The Coriolis Effect: 2Ωv sin φ, and the Scale at Which It Matters

Q: Why does the deflection grow as t² and not as t?

Because it is a sideways acceleration, and a constant acceleration gives a displacement of ½at². The sideways velocity grows as t, and the displacement is the area under that, so it grows as t². This is also why the percentage of the path deflected grows as t: the distance travelled only grows as t, so the ratio grows linearly with the flight time.

The Coriolis effect is not a force. It is what a straight line looks like from a frame that is turning underneath it. Give a speed, a latitude and a time of flight and this returns the sideways deflection, the Coriolis parameter f = 2Ω sinφ that meteorologists use, and the distance scale below which the whole thing is negligible.

Last updated 11 Nov 2026

Deflection from a rotating frame, to your numbers

a = 2Ωv sin φ — an inertial term, not a force

Horizontal speed v (m/s)The speed over the ground, in m/s. Divide km/h by 3.6. Only the horizontal component matters for the horizontal deflection this page reports.

Latitude φ (degrees, negative for the south)Positive north, negative south. The sinφ makes the effect vanish on the equator and reach its maximum at the poles. 51.5 is London; 28.6 is Delhi; −33.9 is Sydney.

Rotation period of the frame (hours)The default is the Earth’s sidereal day, 23 h 56 min 4.09 s — one rotation relative to the stars, not the 24 h solar day. Using 24 understates Ω by 0.27 per cent. Put 0.05 in for a three-minute fairground roundabout and watch what happens.

Time of flight or travel, t (s)How long the object is moving for. The deflection grows as t², so this is the input that decides whether the effect matters at all.

The straight path an inertial observer sees, and the bent path the rotating frame sees. The sideways scale is exaggerated by the factor printed at the bottom, so that a deflection of a fraction of a per cent is visible at all.

5.1362mExample

A hockey ball is struck at 25 m/s in London (latitude 51.5°) and travels for 60 s. How far sideways does the Earth’s rotation carry it, and what does that mean in practice?

The acceleration, the parameter, and the two scales that decide whether it matters

a = 2Ωv sinφ · f = 2Ω sinφ · deflection ≈ ½at² · R = v/|f| · T_inertial = 2π/|f|

Ω: the angular velocity of the rotating frame, in rad/s. For the Earth it is 7.292 115 × 10⁻⁵ rad/s, which is 2π divided by the sidereal day of 86 164.09 s, not by the 86 400 s solar day. The difference is 0.27 per cent and comes from the Earth also going round the Sun
sinφ: picks out the component of the rotation about the local vertical, which is the only component that turns a horizontal velocity within the horizontal plane. It is zero on the equator and one at the poles, and it is why the same planet gives no horizontal deflection in Singapore and a full-strength one in Reykjavik
f: the Coriolis parameter, 2Ω sinφ, defined in the American Meteorological Society’s Glossary as “twice the component of the earth’s angular velocity about the local vertical”. It is the number every atmospheric and oceanic calculation actually uses, and it has units of s⁻¹
½at²: an approximation, good while the deflection is small compared with the distance travelled. It treats the Coriolis acceleration as constant, and it is not: the acceleration is always perpendicular to the velocity, so as the path bends the acceleration turns with it. The exact path is a circle of radius v/|f|, and this formula is its first approximation
what it is not: a force. Nothing exerts it, it has no third-law partner, and it disappears the moment you stop using a rotating frame. It is the second of the two inertial terms — the other is the centrifugal term covered on the centripetal page — and unlike that one it appears only when you MOVE within the rotating frame
the vertical term: there is one, 2Ωv cosφ for eastward motion, and it is ignored here as it is in meteorology. It acts vertically, where it competes with g, so it changes an apparent weight by parts per million rather than bending a path. The horizontal term has no competition and that is why it wins

Worked example

A hockey ball is struck at 25 m/s in London (latitude 51.5°) and travels for 60 s. How far sideways does the Earth’s rotation carry it, and what does that mean in practice?
Angular velocity of the Earth. Ω = 2π/T with T the sidereal day, 86 164.09 s, giving Ω = 7.2921 × 10⁻⁵ rad/s. Using 86 400 s here is the commonest slip and it is wrong by 0.27 per cent.
Coriolis parameter. f = 2Ω sinφ = 2 × 7.2921 × 10⁻⁵ × sin 51.5° = 1.4584 × 10⁻⁴ × 0.78261 = 1.1414 × 10⁻⁴ s⁻¹. Put the latitude into your calculator in degrees or convert by hand — 51.5 × π/180 = 0.8988 rad.
Acceleration. a = fv = 1.1414 × 10⁻⁴ × 25 = 2.854 × 10⁻³ m/s². That is about 1/3400 of g, which already tells you the answer is going to be small.
Deflection. ½at² = ½ × 2.854 × 10⁻³ × 60² = ½ × 2.854 × 10⁻³ × 3600 = 5.14 m, to the right of the direction of play.
Now put it in proportion. In 60 s at 25 m/s the ball covered 1500 m. The deflection is 5.14 m, which is 0.34 per cent of the path. A hockey ball is in the air for about a second, not sixty, and over one second the deflection is 1.4 millimetres — because it grows as t², so shortening the flight sixty-fold cuts the deflection three thousand six hundred fold.
Which is the real lesson. The Coriolis effect is not weak because Ω is small; it is weak because sporting and domestic events are short. Give the same calculation a 500 s shell flight and the deflection is 357 m. Give it a weather system lasting days and the air stops going where the pressure pushes it altogether.

