Talk:Coriolis force/Archive 4

inertial circle
The article states:


 * If an object moves in a rotating system subject only to the Coriolis force, it will move in a circular trajectory called an 'inertial circle'.

This is equivalent to stating:


 * If an object which is stationary with respect to an inertial frame of reference is observed from a rotating frame of reference then the object will appear to move in a circle called an "inertial circle".




 * What are you talking about? The expressions 'inertial oscillation' and 'inertia circle' are sharply defined in meteorology. Only the pattern of motion that is depicted in the picture on the right falls within that definition of 'inertial oscillation'.
 * Anders Persson writes:
 * Coriolis's name began to appear in the meteorological literature at the end of the nineteenth century, although the term "Coriolis force" was not used until the beginning of the twentieth century.


 * Right or wrong, it is a given that in meteorology all wind deflections that cannot be accounted for by pressure gradients are without further ado attributed to 'coriolis effect'. Possibly the meteorological community did not bother to check whether they were using the expression 'coriolis effect' in the same way as in other branches of science. --Cleon Teunissen | Talk 07:43, 31 August 2005 (UTC)


 * This is wrong. In met (as in anywhere else) the Coriolis term is defined by omega cross v. Thats all. William M. Connolley 08:07:25, 2005-08-31 (UTC).

Look how much plainer the second sentence is! No pseudo-physics. A layman can understand. The first sentence, the one in the article, says the same and does so with less clarity. Let us be clear in the article: Nothing can move under the influence of a fictitious force, it can only appear to do so. Invoking the Coriolis force in an explanation without making clear that the force does not exist can only serve to confuse.

Paul Beardsell 02:28, 31 August 2005 (UTC)

The object is stationary (or has constant velocity) so there are no net forces. And surprise, surprise: The forces in balance are both fictitious:  Centrigual and Coriolis! Paul Beardsell 02:35, 31 August 2005 (UTC)


 * I disagree. And so I've restored my version, though rephrased. Your version is wrong: the object doesn't have to be stationary in the inertial system, and indeed usually isn't. As for the "circles" shown on the pic, the object is near-stationary in the rotating system and moving rather fast in the quasi-inertial system. Also, the distinction between real and fictitious forces isn't meaningful or helpful. William M. Connolley 08:16:37, 2005-08-31 (UTC).


 * You pick hairs so I do the same: If the object is NOT STATIONARY then there will be no inertial CIRCLE becuase a circle is defined (in physics as in mathematics [but probably we will be told not in meteorology]) as the locus of points in a plane equidistant from another point.  However, I agree this is overly picky.  Nevertheless, my definition is that of the simplest inertial circle imaginable.  Paul Beardsell 14:38, 2 September 2005 (UTC)

And the explanation in the same section as to why low-pressure systems rotate in the opposite direction is piss poor. From the POV of the rotating observer the Coriolis force prevents the collapse of the low pressure system. The non-rotating observer sees that the rotating earth swings the winds to the right (Northern Hemisphere) which puts the low pressure area on the wind's left, and the wind is sucked to the left. (And here Cleon just doesn't get it: ) And the radius and the wind speed co-depend so that a long lasting system has winds parallel to the pressure gradient. Paul Beardsell 02:41, 31 August 2005 (UTC)

Understanding the mathematics
I copy and past from above:
 * In met (as in anywhere else) the Coriolis term is defined by omega cross v. Thats all. William M. Connolley 08:07:25, 2005-08-31 (UTC).

No that is not all. The problem is that your understanding of the mathematics is insufficient.

In ballistics calculations the coriolis term is present in the formula to account for an observed trajectory that is deformed from what would be seen from a non-rotating point of view. A calculation for a ballistic trajectory as seen from a non-rotating point of view would not contain the coriolis term.

In the equation of motion that is used in meteorology the omega cross v term is present both in the equation of motion for a non-rotating coordinate system and in the equation of motion for a rotating coordinate system. (rotation around the Earth's axis) The omega cross v term that is used in meteorology is invariant under the coordinate transformation. --Cleon Teunissen | Talk 09:07, 31 August 2005 (UTC)


 * Of course it isn't. The omega cross v term *arises* in the course of the coordinate transformation. Omega cross v *is* the coriolis effect. I really don't know how you could possibly write the above. "omega" is the rate of rotation of the respective coordinate systems. The term cannot possibly be coordinate invariant. William M. Connolley 11:15:03, 2005-08-31 (UTC).


 * Let me get this straight William: do you believe that when an observer is watching the motion of inertial wind from an outside, non-rotating point of view, that he will will not see anything of the motion of the inertial wind? What, in your opinion, does the motion of inertial wind look like as seen from a non-rotating point of view? --Cleon Teunissen | Talk 11:51, 31 August 2005 (UTC) By non-rotating I mean a point of view that sees the earth complete a rotation in a sidereal day. --Cleon Teunissen | Talk 12:06, 31 August 2005 (UTC)


 * You are confusing the motion of the winds, with the terms that appear in the equations of motion. If you don't understand this (and it looks like you don't) then... you have a problem. Look back (poss in the archived stuff). I put in the derivation of where the coriolis term comes from. William M. Connolley 15:33:46, 2005-08-31 (UTC).


 * You are avoiding the issue: the derivation that you gave applies in the case of ballistics. That derivation is insufficient in the case of wind motions. The accelerational component of the inertial oscillations is invariant under the coordinate transformation. If you don't understand this invariance (and it looks like you don't) then... you have a problem. --Cleon Teunissen | Talk 16:08, 31 August 2005 (UTC)


 * You still don't understand. The derivation is for the forces that appear as a result of change-of-coordinates. It is pure kinematics (if you rephrase force as acceleration). There is no physics in it. Go away, think about it for a day or two. William M. Connolley 18:46:22, 2005-08-31 (UTC).


 * You still don't understand. You should study the animations by David McIntyre. Study the trajectories he obtains. That is kinematics, he applies it correctly. Go away, and think about these matters. --Cleon Teunissen | Talk 19:13, 31 August 2005 (UTC)

Understanding the mathematics (2)
The procedure that David McIntyre uses to perform a transformation to a rotating coordinate system is correct. His animations represent correctly the coriolis effect that needs to be taken into account in ballistics.
 * N.b. His animations, as he specifies, are on a "spherical earth", not an oblate spheroid (realistic) one. That threw me for a second, especially  on "image 3b, centrifugal". GangofOne 20:58, 31 August 2005 (UTC)
 * Yeah. The animations are illustrations to the article mentioned earlier on this Talk page, which discusses the theoretical case of inertial motion over the surface of a perfect sphere. And yeah, McIntyre mentions that on an oblate spheroid the path of the puck is different. Meteorological modeling takes the oblateness of the Earth into account. --Cleon Teunissen | Talk 02:22, 1 September 2005 (UTC)

In meteorology it is customary to decompose velocity parallel to the geopotential surfaces in the pair (v,u) v is the velocity component parallel to the longitude lines, u is the velocity parallel to the latitude lines.

Here is what McIntyre would do in transforming to a rotating coordinate system, if he would be using the pair u and v:  The velocity component u is used completely, and is transformed to another velocity parallel to the plane of the equator.  The velocity component v is not used completely: only the component parallel to the equator is used. This component is then transformed to another velocity parellel to the plane of the equator.

The transformation that McIntyre performs involves only the component of the motion that is parallel to the plane of the equator. It is rotating transformation around the Z-axis, so the transformation does not involve the component of motion parallel to the Z-axis. --Cleon Teunissen | Talk 13:15, 31 August 2005 (UTC)

Mathematical derivation of the equation of motion for inertial oscillations


The image to the right is genererated by the program code that William has made available here

the crucial lines of code are:
 * u=u+v*f*ts
 * v=v-u*f*ts

where the terms : v*f*ts and  u*f*ts  represent accelerations in a plane tangential to the local geopotential surfaces.

A mathematician will spot immediately that since the terms v*f*ts and  u*f*ts refer to a plane that is not parallel to the plane of the equator they are not related to rotating transformation around the Earth's axis.

The lines of code u=u+v*f*ts and v=v-u*f*ts are interconnecting motion parallel to the plane of the equator and motion parallel to the Earth's axis. On the other hand: rotating coordinate transform around the Earth's axis leaves all components parallel to the Z-axis unaltered.

This section in my version of the coriolis effect article provides a derivation of the underlying math of these lines of code. The derivations are still quite rough around the edges, but the gist is there.

William, you claim that what those lines of code are doing follows from rotating transformation around the Earth's axis. OK, then to prove your claim you should present the mathematical derivation: a rotating coordinate transformation around the Earth's axis (using an orthogonal coordinate system, preferably cartesian) that does what those lines of code are doing: changing the latitude of objects. --Cleon Teunissen | Talk 18:29, 31 August 2005 (UTC)

The distinction between real motion and apparent motion (2)
I copy and paste from above:
 * I define 'real motion' as motion with repect to the local coordinate system. William M. Connolley 21:32:24, 2005-08-30 (UTC).

In the context of newtonian dynamics 'real motion' is the motion with respect to the local inertial frame of reference. An inertial frame of reference is a frame in which the laws of motion hold good. Therefore we have that the motion with respect to the local inertial frame is the real motion (while recognizing the principle of relativity of inertial motion of course).

The equation of motion for a rotating coordinate system incorporates the rotation with respect to the local inertial frame of reference into the equation. That is an "umbilical cord" to the local inertial frame of reference that is never severed. Interestingly, that same "umbilical cord" is also present in the theory of general relativity. If, in GR, you choose a rotating coordinate system for representing the physics taking place then you have the same "umbilical cord". In GR, if you transform to a rotating frame of reference, you incorporate into the formulas the rotation with respect to the local inertial frame of reference.

Those are the reasons that in physics a distinction is recognized between 'real motion' and 'apparent motion'. --Cleon Teunissen | Talk 20:06, 31 August 2005 (UTC)

Revert of sept 2nd
To explain the physics of a motion, or a pattern of motions, you take the motion with respect to the inertial frame of reference. For example, in the case of a satellite, the motion with respect the inertial frame of reference is described by the universal law of gravity. In celestial mechanics the explanation of the motions of planetes and comets around the Sun is always given in terms of the motion with respect to the inertial frame of reference.

To explain the motion of an object as seen from a rotating point of view the observed motion must be decomposed into the motion as seen from a non-rotating point of view, and the other component is (additional) apparent motion, that is explained by the fact that the observer is rotating.

In the case of meteorology the component of the patterns of motion that is the same for both the non-rotating and the rotating observer is prominent.  The physics of the motions that are the same both as seen from a non-rotating and a rotating point of view should be explained.--Cleon Teunissen | Talk 05:34, 2 September 2005 (UTC)


 * Thats no explanation for a revert. I thought we had agreed, above, to think over our comments for a few days. It seems rather off to revert in the middle of that. William M. Connolley 08:30:42, 2005-09-02 (UTC).


 * No agreement of any sort had been reached. You stopped commenting on the talk page. I had stated your version of the artile contradicts itself, and I have specified here in what way your version contradicts itself. You have not commented on that.


 * Your version of the article contradicts itself, and you show no intention to address that issue. That is why I reverted to my version. --Cleon Teunissen | Talk 09:38, 2 September 2005 (UTC)


 * If that was the reason for your reversion, then you should have said so, rather than adding the text above. There is so much text on this page that its hard to reply to everything. I have aproblem with you, in that from my POV you seem to constantly be starting off new things rather than sticking to the point. I'll go back and comment above. William M. Connolley 11:28:12, 2005-09-02 (UTC).


 * I'm trying to elicit responses from you, to find out how your mind works. I wrote: inertial motion over the surface of a great sphere follows a great circle and initially you rejected that, later you corrected yourself. From my POV there are a couple of more warps like that in your thinking, warps that so far are invisible to you. So I am probing all the time, trying to find out how you are getting from A to E. I am aware that flooding the talk page has been counterproductive. --Cleon Teunissen | Talk 13:04, 2 September 2005 (UTC)

The standard practice of explaining observed motion
In astronomy there is the fact that as seen from the Earth the outer planets are from time to time observed to shift to what is called retrograde motion. (Retrogradation) This is explained by pointing out the fact that the Earth is itself rotating around the Sun. (Nasa page with animation) It would be absurd to state that a coriolis force causes Mars to shift to retrograde motion. --Cleon Teunissen | Talk 06:29, 2 September 2005 (UTC)

What has wrapping up got to do with it?
I copy and paste from above: 
 * You present the image to the right as the pattern of motion of inertial oscillations as seen from a co-rotating point of view. --Cleon Teunissen | Talk 14:15, 30 August 2005 (UTC)


 * Sort of. Its actually the 2-d viewpoint wrapped up into a spheroid. William M. Connolley 18:35:44, 2005-08-30 (UTC).

