Presets
Playback
Parameters
The real Earth turns the plane by only about 11 degrees per hour at Paris, so the spin speed-up slider multiplies the rotation rate to make it visible in a minute. The pendulum itself runs at its real period.
Model checks
The science
Rotating frame
The floor is fixed to the Earth, which turns at Ω = 7.292×10-5 rad/s. Only the vertical part of the rotation vector matters for horizontal motion: Ωz = Ω sin φ at latitude φ. With z = x + i y the bob obeys z'' = -ω02 z - 2 i Ωz z' + Ωz2 z, where ω0 = √(g/L). The middle term is the Coriolis force, the last the centrifugal force.
Exact solution
In an inertial frame the small-angle pendulum is a plain oscillator, zin'' = -ω02 zin. The floor turns by Ωz t, so z = e-i Ωz t zin. Released at rest relative to the floor, the bob traces a rosette whose swing plane turns at -Ω sin φ, clockwise in the northern hemisphere. A full turn takes T = 2π / (Ω sin φ) = 23.93 h / sin φ: about 32 h in Paris, 23.93 h at the pole, never at the equator.
The page integrates the equations with a fourth-order Runge-Kutta scheme and compares the result with this exact expression.
Honest simplifications: small swing angles (the bob is a 2D linear oscillator, no amplitude-dependent period), no air drag, no pivot imperfection, and the spin speed-up slider is a visualisation device: at 3000x the Coriolis ratio is still small enough that the plane picture holds, but the numbers are not the real-Earth numbers. Set the slider to 1x to see the real rates in the readouts.
What to try
- Pick North Pole and then Equator: the turn rate goes from one full turn per sidereal day to zero.
- Compare the measured turning points with the yellow predicted line in the first graph.
- Use Real Earth (x1): the readouts show the true rate while the trace looks like a straight line.
- Lengthen the pendulum: the swing period grows as the square root of L but the plane turn rate does not change.