User talk:Cleonis/Archive 2

Re: Chris Hillman's assessment of Kevin Brown

Cleon -

The issue is not Kevin Brown's competence, but rather the usefullness of his writings. When something is so advanced that it is over everyone's head, it is no more informative than if it was written by an idiot.

The work is a set of essays taking in minutia about GR. It is not and probably is not intended to be a web-course in GR, but rather to document the author's thoughts on the subject. Such a thing has its uses, but I do not see its utility as a GR resource.

I have seen Chris Hillman write in much the same style in his regular Mathmatical Musings postings on the USENET. Any one of those can be an education in advanced math, and often also is in GR. However, I notice that in Wikipedia Chris is being very careful to be clear and concise and to directly address the issues related to GR in his writings. Overall, he is succeeding in making some relatively advanced material accessible to people with only a minimal training in the subject.

I do not see Kevin Brown's material as being accessible at all, nor do I find it to be very informative. What is covered in those essays are not the mainstream issues of GR. Instead of aiding prople in obtaining a good understanding of GR, a fairly advanced level of understanding is assumed and required. I can follow that stuff, but it is not easy, and that is with a decade of (admitedly part-time) GR training under my belt. You souldn't need a Ph.D. to comprehend Wikipedia material. That simply is not the intended audience.

So please do not confuse genius with usefullness.

--EMS | Talk 21:57, 11 July 2005 (UTC)[reply]

Interesting. I'm 35 years removed from when I was seriously studying physics, yet I find Kevin Brown's math pages on the subject beatiful and clear; nagging questions other sources don't answer are dealt with head on - I think that's what I like, in particular. I don't know that they are reasonble to first study relativity from, or a reasonable model for wikipedia articles, but they certainly seem reasonable as a web reference on points we may want to cover. --Pallen 04:02, 6 March 2006 (UTC)[reply]

Re: Coriolis effect

Thank you for to have informed me about the article. As soon as possible i would like to integrate my italian translation with the new animations ad new text. I've seen that you uploaded them on the en.wiki, than i've to copy them on commons. The new animations seem to be more clear and simple to understand than the old one, i think. I'll inform you when my work will be done.

PS Please inform me on my italian discussion page if you answer here, i'm a spot on in en.wiki. Bye, Ciao --Guam 14:57, 12 July 2005 (UTC)[reply]

The work is now done, please take a look and thanks for uploading yourself the image on Commons. I have held the section An image of the Coriolis effect because i think is a very impressive and simple description of the origin and relativity of the Coriolis's force. Bye --Guam 11:17, 15 July 2005 (UTC)[reply]

I copy here my answer in my discussion page:

Obviously the term hovercraft is not to be intended literally, but in the same way as a generic frictionless object, as a dry ice punk over a parabolic table. Actually i haven't changed the quick silver table model and the article is still describing it. The section an image of the Coriolis effect, now titled intuitive description (Descrizione intuitiva) is intended for to give an intuitive idea about the origin of the force in a bidimensional rotating model. However the whole article is now under revision by the community. About the chair: the point '1b is very interesting to enhance the relativeness of the rotation and to introduce the mach principle. need to be addend to the article. --Guam 11:52, 18 July 2005 (UTC)[reply]

Sorry, I'm completely unqualified when it comes to the physics of the Coriolis effect, I was just offering stylistic advice.--nixie 10:33, 29 July 2005 (UTC)[reply]

Eugene Shubert

Cleon -

Thanks for the support. I do admit that it is an interesting article. However, I have trouble with things that argue that "X" is not "X", even if they do it as well as this article does. In this case, the argument is made that classical mechanics actually implies relativity. Certainly he has found an elegant way of making the connection, but as early as 1908 Minkowski had shown that classical mechanics was effectively relativity with $c=\infty$ . The Eugene Shubert is little more than an extension of that insight.

At least is not arguing that Newton should have derived relativity himself (and I have seen that done). --EMS | Talk 14:33, 25 July 2005 (UTC)[reply]

All right

I'm waiting for your reply. Bye. BW 10:04, 29 July 2005 (UTC)

Coriolis article

Thanks. I have no technical knowledge on the particular subject, but I am a technical editor, so I feel confident in editing technical subjects because I'm trained enough to know when things are contradictory, confusing, too technical, defy common sense, and so on. I'm generally good at simplifying language and correcting grammar and syntax, which technical articles seem to need especially. That can be my contribution in this case, I hope.

My main area of interest regarding the Coriolis topic is tropical cyclones, but in meteorology there seems to be more going on than Newtonian mechanics (or maybe not?). One problem is when one starts talking about tropical cyclones not forming near the Equator, there are more dynamics involved than the lack of Coriolis force, as I understand it. For one thing, the equator has a zone of high pressure that inhibits convection. And, lately we've had some examples of things that "don't ever" happen actually happening.

As for the approach of the article, I do not know of any standards or references or other articles. It is difficult when some editors feel that two major points need to be made, and other editors want to stick to one major path. In many cases, I think, sectioning and linking can solve some of these problems. I believe in keeping introductory paragraphs simple, and bringing up tangential issues much lower and in their own sections.

Don't know if any of this addresses your concerns, but glad to help where I can. I did do a pretty big edit (meant to only fix some capitalization and wordiness issues, but it was such a long article, I got sucked in). Cheers. DavidH 18:26, July 29, 2005 (UTC)

Hi, saw your comment, and also the rather hypercritical comments from that other guy who's disputing the article's accuracy. Seems like a lot of the argument is over that flow meter, and maybe you are coming to a consensus on that. Unfortunately, he seems to really be the kind of person who "discusses" his concerns with a finger poking you in the chest.

Sorry, though, it would be hard for me to come out one way or the other regarding very technical content. I believe I passed physics in high school, but, man, that was a long time ago. I know that on some of the meteorology discussion boards I frequent, even very smart people start yelling at each other over misunderstanding of the Coriolis effect....<sigh>.

I will keep an eye on the article, and I will try to comment on whether certain statements are understandable. One thing that might help is to remind all editors that the article should be understandable by a person in high-school physics (in my opinion), and points that require much higher math or physics can probably be left out, or dealt with by linking? Good luck! DavidH 22:28, July 29, 2005 (UTC)

William doesn't like you because you're smart. Even if you're not smart, you're trying to learn and teach others while being polite at the same time. William on the other hand has a very different MO.

William does not want to return to addressing his primary question, his MO is to embroil you within minutia (things he should be able to fix himself, if he really knew what he was talking about). As long as he hangs out in your article work (a pet subject of his, if his user page is 1/10 honest) you will become tormented by his dodgy and hypocritical insistence that you answer each and every one of his questions, while he remains impervious to any of yours.

Cleon, I suspect you are talking over the head of Connolley. And Knotts advice for you to address "each and every" thing he says, is sophistry. Think for yourself! Otherwise, I suspect he will rally a troupe of aholes, and see to it that you are driven from Wikipedia. 68.110.237.166 22:33, 29 July 2005 (UTC)[reply]

Laughs. crackpot alert. Theresa Knott (a tenth stroke) 09:06, 30 July 2005 (UTC)[reply]

Well, clearly William M Connolley and I are talking past each other. His effort to misunderstand me as much as possible is palpable. But no matter what William's motives are, a challenge to provide explanation is to be met. That comes with the territory (the wikipedia territory). If you can't take the heat, stay away from the peppers. --Cleon Teunissen | Talk 13:19, 30 July 2005 (UTC)[reply]

I noticed you were asked to improve the opening paragraph. Does this help in any way, give you some new ideas, or a different way to open the topic?

