Talk:Coriolis force/Archive 3

I have archived everything. That may be a bit drastic - apologies if so.

Note to newcomers: this archiving occurred just after the "new version" of 2005/08/22 got put in. William M. Connolley 10:47:05, 2005-08-23 (UTC).

( Coriolis Acceleration of a West Bound Body ) I moved this question to the end, for sanity conservation purposes. GangofOne 05:09, 11 December 2005 (UTC)

Brief Question - Coriolis force vs centrafugal force in Meterology 

Sorry for putting this at the top, seems there is a lot of discussion about this page, and thought it might be more visable till its been resolved....Sorry if this is taking a liberty.

Basically the question is... about the Coriolis in Meteorology section.

you say: Perhaps the most important instance of the Coriolis effect is in the large scale dynamics of the oceans and the atmosphere. In meteorology, Coriolis effects tend to dominate centrifugal effects, because the latter is usually balanced by an ambient pressure gradient (exactly analogously to the slope on a parabolic turntable).

My understanding is that the 'ficticious' centrifugal force would be dominating, by magnitude however it is balanced by the fact that the deformable earth surface has formed into an 'oblate' squashed sphere. Presumably this is what the ambient pressure gradient is???? I was thinking that maybe this could be explained a bit better?

Perhaps also with a small force vector diagram Showing the centrifugal force being balanced??

Oblate

Wideofthemark 18:39, 28 May 2006 (UTC)

Reprise: ballistic vs atmospheric Coriolis effect

Cleon has continued to assert there are two Coriolis effects. He used the (northern hemisphere) examples (both observed from the surface of the Earth) of the floating balloon which, he says, circles to the left because of the "atmospheric Coriolis effect" and a "vomit comet" missile which deviates to the right because of the "ballistic Coriolis effect". He says there are two effects because in one the deviation is to the left and in the other it is to the right.

A thought experiment: (1) We magically stop all the winds (relative to the surface of the Earth). (2) At the centre of a region we magically create a low pressure by removing some of the air or by cooling it down. (3) We now stop the magic and allow physics to take over. (4) We observe the air flowing initially albeit instantaneously directly towards the centre of the region. QUESTION: Is that air deflected to the left or to the right?

Cleon, please answer the question directly. Paul Beardsell 12:13, 23 August 2005 (UTC)

---

Your question is superfluous; it is answered by a pair of images I manufactured weeks ago.

pattern of deflection in the theoretical case of a sudden low pressure area.

First: you have made a mistake with the balloon.

If the floating balloon is moving along with inertial wind then it will follow the motion of the inertial wind: on the northern hemisphere inertial wind flows clockwise: deflection to the right. That is elementary meteorology.

Yes, I agree and I understand although I may have misread something you wrote earlier. But you are still wrong that there are two different Coriolis effects, if that is what you still say. Paul Beardsell 14:47, 23 August 2005 (UTC)

In meteorology inertial wind is defined as: wind that flows without the presence of a pressure gradient force. (Eventually friction dissipates the energy of the inertial wind)

And note that (northern) inertial wind turns right. The deviation in the inertial wind is caused by the Coriolis effect. Paul Beardsell 14:47, 23 August 2005 (UTC)

The thought experiment:
Just look at the image I created weeks ago:
On the northern hemisphere: if by magic a sudden low pressure area is created then air mass that initially is starting to move in east-to-west direction is deflected to the North. That is: deflection to the right. At some point the pressure gradient force becomes overpowering: the transition to cyclonic flow, which on the northern hemisphere flows counter-clockwise.

Yes! And that is what I was saying just before you said I had no understanding of meteorology. That there is some anti-clockwise flow does not mean that the (northern hemisphere) deviation owing to the Coriolis effect is to the left! It is to the right for winds just as it is for missiles. Paul Beardsell 14:47, 23 August 2005 (UTC)

You seem to have trouble keeping track of what is flowing clockwise and what is flowing counter-clockwise. --Cleon Teunissen | Talk 13:41, 23 August 2005 (UTC)

No. Not at all. But I will admit to difficulty in following your sometimes overly long and sometimes not quite to the point replies. Paul Beardsell 14:47, 23 August 2005 (UTC)

What Cleon wrote.

what the direction of inertial oscillation would be if the air mass would move in inertial oscillation

Here I wrote:

On the other hand, when an airship allows itself to be swept along by inertial wind that is at some point in time moving from east-to-west, then that airship, moving along with the wind, is seen to deflect to the north.

Inertial wind is air parcels moving in inertial oscillation. The cycle of inertial wind makes it flow east-to-west regularly. On the northern hemisphere: if you are moving from east-to-west and this east-to-west-motion is deflected to the North, that is deflection to the right.
--Cleon Teunissen | Talk 14:00, 23 August 2005 (UTC)

---

And we get anti-clockwise [i.e. deviation to the left] rotation around low pressure (northern) regions and clockwise [i.e. deviation to the right] rotation around high pressure. But we all KNOW that Coriolis effect causes a deviation to the right. (And this accounts for the less dramatic high pressure weather systems.) I have no problem with this seeming contradiction as, unlike another here, I do not ascribe every atmospheric phenomenum to the Coriolis effect. Does Cleon yet agree there is only one Coriolis effect? Paul Beardsell 14:33, 23 August 2005 (UTC)

Gravitational potential is incorporated in the meteorological models

[...] the turntable: as it clearly states there, the gravitational component is being used to balance the centrifugal (err, and notice this is centrifugal-from-the-axis-of-rotation, not centrifugal-from-the-inertial-circle-centre) force to allow the coriolis dynamics to be seen. In the atmos, the same term does not appear. [...]. William M. Connolley 08:24:15, 2005-08-23 (UTC).

Of course in meteorological models the perpendicular to the axis of rotation component of the Earths gravitation is incorporated: it incorporated in the geopotential height. At the equator the potential gravitational energy is about 108 KiloJoules per kilogram mass higher than at the poles. That is a gravitational potential; the derivative of that potential is the force. It's in the models allright.

Now about that parabolic rotating turntable.
Letting the turntable rotate is helpful in minimizing the small but non-zero friction, but it is the shape that counts.

Let's take a really large parabolic dish: the dish of the Arecibo radio-telescope. Let's say it is strong enough to be driving around on it with a car. Inside the car a plumb line is suspended from the roof.

Given the shape of the dish there is exactly one angular velocity of driving in circles on the dish whereby inside the car the plumb line hangs vertical with respect to the car. Lets call that equilibrium velocity. If the car goes faster than that, then the plumb line will point outward, if the car goes slower than that the plumb line is pulled inward: by gravitation.

Let's say the surface of the dish is almost frictionless, almost no grip with the tires. Then if you want to drive around on it in a car your velocity makes all the difference. More velocity than equilibrium velocity and the car will drift up the slope of the dish. Less velocity than equilibrium velocity and gravitation, redirected by the slope of the surface, will pull the car to the center (but in being pulled to the center the car gains angular velocity (conservation of angular momentum)).

That is what is happening in a thin fluid layer on a rotating parabolic turntable. If you start an inertial oscillation in that fluid layer, then you get an oscillation in the angular velocity of the fluid parcels, together with an oscillation in the distance to the central axis of rotation.

The presence of the Earth's gravitation is an integral part of the inertial oscillatons. There is cancelation for one angular velocity only. Whenever the angular velocity is not exactly the equilibrium velocity, then there is either too little or too much centripetal force and whenever there is unbalance the distance to the central axis of rotation changes. --Cleon Teunissen | Talk 14:49, 23 August 2005 (UTC)

This is not true in the atmosphere in any sense. William M. Connolley 21:41:56, 2005-08-23 (UTC).

