Talk:Foucault pendulum/Archive 1

The California Academy of Sciences
I suggest removing the link: from the External links section.
 * "About Foucault pendulums". California Academy of Sciences, 2003.

The cartoons are gorgeous, but the site presents misleading and wrong information. On the third page two questions about the world are confused. Whether the Earth is flat, and whether the Earth is rotating.

In the community of greek scholars, there was consensus that the Earth is a sphere, since the greek scholars were aware of convincing evidence that the Earth is spherical. All through the centuries it has been recognized by scholars that given the evidence it is clear that the Earth is spherical in shape.

(The american writer Washington Irving concocted the myth that during the middle Ages European scholars believed the Earth to be flat. Unfortunately this myth, started in 1826, is still widespread.)

It is impossible to see whether the Earth is rotating. It was assumed that the Earth is statonary, because given the knowledge available at the time there were compelling reasons to do so.

On the seventh page it is stated:
 * Note that while the Pendulum seems to change its path during the day -- actually it is the floor beneath it that changes, being twisted around by the daily rotation of earth.

This statement is wrong. At latitudes away from the Poles and the Equator the plane of oscillation of the pendulum is rotating both with respect to the Earth and with respect to the stars.  --Cleon Teunissen | Talk 08:47, 21 July 2005 (UTC)

In the past weeks no objection has been raised, so I have removed the link from the external links section. --Cleon Teunissen | Talk 09:49, 12 August 2005 (UTC)

Pendulum Precessing
It appears that it is incorrect to state that the pendulum precesses when it is the surface of the Earth that is turning underneath the plane of the pendulum swing. It is a misleading attribute of the pendulum swing.David Harty 03:11, 25 April 2006 (UTC)


 * You seem to be referring to a polar pendulum. In the case of a polar pendulum, the pendulum does not precess with respect to the fixed stars, the Earth is turning underneath the plane of the pendulum swing.


 * On all other latitudes the plane of swing does precess with respect to the fixed stars.


 * Conversely, if the pendulum swing would always be stationary with respect to the fixed stars, then on all latitudes it would take one sidereal day for the pendulum to precess through a full circle with respect to the Earth. That is not what is observed.


 * It would be erroneous to claim that on latitudes away from the poles the direction of the plane of swing (with respect to the fixed stars) does not change. --Cleonis | Talk 15:07, 26 April 2006 (UTC)

Because the Earth rotates. The obfuscation is dramatic.


 * Well, I'm a messenger here. The pattern of motion that I describe is the motion that is observed. It may go against expectation, but it's the reality. --Cleonis |

Talk 21:33, 26 April 2006 (UTC)

The motion of the plane of the pendulum swing in relation to the Earth has nothing to do with the motion against the stars. You continually distort information to your own benefit as you have above. Previously you wrote,


 * I propose to not merge. The information in the Foucault Pendulum vector diagrams article is very technical, it is on a much different level as compared to the current Foucault pendulum article. If the two articles were to be merged, it would still feel as two distinct articles. --Cleonis | Talk 13:51, 9 January 2006 (UTC)


 * The comment to merge was originally made when there was only one paragraph written for the article. The purpose of the first article is an overview of the Foucault pendulum and I would not like to clutter it up with details that are not pertinent to an overview. I think the first article is fine the way it is. I wrote the second article to explore details not fully explained in books and to attempt to provide diagrams rather than words. If the information gets moved to a Wikibook, that would be fine too, but this was as far as I intended to go at this point. Yes, the current article is Foucault Pendulum Vector Diagrams not Diagrams for Foucault Pendulum. Diagrams for Foucault Pendulum should be deleted. David Harty 05:28, 10 January 2006 (UTC)

Retrieved from "http://en.wikipedia.org/wiki/Talk:Foucault_pendulum"

Now you write to promote your own agenda. Wikipedia should review your actions. Next you will promote the Eotvos effect (again) as the discerning force of the Foucault pendulum. You should read the final comments in the Coriolis Effect Talk if you think you are providing a clear approach to discussion of the subject. You should write your own article and provide links to it rather than continuing to engage in obfuscation.


 * I regret to read that you experience my additions to the Foucault pendulum article as obfuscation. My agenda is to get the physics right. My sources of information are articles that have been published in peer-reviewed journals. I have listed references: the articles by Dale Durran, Anders Persson and Norman Phillips.


