How this instrument works
The Coriolis acceleration is the sideways push a moving object seems to feel because the ground underneath it is turning. Stand at a fixed point on a spinning disc and roll a ball toward the center: to someone standing on the disc, the ball appears to curve away from a straight line, even though no real sideways force touched it — the curve is the observer's own rotation showing up in the arithmetic. On Earth the same bookkeeping gives a = 2ω·v·sin(φ): double the planet's rotation rate, times the object's speed, times the sine of latitude. The factor of two falls straight out of differentiating position twice inside a rotating frame; the cross-product term 2ω×v appears alongside the more familiar centrifugal term, and it is this term, not centrifugal force, that bends the paths of wind, ocean currents, and artillery shells.
The sin(φ) term is doing real geometric work, not decoration. Only the component of Earth's spin axis that points straight up through your position, the local vertical, contributes to horizontal deflection; at the equator that component is zero, so a parcel of air crossing the equator feels no sideways push from rotation at all, while at either pole the full rotation rate acts on it. Long-range naval gunners in both World Wars corrected 16-inch shell trajectories for exactly this drift over ranges past twenty kilometers; meteorologists lean on the same relation, in its geostrophic-wind form, to explain why air spiraling into a low-pressure system curves in toward the center rather than flowing straight there.
What the formula does not capture is worth knowing too. It returns a magnitude for an object moving at speed v at latitude φ; the direction of travel does not enter the arithmetic, which is a simplification of the full vector treatment rather than an error in it. And despite the popular story, this is not why bathwater spins down a drain one way north of the equator and the other way south: at v = 1 m/s even a sink at the pole feels an acceleration of only about 0.000146 m/s², far too small to out-compete the residual swirl left by filling the tub or the exact shape of the drain.
- Enter Velocity — the speed of the moving parcel or object relative to the ground, in m/s, km/h, or mph.
- Enter Latitude — the location on Earth, in degrees north or south (or radians); negative values flip which way the deflection points.
- Leave Earth's rotation rate alone — ω = 7.2921159×10⁻⁵ rad/s is fixed inside the instrument, since it models Earth specifically.
- Read Coriolis acceleration in m/s², or switch its unit to ft/s² for the same figure in imperial terms.
Worked example — a 20 m/s wind at 45° latitude
Take a steady 20 m/s wind blowing across the mid-latitudes, near 45°N, roughly the line through Minneapolis or Milan. Converting 45° to radians gives 0.785398163397448, and the formula a = 2ω·v·sin(φ) becomes 2 × 7.2921159×10⁻⁵ × 20 × sin(0.785398163397448) = 0.00206252184084 m/s². That is the figure the instrument reads back once Velocity is set to 20 m/s and Latitude to 45 degrees.
The number looks negligible next to the wind's own 20 m/s, and for one second it is. But hold that sideways acceleration up over an hour of steady flow — 3600 seconds — and the parcel drifts roughly 13.4 kilometers off the straight-line path it would have taken on a non-rotating Earth. That accumulated drift, not the instantaneous figure, is why storm systems spiral and why transoceanic flight plans and ballistic tables both carry a Coriolis correction term.
Questions
Why does the Coriolis acceleration depend on sin(latitude)?
Because only the part of Earth's spin axis that points straight up through your location, the local vertical, drives horizontal deflection; that part scales as sin(φ). At the equator the spin axis lies flat along the horizon, so sin(φ) = 0 and the horizontal effect vanishes; at either pole the axis points straight down, sin(φ) = ±1, and a given speed produces its largest possible deflection.
Why can't I enter Earth's rotation rate myself?
Because this instrument is built around Earth's Coriolis effect specifically, not a generic rotating-frame problem, so ω is wired in at 7.2921159×10⁻⁵ rad/s, one full turn every 23 hours 56 minutes. Fixing it removes a step you would otherwise have to look up, and it keeps results across different latitudes directly comparable.
What does a negative Coriolis acceleration mean?
It flags the southern hemisphere. Enter a latitude below zero and sin(φ) turns negative, so the acceleration's sign flips too, a direct statement that the deflection reverses: moving objects curve left of their path south of the equator instead of right. The size, 2ω·v·|sin(φ)|, is identical at, say, 45°N and 45°S; only the sense of the turn changes.
Does this explain which way water spins down a drain?
No, and that popular claim is a myth. At v = 1 m/s, even at the pole, the Coriolis acceleration is only about 0.000146 m/s², thousands of times smaller than the swirl a hand, a faucet, or an off-center plug already puts into a sink or tub. The effect only becomes the dominant force over the large distances and long times of wind, ocean currents, and artillery, not basin-sized water over a few seconds.
Is this the force that makes hurricanes spin?
Yes, in origin. The Coriolis effect is what sets a storm system rotating, counterclockwise in the northern hemisphere and clockwise in the southern, once it starts drawing air inward toward low pressure. Hurricane dynamics also fold in pressure gradients, surface friction, and latent heat, so a = 2ω·v·sin(φ) gives the deflection rate for one moving parcel, not a complete storm model.
Why does the answer stay so small even for fast objects?
Because ω itself is tiny: Earth turns through only about 7.29×10⁻⁵ radians each second. Even a jetliner at 250 m/s crossing 45° latitude feels roughly 0.0258 m/s², a small fraction of standard gravity. The effect is real and measurable but stays secondary to the parcel's own speed at any single instant; it matters because it accumulates over the hours a flight or a current runs, which is why navigators correct for it and sprinters never notice it.