---

But as soon as the friction disappears the plumb line hangs (locally) vertically so we can discard the plumb line - it plays no part in the problem anymore. And as soon as we float the surface disappears. We can similarly discard consideration of the shape of the surface when we float! An airplane is unaffected by the sudden appearance of a valley below it. Were the earth a cube there would still be cyclonic phenomena in the atmosphere. Uneven heating will provide low and high pressure zones. The Coriolis effect will deflect air travelling from high to low pressure. If you, Mr Balloonist, (or a packet of air) is floating in the atmosphere then gravity plays NO PART. That is what "floating" means. You can introduce gravity into the calculations, if you like, but only to see it cancelled out. Gravity and the Coriolis effect are unrelated. Paul Beardsell 15:13, 23 August 2005 (UTC)

In other words, please move this discussion to a relevant article. Paul Beardsell 15:14, 23 August 2005 (UTC)

Of course an airplane is unaffected by a valley, an airplane is buoyant! For buoyant objects the slope of the geopotential surface(s) matters. In the case of the Earth the shape of the geopotential surfaces is an oblate spheroid. It was you Paul, who pointed out to me that is is the shape of the geopotential surfaces that matter, not the shape of the solid. --Cleon Teunissen | Talk 16:00, 23 August 2005 (UTC)

You paraphrase me incorrectly. I never used the word "geopotential". Presumably by that you mean gravitational potential. My point is gravity and the Coriolis effect are independent. The "geopotential" can just be neglected when one is floating. (Almost) by definition. Or by Archimedes. Paul Beardsell 16:33, 23 August 2005 (UTC)

Yes, you never used the word geopotential, but you emphasized it wasn't the shape of the solid that matters, but the shape of the surfaces of equal buoyancy, and that put me on the right track. The clearest example of a geopotential surface is the oblate spheriod shape of the Earth's world-wide sealevel: everywhere a plumb line is perpendicular to it.

In general: in orbital mechanics the gravitational potential energy is the primitive of the inverse square force law. The shape of the planetary orbits are fully determined by that gravitational potential.

In meteorology we have that the gravitational potential energy at equator-sea-level is in the order of 108 KiloJoules per kg mass higher than the gravitational potential at polar-sea-level. There are graphs of that gravitational potential as a function of latitude. It's a gravitational well, but not with an inverse square force law. In the case of a parabolic dish the force law is a proportional to radial distance law. The graviational potential energy as a function of latitude is incorporated in the equations of motion of meteorological models, otherwise they wouldn't work.

There are many true things that can be said. But, true or false, the above has nothing to do with the Coriolis effect. Paul Beardsell 17:13, 23 August 2005 (UTC)

Comparison of inertial wind and ballistic trajectories

The ideal case for making the comparison is inertial wind, for in the case of inertial wind there is NO pressure gradient force involved.

On the northern hemisphere: when you start looking at a point in time that the inertial wind is moving exactly in east-to-west direction, then you see the inertial wind curve to the North.

Now a ballistic trajectory that starts in a direction that is east-to-west with respect to the Earth. Let the launching station be situated on 45 degrees latitude (northern hemisphere) There is exactly one great circle that is parallel to the latitude line at the location of the launch site. All projectiles are fired in a direction that is in the plane of that great circle. No matter what velocity the ballistic missile is fired with, if it is launched from 45 degrees latitude, and if its direction on launch is in the plane of the aforementioned great circle then it will never land further away from the equator than 45 degrees latitude.

A missile is in the air for an hour or two. In that time the Earth revolves at most 30 degrees. Just how much Coriolis effect could you actually see? Paul Beardsell 17:46, 23 August 2005 (UTC)

Air parcel in inertial oscillation as seen from non-rotating point of view. This picture could be improved by getting the lines of longitude to rotate with the revolving planet rather than staying stationary w.r.t. to the non-rotating observer.

Inertial wind is an oscillation of air parcels, and those air parcels are oscillating upwards and downwards with a particular latitude line as middle line.

Motion along a great circle that goes all the way around the earth can also be seen as an oscillation. Great circle oscillation always has the Equator als the middle of the oscillation.

This difference between the trajectory of inertial wind and ballistic trajectory is independent of velocity. There is no tweaking of the velocity available that makes the two trajectories coincide.

NONSENSE! If the wind was as fast as the missile, the wind would cross the equator, just like the missile. Paul Beardsell 16:53, 23 August 2005 (UTC)

I agree with PB. The dynamics of inertial wind and ballistic trajectories (projected onto the sfc plane, of course) are the same. All that is different is the speeds involved. William M. Connolley 21:55:34, 2005-08-23 (UTC).

You are being evasive. You have stated several times that if the ballistic motion would be slow enough the trajectories would be the same. But no matter how slow you make it, the trajectories do not coincide. If ballistic trajectory and inertial wind trajectory would be the same physics, then the trajectory would be the same for all velocities.

No, the trajectory is velocity dependent. The circle radius is u/f. William M. Connolley 21:55:34, 2005-08-23 (UTC).

Yes, the same. But there are other NON-CORIOLIS forces at play. The ballistic missile is subject to gravity. Any pocket of air is floating and therefore (by the definition of "floating" or, if you prefer, by Archimedes) is not subject to gravity. Gravity and the Coriolis effect are unrelated. Gravity is far stronger than the Coriolis "force" and Coriolis has a small effect on the missile. Not so for air. But the physics is the same. And there are other forces as strong as Coriolis acting on the air! But the deviation observed between the non-rotating and the rotating frame is described by the same maths term in all cases. ONE CORIOLIS EFFECT. Paul Beardsell 17:39, 23 August 2005 (UTC)

Of course there is only one way to perform a transformation to a rotating coordinate system. No matter what physics you are looking at, the transformation procedure is always the same. --Cleon Teunissen | Talk 17:56, 23 August 2005 (UTC)

And that transformation is 100% the Coriolis effect and the Coriolis effect is 100% that transformation. Paul Beardsell 18:02, 23 August 2005 (UTC)

This is conclusive evidence that the trajectory of inertial wind and ballistic trajectory cannot be described with the same equation of motion. In meteorology, geopotential height is incorporated into the equation of motion.

Notice that you started at the top saying in the case of inertial wind there is NO pressure gradient force involved.. That was correct. Now you've forgotten it. William M. Connolley 21:55:34, 2005-08-23 (UTC).

In ballistics the equation of motion just needs the inverse square law of gravitation. The two equations of motion are indeed different. -Cleon Teunissen | Talk 15:43, 23 August 2005 (UTC)

But the correction required in each example to get from the non-rotating observations to the rotating observations is the same. The correction is the Coriolis effect. The other forces are not the Coriolis effect. This article is about the Coriolis effect. Enough! Paul Beardsell 16:49, 23 August 2005 (UTC)

---

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. The extra term required in the rotating frame is called the Coriolis force and this term is the same in all examples including both ballistic missile and atmospheric wind. Any other phenomenum is not the Coriolis effect. Paul Beardsell 16:43, 23 August 2005 (UTC)

---

Cleon continues to argue with William as though the above ever-so-pertinent point has not been made. Cleon, do you agree with the above paragraph or not? Paul Beardsell 21:53, 24 August 2005 (UTC)

Of oourse there is only one way to perform a transformation to a rotating coordinate system! How many times do I need to assert that my point is not about coordinate transformation. There is precisely one way to perform a transformation to a rotating coordinate system. --Cleon Teunissen | Talk 06:17, 25 August 2005 (UTC)

It looks like a second article is needed

I don't think so. Paul Beardsell 17:30, 23 August 2005 (UTC). Cleon cut'n'pasted the following here:

The other forces are not the Coriolis effect. This article is about the Coriolis effect. Paul Beardsell 16:49, 23 August 2005 (UTC)

Thirty of the articles that link to the coriolis effect article are meteorological articles, and they expect an article that delivers the goods.

Months ago I proposed a split: one article just about the transformation to a rotating coordinate system aspect, and another article that caters to the meteorologists. --Cleon Teunissen | Talk 17:17, 23 August 2005 (UTC)

Possibly in the other 30 articles people are calling things the Coriolis effect which are NOT the Coriolis effect. Or they say things are due to the Coriolis effect when it is only partly the cause. Sounds to me that the 30 articles which refer to this one should be checked for accuracy first. 17:25, 23 August 2005 (UTC)

Cleon, what would the "second article" be called? Paul Beardsell 17:25, 23 August 2005 (UTC)

Well, what that article would be about is what rotational dynamics makes the winds flow parallel to pressure gradients rather than straight towards a low/high pressure area.