 * I agree with you that I have not been self-consistent. In januari I advocated not merging the Foucault Pendulum Vector Diagrams article with the Foucault pendulum article, saying that the diagrams article was very technical in content. What I have recently added to the Foucault pendulum article is quite technical. I sincerely hoped that when I would add the animation and diagrams, they would succeed in convincing you. So yes, when I advocated to not merge, I was in fact trying to buy some time.


 * Of course, the Eötvös effect is not the discerning force of the Foucault pendulum. I made remakes of a diagram designed by Anders Persson. My remakes and the diagram by Anders Persson have the same purpose: to show the nature of the relation between the Coriolis effect as it is taken into account in meteorology, and the Eötvös effect as it is taken into account in taking gravimetric measurements. Image:Coriolis_Eotvos_eastward.png Image:Coriolis_Eotvos_westward.png --Cleonis | Talk 00:28, 28 April 2006 (UTC)

One has to ask the question, "What would be the observed period of rotation of the plane of the pendulum swing if the Earth were a perfect sphere". Since the period of rotation depends on the degree of alignment of the central axis of the pendulum with the axis of rotation of the Earth, there would be no difference in the observed rotation between a perfect sphere and an oblate spheroid. In another article the following statement is made;


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

It may be this is another example of obfuscation.David Harty 21:28, 1 May 2006 (UTC)

perfectly spherical celestial body thought experiment
I copy and paste from above: "One has to ask the question, 'What would be the observed period of rotation of the plane of the pendulum swing if the Earth were a perfect sphere'. David Harty 21:28, 1 May 2006 (UTC)"

That is an interesting thought experiment. (It is not a physically realistic thought experiment, an actual celestial body will assume a shape that matches its rate of rotation, but that does not detract from the thought experiment being interesting.)

The article describes that the pendulum precesses because the centerseeking force is doing work. In the hypothetical case of a perfectly spherical celestial body that is rotating, the poleward force would not be present at all.

In the case of a Foucault pendulum on Earth the center-seeking force pulls the pendulum bob closer to the nearest pole each time the pendulum bob swings from east-to-west. In the case with no presence of the poleward force, a pendulum that swings back and forth in east-west direction would remain swinging back and forth in east-west direction. So in the hypothetical case of a perfectly spherical celestial body, a pendulum that is started in south-north direction will precess somewhat, but once an east-west swing has been reached the pendulum will from then on remain locked in that east-west swing. (East-west with respect to the Earth it is suspended above.)

The perfectly spherical celestial body thought experiment illustrates the importance of the poleward force in shaping the precession of the Foucault pendulum. --Cleonis | Talk 22:56, 1 May 2006 (UTC)


 * Interesting, but one has to ask about the proof of Foucault's Sine Law provided by Louisville in 1851 (as provided in various texts) that "the rate of rotation of the Earth, viewed as a vector, is equal to the sum of two vectors; rotation about the vertical (the center-line of the pendulum) and rotation about a line perpendicular to it and aligned with the local meridian. So the assumption that the two vectors are additive would be invalid and then Foucault's Sine Law would also be incorrect in this particular example of the spherical Earth.  Then the title of Persson's article becomes apropos, "The Coriolis Effect - A Conflict Between Common Sense and Mathematics".David Harty 08:11, 2 May 2006 (UTC)


 * You have raised two issues:
 * 1) whether a pendulum located on a perfectly spherical rotating celestial body would exibit the same precession
 * 2) whether equatorial bulge verses perfect spherical shape matters significantly for dependency of rate of precession on latitude.
 * The second issue cannot be assessed, for in the case of a Foucault pendulum located on a perfectly spherical rotating celestial body, the pendulum will not precess through a whole circle anyway, the cases don't compare.


