Talk:Foucault pendulum/Archive 3

Stereographic animation
I have created a stereographic animation. To obtain the stereo-effect, the animation must be viewed with the eyes looking crosswise: the left eye looking at the image on the right, the right eye looking at the image on the left.

It is a bit uncomfortable that there is no indication that the sphere is turning. The current version of the animation is 26 KB. The animation is so small in Kilobytes because very little changes in the scene. Adding longitude lines to show the rotation of the sphere would result in a much larger file.

The transitions from frame to frame are very jerky. To get smooth motion, I think at least four times as many frames would be necessary.

Generally I prefer to keep the KB's as low as possible. I think there are still a lot of people who only have dial-in connection. --Cleonis | Talk 21:13, 18 March 2007 (UTC)


 * Looks a lot better than the current version. I don't really care for the stereoscopic version, one should be enough imo. I would prefer if the animation also included a vector showing the direction of the plane of swing, I like to keep things to the essentials. But that might just be me and I don't have a strong opinion on that. Personally, I don't see the need to make earth rotate (the pendulum does not care). I still believe that an applet that allows people to rotate earth and look at it from any angle, and that allows people to change the latititude interactively would be more useful. But I still don't know how to inculde java appletts into wikipedia pages... --ShanRen 21:43, 18 March 2007 (UTC)


 * Yeah, an applet would be best. But for the time being animated GIF will have to do. I'll make a non-stereoscopic version (The image caption can present a link to the stereograph.) It'll be a few days, maybe next weekend. --Cleonis | Talk 23:33, 18 March 2007 (UTC)
 * So there is no way to include java applets in wikipedia? Maybe we should just link it then? --ShanRen 23:37, 18 March 2007 (UTC)


 * I anticipate that in the future the wikipedia programmers will add the functionality to include java applets. Until then GIF-animations are the next best thing. By the way, the java applet doesn't seem to work on my PC. I get the sphere made out of wire mesh, and some vectors, but it doesn't move or anything, it just sits there. Probably I need some more software to run the applet, but I don't know which software. The advantage of GIF-animations is the reliability.
 * Hmm. Two windows should open. One with the graphics, that you can spin and rotate with your mouse, and another one with buttons that let you adjust the latitude and let you start and stop the animation (or just go through it step by step). --ShanRen 00:39, 21 March 2007 (UTC)

Version of animation with vector


The animation is smooth now. The number of frames is quite large, 196 frames, but since there is very little change from frame to frame the size in KB's is reasonable: 42 KB --Cleonis | Talk 21:16, 20 March 2007 (UTC)


 * Very nice. --ShanRen 00:07, 21 March 2007 (UTC)
 * I agree. Very nice.  The applet downloaded from above is also extremely helpful.  David Harty 16:34, 23 March 2007 (UTC)


 * Uploaded a final version of the animation. I increased the translucency, to make the vector more constant. Smoothness is increased by increasing the number of frames, plus changing the frame rate from 10 per second to 16 per second. For comparison: the previous smooth version. --Cleonis | Talk 21:30, 23 March 2007 (UTC)

Simpler intuition
It would be nice if a simpler intuition about the pendulum was provided. --128.135.82.146 02:42, 21 March 2007 (UTC)

Foucault Pendulum Thought Experiment
I agree with the previous commenter that a more intuitive example could be provided. The example would be provided for showing how to visualize the forces at work and how the effect is observed. In response to this request an attempt is made through the following thought experiment which could be proposed as a separate caption to the article. However, I think the explanation falls short in trying to explain the observed effect of the Foucault pendulum. I would appreciate any comments, especially related to items 14 and 15. I have replaced the steps below with an updated version but am sure that the description still falls short.David Harty 13:32, 24 March 2007 (UTC)

The following thought experiment is provided to visualize the forces at work and how the effect is observed.
 * 1. Construct a flat turntable in a non-gravity environment such as the space station.
 * 2. Construct a Foucault pendulum on the flat turntable such that vertical support tower is located at the center of the turntable on the axis of rotation of the turntable and has a armature that can be extended horizontally from the top of the tower.
 * 3. Construct a pendulum bob hanging from the end of the armature by a wire.  The pendulum bob has magnetic properties.
 * 4. Construct a strong magnet located at the center point of the turntable.  The bob suspended from the armature is attracted to the center point of the central magnet creating an angle of the plumb line of the pendulum with the vertical support tower.  With the addition of the rotating turntable a new plumb line and angle is established for the rotating system (compare geocentric latitude with geographic latitude).  This angle is defined as "phi".
 * 5. If it is desired to observe the pendulum hanging directly over the axis of rotation of the turntable, then it is necessary to provide a curved support away from the armature such that the pendulum swing is not interfered by the support structure.  However, it is noted that there is an absence of an effect that is observed when the pendulum hangs directly over the axis of rotation.
 * 6. What is important to the construct is the point of attachment of the wire which suspends the pendulum.  This point defines the vertical length "T" of the tower above the turntable (and the same as the distance above the center point of the magnet) and the horizontal displacement "A" from the axis of rotation provided by the armature.
 * 7. The attractive force between the magnets represents the gravitational force provided by the Earth.
 * 8. The rotation of the turntable represents the turning of the Earth.
 * 9. The extension of the armature represents the changing of the latitude from the poles to the equator.  When A is zero the pendulum is located directly on the axis of rotation such as the location of one of the poles of the Earth (cosine phi equals one).  When A is infinite in relation to T the pendulum is located directly perpendicular to the axis of rotation such as the location on the equator of the Earth (cosine phi equals zero).  When A is equal to T then phi is equal to 45 degrees such as the location of 45 latitude (cosine phi equals .707).  When A is equal to 1.732 times the value of T then phi is equal to 60 degrees such as the location of 30 degrees latitude (cosine phi equals .5).  The cosine of phi is equivalent to the sine of the latitude.
 * 10. There are two forces at work on the pendulum.  The force of attraction of the magnets (gravity) and the Coriolis Force generated by the rotation of the turntable (the turning of the Earth).  The resultant force acting on the pendulum bob at all times is defined by the cross product of these two forces.
 * 10a. The Coriolis Force is a force appearing in the equation of motion in a rotating reference and causes the Coriolis Effect. For further information also see the Visualization of the Coriolis Effect.  The presence of the Coriolis Force results in a motion described as inertial circles.
 * 10b. There are two manifestations that could be observed as a result of the rotation of the turntable, the angular velocity and the inertial circle velocity.
 * 11. One more construct is needed.  A support stand needs to be constructed on the turntable underneath the armature with a flat, rigid surface located underneath the pendulum bob.  The support stand needs to be able to extend outwards on the turntable similar to how the armature of the pendulum extends.  The plane of the flat surface needs to be angled such that the surface is perpendicular to the plumb line of the pendulum which is directed towards the center of the central magnet located at the center of the turntable.  The plane of the surface represents the surface of the Earth at the location of the pendulum bob.
 * 12. When the pendulum bob is not swinging the two forces of gravity and Coriolis are still acting on the pendulum bob, however, there is no observed effect in relation to the plane of the flat surface (the flat surface is not affected any differently than the center point of the pendulum bob).
 * 13. When the swinging of the pendulum bob is initiated at the polar location then the plane of swing of the pendulum is established over the rigid, flat surface.  The plane of swing of the pendulum bob is observed to rotate in relationship to the surface and this is the effect observed by the Foucault pendulum.
 * 14. The rotation is observed because the resultant combination of the two forces acting on the bob changes with the change of position caused by the swing of the pendulum.  The plane of the swinging bob is constrained differently than the plane of the rigid, flat surface so that all or part of the rotational velocity can be observed by the change in relationship of one plane to the other.  (Would it be incorrect to say that change that is observed is related to the change in centrifugal force that occurs moving inwards and outwards in relation to the center of the turntable?)
 * 15. For a given location of the Foucault pendulum the magnitude and direction of the gravity term doesn't change significantly over the range of arc of the pendulum swing (though gravity does decrease by the square of the distance between the two objects).  What does change with location (latitude) is the direction of the attractive force supplied by the magnetics (the gravitational attraction of the Earth) in relation to the Coriolis Force suppled by turntable (the rotation of the Earth).  (Would it be correct to say that the change that is observed is related to the increase in the horizontal component of the gravitational force that is acting in opposition to the centrifugal force?  Is it correct to say that the two horizontal forces are 90 degrees out of phase but over the course of a full rotation of the turntable (Earth) the forces cancel out?)
 * 16. The magnitude of the change that is observed between the two planes established is determined by the rotation of the Earth divided by the cosine of phi (the sine of the latitude).
 * 16a. If the central magnet were moved outward along the turntable to the point directly underneath the connection of the pendulum then the pendulum bob would hang perpendicular to the turntable and the full rotation of the Earth could be observed again in relation to the pendulum swing.
 * 16b. When the armature of the pendulum is extended the location of the equator is approached.  At the equatorial location the angular velocity can't be observed by the motion of the pendulum swing because the forces acting on the pendulum cancel out on opposite sides of the pendulum swing.
 * 16c. The Coriolis Force cancels out on opposite sides of the pendulum swing.
 * 17. If there was no rotation of the turntable than the only force acting on the pendulum would be the constant force of gravity and no change would be observed by the swinging pendulum in relation to the surface of the Earth.
 * 18. If there was no attractive force then the only force acting on the pendulum would be the Coriolis force and the pendulum bob would be observed outward and upward(?) to the armature.  It is the action of gravity that is always attracting the pendulum bob to obtain the maximum closeness, i.e., align the connection wire with the plumb line of the pendulum.