Ha! Cause and effect! The reason for this is that winds which do NOT flow parallel to the gradient reduce the gradient causing the winds to reduce and then stop (except for some inertia). Any LONG DURATION WIND MUST BE PARALLEL TO THE GRADIENT. Otherwise there is soon NO WIND! The cause for it goes. Paul Beardsell 17:54, 23 August 2005 (UTC)

Traditionally the meteorologists call that "the coriolis effect". As far as I can tell the meteorological community never really verified whether it was appropriate to call it "the coriolis effect", they probably just went ahead. Now we're stuck with that development.

I guess there is no other option but to name the article that caters to the meteorologists: "Coriolis effect in meteorology".

Personally I would prefer to start using the expression coriolis effect as a sort of umbrella for the different meanings it has in different scientific and engineering communities. --Cleon Teunissen | Talk 17:42, 23 August 2005 (UTC)

I rather rudely hold the opinion that it has the same meaning to all those who are scientifically literate. That has been tested to include lots of engineers and to exclude lots of so-called scientists. Paul Beardsell 17:48, 23 August 2005 (UTC)

The pattern of wind deflection is a theorem of newtonian dynamics

I copy and paste from the previous section:

Ha! Cause and effect! The reason for this is that winds which do NOT flow parallel to the gradient reduce the gradient causing the winds to reduce and then stop (except for some inertia). Any LONG DURATION WIND MUST BE PARALLEL TO THE GRADIENT. Otherwise there is NO WIND! The cause for it goes. Paul Beardsell 17:54, 23 August 2005 (UTC)

This is incredible. You now demonstrate that you haven't got the foggiest idea what meteorology is about.

There is nothing random about the deflection of the winds. The rearrangement that is very very simplified represented in the image always occurs. It is a theorem of newtonian dynamics that such a rearranging occurs, just as in celestial mechanics gyroscopic precession of the Earth is a theorem of newtonian dynamics. --Cleon Teunissen | Talk 18:07, 23 August 2005 (UTC)

Drop as many names and as much scientific terminology as you like. Of course my remark was made slightly tongue in cheek but only slightly. What I said is correct. If the wind was not parallel then the low pressure would be filled. Then there would be no low pressure. No wind. Paul Beardsell 18:10, 23 August 2005 (UTC)

Why, Cleon, doesn't the wind go straight to the low pressure centre, twisting just a little? Becuase at a certain radius the speed is just right for Coriolis to balnace the centrifugal force (both fictitious but you know what I mean). At a certain radius for a certain wind speed the winds are always parallel to the gradient. Other wind speeds and directions are not sustainable because they dissipate the pressure gradient. So they themselves dissipate leaving just the rotating winds and the remaining pressure difference. It's why the pressure differences are small and the winds are slow. Paul Beardsell 18:15, 23 August 2005 (UTC)

Incredibly, it seems you think of meteorological dynamics as a process with analogies to a process of darwinian selection. You seem to think that patterns only last when they happen to be in the right proportions to last. However, the physics involved in meteorology makes it inevitable that winds flow close to parallel to the pressure gradients; it's a mechanical process, just as in celestial mechanics and in thermodynamics. --Cleon Teunissen | Talk 18:35, 23 August 2005 (UTC)

Not for the first time you are putting words into my mouth. Darwin indeed! But there is no arguing with the assertion that things only last if they last! Why do we see only the resonant frequencies in a plucked string? It isn't because there weren't any other frequencies but rather that the other frequencies have been damped. Why is it that the only weather patterns which last for a week have winds parallel to the pressure gradient? Because any weather pattern without this character quickly dissipates. That which quickly dissipates doesn't last for a week. Paul Beardsell 21:25, 23 August 2005 (UTC)

The geopotential height in the equation of motion

I copy and paste from above:

Notice that you started at the top saying in the case of inertial wind there is NO pressure gradient force involved.. That was correct. Now you've forgotten it. William M. Connolley 21:55:34, 2005-08-23 (UTC).

Yes, I must correct myself there. What needs to be retained is the absense of pressure gradient in any direction parallel to the geopotential surfaces. It is inertial wind if there is no pressure gradient in any direction parallel to the geopotential surfaces

Then there is that at the Equator the gravitational potential energy is about 108 KiloJoule per kg mass higher than at the poles. In that sense the surface of the earth is a gravitational well. I had assumed that difference in potential energy is incorporated into the geopotential height. NASA JPL page about measuring altitude Geopotential height is calculated as height above sea level, sea level being the oblate spheroid. In that sense the geopotential height is part of incorporating the oblateness of the Earth into the equation of motion, but it is not a clearcut example so I will no longer present it as evidence.

I should not have claimed that the geopotential height alone does the job of incorporating the difference in gravitational potential. That is not the case. --Cleon Teunissen | Talk 06:11, 24 August 2005 (UTC)

Reprise: inertial wind trajectory and ballistic trajectory

I copy and paste from above:

The dynamics of inertial wind and ballistic trajectories (projected onto the sfc plane, of course) are the same. All that is different is the speeds involved. William M. Connolley 21:55:34, 2005-08-23 (UTC).

In the approximation of taking the Earth as a perfect sphere the prediction is that any motion of a parcel of fluid in a thin fluid layer (in the absence of pressure gradient) will be described with the same equation of motion as a satellite orbit that is perfectly circular. The intersection of the plane of the satellite orbit with the Earth's surface is a great circle

Nonsense. Where *do* you get this from? William M. Connolley 08:42:10, 2005-08-24 (UTC).

In the approximation of taking the Earth as a perfect sphere the only inertial oscillation in the thin fluid layer that is possible is an oscillation in which the parcels of fluid follow great circles, which means that always the equator will be the midline of the oscillation. (Equator remains midline of oscillation in transforming to rotating coordinate system that rotates around the Earth's axis.)

But the equations of motion that are used in meteorology predict correctly that inertial oscillations occur with a particular latitude line as midline.

Throw a small object in east-to-west direction with a velocity of about 10 meters per second, and have a weather balloon be floating along with 10 meter per second inertial wind, flowing in east-to-west direction. Track both of them. The thrown object lands closer to the equator than the latitude it was thrown from, the inertial wind in east-to-west motion has, in the same amount of time, curved to the nearest pole. Same speed, same initial direction: different trajectory, different equation of motion.
--Cleon Teunissen | Talk 06:16, 24 August 2005 (UTC)

---

NB: CT has moved (I've now restored it, so its a copy) this comment of mine from the context above:

Nonsense. Where *do* you get this from? William M. Connolley 08:42:10, 2005-08-24 (UTC).

This is the physics of motion. These are facts of motion that are always observed.

Look at all of the orbits of the satelites that are orbiting Earth. Take for each satellite orbit the vector of its angular momentum. None of those vectors rotates with respect to any of the other vectors. For all satellite orbits we have that the vector of its angular momentum keeps pointing at the same place among the fixed stars.

It is for example impossible for a satellite to be simultaneaously in circumpolar orbit, and trace a path that tracks exactly over the greenwich meridian and the 180 degrees meridian. It is not possible to set up a satellite orbit in such a way that the plane of its orbit rotates with respect to the fixed stars. (Not counting tidal effects for now, that are very small.) It just doesn't happen.

In the case of motion over the surface of a perfect sphere, the only inertial motion over that surface is motion along a great circle. In inertial motion, the plane of that great circle does not rotate with respect to the fixed stars.

These are among the most basic facts of the physics of motion. Anybody who wants to contribute to physics articles should first learn those facts. --Cleon Teunissen | Talk 09:46, 24 August 2005 (UTC)

CT, this is hopeless. You need to learn to write equations, not words. But most importantly, I don't see any of this helping the page. Isn't this all a bit self-indulgent? We're just having a (not very polite) discussion with no obvious purpose. William M. Connolley 11:19:46, 2005-08-24 (UTC).

Evidence is piled up

Clearly, the coriolis effect Talk page is a disaster area now.

William, the problem is that you keep avoiding the issue. The assumption that you have embraced has been falsified, but you refuse to recognize that.

You have stated:

The dynamics of inertial wind and ballistic trajectories (projected onto the sfc plane, of course) are the same. All that is different is the speeds involved. William M. Connolley 21:55:34, 2005-08-23 (UTC).