 * But there is also a purely formal way to address the same question. The equation of motion for motion with respect to a rotating coordinate system has a centrifugal term and a coriolis term. If the poleward force is present (and if the poleward force matches the rotation rate) then the term in the equation for the poleward force and the centrifugal term drop away against each other. Then the motion is described with the coriolis term only.
 * When no poleward force is present, then the centrifugal term will be much larger than the coriolis term. Then the motion is described with the (dominant) centrifugal term and the coriolis term together.
 * But in both cases: poleward force present or poleward force not present, the coriolis term will be the same; the same dependency on latitude.
 * If one chooses to define that the 'coriolis force' and the coriolis term in the equation of motion are one and the same, then it follows that the dependency of the coriolis force on latitude is the same for   a rotating celestial body with equatorial bulge and a rotating celestial body that is perfectly spherical. (More precisely: in the case of the Earth, with its 1/300 flattening, any difference is way negligable.) --Cleonis | Talk 13:38, 2 May 2006 (UTC)

Problems with last three sections in this article
The sections titled "The forces exerted on the pendulum bob", "Comparison with driving on a banked circuit", and "Method of calculation" are full of information that is mostly off the point, misleading, and in many cases wrong. For example, the first sentence of the first section I mentioned says: "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." This statement is false. A Foucault pendulum would precess in essentially exactly the same way on an Earth that was perfectly spherical, but had the same rotation as ours. In general all the information about the centripetal force is misleading, because it really doesn't matter at all. In the coordinate system rotating with the Earth, the centripetal force can be lumped in with gravity, changing the direction of the true gravitational force a bit, and forgotten about. Not so for the Coriolis force, which depends on the motion of the object in question. The Coriolis force is the thing responsible for the precession of the Foucault pendulum. Another falsehood is "The slower the velocity with respect to the Earth, the straighter the path." The Coriolis force is indeed stronger for a faster-moving object, but so too, the momentum of the faster moving object is greater, with the end result that path curvature is greater for a slower moving object, not a faster one. I recommend a major revision of these three sections, or else removal.

Rracecarr


 * Although only reading this comment (I'll do my best later to be more informed), I can already conclude that Rracecarr's objection can't be correct, at least not in full: the "Coriolis force" is a pseudoforce that results from looking at the centripetal acceleration, and thus not be truly causal; and the centripetal force can only be accurately included with the gravitational force and forgotten about, if it is constant. However, that can obvioulsy not be generally true for an object that isn't in perfect circular motion together with the earth surface. Incidently, I vaguley remember a balloon/zeppelin example by Cleon in which he very nicely explained this often overlooked effect (Cleon, do you know what I refer to?). However, I have at this moment no opinion about the claim of the precise origin of the force - it's directly from literature or inferred, and which article in particular? Harald88 21:13, 3 May 2006 (UTC)


 * A geophysics effect that is sideways related is the Eötvös effect. In that article I present an example with an airship. Harald,  you have that Eötvös effect example in mind. Understanding the Eötvös effect is helpful for understandig the coriolis effect as taken into account in meteorology and analysis of the precession of the plane of swing of the Foucault pendulum. --Cleonis | Talk 00:02, 4 May 2006 (UTC)


 * The Foucault pendulum precesses because of the Coriolis force. Read the first sentence of this article (Foucault pendulum).  The centrifugal and Coriolis forces arise from the rotation of the reference frame.  The Coriolis force depends on the motion of an object, but the centrifugal force is proportional to the mass of the object but independent of its motion, and so may be included with gravity, even for objects not at rest relative to the surface of the Earth.  Rracecarr 22:05, 3 May 2006 (UTC)


 * the information I have provided is has been published in peer reviewed articles. Those articles have been listed in the references section. Please study those articles before you proceed. --Cleonis | Talk 06:25, 3 May 2006 (UTC)


 * I have looked at the references. They are good.  However, you have misinterpretted them.  Much of the information you have posted is relevant to so-called inertial oscillations, not to Foucault pendula.  Inertial oscillations occur at the Coriolis frequency, (on the order of 1 cycle per day), and the restoring force for inertial oscillations IS produced by the equatorial bulge.  The geopotetial surfaces of the rotating earth are oblate, but in a non-rotating frame, geopotential surfaces are spherical.  This difference provides the restoring force for inertial oscillations.  It does NOT provide the restoring force for a pendulum--that is provided by gravity acting on the bob as it follows its circular arc around the pivot point.  A pendulum has a much much higher frequency, and all the stuff about poleward forces and cars on banked tracks is irrelevant.  I can see you have put a lot of work into this material.  I think you should create an article on inertial oscillations, or perhaps add your stuff to the page on the Coriolis effect.