The same latitude and speed, at wildly different timescales

Everything in this table is at 25 m/s and 51.5°N; only the time changes. The deflection grows as t² while the distance grows as t, so the percentage column grows as t as well. The last row is marked as a failure deliberately: once the deflection is comparable with the distance, ½at² is no longer the right model and the motion is a circle of radius v/|f|, which here is 219 km.
Situation	t (s)	Distance (m)	Deflection (m)	% of the path
A hockey ball in the air	1	25	0.0014	0.0057%
A long goal kick	4	100	0.023	0.023%
A sniper shot at 1 km	40	1 000	2.28	0.23%
The worked example	60	1 500	5.14	0.34%
A naval shell, long range	500	12 500	357	2.9%
A parcel of air, half a day	43 200	1 080 000	the formula fails	Coriolis dominates

How the effect varies with latitude, for the worked example

The inertial period at the pole is **half a sidereal day**, which is exact and not a coincidence: there f = 2Ω, so 2π/f is half of 2π/Ω. The Sydney row has a negative f and the deflection is to the left; the magnitude is what the page reports, and the hemisphere decides the direction.
Latitude	sin φ	f (10⁻⁴ s⁻¹)	Deflection in 60 s (m)	Inertial period
0° — the equator	0.000	0.000	0.00	no inertial circle
15° — Dakar	0.259	0.377	1.70	46.3 h
28.6° — Delhi	0.479	0.698	3.14	25.0 h
51.5° — London	0.783	1.141	5.14	15.3 h
−33.9° — Sydney	−0.558	−0.813	3.66, to the LEFT	21.5 h
90° — the pole	1.000	1.458	6.56	11.97 h

Why a straight line looks bent, and the bath-drain question answered properly

Start on a roundabout. Stand at the centre of a turning playground roundabout and roll a ball straight out towards a friend on the rim. In the ground frame the ball goes perfectly straight — nothing pushes it sideways. But the rim has moved by the time the ball arrives, so from the roundabout the ball appears to curve away. Film it from above and from a camera bolted to the roundabout and you will see two different paths for one ball. The Coriolis acceleration is the name for the difference.

The Earth is that roundabout. It turns once every sidereal day, so Ω is only 7.29 × 10⁻⁵ rad/s, which is why the effect is so gentle. Only the component of that rotation about the local vertical turns a horizontal velocity within the horizontal plane, and that component is Ω sinφ. Hence the factor sinφ: nothing at the equator, everything at the poles. The deflection is to the right of the motion in the north and to the left in the south, and it does not matter which compass direction you are travelling in — it is always to the right of your direction.

Why it looks like a force and is not one. It has no agent: you cannot finish the sentence “this force is exerted on the shell by …”. It has no third-law partner. It depends on the velocity of the object, which no real force does in quite this way, and it vanishes for an object at rest in the rotating frame. All of that marks it as an inertial term, the price of writing F = ma in a frame where it does not strictly hold. Meteorologists use it anyway, constantly, because working in the Earth’s frame is enormously more convenient than working in a frame fixed to the stars.

The t² is the whole story of what matters and what does not. The acceleration is tiny, but the sideways displacement accumulates as ½at². Over one second the deflection of anything is sub-millimetre. Over a minute it is metres. Over a day it dominates. That single scaling explains why Coriolis is a routine correction in long-range gunnery and in aviation, a controlling influence in meteorology and oceanography, and completely irrelevant to everything that happens in a room.

So: does it decide which way your bath drains? No, and the numbers in this page are why. The water in a bath moves at perhaps 0.1 m/s and drains in a couple of minutes, which gives a Coriolis acceleration of order 10⁻⁵ m/s² and a deflection of millimetres. Any residual swirl from filling the bath, any asymmetry in the plughole, any slope in the basin, is thousands of times stronger. The effect can be demonstrated in a drain, and has been, but only in a symmetrical tank left to stand for a day in a controlled laboratory so that every other motion has died away first. In your bathroom it is swamped.