What is needed to observe the effect of the Foucault pendulum is an attractive force supplied by gravity, a Coriolis force suppled by the rotation of the Earth, a rigid surface for comparison supplied by the Earth's surface and a plane of swing of the pendulum. David Harty 13:32, 24 March 2007 (UTC)

For the case of a reduced attractive force between the magnets, the case of diminishing gravity, one has to be more precise in defining the construct.
 * The angle "phi" is more precisely defined as the angle between two lines; one line created between the center point of the central magnet and the center point of the bob and the other line created by the axis of rotation of the turntable. The values of T and A are more precisely defined in diminished gravity using the center point of the bob in relation to the center point of the central magnet.  With the above construct one can increase or reduce either of the two forces to observe the effect.  The attractive force component can be changed by changing the strength of the magnetic field or changing the distance between the two magnets.  The centrifugal force component can be changed by changing the rotation rate of the turntable.  Other construction parameters can be changed to observe the effect such as changing the center point of the central magnet, changing the location of the vertical support tower in relation to the axis of rotation, or changing the plane of the support stand.  Differences from the Foucault pendulum will be observed.

Please provide comments below related to a numbered step. In particular steps number 14 and 15 where my explanation falls short. I have to give credit to the applet provided above which provided the visualization of the effect that is observed, then using the force diagram to construct a thought experiment to explain the change in the forces which could be observed. I would appreciate any comments and will let the subject digest for awhile while waiting and observing comments. Thanks in advance for all comments. I think an intuitive approach is needed because that is what Foucault was able to envision. David Harty 16:34, 23 March 2007 (UTC)


 * David, the more I think about it, the more I have to agree that the derivation in the article using the Coriaolis force is probably not very helpful. Anyone with the differential equations and physics background needed to understand the computation can easily do it by themselves. And anyone that does not have the background will not understand it. So maybe it should be removed.
 * So the question remains what other explanation could be used to substitute. A thought experiment is a good idea. In that case I would advocate the bicycle wheel as a model. It is a little abstract, but probably the easiest example to understand. The next complicated case would be the Wheatstone-Foucault device. I would favour that over your example. Maybe the simples explanation is that it is not the plane of oscillation that rotates, but the local surface of earth that rotates at a rate given by "the magnitude of the projection of the angular velocity of earth onto the normal direction to earth" (quote from article). This can be best understood (and visualized) using the bicycle wheel. Maybe this train of thought deserves more detail.
 * Some comments about your example:
 * 4. The angle does not model the plumb line but the geocentric latitude. This will get adjusted to the plumb line (geographic latitude) by rotating the disk at the apporpriate speed, so that with the centrifugal force the rest position of the pendulum makes the same angle as the plumb line.
 * 10. The combination of centrifugal force and gravity does not equal the coriolis force! The coriolis force is not directly connected to either one; what it has in common with the centrifugal force is that it arises because the rotating frame is not an inertial frame. Other than that it is not related to the centrifugal force (nor gravity).
 * 14. Answer to you question: Yes, it would be incorrect. See also 10.
 * 15. Partial answer (don't really understand the question) The magnitude of the gravitational force (of any force in the direction of the plumb line) has no effect on the precession of the plane of oscillation.
 * In general, I don't think an intuitive understanding can be established by looking at forces in the rotating frame. The coriolis force is velocity dependent, and that is very difficult to visualize. --ShanRen 17:57, 23 March 2007 (UTC)


