Coriolis Force at work in large-scale geophysical flows

We suggest the reader reference an in-depth discussion of the Coriolis force at work in large-scale, geophysical flows, by James F. Price, Woods Hole Oceanographic Institutionan essay on the Coriolis force

Coriolis Force

See the accompanying sub-section page at Wikipedia, the free encyclopedia

uTube Videos:

1) What’s the Coriolis Effect? - Winds - BBC Four

2) The Coriolis Force

Great Circle routes vs Rhumbar. GE does paths as great circle, while overlays are placed as Mercator projections, which lines show up as Rumbars.

The equation of motion for a rotating coordinate system has two terms to accommodate the rotation of the coordinate system, the centrifugal term and the Coriolis term. Generally the centrifugal term is much larger than the Coriolis term. The earth does not rotate fast enough for the centrifugal term to "throw off" an object from its's surface, it is leveraged in practical use in the case of launching spacecraft: Launching sites closer to the equator are more favorable.

The explanation of this phenomenon is mainly kinematic, i.e., more mathematical than physical.

Coriolis Force and Projectiles

(excerpt here)
Federation of American Scientists, 1725 DeSales Street, NW, 6th Floor,Washington, DC 20036 , email:

Projectiles which travel great distances are subject to the Coriolis force. This is a artifact of the earth's rotation. The local frame of reference (north, east, south and west) must rotate as the earth does. The amount of rotation, also known as the earth rate is dependent on the latitude: earth rate = (2p radians)/(24 hours) sin(latitude). For example, at 30o N, the earth rate is 0.13 radians/hr (3.6 x 10-5 rad/sec). As the frame of reference moves under the projectile which is traveling in a straight line, it appears to be deflected in a direction opposite to the rotation of the frame of reference.

Figure 12. Coriolis force.
In the northern hemisphere, the trajectory will be deflected to the right. A projectile traveling 1000 m/s due north at latitude 30o N would be accelerated to the right at 0.07 m/s2. For a 30 second time-of-flight, corresponding to about 30 km total distance traveled, the projectile would be deflected by about 60 m. So for long range artillery, the Coriolis correction is quite important. On the other hand, for bullets and water going down the drain, it is insignificant!


The angular velocity of earth is 4.17 x 10^-3 deg/sec.

That is 7.27 x 10^-5 sec^-1 (1 deg ~ 0.0175 rad).

If I fire East for example, I'd be at 90 deg to the axis of the earth. But firing NE, I'd have components of both.... If I do have multiple vectors, how do I figure the effects?

Simply split velocity into "north" (angle to Earth's axis depending on latitude) and "east" (always 90 deg to axis) components, and compute and work with Coriolis forces for both. You can do this because vector product is distributive, i.e. (\vec{v}_{north}} + \vec{v}_{east}) \times \omega = \vec{v}_{north}} \times \omega + \vec{v}_{east}} \times \omega.

The "north" force component will push sideways (to east if you fire north, to west if you fire south), and the "east" component will push upwards (that's why they like to blast off eastbound near equator).

Chusslove Illich (Часлав Илић)