 * I completely agree with Rracecarr. The named three sections should be removed from this article. Much of it is present on the Eötvös effect page, where it is more appropriate. --PeR 17:04, 10 November 2006 (UTC)


 * In retrospect, I think the exhaustive explanation I added is overdone for the wikipedia foucault pendulum article. So I won't object to cutting away much of it.
 * What is omportant, I think, is to emphasize the aspect that on any latitude other than the poles the pendulum bob does not keep swinging in the same direction (with respect to the fixed stars). --Cleonis | Talk 14:17, 12 November 2006 (UTC)

References for the Foucault pendulum article

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


 * Persson, A.
 * How do we Understand the Coriolis Force? Bulletin of the American Meteorological Society 79, 1998, 1373-1385.
 * The Coriolis Effect: Four centuries of conflict between common sense and mathematics, Part I: A history to 1885 History of Meteorology 2 (2005)


 * Norman A. Phillips
 * An Explication of the Coriolis Effect, Bulletin of the American Meteorological Society: Vol. 81, No. 2, 2000, pp. 299–303.
 * What Makes the Foucault Pendulum Move among the Stars? Science and Education, Volume 13, Number 7, November 2004, pp. 653-661(9)

Specifically, the Bulletin of the Americal Meteorological Society (BAMS) and the Meteohistory.org site count as peer-reviewed journals. You should not oppose information that comes from articles that have passed the rigours of getting it published in a peer-reviewed journal.

If you do a google search of the respective authors, you will find they are well respected experts in the field of metorology, specialized in the mathematics of modeling the motions of air over the Earth. Understanding the mechanics of the coriolis effect is the very core of their expertise. --Cleonis | Talk 06:25, 3 May 2006 (UTC)

Addendum: The expertise of meteorologists is the coriolis effect as it is taken into account in meteorology. The connection with the Foucault pendulum is that both in meteorology and in the case of the Foucault pendulum the mass is constrained to motion that is parallel to the Earth's surface. In ballistics, motion is not constrained to the surface, and because of that the physics of ballistics and the physics of meteorology are not analogous. --Cleonis | Talk 15:01, 3 May 2006 (UTC)

The swing of the pendulum; and the precession of the plane of swing
I copy and paste from above: "I have looked at the references. They are good. However, you have misinterpretted them.  [...] I can see you have put a lot of work into this material.  I think you should create an article on inertial oscillations, or perhaps add your stuff to the page on the Coriolis effect. Rracecarr"

No, I have not misinterpreted the references. About the frequency of inertial oscillation: close to the poles, the frequency is one cycle per 12 hours. Further away from the poles the cycle takes longer. At every latitude, the period for completion of one inertial oscillation is half the amount of time of completion of a precession cycle of a foucault pendulum. It is unclear what you mean by "the bob as it follows its circular arc around the pivot point". One of the characteristics of a Foucault pendulum is that it is released with great care. The swing of the pendulum, as seen from the side, must be strictly planar. The pendulum bob does not follow a circular path with respect to the suspension point; the pendulum bob must only swing back and forth, otherwise you don't see anything special. As far as I can tell, you have confused on one hand the swing of the pendulum, which, in for example the original demonstration with a 67 meter wire, has a period in the order of tens of seconds, and on the other hand the period of the precession of the plane of swing.

Writing an article about inertial osciallations, with one or two animations, is on my to do list. For now, please check out the following animated GIF's: Image:Coriolis_effect06.gif Image:Coriolis_effect07.gif Image:Coriolis_effect15.gif Image:Parabolic_dish_ellipse_oscill.gif An article that is sideways related to inertial oscillations is the Rotational-vibrational_coupling article. There are three animated GIF's in that article.