It is unclear what you mean by 'wrapping up'. Take a globe, draw circles on it, smal radius close to the poles, big radius close to the equator, take a picture of that globe with a photocamera, and it will look just as Anders Persson has drawn it. If you take two pictures from a slightly different angle then that can serve as a stereoscopic image. An animation studio can make a 3D animation featuring those inertial winds.

It is unclear how a 'wrapping up of a 2D viewpoint' is supposed to figure in this. That image presents a 3D viewpoint, without any intermediate 2D viewpoint in the process of making it. --Cleon Teunissen | Talk 12:35, 2 September 2005 (UTC)

Inertial circles again
My earlier simplification of the definition of an inertial circle (reverted by Cleon) needs to be re-instated.


 * Your "simplification", which was in error, was reverted by William M Connolley. It is often that an expression has a specific meaning in a particular branche of science. 'Inertia circle' is an example of that. It has a specific meaning, quite different from what you seem to have in mind. --Cleon Teunissen | Talk 13:13, 2 September 2005 (UTC)


 * Cleon, I'm sorry to have taken your name in vain. But on the issue of what an inertial circle is you are wrong.    For you an intertial circle requires gravity, rotating spheres, etc.  You want to describe inertial circles as a consequence of the Coriolis effect!  They are simply two terms for the same thing.  There is a lot of conflation of concepts happening in this article and on this talk page.  Paul Beardsell 13:25, 2 September 2005 (UTC)


 * Paul: see for my explanation of why I thought your change was wrong. Note that I only removed part of your change. William M. Connolley 13:49:49, 2005-09-02 (UTC).


 * OK, but here is my refutation. Paul Beardsell 14:43, 2 September 2005 (UTC)


 * I disagree. Objects non-stationary in the inertial frame nonetheless move in (on the earth, quasi-)circles in the rotating frame. William M. Connolley 16:01:40, 2005-09-02 (UTC).

Any object in the inertial frame of reference appears to move in a circle when observed from the rotating frame. That is the simplest inertial circle. To not at least acknowledge that, in the article, means we make the concept difficult to understand. Any freely moving object's motion will appear to be a precessing circle or a wavy line. The Coriolis effect is now fully described. Nothing more really needs to be said.

An object on the surface of the Earth is NOT, repeat not, freely moving.


 * Depends, if you are thinking of real physics, then yes. If you're thinking of a 2-d coords on a sphere, then no. William M. Connolley 16:01:40, 2005-09-02 (UTC).

An object affected by gravity (a satellite or a ballistic missile) is not freely moving. A object on the surface and stationary w.r.t. to (only!) the surface of the Earth and, in observing a point in the Inertial frame, sees a Coriolis effect. An object moving on the surface of the Earth has a more complex motion and sees SUPERIMPOSED Coriolis effects when observing an object in the inertial frame. An object stationary w.r.t. (only!) the surface of the Earth, when observing a non-freely moving object such as a satellite, a missile or an object moving on the surface of the Earth such as a wind or a puck, sees SUPERIMPOSED Coriolis effects.

Arguments presented here often make the mistake of assuming that something on the surface of the Earth or a gravity affected object is in the inertial frame. Such an object, in observing another such object, is going to see SUPERIMPOSED Coriolis effects. The very simple maths requires the vector addition of two rotating motions if we are going to continue doing this. Two Coriolis terms!

Paul Beardsell 12:49, 2 September 2005 (UTC)

Babylonian confusion
I copy and paste from above:
 * Cleon [...] You want to describe inertial circles as a consequence of the Coriolis effect!   Paul Beardsell 13:25, 2 September 2005 (UTC)

I most certainly do not want to describe the phenomenon of inertial oscillations as a consequence of the coriolis effect! (NB I am now exclusively referring to the meaning of 'inertial oscillation' as it is current in meteorology (and oceanography)).

In an earlier version I was still using the expression 'coriolis effect' in a meaning that I had learned in the past. As it turns out most people do not know the expression in that meaning, so I stopped using it in that way.

Since I have to give it some kind of name I am in my version of the article consistently naming 'rotational dynamics' as general causative agent. I now regard the expression 'coriolis effect' as useless for communication, for it is unpredictable what meaning the reader will have in mind.

Also, I now know that I cannot use the expressions 'inertial oscillation' an 'inertia circle' in communication with you, for the meaning of that expression that you seem to have in mind is different from the standard in meteorology. --Cleon Teunissen | Talk 13:46, 2 September 2005 (UTC)

Cleon you can use whatever terms however you like as long as you define your terms and in so doing you do not claim some kind of exclusivity over them. Of course one can read that "a pattern of air movement is an inertial circle" but that does not mean that an inertial circle is a pattern of air movement! A "dog is an animal" does not mean all animals are dogs. A particular meteorological phenomenum is an example of inertial circles. Dogs are examples of animals. Paul Beardsell 14:28, 2 September 2005 (UTC)

When you say "NB I am now exclusively referring to the meaning of 'inertial oscillation' as it is current in meteorology (and oceanography)" I need to point out that this is NOT an article "exclusively" on those subjects. So you must take the general case into account in your explanations. Paul Beardsell 14:28, 2 September 2005 (UTC)


 * Good grief Paul, in the article the specific-to-meteorology-meaning of 'inertial oscillation is introduced with proper explanation and images, and it is only used in the last section, and usually only the really interested readers read all the way to the last sections. This is the talk page, it would be hopelessly cumbersome to introduce and explain the expression 'inertial oscillation' again and again in comments on the Talk page. --Cleon Teunissen | Talk 15:04, 2 September 2005 (UTC)


 * Good grief yourself. The inertial circle section appears early in the article.  The non-meteorological meaning was one I was attempting to introduce - not one that present but one that should be!  The really interested reader is not necessarily interested in meteorology.  But that the Coriolis effect is only of interest to meteorologists is obviously I point I continue to miss.  That all animals are dogs is not an impression you have to continue to make (not that I say you think that all animals are dogs) but the language used in the article and here continues to have that appearance.  For me it is as if you were insisting on explaining addition in metereolgical terms, (mis)appropriating arithmetical terms, two clouds plus two clouds is four except sometimes they merge and then there are three clouds.  No, say I:  Addition is not only part of meteorology but part of arithmetic.  And the remarkable phenomenum of addition is best explained in arithmetical terms not metereological.  Substitute "Coriolis effect" for "addition"; "physics" for "arithmetic".  "Inertial circles" for "two plus two".  Paul Beardsell 15:36, 2 September 2005 (UTC)

Observers/Perceived
I removed the addition of "observers", which I think is unnecessary and carries unfortunate connotations of QM measurement, and also ties us to real planets too much. The effect exists in pure maths too with no physics. I also removed "perceived" for much the same reasons, and because there is nothing wrong with calling the velocity in the rotating system real, and the article shouldn't say otherwise. Velocity is meaasured with respect to a coorindate system. Thats all. William M. Connolley 16:04:26, 2005-09-02 (UTC).


 * Angular velocity is measured with respect to space. There is only one real angular velocity and that is the measured angular velocity with respect to space. --Cleon Teunissen | Talk 16:17, 2 September 2005 (UTC)


 * Good to see that there still exist believers in the good old "absolute space", but I'm not one of them. And anyway, thats physics, which the coriolis effect exists indepentently of, since its pure kinematics. Angular velocity is just the relative rate of rotation of two coordinate systems. William M. Connolley 20:55:56, 2005-09-02 (UTC).


 * Whoever wrote that should know better. Kinematics is physics in the same way arithmetic is mathematics.  Paul Beardsell 20:37, 2 September 2005 (UTC)


 * Twas me. I defend it. Coriolis can be made to arise purely from coordinate transforms, which has no physics in it at all. William M. Connolley 20:55:56, 2005-09-02 (UTC).

No I am not referring to Newtonian absolute space, if that is what you mean. I am talking about measuring rotation with respect to Einsteinian space. General relativity is more counterintuitive than you think. General relativity describes what has been known for centuries: you can measure rotation with respect to the local inertial frame of reference. Ask EMS, he will confirm that a ring laser interferometer setup measures rotation with respect to space, as do many other devices.

EMS defines 'absolute rotation' as follows: 
 * Absolute rotation occurs with respect to the local inertial frame of reference. [...] --EMS | Talk 17:23, 3 August 2005 (UTC)


 * There always is a local inertial frame of reference. And one always knows if one is rotating in relation to it.  "Local" doesn't mean "arbitrary". Paul Beardsell 20:40, 2 September 2005 (UTC)

The example of the ring laser interferometer is particularly striking since it doesn't require prior calibration. EMS wrote on his talk page:
 * That is a very good example. The use of radio signals and a beat frequency is different than what I was thinking of, and I do see that it works and why it works. (In a sense, the device is self-calibrating, but that is almost semantic issue. You really do make you point here, and effectively.)  --EMS | Talk 17:49, 3 August 2005 (UTC)


 * You can find those EMS quotes in his Archive no. 2 

Somebody has bamboozled you into believing that Angular velocity is just the relative rate of rotation of two coordinate systems. And now you believe that with religious ardour, and any facts contrary to that belief sytem are ignored by you. --Cleon Teunissen | Talk 19:41, 2 September 2005 (UTC)

Be careful about taking EMS's name and thoughts in vain. The *local* inertial frame he means isn't the one your are thinking of. William M. Connolley 20:55:56, 2005-09-02 (UTC).


 * The expression 'inertial frame of reference' is unambiguous in physics. You release a number of test masses, and you observe their motion. If you zoom in to the part of space very close around those test masses then you are observing the motion with respect to the local inertial frame. There is exactly one point of view from which you do not see the test masses spiral around each other in some way. That is the one non-rotating frame: the local inertial frame.


 * With 'local', I mean the following sequence of larger and larger spaces. For the astronauts onboard a space station, the frame co-moving with the station's center of mass is their local inertial frame. For all of the satellites orbiting Earth, the frame co-moving with the center of mass of the Earth is the local inertial frame. For all of the planets and comets etc. of the solar system, the frame co-moving with the center of mass of the Solar system is their local inertial frame. For the stars of our Galaxy the center of mass of the Galaxy is the "local" inertial frame.


 * All those inertial frames are constantly accelerating with respect to each other but none of those inertial frames rotates with respect to the other inertial frames. --Cleon Teunissen | Talk 09:06, 3 September 2005 (UTC)


 * I think you need to ask EMS about inertial frames. Or read the article. How about Einstein’s general theory removes the distinction between nominally "inertial" and "noninertial" effects, by replacing SR’s "flat", Euclidean geometry with a curved non-Euclidean metric.


 * What GR does is that it explains what was an observation in newtonian dynamics: the acceleration caused by gravitation is not registered by accelerometers. But GR does not say you can't measure rotation with respect to space. On the contrary, GR confirms that you can measure (with a strictly local measurment) rotation with respect to space.


 * The field equations of GR are written in a mathematical language that operates on a higher level of abstraction than geometry. In working with the field equations you are uncommitted to any choice of coordinate system. First the equations are solved, then you decide to what kind of coordinate system you are going to map the obtained solution. If you map the obtained solution to a rotating frame of reference, then you must incorporate the rotation with respect to the inertial frames of reference, just as in a transformation in newtonian dynamics.


 * But as long as you haven't mapped to any particular choice of coordinate system the laws of physics are the same for any situation. This is called general covariance. That is one of the brilliant achievements of GR.


 * Indeed, this is why calling the coriolis force is fictitious is meaningless. William M. Connolley 19:37:58, 2005-09-03 (UTC).


 * In GR, just as in Newtonian dynamics the concept of inertial frame is defined by what test masses do when you release them. The concept of inertial frame is unambiguous in physics. 
 * The real GR is a brilliant achievement, your interpretation of it is a silly caricature. --Cleon Teunissen | Talk 11:18, 3 September 2005 (UTC)


 * Your understanding of it is deficient. William M. Connolley 19:37:58, 2005-09-03 (UTC).


 * Well, I shall have to come back to this issue, later. But first things first. --Cleon Teunissen | Talk 21:05, 3 September 2005 (UTC)

Babylonian confusion about the word kinematics?
The wikipedia article on kinematics reads: 
 * In physics, kinematics is the branch of mechanics concerned with the motions of objects without being concerned with the forces that cause the motion. In this latter respect it differs from dynamics, which is concerned with the forces that affect motion.


