Talk:Coriolis force/Archive 4

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

message: The image to the right is genererated by the program code that William has made available here
the crucial lines of code are:
:u=u+v*f*ts
:v=v-u*f*ts
where the terms : v*f*ts  and  u*f*ts  represent accelerations in a plane tangential to the local geopotential surfaces.
A mathematician will spot immediately that since the terms v*f*ts  and  u*f*ts refer to a plane that is not parallel to the plane of the equator they are not related to rotating transformation around the Earth's axis. 
The lines of code u=u+v*f*ts and v=v-u*f*ts are interconnecting motion parallel to the plane of the equator and motion parallel to the Earth's axis. On the other hand: rotating coordinate transform around the Earth's axis leaves all components parallel to the Z-axis unaltered.
This section in my version of the coriolis effect article provides a derivation of the underlying math of these lines of code. The derivations are still quite rough around the edges, but the gist is there.
William, you claim that what those lines of code are doing follows from rotating transformation around the Earth's axis. OK, then to prove your claim you should present the mathematical derivation: a rotating coordinate transformation around the Earth's axis (using an orthogonal coordinate system, preferably cartesian) that does what those lines of code are doing: changing the latitude of objects. --Cleon Teunissen | Talk 18:29, 31 August 2005 (UTC)
thread_title:  Mathematical derivation of the equation of motion for inertial oscillations 

message: I copy and paste from above:<BR>
: I define 'real motion' as motion with repect to the local coordinate system. William M. Connolley 21:32:24, 2005-08-30 (UTC).
In the context of newtonian dynamics 'real motion' is the motion with respect to the local inertial frame of reference. An inertial frame of reference is a frame in which the laws of motion hold good. Therefore we have that the motion with respect to the local inertial frame is the real motion (while recognizing the principle of relativity of inertial motion of course).  
The equation of motion for a rotating coordinate system incorporates the rotation with respect to the local inertial frame of reference into the equation. That is an "umbilical cord" to the local inertial frame of reference that is never severed. Interestingly, that same "umbilical cord" is also present in the theory of general relativity. If, in GR, you choose a rotating coordinate system for representing the physics taking place then you have the same "umbilical cord". In GR, if you transform to a rotating frame of reference, you incorporate into the formulas the rotation with respect to the local inertial frame of reference.
Those are the reasons that in physics a distinction is recognized between 'real motion' and 'apparent motion'. --Cleon Teunissen | Talk 20:06, 31 August 2005 (UTC)
thread_title:  The distinction between real motion and apparent motion (2) 

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

message: I copy and paste from above: <BR>
:You present the image to the right as the pattern of motion of inertial oscillations as seen from a co-rotating point of view. --Cleon Teunissen | Talk 14:15, 30 August 2005 (UTC)
replies: text:  Sort of. Its actually the 2-d viewpoint wrapped up into a spheroid. William M. Connolley 18:35:44, 2005-08-30 (UTC).
It is unclear what you mean by 'wrapping up'. Take a globe, draw circles on it, smal radius close to the poles, big radius close to the equator, take a picture of that globe with a photocamera, and it will look just as Anders Persson has drawn it. If you take two pictures from a slightly different angle then that can serve as a stereoscopic image. An animation studio can make a 3D animation featuring those inertial winds. 
It is unclear how a 'wrapping up of a 2D viewpoint' is supposed to figure in this. That image presents a 3D viewpoint, without any intermediate 2D viewpoint in the process of making it. --Cleon Teunissen | Talk 12:35, 2 September 2005 (UTC)
thread_title:  What has wrapping up got to do with it? 

