The Coriolis Effect, Explained
After reading this you will be able to predict which way a moving object appears to deflect on a spinning platform, compute the size of that deflection from the rotation rate and speed, and explain why cyclones obey the Coriolis effect while your sink does not.
What the Coriolis effect actually is
Roll a marble straight across a spinning turntable. To someone standing beside the table, the marble travels in a perfectly straight line. To a camera bolted to the turntable and spinning with it, the same marble curves. Nothing pushed it sideways. The path bends only because the observer is turning underneath it.
That is the whole idea. The Coriolis force is not a force in the Newtonian sense. It is a bookkeeping term you must add when you insist on measuring motion from a rotating frame. The simulator shows both views at once: on the left, the inertial frame, where a force-free ball goes dead straight; on the right, the co-rotating frame, where the identical motion sweeps into a curve.
The hook example is the artillery shell. A shell fired 30 km north in the northern hemisphere lands noticeably to the right of its aim. Gunners in 1914 learned to correct for this. The shell never felt a sideways push. The Earth simply rotated a little during the shell's flight time, so the target moved out from under it.
When the effect matters and when it does not
The rule of thumb is time. The Coriolis acceleration is tiny, so it only accumulates into a visible curve when the motion lasts a long time over a large distance. The relevant comparison is between the travel time T and the rotation period 2\pi/\Omega.
On Earth, \Omega \approx 7.29 \times 10^{-5} radians per second (one turn per sidereal day). A parcel of air crossing 500 km at 10 m/s spends about 50,000 seconds in transit. That is long enough for the deflection to steer whole storm systems. Water in a sink spends 5 to 10 seconds draining. In that window the Coriolis effect is swamped by the shape of the basin, leftover swirl from the tap, and the direction you pulled the plug.
The draining-sink legend does not survive the arithmetic. At mid-latitudes the horizontal Coriolis acceleration on water moving 0.1 m/s is about 1e-5 m/s², roughly a millionth of gravity. Any real sink has residual currents thousands of times stronger. The hemisphere your drain is in tells you nothing about its swirl.
The formula and where each term comes from
In a frame rotating with angular velocity vector \vec{\Omega}, a free particle appears to accelerate even though no real force acts on it. The apparent acceleration has two pseudo-force terms:
Here \vec{v} is the velocity measured in the rotating frame and \vec{r} is the position from the rotation axis. The first term is the Coriolis acceleration. The second is the centrifugal acceleration, which points outward and has magnitude \Omega^2 r.
For horizontal motion on a flat disk spinning about a vertical axis, the Coriolis term simplifies. Its magnitude is 2\Omega v and it always points at a right angle to the velocity. The sign of \Omega sets the direction: for a counterclockwise spin (positive, northern-hemisphere convention) the deflection is to the right of motion; for clockwise (negative, southern hemisphere) it is to the left.
Two facts fall out immediately. Doubling the speed doubles the Coriolis acceleration. And because the deflection is always perpendicular to \vec{v}, it changes direction but never speed. In the rotating frame a free ball traces a circle called an inertial circle, of radius r_i = v/(2\Omega).
A worked example matching the demo
One ball, two frames, with the default settings
Use the demo defaults: rotation rate \Omega = 0.5 rad/s (positive, so counterclockwise), launch speed v = 2 m/s, launched straight across the disk from the center. Work in the rotating frame.
- Coriolis acceleration magnitude: 2\Omega v = 2 \times 0.5 \times 2 = 2.0 m/s². It points 90 degrees to the right of the velocity because \Omega is positive.
- Inertial-circle radius: r_i = v/(2\Omega) = 2/(2 \times 0.5) = 2.0 m. In the rotating view the ball rides a circle 2 m across.
- Time to complete one inertial circle: T = 2\pi/(2\Omega) = \pi/\Omega \approx 6.28 s. That is half the disk's own rotation period of 2\pi/0.5 \approx 12.57 s.
- Deflection after a short time t = 1 s: to first order the sideways offset is \tfrac{1}{2} a_{\text{Cor}} t^2 = 0.5 \times 2.0 \times 1^2 = 1.0 m, while it has traveled v t = 2 m forward.
Check this against the inertial frame. There the ball goes straight, covering v t = 2 m in 1 s with zero sideways motion. Both descriptions are correct. They disagree only because one camera is spinning.