That statement has been shown to be in error.

I'm trying to explain these matters you, but you are systematically looking the other way. I have provided mathematical derivations in the version I wrote, Dale R Durran provides a mathematical discussion, Norman A. Phillips provides a mathematical discussion. Evidence is piled up, but you keep looking the other way.

So yes, the situation looks hopeless. --Cleon Teunissen | Talk 12:20, 24 August 2005 (UTC)

Yes I've said that, no you haven't shown it to be in error. I say the dynamics is that the acceleration (horizontally) is given by f*k cross u, in the case of (inertial) wind or ballistics. What do you say the dynamics is? No handwaving words, equations please. William M. Connolley 12:35:33, 2005-08-24 (UTC).

OK. Inertial motion following a perfect sphere: a parametric equation for inertial motion as a function of time, along the equator, in cartesian coordinates, non-rotating point of view:

$x=R_{e}*\cos {(qt(2\pi /24))}$

$y=R_{e}*\sin {(qt(2\pi /24))}$

Where $R_{e}$ is the radius of the Earth. x and y in the equatorial plane. t stands for time in hours and q is a factor for the velocity: for q = 1 the great circle is traveled in 24 hours; for q = 2 the great circle is traveled in 12 hours.

To tilt that great circle 45 degrees with respect to the equator (x-axis as axis of the tilting:

$x=R_{e}*\cos {(qt(2\pi /24))}$

$y={\sqrt {2}}*R_{e}*\sin {(qt(2\pi /24))}$

$z={\sqrt {2}}*R_{e}*\sin {(qt(2\pi /24))}$

A transformation to a rotating coordinate system (rotating around the Earth's axis) will affect the x- and y-coordinate, but it will leave the z-coordinate unaffected.

Any great circle on a perfect sphere remains an oscillation with the equator as the midline under Earth-axis coordinate transformation to rotating coordinate system.

In the case of inertial motion along the surface of a perfect sphere there is a centripetal acceleration: acceleration towards the center of the sphere.

$a=\omega ^{2}r$.

Where $\omega$ is the angular velocity.

That inertial motion over the surface of a perfect sphere is always a great circle is an elementary theorem of mechanics. (Of course an accelerometer will register the presence of gravitation. In moving over a surface 'inertial motion' refers to not measuring any acceleration in the 2 degrees of freedom that are left: moving over the surface.)
--Cleon Teunissen | Talk 14:46, 24 August 2005 (UTC)

Great circles David McIntyre discusses at great length that in the case of motion over the surface of a perfect sphere the inertial motion is uniform motion along a great circle
McIntyres animations
--Cleon Teunissen | Talk 16:39, 24 August 2005 (UTC)

I think you've misunderstood my question. I asked about the dynamics: ie, the equation of motion. Not the *path* of motion. Thats what I gave: the acceleration is k*f cross u. You haven't given the acceleration/force term. An analogy, if it helps: under gravity, the *acceleration* is g. But the path is y=t^2. If you think that ballistics and inertial wind have different dynamics, then their equations of motion are different. If so, that means you know what the equations are, and how they differ. No? William M. Connolley 16:52:19, 2005-08-24 (UTC).

And... AFAIK I have no objection to the idea that inertial circles on a sphere are great circles (from a non-rotating viewpoint). It seems quite reasonable. What have I ever said that makes you think I doubt it? And what has it got to do with the difference between inertial circles and ballistics? William M. Connolley 16:58:54, 2005-08-24 (UTC).

A pic: inertial circles

Inertial circles

I don't know if this helps, but I drew a picture of various inertial circles. The program to draw it (in IDL) is at http://www.antarctica.ac.uk/met/wmc/cor.pro. Disclaimer: done is a hurry, uses crude time-stepping, etc etc. William M. Connolley 12:29:38, 2005-08-24 (UTC).

Nice pic. And the red line showing the $\beta$-effect. (Anders Persson describes the $\beta$-effect as: 'at the higher latitude the coriolis effect is stronger, so you get stronger curving there, resulting in an overal westward transportation.)

I have no programming experience; I use Excel and Paint shop Pro. To make an animation I imagine the entire animation in my head, then set up sines and cosines in Excel to get the coordinates for each consecutive frame, and then I manufacture the frames.

Of course, in actual meteorology winds are not swept up to the velocities that would hurl them all the way to the other hemisphere. The interesting velocity is typical wind velocity: the red line. -Cleon Teunissen | Talk 14:23, 24 August 2005 (UTC)

Perfect spheres and great circles

I copy and paste from above:

[...] I have no objection to the idea that inertial circles on a sphere are great circles (from a non-rotating viewpoint). It seems quite reasonable. [...] William M. Connolley 16:58:54, 2005-08-24 (UTC).

Your caution is somewhat hilarious. It is an elementary theorem of the physics of motion, as elementary as, say, Pythagoras' theorem and you very very carefully say that "it seems quite reasonable".

Anyway, this is good news. Effectively you concede that the equation of motion that is used in meteorology is tailored to take the oblateness of the Earth into account. The equation of motion that is used in meteorology predicts correctly that on each hemisphere inertial oscillations are confined to that hemisphere.

Inertial motion over a perfect sphere (uniform motion along a great circle) always crosses to the other hemisphere. It's not a matter of how much velocity, for the shape of the great circle is independent of velocity. For every velocity, slow or fast, the inertial motion on the surface of a perfect sphere oscillates from hemisphere to hemisphere. --Cleon Teunissen | Talk 06:20, 25 August 2005 (UTC)

The position on the Z-axis is invariant under the coordinate transformation

I copy and paste from above:

[...] I say the dynamics is that the acceleration (horizontally) is given by f*k cross u, in the case of (inertial) wind or ballistics. What do you say the dynamics is? [...] William M. Connolley 12:35:33, 2005-08-24 (UTC).

I assume that you mean by horizontal: the plane tangential to the local surface. The acceleration In the meteorological equation of motion: $a=f*(\omega \times v)$

Let's take a look at the coordinate transformation again: the axis of rotating transformation is the Earth's axis. Let the equatorial plane be the x-y-plane, let the Earth's axis be the z-axis. The coordinate transformation is around the z-axis, so the coordinate transformation will only affect the x- and y-coordinates, the position of the air mass on the z-axis is invariant under the coordinate transformation.

So how come the meteorological equation of motion contains a term that correlates a velocity parallel to the plane of the equator to motion along the z-axis? The position on the z-axis is invariant under the coordinate transformation.

The equation of motion says: on the northern hemisphere, when inertial wind is moving from east-to-west with respect to the Earth, it will proceed to curve to the north.

Inertial oscillation on oblate spheroid as seen from non-rotating point of view

The meteorological equation of motion for the rotating coordinate system has that compact vector cross product. The component of the outcome of that vector cross product parallel to the equator is consistent with the coordinate transformation, the component of that vector cross product that is a vector for acceleration up the z-axis is not related to coordinate transformation. That component is there to take the oblateness of the Earth into account. That acceleration is the centripetal pulling action of the centripetal component of gravitation. That is why the inertial oscillation on the oblate spheroid is not a great circle.

That brings me to the original question: what will the equation of motion for inertial motion over a perfect sphere look like? For one thing: it will not feature that vector cross product, for on a perfect sphere there is only the spherically symmetrical force of gravitation, that is completely canceled by the normal force.

As seen from a non-rotating point of view, there is no acceleration parallel to the local surface. An equation for acceleration parallel to the local surface would look something like: $a=0$. The motion along the great circle is exhaustively described by giving the position as a function of time; the shape is independent of the velocity.

You can transform that function to a rotating coordinate system. That will affect the x- and y- coordinate as a function of time, the z-coordinate is invariant under that rotating transformation. --Cleon Teunissen | Talk 06:27, 25 August 2005 (UTC)

I recommend...

