I’m trying to refresh my Physical Geography knowledge and ran into a problem I don’t remember the solution to, and googling doesn’t help. The problem is to do with the effect of the Coriolis force on movements that are undertaken along a parallel. I understand the deflection of moving mass has to do, basically, with the different speeds at which masses travel in different latitudes, so that a mass of air, say, which comes north from the aquator, has a stronger movement component to the right than a mass of air that is at the tropic of the cancer, or at the polar circle, and accordingly, it is deflected to the right; but although I know masses that travel parallel to a parallel are similarly deflected, I can’t remember why. Nor do my books tell me; maybe the Dope can help out?
But, they are not. The magnitude of Coriolis deflection is it at maximum for motion along the longitude, and zero for motion perpendicular to it.
Are you talking about masses moving parallel to each other, but at different speeds?
(Just trying to clarify.)
Earth has an angular velocity of, well, let’s just call it 24h/revolution… And everything **on **Earth is traveling along for the ride. Now at the Equator, this translates to 40000 Km/24 h, or roughly 1660 Kph. As you travel North, the length of each consecutive line of Longitude is slightly less – by the time you’re up to the 60th (? I think it’s Cosine) parallel, it’s length is only 20000 km. At the Pole, of course, the length of the 90th “parallel” is zero. So, somewhere, relatively near the equator, the length of a particular line of Longitude (“Parallel”) will be 38400 km, and points at this latitude will all have a linear velocity of 1600 kph (38400/24).
Now your air-mass, traveling North, starts out with a linear velocity going Eastward at 1660 kph. When it reaches the 38400-km long parallel, it is still traveling East at 1660 kph. Except all the points above which it is going are only doing 1600 kph Eastward… so you would (in a theoretical, friction- and other-atmosphere-free, of course) world have your air-mass traveling 60 kph East relative to the ground.
Various effects (such as friction) modify and moderate this, of course, but this still is a real effect in the real world. So, a Southernly wind (going North from the Equator) will tend to “curl” East, while a Northernly wind (traveling South toward the Equator) will “curl” West (unless I got my directions wrong and it’s the other way around? :o)
In the Southern hemisphere, of course, the directions are reversed.
See, that was what I thought it had to be; but I’ve been rather specifically told by all my sources that all movements are deflected to the right in the northern hemisphere, even those movements that are perpendicular to the longitude (that is, parallel to latitude). Even movements the go east-west, then are deflected to the right; but I can’t understand why.
I’ve been unclear about what I need to know; the general idea is clear to me, with regard to north-south movements. I’m wondering about why an air-mass moving ON the 34th (lets say) parallel, not perpendicular to it, would be deflected, precisely because of the reasoning you offer. If my air mass does not change into the influence of a different rotational speed, there should be no deflection.
Sorry, my misunderstanding – I didn’t read your OP closely enough :o
To address your original question: My WAG is, the linear velocity **gradient **of the “stationary” (WRT Earth) air-wall standing in the path of your moving air-mass will cause a wedge-like effect, deflecting the air-mass in the direction in which the “stationary” air is traveling at a velocity closer to that of the moving air-mass – so an East-moving wind in the Northern hemisphere (linearly “faster” than the surrounding air) will tend to be deflected South, toward stationary air traveling at a velocity closer to the wind’s.
But this is just a WAG this time. I’m not even sure I got the directions right this time either (I’m working on 25±Y.O. memories from High School.)
Your sources are wrong. The coriolis force for the earth, for example, is proportional to the cross product of the angular velocity of the earth and the velocity of the particle in question. The earth’s velocity is a vector parallel to the earth’s axis of rotation and so is the velocity of a particle moving along a parallel of latitude. The two vectors being parallel have a cross product of zero.
After thinking on this some more I think that the above is wrong. Ignore it for the time being and I’ll be back.
Your sources are correct (how’s that for a switch?). At some spot on the earth’s surface of latitude ß erect a coodinate system with the x axis due east, the y axis northward tangential to the surface and the z axis radially outward.
The angular velocity W[sub]s[/sub] of a point on the surface in this system will be a vector having the components:
**
0
W[sub]s[/sub] = Wcosß
Wsinß**
the velocity of the particle in question will have the components in this coordinate system of:
**
vx
V = vy
vz**
The coriolus vector is equal to -2W[sub]s[/sub] x V
**
▌x y z ▌
-2*▌0 Wcosß Wsinß▌
▌vx vy vz ▌**
This is the vector:
**
2vyWsinß - 2vzWcosß
-2vxWsinß
2vxWcosß**
When the air is moving due east vy and vz are zero and the corilous effect is a vector south and up. Moving west the vector is north and down. When the wind is moving north the vector is east and when moving south it is west.
I used Wikipedia as a refresher course.
And I would assume the relative motion of the particle is right and up because the spot on the earth in the northern hemisphere is moving down and left relative to a line tangential to the parallel of lattitude except at the equator.
Well, I do believe I get what you’re saying and thank you kindly for it!
Math is so cool.
Tris