"One of our more common interaction with this term and the effect it describes can be seen in the patterns of large weather systems, particularly those with a cyclonic circulation. In the northern hemisphere deflection is to the right, causes spin in a counter-clockwise direction, while southern latitudes generate clockwise rotation and deflect to the left. In the case of atmospheric patterns, Coriolis effect is a combined product of the atmosphere interacting with surface friction of our rotating Earth, and gyroscopic precession. Without precise laboratory conditions, one may have mistakenly observed Coriolis effect in the spin of a draining bath tub of toilet bowl."

Let me know how you would change it, and/or if you would like something like this placed in the introduction. Take care... TTLightningRod 03:50, 30 July 2005 (UTC)[reply]

Er no! you can't see the Coriolis effect in a bath or toilet. Theresa Knott (a tenth stroke) 09:06, 30 July 2005 (UTC)[reply]

So easy to fix, right? But I agree with Cleon, that none of this need appear so early in the article. TTLightningRod 13:05, 30 July 2005 (UTC)[reply]

Error in edits

Sorry that by shortening it I made it incorrect.

Original:

What happens to air flow that initially starts flowing from West to East...

my edit:

What happens to air flow that initially flows from west to east...

"Initially starts flowing" seems convoluted because "initial" and "starts" both refer to a beginning. Something either flows, or it doesn't.

In any case, the directions west, east, north, south are not capitalized unless it's in a name like the North Pole or the South China Sea.

Original:

Because of the extra velocity on top of the normal velocity from co-rotating with the Earth as whole...

My edit:

Because of the extra velocity from co-rotating with the Earth as whole...

I see your point, the increase isn't from co-rotating, but it's also not explained. "Extra" and "on top of" are the same type of qualifiers. Something that's "extra" is something that's "on top of." It needs to focus on the increase, not what it's already doing (co-rotating). That's what led to my over-simplification.

I think this is closer to what it needs to say:

A pressure gradient increases the velocity of an air mass, which forces the flow away from the Earth's axis of rotation. For example, if air flowing west to east accelerates, the velocity increase turns the flow northward. (if that's right, direction-wise?)

This is somewhat like a car driving around a curve. If it accelerates, the car is forced outward. If the curve is banked like a race track, as the car goes faster, it climbs higher on the bank.

If you can use any of that, do. Good luck, I appreciate your hard work on the article. DavidH 10:07, July 31, 2005 (UTC)

toward/s

Either is OK, but in AP style no "s" is preferred, so that's what I do. I go forward, I go backward, I go toward. Or landward, shoreward, seaward, northward, upward, downward, and so on.

I tell you, the banked car track is something most people understand (NASCAR racing so big in US now), so if it applies, it could help. But since the track isn't rotating, I guess that's only simple angular momentum, not Coriolis at all. DavidH 11:04, July 31, 2005 (UTC)

Re: The mathematics of coordinate transformation

Cleon -

I doubt that I can be on you side on this issue. This is an area where I have noticed that you have ongoing difficulties.

I have added a response to your posting on my talk page. Please see it. I will also ask William to comment. --EMS | Talk 15:20, 3 August 2005 (UTC)[reply]

Centrifugal Force

Cleon,

Thanks for the replies on Talk:Centrifugal force. I think we've gotten onto the same page, conceptually. Can you now see any beneficial edits we might make to the article, or should I take a whack at it? --Jeepien 15:41:48, 2005-08-08 (UTC)

Coriolis and Meteorology

I am a very new Wikipedian, so I really appreciate the feedback. I am also very grateful that you found and corrected my error about Hadley and Coriolis.

I really enjoyed your version of the Coriolis Effect article, and I found it to be clear and educational. I would be sad to see any of your excellent animations go. The article seems to load plenty fast for me, but I am not experienced enough (or allowed, maybe) to know if the article is relatively large.

I am in total agreement with you about this article's importance. I think coherency in science articles is vital.

You should know that my continuous math skills are poor (working on that...thank you wikibooks!), so any help I provide in that capacity will be limited. However, I am passionate about science and the 'learn free' ideal and I have nothing but time at this stage in my life.

Meteorology is the latest project I've been working on, but my first baby on this wiki was the article on supercells. I would be honored if you could take a look at it. Here's to improving the quality of science articles on the wikipedia.--demonburrito 07:29, 14 August 2005 (UTC)[reply]

Demonburrito's understanding of George Hadley's work

I'm curious: have you learned what it was exactly that Hadley had in mind, and how he was later corrected? Hadley can't be blamed for his mistake, even Laplace, the foremost expert in mechanics of his time overlooked it. --Cleon Teunissen | Talk 08:12, 14 August 2005 (UTC)

Well, I have to admit that my knowledge on this subject is rather sparce. The work I did on meteorology was hurried because I felt that it was harmfully inaccurate (in searching for sources for claims in the article, I was horrified to find thousands of commercial sites using the text). Much of my research was from sites like this: [1]. What I gleaned from it was this:

Hadley pointed out that equatorial solar heating would make tropical air lighter than air at high latitudes. Tropical air would thus rise, high latitude air would sink, winds at the surface would blow towards the equator, and winds aloft would blow towards the poles. Hadley realized that the rotation of the earth would cause the surface wind blowing toward the equator to veer towards the west, thus producing the system of trades familiar to the mariners of his day.

I expect my understanding to improve drastically as I start to thoroughly cite original sources, and read them in entirety.

Actually, the peak of my understanding was when I read your version of Coriolis a few hours ago. I would love a more in-depth explanation of the history of the concept; from Halley to LaPlace, to Hadley, to Coriolis and to Rossby.

Concerning your cause: As this is a subject that I am fascinated by, I would like to spend a few days sussing this out. Links to accessible scientific papers would be appreciated (I don't have a subscription to any of the online libraries). Perhaps you could describe the conflict for me on another userpage?

So, in short, hit me.--demonburrito 08:54, 14 August 2005 (UTC)[reply]

Laithwaite and gyroscopes

Cleon, you seem to know something about gyroscopes. A while ago, I did some editing on Eric Laithwaite who was obviously just plain wrong on the subject. However, when I read accounts from experts about the nature of his mistake I'm none the wiser. They talk about fast tops and slow tops and are obviously refering to a body of knowledge that I've never managed to find. Can we have some more content here? An account of where Eric Laithwaite went wrong would be really useful. Cutler 13:39, August 16, 2005 (UTC)

J. D. Norton link

Hi Cleon. Yeah, his massive review is rather lengthy - I still haven't read it! Thanks for the great link to J. D. Norton's home page. Much appreciated. Feel free to contribute to my sandbox article on general covariance. ---Mpatel (talk) 15:17, August 27, 2005 (UTC)

All apologies

Cleon, I'm so sorry that I haven't responded in so long. My nomadic lifestyle is sometimes not conducive to study.

Also, I was wondering if you have been following the Venus Express program. There is little on the wikipedia on the venusian atmosphere, but I think it may interest you. For a taste, if you are not familiar: Venus has a weak magnetic field, and yet the surface pressure is 100 bars. Venus has a retrograde rotation, and has massive steady-state cyclones over the poles.

The BBC World Service radio has last week's [Science in Action] archived, and this includes a nice introduction.