This: http://www.physics.oregonstate.edu/paradigms/Publications/GreatCircles.html. I think CT may have used it. Just before section IV:

discuss the consequences of the oblateness of the real earth... the shape of the earth is such that particles at rest with respect to the rotating earth remain at rest. For the earthbound observer, this means that there is no net force on the stationary hockey puck, and hence the centrifugal deflections shown in Fig. 3 are not present on an oblate earth. For the inertial observer, the net force on a hockey puck at rest with respect to the oblate earth is toward the axis of rotation, causing the puck to travel in a circle along its line of latitude; in contrast to the net radial force on a spherical earth and the resultant great circle path... In the rotating frame, the deflection of the puck from its intended target is due only to the Coriolis force. This is why discussions of the effects of rotation upon the weather, ocean currents, and rivers on our oblate earth invoke only the Coriolis force.

There is only one way to perform a transformation to a rotating coordinate system

What McIntyre writes about the oblateness is the same as what Dale R. Durran, Norman A Phillips and Anders Persson write.

For the inertial observer, the net force on a hockey puck at rest with respect to the oblate earth is toward the axis of rotation, causing the puck to travel in a circle along its line of latitude; in contrast to the net radial force on a spherical earth and the resultant great circle path. The net centripetal force on the puck is matched to the rotation frequency omega of the earth, but is not matched should the puck rotate at a different frequency. This concept provides a common explanation of the Coriolis force from the inertial viewpoint. A hockey puck launched to the east has an inertial velocity faster than that of the local surface of the earth. The net centripetal force that kept it at rest before is now not sufficient to keep the puck traveling in a circle at this speed, so the puck moves to a larger radius where there is a smaller centripetal acceleration $v^{2}/r$. The eastward-launched puck thus moves southward, explaining the rightward Coriolis deflection (in the Northern Hemisphere). A puck launched to the west is traveling too slowly and moves to a smaller radius where there is a larger centripetal acceleration. The westward-launched puck moves northward, again to the right (in the Northern Hemisphere). In the rotating frame, the deflection of the puck from its intended target is due only to the Coriolis force.

What we have that there is a difference in the physics of frictionless motion over the surface of a perfect sphere on one hand, and over an oblate spheroid on the other hand.

In the case an oblate planet (a rotating planet) there is no spherical symmetry but cilindrical symmetry instead. In the case of an oblate spheroid there is for buoyant objects a component of gravitation perpendicular to the axis of rotation of the oblate planet. (For objects in any kind of orbit the decomposition does not apply).

The last remark of David McIntyre is clearly a mistake; effectively he claims there are two different coriolis effects, which is wrong.

Transformation to a rotating coordinate system preserves all relative positions, preserves all relative velocities, preserves all relative accelarations. Any force that is present is present in all coordinate systems, the only thing that may change under coordinate transformation is the representation of the force.

For example, an observer is standing on a merry-go-round, holding on to a pole. The pole bends somewhat. The pole is exerting a centripetal force on the observer, maintaining the circular motion of the observer. The amount of centripetal force that the pole exerts on the observer can be seen from the amount of centripetal acceleration. The observer is exerting a force in centrifugal direction on the pole, bending the pole. The amount of force that the pole exerts on the observer, and the amount of force that the observer exerts on the pole is the same (Newtons third law.)

The difference between the observer and the pole is that the observer is not attached to anything else so the force exerted on the observer pulls him in circular motion. The pole on the other hand is attached to the big and heavy merry-go-round; the pole bends a little, but that's it.

No matter what transformation you apply, the amount of bending of the pole is always the same: the same force is present in all coordinate systems.

In the case of motion over the surface of an oblate spheroid, the component of gravitation perpendicular to the axis of rotation is the same in all coordinate systems, the only thing that changes under coordinate transformation is the representation of the force.

McIntyre's mistake is that he effectively claims there are two different coriolis effects. That makes no sense; there is only one way to perform a transformation to a rotating coordinate system. --Cleon Teunissen | Talk 00:51, 26 August 2005 (UTC)

In the case of meteorology on Earth: the shape of the Earth and hence the component of gravitation perpendicular to the Earth's axis are cilindrically symmetrical with respect to the Earth's axis. That is convenient, for the transformation to the rotating coordinate system is around the Earth's axis too. So in this case the shape of the gravitional potential is invariant under the coordinate transformation that is performed. --Cleon Teunissen | Talk 01:08, 26 August 2005 (UTC)

Reprise (2): inertial wind trajectory and ballistic trajectory

I copy and paste from above:

[...] I have no objection to the idea that inertial circles on a sphere are great circles (from a non-rotating viewpoint). [...] And what has it got to do with the difference between inertial circles and ballistics? William M. Connolley 16:58:54, 2005-08-24 (UTC).

It is unclear why you ask the question about the difference between ballistics, and inertial oscillations over the surface of an oblate spheroid. The answer to that is so elementary that any mathematician will immediately see it.

From a glossary
Orbit
The path of a body acted upon by the force of gravity. Under the influence of a single attracting body, all orbital paths trace out simple conic sections. Although all ballistic or free-fall trajectories follow an orbital path, the word orbit is more usually associated with the continuous path of a body which does not impact with its primary.
[end quote]

Ballistic trajectories are part of the same mathematical group of trajectories as satellite orbits. The shape of any ballistic trajectory is an ellipse (conic section), with the center of gravitation of Earth at one focus of that ellipse. The plane of the trajectory does not rotate with respect to the inertial frame of reference

If you intersect the plane of the ballistic trajectory with the surface of the Earth you get a great circle.

I have credited Paul Beardsell for pointing out the background of the difference. Objects that are in any form buoyant are moving parallel to a geopotential surface. By reasonable approximation, buoyant objects are free to move in two dimensions only. Ballistic trajectory is freedom in all three dimensions of space.

For motion parallel to a geopotential surface of a rotating oblate planet the shape of the gravitational potential is cilindrically symmetrical. For orbital motion the shape of the gravitational potential is (by good approximation) spherically symmetrical. --Cleon Teunissen | Talk 06:33, 27 August 2005 (UTC)

See also:
Going Ballistic. By Neil deGrasse Tyson, From Natural History magazine, November 2002

In physics there is one inertia

In his version of the coriolis article WMC has written:

In an inertial circle, the force balance is between the centrifugal force (directed outwards) and the Coriolis force (directed inwards).

I commented:

So what is the story of inertia tugging outwards (centrifugal) and inwards (coriolis) simultaneously? (in the case of inertial oscillation) Does William M Connolley believe that inertia is tugging at the parcel of fluid in two opposite directions simultaneously? Does William M Connolley believe there are two kinds of inertia that can oppose each other?--Cleon Teunissen | Talk 14:07, 22 August 2005 (UTC)

WMC replied:

Its what the equations say. [...] William M. Connolley 14:24:31, 2005-08-22 (UTC).

William, you are here effectively proposing a novel theory of physics. There is no reason to suggest there is more than one inertia. In the equation of motion for a rotating coordinate system the acceleration term, coriolis term and the centrifugal term together represent the direction of inertial motion. In the equation of motion it is the factor $m$ that represents the inertia. The vectors do not represent the inertia, they represent direction. There is no such thing as the coriolis force and centrifugal force opposing each other, there is always only one inertia.

Let there be two coordinate systems (1) and (2) rotating relative to each other, with corresponding expressions for F(1) and F(2).

$m\mathbf {a} _{\mathrm {1} }=m\mathbf {a} _{\mathrm {2} }-m{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \mathbf {r} )-2m{\boldsymbol {\omega }}\times \mathbf {v} _{\mathrm {1} }$

which means:

$m\mathbf {a} _{\mathrm {1} }=m{\Big (}\mathbf {a} _{\mathrm {2} }-{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \mathbf {r} )-2{\boldsymbol {\omega }}\times \mathbf {v} _{\mathrm {1} }{\Big )}$

Inertia is the tendency to keep moving in the same direction, and that is always a single direction. Look at what the equation says: for a rotating coordinate system you add the vectors of the various term, and the resultant vector gives that single direction. --Cleon Teunissen | Talk 16:41, 27 August 2005 (UTC)

Revert of august 28th.

I have reverted the article to a somewhat rewritten version of my august 18th version.

My version of the coriolis effect article is based on the work of the meteorologists mentioned in the references section.
--Cleon Teunissen | Talk 19:38, 28 August 2005 (UTC)

I've reverted to mine. What did you expect? We seem to spend a lot of time talking at cross purposes, but there was clearly no agreement to revert to yours, and at least some other people liked mine [1] a lot.