The article by Anders Persson for meteohistory.org is a shortened version of the following article: The coriolis effect PDF-file 800 KB) The longer version is much more informative. --Cleonis | Talk 23:53, 3 May 2006 (UTC)


 * I do not want to get into a protracted debate. You have indeed misinterpretted the papers.  The circular arc I'm refering to is in the vertical plane.  The reason the pendulum oscillates is that as it moves to the side (horizontally), it also moves UPWARD, following a circular arc, and gravity pulls it back toward its equilibrium position.  This is different from inertial oscillations.  At the north pole, if you give a velocity to a parcel of water, it will undergo inertial oscillations--in the earth frame, these oscillations appear circular with a period of 12 hours.  In a nonrotating frame fixed relative to the stars, the oscillations are linear and have a period of 24 hours.  The restoring force which drives this linear oscillation is gravity pulling the water toward the pole because of the slightly flattened shape of the earth.  This restoring force, and all the stuff you wrote, has nothing whatever to do with Foucault pendula.  Please take me seriously.  I am a physical oceanographer, and motion on a rotating sphere is in my area of professional expertise.  If you like, send an email to the authors of the articles you keep referring to.  They will probably respond. I do not want to simply delete your work. It belongs in another article.


 * I repeat: the swing of the pendulum, with its period in the order of tens of seconds, is 'not analogous to the inertial oscillations as recognized in physical oceanography.
 * The fact that the plane of swing of the Foucault pendulum precesses involves the same force as the force that is involved in inertial oscillations (oceanography)': the poleward force.
 * If you write to Norman Phillips and Anders Persson they will probably respond; they responded to me when I wrote them. I notified Anders Persson when I had uploaded my additions to the Foucault pendulum article. --Cleonis | Talk 00:32, 4 May 2006 (UTC)
 * Addendum: my first email-contact with Anders Persson was in july 2005, my first email-contact with Norman Phillips was in october 2005. --Cleonis | Talk 01:03, 4 May 2006 (UTC)

At 30 degrees latitude, a complete cycle takes two days
I copy and paste from above: "The Foucault pendulum precesses because of the Coriolis force. Read the first sentence of this article (Foucault pendulum). The centrifugal and Coriolis forces arise from the rotation of the reference frame. [...] Rracecarr 22:05, 3 May 2006 (UTC)"

There is a serious problem with your claim.

A foucault pendulum located at 30 degrees latitude (Casablanca), takes TWO DAYS to complete a full precession cycle.

By comparison, a Foucault pendulum located at the equator precesses with respect to the fixed stars in ONE DAY.

How do you propose to explain the difference in amount of time that the precession cycle takes, between the equator and 30 degrees latitude. Why does it take twice as long on 30 degrees latitude? --Cleonis | Talk 00:18, 4 May 2006 (UTC)


 * Simple--at any location, the local rotation rate (the rate at which the local reference frame rotates about the local vertical) is one cycle per day times the sine of the latitude. At 30 degrees, that means the local rotation rate, relative to the earth, is 0.5 cycles per day, for a Foucault pendulum period of 2 days.  Since the rotation rate of the earth is 1 cycle per day, the rate of precession relative to the stars is 1-.5=.5 cycles per day.  At the equator, local rotation is 0 cycles per day (no rotation), which means the pendulum does not precess relative to the earth.  Which means it returns to the same orientation once per day relative to the stars.


 * What do you mean by 'the local rotation rate'?
 * Are you claiming that at 30 degrees latitude the Earth is rotating around its axis slower than at the Equator? --Cleonis | Talk 18:29, 4 May 2006 (UTC)


 * You have introduced something new: a concept of 'local reference frame'. However, there are only two reference frames that matter: the inertial frame, and the frame that is co-rotating with the Earth. Both those frames are global frames. --Cleonis | Talk 18:37, 4 May 2006 (UTC)


 * Once you have lumped the centrifugal force in with Newtonian gravity, forming an effective gravity, you can forget about the component of the rotation vector parallel to the surface of the earth. Angular velocity is represented by a vector: the magnitude is the angular speed, and the direction is parallel to the axis of rotation.  This vector is the sum of two components: one parallel to the surface of the earth, and one perpendicular.  Thus, if you are at 30 degrees latitude (standing still) you are rotating once per day about the earth's axis.  But, equivalently, you are rotating once every two days (sin 30 = .5) about an axis running from your head to your feet, and at the same time once every 2/sqrt(3) days about an axis parallel to the ground and pointing toward the pole.  The sum of these two simultaneous rotations gives you your once per day rotation about the earth's axis, but the horizontal component doesn't matter for calculating motions along the earth's surface.  That is the reason for defining a local reference frame.  It is the reason the Corilis force, the precession rate of a Foucault pendulum, and the frequency of inertial motions change with latitude.  The absolute rotating remains the same everywhere, but it is only the rotation about the local vertical that matters.