Doe voorzichtig!--demonburrito 18:10, 27 October 2005 (UTC)[reply]

Username change

Your request has been completed. Regards — Dan | talk 00:39, 19 December 2005 (UTC)[reply]

Foucault Pendulum Velocity Diagrams

Dear Sir,

I hope you will look at the article one more time. I have made some changes to it since our last discussion. In particular, I rewrote the section about the Foucault pendulum connections and would like to know if it is now better written. It was very poorly written to begin with and was only intended to get a more detailed article written about the connections of the pendulum. I think it is I that owe you an apology because my usage of the 'common terminology' is not precise nor am I an expert. However, I was motivated by my trip to Paris and my observation and reading about the Foucault pendulum.

I had debated about whether to place the article in Wikipedia and have considered removing the article and the links to it. I think maybe the surface velocity vectors is a mathematical contrivance and not really helpful to understanding the forces involved. The intention was to provide visual diagrams of the orientation not a detailed mathematical proof. I would welcome your thoughts in this regard and would willingly remove the article. I fully expected edits to be made to the article as initially written.

I'm still not sure that I fully understand all of the common terminology that is used and was not clear about all the comments in the article discussion. For example, I'm still not sure about the usage of the terms in the phrase "precession of the plane of oscillation".

I also added a further statement to one of the early comments as I understood it better, as follow:

The swing of the pendulum establishes a vertical plane above the surface of the Earth that is independent of the rotation of the Earth. The initial conditions imparted to the pendulum swing, however, are dependent upon the location of the support structure on the Earth.

Please feel free to comment to me at my user talk, in the article discussion, or even by editing the article. However, I am a layman to the subject and fairly new to Wikipedia so it might not be time well spent.

Thank you for your comments and I would welcome any feed back that you would provide.

David Harty 13:57, 11 January 2006 (UTC)[reply]

It is apparent to me that I still have got it wrong about the independence of the swing to the rotation of the Earth. Let me try again with the following:

The pendulum swing is dependent upon the motion of the support where it is located on the surface of the Earth. The support moves with the surface velocity vectors of the Earth. At the equator the support moves with the equatorial rotation of the Earth and moves the pendulum swing along with this rotation. At the poles the support-point is located on the axis of the Earth so the support-point rotates but does not have a horizontal velocity component as it does at the equator (and progessively less with increasing latitude). The pendulum swing at the poles remains aligned toward a star if not forced to rotate with the support.

Maybe this is better but I think my terminology is still lacking. David Harty 15:13, 11 January 2006 (UTC)[reply]

I wanted to say thank you, again, for your previous comments. After some thought, I added a brief paragraph about the Coriolis effect and referred to the article in Wikipedia. I hope that this is helpful to clarifying the information as this is the effect that is at the heart of the behavior of the Foucault pendulum. I think the Foucault pendulum is elucidating to the Coriolis effect. It seems to me that a precise understanding of terminology is needed in order to gain a full understanding of the observations. Often times, it seems that the terminology seems to be used or interpreted with different meanings of the word. Anyway, thanks a third time for your previous comments and to keep me thinking about the 'interpretations' of the 'observations'. My hope is that the article keeps getting better and is useful even for a layman.David Harty 15:21, 22 January 2006 (UTC)[reply]

WMC adminship

FYI. thought you might like to vote on Wikipedia:Requests_for_adminship/William_M._Connolley_2 --GangofOne 20:50, 12 January 2006 (UTC)[reply]

centripetal/centrifugal force

I did an attempot to improve a phrase in the centripetal force article. NOw it reads:

Centripetal force must not be confused with centrifugal force. In an inertial reference frame (not rotating or accelerating), the centripetal force accelerates a particle in such a way that it moves along a circular path. In a corotating reference frame, a particle in circular motion appears to have zero velocity, if the rotation is not accounted for. The centripetal force is exactly cancelled by a centrifugal reaction force that in this approach appears as a pseudo gravitational force. Centripetal forces are according to Newtonian mechanics true causal forces, while centrifugal forces only appear relative to rotating frames.

Do you agree that this is an improvement? Is it good enough do you think? Harald88 23:21, 26 January 2006 (UTC)[reply]

I know that you are phrasing it in the usual terminology. Unfortunately, that terminoloty rubs me the wrong way.

I feel very uncomfortable with the expression 'in a rotating frame of reference'. It's just so wrong. I usually write: 'motion with respect to a rotating coordinates system', or 'motion as mapped in a rotating coordinate system'.

And I hate the expresssion 'pseudo gravitational force' so bad I can taste it.

for newtonian dynamics, the way I would formulate it (with the disadvantage of being unusual terminology) is a bit as follows:

Centripetal force must not be confused with centrifugal force.
In order to maintain motion along a circular path, a force directed towards the center of rotation must be exerted continuously. This center-directed force is referred to as the centripetal force. When the motion of a circling object is mapped in an inertial coordinate system there is centripetal acceleration of the circling object.

Alternatively, motion of a circling object can be mapped in a co-rotating coordinate system. A rotating coordinate system has the property that each point of that coordinate system is accelerating in centripetal direction with respect to an inertial coordinate system. Conversely, any inertial motion corresponds to centrifugal coordinate acceleration with respect to a rotating coordinate system. The centrifugal coordinate acceleration with respect to a co-rotating coordinate system is equal in magnitude to the earlier mentioned centripetal acceleration with respect to an inertial coordinate system.

Maintaining a circular path requires a centripetal force. Let a very heavy object A be exerting a centripetal force, causing centripetal acceleration, on object B. Then object B will in return exert a reaction force on object A, as described by Newton's third law. In the case of motion along a circular path the reaction force acts in centrifugal direction. If object A is much heavier than object B the, then A will not be noticably moved by the reaction force.

It should be noted that the reaction-force-in-centrifugal-direction and the centrifugal coordinate acceleration are quite different concepts. The reaction-force-in-centrifugal-direction is a physical force, which can have physical consequences, the centrifugal coordinate acceleration is a calculational device, with no physical counterpart.

I just cannot express hard enough my rejection of expressions such as 'fictitious force' and 'inertial force' and 'pseudo force'. I can read between the lines of course, I know what is meant when those expressions are used. But it is such a stupid stupid idea to allow such expressions into the physics vocabulary, and I won't have any of that. --Cleonis | Talk 01:23, 27 January 2006 (UTC)[reply]

As it happened, I did not entirely rewrite it. I think we fully agree that "using" a rotating frame or "mapping to " a rotating frame would be more accurate. We both hate pseudo nonsense, but as Wikipedia editors we must fairly and accurately explain how others use/abuse Newtonian physics. As soon as one regards a rotating frame as non-rotating, then such things appear, that's undeniable. The problem is that some people scrap that explanation, and then the mess is complete... BTW, I think that you are mistaken about "inertial force": I understand it to mean: a (real) reaction force due to inertia. But I should check the literature to be sure! Harald88 23:07, 27 January 2006 (UTC)[reply]

About reaction force and how it relates to the expression 'inertial force'. I come again with the example of the Pilot Training Centrifuge (PTC). Let a weighing scale be positioned on the floor of the pod of the PTC. Let a hard object, such as a bowling ball, be put on top of the weighing scale. When the PTC is pulling G's, then the weighing scale will show how much reaction force is exerted by the ball on the weighing scale.

When the PTC is pulling G's the weighing scale is exerting a centripetal force on the ball, and in reaction to that the ball exerts a reaction force on the weighing scale.

So if you draw a force-vector diagram, whith the base of each arrow positioned at the actual point of grip, then the force-vector of the reaction force will be positioned at the contact point between ball and weighing scale.

