The sidereal day 86164.1 s (one inertial Earth rotation) sets Omega_earth = 2*pi / T. The plane appears clockwise in the northern hemisphere and counterclockwise in the southern.
center = directly below pivot / blue solid = current swing axis / gray dashed = swing axis at t=0 / orange arc = precession angle Δθ / N/E/S/W = compass
x = elapsed time t (h) / y = precession angle Δθ (°) / blue = current latitude / gray = equator (0°), 35°, pole (90°) / yellow dot = current (t, Δθ)
The Earth's spin vector Omega contributes only the component along the local vertical at latitude phi. That single component drives the rotation of the Foucault swing plane.
Simple-pendulum swing period (small amplitude):
$$T_{\mathrm{osc}} = 2\pi\sqrt{\dfrac{L}{g}}$$Precession angular rate at latitude $\varphi$:
$$\Omega_{\mathrm{pre}} = \Omega_\oplus \sin|\varphi|,\qquad \Omega_\oplus = \dfrac{2\pi}{86164.1\,\mathrm{s}} \approx 7.292\times10^{-5}\,\mathrm{rad/s}$$Full-rotation time and accumulated angle after time $t$:
$$T_{\mathrm{pre}} = \dfrac{2\pi}{\Omega_{\mathrm{pre}}},\qquad \Delta\theta = \Omega_{\mathrm{pre}}\,t$$$L$ is the pendulum length [m] and $g$ is the local gravity [m/s²]. The precession depends only on $\sin\varphi$ — not on $L$ or mass — so the rotation is clockwise in the northern hemisphere, counterclockwise in the southern, and zero on the equator.