 * Thank you for your comments. They have been very helpful.  All I am trying to do is identify the why, then refer to other articles such as the Coriolis Effect article for more detailed development.  It seems to me that there are two manifestations that have to be identified and that manipulating one manifestation provides elucidation to the other manifestation.  That is all that I am trying to do and bring to the article.  The two manifestations that need to be compared are the angular velocity of the Earth and the velocity of the inertial circles as caused by the Coriolis Force.  That is why I chose the construct that I did, rather than have to convert other constructs into the Foucault pendulum reference frame.  Identifying then changing the parameters in the construct allows one to compare how the observations change and what causes it.  My intuition tells me that the inertial circle mechanism cannot be observed by the Foucault pendulum construction because the swings on opposite sides of the plumb cancel out the effect.  If the pendulum were mounted on a fluid that would be a different construct... but a parameter which could be manipulated in the construct and the effect observed.  I have replaced the section above and added comments and strikeouts as it is noted that this talk page is already large.  Your comments are well received and it has occurred to me that it is not possible for me to provide a concise and accurate description of the interaction of the two manifestations.  David Harty 13:24, 24 March 2007 (UTC)
 * In commenting on my own thought experiment it has occurred to me that I don't have a complete understanding of the forces at work and the resulting effects. Several of the statements above are actually intended to be questions, for example, 16a, 16b, 16c, 17, 18.
 * It is also noted that the surface velocity vector decreases as one moves inwards towards the axis of rotation. This surface velocity vector decreases from the equator to the pole with the cosine of the latitude since the projection of the latitude onto the equatorial plane is equivalent to the radius from the axis of rotation.  The angular velocity remains the same so the ratio of the surface velocity to the angular velocity decreases with cosine of the latitude (or, the ratio of the angular velocity to the surface velocity increases with the sine of the latitude).  So maybe 16a is incorrect in that it is the alignment of the angular motion to the axis of rotation that is important rather then with the force of gravity, and the pendulum swing would still be observable in 16a (basic question 1).  I don't know the answer to this question posed by the experiment.
 * Related to this is a comment about the vector diagram provided with the applet. I think the white vectors represent the force of gravity (directed inwards) and the angular velocity along the rotating surface of the Earth, which is constant.  Perhaps it may be helpful to depict the surface velocity vector, as well, along with the angular velocity, since the surface velocity changes with latitude.  Thus, one will get the sense that the surface velocity is changing with latitude w.r.t the angular velocity, as well as, the direction of the force of gravity is changing with latitude w.r.t the angular velocity.  This may provide a better representation of the basic premise.
 * One last fundamental question regarding cause and effect. Are the basic forces involved with gravity and angular momentum resulting in an effect that results in the observed motion of the pendulum swing, or do the basic forces give rise to an additional force, the Coriolis force, that than results in the observed motion of the pendulum swing, or is the Coriolis force just another effect of the basic forces (basic question 2).  The article should be made clear in this regard.  Your comments are well received. David Harty 09:10, 28 March 2007 (UTC)

David, some quick answers. The single important quantity for the Foucault pendulum and the related systems discussed here is the angle between the angular momentum of earth (or the disk) and the plumb line (equilibrium of the pendulum). The pendulum is assumed to oscillate in a plane, so it passes through the equilibrium point with every swing. (If not, other effects of comparable size may appear. For example, one can build a pendulum that traces out an ellipse, but the ellipse does not precess at all. This is because the period of a physical pendulum depends on the amplitude, and the mayor and minor axis precess just because of the difference in amplitudes. One can engeneer this to exactly counteract the Foucault effect (or double it if one wishes to do so.) From the perspective of an earthbound frame, the driving mechanism is the Coriolis force. It is difficult to visualize how it acts, because it depends on the velocity of the pendulum bob. From an inertial frame outside of earth the motion of the plane of swing can be visualized as follows. Model the plane of oscillation by its normal vector. Note that it is tangent to earth at all times. To understand its motion, take the vector and move it in 3-space along a path of fixed latitude for some distance. At it's new position it is not tangent to earth any more, but sticks out (or in) a little. Correct this by projecting it back onto the tangent plane. Repeat. In the limit of infinitesimal movements, this is exactly what the Foucault pendulum does. In the demo, the white vectors merely indicate the earth-bound coordinate system given by north and east directions. View this as axes drawn on the floor underneath the pendulum. The red vector gives the direction of swing of the pendulum. For 16a, this is (almost) the same as a Foucault pendulum just a meter or so away from the north pole. During the course of a day the suspension point traces out a circle, but the pendulum comes back to the same (almost) position after one day. --ShanRen 13:43, 28 March 2007 (UTC)


 * Thank you for your comments and the time it takes to prepare a response. The comments have been very helpful as has been the recent discussions.  First off, I must apologize for my inaccurate portrayal of the white arrows on the applet.  There really is no excuse for my inattention to that detail.
 * What I was thinking about was a force and velocity diagram to enhance the representation shown on the applet. The gravity force vector is always directed towards the center of the earth.  The centripetal force vector is always directed inward and perpendicular to the axis of rotation in the plane of constant latitude (parallel to the equatorial plane).  The centripetal force is reduced with increasing latitude because the radius of the rotating plane (defined by constant latitude) is reduced (decreases with the cosine of the latitude).  Although the angular velocity remains the same with changing latitude, the surface velocity vector decreases with latitude.  Thus, I thought it would be helpful to define the system of force and velocity vectors in which the pendulum is operating.
 * In regards to the bicycle analogy, I made another different interpretation in that I let the bicycle wheel represent the constant latitudinal plane (the armature indicated above) and then I constructed the magnets to represent gravity, then attached a pendulum to the rotating wheel. I did this only to visualize the construction.  However, I see that angling the bicycle wheel to the disk (the equatorial plane) provides a similar effect by changing the angle between the centripetal force and the gravitational force.  However, the plane of the bicycle wheel in this case represents the plane of the coordinate system that is shown in the applet diagram, not the plane of the pendulum swing.  Since the Earth can't change like the coordinate system, it is necessary for us on Earth to have a pendulum which has the freedom of motion.
 * At this point pictures are worth a thousand words which is why I appreciate the representation in the applet diagram.
 * So I can visualize the gravitational force, the centripetal force, the angular velocity and the surface velocity. This system gives rise to the Coriolis force.
 * The solid Earth cannot show this change in relationship of the forces so a mechanism is needed which has freedom of motion to show this change as the latitude changes. The oceans have freedom of motion to show this change and gives rise to the inertial circles  that are present in the ocean.  Adding a Foucault pendulum provides an earth-bound reference for comparing the change in relationship of these forces which shows that the Earth is turning.
 * So from my simplified viewpoint it appears to me that the motion observed by the plane of swing of the pendulum is similar to how the Coriolis force is observed, in that they arise as a result of the forces and rotational velocity of the system. So maybe it would better to explain this system of forces as the cause, rather than saying that "the precession of the pendulum is explained by the Coriolis force", and also provide a link to the centripetal force article.