On the other hand, what you sometimes see in diagrams that the base of some vector is positioned at the center of mass of the ball, (and pointing in centrifugal direction), and next to that arrow the words: 'centrifugal force'. That's no good, because inertia does not fall in the category of force. In a force-vector-diagram, you just don't insert a vector for inertia. --Cleonis | Talk 23:37, 27 January 2006 (UTC)[reply]

I agree with that. Nevertheless, gravitation isn't a force either, not even according to Newton: instead it's the cause of a force. No gravitation, no attraction to bodies; and no inertia, also no reaction force. Simple as that! ;-)

BTW, I now did an attempt to correct the "rotational frame" mess in centrifugal force, as it's also a redirect from centrifugal acceleration. Harald88 11:32, 28 January 2006 (UTC)[reply]

Just for my own sanity, b/c this distinction between the two forces is maddening, is the centrifugal force simply the opposing component of the original motion vector? Is that essentially the name given to that force, since centripital acc. pulls the object in and alters the original single vector into a composite one? I'm sure I'm not phrasing this properly, and if I had a diagram it'd be a lot easier. In fact, I think that a graphic is very much in order, as someone coming to this article with little knowledge of physics will be utterly confused by these discussion and formula. - IstvanWolf

Hi Cleonis, by a strange quirk of fate I had never actually taught our (rigid body) dynamics classes until last year. I am a big fan of d'Alembert's principle (add the "inertial forces" and "inertial torques" to the system and then treat it as a statics problem - i.e. put one's self in a reference frame attached to the rigid body, which is undergoing linear accleration as well as angular accleration). I notice some of the text-books express a similar idea to yours about inertial forces. They really hate them. I set an exam question where I said "hint: use d'Alembert's principle". The exam checker, who is much more adept at this than I am, checked the question using Lagrange's equations and made the comment that the students will never solve it in the time required (he ghad covered many pages with calculations). I said, that is why I said use d'Alembert's principle - it takes about 5 mins. Now dynamics is his speciality, yet he had never even heard of d'Alembert's principle (or had forgotten it). It seems a shame that such a powerful tool is unknown because of a philosophical objection. And of course, d'Alembert's principle promted Einstein to propose the equivalence principle and get the very simple prediction of bending of light by gravity - so it is very useful at least. E4mmacro 22:04, 1 February 2006 (UTC)[reply]

Is gravitation a force?

I copy and paste from above:

[...] gravitation isn't a force either, not even according to Newton: instead it's the cause of a force. No gravitation, no attraction to bodies; and no inertia, also no reaction force. Simple as that! ;-) [User:Harald88|Harald88]] 11:32, 28 January 2006 (UTC)

I am unaccustomed to making a distinction between 'gravitation' and 'gravitational force', but I see how it can be put to use. For me generally, the two expressions are inter-exchangable. Effectively, my definition of 'force of gravity' is: that which is described by Newton's inverse square law of gravity.

My definition of inertia is: that which is axiomatically described by Newton's three laws.

I need to arrange my definitions in such a way that they also apply in the context of the general theory of relativity. As I often point out: in general relativity, the description of gravitation and the description of inertia have been unified into a description of properties of a single entity.

I think I like the following interpretation of newtonian dynamics: (an interpretation with general relativity in mind):

When I am standing somewhere on Earth, the force of gravity is being exerted on each atom of me. When I am standing on a weighing scale, then the end result of this distributed force is that the soles of my feet compress the spring of the weighing scale. So there is a force-vector associated with my center of mass, and there is a force-vector associated with the contact surface between my feet and the weighing scale. Because I am exerting a force on the weighing scale, the weighing scale exerts a Normal force on me.

There is an interesting role reversal as compared to the Pilot Training Centrifuge (PTC) story

In the case of the PTC, the causal sequence starts with the weighing scale exerting a centripetal force on the ball, and because of inertia there is a reaction force exerted by the ball on the weighing scale. In the case of me standing on a weighing scale, because of gravitation my feet are exerting a force on the weighing scale, and in response the weighing scale exerts a Normal force on me. --Cleonis | Talk 13:54, 28 January 2006 (UTC)[reply]

Addendum: I follow the modern interpretation of general relativity, in which the concept of gravitation is associated exclusively with curvature of space-time. In flat space-time, there is by definition no gravitational field. This is a matter of definition, not a matter of physics. --Cleonis | Talk 14:01, 28 January 2006 (UTC)[reply]

Simultaneity

Cleonis. Since you ask, I do think your discussion of Poincare's clock synchronisation is probably historically wrong. On that talk page I discuss it. What you put is interesting of course, but maybe a little complicated. Can I see your re-write before you post it? Thanks. P.S. I actually created (not just edited) the page with the idea of showing one thing that I thought was not well known. That relative simultaneity is not necessarily the same thing as time dilation. So I feel "protective" about the page and that point. Would not like to lose sight of that. E4mmacro 21:45, 1 February 2006 (UTC)[reply]

what does simultaneity do or is that even a word E-Series 19:52, 17 February 2006 (UTC)[reply]

Fictituous forces

See the talk page on centrifugal force : For some reason I am not good in explaining why the use of fictituous forces has no benefit whatever (at least, none of the claimed advantages of given examples hold). Perhaps you are better in explaining that appropriate mapping takes care of everything? Thanks, Harald88 11:59, 6 March 2006 (UTC)[reply]

Concentrating the gravitational mass of the reference ellipsoid in a point

GangOfOne wrote on the science reference desk:

[...] you say "for an object resting on the equator, the point of grip of the Earth's gravitation is about 10 kilometers away from the Earth's center", how did you compute that? [...] GangofOne 22:12, 21 March 2006 (UTC)

Recapitulating: I am looking for the point where all the mass would have to be concentrated to exert the same gravitational force as the reference ellipsoid does (in the case of an object that is at rest on a non-spinning reference ellipsoid).

Approximate calculation for object at rest on the equator:
An object at rest on the equator of a non-rotating celestial body with the shape of the reference ellipsoid, has a higher gravitational potential energy than the same object at rest on one of the poles. I will call the difference in gravitational potential between those two locations the 'pole-equator potential energy'.
An object at rest on the equator of the spinning Earth has a velocity of about 465 m/s. This corresponds to a kinetic energy of about 1.1 x 10^5 joule per kg of mass.
It is a theorem of the dynamics of rotation that in the case of perfect circular motion the kinetic energy and the potential energy are equal in magnitude. So in this case the pole-equator potential energy equals the kinetic energy.
Approximating the gravitational acceleration as 9.8 m/s^2 for every latitude, the 1.1 x 10^5 joule per kg of mass corresponds to a difference in height of about 10 kilometers.
This puts the theoretical point 'where all the mass would have to be concentrated' at about 10 kilometers away from the geometrical center of the Earth. Of course, this theoretical point is not a single point; its éxact location is a function of the precise geometrical locations of the objects that are involved.

About that 'equal in magnitude' of kinetic energy and potential energy. Here, the potential energy is defined as follows: the integral of the total amount of work that the centripetal force would do if it would be allowed to pull an object all the way to the axis of rotation. --Cleonis | Talk 14:28, 22 March 2006 (UTC)[reply]

You may also be interested in the equations near the bottom of the latitude page, geodetic vs. geocentric vs reduced latitude. May help w/ finding the direction of the vectors you are interested in. Also the conversion from geodetic to earth centered, earth fixed coordinates on the colorado.edu page may help. As far as the magnitude of the vectors go, i can't be much help. You have caused me confusion about the geoid's relationship to the reference ellipsoid. I know these two surfaces match very closely, but never considered why or what effect the rotation of the earth has on the shape of the geoid.