message: I copy and paste from above:
: Cleon [...] You want to describe inertial circles as a consequence of the Coriolis effect!    Paul Beardsell 13:25, 2 September 2005 (UTC)
replies: text: Good grief Paul, in the article the specific-to-meteorology-meaning  of 'inertial oscillation is introduced with proper explanation and images, and it is only used in the last section, and usually only the really interested readers read all the way to the last sections. This is the talk page, it would be hopelessly cumbersome to introduce and explain the expression 'inertial oscillation' again and again in comments on the Talk page. --Cleon Teunissen | Talk 15:04, 2 September 2005 (UTC)
replies: text: Good grief yourself.  The inertial circle section appears early in the article.  The non-meteorological meaning was one I was attempting to introduce - not one that present but one that should be!  The really interested reader is not necessarily interested in meteorology.  But that the Coriolis effect is only of interest to meteorologists is obviously I point I continue to miss.  That all animals are dogs is not an impression you have to continue to make (not that I say you think that all animals are dogs) but the language used in the article and here continues to have that appearance.  For me it is as if you were insisting on explaining addition in metereolgical terms, (mis)appropriating arithmetical terms, two clouds plus two clouds is four except sometimes they merge and then there are three clouds.  No, say I:  Addition is not only part of meteorology but part of arithmetic.  And the remarkable phenomenum of addition is best explained in arithmetical terms not metereological.  Substitute "Coriolis effect" for "addition"; "physics" for "arithmetic".  "Inertial circles" for "two plus two".  Paul Beardsell 15:36, 2 September 2005 (UTC)
thread_title:  Babylonian confusion 

message: I copy and paste from above:
:The derivation is usually done without an explicit coordinate system: I gave it here  for reference. [...] William M. Connolley 10:41:48, 2005-09-04 (UTC).
Yes, that is de derivation for the coriolis effect as it is taken into account in ballistics. That formula is sufficient in the case of inertial motion in the inertial frame. In the case of ballistics the fictitious coriolis force that is in the formula acts parallel to the plane of the equator, as is illustrated by David McIntyre writing in his article about Inertial motion over a perfect sphere:
:We work primarily with the latitude–longitude coordinate description of the motion, which makes the transformation from one frame to the other simple—only the longitudinal difference  \omega t  due to the rotation is required.
The latitudinal position as a function of time is independent of the point of view of the observer.
It is wrong to suggest that all of the motions of winds (that is:those wind motions that cannot be attributed to pressure gradient alone as cause) can be accounted for in terms of the fictitious coriolis force. --Cleon Teunissen | Talk 16:03, 5 September 2005 (UTC)
thread_title:  The fictitious coriolis force acts parallel to the plane of the equator 

message: Imagine you are a high school student and you look up this article.  Its unreadability would put you off physics for life.  You all seem to be trying to get to an intellectally perfect description and have lost sight of your potential users.JMcC 19:30, 26 October 2005 (UTC)
replies: text: The first paragraph 'simple examples' gives in my opinion a simple enough explanation, that does not contain complex mathematical formulas, and looks well readable. Things that can be considered are: (1) the possibility of adding pictures to the two examples; (2) the possibility of having the caroussel as the first example instead of as the second. Bob.v.R 01:15, 27 October 2005 (UTC)
replies: text: I am glad you liked it.  After writing the paragraph above calling for a simple explanation, I wrote it.JMcC 08:43, 27 October 2005 (UTC)
replies: text: I see! And what do you think of my two suggestions? Bob.v.R 19:07, 28 October 2005 (UTC)

text: Quite correct. Only a moving object experiences the Coriolis effect.  An object at a pole isn't moving.  An object moving towards a pole would of course experience the effect.JMcC 16:39, 27 October 2005 (UTC)
replies: text:  No, an object (instantaneously) at the pole can indeed be moving. But the "simple examples" text unfortunately doesn't tell you that the effect is proportional to the speed of the object, and it seems to me to imply no effect at the pole. The "consider a missile..." bit I find unhelpful and handwavy (and, I think, if you work it out it gets the wrong numbers). William M. Connolley 18:26, 27 October 2005 (UTC)
replies: text: I agree that an object passing over the pole would experience the Coriolis effect, but I don't think the article implies that it wouldn't.  It should be possible to explain clearly the Coriolis effect to a layman, which the article previously failed to do.  I had a stab, and may not have fully succeeded but the way that Wikipedia works is that somebody should now be able to make it even clearer.  If you want to make the point about proportionality to speed, you merely need to say that if the missile travelled faster from the North Pole, the point at the equator would have moved less by the time the missile was due to arrive. I suppose it is handwavy, but it is closer to an understandable explanation, which is what we are all trying to achieve.  Having had a go, I will now bow out from the article.  Good luck JMcC 23:24, 27 October 2005 (UTC)
replies: text:  I will have a go, but not tonight. William M. Connolley 19:26, 28 October 2005 (UTC)
thread_title: We need a simplified explanation first