An interactive to feel the two terms
Reading and interpreting the results
Three numbers on screen tell the story. The Coriolis acceleration 2\Omega v sets how sharply the path bends per unit time. The inertial-circle radius v/(2\Omega) sets the size of the loop. The centrifugal term \Omega^2 r grows with distance from the axis and pushes outward, which is why a ball launched near the rim spirals rather than tracing a clean circle.
Watch the sign convention. Positive \Omega (counterclockwise, northern hemisphere) deflects motion to the right. That is why northern-hemisphere cyclones spin counterclockwise: air rushing inward toward a low-pressure center is deflected right, and the balance between the inward pressure pull and the rightward deflection sets up a counterclockwise circulation. Flip to negative \Omega and everything mirrors: deflection to the left, clockwise cyclones.
| Speed v (m/s) | 2Ωv (m/s²) | Inertial radius v/(2Ω) (m) | Loop period π/Ω (s) |
|---|---|---|---|
| 0.5 | 0.5 | 0.5 | 6.283 |
| 1.0 | 1.0 | 1.0 | 6.283 |
| 2.0 | 2.0 | 2.0 | 6.283 |
| 4.0 | 4.0 | 4.0 | 6.283 |
Notice the loop period is the same in every row: \pi/\Omega \approx 6.283 s. A faster ball rides a bigger circle but completes it in the same time, because the acceleration scales up in step with the speed.
Common mistakes
The first mistake is treating the Coriolis force as real. It has no source object. It appears and vanishes as you switch frames. In the inertial view of this simulator, no such force exists and the ball goes straight.
The second mistake is thinking speed changes. The Coriolis term is always perpendicular to velocity, so it does no work. The ball on the disk keeps its exact launch speed forever; only its direction rotates.
The third mistake is scale confusion, the sink myth again. People assume any rotating frame produces a visible curl on any object. The curve needs travel time comparable to the rotation period, or at least a large fraction of an inertial circle. Sink water completes a negligible arc before it drains.
To decide whether the Coriolis effect matters, compute the Rossby number: Ro = v/(L\,f), where L is the length scale and f = 2\Omega\sin(\text{latitude}). When Ro \ll 1 the effect dominates (weather systems). When Ro \gg 1 it is negligible (your sink, a thrown baseball).
Related tools
The Coriolis effect is a rotating-frame illusion, so tools about rotation and reference frames pair naturally with it. The Gyroscope Precession simulator shows another rotation surprise, where a spinning wheel circles sideways instead of falling. The Tides Simulator explains a related geophysical rhythm driven by the rotating Earth. For the sideways force that really is a push, unlike Coriolis, see the Magnus Effect, where spin bends a ball through the air. If you want to see how a completely different pseudo-force reasoning works, the Special Relativity Visualizer shows how observers in different states of motion disagree about lengths and times.
Frequently asked questions
Does the Coriolis effect make water drain the other way in the southern hemisphere?
No, not in any ordinary sink. The Coriolis acceleration on draining water is around a millionth of gravity and acts for only a few seconds, far too weak to overcome the basin shape and residual currents. Careful laboratory experiments with perfectly symmetric tanks left still for a day can just detect it, but no real bathroom sink does.
Why do cyclones spin opposite ways in the two hemispheres?
Air flowing toward a low-pressure center is deflected right in the northern hemisphere (counterclockwise circulation) and left in the southern (clockwise). The sign flips because the effective rotation rate f = 2\Omega\sin(\text{latitude}) is positive north of the equator and negative south of it, matching the positive and negative \Omega in the simulator.
Is the Coriolis force a real force?
No. It is a pseudo-force that appears only when you describe motion from a rotating frame. Switch to the inertial frame and it vanishes entirely, along with the centrifugal force. The ball moves in a straight line at constant speed in that frame, as Newton's first law requires.
Why is the deflection always sideways and never forward or backward?
Because the Coriolis acceleration is the cross product -2\vec{\Omega} \times \vec{v}, which is perpendicular to the velocity by construction. A perpendicular acceleration turns the path without changing the speed, so the ball curves but never slows down or speeds up.
How big is the effect for a long-range rifle shot?
Take a bullet at 800 m/s over a 1 km shot at mid-latitude, where f \approx 1.0 \times 10^{-4} per second. Flight time is about 1.25 s, and the sideways drift is roughly \tfrac{1}{2} f v t^2 \approx 0.5 \times 10^{-4} \times 800 \times 1.25^2 \approx 0.06 m, about 6 cm. Small but real, and target shooters correct for it.
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