 * So, despite Williams assertions to the contrary above, Kinematics is physics. As your quote says: "In physics...".  Paul Beardsell 03:02, 3 September 2005 (UTC)


 * Do you really believe you can source things that way? In that case, the current Coriolis article is manifestly correct, because it says so. Kinematics is physics-independent. William M. Connolley 10:01:45, 2005-09-03 (UTC).

I am aware of two different ways to interpret the above.  For example, I read someone who described that if in calculation exclusively the principle of conservation of momentum is used to figure out an outcome, then the calculation does not explicitly describe a force being exerted, so that kind of calculation would count as ''applying kinematics'.

On the other hand, WMC uses the expression 'kinematics' for mathematical operations that in themselves are not about physics taking place. Transforming to another coordinate system is not a description of physics taking place.

There are multiple meanings around for the word kinematics, and it is unpredictable what meaning which reader is familiar with. --Cleon Teunissen | Talk 21:30, 2 September 2005 (UTC)


 * But we all should be agreed that the Coriolis effect is just that, an effect. And the Coriolis force is fictional:  It does not exist.  It is used/invented by those who choose to ignore that they are rotating to explain motion they see which does not obey Newton I.  Amazingly I am not confident that we all do agree this.  Paul Beardsell 02:49, 3 September 2005 (UTC)


 * When we transform from one coordinate system to another no real extra velocity / acceleration / force is introduced. I want to think we all believe that too.  But then an inordinate (no pun intended) amount of time is being spent describing motion relative to an observer on the rotating surface of the Earth as if the observer is not rotating.  Paul Beardsell 02:56, 3 September 2005 (UTC)


 * Yes, all three of us assert that coordinate transformation only changes the representation of the physics taking place, that is as obvious as can be. One can convert from a cartesian (non-rotating) coordinate system to a system of (non-rotating) spherical coordinates; change of the representation of the physics taking place. This (java-app) animation shows a nice example of coordinate transformation.


 * the coriolis term and the centrifugal term in the equations of motion for a rotating coordinates system represent the inertia of the objects under observation.


 * $$m \mathbf a_{\mathrm{Rot}} = m \Big( 2\boldsymbol\omega\times\mathbf v_{\mathrm{Rot}} +\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)\Big) $$


 * Where m represents the inertia of the object. The vector addition of the vectorial expresssions represents the apparent acceleration of an object as seen from a rotating point of view. --Cleon Teunissen | Talk 06:33, 3 September 2005 (UTC)

Repeating the question about the North/south component


WMC has written about the motion patterns depicted in the image: 
 * The inertial oscillation (ie, the thing moving in a quasi-circle) *is* due to the coriolis force (and it alone, since the centrifugal force is absent/balanced). [...] William M. Connolley 13:58:38, 2005-09-02 (UTC).

I now repeat the question I asked earlier, because WMC hasn't answered the question: I take it you agree with me that the North/South component of the depicted oscillations is identical as seen from either the rotating or the non-rotating point of view. (Rotation around the Earth's axis)


 * You write so much. How can you expect anyone to read it all? Yes, it would seem to be true. But I still fail to see what this has to do with reverting the article, which makes no statements about these as seen from the inertial frame. William M. Connolley 19:46:15, 2005-09-03 (UTC).

So what, according to you, is the coriolis force doing to the North/South component of the depicted oscillations? --Cleon Teunissen | Talk 15:31, 2 September 2005 (UTC)


 * Can't say I've ever thought about it. Of the top of my head, the answer is: when its at the "top" (or "bottom") of the circles then the force, being omega cross v, is pushing it equatorward (or poleward). Why do you ask? William M. Connolley 19:46:15, 2005-09-03 (UTC).


 * The standard definition of the coriolis effect (as applied in ballistics) is as follows: The coriolis term (together with the centrifugal term) is used to handle...


 * No. The std definition is exactly that given in the "formula" section: omega cross v. Not a pile of words. We have more than enough of those already. William M. Connolley 21:17:14, 2005-09-03 (UTC).


 * the difference between what is seen from a rotating point of view, and what is seen from a non-rotating point of view. (I noticed that in your version of the article you do not explicitly mention that 'difference' bit). Anyway, the standard definition is that the coriolis/centrifugal terms cover exclusively the difference.


 * The North/South component of the depicted oscillations is identical as seen from either the rotating or the non-rotating point of view. (Rotation around the Earth's axis). So the north/south component of the depicted oscillation cannot be attributed to the "coriolis force": it is identical as seen from either point of view.


 * The vector of the coriolis force that is associated with the rotating coordinate transformation points parallel to the plane of the rotation. In the case of meteorology, the rotating coordinate transformation is around the Earth's axis: the coriolis force associated with that rotating coordinate transformation points in a direction parallel to the equator. It cannot point in a direction perpendicular to the equator.


 * Possibly the definition that you have in mind is different from the standard definition. Your personal definition being different from the standard one would explain a number of things. --Cleon Teunissen | Talk 20:46, 3 September 2005 (UTC)


 * Formula, not words. One day you'll learn that. William M. Connolley 21:17:14, 2005-09-03 (UTC).

William, it is now clear that you are unaware of the standard definition of the coriolis effect as it is applied in ballistics, and you are using some other definition instead.


 * Cleon, it becomes painfully obvious that you still ahven't realised that there is only one Coriolis effect. the confusion is because you can't cope with the maths. William M. Connolley 10:41:48, 2005-09-04 (UTC).

The derivation of the coriolis term and centrifugal terms is usually done in polar coordinates: $$ r $$ and $$ \omega $$, and it is always a derivation for motion in a plane. That derivation applies to motion in two-dimensional space!  In ballistics, the plane of rotation is the equator and and everywhere the direction of the fictitious coriolis force is parallel to the plane of the equator.


 * Twaddle. The derivation is usually done without an explicit coordinate system: I gave it here for reference. I suggest you read and understand it. William M. Connolley 10:41:48, 2005-09-04 (UTC).

For a cartesian coordinate system:  Let there be a cartesian coordinate sytem C, Z-axis coinciding with the Earth's axis, and a rotating coordinats sytem C', Z-axis coinciding with the Earth's axis. Let $$ f(t) $$, $$ g(t) $$ and $$ h(t) $$ be functions for a parametric equation of motion.

A parametric equation of motion.


 * $$ x = f (t) $$
 * $$ y = g (t) $$
 * $$ z = h (t) $$

That parametric equations of motion can be transformed to a coordinate system that is co-rotating with the Earth.

The parametric equations for the rotating coordinate system C' are as follows:


 * $$ x' = f' (t) = f (t) * cos (\omega*t) +  g (t) * sin(\omega*t) $$
 * $$ y' = g'(t) = g (t) * cos(\omega*t) -  f (t) * sin(\omega*t) $$
 * $$ z' = h (t) $$

That is in mathematical language the structure of the coordinate transformation as it is applied in ballistics.<BR>

No matter what coordinate system is used (cartesian, cilindrical or spherical), the coordinate transformation (rotation around the Earth's axis), involves exclusively the component of the motion parallel to the equator. --Cleon Teunissen | Talk 22:14, 3 September 2005 (UTC)


 * You say: "everywhere the direction of the fictitious coriolis force is parallel to the plane of the equator." This is true. Omega is a vector pointing to Polaris, the cross product of omega and any vector is perpendicular to omega, so that cross product is parallel to the plane of the equator. But it's not clear what you're otherwise getting at. The coordinate-free way of thinking about it suggested by WMC is cleaner. GangofOne 21:11, 4 September 2005 (UTC)


 * The fictictious coriolis force is always parallel to the plane of the equator. On the northern hemisphere: wind that is flowing from east-to-west will proceed to bend towards the north (unless a pressure gradient prevents that). The bending towards the north is the same both as seen from a non-rotating and from rotating point of view. So in two ways it is inconsistent to attribute the bending to the north to the fictitious coriolis force: (1) the fictitious coriolis force only acts parallel to the plane of the equator. (2) the north/south component of the motion is identical as seen from a non-rotating and rotating point of view.


 * No matter what coordinate system you are using: the component of the motion that is paralell to the Earth's axis is identical as seen from either a non-rotating or a rotating point of view. --Cleon Teunissen | Talk 15:43, 5 September 2005 (UTC)

Paul Beardsell wrote about the standard definition:<BR>
 * There is a "correction" required to the force equations to describe the difference between the forces apparent in the rotating frame of reference to the forces apparent in the non-rotating frame of reference. And this correction is called the Coriolis effect. [...] Paul Beardsell 16:43, 23 August 2005 (UTC)

The fictitious coriolis force acts parallel to the plane of the equator
I copy and paste from above:
 * The derivation is usually done without an explicit coordinate system: I gave it here for reference. [...] William M. Connolley 10:41:48, 2005-09-04 (UTC).

Yes, that is de derivation for the coriolis effect as it is taken into account in ballistics. That formula is sufficient in the case of inertial motion in the inertial frame. In the case of ballistics the fictitious coriolis force that is in the formula acts parallel to the plane of the equator, as is illustrated by David McIntyre writing in his article about Inertial motion over a perfect sphere:
 * We work primarily with the latitude–longitude coordinate description of the motion, which makes the transformation from one frame to the other simple—only the longitudinal difference $$ \omega t $$ due to the rotation is required.

The latitudinal position as a function of time is independent of the point of view of the observer.

It is wrong to suggest that all of the motions of winds (that is:those wind motions that cannot be attributed to pressure gradient alone as cause) can be accounted for in terms of the fictitious coriolis force. --Cleon Teunissen | Talk 16:03, 5 September 2005 (UTC)

Rv
CT reverted to his version; I've reverted back. Cleon, despite endless words, you still haven't told me (or indeed I suspect anyone else) what you think is wrong with my version (which is also the version that others have OK'd, and so has Anders Persson (After a first, quick glance does not invite any immediate protests I have from him). I don't know if it is


 * vital things are missing
 * something is wrong with it

If its (1), then name the first thing that you consider to be missing. If its (2), please quote the first sentence that you regard as objectively false, or state clearly that nothing is objectively false but your objections are more subtle that that. William M. Connolley 16:11:14, 2005-09-05 (UTC).

The opinion of Anders Persson
Anders Persson will confirm that the oblateness of the Earth and the role of gravitation ought to be mentioned.


 * So you say. But in fact, he didn't. If you can persuade him to comment further, that would be splendid, but please don't put words into his mouth. William M. Connolley 19:16:42, 2005-09-05 (UTC).


 * I think Anders Persson's articles are sufficient. --Cleon Teunissen | Talk 20:11, 5 September 2005 (UTC)

Anders Persson will certainly disagree with a statement that the Earth's gravitation does not play a part in the motion patterns of winds.


 * That statement is not made, so why do you bring it up? William M. Connolley 19:16:42, 2005-09-05 (UTC).


 * Then it is confirmed that the Earth gravitation plays a part in the motion patterns of winds. --Cleon Teunissen | Talk 20:11, 5 September 2005 (UTC)

In Persson's words:
 * For an eastward movement the centrifugal force is increased and the body is deflected toward the equator, to the right of the movement. For a westward movement, the centrifugal force is weakened and can no longer balance the gravitational force, which is the physical force that moves the body in the poleward direction, to the right of the movement.

My choice of words is different from that of Anders Persson, but the essential element is gravitation. It is the Earth's gravitation (a component of it) that pulls east-to-west moving air to the nearest pole.<BR>

When Anders Persson is using the expression 'coriolis force', he is not referring to the fictitious coriolis force associated with coordinate transformation. From his writing it is clear that Anders Persson uses the expression 'coriolis force' in the umbrella meaning: that which deflects the winds (other than pressure gradient). --Cleon Teunissen | Talk 17:33, 5 September 2005 (UTC)


 * Once again, you are putting words into his mouth. Don't do this. William M. Connolley 19:16:42, 2005-09-05 (UTC).


 * The quote I gave is from the Bulletin of the American Meteorological society article. (Page 1375, second column) The quote is unambiguous. Air mass that is moving from east-to-west with respect to the Earth will be pulled to the nearest pole by a component of the Earth's gravitation. <BR> --Cleon Teunissen | Talk 20:11, 5 September 2005 (UTC)

What is wrong with WMC's version of the article.
Here are three important omissions


 * It fails to mention that the fictitious coriolis force only acts parallel to the plane of rotation.
 * This is incorrect. It says quite clearly The formula implies that the Coriolis force is perpendicular both to the direction of the velocity of the moving mass and to the rotation axis. Perpendicular to the rotation axis is "in the plane of rotation" in your terms. William M. Connolley 19:13:05, 2005-09-05 (UTC).