I was Paul Beardsell who wrote that. Let me copy and paste some things that Paul Beardsell wrote, to give you an idea of his level of understanding of meteorology:

Ha! Cause and effect! The reason for this is that winds which do NOT flow parallel to the gradient reduce the gradient causing the winds to reduce and then stop (except for some inertia). Any LONG DURATION WIND MUST BE PARALLEL TO THE GRADIENT. Otherwise there is NO WIND! The cause for it goes. Paul Beardsell 17:54, 23 August 2005 (UTC)

Why, Cleon, doesn't the wind go straight to the low pressure centre, twisting just a little? Becuase at a certain radius the speed is just right for Coriolis to balnace the centrifugal force (both fictitious but you know what I mean). At a certain radius for a certain wind speed the winds are always parallel to the gradient. Other wind speeds and directions are not sustainable because they dissipate the pressure gradient. So they themselves dissipate leaving just the rotating winds and the remaining pressure difference. It's why the pressure differences are small and the winds are slow. Paul Beardsell 18:15, 23 August 2005 (UTC)

It is quite clear that Paul Beardsell doesn't understand geostrophic adjustment, that he doesn't know what meteorology is about. In my judgement Paul's preference for your version doesn't carry weight.

I told you an elementary consequence of the physics laws of motion: that inertial motion over the surface of a perfect sphere always follows a great circle, and your initial response was: Nonsense. Where *do* you get this from?

I did. I was wrong. Sorry. On a sphere, particles moving in the absence of forces other than those normal to the sfc move in great circles. William M. Connolley 19:35:13, 2005-08-29 (UTC).

No need to say sorry. The next steps are to follow the ramifications.

The logical inference is that in the case of inertial oscillations a force is involved, making the air parcels follow a non-great-circle trajectory. That other force is gravitation, in this case a component of gravitation, providing a centripetal force. --Cleon Teunissen | Talk 21:08, 29 August 2005 (UTC)

That illustrated sharply that there are things that you still need to learn. If it would be about a difference of opinion I would yield. But this is about logic, the logic of the laws of motion. I choose to uphold logic.

I shall have to be patient. I hope that at some point in time some other physicists will get involved, and then hopefully the matter can be settled. --Cleon Teunissen | Talk 19:00, 29 August 2005 (UTC)

The laws of rotating motion applied to a motion over a Velodrome

This website offers picturs of a velodrome

A velodrome is a racetrack specifically for bicycle races. A velodrome with a track length of 250 metres usually has straights of about 60 metres long, and U-turns with an inner radius of about 20 metres, and about 65 metres in length. The banking of the turns is 42 degrees.

The motion of the cyclists in taking the U-turns is described by the equations of rotating motion. There is a close analogy between the rotating motion in the curves of a velodrome and motion of fluid parcels of a thin fluid layer on a parabolic rotating turntable.

First calculation: at what velocity are the cyclists perpendicular to the banking of the curve, if they follow a curve with a radius of 24 metres?

The cosine of 48 degrees is 0.67, so for every kilogram of mass there will be 0.67 * 9.8 = 6.6 Newton of centripetal force. This amount of centripetal force is a given, it is exclusively determined by the slope of the banked curve.

$a=v^{2}/r$
$\Leftrightarrow v={\sqrt {a*r}}={\sqrt {158.4}}$ = 12.6 m/s, about 45 km/h (28 miles/h)

Theoretically it is possible to go around the track of the velodrome with a smal hovercraft, but not at any velocity. The only way a hovercraft can make it to the other end of the U-turn is to go at exactly that velocity of 12.6 meters per second. Slower than that and gravitattion will pull the hovercraft down the banked curve, faster than that and the hovercraft will swing wide, climbing up the banked curve.

Second calculation: suppose the hovercraft can exert a sideways thrust, how much sideways force must the hovercraft exert to stay on the 24 meter radius while going either faster or slower than the equilibrium angular velocity?

$F_{v}$ stands for the total centripetal force, that must be exerted to maintain circular motion with radius $r$ and velocity $v$.
$F_{e}$ stands for the centripetal force from the Earth's gravitation. $v_{e}$ stands for the equilibrium velocity. $\omega _{e}$ stands for the corresponding angular velocity. $v_{e}$ stands cor the equilibrium velocity for staying on the 24 metre radius all through the curve. $\Delta v$ stands for the difference between the actual velocity and the equilbrium velocity.

$F=F_{v}-F_{e}=-m(v_{e}+\Delta v)^{2}/r-(-mv_{e}^{2}/r)$

$=-m((v_{e}+\Delta v)^{2}-(v_{e})^{2})/r$

$=-m(v_{e}^{2}+2v_{e}*\Delta v+(\Delta v)^{2}-v_{e}^{2})/r$

$=-2m(v_{e}/r)*\Delta v-m(\Delta v)^{2}/r$

$=-2m(\omega _{e}*\Delta v)-m(\Delta v)^{2}/r$

This formula also indicates with what radial acceleration the hovercraft will slide down the bank or climb up should the supporting sideways thrust suddenly fail.

This calculation also applies in the case of rotating motion over the surface of a parabolic dish. On a parabolic dish $\omega _{e}$ is the same for every radius of motion. In the case of motion of a parcel of fluid of a thin fluid layer on a rotating turntable: if the angular velocity (with respect to the main axis of rotation) is faster than equilibrium velocity the fluid parcel will slide up the slope, if the angular velocity is slower than the equilibrium velocity gravitation will pull the fluid parcel down. --Cleon Teunissen | Talk 19:13, 29 August 2005 (UTC)

What does all that add to the existing page text To demonstrate the Coriolis effect, a turntable can be used. If the turntable is flat, then the centrifugal force, which always acts outwards from the rotation axis, would lead to objects being forced out off the edge. If the turntable has a bowl shape, then the component of gravity tangential to the bowl surface will tend to counteract the centrifugal force. If the bowl is parabolic, and spun at the appropriate rate, then gravity exactly counteracts the centrifugal force and the only net force (bar friction, which can be minimised) acting is then the Coriolis force. William M. Connolley 19:41:16, 2005-08-29 (UTC).

Clearly you prefer your version to the current, my, version. Do you think that there are actually any *errors* in the current version, and if so, what is the *first* error? William M. Connolley 19:41:16, 2005-08-29 (UTC).

What I see as the principle error in WMC's version

Observation:
The hovercraft going through a banked velodrome curve, being pulled down if it is too slow is a real deflection. Note that it is a deflection in the up-down direction.

Think of it in cartesian coordinates, with the x-y-plane in the plane of the velodrome, and the Z-axis vertical. perform a transformation to a rotating coordinate system around the Z-axis. That transformation does not alter the position on the z-axis. That transformation only alters x and y-coordinate values.

WMC's version of the of the coriolis effect article is contradicting itself. In the opening section it is stated that coriolis effect is about apparent deflection. More specifically: in the case of the examples discussed it refers exclusively to apparent deflection of position parallel to the x-y-plane. (the rotation being around the Z-axis) But in the sections about the rotating parabolic turntable and the rotating Earth WMC's version is suddenly talking about deflections up and down!

I dislike that bit about apparent in the intro, its only there to keep JB [sorry: I mean PB] happy. As far as I'm concerned, the deflection in the rotating system is real, as measured in the coordinates of that system. I'm still having trouble understanding what you think is wrong: clearly, something in the parabolic turntable section. Please quote the specific text that you think is wrong. William M. Connolley 08:56:21, 2005-08-30 (UTC).

To clarify that: PB and I agree on the physics, we just disagree slightly over the wording. Its no big thing.

William, you are avoiding the issue. The up/down deflection is seen both from a non-rotating point of view and from a rotating point of view. The up/down deflection is invariant under the coordinate transformation.

The the word apparent applies when an obvserver viewing from a non-rotating perspective sees something different from what an observer viewing from a rotating perspective sees. Both observers see the same amound of up/down deflection'; the up/down deflection is real. --Cleon Teunissen | Talk 10:49, 30 August 2005 (UTC)

Soprry, still don;t know which bit you mean. Can you actually quote the bit of text from the parabolic turntable section that you think is wrong, please, as requested above. William M. Connolley 12:40:29, 2005-08-30 (UTC)

Also: I asked Anders P to look over the article. It was OK by him. William M. Connolley 10:21:09, 2005-08-30 (UTC).

