 4. Sanity check against Paris: $9.7°/\mathrm{h}$ at $40°$ versus $11.3°/\mathrm{h}$ at $48.85°$ — smaller latitude, slower turn, as the sine demands.

2 · The pendulum as a latitude meter (exam level)

An expedition pendulum, latitude unknown, is timed through one full turn of its swing plane: $T_F = 38.0 \pm 0.3\ \mathrm{h}$ . (a) Find the latitude and its uncertainty. (b) The expedition's junior member uses the 24-hour solar day instead of the sidereal day — what systematic error does that inject? (c) Could the pendulum tell which hemisphere it is in?

Step 1. Invert (2): $\sin\varphi = T_{\mathrm{sid}}/T_F = 23.93/38.0 = 0.6298$ , so $\varphi = \arcsin(0.6298) = \boxed{39.0°}$ .

Step 2. Propagate the timing uncertainty. Since $\sin\varphi \propto 1/T_F$ , the fractional errors match: $\delta(\sin\varphi) = 0.6298\times(0.3/38.0) = 5.0\times10^{-3}$ . Convert through the derivative $\delta\varphi = \delta(\sin\varphi)/\cos\varphi = 5.0\times10^{-3}/0.777 = 6.4\times10^{-3}\ \mathrm{rad} = \boxed{0.37°}$ — about 41 km of north–south position from a wire and a stopwatch.

Step 3. The solar-day mistake: using $24\ \mathrm{h}$ gives $\sin\varphi = 24/38.0 = 0.6316$ and $\varphi = 39.17°$ — a systematic shift of $\boxed{+0.13°}$ , comparable to the entire random error budget. The 0.27% clock error survives undiluted into the sine. At high latitude it is worse: $\cos\varphi$ in the denominator of Step 2 amplifies it.

Step 4. Hemisphere: yes — from the sense of rotation, not the rate. Clockwise from above means $\varphi > 0$ (north); counterclockwise means south. The magnitude $\lvert\sin\varphi\rvert$ is blind to the sign, but the direction of the creep is not. An expedition at $39.0°$ S would time the same 38.0 hours with the plane turning the other way.

Origins · 2 minLéon Foucault and how this was discovered

Léon Foucault

1819–1868 · Paris Observatory

Foucault never held a physics degree — he abandoned medicine (he could not bear blood) and made himself into the finest experimental physicist in France from a home workshop. The pendulum idea struck him in January 1851 while machining a steel rod in a lathe: set vibrating, the rod's swing plane held its orientation as the chuck slowly turned. Within weeks he scaled the effect from a lathe to the Panthéon, and announced that the turning rate followed the sine of the latitude — a law he stated from experiment and physical reasoning, leaving the full derivation to others, which scandalized the Academy's mathematicians and delighted everyone else. The following year he invented and named the gyroscope as a second, cleaner rotation detector; he went on to measure the speed of light in a Paris basement to 0.6% with a rotating mirror, discover eddy-current damping, and build the telescope mirrors that became the modern standard. The Foucault pendulum remains what it was in 1851: the only proof of the Earth's rotation you can watch with your own eyes in an afternoon.

History

1661

— Vincenzo Viviani in Florence notes that a swinging pendulum's plane slowly drifts, but records it only as a nuisance to be suppressed with a second wire — the effect goes unexplained for 190 years.

1835

— Gaspard-Gustave de Coriolis publishes the rotating-frame equations of motion containing the velocity-dependent force that will eventually carry his name.

1851

— Foucault demonstrates the pendulum in his cellar on 3 January, at the Paris Observatory in February, then publicly under the Panthéon dome with a 67 m wire; he announces the law of the sine, and replications at other latitudes confirm it within months.

1879

— Heike Kamerlingh Onnes, in his doctoral thesis, gives the definitive analysis of the pendulum's error sources — the elliptical precession that plagues short pendulums — turning the demonstration into a precision instrument.

2001

— A Foucault pendulum rigged in a six-storey stairwell at the Amundsen–Scott South Pole Station completes its turn in approximately 24 hours, confirming the sine law at the last untested latitude on Earth,

\varphi = -90°

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 10 (Accelerating Frames of Reference) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 4-5 · Landau & Lifshitz, Mechanics, Vol. 1, §39 · Taylor, Classical Mechanics, Ch. 9

Provenance. Verifiers should recompute: (1) $\boldsymbol{\Omega}$ decomposition giving vertical component $\Omega\sin\varphi$ ; (2) the coupled equations $\ddot x-2\Omega_z\dot y+\omega_0^2 x=0$ and the complex solution $\xi=e^{-i\Omega_z t}(Ae^{i\omega_0 t}+Be^{-i\omega_0 t})$ ; (3) numbers: $\Omega=2\pi/86164\,\mathrm{s}=7.292\times10^{-5}\,\mathrm{rad/s}$ , Paris $\varphi=48.85^\circ$ gives $T_F\approx31.8\,\mathrm{h}$ and $11.3^\circ/\mathrm{h}$ ; (4) holonomy $2\pi(1-\sin\varphi)$ via cone unrolling. Cross-check against Morin Ch. 10, Goldstein Ch. 4-5, Landau-Lifshitz Vol. 1 §39, and Taylor Ch. 9. Worked-example numbers on this page (museum pendulum $9.7°/\mathrm{h}$ , $37.2\ \mathrm{h}$ , $7\ \mathrm{cm}$ per 10 min at $\varphi=40°$ ; latitude meter $\varphi = 39.0°\pm0.37°$ from $T_F = 38.0\pm0.3\ \mathrm{h}$ ; sidereal day $86{,}400\times365.2422/366.2422 = 86{,}164\ \mathrm{s}$ ) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Coriolis force · Time derivatives in rotating frames · Angular velocity vector of a rotating frame · Effective gravity on the rotating Earth. Leads to: Centrifugal force · Effective gravity on the rotating Earth.