David Harty 12:43, 29 March 2007 (UTC)
 * David, the centripetal force has little to do with the Foucault precession. If you neglect it from the computation, the result changes only very slightly. I recommend you don't consider the centripetal force at all, since it seems to distract you from what is going on. If you would like to visualize this, consider earth (or your disk) rotating very slowly.
 * If you mount the bicycle wheel the way you said, it does actually represent the motion of the pendulum. Mount a bicycle wheel underneat the pendulum, with axis in the direction of the plumb line. Then start the swing of the pendulum, and mark on the bicycle wheel the initial plane of oscillation. Then over time, the marker on the bicycle wheel will always coincide with the plane of oscillation, the bicycle wheel turns exactly like the pendulum. --ShanRen 14:14, 29 March 2007 (UTC)
 * I thought I was referring to the bicycle wheel as the constant longitudinal plane rotating like the Earth, not angled like the surface of the Earth is to the axis of rotation. Unfortunately, the surface of the Earth (as shown by the coordinate system on the applet) can't rotate like the angled bicycle wheel.  North is always North.  So I agree with you, the angled bicycle wheel (axis in the direction of the plumb line) turns like the pendulum.  I think that is what you mean and I appreciate your thoughts on the matter.   70.58.144.193 18:03, 29 March 2007 (UTC)  Nevermind, I see what you mean.  I am just adding extra steps that don't do anything and angling the bicycle wheel the wrong way.  Thank you for your patience.  70.58.144.193 22:12, 29 March 2007 (UTC)

The Coriolis force is the cause of the precession of the pendulum swing, just like it is the cause of the inertial circles. It appears to me that it would be appropriate in the first diagram in the Foucault pendulum article to show the precession of the pendulum swing due to the Coriolis force. This would result in the curve being zero at the South and North Poles and -2pi at the equator. It is the pendulum that is precessing with respect to the surface of the Earth as caused by the Coriolis force, not the surface of the Earth that is turning underneath the plane of swing of the pendulum. Of course, the situation arises because the Earth is rotating and has a gravitational field. David Harty 08:14, 30 March 2007 (UTC)
 * Actually, the centripetal force becomes very important because it is the force that defines the angular velocity which is fundamental to the equations, even though the magnitude may be small. The centrifugal force at the outside of the wheel decreases with the radius from the axis of rotation.  I think you missed the point and the force diagram is helpful in defining the motion of the Foucault pendulum.David Harty 11:26, 31 March 2007 (UTC)

If the Coriolis force were to be shown on the force vector diagram a lot could be explained from that one diagram.David Harty 14:58, 31 March 2007 (UTC)

Combination Diagram with Force Vectors
From Wikipedia the following statement is provided;
 * "The centripetal force is the external force required to make a body follow a circular path at constant speed. The force is directed inward, toward the center of the circle. Hence it is a force requirement, not a particular kind of force. Any force (gravitational, electromagnetic, etc.) can act as a centripetal force. The term centripetal force comes from the Latin words centrum ("center") and petere ("tend towards").
 * The centripetal force always acts perpendicular to the direction of motion of the body. In the case of an object that moves along a circular arc with a changing speed, the net force on the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal force), and a parallel, or tangential component, that changes the speed."

The Coriolis force acts like a centripetal force. I would like to propose a diagram that shows the forces acting on the surface of the Earth and the forces acting on the pendulum. I think this would be a combination of the applet provided above and the animated gif diagram provided in the article. The applet is excellent because it is able to move to a position on the entire globe and the animated gif diagram is excellent because it shows the pendulum and shows the change in the plane of swing of the pendulum. By showing the force vectors on the combined diagram many questions are answered, definitions provided and the Foucault pendulum becomes a basic demonstration of the Coriolis force.David Harty 16:01, 31 March 2007 (UTC)


 * Let me say it again. The coriolis force does not act like a centripetal force. Forget about the centripetal force, it is not important in this setting. Drawing force vectors for the coriolis force is difficult since the force depends on the velocity of the pendulum bob. It is not a force like gravity, or the centrifugal force in the case of earth for that matter, that are vector fields (as a function of space). One could include the actual movement of the bob in the diagrams, and then show how the coriolis force acts at each particular time. But this is quite cumbersome and not very illuminating. It takes many osciallations of the pendulum to see even the slightest effect. One could of course artificially slow down the period of the pendulum as has been done in the animations below, but there are other problems with this as these animations are generally unphysical. As I explained before, if the motion of the penudulum is not (to very good approximation) planar but elliptical, then other effects dominate the Foucault effect.
 * Also, the Foucault pendulum does not precess "because" of the coriolis force. The coriolis force is merely a tool to compute the precession rate in an earthbound frame. In an intertial frame, there is no coriolis force. --ShanRen 19:53, 31 March 2007 (UTC)

Visualizing the forces


In my opinion, the device described by Wheatstone, as shown in the current version of the Foucault pendulum article, is most suited for visualizing the forces. Actually, I prefer an adapted design, aimed specifically at being a tabletop model for the terrestrial mechanics of the Foucault effect. When the bob is displaced from the rest position the springs extend and/or contract, depending on the direction of displacement. Visualizing the extension of the springs gives a visceral impression of the forces that are acting. --Cleonis | Talk 22:13, 23 March 2007 (UTC)

Possible experiment
In the coriolis effect article it is described that the coriolis effect that is taken into account in meteorogy is the same as the effect that causes the Foucault pendulum to precess.

The following thought experiment is aimed at allowing for the connection with meteorology: The most level surfaces on the Earth are salt flats that are flooded yearly. If memory serves me, the bonneville salt flats are flooded yearly with a thin layer of water, which then slowly evaporates, leaving a level surface that stretches for kilometers.

A hovercraft that hovers above a surface that is not perfectly level will tend to float down the incline. On the Bonnevill salt flats, a hovering hovercraft has no such tendency: perfect equilibrium. What happens if a hovercraft driver starts going back and forth? Let the hovercraft driver take on the role of providing the equivalent of the restoring force. A spot is marked, and the driver is orienting the thrust of the hovercraft in such a way that the restoring force is at all times directed towards that particular spot. The hovering of the hovercraft is analogous to the fact that air mass tends to remain at the same altitude (height above the Earth's surface) when it is set in motion.

It is also interesting to consider whether designers of vehicles for breaking land speed records need to be aware of the coriolis effect. An object that is co-rotating with the Earth at the latitude of the Bonneville salt flats and Black rock desert is circumnavigating the Earth at about 300 meters per second, which is about the velocity of Thrust SSC. When Thrust SSC was going in west-to-east direction, it was circumnavigating the Earths axis at a velocity of about 600 meters per second. When Thrust SSC ran the course in the opposite, east-to-west direction, it was close to being at a standstill with respect to the Earth's axis. In both directions the car will have had a tendency to veer to the right. --Cleonis | Talk 18:53, 24 March 2007 (UTC)

Vincenzo Viviani's pendulum
It has been mentioned that Vincenzo Viviani observed a veering of a pendulum. It is unlikely that the crude pendulum of that demonstration actually showed the Foucault precession, for the following reasons.