The 'equatorial bulge' sentence must be changed.
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".

I also posted this statment on Cleon's talk page. If Cleon does not change the statement this week, I will.

Rracecarr 19:39, 4 May 2006 (UTC)


 * Rracecarr, that makes sense to me: the flattened earth is important for problems that relate to the earth's surface. However, the pendulum's "absolute" motion depends on the centre of mass of the earth and the earth's rotation at the point from which the pendulum is suspended. Thus the bulging of the earth's surface doesn't really matter, except for a slight displacement of the centre of gravity. That displacement is minor, I can't see how it would make a difference in cause and effect; nor did I spot a corresponding claim in the references. Harald88 19:04, 6 May 2006 (UTC)


 * It is glaringly obvious that the equatorial bulge is the cause of the poleward force. (in the case of poleward force as defined in the article.)
 * Evidence is abundant:
 * In Geophysics the Eötvös effect is taken into account when gravimetric readings are taken. The Eötvös effect must be taken into account when the measuring instrument has a velocity relative to the Earth. (This is often the case, for many gravimetric surveys are done onboard moving ships and aeroplanes)
 * Rracecarr is disregarding clear evidence. --Cleonis | Talk 20:05, 4 May 2006 (UTC)

Actually, the opposite is glaringly obvious. The "poleward force" is really just the centripital force that keeps things on the earth going in circles. Said in another way, it is the difference between the true gravitational force (pointing toward the center of the earth) and the effective gravitational force, which adds in the centrifugal force, and is what is 'felt' at the surface of the earth. For the third time, a Foucault pendulum on a perfectly spherical earth would behave, to a very good approximation, exactly the same as one on the real earth. It is silly and stubborn to refuse to acknowledge this fact.

On a spherical earth, your picture with the red, green and blue arrows would look just the same, except the black oval would be a circle, and the green "normal force" would no longer be perpendicular to the black line. The red, green, and in particular blue arrows would remain the same. The "normal force" on the spherical earth is still provided by the wire. The only difference is that the wire hangs, at equilibrium, slightly toward the equator at the bottom, rather than perpendicular to the surface of the earth. Rracecarr 21:06, 4 May 2006 (UTC)

OK to copy a list?
The list of north American Foucault pendulum locations recently added by 157.203.42.40 appears to have been copied from http://www.calacademy.org/products/pend3.html

Armm... there is a novel wirte by Umberto Eco named "The Foucault's pendulum". It's pretty broing, but i like it ^__^

The spikyness of the extremes of the swing
I have reverted a diagram being replaced with an animation. The animation: Image:Foucault pendulum animated.gif is correct for the following situation: A foucault pendulum located on the southern hemisphere that swings so slowly that it takes 24 hours to complete 8 swings.

The diagram that I made Image:Foucault pendulum precession2.png is strictly speaking physically unrealistic: my diagram shows on one hand the 'spikyness' that applies in the case of a Foucault pendulum performing thousands of swings in 24 hours. On the other hand, my diagram shows a large angle from one swing to the next, which physically corresponds to the case of a pendulum that swings in the order of ten times in 24 hours. Strictly speaking, my diagram is a hybrid, moving away from what is physically realistic in order to bring particular features more into focus. Actual Foucault pendulums swing thousands of times in 24 hours. Clearly, it is not possible to produce a transparent diagram or animation that remains true to that.

In Alonso and Finn Fundamental University Physics (second edition) Volume 1, section 6.4 (page 121), the diagram representing the swing of Foucault pendulum is the same as my diagram. Evidently Alonso and Finn opted for the same compromise: the spikyness at the extremes is retained, only the amount of curving to the right (on the northern hemisphere) is exaggerated. --Cleonis | Talk 14:56, 11 September 2006 (UTC)

