Compton generator

Analysis

Assume the diameter of the glass tube is much smaller than the diameter of the ring, and ${\displaystyle R}$ is the radius of the ring, ${\displaystyle \omega }$ is the Earth rotation rate and ${\displaystyle \lambda }$ is the latitude.

Initially the ring is horizontal and the water is stationary. Second the ring is then quickly rotated by 180° around its East-West diameter and stopped, such that it again lies flat on the table surface, which is horizontal. At this time, the velocity ${\displaystyle v_{th}}$ of the water in the tube is given by
${\displaystyle v_{th}=2\omega R\sin \lambda }$

Note that a rotation from the vertical to the vertical position produces the velocity ${\displaystyle v_{tv}}$ of the water in the tube is given by
${\displaystyle v_{tv}=2\omega R\cos \lambda }$

This is derived by first integrating the torque due to the Coriolis force around the ring, then integrating the torque over the time it takes for the ring to flip, to obtain the change in angular momentum.

With these two equations, one can solve for both ${\displaystyle \omega ,\lambda }$, thus finding both the rotation speed of earth and the latitude of the apparatus.

Experimental verification

Compton used this measured drift velocity to determine his latitude to within 3% accuracy. He also used it to measure the rotational period of earth to an accuracy of 16 minutes per day (accuracy of 1%). By careful methods he could observe the effect in a ring with a radius of only 9 inches (23 cm).

Earth's rotation is 7.3 × 10⁻⁵ radians/second. In the original report, Compton used a ring of 1 meter in radius at the College of Wooster (latitude 41 degrees). This would translate to a velocity of about ${\displaystyle v_{th}\approx 0.1\mathrm{m}\mathrm{m}/\mathrm{s}\mathrm{e}\mathrm{c}}$.