Where this goes wrong in exams and arguments. Four ways. One: using the 24-hour solar day for Ω instead of the 86 164 s sidereal day. Two: putting the latitude into sin as though the calculator were in radians when it is in degrees, or the reverse — and note this page converts by hand with φ × π/180 because the engine has no degree function. Three: calling it a force in a sentence that also mentions a free-body diagram. Four: quoting the deflection without the distance, which hides the only thing that matters — 5 m sounds impressive until you learn the ball travelled 1.5 km.

Frequently asked questions

Is the Coriolis effect a real force?

Not in the Newtonian sense. It has no agent, no third-law partner, and it vanishes in an inertial frame. It is one of the two inertial terms you must add to make F = ma work in a rotating frame, and in that frame it is entirely real in its consequences — the shell does land somewhere else. Both statements are true at once, and arguing about which is “really” right is arguing about a choice of frame.

Why sin of the latitude?

Because only the part of the Earth’s rotation that is about your local vertical can turn a horizontal velocity within the horizontal plane. At the pole the rotation axis is vertical and sinφ = 1. On the equator the axis is horizontal, there is no vertical component at all, and sinφ = 0. The Earth is turning just as fast in both places; what changes is the geometry of your local horizontal plane relative to it.

Does it make bathtubs drain the other way in Australia?

Not in practice. The deflection over a draining bath is of order millimetres, and the residual swirl left over from filling it is thousands of times larger. Careful experiments in symmetrical tanks, left standing for a day so every other motion dies, do show the predicted sense in both hemispheres — but a real bath is nothing like that. The common demonstration offered to tourists at the equator is a conjuring trick.

Why does the deflection grow as t² and not as t?

Because it is a sideways acceleration, and a constant acceleration gives a displacement of ½at². The sideways velocity grows as t, and the displacement is the area under that, so it grows as t². This is also why the percentage of the path deflected grows as t: the distance travelled only grows as t, so the ratio grows linearly with the flight time.

What is the inertial circle for?

It is the natural scale of the effect. If no other horizontal force acted at all, a body moving at speed v in a frame with Coriolis parameter f would travel a circle of radius v/|f|, once every 2π/|f|. At 25 m/s in London that is a 219 km circle taking 15.3 hours. Compare your own problem with those numbers and you have your answer immediately: if your path is kilometres and your time is minutes, Coriolis is a rounding error.

Is this on my syllabus?

Usually not as an examinable formula. Neither the CBSE Class XI–XII curriculum nor Cambridge 9702 lists the Coriolis term among its learning outcomes, and the two of them cover rotating frames only through the centripetal requirement. It appears in geography and in earth-science courses, in university mechanics when non-inertial frames are introduced properly, and in almost every argument anyone ever has about bathtubs. It is here because it is the clearest example there is of an effect that is real, measurable and still not a force.

Related calculators

References

International Earth Rotation and Reference Systems Service, “Useful Constants” (hpiers.obspm.fr/eop-pc/models/constants.html, table maintained by C. Bizouard, read 11 November 2026). Gives the mean angular velocity of the Earth as Ω = 7.292 115 0(1) × 10⁻⁵ rad s⁻¹, the nominal value Ω_N = 7.292 115 146 706 4 × 10⁻⁵ rad s⁻¹ for epoch 1820, and the sidereal day as 86 164.090 530 832 88 s. The page cites IERS Numerical Standards (IAG 1999) for Ω and IERS Conventions (2003) for the stellar-day ratio. This is the figure the page uses, and the reason the default rotation period is 23.934 470 h and not 24.
American Meteorological Society, Glossary of Meteorology, entry “Coriolis parameter” (read 11 November 2026 from the glossary wiki). Defines it as “twice the component of the earth’s angular velocity about the local vertical, 2Ω sinφ, where Ω is the angular speed of the earth and φ is the latitude”, and notes that for a parcel of horizontal speed V the product fV is the horizontal Coriolis force per unit mass — which is exactly the acceleration this page reports. The entry gives no numerical values for Ω, which is why the constant above is cited separately.
BIPM, The International System of Units (SI), 9th edition (2019). The radian is the coherent SI unit of plane angle and has dimension one — which is why rad/s and s⁻¹ pass the same unit check, why the radian can appear and vanish from a dimensional analysis without anything being wrong, and why every rotational formula on this site is written in radians and not degrees. Table 4 gives the newton as kg m s⁻².
PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Unit III, Laws of Motion, reads in part: “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. The parenthesis “no derivation” is why this page gives you the standard results rather than the integrals behind them. Quoted to establish what is in scope, never to claim any endorsement by the board.