GangOfOne, if you stop by, feel free to visit the Datum page and change datums to data if you need to get it out of your system. EricR 15:57, 22 March 2006 (UTC)[reply]

"Datums" seems to specialized terminolgy of that field, so I will accommodate. (I studied Latin, so I am sensitive to Latin-abuse.) GangofOne 22:26, 22 March 2006 (UTC)[reply]

Cleonis, about your derivation, I don't know if it's right, need to think about it. You generally use a round-about reasoning process, which may or may not lead to the right answer (depending on whether every step is valid). That's good, being creative about it. But I don't see how bringing in the rotation of the earth can have anything to do with figuring out the static case, (except possibly because the elipsoid is an equilibrium shape, ie. special). However, it would seem that the direct answer to the question of 'What is the position of a point particle of the same mass, that gives the same grav field as some arbitrary shaped lump of matter.' has a conceptually simple answer. (For simplicity I will assume that the masses I mention are spatially homogeneous. And Newtonian gravity, not Einsteinian GTR) The gravity field at the test position is just

$\int {{\vec {r}}dm/r^{3}}\$ where r is distance to the point from dm. So once you know the Force_gravity, and the total mass (which is just $\int {dm}\$ ), then you just compute where to put a point particle that gives you that Force. So now the problem has been reduced to a calculus problem. Of course, in the special case of a spherical distribution of mass Newton proved it acts as if all the mass were at the center. Anything else, I have to calculate about to convince myself. GangofOne 23:13, 22 March 2006 (UTC)[reply]

The only purpose of making use of the kinetic energy associated with the rotation is that it gives a good approximation while minimizing the computational load.

What I am certain about is that at every latitude the gravitational potential energy (as compared to the gravitational potential energy of an object at rest on the pole) is equal in magnitude to the kinetic energy.

Anyway, there are several questions that I am very curious about:

In the case of an object resting on a pole of the reference ellipsoid, does that mass distribution act the same as all the mass concentrated at the geometrical center?
The more general case: for an object at rest at any latitude, what is for each configuration the position of the corresponding point mass? Specifically, is the location of the corresponding point mass always in the plane of the equator of the reference ellipsoid?

I don't have enough mathematical skill to perform the static case calculus myself. I do have sufficient calculus knowledge, I think, to follow the steps of an actual calculation of the static case.

By the way, I don't intend to use any result of this enquiry in any wikipedia article, I'm just very curious. --Cleonis | Talk 00:32, 23 March 2006 (UTC)[reply]

"I'm just very curious." Excellent. First, do we know the oblate spheroid is an elipsoid? Ie do we know the cross section through the poles is an ellipse? (that is , taking the case of a perfectly smooth fluid planet that has rotation of one rev. per day.) How can be prove this before we continue? GangofOne 06:40, 24 March 2006 (UTC)[reply]

How to prove that the cross section of the Earth is an ellipse?

All sources that I reviewed state that in the simplification of a body of mass with uniform density, the equilibrium shape for a spinning celestial body is an ellipsoid. (Of course, in the case of the reference ellipsoid for the actual Earth the fact that the iron core is much denser is also taken into account.)

The possibility must be faced that all writers on the subject of the shape of the Earth have merely copied from each other, without independent check. I assume that in the past some mathematician has given a rigorous proof, but I do not remember seeing such a proof, so I'm taking the statement 'The earth is to first approximation an ellipsoid' on good faith.

I would be interested in mathematical proof that the shape of a spinning celestial body is an ellipsoid. But I'm also contend with taking that shape on good faith. I'll try some googling, but I don't have much hope.

I prefer to take it as granted that the shape of a spinning celestial body is an ellipsoid. --Cleonis | Talk 08:18, 24 March 2006 (UTC)[reply]

I was a bit to pessimistic about finding a derivation, I think. For example, this PDF-document (125 KB) by David T.Sandwell about 'the shape of the Earth and its gravity field'. Sandwell refers to recent textbooks in which the full derivation is given. --Cleonis | Talk 08:58, 24 March 2006 (UTC)[reply]

MacCullagh's formula

Judson L. Ahern of the School of Geology & Geophysics, University of Oklahoma gives a derivation.

MacCullagh's Formula The gravitational potential of a near-spherical oblate spheroid. This formula can be used to calculate the gravitational potential of an object at rest on the surface of a stationary oblate spherroid.

Shape of a self-gravitating, rotating fluid body Ahern concludes: "Therefore, we find excellent agreement between the shape of the Earth and the shape predicted for a hydrostatic ellipsoid, arguing strongly for a fluid Earth on long time scales. In fact, the discrepancy (297.811 vs. 298.25) comes primarily from dropping higher order terms in our derivation."
--Cleonis | Talk 13:52, 24 March 2006 (UTC)[reply]

Connection with spherical harmonics

A comprehensive treatment: The Earth’s Gravitational ﬁeld (Available as part of the MIT open courseware project)

It is shown that the shape of the reference ellipsoid can be written as as a series of spherical harmonics. The J₂ term corresponds to the oblateness. In the MIT document it is stated that 'The higher order terms J₄, J₆ are smaller by a factor 1000. The higher order terms are incorporated in the reference spheroid.'
--Cleonis | Talk 10:31, 25 March 2006 (UTC)[reply]

centrifugal force

Hi Cleon, please have a look at Claim of "usefulness" of centrifugal force's "potential energy" on [[2]]. It may be useful if it's not only me explaining them how to map frames without invoking fictituous forces. :))

Cheers, Harald88 06:50, 21 April 2006 (UTC)[reply]

Hi Harald, my view on education of mechanics is along the following lines:

As you know, I think that a gut-feeling explanation of what physics is taking place should never invoke fictitious forces, only actual forces. On the other hand, it is also good physics education to teach resourcefulness in calculational approach, and having read the contributions by Henning Makholm and Yevgeni Katz I see that there are examples where invoking a potential of a centrifugal force leads to the shortest calculation.

Some people blindly assume that any calculational strategy that yields the correct answer corresponds to actual physics. (Some even go a step further, believing that the most economic calculation inherently corresponds to actual reality.)

That subtle distinction is hard to communicate on a wikipedia talk page. To some people it is a totally novel (and unthinkable) concept that a particular approach can be powerful calculational strategy, and nonetheless bad physics thinking.

So I'm reluctant to step in on the centrifugal force talk page. That which I would love most to communicate is so very hard to communicate. --Cleonis | Talk 00:08, 22 April 2006 (UTC)[reply]

Above I used the expression 'gut-feeling explanation', but a lot of people have an (erroneous) gut-feeling that there actually is a centrifugal force, so that expression is ambiguous. Basically I am referring to a principle. A principle that states: "any theory of physics must exclusively invoke physical interactions to account for effects that occur". Centrifugal force is not an interaction between (pairs of) objects, hence it is inadmissable as a physics explanation.

However, not all people adhere to that principle. Their reasoning goes as follows: "Our day to day experience is that when we release an object it accelerates towards the Earth's surface. We have invented the concept of "gravity" to account for that acceleration; the concept of gravity is a human invention. Likewise, we are perfectly entitled to invent a centrifugal force. If calculations that invoke that centrifugal force never lead to an inconsistency then centrifugal force is a perfectly valid physics concept." I think that is rubbish, but there is no arguing with that kind of reasoning. --Cleonis | Talk 08:33, 22 April 2006 (UTC)[reply]

I know, and it's even the reasoning that Einstein had in 1918. However, the example that is now discussed (likely to be next transfered to the article space) is simply a comparison between a calculation with and without pseudo forces - apparently many people have not received the mechanics education that we had, or they misunderstood it, making them believe that one has to use pseudo forces in order to make calculations simple. See also my new comment on the pseudo force talk page, it's very much related. Thus what's needed is not a debate, but an explanation of how the same problems are calculated in classical mechanics - isn't it astonishing? Harald88 10:52, 22 April 2006 (UTC)[reply]

I don't know the content of Einstein's thinking. Historians of science describe that Einstein abandoned Mach's principle in the early twenties. (More precisely, he abandoned Mach's principle as a way of interpreting/underpinning general relativity). But that is another matter.