I have introduced the velodrome example to make comparisons. In the case of a cyclist following through the banked curve at 45 km/h (28 miles/h) he is perpendicular to the banking. When the cyclist is perpendicular to the banking of the curve, gravitatian is providing all the necessary centripetal force, the tires do not have to provide any sideways grip then.

Next I shifted to virtually frictionless motion over the velodrome: the whacky idea of going around it in a single-person hovercraft.

The logical next step is to cut out the straights that connect the curves, and making the banked curves parabolic, and then you have a parabolic dish.

These shifts are inconsequential for the physics taking place. Obviously the same physics must be going on in the various compared examples.

If the angular velocity of the hovercraft drops then gravitation will instantly pull it down, and that real deviation is unrelated to coordinate transformation. The larger the deviation from equilibrium velocity, the stronger the tendency for up/down deflection. --Cleon Teunissen | Talk 21:14, 29 August 2005 (UTC)

Motion that is the same as seen from rotating and non-rotating poing of view

I copy and paste from above:

Sorry, still don't know which bit you mean. Can you actually quote the bit of text from the parabolic turntable section that you think is wrong, please, as requested above. William M. Connolley 12:40:29, 2005-08-30 (UTC)

In the article it is stated that the center of the turntable that is used is a centimeter deeper than the rim. So in moving away from or toward the center the position of the puck with respect to the Z-axis changes.

OK, that sounds about right. Of course when looking at the motion (in the rotating frame) we're thinking in a 2-d system. William M. Connolley 18:35:44, 2005-08-30 (UTC).

You present the image to the right as the pattern of motion of inertial oscillations as seen from a co-rotating point of view.

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

The motion depicted on the picture also has a component parallel to the Z-axis, the Earth's axis. The amount of motion with respect to the Z-axis is the same both as seen from a non-rotating point of view and as seen from a rotating point of view, which means that it is a real motion, not an apparent motion in the sense of transformation to a rotating frame of reference. --Cleon Teunissen | Talk 14:15, 30 August 2005 (UTC)

Could be. I'm still not sure what you think is actually *wrong* with the page. Are you disliking the use of "real" and "apparent"? Is that all?

Teaching CT very elementary politeness

Do make an effort, old fruit: deliberate impoliteness isn't going to help you. William M. Connolley 18:35:44, 2005-08-30 (UTC).

Yeah, I know, it doesn't help. It's part of the choreography between us. You avoid answering questions of mine to you, I keep answering every question you ask, and you keep asking me questions that you can easily figure out yourself. You keep asking trivial question, and the result is that the real issues get buried under piles of words. --Cleon Teunissen | Talk 20:24, 30 August 2005 (UTC)

Let there be a cartesian coordinate system C, non-rotating, with the z-axis coinciding with the Earth's axis. Let there be a rotating cartesian coordinate system C', with the Z-axis coinciding with the Earth's axis. The angular velocity of C' with respect to C is $\omega$

Let the following three functions be a parametric equation of motion for coordinate system C of an object that moves over the surface of the Earth. (Like David McIntyre uses parametric equations to describe motion along a great circle)

$x=f(p)$

$y=g(p)$

$z=h(p)$

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

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

$x'=f'(p)=f(p)*cos\omega +g(p)*sin\omega$

$y'=g'(p)=g(p)*cos\omega -f(p)*sin\omega$

$z'=h(p)$

Oh great and wise maths teacher, should there not be some form of time dependence in these equations, since C' is *rotating*? cos(omega) is appropriate to a *fixed* rotation. I think you meant cos(omega*t). Then you need to relate t and p. But even with some such correction, the point of this is not at all obvious. William M. Connolley 18:35:44, 2005-08-30 (UTC).

I had specified that $\omega$ is to be read as an angular velocity. So it a stretch to suggest that the formula given appears to represents a single rotation event. If only a factor $\Omega$ would have been declared (as a particular angle) and if that would be the only factor in the formula, then the formula appears to refer to a single rotation event.

And yes, of course I should not have used $\omega$ in that parametric equation, but instead a factor $(\Omega *t)$; a particular angle multiplied with time. You could have easily made the correction yourself. From the context it is obvious that I am talking about a transformation to a rotating coordinate system, not a one-event rotation. It is petty to avoid the physics and to start picking on an easily corrected editing error. --Cleon Teunissen | Talk 20:24, 30 August 2005 (UTC)

But there is no physics, its just a coordinate transform. You made an elementary mistake, which is embarassing when you are pretending to try to teach elementary matters. William M. Connolley 20:45:15, 2005-08-30 (UTC).

The position with respect to the Z-axis is invariant under this coordinate transformation. For every moment t in time we have that z = z' = h(p).
When the word 'apparent' is used it refers to the fact that in general $x\neq x'$ and $y\neq y'$ It is this transformation to a rotating coordinate system (transformation around the Earth's axis) that David Mcintyre is depicting in his animations.

William, the above discussion of coordinate transformation is correct. Let's proceed from there. --Cleon Teunissen | Talk 14:15, 30 August 2005 (UTC)

It doesn't seem to be relevant. Notice that in the discussions of motion on a sphere, people use spherical coords, not cartesian, because its far more convenient. William M. Connolley 18:35:44, 2005-08-30 (UTC).

You are avoiding the issue. I was not discussing what coordinate system is used, I was discussing the type of transformation that is used by David McIntyre. The kind of transformation that he performs is a cilindrical transformation, so that the z-coordinate is invariant under the coordinate transformation. That is most clearly demonstrated with cartesian coordinates. --Cleon Teunissen | Talk 20:24, 30 August 2005 (UTC)

I'm telling you that I have no idea what issue you are trying to discuss. It appears to have no relevance to the page. Its was almost correct, but even when corrected appears irrelevant. Can you please get to the point, if you have one. William M. Connolley 20:45:15, 2005-08-30 (UTC).

The distinctiion between real motion and apparent motion

I define the expression 'real motion' as follows: the motion with respect to the local inertial frame of reference.

Ah, definitions, how splendid. You can define black as white if you like, but it will confuse people. I define 'real motion' as motion with repect to the local coordinate system. But this has very little to do with the article.

(Of course, both newtonian dynamics and relativistic dynamics recognize the principle of relativity of inertial motion (and for non-relativistic velocities the transformation between inertial frames of reference of special relativity reduces to the galilean approximation).

I define the expression 'apparent motion' as follows: apparent motion is motion as seen from a rotating point of view.

These definitions are unambiguous, for a rotating observer can always perform a local measurement, measuring his own angular velocity with respect to space.

The relevant 'local inertial frame of reference' for the purpose of meteorology is the inertial frame of reference that is co-moving with the center of mass of the Earth.

Although that system isn't inertial, of course, only approximately so. William M. Connolley 21:32:24, 2005-08-30 (UTC).

It should be noted that the equation of motion for a rotating coordinate system is itself referring to another coordinate system. The equation of motion for a rotating coordinates system contains a factor, usually represented with the letter $\omega$, that 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. There are no exceptions to this: the local inertial frame of reference is the true frame of reference, hence motion with respect to the local inertial frame is the real motion (while all the time observing the Principle of Relativity of Inertial Motion, of course.)
--Cleon Teunissen | Talk 20:50, 30 August 2005 (UTC)

---

I copy and paste from above:

I'm telling you that I have no idea what issue you are trying to discuss. It appears to have no relevance to the page. Its was almost correct, but even when corrected appears irrelevant. Can you please get to the point, if you have one. William M. Connolley 20:45:15, 2005-08-30 (UTC).

I did come to the point, months ago, but you kept asking trivial question, and the result was that the main issues got buried under piles of words.

Ooooooooooooooooooohhhh cutting. Check the pages: the piles of words are yours, not mine. William M. Connolley 21:34:35, 2005-08-30 (UTC).

My point: The inertial oscillations as recognized in meteorology are real motion, not apparent motion.

(Note that this image would be improved if the longitude lines rotated with the planet and did not stay stationary relative to the non-rotating observer.)

