As air moves from high to low pressure in the northern hemisphere, it is deflected to the right by the Coriolis force

They use an example of a merry-go-round, which makes sense intuitively. An object pushed from the outer edge inward is deflected to the right if the merry go round is spinning counter-clockwise.

Applying that to winds converging on a low pressure system on the surface of earth: we are told that Coriolis Force causes the winds to deflect to the right.

Question: what about winds moving from north to south in this system? This seems like it would be the 2D equivalent of pushing a ball from the inside to the outside of the merry go round, which if spinning counter-clockwise, would deflect to the left? (a ball pushed from inside to outside still deflects to the right, relative to co-rotating observer.)


I never did like the merry-go-round explanation for the Coriolis Effect. It may be intuitive for some, but I never could wrap my head around it. (And it's properly the Coriolis Effect, not Force. It's no more a "force" than the "force" pushing you against your car seat as your car accelerates.)

This works better for me: Imagine you plan on launching a projectile from the equator towards the North Pole. Your projectile will have an easterly velocity 1,039 mph, but let's, for the sake of simplicity, say it's 1,000 mph. As your projectile crosses, say, 20° north latitude it will still have that 1,000 mph eastward velocity, but the ground at 20° north latitude will have an eastward velocity of 940 mph (cosine 20° = .93969). IOW, the projectile will be travelling about 60 mph to the east faster than the ground. At 45° north latitude the ground will be moving at 707 mph, so your projectile will be traveling to the east almost 300 (293) mph faster than the ground.

Now imagine the opposite situation, launching a projectile from, say 40° north towards the equator. Your projectile will have an eastward velocity of 766 mph. As it crosses 20° north it will still have that eastward velocity of 766 mph, the ground will be travelling at 940 mph. IOW, the ground will be moving eastward 174 mph faster than the projectile, which will appear to deflecting towards the west.

In both cases the projectile will be appearing to deflect towards the right. From the equator that would be towards the east, but from north of the equator towards the south, the deflection would appear to be towards the west.

From a stationary observer above the earth, the projectile would appear to travel in a straight line while the earth rotated underneath it, which is in reality what is happening.

In the case of an air mass moving into a low pressure center, that air mass gets deflected towards the right as it enters the influence of the low, so it always enters the low toward its right. It then has no other direction to turn but towards the left.

  • $\begingroup$ Exactly the answer I was looking for. Took me for ever to realize that this is, as you said, not really a force but an effect. Also: I feel you should change 60mph faster than the ground to "relative to the ground the object is moving x mph etc. etc." $\endgroup$ – OneChillDude May 11 '16 at 15:44
  • $\begingroup$ FWIW, we refer to the centrifugal force. I've never heard the term centrifugal effect. Well, I agree with you, but language is a soft kind of thing. Usage defines it. $\endgroup$ – garyp May 11 '16 at 15:51
  • $\begingroup$ @OneChillDude, I tried to incorporate your suggestion into my answer, but just couldn't get the verbiage to work; it sounded awfully forced. $\endgroup$ – BillDOe May 12 '16 at 0:00
  • $\begingroup$ @garyp, I learned the Coriolis Effect as it being an effect, and my professor was really adamant about it. I kinda carried that on. It seems less confusing. $\endgroup$ – BillDOe May 12 '16 at 0:00
  • $\begingroup$ @BillOer eh no worries. Thanks for trying though! $\endgroup$ – OneChillDude May 12 '16 at 21:28

In physics, the Coriolis effect is a deflection of moving objects when they are viewed in a rotating reference frame. In a reference frame with clockwise rotation, the deflection is to the left of the motion of the object; in one with counter-clockwise rotation, the deflection is to the right.

In the Northern hemisphere the Coriolis force ALWAYS pushes the moving particle to the right, no matter if it is moving N or S or E or W. In the Southern hemisphere the Coriolis forces ALWAYS pushes the moving particle to the left, no matter if it is moving N or S or E or W.

Moreover, near the equator the effect should be zero if you are traveling to the N or to the S because the velocity and earth's rotation vectors are parallel, and if you travel east to west on the equator, the Coriolis force is actually pointing inwards! That is why hurricanes rarely form at these latitudes.

  • $\begingroup$ Hmm. Viewed from the rotating earth, wouldn't the deflection always be to the west in the northern hemi? Also, I think you have to add to your description the curvature of the earth. The effect is greater as you go further north (otherwise no rotation and no hurricanes). So wind coming from the north dominates that from the south, is deflected west, and results in counter-clockwise wind pattern. I readily admit that I'm making this up as I go along, and I may be dead wrong. If so, please help! No doubt: the Coriolis force is poorly described in most places. $\endgroup$ – garyp May 10 '16 at 18:24
  • $\begingroup$ Also, I imagine that the dragging of the atmosphere over the surface (which is not the Coriolis force, and would deflect to the east) must complicate things. I concur about sending this to Earth Science. $\endgroup$ – garyp May 10 '16 at 18:27
  • $\begingroup$ Sure, the boundary layer drag adds forces to the force balance. But still, the Coriolis force is going to be present. Causing particles to deflect to the right in the N. hemisphere. I think you are over complicating things. Basically if it is a high pressure then wind coming out of the center of the high pressure will veer right, resulting in a clockwise wind pattern. If it is a low pressure then wind going into the center of low pressure veers to the right causing a counter-clockwise wind pattern around the center of low pressure. $\endgroup$ – Isopycnal Oscillation May 10 '16 at 18:36
  • 1
    $\begingroup$ BTW this website does a pretty good job at explaining Coriolis stratus.ssec.wisc.edu/courses/gg101/coriolis/coriolis.html $\endgroup$ – Isopycnal Oscillation May 10 '16 at 18:43
  • $\begingroup$ Yes it does. Thanks. (One of) my error(s) was neglecting the eastward component of the velocity. $\endgroup$ – garyp May 10 '16 at 20:07

In the northern hemisphere winds moving from north to south do deflect to the right i.e. towards the west. Two an example of this would be:

- In the northern hemisphere winds in hurricanes circle in a clockwise sense so on the eastern side of the storm winds a moving towards the south and deflected west - to the right.

  • In the Hadley cell in the northern hemisphere. Hot air at the equator rises, cooler air from the north flows south to replace it and is deflected right - to the west on the way. This gives rise to the trade winds.
  • $\begingroup$ Uh... hurricanes in the north rotate counterclockwise $\endgroup$ – garyp May 10 '16 at 18:25
  • $\begingroup$ Right - I'll just take that out of my answer. $\endgroup$ – M. Enns May 10 '16 at 19:19

OK, so I figured out why I was so confused: The keywords that helped me have the "aha" moment were on wikipedias page for Coriolis Force.

the Coriolis force is an inertial force (also called a fictitious force)

In my head I was imagining Coriolis Effect being caused by fluid friction, but it's actually caused by inertia with not enough fluid friction to accelerate air as it moves up and down the globe. This video also helped my understanding: https://www.youtube.com/watch?v=i2mec3vgeaI

  • $\begingroup$ Coriolis arises because particles are moving in a rotating frame of reference. i.e. the earth is rotating. $\endgroup$ – Isopycnal Oscillation May 11 '16 at 18:18

protected by Qmechanic May 10 '16 at 19:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.