message: I copy and paste from above:
:129.92.250.41 wrote:<BR>[...] Let's decouple the coriolis effect from the great circle issue by considering a body moving east or west along the equator.  How does the rotation of the earth cause an acceleration?
Let's look into that matter a little closer, and do some calculations. <BR>
There are plans for huge cargo-lifting zeppelins, the weight of the zeppelin plus the cargo will be around 500 tons. The cruising velocity of these cargolifters will be in the order of 25 meters per second.
What does it take for the cargolifter to be neutrally buoyant when it is stationary with respect to the Earth? The fact that the Earth rotates must be taken into account. At the equator the velocity of Earth's surface is 460 meters per second. The amount of centripetal force required to cause a mass to move along a circular path with a radius of 6370 kilometer (the Earth's radius), at 460 m/s, is about 0,0336 newton per kilogram of mass. For the 500 ton cargolifter that becomes about 16800 newton. The amount of buoyancy force required is the weight of the cargolifter minus those 16800 newton.
When cruising at 25 m/s due East the total velocity becomes 460 + 25 = 485 m/s, which requires a centripetal force of 18460 Newton. Cruising at 25 m/s due West: 14850 Newton. So if the cargolifter is neutrally buoyant while cruising due East, it will not be neutrally buoyant anymore after a U-turn; the airship will have to be re-trimmed.
=== Derivation of the formula ===
Derivation of the formula for motion along the equator.<BR>
Notation:<BR>
 \Delta \omega  is the angular velocity of the cargolifter relative to the angular velocity of the Earth. <BR>
 \omega_e  is the angular velocity of the Earth: one revolution per 24 hours. <BR>
(Hence the total angular velocity of the cargolifter is  \omega_e + \Delta \omega ) <BR>
 F_e  is the centripetal force for co-rotating with the Earth. <BR>
 F_v  is the total centripetal force for an object moving along the surface of the Earth.<BR>
 (\Delta \omega) * r = v_c  is the cargolifter's velocity (velocity relative to the Earth)
How much does the buoyancy of the cargolifter need to be retrimmed when the cargolifter's velocity changes?
: F = F_v - F_e  = - m (\omega_e + \Delta \omega)^2 r - (- m (\omega_e)^2 r) 
: F = - m r ((\omega_e + \Delta \omega)^2 - (\omega_e)^2) 
: F = - m r (\omega_e^2 + 2 \omega_e \Delta \omega + (\Delta \omega)^2 - \omega_e^2) 
: F = - 2 m \omega_e (\Delta \omega) r - m (\Delta \omega)^2 r 
: F = - 2 m \omega_e v_c - m (\Delta \omega)^2 r 
When  \Delta \omega  is very small compared to  \omega_e  then the following shorter version is a good approximation:
: F = - 2 m (\omega_e * v_c) 
This derivation explains why the formula (2 * omega cross v) works for east-to-west and west-to-east motion.
This particular derivation was for motion along the equator. A more general derivation explains why a cargolifter cruising due west at say 45 degrees latitude will tend to veer towards the nearest pole. The buoyancy is a crucial element. --Cleon Teunissen | Talk 11:14, 15 December 2005 (UTC)
thread_title:  The physical consequences of making a U-turn 