The component of the motion of the air parcels in north/south direction is invariant under the coordinate transformation, it is real motion.

The big difference between the inertial oscillations, and what is represented in the animations by Mcintyre, is that in McIntyres animations the great circles are the real motion (the motion with respect to the inertial frame of reference) and any apparent distortion of that great circle shape as a consequence of coordinate transformation to a rotating coordinate system represents addition of apparent motion.

The real motion of the air parcels moving in inertial oscillation is represented in the animation I made. That animaiton shows the motion as seen from a non-rotating point of view. --Cleon Teunissen | Talk 21:15, 30 August 2005 (UTC)

OK... so now your point is not the parabolic turntable, you've shifted: your point is that the inertial osc are real motion. Errr... fine. So what? Does the page say any different? I don't see any mention of this "real motion" stuff in the inertial circles section. Is your point that omitting this represents a crucial error? I wouldn't agree, if that is your point. William M. Connolley 21:32:24, 2005-08-30 (UTC).

I appears you now suddenly fault me for trying to reduce the flow of words. I did not explicitly mention parabolic turntables,

Ooooooooooooooooooohh yes you did: look at The laws of rotating motion applied to a motion over a Velodrome and following. William M. Connolley 22:20:35, 2005-08-30 (UTC).

and now you suddenly jump to a wrong assumption. You are giving a strong signal to me that I should not try to be economic with words, for you will immediately draw a wrong conclusion should I fail to mention any mere detail.

No, you certainly should be terse, because if you create huge piles of words people will miss things you consider important. William M. Connolley 22:20:35, 2005-08-30 (UTC).

The analogies between motion on a parabolic rotating turntable and motion on the rotating oblate spheroid that Earth are profound. That is why the rotating parabolic turntable is a good educational tool. In both cases the same logic applies.

Your version of the coriolis effect article gives a description that implies that as seen from a non-rotating point of view the observer would not observe a deflection of winds. However, such a thing is only the case for ballistic motion. In ballistic motion (not counting air resistence for now) the real motion is planar, and the trajectory is elliptical, with the center of mass of the Earth at one focus. Any apparent deviation from that motion in a plane is apparent deflection and this apparent deflection is described mathematically with the coriolis term and the centrifugal term of the equation of motion for a rotating frame of reference.

The last section of your version of the article contradicts the opening paragraphs. The inertial oscillations are real motion. Especially the component of the oscillation in north/south direction is the same both as seen from a non-rotating point of view and rotating point of view.

Despite this fundamental difference, you refer to both of those situations as 'coriolis effect'. --Cleon Teunissen | Talk 22:03, 30 August 2005 (UTC)

Please, can you actually quote the bits you don't like. I've asked this so many times now. Don't give me your paraphrases of whats there, quote. The last section says: 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'.. Do you think that is wrong? If not that, then which bit? I think I'm beginning to see your well-hidden point re the ballistics bit: which is (in my language) that ballistic motion needs the centrifugal term in too. William M. Connolley 22:20:35, 2005-08-30 (UTC).

William, that is not correct. The object must be subject to other forces otherwise it would travel in a straight line relative to the stationary observer. Gravity is the main real force. The Coriolis force is not a real force! Paul Beardsell 23:43, 30 August 2005 (UTC)

Ah, sorry, this is a terminology confusion. By this stage, we're in a 2-d coordinate system on the surface of the sphere. The equations of motion have been written in this 2-d system, and in those equations a term appears which is the coriolis force. There is no third dimension at this point. William M. Connolley 08:02:03, 2005-08-31 (UTC).

Apparent vs real

As has been said before:

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. The extra term required in the rotating frame is called the Coriolis force and this term is the same in all examples including both ballistic missile and atmospheric wind. Any other phenomenum is not the Coriolis effect. Paul Beardsell 16:43, 23 August 2005 (UTC)

Now, as the observed difference between the two observations is a bending or an acceleration this must be accounted for by the rotating observer as an extra force. We know the force is not really there and hence the use of the word "apparent". Paul Beardsell 23:36, 30 August 2005 (UTC)

Observing the inertial oscillations

I copy and paste from above:
[...] can you actually quote the bits you don't like. William M. Connolley 22:20:35, 2005-08-30 (UTC).

The current article states:

In physics, a Coriolis effect is the apparent deflection of a freely moving object as observed from a rotating frame of reference. The effect can be accounted for in the rotating frame by the introduction of the Coriolis force which then balances the the equations of the apparent motion.

Note that it wouldn't change that much if the word 'apparent' is replaced with a more neutral word.
You write: as observed from a rotating frame of reference
You state that what makes the difference is from what point of view you are observing the motion.

That applies in a straightforward way for ballistics, but in the case of meteorology there is more to the story.

Especially the north/south component of the motions depicted in the image on the right are the same both as seen from a rotating and a non-rotating point of view.

That contradicts the assertion that it will only be seen as observed from a rotating frame of reference. Both observers, rotating and non-rotating, see the oscillation. --Cleon Teunissen | Talk 23:58, 30 August 2005 (UTC)

But they observe an apparently different oscillation. One of them sees a wave, the other sees circles. Paul Beardsell 00:04, 31 August 2005 (UTC)

Both observers see the same pattern of oscillations. As seen from a non-rotating perspective the the component of the motion parallel to the latitude lines isn't a constant velocity. As seen from a non-rotating perspective the velocity oscillates between, say, 320 and 340 metres per second. As seen from a co-rotating perspective the velocity with respect to the Earth oscillates between moving with 10 metres per second to the west and 10 metres per second to the east. Both observers see the same accelerational pattern. The accelerational pattern of the inertial wind is invariant under the coordinate transformation. --Cleon Teunissen | Talk 07:23, 31 August 2005 (UTC)

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It seems that CT regards the problem he is describing here as severe enough to warrant reversion to his version. I disagree. Now if we're talking about:

In physics, a Coriolis effect is the apparent deflection of a freely moving object as observed from a rotating frame of reference

I agree that this is somewhat carelessly phrased, because there are both coriolis and centrifugal terms, unless something else is acting to counter the centrifugal. But that seems more a case for rephrasing the wording than wholesale reversion. I've done a minor rephrase, replacing *the* with *an*, which now makes is technically correct, though perhaps ambiguous.

No. There is no centrifugal force on a freely moving object. I can only think you refer to a body on the surface of the Earth. That body is not freely moving. Paul Beardsell 12:25, 2 September 2005 (UTC)

CT also objects:

That contradicts the assertion that it will only be seen as observed from a rotating frame of reference.

I don't see which text you're referring to. Where does it say that motion is *only* observed in a rotating system?

William M. Connolley 11:38:37, 2005-09-02 (UTC).

---

The first paragraph of the opening section states:

In physics, the Coriolis effect is an apparent deflection of a freely moving object as observed from a rotating frame of reference. The effect can be accounted for in the rotating frame by the introduction of the Coriolis force which then balances the equations of the apparent motion.

This text refers to the following: as seen from a rotating point of view, you see a different motion as compared to what you see from a non-rotating point of view. The correction is applied in the equation of motion for the rotating coordinate system.
The component of the observed motion that is attributed to the coriolis force is the component of the motion that is seen from a rotating point of view, but not from a non-rotating point of view. That is what McIntyre's animations show: uniform motion along a great circle as seen from a non-rotating point of view, distorted motion as seen from a rotating point of view.

The picture on the right displays a motion pattern. A pattern of oscillations. Let's concentrate on the component of those oscillations that is parallel to the Earth's axis. That component is the same both seen from a non-rotating and a rotating point of view. Nonetheless the text in the final section claims that the inertial oscillation is to be attributed to the coriolis force. That is where the opening section and the last sections contradict each other. --Cleon Teunissen | Talk 12:38, 2 September 2005 (UTC)

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). If your objection is that the thing would be moving in the inertial system, then I agree you're right, but I don't agree that its a flaw in the current page. William M. Connolley 13:58:38, 2005-09-02 (UTC).

I take it you agree with me that the North/South component of the Inertial oscillation is identical as seen from either the rotating or the non-rotating point of view. (Rotation around the Earth's axis)

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