Observing his pendulums on long runs, Léon Foucault became aware of perturbations from the theoretically expected precession.

In his book, William Tobin describes that Foucault tried to understand the perturbations by returning to the demonstration that led him to the pendulum in the first place. Foucault had clamped a metal rod in the chuck of a lathe, and he noticed that the direction of the (immaterial) plane of vibration remained the same when slowly turning the lathe.

However, when the lathe did not turn at all, the plane of swing would slowly veer, with a cycle time in the order of tens of seconds. William Tobin writes that Foucault reasoned as follows. Suppose that the metal rod is not perfectly uniform. Then the natural frequency of vibration in one direction will be different from the natural frequency in a direction perpendicular to it. Non-uniformity of the rod gives a directional bias in the rod. When a vibration of the rod is started, this induced vibration is unlikely to be perfectly aligned with the directional bias of the rod itself. Gradually, the energy of the vibration will transfer from one direction of swing to another.

Foucault reasoned that if the lathe rotates slowly, say one revolution per second, then any directional bias in the rod is averaged out. In fact Foucault contemplated incorporating this design feature in an actual pendulum, in order to average out bias in the suspension wire, but eventually he did not do so.

It is likely that what Vincenzo Viviani noticed was veering due to non-uniformity of the suspension wire. (or a non-uniformity of the upper attachment point of the wire. --Cleonis | Talk 08:39, 24 March 2007 (UTC)


 * Cleonis, you seem to have some more information on Viviani's experiment. I could not find anything, could you share your sources? --ShanRen 15:12, 24 March 2007 (UTC)


 * My comment was a general one. In Galileo/Viviani's time, no special effort was made to exclude effects other than the Foucault effect. On that ground it is reasonable to assume that in Viviani's time no Foucault effect was (accidentally) observed. You have pointed out that attempts to trace the origin of the Viviani story have been unsuccesful. So it's possible that the Viviani story is just a mystification. --Cleonis | Talk 18:22, 24 March 2007 (UTC)


 * As you have no additional information I have to conclude that your reasoning is bogus. You claim that Viviani's pendulum was "crude" without having any information to back this up. Contrary to what you claim, Galileo is known to have been a very capable experimentor. The reasons that you cite as a possibility for the observation, namely problems with the wire or the suspension point, are extremely unlikely to cause a veering if the experiment was set up somewhat carefully (which is quite safe to assume, looking at descriptions of experiments at the time). The problems that other people has with the wire and the suspension point are during long-term observations, they certainly don't come up in short-term observations. The fact that Viviani was not looking for the Foucault effect does not mean that he could not have done a careful experiment. From what we know so far, he paid attention to the most important factor in the experimental setup, namely that the pendulum was set into planar motion.
 * Of course, verifyability of the source is necessary for this to be included in the article. Once we have all the relevant information we may get a better idea what Viviani observed, and how relevant it is. And most importantly, we may be able to give an informed opinion. --ShanRen 19:21, 24 March 2007 (UTC)

Bicycle wheel mounted on a disk
The following passage from the section 'related physical systems' needs clarification:

"a non-spinning perfectly balanced bicycle wheel mounted on a disk so that its axis of rotation makes an angle $\phi$ with the disk. When the disk undergoes a full clockwise revolution, the bicycle wheel will not return to its original orientation, but will have undergone a net rotation of $-2\pi\, \sin(\phi).$"

A rewording is necessary because the current formulation contains a selfcontradiction. When the bicycle wheel is non-spinning, then its angular velocity with respect to the disk is $$-\Omega$$; the reverse of the angular velocity of the disk. After the bicycle wheel has been tilted, its angular velocity (with respect to the disk) is no longer the reverse of the disk's angular velocity.

I propose the following:
 * A perfectly balanced bicycle wheel is mounted on a disk in such a way that the direction of the bicycle wheel's axis is adjustable. In the initial state the axis of the bicycle wheel is parallel to the central axis (the axis of the disk). The disk is rotating with angular velocity $$\Omega$$, and the bicycle wheel is initially non-spinning. Then, without touching the wheel, the wheel's axis is tilted so that it has an angle with respect to the disk. After that the bicycle wheel will have an angular velocity with respect to the disk: after each full revolution of the disk the bicycle wheel will not return to its original orientation, but will have undergone a net rotation of $$-2\pi\, \sin(\phi).$$

The 'bicycle wheel mounted on a disk' is quite a neat example of the sine law in rotating systems with angled motion. I think I will make an animation that shows the above described process. --Cleonis | Talk 21:33, 26 March 2007 (UTC)


 * Let me clarify the system, it does not quite work the way you are describing it. The system you are descibing will not exhibit that exact phase shift. Here is a clearer version of what I meant to say:
 * a non-spinning perfectly balanced bicycle wheel mounted on a disk, which initially is also fixed, so that its axis of rotation makes an angle φ with the disk. When the disk is then rotated so that it undergoes a full clockwise revolution and is returned to its initial orientation, the bicycle wheel will not return to its original orientation, but will have undergone a net rotation of $$-2\pi\, \sin(\phi).$$
 * --ShanRen 00:24, 27 March 2007 (UTC)


 * The detail of the disk not rotating initially introduces an element that is not present in the Foucault setup. Interestingly, it is not necessary for the disk to be non-rotating initially.


 * Let the disk be rotating with a constant angular velocity troughout. The angle of the wheel with respect to the disk, $$\phi$$, is kept the same throughout. Initially, the wheel is driven (a torque is exerted) in such a way that the wheel counterrotates with respect to the disk, and after each full rotation of the disk the wheel is back in the same orientation. Then the wheel is released to rotate freely around its own axis. After the wheel has been released, it will undergo a net rotation of $$-2\pi\, \sin(\phi)$$ for each full circle of the disk.


 * The analogous procedure for a pendulum is as follows: a pendulum can be constructed in such a way that the swing is forced to go through a full circle (with respect to the Earth) in one sidereal day. That constraint of the pendulum requires a torque. When the pendulum is released to free swing, the plane of swing will undergo a net rotation of $$-2\pi\, \sin(\phi)$$ for each full circle of the disk.