( Yet another sideline: the historian of science John Norton has pointed out that Mach himself has never proposed Mach's principle. In one of his writings Mach sketches something that from a modern point of view can be regarded as Mach's principle, and Mach then proclaims that such a concept is so speculative in nature that it falls outside the scope of proper conduct of science. Source: How Hume and Mach Helped Einstein Find Special Relativity '(PDF-file 435 KB) (the remark about what later became known as Mach's principle is on page 28) )

Interesting reference, thanks I plan to read it. :) Harald88 10:47, 23 April 2006 (UTC)[reply]

About calculations: let a test-tube, 1 cm in diameter, 10 cm in length, be mounted on a centrifuge wheel with a large diameter, say a meter. When that centrifuge is rotating at a constant rate, pulling G's, then obviously the simplest approach is to treat the system as if there is a gravitational potential over the length of th test tube. So it is obvious to me that there are situations where using a fictitious force (with a corresponding potential) is the most efficient calculational strategy. In other situations it is a matter of taste whether to use a rotating coordinate system or to use an inertial coordinate system.

That is the question: until now there only have been vague claims, while every time that I checked it, it appeared that approximately the same calculations are required in both cases. The only way that I see to make it a little simpler is if G is given (measured) and not omega. And the claim that classical mechanics requires one to stick to an inertial frame of reference are bogus, but again repeated as argument on the corresponding talk page. We are now in the process of working out an example on that talk page that when ready should be included in the article. However, I'm not sure to correctly interpret the example. Harald88 10:47, 23 April 2006 (UTC)[reply]

I have used the animations that I made for the Rotational-vibrational coupling article sevreal times as an illustration. I believe the the rotational-vibrational coupling article shows that looking at the motion from an inertial point of view is the most simple approach. I contribute articles like rotational-vibrational coupling, and hopefylly the style of thinking of that article will eventually rub off. My experience is that arguing on talk pages is non-productive (and sometimes even counterproductive)

Please watch the Foucault pendulum article. I am working on an animation for that article. Very few sources mention that the Foucault pendulum precesses with respect to the fixed stars. --Cleonis | Talk 18:10, 22 April 2006 (UTC)[reply]

OK, keep up the good work! Harald88 10:47, 23 April 2006 (UTC)[reply]

PS now that I tried to understand the last example on the centrifugal force talk page, I have the impression that it is in fact erroneous - which is quite possible as one easily makes mistakes when using that concept, it's one of the reasons why I dislike that approach. Still, several people seem to think it's correct! Thus, could you please at least have a look at it and give me your opinion?

I haven't made any calculation yet, but the way I understand it, one may choose different paths in which the caroussel imparts different amounts of kinetic energy to the object (e.g. nothing at all or a straight line relative to the rotating frame) - which should result in different end speeds, right? Harald88 11:33, 23 April 2006 (UTC)[reply]

OK, I now understand that of course the question was about speed relative to point B - thanks for your clarification there. :)

Response from Rracecarr about Foucault pendulum

You are right that Anders Persson supports your claims--but he is wrong. If you have a physics background, you will recognize that the claim that precession relative to the stars requires that a force does work is a false statement. If a force is doing work, it must increase the energy of the system, but the amplitude of the swing in actuality does not increase. No force does any net work on a Foucault pendulum. The many typos and grammatical errors also indicate that the Anders Persson article is not a peer-reviewed paper from a reputable journal. I am sorry that some of the references on the web have steered you wrong. It is a shame that there is so much misinformation out there. But let's not make Wikipedia another example of this.

Here is a reference that is actually right-- http://www.physics.orst.edu/~mcintyre/COURSES/ph429/foucault.pdf They mention Marion Thornton. I own the 4th edition of this classic text, and sure enough, on pages 398-401 the precession of a Foucault pendulum is carefully worked out. Nowhere does the oblate shape of the earth enter the calculation. All that matters is the local vertical component of the rotation of the earth. That is, the rotation of the earth may be represented as a vector parallel to the earth's axis, with a magnitude of one cycle per day. At any location on the earth, the locally vertical component of this vector is one cycle per day times the sine of the latitude. It's simple trigonometry. The explanation of the precession of a Foucault pendulum is really very simple: the earth turns under the pendulum (in the rotating frame of the earth, an equivalent explanation invokes the Coriolis force). At the poles, this is easy to see, but at other places on the earth, it is still true--only the pendulum only sees the local vertical component of the rotation. This means that at other locations besides the poles, after one day, the pendulum will have made less than a full precession cycle, even though it has returned to the same postion on earth relative to the stars. Thus it has precessed relative to the stars. The cause of this precession is the change in the direction of gravity as the earth spins: gravity always pulls toward the center of the earth, but as the earth turns, that direction changes relative to the stars.

The following statement in Foucault pendulum must be revised, because it is flat wrong: "The origin of the force that is involved in the precession of the Foucault pendulum with respect to the fixed stars is in the equatorial bulge of the Earth." Here is the deal: it is pretty easy to understand the precession of the pendulum relative to the earth--the earth literally turns under the pendulum. The reason for the precession relative to the stars is not so easily grasped, but has to do with the changing direction of gravity (relative to the stars) with time. This changing direction can be described in terms of a "poleward force". This "poleward force" is the same thing that causes the equilibrium shape of the earth to be flattened from a perfect sphere (the so-called equatorial bulge). HOWEVER, IN NO WAY is the equatorial bulge the SOURCE of the poleward force, or of precession relative to the stars.

To reiterate, take a hypothetical, perfectly spherical earth, which due to its rigidity resists the flattening influence of rotation--a huge ball bearing. A Foucault pendulum on this earth would precess relative to the stars in the same way as one on the real earth, and due to the same "poleward force".

You misunderstand. In the hypothetical case of a perfectly rigid, perfectly spherical rotating celestical body, the poleward force would be absent, and the Foucault pendulum would precess differently. Most importantly, the foucault pendulum would not precess through a complete circle. Instead, once the pendulum would reach a state of east-west swing, the pendulum would remain locked in that east-west swing.