message: Derivation of the formula for motion along the equator.<BR>
Notation:<BR>
 \Delta \omega  is the angular velocity of the cargolifter relative to the angular velocity of the Earth. <BR>
 \omega_e  is the angular velocity of the Earth: one revolution per 24 hours. <BR>
(Hence the total angular velocity of the cargolifter is  \omega_e + \Delta \omega ) <BR>
 F_e  is the centripetal force for co-rotating with the Earth. <BR>
 F_v  is the total centripetal force for an object moving along the surface of the Earth.<BR>
 (\Delta \omega) * r = v_c  is the cargolifter's velocity (velocity relative to the Earth)
How much does the buoyancy of the cargolifter need to be retrimmed when the cargolifter's velocity changes?
: F = F_v - F_e  = - m (\omega_e + \Delta \omega)^2 r - (- m (\omega_e)^2 r) 
: F = - m r ((\omega_e + \Delta \omega)^2 - (\omega_e)^2) 
: F = - m r (\omega_e^2 + 2 \omega_e \Delta \omega + (\Delta \omega)^2 - \omega_e^2) 
: F = - 2 m \omega_e (\Delta \omega) r - m (\Delta \omega)^2 r 
: F = - 2 m \omega_e v_c - m (\Delta \omega)^2 r 
When  \Delta \omega  is very small compared to  \omega_e  then the following shorter version is a good approximation:
: F = - 2 m (\omega_e * v_c) 
This derivation explains why the formula (2 * omega cross v) works for east-to-west and west-to-east motion.
This particular derivation was for motion along the equator. A more general derivation explains why a cargolifter cruising due west at say 45 degrees latitude will tend to veer towards the nearest pole. The buoyancy is a crucial element. --Cleon Teunissen | Talk 11:14, 15 December 2005 (UTC)
thread_title:  Derivation of the formula 

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

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

message: I swapped the pic of the hurricane for a (featured) pic of an icelandic low. Because... I'm not sure how important CF is in hurricane dynamics. But it certainly is in low dynamics. William M. Connolley 22:34, 24 January 2006 (UTC)
replies: text: Actually, there was an article publiched in a British journal (possibly the "Journal of Fluid Mechanics" many yearts ago, that demonstrated the bathtub effect. The authors put a few drops of ink into the water in the tub, and waited till all the swirling ceased, i.e. until the viscosity of the water totally dampened the vorticity induced by the inflow. Then, if they opened the bottom drain v-e-r-y slowly, so as not to induce any vorticity, the flow of ink in the outflow clearly demonstrated a counter-clockwise swirl in England. However, they had to wait at least one week for the initial vorticity to dissipate, and it is this that vitiates all simple experiments---the initial conditions of the water are not mathematically curl (i.e vorticity)-free.
this is fair enough (though it reads more like a talk comment) - but its already mentioned. It would be nice to have the exact ref, though. William M. Connolley 21:34, 5 February 2006 (UTC)
thread_title:  New pic 

message: The meteorologist Brian Fiedler has manufactured some physlets, one physlet illustrating motion over a flat surface  and a physlet illustrating motion over a parabolic surface. These physlets use the system of showing the motion both seen from a non-rotating point of view and from a rotating point of view. If something like that would be possible for wikipedia I would make something like that.
The big advantage of those physlets is of course that you can adjust a parameter, you can "fiddle around" with it. GIF-animations just grind on, but for wikipedia a GIF-animation will have to do. --Cleonis | Talk 17:30, 22 February 2006 (UTC)
thread_title:  Physlets by Brian Fiedler. 

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

message: It is very common to see in meteorological texts statement like: 'because of the Coriolis effect, air tends to keep flowing around pressure areas'. Dozens of wikipedia articles about winds and weather in general mention a similare 'bevcause of the coriolis effect' and refer to the Coriolis effect article, clearly expecting that the Coriolis effect article will deliver the goods, that the Coriolis effect article will explain what physics is at play in affecting the winds.
The central theme is of course the expression fk*u
The article about rotational-vibrational coupling is designed to describe the simplest possible example of Coriolis effect. The spring exerts a center-seeking force, the strength of the center-seeking force is proportional to the displacement from the center of rotation. The vector for the center-seeking force and the vector of the centrifugal term of the equation of motion are at all times opposite in direction and equal to each other in magnitude. The motion depicted in the rotational-vibrational coupling animation has the property that the motion with respect to the rotating coordinate system is described by the coriolis term: fk*u. 
Conclusion: it is an example of the pure form of Coriolis effect if and only if the strength of the center-seeking force is proportional to the displacement away from the center of rotation. 
Coriolis effect is occuring when the center-seeking force is doing work. More precisely: when there is an oscillation (as discussed it the rotational-vibrational coupling article) then the center-seeking force oscillates between doing work and doing negative work. The aim of physics education is to show what forces are doing work. Showing what forces are doing work shows the cause-to-effect relations.
It is common in physics to use the expression effect in referring to an example a particular recognizable effect, and there can be a class of examples, related by analogy. Examples: the Casimir effect and the nautical Casimir effect. discussing the nautical Casimir effect in the Casimir effect article is appropriate.
An area of study in molecular physics is rotational-vibrational spectroscopy. In molecular physics, coupling between rotational and vibrational energy-levels is considered to be an example of a Coriolis effect, hence it is widely referred to as coriolis-coupling. (16.000 Google hits for "coriolis coupling" plus "molecule") --Cleonis | Talk 21:53, 12 April 2006 (UTC)
thread_title:  What is expected when other articles refer to the coriolis effect article 