 * In the two analogous cases the direction of the torque prior to release is the same. --Cleonis | Talk 15:19, 27 March 2007 (UTC)


 * Cleonis, having the bicygle wheel that is spinning initially just complicates the situation. It makes it difficult to read off the phase shift and it suggests the need for a relation between the initial frequency of the wheel and the frequency of the roation of the disk, which is totally artificial.
 * The setup you describe does not have the outcome you describe. If the wheel is initially "corotating" with the disk (i.e. returns to its original position after each turn of the wheel), then it will continue to do so -- if no external forces are applied.
 * As a side note, in your arguments you need to explain why the gyroscopic force (when you change the direction of the angular momentum of the wheel, or the conical pendulum) has no effect. This especially applies to your "analogy" using a conical pendulum (I assume that's what you mean). Anyway, there is something wrong with the pendulum you describe. You get into trouble if you have a period of 1 day. Just imagine it at the north pole!
 * Lastly, the initially non-rotating bicycle wheel relates to the Foucault pendulum in the following way. It simply defines an inertial frame along the path of fixed latitude, so it shows at what rate the ground rotates underneath the pendulum. --ShanRen 16:27, 27 March 2007 (UTC)


 * Shanren, do note that what I wrote is as follows: initially, the wheel is driven (a torque is exerted) in such a way that the wheel counterrotates with respect to the disk, and after each full rotation of the disk the wheel is back in the same orientation.
 * You, on the other hand, are describing a setup in which the wheel is initially co-rotating with the disk, which obviously is an uninteresting initial condition.


 * When the wheel is at an angle with respect to the disk and the wheel counterrotates with respect to the disk, then the wheel is not motionless with respect to the inertial frame. Only if the wheel's counterrotation is an exact match for the disk's rotation, does the wheel return to the same orientation after every full turn of the disk. --Cleonis | Talk 21:33, 27 March 2007 (UTC)


 * Cleonis, ok you said "counterrotating". But what's your point? Either way, if your wheel is rotating (co or counter) in a way so that it initially comes back to its original orientation after each turn of the disk, it will continue to do so if there are no external forces. In short, the sytem that you describe does not exhibit the behavior you assert. So please focus on the issue instead of irrelevant co or counter discussions.
 * Also, what's with the "wheel is not motionless with respect to the inertial frame" business? The axis of the wheel is changin it's direction, so of course it is not motionless -- regardless of how it rotates. --ShanRen 01:14, 28 March 2007 (UTC)

How to define an inertial frame along a path of fixed latitude
I copy and paste from above: "Lastly, the initially non-rotating bicycle wheel relates to the Foucault pendulum in the following way. It simply defines an inertial frame along the path of fixed latitude, so it shows at what rate the ground rotates underneath the pendulum. --ShanRen 16:27, 27 March 2007 (UTC)"

Shanren, you need to clarify what you mean by 'inertial frame' in this context

Recapitulating: Inertial motion over the surface of a sphere follows a great circle. This is of course elementary spherical geometry. In euclidean 2-space inertial motion follows a straight line, and in spherical geometry this translates to motion along a great circle. It is appropriate to refer to motion along a great circle as 'inertial motion' for although there is a force perpendicular to the surface, confining the motion to that surface, there is no force in a direction tangent to the surface.

If it is assumed that the bob of the Foucault pendulum is at all times in inertial motion, then it follows that at all times the bob must be moving along a great circle.

We agree that the equation of motion for the Foucault pendulum is as follows:

$$ \left\{ \begin{array}{ll} \dfrac{d^2x}{dt^2} = -\omega^2 x + 2 \Omega \dfrac{dy}{dt} sin(\phi)\\ \dfrac{d^2y}{dt^2} = -\omega^2 y - 2 \Omega \dfrac{dx}{dt} sin(\phi) \end{array}\right.\qquad\qquad(A) $$

In the image on the right, Great circle and coriolis motion, the red circle represents a great circle that is tangent to the latitude line of 30 degrees northern latitude, and the blue line represents the motion of the pendulum bob during the swing in which it swings in east-to-west direction.





Motion along a great circle that is tangent to a latitude line has the following property: no matter the direction (west-to-east or east-to-west): the motion always proceeds towards the equator. It cannot move north of the tangent latitude line On the other hand: motion of the pendulum bob as described by the coriolis term has the following property: motion in east-to-west direction that starts tangent to a latitude line proceeds towards the nearest pole.

It follows that the pendulum bob is not in inertial motion. Hence a frame of reference that is co-rotating with the plane of swing of a pendulum is not an inertial frame of reference.

Shanren, that is why you must clarify what you mean by 'It simply defines an inertial frame along the path of fixed latitude'. --Cleonis | Talk 22:23, 28 March 2007 (UTC)


 * To answer your question. We are interested in observing a direction that is tangent to the sphere along a path of fixed latitude, namely the orientation of the plane of oscillation. To quantify what happens, we can define a moving frame along the path.
 * Let's look first at a simple example of a pendulum in the plane. In the plane the pendulum can act like a compass. If you take a planar pendulum on a car, and slowly drive along some path in the plane, the plane of oscillation does not change. (Lots of "plane"s here, hoepfully not too confsing.) The pendulum always swings in the same plane. Now suppose you drive along a circle and wish to understand the motion of the direction of the plane of oscillation. To record the motion of the pendulum, you can define a moving frame along then path. There are many choices to do so, and some are better or more natural than others. I want to draw your attention to two such choices. Firstly, in the plane we have an absolute notion of direction, say along the x and y axis (just make a choice). If we record the motion of the pendulum in these axis, the pendulum does not turn. The reason is that there are no forces that can make the direction of the pendulum turn, and the frame is an intertial frame. But we could also choose a different frame, namely one attached to the car, that is the tangent to the circle and the perpendicular direction. If we record the motion of the plane of oscillation in that frame, the pendulum rotates. Of course, there still aren't any external forces that make the pendulum turn, but the frame we chose was not an intertial frame.
 * Observe that while the path along which we took the pendulum was not a straight line, we could still define an inertial frame along the path. If we take a straigh-line path, then the two definitions would coinside.
 * Now for the Foucault pendulum, that is a pendulum on earth that moves on a circle of fixed latitude, there are also many choices to write down a moving frame along the path. One such choice is given by an earthbound frame, given by "east" and "north" (or by the tangent to the path and it's perpendicular). But there are many other choices. Some frames are more useful than others for doing physics. The earthbound frame has the problem that it is not an inertial frame along the path in the following way. (This is of course related to the fact that the circle of fixed latitude is not a straight line, except the equator.) If we observe a direction that does not feel any phsyical forces that could compel it to turn -- like the bicycle wheel -- then this direction will turn in the earthbound frame. However, just like in the plane, we can define an inertial frame along the path, that is a frame in which directions stay fixed, if there are no external forces that make it turn. The bicycle wheel defines such a frame. Of course we can give an intrinsic definition of such a frame. In physics this is called a parallel frame, and it is defined by the condition that the derivative of the frame in the direction of the path vanishes. (See parallel transport, although that wikipedia article is not very good.)
 * Hope that helps. --ShanRen 02:42, 29 March 2007 (UTC)


 * It appears that the definition of inertial frame that you are applying is different from the standard one. Standard in newtonian dynamics is the following: inertial motion is motion along a straight line.