I propose that we declare total failure of communication. I am unable to reach you. You disregard the evidence that is presented to you. --Cleonis | Talk 19:51, 4 May 2006 (UTC)[reply]

I have gone ahead and edited the article. I have tried to leave as much of your work undisturbed as possible, while removing the most important fallacy. You are incorrect that the poleward force would be absent on a spherical earth. You are incorrect that a Foucault pendulum would remain in an east-west swing on a spherical earth. I will agree to your "failure to communicate" proposal, and won't leave you any more messages after this. I can't resist a final attempt though. Please forget that it is me talking (I know you don't like me) and consider this: do you think that a Foucault pendulum at the north pole of a spherical earth would not precess relative to the earth? Of course it would--it would precess once per day, as the earth turned under it. Ok, now put a Foucault pendulum a mile away from the north pole (still on the spherical earth), and swing it in the east-west direction. Would it precess relative to the earth? Sure, and at very nearly once per day. Where is the magical cutoff as you move away from the pole where a the precession of a Foucault pendulum swinging east-west will suddenly drop to zero? Rracecarr 20:08, 9 May 2006 (UTC)[reply]

Sagnac-effect

Hello Cleonis, I like the animation you made. However, at the moment I don't have the time to incorporate it in the article. If you have the time, maybe you can do it yourself, or wait a few weeks. BoH 22:40, 29 July 2006 (UTC)[reply]

:Image:Foucault_pendulum_precession2.png

the animation by DemonDeLuxe
Hi Cleonis,

I see that you made that illustration showing the precession of the Foucault pendulum. I have done a little animation showing the pendulum and are a little bit baffled as far as the precession is concerned. As you can clearly see, our graphs are different insofar as yours is "pointy", which I, honestly, can't understand. The pendulum's motion gets slower the farther the weight swings, so there's more "earth turning underneath" per centimeter (or second) of swinging than more to the center of the motion. The result, as far as I can see it, should be a graph like is shown in my animation. In your graph, however, it looks as if the pendulum would swing at constant speed, until it abruptly turns around and instantly swings back at the same speed! But seeing that you seem to have a profound understanding of physics (which I don't - my graph simply comes from telling the rendering program to turn the whole pendulum around by 360 degrees) there might be something I have missed.

Which graph is correct? --DemonDeLuxe 19:18, 13 August 2006 (UTC)[reply]

Thank you very much for your very detailed answer. "But" *g*. The obviously quite common image of a "spiked" graph indeed seems to come from the pendulum swinging through fine sand, which would explain the sharp points (since such a graph would typically exclude the outside parts). All physical observations aside, here's what I actually told the rendering program to do:

a) swing the pendulum back and forth 8 times, with the speed curve being a sine curve (AFAIK it is not too easy to calculate the exact formula but I was told a sine curve would be a sufficient approximation).

b) in the course of those 8 double swings, turn around the whole pendulum by 360 degrees (it shouldn't matter whether you turn the pendulum or the earth underneath, right?)

Other than with most real pendulums, the graph is drawn over the whole movement. It mimicks a pendulum filled with sand (or any other more or less fluid substance) - in the 3D program, it's an eruption of particles coming from the tip of the pendulum, going from there in a straight line elongating the pendulum axis and stopping as soon as they hit the floor.

Still, I think that, seeing that the pendulum's movement must get slower towards the extremes of it's swings (otherwise a swing would never end), while earth continues it's spin at constant speed, the relative sideways shift per cm of "graph path" must be greater than in the middle of a swing at the pendulum's maximum speed.

As for the graphics and the sheer data volume: I've been dabbling in 3dsmax for a couple of months now. While I am a web guy by profession (and in fact AM what you could call a "bandwidth nazi" fighting for every byte normally), the aim of this animation was to actually show a full turn while still having some visual appeal (thus the mirror image of the pendulum on the reflective surface of Foucault's portrait). I may actually have gone a bit overboard with that one *g* Have a look at Image:Newtons_cradle_animation_book.gif, that's more my normal way of a compromise between style and bandwidth ;O) Regards, --DemonDeLuxe 07:54, 18 August 2006 (UTC)[reply]

Equivalentie principe

Hallo Cleon; in the dutch wikipedia "Equivalentie principe" was still absent so I tried a translation of the english one. Maybe you could have a look and see if i did not introduce too many stupidities ? From the discussion page on the english version i understand that you are in favour of inserting General Relativity points in the explanation of the "Einstein equivalence principle". From a didactic point of view (i am a teacher) i would oppose that: i think the Einstein Equivalence principle should be explained without referring to more complex theories. But please have a look. Sjoerd22 12:04, 20 August 2006 (UTC)[reply]

Your comment was that you do not approve of the Equivalence principle article as it stands. My opinion is that the "principle" is maybe not a major physical law or even insight, but from a didactic point of view it could be valuable to introduce relativity to someone who understands Newtonian mechanics. So unless you can be more clear about at what point the "story" is incorrect i will let it stand as is. Thanks for your comments.

Sjoerd22 20:23, 20 August 2006 (UTC)[reply]

GR sources

Hi Cleonis. Thanks very much for the sources. By the way, I didn't realise that there was an article on the Eötvös effect - thanks! MP (talk) 12:18, 30 October 2006 (UTC)[reply]

Bad link

Cleon - Please check the last article link in General_relativity_resources#Special_topics_2. I cannot get it to work. --EMS | Talk 04:46, 2 November 2006 (UTC)[reply]

Thanks for fixing it. As for why the "www" was not approrpiate: You need to look at the first part of a URL as a computer name or alias. "www" by convention identifies the primary world-wide-web host for a given site. However, as long as the right machine can be found by a DNS lookup, a URL can start with anything that you like. (In fact, if the URL starts with numbers and dots, the computer is being identified with an IP address, but that is fairly unusual these days.) --EMS | Talk 15:14, 2 November 2006 (UTC)[reply]

Centrifugal force again

I did a little clean-up of the intro, based on the last discussion on its Talk page. Could you check it and help out if needed? Harald88 11:00, 11 November 2006 (UTC)[reply]

Twin paradox and GRT "explanation"

Please comment on Twin paradox, as there some people seem to want to suppress the modern and notable view that GRT is a theory of gravitation and that Einstein's 1918 equivalent "gravitational field" on an accelerated body in free-space is a pseudo-field, and not a "real" field as he claimed at the time. (it doesn't really matter if you or I agree or disagree, this a NPOV matter; but I am sure that you understand the topic). Harald88 00:02, 26 November 2006 (UTC)[reply]

Image:Foucault pendulum plane of swing.gif

Good job on the animation Image:Foucault_pendulum_plane_of_swing.gif. It could be improved, in my opinion, by removing the text "Day ." at the top. The benefits would be that (1) the picture can more easily be included in other language wikipedias and (2) half of the frames can be removed as the pictures for "Day 2" are identical to the ones for "Day 1" at the same hour. Cheers. --PeR 20:36, 3 January 2007 (UTC)[reply]

I designed that animation the way I did because I wanted to emphasize that it takes a foucault pendulum located at 30 degrees latitude 48 hours to precess a complete cycle. With a 24 frames version this could easily be overlooked. So I very much prefer a 48 frames version over a 24 frames version. (The animation is 50 KB in size, so it's not a bandwith guzzler.)

However, I agree that a language-independent version would be better. It occurred to me that a marker can be placed at the perimeter of the plane of swing, just as a marker could be moved along the perimeter of a foucault pendulum display, keeping it aligned with the plane of swing. That would vividly show that precessing through a complete cycle takes two days. I will see what I can do. A possible name for a thus reworked version could be Foucault_pendulum_plane_of_swing_marker. --Cleonis | Talk 01:02, 4 January 2007 (UTC)[reply]

The fact that it takes two days for one full revolution is something that is best pointed out in the caption, imo. A plane of swing is symmetric in itself, so I see no reason to depict it as asymmetric. For the extra bandwidth I'd rather have a smoother version with one frame every half hour. But that's just me. --PeR 08:38, 4 January 2007 (UTC)[reply]

POV-Ray

Hey man, you have some good 2D animations out there! Nice to see you're interested in POV-Ray, it's quite a nice program, especially for those into mathematics and programming. :)

Go to POV-Ray's website and get it if you haven't. There's MegaPOV out there as well, which has a lot of nice advanced features (like physics simulation and whatnot.) I don't quite like it, though. Kind of misses the point (there's a lot of better programs out there for this sort of advanced motion thing)

The thing is, POV-Ray is a lot more low-level than most people realize. You have to get your hands in a lot of math to make things move nicely and look nicely, which is exciting and challenging. But since we're talking mostly about diagrams and models here, most animations won't be too complex. A trick I found is that you shouldn't try to make everything at once, as in, figure the equations to move everything as it should all at once. That's usually too complicated to bother. Instead, divide the animation into steps and just set the code for each part of the animation.