 * I stand corrected. You do say that the vector of the coriolis term in the equation of motion for a rotating system is always in the plane of rotation.


 * It does not mention the oblateness of the Earth as a key factor in meteorology.
 * The article is about Coriolis effect. Its not really clear that a lot of met stuff belongs. This issue is touched on in the "Visualisation of the Coriolis effect" section (but analogously, for the rotating turntable). The matter was mentioned quite well in an earlier version but I removed it on revising the page, as being too far away from a pure Coriolis effect page. There should be a flow-on-planets page, though. William M. Connolley 19:13:05, 2005-09-05 (UTC).


 * It fails to mention explicitly that the Earth's gravitation is a key aspect of the physics of wind motions. Air mass that is flowing from east-to-west with respect to the Earth is pulled towards the nearest pole by a component of the Earth's gravitation. At every latitude, this component acts perpendicular to the Earth's axis.
 * This is essentially the same point as (2). Put the way you have, I think its wrong. Essentially, oblateness and centrifugal force cancel out, so one is left only with Coriolis force (which is what this article is about). Thats explained on.

--Cleon Teunissen | Talk 17:14, 5 September 2005 (UTC)

In short: for point 1, I think you're wrong. For points 2 and 3, I think its (a) not really appropriate for this article and (b) two effects cancelling.

Can we start off with point 1, which is I hope most easily sorted out. Do you hold to your objection? William M. Connolley 19:13:05, 2005-09-05 (UTC).


 * OK. Point 1, it turns out, is not in dispute: the fictitious coriolis force acts parallel to the plane of rotation.


 * That hammers in point 2/3. The fictitious coriolis force does nothing to explain the deflection pattern of wind, its the Earth's gravitation, and the fact that the system as a whole is rotating.


 * Next step: your theory of mutual cancelation of the perpendicular to the Earth's axis component of gravitation and centrifugal force. That is covered by my discussion of motion over a velodrome track. There is equilibrium for one angular velocity only. Only for objects that co-rotate with the Earth is the centripetal force just right. For any other angular velocity around the Earth's axis (air mass flowing east-to-west or west-to-east with respect to the Earth) there is no equilibrium, and therefore only partial "cancelation". --Cleon Teunissen | Talk 20:49, 5 September 2005 (UTC)

Decomposing the fictitious coriolis force
I have not followed all these lengthy discussions, but I wonder why you find it so important that the Coriolis force is perpendicular to the axis. That is a clear fact, that has been in all versions of the article. However you should decompose it into (locally) vertical and horizontal components.

&minus;Woodstone 21:11:13, 2005-09-05 (UTC)
 * The vertical component (perpendicular to the surface or better to a geopotential plane) is counteracted by gravity (and centripetal force) in an equilibrium. The air cannot massively rise, because that would create a vacuum near earth, or fall, creating overpressure (unlike a bullet).
 * That leaves the horizontal component to deflect the air flow over the surface. On the northern hemisphere this component is always to the right of the air flow velocity vector.
 * For inertial flow, it just keeps curving to the right, completing a circle clockwise
 * For flow on a pressure gradient, it cannot deflect further than 90&deg; from the gradient, so starting to deviate to the right, it reaches a balance where it finally needs to circle around anticlockwise, to stay at less than 90&deg; from the gradient.


 * Hi Woodstone. About the fictitious coriolis force: there is no such thing as deflecting it; it's not a force. --Cleon Teunissen | Talk 21:30, 5 September 2005 (UTC)
 * And the only directions the fictitious coriolis force can be decomposed in are other directions that are also parallel to the plane of the equator. --Cleon Teunissen | Talk 21:39, 5 September 2005 (UTC)

It acts like a force in the rotating coordinate system. It deflects the air flow relative to the rotating coordinate system. I'm not saying the force is deflected, but that the air flow is deflected. In a non-rotating frame of reference there is of course only inertia.

And yes, of course you can decompose outside of the parallel plane! Just project the Coriolis vector onto the local vertical and the locally horizontal plane. When you decompose these new component vectors back onto the parallel plane and axis directions, you will find that the axial directions of the two vectors are equal and of opposite sign. Nothing mysterious. &minus;Woodstone 21:56:38, 2005-09-05 (UTC)


 * It seems to me that you just confirmed what I wrote. You can go through elaborate decomposing, and any component vectors perpendicular to the plane of rotation come in a pair that cancels each other: the vector of the coriolis term never gets away from the rotational plane.


 * If you have a satellite that is in perfectly circular circumpolar orbit, with a period of days, then as seen from a non-rotating point of view the orbit is in a plane (as satellite orbits are always planar). As seen from a point stationary with respect to the Earth that orbit looks to follow a rather corkscrew-like trajectory. That satellite is nowhere climbing higher or descending. It is pointless to decompose the fictitious coriolis force, any pair of components perpendicular to the plane of rotation will mutually cancel anyway. And the fictitious coriolis force is not acting as a force then, the only force on the satellite is the Earth's gravitation, directed towards the center of gravitation of the Earth. --Cleon Teunissen | Talk 06:40, 6 September 2005 (UTC)

There is certainly a point to decomposing in convenient components. The vertical component can be discarded from consideration, because it will be balanced by the boyancy of the air. This leaves exposed very clearly the only component (in the horizontal plane) relevant for explaining the cyclonic or inertial flows. &minus;Woodstone 19:48:11, 2005-09-06 (UTC)


 * There is no such thing the fictitious coriolis force being "balanced" by a real force; the fictitious coriolis force isn't a force, it's a calculational tool, it "exists" only on paper. An example of that is the retrograde motion of Mars. The real orbit of Mars is an ellipse around the Sun with the Sun at one focus. The occasional regrograde motion of Mars is an apparent retrograde motion, that is not caused by a force; the Earth is moving. (NASA page with animation Another nice animation Likewise the satellite orbit. The satellite is following a planar orbit in space, and the Earth is rotating underneath it.


 * In transforming to a rotating coordinate system the picture as a whole is rotated. It is like watching the images taken by a videocamera that is itself rotating. The picture as a whole is rotated: what is moving parallel in the non-rotating coordinate system is still moving parellel in the rotating system. You seem to be describing a force that can rearrange wind pattern. The fictitious coriolis force "affects" all objects in a situation simultaneously, and rotates the whole picture, preserving all relative positions, all relative velocities, and all relative accelerations. -Cleon Teunissen | Talk 21:45, 6 September 2005 (UTC)
 * In your last sentence, if you replace "The fictitious coriolis force" with "The coordinate transformation", then I won't be confused by what you say.GangofOne 05:56, 7 September 2005 (UTC)
 * Yes, that is a good replacement: the coordinate transformation rotates the whole picture. --Cleon Teunissen | Talk 07:45, 7 September 2005 (UTC)

The concept of fictitious force has no meaning and no use. You're welcome to your own view on this, of course. William M. Connolley 22:07:29, 2005-09-06 (UTC).


 * Here is my view: the coriolis term and the centrifugal term in the equation of motion for a rotating coordinate system are useful calculational tools, and nothing beyond that. For the purpose of explaining the physics taking place the concept of fictitious force is of no use. To explain the physics taking place you take the motion with respect to the inertial frame of reference. Then inertia is simple plain inertia. --Cleon Teunissen | Talk 07:45, 7 September 2005 (UTC)


 * In an inertial coordinate system, F=ma. A rotating c. s. is NOT an inertial c. s., and F NOT= ma. However, if you add 2 well-defined fictitious "forces" to F then (F + centrifugal + coriolis) DOES = ma. And then everything you learned in school about the kinematic equation F=ma is applicable. GangofOne 05:52, 7 September 2005 (UTC)

We need a simplified explanation first
Imagine you are a high school student and you look up this article. Its unreadability would put you off physics for life. You all seem to be trying to get to an intellectally perfect description and have lost sight of your potential users.JMcC 19:30, 26 October 2005 (UTC)


 * The first paragraph 'simple examples' gives in my opinion a simple enough explanation, that does not contain complex mathematical formulas, and looks well readable. Things that can be considered are: (1) the possibility of adding pictures to the two examples; (2) the possibility of having the caroussel as the first example instead of as the second. Bob.v.R 01:15, 27 October 2005 (UTC)


 * I am glad you liked it. After writing the paragraph above calling for a simple explanation, I wrote it.JMcC 08:43, 27 October 2005 (UTC)


 * I see! And what do you think of my two suggestions? Bob.v.R 19:07, 28 October 2005 (UTC)

Unfortunately the new para implies that the coriolis effect is zero at the poles. It isn't. William M. Connolley 12:17, 27 October 2005 (UTC).


 * Quite correct. Only a moving object experiences the Coriolis effect. An object at a pole isn't moving.  An object moving towards a pole would of course experience the effect.JMcC 16:39, 27 October 2005 (UTC)


 * No, an object (instantaneously) at the pole can indeed be moving. But the "simple examples" text unfortunately doesn't tell you that the effect is proportional to the speed of the object, and it seems to me to imply no effect at the pole. The "consider a missile..." bit I find unhelpful and handwavy (and, I think, if you work it out it gets the wrong numbers). William M. Connolley 18:26, 27 October 2005 (UTC).


 * I agree that an object passing over the pole would experience the Coriolis effect, but I don't think the article implies that it wouldn't. It should be possible to explain clearly the Coriolis effect to a layman, which the article previously failed to do.  I had a stab, and may not have fully succeeded but the way that Wikipedia works is that somebody should now be able to make it even clearer.  If you want to make the point about proportionality to speed, you merely need to say that if the missile travelled faster from the North Pole, the point at the equator would have moved less by the time the missile was due to arrive. I suppose it is handwavy, but it is closer to an understandable explanation, which is what we are all trying to achieve.  Having had a go, I will now bow out from the article.  Good luck JMcC 23:24, 27 October 2005 (UTC)


 * I will have a go, but not tonight. William M. Connolley 19:26, 28 October 2005 (UTC).

I've now done a first stab at my promised revamp. Firstly, the new section was called "simple examples" but was clearly an attempt at a simple *explanation* so I've renamed it. Second, I needed to refer to the formula, so I moved that up. Formula first, *explaining* it next. Third, there are (as I've now written) 2 explanations: the one everyone actually using the CE uses; and the one people who don't use it seem to desire. I've separated the two cleanly. William M. Connolley 17:16, 11 November 2005 (UTC).

Coriolis frequency
I redirected coriolis acceleration and coriolis parameter here, as they are talked about in the article. I made a redirect for coriolis frequency as well, but I'm not sure the article is talking about the same thing as this article. Is the equation just in a different form? If it's not the same thing, it would be great if someone could mention the actual coriolis frequency in this article or its own article, if necessary. Thanks, Kjkolb 09:15, 25 November 2005 (UTC)


 * Interesting... the scienceworld page bit is essentially done in the inertial circles section. v^2/r=2.omega.v == v = 2.omega.r, which is the same as the sections r = v/f, if you understand f=2.omega(.sin latitude). The formula is more interesting on the sfc of a planet (or other spheroid) cos you get f rather than 2.omega. Puzzle: why little-omega = 2 big-omega? William M. Connolley 10:35, 25 November 2005 (UTC).

Coriolis Acceleration of a West Bound Body
I moved this question from the beginning, for sanity conservation purposes. (GangofOne)

Using the Physical explanation of Coriolis Acceleration, there would be no acceleration on an East or West Bound Body. However, if you apply omega cross v, you would get an acceleration. Can anyone explain this? (unsigned by 12:09, 10 December 2005 129.92.250.41)

Part of an answer: omega cross v is true, if the explanation leads you to believe otherwise, it is a bad explanation. I believe it was intented to simplified and understandable, but maybe it's not adequate. The "Physical explanation" as writen only discusses north-south motion, but it is misleading, there is a Coriolis force whenever omega cross v is not zero. The way to begin to visualize the answer to think about a nonrotation planet first, and realize that a projectile launched due west does not follow the latitude line, but follows a great circle. Then realize that a rotating planet will cause a projectile to move more northernly from that great circle (in the N. hemisphere, launched due west). Think about it, then ask more, or someone else can chime in. GangofOne 05:09, 11 December 2005 (UTC)


 * Thanks for the response, I have been thinking about it as you suggested. Let's decouple the coriolis effect from the great circle issue by considering a body moving east or west along the equator.  How does the rotation of the earth cause an acceleration?


