Slide a frictionless puck across a spinning disc. In the room it goes perfectly straight; on the disc it curves. Where does the "Coriolis force" come from?
Each preset changes one idea. Blue = path on the disc, orange = path from the closed form, grey dashed = straight line without rotation.
The puck starts at distance r on the left of the centre; the aim angle is measured on the disc (0° = through the centre). Speed is the speed on the disc at launch. Changing anything relaunches.
Turntable: the closed form is a straight line in the room frame, rotated back onto the disc. Dish: w = ux + i uy obeys dw/dt = −2iΩw. Neither uses the integrator.
Launches from the centre at your speed and measures the angle by which the path has curved when it reaches r = 0.8 m, for seven spin rates, then fits a line. Prediction: θ = Ω r / u (the disc simply turns by Ω times the crossing time r/u).
In a frame rotating at angular velocity Ω a free body obeys a = −2 Ω × v − Ω × (Ω × r). The first term is the Coriolis acceleration (perpendicular to the velocity, does no work); the second is the centrifugal acceleration, Ω² r outward. Neither is a real force: they are what the straight, constant-velocity motion of a free body looks like to an observer who spins.
Turntable. A frictionless puck leaves with room-frame velocity u + Ω × r₀ and moves in a straight line, so its position on the disc is the room-frame line rotated by −Ωt. This closed form is what the page compares with the RK4 integration. The Jacobi energy ½|v|² − ½Ω²r² is conserved.
Dish. On a paraboloid z = Ω² r²/(2g) the tilt of the surface cancels the centrifugal term exactly (the principle of a liquid-mirror telescope), leaving only the Coriolis term. Then dw/dt = −2iΩw, so the speed is constant and the puck circles with radius u/(2Ω) and period π/Ω. This is the mechanical model of the inertial oscillation seen in the ocean and atmosphere, where gravity and the flattening of the Earth play the role of the dish. The dish is drawn flat for clarity.
Earth. At latitude φ the vertical component of Earth's rotation sets f = 2ΩE sin φ with ΩE = 7.292×10−⁵ rad/s. For a short trip D at speed v the sideways drift is ≈ f v t²/2 = ΩE sin φ · D²/v and the inertial radius is v/f. The calculator ignores the horizontal component of Ω (it matters near the equator for vertical motion), air drag, the Earth's curvature and the east-west Eötvös effect.
Simplifications: no friction or rolling resistance, point-like puck, a perfectly rigid disc spinning at constant Ω. The puck is stopped when it reaches the rim. This model does not predict which way a bathtub drains: at that scale f is far too small compared with other effects.