Take the Villarceau circles animation, for example. This image has 8 different steps (If you check the POV-Ray source included, you'll see all the equations for these commented out):

Fade in plane
Move plane for slice motion
Fade out plane and half-torus while moving the half-torus away
Fade in circles
Rotate everything
Fade out half-torus
Fade in full torus
Fade out circles

I've done the same thing with the gyroscope, coffee mug to torus morph and several other animations not for Wikipedia. It's a pretty good approach.

Now, POV-Ray comes with an excellent documentation already, so just follow the tutorial, play with it a bit and then give it a shot on animations. It's actually quite easy, really.

On the top of the editor window, you'll find a little edit box. You just have to type in here +kffN and render, and most things will be set for animation already. +kff means "clock final frame", and the number in front of it is the number of frames to render.

The keyword clock (lowercase) is used in the POV-Ray source code to represent a time variable, the single one you'll use when rotating, translating, changing colors and whatever else you want. By default, it goes from 0 (first frame, #1) to 1 (last frame, specified after +kff.) You can change these, of course:

+ki2 +kf5 — clock will start at 2 (+ki = clock initial) and end at 5 (+kf = clock final)
+kfi11 +kff20 — first frame will be 11, last frame will be 20 (with clock running from 0-1 or whatever else you specify instead)

Setting different start and final frames is very useful for the per-part rendering I told you about. You want to render, say, 10 frames in a fade in? Use +kff10, and use your equations so when clock = 1, that part is over (frame = 10.) Wanna render part 2 now, in 20 frames? Use +kfi11 +kff30. clock will, again, run from 0 to 1 throughout these frames.

Mug morphing to a torus
I find it very useful that clock runs from 0 to 1 by default. It's good because you usually don't have to bother thinking about clock in any other range, since you can use simple functions with that to interpolate things non-linearly (what usually looks like crap.) My personal favourite is pow(sin(clock*pi/2),2) ( $\sin ^{2}clock{\frac {\pi }{2}}$ ), since it creates a smooth and very natural-looking acceleration and deceleration movement (not surprisingly, of course) like the one in this image with the coffee mug.

Now, for the sort of thing you're trying to do, like the vibrating tubes of a mass flow meter, if you're looking for actual "bending" of shapes you might need a few more advanced objects, like isosurfaces or parametrics. The problem is that these are usually very slow to render in POV-Ray (the torus to mug animation had to use isosurfaces and took a while...) The reason they're slow to render is that POV-Ray, by design, deals with pure mathematical objects. A sphere in POV-Ray isn't a triangle mesh, it's an actual, perfectly mathematical sphere. So isosurfaces and parametrics have to be approximated during rendering, what usually takes a lot of processing and can give awful results if not properly set. There's a nice tutorial on them over this website, which also includes links to a few POV-Ray scripts that may save you a lot of trouble by converting your isosurfaces into meshes, which are a whole lot faster to render. But ideally, you should try dealing with basic primitives before you use that sort of thing.

Well, I'm kind of throwing all hints I have here in order to be helpful, so forgive me if I'm completely missing your point when you asked for help. In any case, just ask whatever else you want and I'll do my best to help.

Good luck, and don't hesitate to ask for further help! — Kieff 05:12, 15 January 2007 (UTC)[reply]

Reply about the two images

Cleon, I replied on my own talk page. Cbdorsett 20:30, 30 January 2007 (UTC)[reply]

Mass flow meter

Thanks for the tip! Great animations too! --Frokor 22:31, 2 February 2007 (UTC)[reply]

Deleting images

Hello and thank you for your contributions to Wikipedia. I know this happened over a year ago and you've probably learned better by now, but I thought I'd tell you that you could have tagged Image:Twins paradox diagram02.png with {{db-author}} instead of {{ifd}}. —Remember the dot (t) 20:57, 7 February 2007 (UTC)[reply]

Een vraagje

Hoi! Kun jij misschien proberen de engelse wikipage coriolis effect te updaten? Het is daar zeg maar een zootje. Qwestie van twee dingen onderscheiden: (1) de eigenlijke corolis effect waardoor een bal van de rechte lijn afwijkt (zie film van een bal op een rad; link staat onderaan) en (2) de visuele illusie van een in een recht lijn lopende bal. Maar ik vermoed dat je bete gescikt ben om het te updeten. Kun jij da doen alsublieft? Ik heb al daar een disussie geopened bij de "talk" pagina van de artikel. En -bij de wey- de coriolis animatie is ook maar eenzijdig. Ik heb een beetje rond gezien dat je ook een beetje "animatieknobbel" hebt. Dus is er dringend een andere animatie nodig van de andere (aktuele) coriolis effect zoals dat hier in een mpeg onderaan op deze pagina mooi staat afgebeeld: http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/crls.rxml Dus als het je geen moeite kost, aan de slag zou ik zeggen. Alvast bedankt. Benua 13:38, 27 February 2007 (UTC)[reply]

Foucault Pendulum

Explaining the Foucault pendulum through parallel transport is not my theory, but became popular in the mid-80s and is pretty much standard knowledge in physics by now. I learned about this from comments by Frank Wilczek in the book cited in the article.

You do a lot of talking about how the poleward force is responsible for the pendulum. How about actually writing down a derivation for the motion of the pendulum? Or pointing me to a place where I can find one? All I have seen so far is a bunch of claims and some diagrams explaining the poleward force. Nowhere have I seen the corresponding differential equations been written down.

The poleward force and the shape of earth have an influence on the direction of the plum line. This influence is not very big, and is commonly absorbed in the choice of which definition of latitude one works with. I claim (and my derivations show) that the direction of the plum line is really all that matters to the Foucault pendulum.

You claimed that Wheatstone's device works like a Foucault pendulum. We agree on this. Do you believe that an analogue of the "poleward force" is also responsible for the precession in Wheatstone's device?

--ShanRen 13:53, 3 March 2007 (UTC)[reply]

I responded on my tak page. --ShanRen 18:31, 3 March 2007 (UTC)[reply]

And another response on my talk page. --ShanRen 23:25, 3 March 2007 (UTC) (thanks for catching the spelling mistake.... )[reply]

Responded. And I have no idea what you mean by "The challenge is to set up the equations for motion in geometrically flat 3-space". But I guess neither do you. --ShanRen 03:38, 4 March 2007 (UTC)[reply]

The usual approach is to set up the equations for motion in curved 2-space (either a curved configuration 2-space or curved physical 2-space; the two representations are isomorphic)

I want to explore the representation where the motion of the pendulum bob is represented with respect to standard 3-space. Just good old euclidean 3-space. I think there are certain advantages to representing it that way. --Cleonis | Talk 09:40, 4 March 2007 (UTC)[reply]

Do you have no idea what you are talking about, do you?

Response on my page. --ShanRen 15:19, 4 March 2007 (UTC)[reply]

Response on my page. --ShanRen 02:57, 14 March 2007 (UTC)[reply]

Time to open your phyisics book. --ShanRen 02:57, 14 March 2007 (UTC)[reply]

Evidence for what? Why are you writing this? --ShanRen 17:18, 27 March 2007 (UTC)[reply]