 * There is no Coriolis effect on the equator (err, at least no effect in the plane tangent to the sfc). William M. Connolley 21:44, 21 December 2005 (UTC).


 * Under those circumstances a geophysical surveyer must take the Eötvös-effect into account. Let a ship cruise at a constant moderate speed along the equator. The ship has extremely sensitive instruments onboard to measure the local gravitational acceleration. Readings obtained when sailing to the east are different from readings when sailing to the west. The geophysicist must know how to correct the readings, to get at what would have been measured if the ship would have been co-rotating with the Earth during the measurement run. Cleon Teunissen &#124; &#91;&#91;User talk:Cleon_Teunissen&#124;Talk]] 21:32, 13 December 2005 (UTC)

Hi GangofOne,<BR> It's been a while since I visited Wikipedia. The question of the predicted trajectory of a projectile launched due west is an interesting one.<BR> We have a projectile that is launched due west, horizontal to the ground, from 45 degrees latitude. The shape of the trajectory will be a keplerian orbit. It will be a planar orbit, with the center of mass of the Earth at one focus of the ellipse. As you describe, the line of intersection of that keplerian orbit and the surface of the Earth is a great circle.

So when a projectile is launched due west it will on hitting the ground be closer to the equator than the latitude it was launched from. Geometrically: a great circle that is tangent to the 45 degrees latitude line is situated in the band between 45 degrees latitude North and 45 degrees latitude South.

Of course, since armies want to fire shells accurately over tens of kilometers, those ballistics issues have been worked out long ago. In ballistics it is well known that no matter what the launch velocity is, a projectile launched due west will never proceed towards the nearest pole. The projectile may possibly land on the same latitude as it was lauched from, but generally it will land closer to the equator than it was launched from.

This is related to the caveat that William M Connolley added to the "simple explanation"<BR> "Some care is needed in formalising this argument; a direct formalisation leads to a Coriolis effect of only half the correct magnitude. This is not true at all."

In the case of motion due west the "simple explanation" predicts the opposite of what it is supposed to explain. Cleon Teunissen &#124; &#91;&#91;User talk:Cleon_Teunissen&#124;Talk]] 18:22, 13 December 2005 (UTC)


 * Note: I didn't write This is not true at all and have just removed it... William M. Connolley 21:44, 21 December 2005 (UTC).


 * Thank you for pointing that out. Anyway, it is your caveat that matters. It is clear that the simplified example, using a cannonball, is wrong for meteorology, for the reason that you state in the caveat: if you derive the mathematics for the example of the cannonball, you end up with a term that is half of the coriolis term. The proper mathematical derivation of the coriolis term for motion of west or eastbound buoyant mass, is indicated in the section below, about making a U-turn, in which the buoyancy of zeppelins moving relative to the earth is discussed.
 * By the way, my signature is changed, I have been granted a change of user name, from 'Cleon Teunissen' to Cleonis. --Cleonis | Talk 00:52, 31 December 2005 (UTC)

The physical consequences of making a U-turn
I copy and paste from above:
 * 129.92.250.41 wrote:<BR>[...] Let's decouple the coriolis effect from the great circle issue by considering a body moving east or west along the equator. How does the rotation of the earth cause an acceleration?

Let's look into that matter a little closer, and do some calculations. <BR> There are plans for huge cargo-lifting zeppelins, the weight of the zeppelin plus the cargo will be around 500 tons. The cruising velocity of these cargolifters will be in the order of 25 meters per second.

What does it take for the cargolifter to be neutrally buoyant when it is stationary with respect to the Earth? The fact that the Earth rotates must be taken into account. At the equator the velocity of Earth's surface is 460 meters per second. The amount of centripetal force required to cause a mass to move along a circular path with a radius of 6370 kilometer (the Earth's radius), at 460 m/s, is about 0,0336 newton per kilogram of mass. For the 500 ton cargolifter that becomes about 16800 newton. The amount of buoyancy force required is the weight of the cargolifter minus those 16800 newton.

When cruising at 25 m/s due East the total velocity becomes 460 + 25 = 485 m/s, which requires a centripetal force of 18460 Newton. Cruising at 25 m/s due West: 14850 Newton. So if the cargolifter is neutrally buoyant while cruising due East, it will not be neutrally buoyant anymore after a U-turn; the airship will have to be re-trimmed.

Derivation of the formula
Derivation of the formula for motion along the equator.<BR> Notation:<BR> $$ \Delta \omega $$ is the angular velocity of the cargolifter relative to the angular velocity of the Earth. <BR> $$ \omega_e $$ is the angular velocity of the Earth: one revolution per 24 hours. <BR> (Hence the total angular velocity of the cargolifter is $$ \omega_e + \Delta \omega $$) <BR> $$ F_e $$ is the centripetal force for co-rotating with the Earth. <BR> $$ F_v $$ is the total centripetal force for an object moving along the surface of the Earth.<BR> $$ (\Delta \omega) * r = v_c $$ is the cargolifter's velocity (velocity relative to the Earth)

How much does the buoyancy of the cargolifter need to be retrimmed when the cargolifter's velocity changes?


 * $$ F = F_v - F_e = - m (\omega_e + \Delta \omega)^2 r - (- m (\omega_e)^2 r) $$
 * $$ F = - m r ((\omega_e + \Delta \omega)^2 - (\omega_e)^2) $$
 * $$ F = - m r (\omega_e^2 + 2 \omega_e \Delta \omega + (\Delta \omega)^2 - \omega_e^2) $$
 * $$ F = - 2 m \omega_e (\Delta \omega) r - m (\Delta \omega)^2 r $$
 * $$ F = - 2 m \omega_e v_c - m (\Delta \omega)^2 r $$

When $$ \Delta \omega $$ is very small compared to $$ \omega_e $$ then the following shorter version is a good approximation:


 * $$ F = - 2 m (\omega_e * v_c) $$

This derivation explains why the formula (2 * omega cross v) works for east-to-west and west-to-east motion. This particular derivation was for motion along the equator. A more general derivation explains why a cargolifter cruising due west at say 45 degrees latitude will tend to veer towards the nearest pole. The buoyancy is a crucial element. --Cleon Teunissen | Talk 11:14, 15 December 2005 (UTC)

What is K-12?
Someone added in a reference: "Targeted towards K-12 teachers and students". That writer does not realise that "K-12" has no meaning outside his country. Can someone convert this to an age group? &minusWoodstone 08:52, 16 December 2005 (UTC)
 * 5 years (Kindergarten)-18 years (12th grade) ie anything before college GangofOne 08:57, 16 December 2005 (UTC)

Coriolis acceleration of a west bound object (2)
I copy and past from above:
 * [...] a projectile launched due west does not follow the latitude line, but follows a great circle. Then realize that a rotating planet will cause a projectile to move more northernly from that great circle (in the N. hemisphere, launched due west). Think about it, then ask more, or someone else can chime in. GangofOne 05:09, 11 December 2005 (UTC)

The statement above about causing a west bound projectile to proceed northernly is incorrect. To see this, try out Gavin Chung's orbit simulator. This Java applet orbit simulator shows the surface trace of a satellite, orbiting earth along a circular orbit. The trace of any orbit over the surface of the Earth is a great circle.<BR> Satellites in geosynchronous orbit are usually orbiting in the earth's equatorial plane. If the geosynchronous orbit is tilted with respect to the Earth's equatorial plane then the orbit's trace over the surface of the earth traverses from hemisphere to hemisphere. On each hemisphere, there is one point where it is tangent to a latitude line. Of course, to be moving tangent to a latitude line is to be moving either due east or due west.

Try the following settings for Gavin Chung's orbit simulator. Altitude: maximum (39000 kilometer) Inclination: 10 degrees. When a a satellite is at 39000 kilometer altitude, it's orbital period is more than 24 hours: hence the satellite's trace travels westward with respect to the Earth's surface. The orbit simulator shows that when there is orbital motion due west, then the object will proceed to move closer the Equator.

The same is valid for any projectile that is launched due west: there is no way a projectile will veer to the nearest pole. Generally a projectile that is launched due west will land closer to the Equator than the latitude it was launched from. --Cleonis | Talk 15:02, 31 December 2005 (UTC)

New pic
I swapped the pic of the hurricane for a (featured) pic of an icelandic low. Because... I'm not sure how important CF is in hurricane dynamics. But it certainly is in low dynamics. William M. Connolley 22:34, 24 January 2006 (UTC).

Someone added:


 * Actually, there was an article publiched in a British journal (possibly the "Journal of Fluid Mechanics" many yearts ago, that demonstrated the bathtub effect. The authors put a few drops of ink into the water in the tub, and waited till all the swirling ceased, i.e. until the viscosity of the water totally dampened the vorticity induced by the inflow. Then, if they opened the bottom drain v-e-r-y slowly, so as not to induce any vorticity, the flow of ink in the outflow clearly demonstrated a counter-clockwise swirl in England. However, they had to wait at least one week for the initial vorticity to dissipate, and it is this that vitiates all simple experiments---the initial conditions of the water are not mathematically curl (i.e vorticity)-free.

this is fair enough (though it reads more like a talk comment) - but its already mentioned. It would be nice to have the exact ref, though. William M. Connolley 21:34, 5 February 2006 (UTC).

New animations uploaded
I have uploaded three animations. <BR> The animation is inspired by Robert Bryll's Java applet of a rotating fountain







My intention is to add these animations to the coriolis effect article, along with other explanatory animations. --Cleonis | Talk 16:59, 19 February 2006 (UTC)


 * I have a problem with this. Namely, since the particles move in  a straight line, it's pretty clear that this must be on a flat turntable.  That means that any apparent deflections, viewed from the rotating camera, observed are the result of both Coriolis and Centrigugal terms. That's not terribly helpful in explaining Coriolis. -- GWO

Sure, a parabolic dish is more relevant; those animations were the first I manufactured. They are very schematic renderings of this MIT fluid dynamics lab demonstration : Trajectory of an object moving frictionless over a parabolic dish and Trajectory of an object moving frictionless over a parabolic dish as seen from co-rotating point of view I intend to make new versions of those animations, in which a rotating dish with a fluid layer on it is represented.

The rotating fountain animations are designed to illustrate why the coriolis force is referred to as a fictitious force. The coriolis effect article must be very clear on the following point: the fictitious forces (the centrifugal term and coriolis term) are not physical causes. Fictitious forces play no part in the physics taking place. --Cleonis | Talk 14:15, 20 February 2006 (UTC)


 * I think the best way to treat this would be as 2 pairs of animations... the two animations in each pair are shown side-by-side looking at the same trajectory on a parabolic dish in the two frames. The shaded quarters on the disc would be great. Something like this...


 * And... please leave out the "fictitious" stuff. Its not necessary or helpful. William M. Connolley 14:58, 20 February 2006 (UTC).

Luckily, an animation of a object in perfect circular motion is comparatively simple. I don't know how much time making those animations is going to take. I'll be back when they are finished. As you mention, GWO, the natural way of seeing the inertial oscillation is as a perturbation away from circular motion. The circular motion, perturbed or unperturbed, is sustained by a centripetal force. See also the article about Rotational-vibrational coupling --Cleonis | Talk 15:18, 20 February 2006 (UTC)

It seems to me that this : shows one exactly what an inertial circle is... Along with the circular one for comparison...  Most people, once told the table is curved, can readily see how the left hand case works (especially if they've played "Beat The Black Ball"?), and the difference between the two animations is clearly just the rotating point of view. Since we're neglecting friction, we can even treat the bowl itself as stationary, as long as the camera is spinning at the correct rate.

More confusion and incorrectness
''Suppose the cannon is at, say, 45 degrees north, and is fired facing due north. The earth is rotating towards the east; at the instant of firing, the cannon ball shares in this rotation. Further north, the earth is rotating more slowly (the angular velocity is the same, but the rotation rate of the surface is the angular velocity times the distance from the polar axis, and hence decreases from equator to pole).''

That's true, but that's *not* Coriolis. Coriolis doesn't care if the Earth is curved, or whether you're travelling N-S or E-W. Any description of Coriolis force that relies on either of these two facts is quite simply incorrect...

Suppose that was how Coriolis worked .... how can you possibly explain the southwards deflection of a mid-latitude shell fired directly eastwards?


 * The confusion arises, I think, because, the centrifugal term and the coriolis term are involved in calculating cannonball trajectories, for example in the animations by David McIntyre of motion along great circles over the surface of the Earth. In order to calculate those trajectories David McIntyre performs a transformation to a rotating coordinate system.