If the Earth rotates once a day, why doesn't every Foucault pendulum's plane also turn once a day?

Because what the pendulum tracks is not "the Earth's rotation" as a lump but the rotation of its own local floor about the local vertical — and that is only the component $\Omega\sin\varphi$ of the planet's spin. At the pole the floor twirls at the full rate and the plane laps in 23.9 h. At latitude $40°$ the floor twirls at 64% of the rate, so the lap takes 37 h. On the equator the floor tilts through space but never twirls, and the plane holds still forever. A once-per-day period everywhere would actually require the swing plane to be nailed to the stars — which, off the poles, it is not.

Is the swing plane fixed relative to the stars, with the Earth just rotating underneath?

Only at the poles. Everywhere else this appealing picture is quantitatively wrong. At latitude $\varphi$ the plane turns at $\Omega\sin\varphi$ relative to the ground, and the ground turns about the local vertical at that same $\Omega\sin\varphi$ relative to the stars — but these do not stack up to a star-fixed plane, because the local vertical itself sweeps around a cone daily. After one sidereal day the plane returns rotated by $2\pi(1-\sin\varphi)$ in inertial space. That leftover angle is a geometric effect — the holonomy of carrying a direction around a latitude circle on a sphere — and is the subject of the parallel-transport article.

At 30 degrees latitude the horizontal component of the Earth's spin is bigger than the vertical one. How can the smaller piece dominate the physics?

Because the competition is not between sizes but between persistence. The vertical component pushes the bob always to the same side of its velocity — a bias that never cancels and compounds over thousands of swings into degrees per hour. The horizontal component couples east–west force to the bob's vertical velocity, which oscillates up-down-up twice per swing with zero average: every push east during the descent is refunded west during the rise. Its whole legacy is a closed micrometre-scale wiggle at twice the swing frequency. A tiny secular effect always outruns a large oscillatory one, given time — the same reason a weak tide can move a harbor while big waves just slosh it.

Does the precession rate depend on the pendulum's length, mass, or amplitude?

Not to leading order — and this universality is the law's signature. In the derivation, the rotating prefactor $e^{-i\Omega_z t}$ carries the rate $\Omega\sin\varphi$ with no reference to $\omega_0 = \sqrt{g/\ell}$ , the mass, or the amplitude; those parameters live only inside the fast oscillation. Length and mass matter practically: a long wire and heavy bob slow the swing, extend coasting time against drag, and shrink the spurious elliptical (Airy) precession that competes with the Earth's signal on short pendulums — which is why Foucault used 67 m and 28 kg, and why tabletop versions disappoint. The apparatus battles are covered in the Foucault pendulum article.

Could I detect the Earth's rotation at all if I lived on the equator?

Yes — just not with a Foucault pendulum's precession, which reads exactly zero there. The pendulum measures one component of $\boldsymbol{\Omega}$ , and on the equator that component vanishes. But the horizontal component is at full strength: drop a stone down a deep shaft and it lands to the east by $\tfrac13\omega g t^3\cos\varphi$ — maximal at the equator, about 2 cm for a 100 m drop — as computed in the Coriolis force article. A ring-laser gyroscope oriented with its axis along the local north also reads the full $\Omega\cos\varphi$ . Nature offers several rotation meters; Foucault's happens to be the one that projects onto the vertical.

Why 23.93 hours and not 24 — the Sun comes back overhead every 24 hours, doesn't it?

It does, but the Sun is a moving target. While the Earth spins, it also advances about $1°$ along its orbit each day, so it must rotate that extra degree — about 3 min 56 s — to bring the Sun back overhead. Relative to the stars, one rotation takes only 23 h 56 min 4 s = 86,164 s. The pendulum answers to the stars: its Coriolis dynamics is generated by rotation relative to inertial space, not relative to the Sun. Over a year the two counts differ by exactly one full turn (366.24 stellar rotations versus 365.24 solar days). Using 24 h in the formulas injects a 0.27% error — small, but three times the random error of a careful latitude measurement.

If I timed a full turn of the plane, how accurately could I locate myself in latitude?

Surprisingly well. From $\sin\varphi = T_{\mathrm{sid}}/T_F$ , a timing good to $0.8\%$ (say $\pm0.3$ h in 38 h) gives latitude to about $0.37°$ — roughly 41 km north–south, from a wire, a bob, and a clock. The sensitivity degrades near the poles, where $\cos\varphi \to 0$ amplifies errors in the arcsine, and near the equator, where the turn takes days and drag kills the swing before a lap completes. Mid-latitudes are the sweet spot. You get the hemisphere for free from the turning sense: clockwise from above is north, counterclockwise is south. This is a genuine inertial navigation principle — measuring where you are from rotation alone, no sky required.

My physics teacher says the precession rate should have a factor of 2 like the Coriolis force. Where does the 2 go?

Careful — two different quantities. The Coriolis force strength on horizontal motion is set by $f = 2\Omega\sin\varphi$ (the meteorologists' Coriolis parameter), with the honest factor of 2 from the acceleration transformation. The precession rate of a Foucault pendulum is $\Omega\sin\varphi$ — no 2. The resolution: a coupling of strength $2\Omega_z$ in the equations of motion is removed by co-rotating at only $\Omega_z$ , because the substitution $\xi = e^{-i\Omega_z t}\xi_0$ regenerates the factor of 2 when differentiated twice. The full bookkeeping is worked step-by-step in the Coriolis-force article's Foucault derivation. Quoting $2\Omega\sin\varphi$ as the precession rate — a 12-hour Paris period instead of 31.8 h — is a classic gradebook error.