Talk:Coriolis force/Archive 3

message: 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)

message: 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. <BR> --Cleon Teunissen | Talk 14:00, 23 August 2005 (UTC)
thread_title:  What Cleon wrote. 

message: I copy and paste from above:<BR>
: 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)
thread_title:  The geopotential height in the equation of motion 

message: 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)
thread_title:  A pic: inertial circles 

message: 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)
thread_title:  Perfect spheres and great circles 

message: 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. 
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)
thread_title:  The position on the Z-axis is invariant under the coordinate transformation 

message: 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)
thread_title:  I recommend... 

message: 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)
thread_title:  There is only one way to perform a transformation to a rotating coordinate system 

message: I copy and paste from above:<BR>
:[...] 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 <BR>
Orbit <BR>
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. <BR> [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)
thread_title:  Reprise (2): inertial wind trajectory and ballistic trajectory 

message: 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)
thread_title:  Apparent vs real