Dependence on how the pendulum bob is released.
In his demonstration in the Pantheon, Foucault used a method of release that was technically good as well as theatrical: he would pull the pendulum to the side with a thin cord, wait until any rocking and swinging of the pendulum had subsided, and then he burned the cord. Let a very long Foucault pendulum be located exactly above the geometric south pole. Assume that the pendulum is a perfect pendulum. With this south pole pendulum the same release method is used as was used by Foucault. After being released in this way, the trajectory of the pendulum bob with respect to inertial space is not planar, it is an ellipse with one axis much longer than the other. The long axis is the swing of the pendulum between its outermost points. The non-zero-ness of the short axis is due to the fact that a an object that is stationary with respect to the Earth, has a tangential velocity with respect to the Earth's axis. By contrast, in order for the pendulum swing to be planar with respect to inertial space, the release mechanism must compensate for the rotation of the Earth. This can be achieved with a mechanism similar to what is used for telescopes: motors and gears in the supporting rig make the telescope rotate with respect to the Earth in such a way that the telescope remains pointing at the same star. The release method that Foucault used does not impart a pure planar swing to the pendulum bob. The animation Image:Foucault pendulum animated.gif is based on assuming a swing that is planar with respect to inertial space. --Cleonis | Talk 15:29, 11 September 2006 (UTC)

Am I the only person who spotted this?
In the article, and I quote here, "In summary: at 45 degrees latitude, an object with a weight of a 1000 kilogram that is buoyant, is subject to a poleward force of 1.7 kilogram of force. Of course, this poleward force does not apply in the case of an object in free fall."

kilogram is not a unit of force / weight. Please re-write.


 * In this text the objective was to illustrate what percentage of the weight the poleward force is. If an object has a mass 1000 kilogram, and it is buoyant, then the poleward force is about 17 newton (at 45 degrees latitude). To a first approximation 10 Newton is 1 kilogramforce.


 * See this Dictionary of Units of Measurement on the website of the University of North Carolina. "kilogram force (kgf) a unit of force equal to the gravitational force on a mass of one kilogram." I don't think there is a need for a rewrite. --Cleonis | Talk 19:25, 19 October 2006 (UTC)


 * The actual weight of the object (w.r.t Earth's gravity) was completely irrelevant to the airship example. The poleward force was calculated based on the MASS of the airship and its angular velocity.  One kilogram could weigh 5 Newtons instead of 10, and the centripetal force calculation in the example would not be affected.  That's why in my opinion a rewrite was necessary.


 * On second thoughts: both values should be stated in terms of force: a buoyant object located at 45 degrees latitude with a weight of 1000 kilogramforce is subject to a poleward force of about 17 kilogramforce. I will rewrite accordingly. --Cleonis | Talk 08:28, 20 October 2006 (UTC)


 * Thank you.

How about this image
What do you think of this animation, I was going to put it in the article but I didn't know where to exactly put it and I didn't want to delete any other images. Any Takers-- Seadog Talk |undefined 00:05, 27 October 2006 (UTC)


 * I think that animation should not be inserted. It does not represent the motion of the pendulum correctly, which is explained in this section that is two section above this one. This image Image:Foucault_pendulum_precession2.png shows the pattern of motion correctly.


 * Also, there is the sheer size of the animation: 1047 KiloBytes. There are still a lot of people with dial-in connection to internet. Downloading 1074 KiloByto over a dial-in connection takes five minutes. Now that this animation is on the talk page, anyone downloading the talk page gets to download that 1 Megabyte animation.


 * I make GIF-animations too, I try hard to keep them under 100 KB. I think anything bigger than 100 KB counts as HUMONGOUS. --Cleonis | Talk 16:15, 29 October 2006 (UTC)

Globalize
Today I tagged the article with the template for the following reason: The following is indicative of this lack of a global perspective:
 * Lack of description of behaviour of pendulum when observed from the Southern Hemisphere
 * The red line shows the precession with respect to the Earth of a Foucault pendulum located anywhere on the northern hemisphere. At the north pole the pendulum precesses (with respect to the Earth) through an entire circle in one sidereal day. - There is no description of the behaviour as seen from the Southern Hemisphere.
 * A Foucault pendulum located on the northern hemisphere at 30 degrees latitude will take two days to precess through an entire circle with respect to the Earth, precessing clockwise with respect to the Earth at a rate of 7.5 degrees per hour. In those two days the pendulum also precesses counterclockwise through a full circle with respect to the fixed stars at a rate of 7.5 degrees per hour. - There is no description of the behaviour as seen from the Southern Hemisphere.