 * No. The argument explicitly appeals to the curvature of the Earth as the mechanism for apparent deflection.  That means that it isn't Coriolis, and it's not what most would understand as centrifugal either.  It's the variation of f with latitude.  That's close to the beta effect, but as the major contribution comes from the &omega; x r centrifugal term, it isn't really that, either.  In any case, these ballistic discussions are not relevant to discussion of the Coriolis effect, and do not belong in this article.  -- GWO


 * I support your statement about ballistics discussions. Ballistics isn't relevant for the coriolis effect article. --Cleonis | Talk 15:52, 21 February 2006 (UTC)


 * Those cannonball trajectories are unrelated to wind trajectories; David McIntyre isn't trying to explain wind trajectories with those animations. --Cleonis | Talk 21:03, 20 February 2006 (UTC)


 * OK. I've cut the two ballistics sections.  They were both somewhat misleading, if not actively wrong.

The rotating parabolic dish and modeling balances
GWO wrote in the article:
 * In meteorology, Coriolis effects tend to dominate centrifugal effects, because the latter is usually balanced by an ambient pressure gradient (exactly analagously to the slope on a parabolic turntable).

I think it is worthwile to indicate why a fluid layer on a parabolic dish is an informative model of the atmospheric layer on the Earth.

To manufacture a parabolic dish, a resin is poured on a rimmed rotating turntable. Due to the rotation, the resin automatically assumes the parabolic shape, (solidifying in that shape). If you spin the turntable at exactly that rate, a fluid that is poured on the parabolic dish will spread out into a fluid layer with equal thickness everywhere.

In the case of the Earth: Earth started as an accretion disk. As the spinning proto-Earth formed, it automatically assumed the ellipsoidal shape, and solidified in that shape. Because of that, the Earth's atmosperic layer is equally thick everywhere, even though the Earth's poles are 21 kilometers closer to the Earth's center than the equator. At the poles, g = 9.83 m/s^2, at the equator, g = 9.78 m/s^2. --Cleonis | Talk 01:02, 22 February 2006 (UTC)


 * I disagree. Firstly, we're discussing frictionless pucks on a solid parabolic dish, not a fluid layer.  An introduction of a layer, whether of uniform thickness or not would only serve to introduce extra (pressure) terms to the equation of motion.  A sliding puck experiences only Coriolis, and therefore nothing else is necessary to describe it's effect.
 * The reason this article so badly needly cleaning was because it tried to encapsulate all facets of all kinds of dynamics on a rotating sphere. Clarity can only be achieved by focussing explicitly on the effects of $$f{\mathbf k} \times {\mathbf u}$$ and omitting everything that is not dependent on those things.
 * Secondly, there is no need to discuss the formation of the Earth in an article on the Coriolis effect. A parabolic dish is good for showing Coriolis in isolation, because we've arranged for centrifugal forces to be countered. The atmosphere shows Coriolis in isolation because centrifugal forces can be balanced (by pressure and gravity).  That's what the article says, and that's all that need be said.  Further commentary about the earth's curvature and its creation through accretion will only serve to confuse the issue here.  It may be relevant to History of the Earth.   -- GWO

I prefer a demonstration with a parabolic dish with a fluid layer on it, and a tiny buoy suspended in that fluid layer. In oceanography, inertial oscillations are recognized as quite a common phenomenon, and the inertial oscillations are detected by monitoring buoy positions. I've tried to introduce the dry ice puck animation before, and several people didn't recognize the relevance for fluid dynamics.
 * Well, if you're interested in buoys in a fluid layer, you don't need a parabolic rotating table. A flat one will do just fine.

Can you specify how you envision a centrifugal force being balanced by pressure and gravity? At the poles, g = 9.83 m/s^2, at the equator, g = 9.78 m/s^2. If you compare two columns of air, one at the north pole and one at the equator, both containing the same number of atoms, then the polar column is heavier, corresponding to a larger air pressure. Hence, you expect air pressure at the poles to be larger than at the equator, even when not counting temperature difference. How does that larger polar pressure fit into your picture of a centrifugal force being balanced? --Cleonis | Talk 15:18, 22 February 2006 (UTC)


 * In purely pragmatic terms, when solving the equations of motion for an atmosphere, we can "hide" the centrifugal terms in the pressure, since they're dependent solely on position. Have you ever written down the equations
 * $$\frac{\partial u}{\partial t} - f v = -\frac{1}{\rho}\frac{\partial p}{\partial x}$$,
 * $$\frac{\partial v}{\partial t} + f u = -\frac{1}{\rho}\frac{\partial p}{\partial y}$$?
 * That's the basic equations of fluid flow on an f-plane. Ever wonder where the centrifugal terms went?  They're built into the pressure.
 * Your discussion concerns only hydrostatic pressure which has already got this balancing pressure built in. Similarly, when we talk about pressure gradient force, and the such like, then we're talking about pressure after adjustment to balance Centrifugal terms.
 * And again, you're discussing curvature.  That's fascinating, but it's nothing to do with the Coriolis effect. -- GWO


 * Also, the shape of the earth has nothing to do with its formation: its just not solid at the very large scale William M. Connolley 16:06, 22 February 2006 (UTC)


 * Of course: the Earth is ductile, otherwise there wouldn't be any continental drift. In its 4 x 10^9 years of existence the rotation rate of the Earth has decreased, and the shape of the Earth has of course followed that.
 * You are indicating to me, William, that for you my talk page messages are way too short. You are indicating to me that I should discuss all the fine points of the physics (mentioning the Earth's ductility right away), rather than limiting myself to discussing the general idea. --Cleonis | Talk 16:43, 22 February 2006 (UTC)
 * No, excess verbosity is not needed. I was pointing out that Earth started as an accretion disk is irrelevant; adding more words won't make it any more relevant. "The earth is not solid" is all you need. William M. Connolley 18:37, 22 February 2006 (UTC).

Physlets by Brian Fiedler.
The meteorologist Brian Fiedler has manufactured some physlets, one physlet illustrating motion over a flat surface and a physlet illustrating motion over a parabolic surface. These physlets use the system of showing the motion both seen from a non-rotating point of view and from a rotating point of view. If something like that would be possible for wikipedia I would make something like that.

The big advantage of those physlets is of course that you can adjust a parameter, you can "fiddle around" with it. GIF-animations just grind on, but for wikipedia a GIF-animation will have to do. --Cleonis | Talk 17:30, 22 February 2006 (UTC)

Animations: parabolic dish with shaded quarters


Here are the shaded quarter versions of the parabolic dish with frictionless puck animations. I have also uploaded 256x256 pix versions of each left and right side of the animation, the links are on the description pages of the animations. I decided to make 256x128 pix versions in order to keep the amount of KB's down.

It is often claimed that the parabolic dish 'balances the centrifugal force'. That is correct only for the case of motion with a constant distance to the overall axis of rotation. In the case of the ellipse-shaped trajectory, the centrifugal force is not balanced; there is an oscillation. When the puck has reached the point closest to the overall axis of rotation, then it moves away from the axis of rotation again, because there is not enough centripetal force there to maintain that distance to the center of rotation. --Cleonis | Talk 14:43, 27 February 2006 (UTC)


 * The diagrams are absolutely fantastic. GWO
 * Sentences like : because there is not enough centripetal force there to maintain that distance to the center of rotation are less so. It's true for some definitions of what we mean by centripetal force but it's not true for others.
 * It is entirely, 100% true to say that on the parabolic surface gravity balances the centrifugal terms ($$\omega^2 r$$) induced by the rotation of the frame. If the particle has angular velocity in addition to that of the frame, then all bets are off.  That's why we need to be very careful about using the phrase 'centrifugal force'.  Do we mean the fictitious force that arises from the change of frame, or the (possibly absent) physical force that would necessary to hold the particle to its present radius?  IMHO, if we limit ourselves to the former usage, everything is much, much clearer.  The fictitious centrifugal acceleration related to the motion of the frame is always counterbalance by the sloping surface.  The remainder is ehat we call Coriolis.  -- GWO

I agree that the demarcation that you describe is the most practical. Personally, I like to use the expression coordinate acceleration for the acceleration of a non-inertial coordinate system with respect to an inertial coordinate system. My policy is to discuss physics and coordinate acceleration separately. From here on I will use the expression 'centrifugal force' exclusively in the meaning of coordinate acceleration, with 'centrifugal force' associated exclusively with the centrifugal term of the equation of motion for motion with respect to a rotating coordinate system. If you would like to incorporate the parabolic dish animations into the coriolis effect article, please do so. For the time being, my attention is focused on other animations I want to make. (For example for the coriolis flow meter article.) --Cleonis | Talk 17:48, 27 February 2006 (UTC)

The Coriolis effect (meteorology) varying with latitude
Myrtone has written in the section Coriolis_effect: "The Coriolis effect does not depend on the curvature of the Earth, only on its rotation. However, the strength of the Coriolis force varies with latitude, and that is due to the Earth being slightly oblate."

While the fact that the Earth is an oblate spheroid is implicitly taken into account in meteorology, the oblateness is not significantly involved in the latitudinal varation of the strength of the coriolis effect (for the Coriolis effect as taken into account in meteorology).

Close to the poles the Earth's surface is close to perpendicular to the Earth's axis of rotation. Objects that are near to the equator and that move over the Earth's surface are moving almost parallel to the Earth's axis, hence a diminished Coriolis effect.

It is incorrect to state that the Earths oblateness is involved in the strength of the Coriolis force varying with latitude. --Cleonis | Talk 10:57, 4 April 2006 (UTC)


 * You're right.  Be bold.  Specifically, fh changes due to sphericity, and oblateness can be  neglected.  -- GWO


 * As I said, implicitly the oblateness is taken into account (in meteorology). Consider the following thought experiment: a celestial body, perfectly spherical, that is rotating. Motion over the surface of that sphere will not be the same as motion over the surface of an oblate spheroid celestial body. The equations that are applied in meteorology rely on the oblateness of the Earth being in equilibrium with its rate of rotation. That is: those equations yield good results exclusively for the case of the oblateness being in equilibrium with the rate of rotation.
 * This is important: the oblateness of the Earth is not neglected (in meteorology), it is just that it is seldom pointed out explicitly that the equations take it into account. To show those things, I need to develop some more animations. Then I will proceed to edit the coriolis effect article.


 * Anyway, the original subject of this thread was latitudinal varying. Summarizing, fh just depends on the angle between the normal force that is exerted by the Earth's surface and the Earth's axis. --Cleonis | Talk 12:50, 4 April 2006 (UTC)

"The Coriolis effect does not depend on the curvature of the Earth, only on its rotation. However, the strength of the Coriolis force varies with latitude, and that is due to the Earth being slightly oblate."

The mention of oblateness is there for accuracy. Let's face it, the Earth is shaped like a slightly oblate spheroid, so this was why I rephrased it.Myrtone (the strict Australian wikipedian):-(


 * It is, but the oblateness is (basically) irrelevant to this, and mentioning the oblateness makes it sound relevant. Which it isn't.  And, given the number of misapprehensions about Coriolis, its more important not to include irrelevant (and hence possibly misleading) information (i.e. oblateness).  "Approximately spherical" is (i) accurate and (ii) does not induce the reader to believe the deviation of the earth from a sphere is any way important (it isn't). Do you know anyone doing weather forecasting who doesn't pretend the Earth is spherical?  -- GWO

It being irrelevent runs counter to the wikipedia aim on being highly accurate, either it is irrelevent or highly accurate (as wikipedia favours), but *not* both.Myrtone (the strict Australian wikipedian)(talk):-(
 * The earth's surface is also 2/3rds water, and appears blue/green from space. That too is highly accurate but it's not relevant to the effect of Coriolis in meteorology.   Furthermore, the capital of Canada is Ottowa.  That too is highly accurate, but not relevant to the effect of Coriolis in meteorology.  GWO  (This would make more sense if I'd spelt Ottawa correctly).