 * Recapitulating: In terms of newtonian dynamics, the motion of the planets around the Sun is considered to be non-inertial motion. In terms of newtonian dynamics the force exerted by the Sun is inferred from the fact that the planets do not move in straight lines. More generally, in newtonian dynamics, the equivalence class of inertial frames of reference can be singled out operationally in the following manner: observe a large number of test masses, and if they all are in inertial motion, then their velocity relative to each other can be represented as a linear relation. By contrast:
 * If two of the testmasses exert a force on each other (attracting or repelling each other) then both of these test masses have motion that is not linear with respect to the other, inertially moving test masses.
 * If the motion of the entire set of test masses is mapped in an accelerating frame, for example a rotating frame, then their relative motion appears to be non-linear. Choosing an inertial frame to map the motion in removes the appeareance of non-linear motion.


 * The very definition of the concept of inertial frame (in terms of newtonian dynamics) rules out the possibility of defining an "inertial frame" that follows a path of fixed latitude.




 * Let vt,e be the total velocity, in tangential direction, of a plumb line, and vt the total velocity of the east-to-west (relative-to-the-Earth) moving bob. Then the surplus of centripetal force (that causes acceleration in radial direction) is given by:


 * $$ a_r = \frac{v_{t,e}^2}{r} - \frac{v_t^2}{r} $$


 * This can also be expressed as an equation that shows how the radial acceleration depends on the tangential velocity relative to the rotating system. Let vt,r be tangential velocity, relative to the rotating system. We have: vt = vt,e + vt,r


 * $$ a_r = \frac{v_{t,e}^2}{r} - \frac{(v_{t,e} + v_{t,r}^2)}{r} $$


 * $$ a_r = \frac{v_{t,e}^2 - (v_{t,e} + v_{t,r})^2)}{r} $$


 * $$ a_r = \frac{v_{t,e}^2 - v_{t,e}^2 - 2v_{t,e}v_{t,r} - v_{t,r}^2}{r} $$


 * $$ a_r = \frac{- 2v_{t,e}v_{t,r} - v_{t,r}^2}{r} $$


 * When the velocity relative to the rotating system is small the above expression simplifies to:


 * $$ a_r = - 2 \omega v_{t,r} $$


 * This provides a straightforward explanation of why the coriolis term is proportional to the velocity relative to the Earth. --Cleonis | Talk 23:27, 31 March 2007 (UTC)


 * Cleonis, I understand that the concept of inertial frames is difficult and can be confusing. An inerital frame is simply a frame in which newtons second law holds. In euclidean 3-space, it is fairly straightforward to understand what is meant by that, since there is a global inertial frame. In our case we have a system that is constraint to tangent directions on the sphere, and there things are admittedly a little harder to understand. There is no global frame for tangent directions on the sphere, let alone a global inertial frame. There is not even a local inertial frame. But there is an inertial frame along any path (not just along geodesics), that is a frame in which Newton's second law holds for the constrained system. That is there is a frame along each path, so that a physical system that defines a direction in that frame will not change its direction in that frame as the system is moved along the path, as long as no external forces are present. If this is too abstract for you, look at a physics book and specifically look at the part where constrained motion is discussed.
 * While I am at it, you seem to confuse the notion of inertial frame and inertial motion. Paths of fixed latitude don't arise from inertial motion, but we can still study (inertial) motion of a tangent direction along that path. --ShanRen 00:34, 1 April 2007 (UTC)


 * More generally, in newtonian dynamics, an inertial frame is a frame in which all of newton's laws hold good, the three laws of motion and the law of gravity.
 * Indeed it is interesting to figure out how that translates to spherical geometry, for in the case of motion constrained to the surface of a perfect sphere, there is no global inertial frame.
 * In newtonian dynamics, gravitation is classified as an external force. You state the following condition: 'as long as no external forces are present'. The Foucault pendulum setup doesn't meet that condition, for in the case of a Foucault pendulum, there is an angle between the plumb line and the direction of newtonian gravity. For example, at 45 degrees latitude, that angle is 0.10 of a degree. The fastest way to obtain that angle is as follows:
 * Being at 45 degrees latitude corresponds to being at 4510.0 kilometers away from the Earth's axis, which, given the Earth's angular velocity, corresponds to having a tangential velocity of 328.9 meters per second. The amount of centripetal acceleration to sustain that circumnavigating motion is 0.024 m/s². Resolved into the direction tangent to the local surface that is 0.017 m/s². For the perpendicular to the surface component I take the average of polar and equatorial effective gravity: 9.81 m/s². The angle betweeen the plumb line and the direction of newtonian gravity is obtained from the ratio of 0.017 m/s² to 9.81 m/s². That yields 0.10 of a degree.
 * The angle between the plumb line and the direction of newtonian gravity is determined by the rotation rate of the rotating system, in this case the Earth, and by the the distance to the central axis of rotation. --Cleonis | Talk 07:11, 1 April 2007 (UTC)
 * Addendum: in the case of the Wheatstone-Foucault device, the motion is not rigidly constrained to motion along a spherical surface. The constraining force first needs to build up in the form of extension and/or contraction of (various parts of) the helical spring; which can be extension in any direction. A more general approach handles the Wheatstone-Foucault device just as well as the case of perfectly rigid constraint to motion along a spherical surface. --Cleonis | Talk 07:34, 1 April 2007 (UTC)


 * In the case of the Focuault pendulum, the condition "tangent to earth" means perpendicular to the plumb line. The question whether or not the direction given by the Foucault pendulum feels any forces is key of course, although that has nothing to do with the existence of an inertial frame. The point you raise about the angle between gravity and the plumb line is really a question about the definition of latitude but I guess I am repeating myself for the 10th time now. If you want to understand how this relates to spherical geometry, and how the Wheatstone device comes in, you need to understand the Gauss map. Geographic latitude has the Gauss map build in already.
 * And while I am at it, an inertial frame is really defined to be a frame where Newton's second law holds, it does not make much sense to sharpen the definition further. You should consult some physics books on this. --ShanRen 14:42, 1 April 2007 (UTC)

Traces of the motion
Thank you to Cleonis for his image which has been used in a discussion of the French page. In fact, the very nice image (B) from a German Wikipedian de:User:DemonDeLuxe (who died recently, I read)) was unrealistics in the sense that such a pendulum motion can't be easily obtained (see below). Foucault himself in order to launch the pendulum with a zero speed, bended the pendulum (on the East side for example), waited until the cable had no more oscillations and then burned a cord which retained the pendulum.