message: I have read your article because my boy asked why the hurricanes and tornados are rotating? Why the air does not go in a straight line from the area of high pressure to the area of low pressure - as common sense suggests?
I found the introduction - beginning with the mathematical formulae 
based on the vector product a bit "steep". Being honest I did not understand the coriolis effect when I read this formulae.
Well, I have studied some vector analysis - but this is 
20+ years ago. And I do not think that any of my colleagues
(all of which are engineers and computing people) would understand this formulae.
This raises the question: for whom is this introduction written?
What is the target audience?
Obviously not students and professors of physics - because they need a much more in-depth understanding of the issue.
For us other lesser mortal? Probably not - because we do not
understand it. Notice that I liked math when I was at university and I was among the best in my year. What about those who did not like math or have never heard of the deeper secrets of vector analysis?
My suggestion: start the article with a more "common sensical" introduction
I suggest the following explanation (however I do not know if it
is correct from a physical point of view).
A wind of 50 miles per hour is blowing in New York City exactly from
South to North. 
: If you think the Coriolis effect knows the difference between North/South and East/West, you do not know what the Coriolis effect is. -- GWO
This wind takes a feather of a bird (or a cloud of feathers) with it. After a while the cloud of feathers arrive somewhere
near Montreal (which is North of NY). To this point an observer on the earth agrees with an observer from space - both of them see the feathers near Montreal.
The point of view of the observer from space however is: since the earth is turning to the east NY City moves eastwards with x miles per hour.
(I do not have the formlae at hand to calculate x - but I "guestimate" it should be a few thousand miles per hour). For this reason the observer from space sees the cloud of feathers NOT moving to the north - but with x miles to the east and 50 miles to the north. (Since x is much faster than 50 miles per hour the observer from space sees the cloud moving almost exactly to the east with a slight tendency to the north).
Well after a while the cloud arrive somewhere near Montreal. The 
feathers and the surrunding air have some inertia. For this reason they
have the tendency to continue their journey in a straight line. And 
this is what the observer from space is seeing: the feathers travel with x miles to the east and 50 miles per hour to the north.
However Montreal is closer to the Northpole than NY. The 
surcumfence of the earth at Montreal is smaller than at NY. Since both Montreal and NY need 24 hours for one rotation Montreal is moving slower to the East than NY. This means the eastward speed of Montreal
is less than x miles per hour. The feathers and the surrounding air
however continue to fly with x miles per hour to the east and 50 miles per hour to the North. (From the point of view of the observer in space).
For this reason an observer in Montreal will see the feathers flying NOT exactly to the north. Instead he sees the feathers flying somewhere to the North-East. This (appearent) change of direction is the coriolis effect.
I do not regularily read the discussions of that article. So: if you want to answer something it might be better to send it to johann_rost@yahoo.com.
Johann
: I think the problem with this is that it mixes up geometry with Coriolis, as people so often do. It would be desirable for a simpler explanation, but yours isn't it. Maybe a picture in that section would help. Really its very simple: the acceleration is proportional to the rotation and the velocity, and perpendicular to both. William M. Connolley 15:27, 19 April 2006 (UTC)
thread_title:  Formula of coriolis effect