 * I'm with GO here. The oblateness is unimportant *for the coriolis effect*. It is also, effectively, unimportant for the dynamics, too, as it can be hidden and it is as if it isn't there William M. Connolley 09:10, 5 April 2006 (UTC)

"The earth's surface is also 2/3rds water, and appears blue/green from space. That too is highly accurate but it's not relevant to the effect of Coriolis in meteorology. Furthermore, the capital of Canada is Ottowa. That too is highly accurate, but not relevant to the effect of Coriolis in meteorology." What has all this, especially the latter fact got to do with the shape of the Earth.Myrtone (the strict Australian wikipedian)(talk)
 * Almost exactly as much as the Earth's slight deviation from spherical has to do with the variation of Coriolis effect with latitude. Practically nothing. -- GWO
 * I sopose that the Capital of Canada (Ottawa) is more off the topic than the earth's slight deviation from spherical. Capital cities have nothing to do with the effect, but the precise shape of the Earth does. If the Earth were undeviated from spherical, the Coriolis effect (meteorology) varying with latitude would differ somewhat.Myrtone (the strict Australian wikipedian)(talk)
 * "Somewhat" in the sense of "a completely negligible amount." I've never known a meteorologist who considers the earth's deviation from spherical to terribly be important in weather or climate dynamics.  Suffice to say, it doesn't come up at conferences very often. -- GWO

I removed "slightly oblate" leaving just "shape". I did this because otherwise there is a danger of people thinking it is the slight oblateness that is the important bit, rather than the near-sphericalness William M. Connolley 08:18, 23 April 2006 (UTC)

The Eötvös effect
One of the ways that the meteorologist Anders Persson uses to address the confusion surrounding the Coriolis effect as taken into account in meteorology is to point out ( in this article ), the close relationship with the Eötvös effect.

Eötvös was the first to describe it, and in 1915 he constructed an instrument to demonstrate the Eötvös effect. See also: the gravity field (PDF-file, 1.6 MB, University of Sydney, school of geosciences)

The way that in which geophysicists take the Eötvös effect into account is closely analogous to the way the Coriolis effect is taken into account in meteorology. The Coriolis effect as taken into account in meteorology, and the Eötvös effect, are complementary to each other, in the way the sine and the cosine are complementary to each other.

Recognizing the close relationship with the Eötvös effect goes a long way in understanding the Coriolis effect as taken into account in meteorology. --Cleonis | Talk 09:47, 6 April 2006 (UTC)
 * I disagree. The Eotvos (excuse my umlaut scarcity) is a correction for the centripetal terms.  If there's one thing that must be agreed on, it's that any.  They're complementary, but they're utterly distinct.  The Eotvos effect, rather than intrinsically linked to Coriolis, is just another thing that the Coriolis force is not. There's no Eotvos on a parabolic dish.   We've already gone to great length to get centripetal effects out of this article, as they're the basis for much confusion.  Let's not reintroduce them, just because geophysical centripetal forces have a cool name.  -- GWO

The Eötvös effect in the case of motion over the surface of a parabolic dish
A thought experiment: a very large parabolic dish, rotating, the parabolic curvature of the dish matches the rotation rate. Let a gravimeter rest on the parabolic dish, at rest with respect to the parabolic dish. Let r denote the distance of the gravimeter to the main axis of rotation. The gravimeter will measure a particular acceleration in the direction perpendicular to the local surface. Now consider the case of the same gravimeter, circling the main axis of rotation a with an angular velocity that is different from the angular velocity of the parabolic dish. The gravimeter will then measure a slightly different acceleration (in the direction perpendicular to the local surface), as compared to the co-rotating with the dish situation. This shows that in the case of motion with respect to a rotating parabolic dish, there is an Eötvös effect, analogous to the Eötvös effect for a gravimeter in motion over the surface of the Earth. I emphasize that the operative factor in the case of Eötvös effect is velocity relative to the rotating body that the gravimeter is resting on (either the parabolic dish or the Earth). Obviously, when the velocity of the gravimeter relative to the parabolic dish changes, the centripetal force does not change, for the centripetal force depends only on the slope of the parabolic dish. This shows that the correction for the Eötvös effect is not a correction for a changed centripetal force. --Cleonis | Talk 18:03, 6 April 2006 (UTC)
 * Obviously, when the velocity of the gravimeter relative to the parabolic dish changes, the centripetal force does not change, for the centripetal force depends only on the slope of the parabolic dish. 
 * Only because you're confusing definitions of centripetal force, again. The inward component of the reaction force between dish and particle depends only on shape.
 * You can do all the thought experiments you like, a particle on a parabolic dish follows the equations
 * $$ \frac{\partial u}{\partial t} - fv = 0, \quad \frac{\partial v}{\partial t} + fu = 0$$
 * (i.e. it moves under the influence of the Coriolis force alone) does not experience the Eotvos effect. Similarly  the Eotvos effect is apparent at the equator, where there is no Coriolis effect at all.
 * Therefore, the Eotvos effect is not the Coriolis effect, and the Coriolis effect is not the Eotvos effect. -- GWO


 * In geophysics, the Eötvös effect is corrected for at all latitudes. Check the formula in the Eötvös effect article, it contains $$\phi$$; the latitude where the measurements are taken. The magnitude of the Eötvös effect is proportional to the cosine of the latitude; strongest at the equator, weakest at the poles.
 * I stated that the Coriolis effect and the Eötvös effect are complementary to each other, as sine and cosine are complementary. Obviously that implies that the Coriolis effect and the Eötvös effect are not the same. --Cleonis | Talk
 * Then why are you trying to introduce a discussion of Eotvos into an article on Coriolis? They're two separate facets of rotating motions. It's a See Also at best.  PS: I never said that Eotvos was apparent only at the equator.  I was using the equator to point out that you can have Eotvos without Coriolis, thereby ruling out one being an artifact of the other. PPS: Has your thought experiment showed you how to derive the Eotvos formula from the equations given above, yet?  -- GWO


 * The Coriolis effect as taken into account in meteorology and the Eötvös effect are closely related. Understanding the physics of the Eötvös effect is a great help in understanding the physics of the Coriolis effect as taken into account in meteorology.
 * In the Rotational-vibrational coupling article that I wrote the explanation can be found why the following equations hold good (to a good approximation) for motion over the surface of a parabolic dish (motion as mapped in a co-rotating coordinate system)
 * $$ \frac{\partial u}{\partial t} - fv = 0, \quad \frac{\partial v}{\partial t} + fu = 0$$
 * In the rotational-vibrational coupling article it is shown that in the presence of a center-seeking force that is proportional to the distance to the center, the moving mass follows exactly the same pattern of motion as inertial oscillations. animation: rotational-vibrational coupling animation: rotational-vibrational coupling as mapped in rotating coordinate system animation: motion over the surface of a parabolic dish, both perspectives
 * My approach to writing for wikipedia is that I want to provide explanations to interested non-experts. My emphasis is on visualisation, rather than on formal mathematics. I'm not writing for people who are already experts on the subject. --Cleonis | Talk 19:37, 6 April 2006 (UTC)
 * If it's not in the mathematics, it doesn't exist. The explanation doesn't have to be mathematical, but if the explanation is contrary to the mathematics, the explanation is wrong.
 * The Eotvos article is excellent. But by your own admission its nothing to do with the Coriolis effect -- except by a rather tenuous analogy -- so this argument is moot. -- GWO
 * The physlets by Brian Fiedler show clearly what is in the mathematics. motion over a flat surface motion over a parabolic surface. The presence of a center-seeking force is the operative factor.
 * The over-arching aim in physics is to provide an explanation that is frame-independent. In the physlets by Brian Fiedler the coriolis term and centrifugal term are part of the mathematics only in the case of representing the motion with respect to a rotating coordinate system. On the other hand, what Brian Fiedler refers to as the tangent gravity is frame-independent; it is taken into account in both representations. The purpose of the centrifugal term and the coriolis term is to effectuate the coordinate transformation; the centrifugal term and coriolis term are not part of the physics taking place. Only the tangent gravity is physics. --Cleonis | Talk 14:17, 8 April 2006 (UTC)
 * Yes. I know.  But that does not mean that everything other than gravity is the Coriolis effect.  The Coriolis effect is one, specific term in that frame transformations : It is the fk*u term.  Anything that isn't fk*u is not Coriolis.  OK? It might be interesting, and real and important and cool and exciting and BUT IT ISN'T CORIOLIS.  This article is exclusively about the Coriolis Effect (the clue is in the title), therefore, things that are not Coriolis do not belong in this article. Sorry for shouting, but I'm getting heartily sick of repeating myself to someone who seems to think the Coriolis effect is a catch all term for the physics of rotating systems. -- GWO

This article is about the coriolis effect, not the physics of motion in rotating frames. So let us, as GWO says, keep things that aren't coriolis out of it, as far as possible. William M. Connolley 14:22, 8 April 2006 (UTC)

Coriolis Force acting on a Foucault Pendulum
It might be worthwhile to identify the Coriolis force related to the observed motion of the Foucault pendulum. There is a simple difference in the forces acting on a pendulum bob before it is released compared to after it is released. This is why there is a change in the motion of the pendulum swing in relation to the surface of the Earth and why there is a change in effect with latitude. If this is elucidating to the discussion of the Coriolis force it might be identified in the article. The Foucault pendulum also has important historical significance. The magnitude of the effect is related to the degree of alignment of the central axis of the pendulum to the Earth's axis of rotation.David Harty 12:11, 7 April 2006 (UTC)

Coriolis effect (in meteorology), Eötvös effect, and Foucault pendulum
At 45 degrees latitude, the Coriolis effect (as taken into account in meteorology) and the Eötvös effect are equal in magnitude. (Sine and cosine of 45 degrees)

In the Eötvös effect article I used the example of the amount of buoyant force that a zeppelin needs to maintain height above the surface of the Earth. When a fast-moving zeppelin makes a U-turn, going from eastward to westward motion (with respect to the Earth), it must retrim its buoyancy.

During eastward travel the zeppelin will tend to veer to the south equator, during westward travel the zeppelin will tend to veer to the nearest pole. The Coriolis effect (as taken into account in meteorology) and the Eötvös effect are perpendicular components of one and the same effect, that is why they both have the same dependency on velocity with respect to the Earth, with that typical factor 2.

Thís mechanism also explains the precession of the plane of swing of a Foucault pendulum. The trajectory of the pendulum bob is not a ballistic trajectory, for the pendulum bob is suspended. Let a Foucault pendulum be located at 45 degress latitude. When the pendulum bob swings in eastward direction, it veers towards the south equator, on the opposite swing, to the west, it veers to the nearest pole. That is part of the mechanism that makes a Foucault pendulum precess though an entire circle. For a Foucault pendulum located at 45 30 degrees latitude, precessing through an entire circle takes 2 sidereal days, and the rate of precession is fairly constant throughout. --Cleonis | Talk 14:47, 8 April 2006 (UTC)


 * It is not clear that the Eotvos effect is the correct term to use in discussion of the motion of the pendulum bob in relation to the surface of the Earth. The Eotvos effect (per the definitions) is a vertical deflection that changes the gravity and with it the weight of the body for an object that is moving.  It is not clear that this effect is what causes the motion of the pendulum bob to change with respect to the surface of the Earth (as compared to a change in the vertical or gravitational force or buoyancy force).  There is a 3-dimensional deflection system for which the Coriolis effect is only one part and the Eotvos effect is one part.  In many articles and books there is an inconsistency in the usage of the 3-dimensional gyroscopic terms and attributions to the Coriolis effect as well as the commonly known ballistic effects.  These 3-dimensional gyroscopic terms are being clarified in this discussion and Coriolis effect article.  The point of connection of the pendulum bob is not moving with respect to the surface of the Earth and forces acting on the pendulum bob are acting through the point of connection of the pendulum apparatus.  Since the point of connection is not moving with respect to the surface of the Earth it is not clear that the Eotvos effect is applicable to the observed change in motion of the Foucault pendulum.  David Harty 10:40, 10 April 2006 (UTC)
 * It is not clear that the Eotvos effect is the correct term to use in discussion of the motion of the pendulum bob in relation to the surface of the Earth.. In fact, it is readily apparent that the Eotvos effect is not the correct term to use in discussion of the motion of the pendulum bob in relation to the surface of the Earth.  However, Cleonis will continue to mention about the Eotvos effect (because he cannot distinguish it from the Coriolis effect) even thought Foucault's pendulum will work beautifully at the poles (with period 24 hours), where the Eotvos effect vanishes, and the Coriolis effect doesn't...  -- GWO
 * It appears to me that the point of the Anders Perssons article is to describe all of the three dimensional deflective mechanisms in a rotating system. The title becomes somewhat misleading since the paper deals with the 3-dimensional system not just the Coriolis Effect.  He describes horizontal deflection of vertical motion, vertical deflection of horizontal motion (Eotvos Effect), and the horizontal deflection of horizontal motion of which there are two parts which he calls the common centrifugal force and the compound centrifugal force (the Coriolis Force.  These deflective mechanisms are apparently all present together because they are all part of the rotating system.  The purpose for bringing up the Foucault pendulum is that it may isolate certain forces and be helpful in defining which forces cause the observed period of rotation of the pendulum.  It seems that the force which determines the rotation is horizontal deflection of horizontal motion (Coriolis force) and the vertical deflection of horizontal motion (Eotvos) is not contributing to the observed period of rotation.  If the pendulum wire were a spring rather than a ridgid wire would the bob move up and down due to the Eotvos Effect?  David Harty 16:10, 12 April 2006 (UTC)

The Coriolis effect (meteorology)/Eötvos effect relation






The Coriolis effect as taken into account in meteorology and the Eötvös effect (for eastward and westward motion) are perpendicular components of one and the same effect. It is not necessary to describe the Eötvös effect in the Coriolis effect article, but it is necessary for anyone writing about the Coriolis effect as taken into account in meteorology to grasp the relation with the Eötvös effect. This decomposition in Coriolis-component and Eötvös-component applies only in the case of geosciences; motion that follows the Earth's surface is motion in 3 dimensions of space.