--B.d.mills 03:05, 3 November 2006 (UTC)


 * You'd have to create a parallel article with everything mirrored to create a southern hemisphere version. What wouldn't work, I think, is constantly flipping from northern to southern hemisphere perspective within a single article.


 * I think that anyone who is willing and able to follow the complete description will be smart enough to do the necessary mirroring mentally. I vote to not create a parallel article and to remove the template. --Cleonis | Talk 03:24, 3 November 2006 (UTC)


 * A better approach would be to do one of the following:
 * Include appropriate descriptions of the behaviour as seen from the Southern Hemisphere.
 * Describe the behaviour in a hemispheric-neutral manner (perhaps by using the Equator as a reference plane), describing the Coriolis motions in relation to the Equator, and then translating that motion into the pendulum motions that are observed in the northern and southern hemispheres.
 * Surely that's not too hard to do?
 * The boreocentrically-biased approach that is all too common in Wikipedia is to describe such matters from the Northern hemisphere only, and then expect people to infer the correct motions in the other hemisphere. Many people do not understand mathematics and science to be able to make that logical inference with sureness and accuracy. Imagine if the article only described motions from the southern hemisphere - would you be sure of getting the northern hemisphere behaviour correct? Would you accept an encyclopedia article worded in such a manner? This is meant to be an Encyclopedia with a global perspective. Do not make the common boreocentric mistake of assuming the globe ends at the Equator. --B.d.mills 03:47, 3 November 2006 (UTC)


 * I have moved the Globalize/Northern template to the section in question. Clarification of the pertinent section may be all that is needed to fix the lack of global perspective. --B.d.mills 03:58, 3 November 2006 (UTC)


 * In my opinion this is not about globalization. As it happens, there is a permanent station located smack on the geographic South pole, with a crew of scientists numbering somewhere around a hundred. A couple of them actually set up a Polar Foucault pendulum A funny detail of their story is that initially their pendulum precessed the wrong way (due to a bias in the suspension or so) but initially they didn't notice the mismatch because they had the direction of precession of a northern hemisphere pendulum in mind. So yeah, its tricky; you do need to pay attention.


 * If one desparately wants to give the southern hemisphere a piece of the action, one can emphasize that the only polar Foucault pendulum was on the south pole. But really, while Northern hemisphere bias can be a genuine issue in articles that deal with politics or economics or history, addressing "northern hemisphere bias" in a physics article is stretching political correctness to absurdity.


 * The southern hemisphere point of view is a mirror image of the Northern hemisphere point of view. I think it would be rather silly to have a parallel version of a particular section that is a verbatim copy of the preceding section, with the words 'clockwise' and 'counter-clockwise' swapped. --Cleonis | Talk 13:29, 3 November 2006 (UTC)


 * Political correctness? It is not an issue of political correctness. It is an example of systemic bias in Wikipedia, and such systemic bias is not desirable in an encyclopedia that is supposed to have a global scope. Your world view seems to end at the equator, as do too many articles in Wikipedia. You have presented no compelling reason why the article and diagrams cannot be amended to include the behaviour as seen from the Southern Hemisphere.


 * All that's needed is a paragraph and two clearer diagrams. That's not too much to ask. I don't understand why you feel it's such a burden to remove this example of boreocentric bias from Wikipedia and why you feel that acknowledging the existence of the Southern Hemisphere in a physics article is so undesirable.


 * I will point this out again because it is very important - not everyone understands physics well enough to be able to deduce the correct behaviour of a Foucault pendulum in the Southern Hemisphere just by reading that article. Imagine you were a Southern Hemisphere resident who wants to know the correct behaviour and who does not know enough physics to be able to make the appropriate inference with confidence. For such a person, the article would be useless. I would rather make the article more useful. --B.d.mills 22:40, 9 November 2006 (UTC)


 * Well, I suppose that our points of view have been stated, and unfortunately there is no consensus in sight. You know my point of view: a reader who is astute enough to follow through the reasoning for the northern hemisphere case is astute enough to perform the necessary mirror operation mentally. In fact, I think that following through the physical reasoning is quite a bit more challenging than mentally performing the necessary mirror operation. --Cleonis | Talk 00:20, 10 November 2006 (UTC)