Coriolis Deflection

Equivalent forms

\vec{F}_{\text{Cor}} = -2m\,\vec{\omega}\times\vec{v}

a_{\text{Cor}} = 2\omega v \sin\phi

It is not a real force at all — just the price of insisting that a spinning frame is 'at rest.'

Symbol	Name	SI unit	Dimension	Range
$a$	Coriolis accelerationoutput Apparent sideways acceleration in the rotating frame	m/s^2	LT^-2	0 – 50
$ω$	Angular velocity Rotation rate of the frame	rad/s	T^-1	0 – 5
$v$	Speed in frame Speed of the body relative to the rotating frame	m/s	LT^-1	0 – 10
$φ$	Latitude Latitude angle (sets the vertical component of ω on Earth)	rad	1	0 – 1.5708

Unit systems

Symbol	SI	CGS	Imperial
$a$	m/s^2	cm/s^2	ft/s^2
$ω$	rad/s	rad/s	rad/s
$v$	m/s	cm/s	ft/s

Where it holds

Any non-inertial rotating reference frame; on Earth

\omega = 7.29\times 10^{-5} rad/s

, so the effect is tiny over short ranges but dominant for large-scale, long-duration motion (weather, ocean gyres, long-range artillery).

Dimensional analysis

[LT^{-2}]

[T^{-1}][LT^{-1}] = [LT^{-2}]

2\omega v

has units

(rad/s)(m/s) = m/s^{2}

; rad is dimensionless.

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Discovery

Gaspard-Gustave de Coriolis · 1835

Coriolis derived the supplementary 'compound centrifugal force' in 1835 while studying energy transfer in rotating machines like waterwheels. Decades later meteorologists realized it governs the rotation of cyclones; Foucault's 1851 pendulum made the Earth's rotation visible through the same effect.

Try this

Fire a cannon due north and the shell lands to the east of its target — without any wind. The ground turned under it.

On a spinning platform a body moving in a straight line (as seen from space) appears to curve sideways. The faster the spin or the body, the harder the apparent push.

Research status: stable

Real-world applications

Cyclone and trade-wind circulation in meteorology
Ocean gyres and the Ekman spiral
Long-range artillery and ballistic-missile targeting
Coriolis mass flow meters in industry

Common misconceptions

The Coriolis force drains your bathtub or toilet a fixed way — far too weak at that scale; basin shape dominates
It is a real force — it is fictitious, appearing only because the frame rotates
It speeds objects up or down — being perpendicular to v, it does no work and only changes direction

Experimental verification

Foucault's pendulum (1851) precesses

at \omega \sin \varphi per day

, directly visualizing the effect; weather satellites show cyclones rotating counter-clockwise in the North and clockwise in the South; long-range naval gunnery tables include Coriolis corrections.

Derivation

Transforming Newton's law to a frame rotating

at \omega

, the acceleration acquires three fictitious terms: a_

rot = a

_

in - 2\omega \times v_rot - \omega \times (\omega \times r) - (d\omega /dt)\times r

.

The velocity-dependent piece

-2\omega \times v

is the Coriolis acceleration.

For horizontal motion on Earth only the vertical component

\omega \sin \varphi

deflects it, giving magnitude

2\omega v \sin \varphi

, to the right in the Northern Hemisphere.

Limiting cases

v parallel

to \omega

⟶

a = 0

The cross product vanishes — motion along the rotation axis feels no Coriolis push.

\varphi = 0

(equator)⟶ No horizontal deflection of horizontal motionThe vertical component of Earth'

s \omega is

zero at the equator, so cyclones can't spin up there.

\omega \to 0

⟶

a \to 0

A non-rotating frame is inertial; straight lines stay straight.

What if…

What if you walk toward the center of a fast merry-go-round?

You feel a strong sideways shove — the Coriolis force trying to conserve your angular momentum.

What if Earth spun ten times faster?

Coriolis effects scale with $\omega$ , so storms would be far tighter and faster-rotating.

What if you throw a ball straight up at the equator?

It lands slightly west, because vertical motion couples to the eastward part $of \omega$ .

1

Artillery shell at mid-latitude

Given ·

v:: 800
ω:: 0.0000729
φ:: 0.785

Find · Coriolis acceleration

Steps

$a = 2\omega v \sin \varphi$
$= 2\times 7.29e-5\times 800\times 0.707$
$\approx 0.082\,\mathrm{m/s}^{2}$ — small, but over 30 s of flight it shifts impact tens of metres

Result ·

a = 2\times 7.29e-5\times 800\times \sin 45^{\circ} \approx 0.082\,\mathrm{m/s}^{2}

2

Foucault pendulum precession

Given ·

φ:: 0.785

Find · precession period

Steps

Precession rate $= \omega \sin \varphi$
Period = (sidereal $day)/\sin \varphi$
$= 24/0.707 \approx 33.9\,\mathrm{h}$

Result ·

T = 24\,\mathrm{h} / \sin 45^{\circ} \approx 33.9

hours per full turn

Centripetal Acceleration→Angular Momentum→Newton's Second Law→Simple Pendulum Period→