To obtain an example without Eötvös effect, I made the Rotational-vibratational coupling animation. There the motion is strictly planar, rather than only approximately planar. In the case of strictly planar motion there is only a Coriolis effect.

A general remark about writing for an encyclopedia: the over-arching aim of education is to unhide anything that has remained hidden. Hiding aspects of the physics taking place is bad physics education. --Cleonis | Talk 14:20, 11 April 2006 (UTC)


 * Holy !*$%*!*%*#*%! You're appealing to the sphericity of the earth again. Do you really believe that the Coriolis effect is anything to do with the curvature of the Earth?
 * For the umpteenth time -- this article is about the Coriolis effect. This article is not a general discussion of the role of rotation in geophysics (of which the Coriolis effect is one specific feature).
 * Those two things are not the same, and yet you continue to fail to grasp that very simple fact.
 * No-one is suggesting the Eotvos effect isn't important in some applications, buy things that are not the Coriolis effect do not belong in articles about the Coriolis effect, any more than the text of this article belongs in Eotvos effect. -- GWO

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What is expected when other articles refer to the coriolis effect article
It is very common to see in meteorological texts statement like: 'because of the Coriolis effect, air tends to keep flowing around pressure areas'. Dozens of wikipedia articles about winds and weather in general mention a similare 'bevcause of the coriolis effect' and refer to the Coriolis effect article, clearly expecting that the Coriolis effect article will deliver the goods, that the Coriolis effect article will explain what physics is at play in affecting the winds.

The central theme is of course the expression fk*u The article about rotational-vibrational coupling is designed to describe the simplest possible example of Coriolis effect. The spring exerts a center-seeking force, the strength of the center-seeking force is proportional to the displacement from the center of rotation. The vector for the center-seeking force and the vector of the centrifugal term of the equation of motion are at all times opposite in direction and equal to each other in magnitude. The motion depicted in the rotational-vibrational coupling animation has the property that the motion with respect to the rotating coordinate system is described by the coriolis term: fk*u.

Conclusion: it is an example of the pure form of Coriolis effect if and only if the strength of the center-seeking force is proportional to the displacement away from the center of rotation.

Coriolis effect is occuring when the center-seeking force is doing work. More precisely: when there is an oscillation (as discussed it the rotational-vibrational coupling article) then the center-seeking force oscillates between doing work and doing negative work. The aim of physics education is to show what forces are doing work. Showing what forces are doing work shows the cause-to-effect relations.

It is common in physics to use the expression effect in referring to an example a particular recognizable effect, and there can be a class of examples, related by analogy. Examples: the Casimir effect and the nautical Casimir effect. discussing the nautical Casimir effect in the Casimir effect article is appropriate. An area of study in molecular physics is rotational-vibrational spectroscopy. In molecular physics, coupling between rotational and vibrational energy-levels is considered to be an example of a Coriolis effect, hence it is widely referred to as coriolis-coupling. (16.000 Google hits for "coriolis coupling" plus "molecule") --Cleonis | Talk 21:53, 12 April 2006 (UTC)

Formula of coriolis effect
I have read your article because my boy asked why the hurricanes and tornados are rotating? Why the air does not go in a straight line from the area of high pressure to the area of low pressure - as common sense suggests?

I found the introduction - beginning with the mathematical formulae based on the vector product a bit "steep". Being honest I did not understand the coriolis effect when I read this formulae. Well, I have studied some vector analysis - but this is 20+ years ago. And I do not think that any of my colleagues (all of which are engineers and computing people) would understand this formulae. This raises the question: for whom is this introduction written? What is the target audience? Obviously not students and professors of physics - because they need a much more in-depth understanding of the issue. For us other lesser mortal? Probably not - because we do not understand it. Notice that I liked math when I was at university and I was among the best in my year. What about those who did not like math or have never heard of the deeper secrets of vector analysis?

My suggestion: start the article with a more "common sensical" introduction I suggest the following explanation (however I do not know if it is correct from a physical point of view).

A wind of 50 miles per hour is blowing in New York City exactly from South to North.
 * If you think the Coriolis effect knows the difference between North/South and East/West, you do not know what the Coriolis effect is. -- GWO

This wind takes a feather of a bird (or a cloud of feathers) with it. After a while the cloud of feathers arrive somewhere near Montreal (which is North of NY). To this point an observer on the earth agrees with an observer from space - both of them see the feathers near Montreal. The point of view of the observer from space however is: since the earth is turning to the east NY City moves eastwards with x miles per hour. (I do not have the formlae at hand to calculate x - but I "guestimate" it should be a few thousand miles per hour). For this reason the observer from space sees the cloud of feathers NOT moving to the north - but with x miles to the east and 50 miles to the north. (Since x is much faster than 50 miles per hour the observer from space sees the cloud moving almost exactly to the east with a slight tendency to the north). Well after a while the cloud arrive somewhere near Montreal. The feathers and the surrunding air have some inertia. For this reason they have the tendency to continue their journey in a straight line. And this is what the observer from space is seeing: the feathers travel with x miles to the east and 50 miles per hour to the north. However Montreal is closer to the Northpole than NY. The surcumfence of the earth at Montreal is smaller than at NY. Since both Montreal and NY need 24 hours for one rotation Montreal is moving slower to the East than NY. This means the eastward speed of Montreal is less than x miles per hour. The feathers and the surrounding air however continue to fly with x miles per hour to the east and 50 miles per hour to the North. (From the point of view of the observer in space). For this reason an observer in Montreal will see the feathers flying NOT exactly to the north. Instead he sees the feathers flying somewhere to the North-East. This (appearent) change of direction is the coriolis effect.

I do not regularily read the discussions of that article. So: if you want to answer something it might be better to send it to johann_rost@yahoo.com.

Johann


 * I think the problem with this is that it mixes up geometry with Coriolis, as people so often do. It would be desirable for a simpler explanation, but yours isn't it. Maybe a picture in that section would help. Really its very simple: the acceleration is proportional to the rotation and the velocity, and perpendicular to both. William M. Connolley 15:27, 19 April 2006 (UTC)

References: meterologists who point out that Earth's oblateness matters
I copy and paste from above: "I've never known a meteorologist who considers the earth's deviation from spherical to terribly be important in weather or climate dynamics. Suffice to say, it doesn't come up at conferences very often. -- GWO"

Please study the following references:


 * Durran, D. R., 1993: Is the Coriolis force really responsible for the inertial oscillation?, Bull. Amer. Meteor. Soc., 74, 2179–2184; Corrigenda. Bulletin of the  American Meteorological Society, 75, 261


 * Durran, D. R., and S. K. Domonkos, 1996: An apparatus for demonstrating the inertial oscillation, Bulletin of the American Meteorological Society, 77, 557–559.


 * Persson, A., 1998 How do we Understand the Coriolis Force? Bulletin of the American Meteorological Society 79, 1373-1385.


 * Norman A. Phillips., 2000 An Explication of the Coriolis Effect, Bulletin of the American Meteorological Society: Vol. 81, No. 2, pp. 299–303.

You will find that those meteorologists point out that the Earth's oblateness forms a constant background factor that is in fact taken into account in the equations of meteorology. Of course, being constant, like gravity itself, the Earth's oblateness is not a variable in the equations, so while it has been incorporated in the equations, it is not a subject of meteorological study. --Cleonis | Talk 15:33, 26 April 2006 (UTC)


 * All this is besides the point. Std met textbooks for the equations of motion (e.g. the recent Cullen book; but Gill will do as well) point out that you end can end up in spherical coords to a very good approximation William M. Connolley 15:55, 26 April 2006 (UTC)


 * ''I see lots of people who note oblateness. Of course, they're not idiots.  I still see no-one who considers it terribly important.  Can you give me a quote in which its terrible importance in meteorology is mentioned?
 * Durran's article is particularly odd. If you add the non-vertical gravity to his first two equations, then eliminate p, you get exactly the same oscillatory motion. The same is true of all uniform body forces. And if you say g is latitude dependent, then you really need to put planetary beta in too. C-.  Everyone else refers back to Durran, and at the moment I don't think he's right.  The idea that the poleward component of g is magically balanced by w^2r.  I mean, is this supposed to be altitude dependent.  g decreases with altitude, w^2r increases.  Do they always magically balance?  That's pretty fortunate, if true.  '' -- GWO


 * About Durran's article, I mention it because it kick-started Anders Persson's thinking about the Coriolis effect. I find Durran's article rather inaccessable, not suitable for helping to understand the physics.
 * About the poleward force as described by Durran, Persson and Phillips: for an atmosphere without any flow or convection the following is valid: at every latitude and at every altitude the poleward force povides precisely the amount of centripetal force that is required. Of course, that does not involve magic or fortune. The state with that equilibrium is the state with the lowest attainable energy. Any system that can flow will inexorably end up in that state where all the energy that could dissipate is dissipated.
 * Here is an example: a heavy dish is filled with water. Quickly, the dish is started spinning, after that no force is exerted on the dish, it must continue spinning on its own rotational kinetic energy. Friction causes the water in the dish to spin up. Due to the friction kinetic energy is transferred from the dish to the water (and in the process some energy dissipates). Inexorably the water will reach a stage of solid body rotation. When the water is in solid body rotation there is no more convection, hence no more dissipation. When a fluid is in solid body rotation, then at all points the required centripetal force is being exerted. --Cleonis | Talk 01:59, 4 May 2006 (UTC)

I've just read Durran, D. R., and S. K. Domonkos, 1996: An apparatus for demonstrating the inertial oscillation, Bulletin of the American Meteorological Society, 77, 557–559. Interesting, but irrelevant to the point at hand. Why did you ask us to read it? Also note the bit at the end: Once we actually began using the apparatus... This demonstrates a certain lack of thought on Duggans part: I've realised this without haveing to make anything William M. Connolley 21:19, 3 May 2006 (UTC)


 * William, you ask me why I have included the Durran/Domonkos article in the references. Surely, that is a no-brainer. Durran's purpose is to show that in order for inertial oscillations to occur, at all points a center-seeking force must be exerted on the puck that is equal in magnitude and opposite in direction to the $$ w^2 \vec{r} $$ term. Analogously: in order for the terrestrial inertial oscillations to occur the poleward force must at all latitudes be equal in magnitude and opposite in direction to the $$ w^2 \vec{r} $$ term.


 * In january I manufactured and uploaded an animation that depicts a terrestrial inertial oscillation. That animation is unsuitable for an article, being a whopping 600 KB in size. I hope to find a way to make an animation that conveys the same idea, while not exceeding 50 KB. Most of my animations are strictly 2-dimensional. This one is closer to a 3-D animation than to 2-D. Image:Inert_oscill_rotat_planet.gif --Cleonis | Talk 07:39, 4 May 2006 (UTC)
 * You are refered to Pedlosky "Geophysical Fluid Dynamics", Eq 1.6.6. or Gill "Atmosphere-Ocean Dynamics" Eq 4.5.10. Both of those seminal works (have you read either?) describe how the centripetal and gravitation forces (never assumed to be vertical) can be absorbed into a single geopotential, and thus effectively hidden in the hydrostatic pressure (in as much as whenever we eliminate the pressure, as Durran does) the gravitational and centripetal terms vanish also.  Nowhere is it suggested that they must balance; that's a nonsense of Durran's invention. PS: "I made an animation once" is not a grown-up scientific argument. -- GWO