Thus, I started with the equations (A) above and tried to give an understandable solutions (see equation (2) of the French page). The nice German image corresponds to the initial conditions $$z_0=0$$ and $$\dot{z}_0=V_0$$ which are hard to get!

I tried to give two views of the pendulum: a standard view from the Panthéon and a second from the rotation plane (unfortunately the thumb of the second does not come). Would the addition of the speed in red, the projection on Earth (green) and of the trace in blue helps understanding the motion and the ellipse? That's my hope. Animations are done using Gnuplot, a freely available software on any operating system. The source code of the drawings are licensed under GPL and can be improved by anyone; the images are under GFDL and CC as usual. Images and sources are on the same address of Commons .--Nbrouard 14:29, 30 March 2007 (UTC)


 * Hi Nbrouard. The image with eight consecutive swings is a still image that represents the sucessive halfswings that are representes in the animation motion of the bob.


 * As you point out, the animation by DemonDeLuxe represents a pattern of motion that is difficult to obtain. Let an astronomical observatory be build smack on the South pole. Like all telescopes, the suspension of the mirror has motors that make the mirror remain stationary with respect to the stars. A polar pendulum that is attached to the star-stationary frame will on release follow a purely planar path.
 * On the other hand, when a polar pendulum is released from a point that is co-moving with the Earth, then the motion of the pendulum bob will be slighly ellipse-shaped. This deviation is imperceptible in the case of a real Foucault pendulum, but in schematic representations, with a ratio of swing frequency to rotation frequency in the order of 10 to 1, it makes a lot of difference. The animation by DemonDeLuxe assumed release from a star-stationary point, my animation assumes release from a point that is co-rotating with the rotating system.
 * If Foucault would have tried for the utmost, he would have made a rig to compensate for the rotation of the release point with respect to the suspension point. --Cleonis | Talk 12:15, 31 March 2007 (UTC)

I really like the animation (C), the one with the green trace and rotating blue ellipse. I have never seen one that so clearly demonstrates the motion in inertial and rotating frames simultaneously. I would definitely be in favor of including it in the article (perhaps with the minor change of English labels for the cardinal directions). Rracecarr 16:34, 2 April 2007 (UTC)

The concept of parallel transport
Shanren, you seem to be arguing the following:

You are discussing a setup with two constraining factors that play a separate, consecutive role: the overall constraining factor is that the motion is confined to motion along the surface of a perfect sphere. A subsequent (optional) constraint is that the motion is confined to motion along fixed latitude.

The path along fixed latitude is a non-inertial path. When a local physical system (such as a Foucault pendulum) is in motion along a fixed latitude, then the overall acceleration vector can be decomposed in a linear component and a rotational component. Being constrained to fixed latitude, the Foucault pendulum is barred from yielding to the linear component of the acceleration vector, but it is free to follow the rotational component.

This, I surmise, is what you are driving at by using the highly abstract expression 'parallel transport'.

Another way of putting it: if motion is confined to motion parallel to the surface of a perfect sphere, then the remaining two degrees of spatial freedom can meaningfully taken to represent inertial motion. So after constraining the motion even further (to the point that only rotation is left free), then that remaining freedom can still be taken to represent inertial motion of that very last degree of freedom.

If that is what you mean then I agree that that is a self-consistent way of representing the setup. Not suitable for novices however, it is necessary to understand the setup first in terms of basic principles before embarking on such abstractions. --Cleonis | Talk 08:49, 1 April 2007 (UTC)


 * No, that't not what I meant and what you say (in regard to your understanding of parallel transport) makes no sense. I agree that parallel transport is in general an abstract concept, that's why the article merely mentions it so that someone familiar with parallel transport can grasp immediately what goes on. For everyone else, the article tries to provide an explanation without direct reference to that term. In the end of course all explanations boil down to parallel transport. For example, the coriolis force shows up as the Christoffel symbols in the frame given by latitude and longitude. --ShanRen 14:56, 1 April 2007 (UTC)

The angle between newtonian gravity and the plumb line.
I copy and paste from above: "The point you raise about the angle between gravity and the plumb line is really a question about the definition of latitude. --ShanRen 14:42, 1 April 2007 (UTC)"

Shanren, the question of latitude convention is where you are wrongfooting yourself really badly. The angle between newtonian gravity and the plumb line is a function of rotation rate of the rotating system and distance to the central axis of rotation. Latitude comes in only indirectly; the total required centripetal force is resolved in the direction tangent to the local surface, and the closer to the Equator the smaller the component tangent to the local surface.

So let's do some calculations: I will compare two possible conventions of numbering latitude.
 * Geometric latitude, the angle between the plane of the equator and the line from the Earth's geometrical center to a point on the Earth's surface.
 * Geographically measured latitude, that I will count in kilometers. 5000 kilometer northern latitude is 5000 kilometers from the equator. (the distance from the Equator to the poles is 10000 kilometers)

(Another possibility is geodetic latitude, 45 degrees geodetic latitude is where the actual Earth's surface is at an angle of 45 degrees to the plane of the equator.) The polar radius of the Earth is 6356.752 kilometers, the equatorial radius is 6378.137 kilometers. The cross section of the Earth can be approximated with the following parametric representation. Let the x-direction be the equatorial diameter and let the y-direction be the distance from pole to pole.

\begin{cases} x = 6378.137*cos(t)   \\ y = 6356.752*sin(t) \end{cases} $$

I used a computer program to calculate line integrals. The geographical latitude of 5000 corresponds to the value 0.25022*&pi; of the parameter t. Geometric latitude of 45 degrees corresponds to the value 0.25053*&pi; of the parameter t

The difference between geometrical latitude and geographical latitude amounts to about 6 kilometers along the surface of the Earth, which corresponds to an angle of about 1/20th of a degree.

At 5000 kilometers latitude geographically, the distance to the center of the Earth is 6378.137*cos(0.25022*&pi;) = 4506.9 kilometers. At 45 degrees geometrical latitude, the distance to the center of the Earth is 6378.137*cos(0.25053*&pi;) = 4502.5 kilometers. The difference between those two is less than 1/1000th. Obviously, what latitude convention is used is insignificant compared to other factors. --Cleonis | Talk 16:15, 1 April 2007 (UTC)


 * Cleonis, please read before spamming this page with your irrelevant rambling. I am getting really tired of this. Nobody claimed that the choice of which latitude is being used makes a big difference, actually quite the opposite is true.
 * The defnition of latitude simply corresponds to the definition of "horizontal". And what counts for the Foucault pendulum is not the horzontal component of the centripetal force as you seem to suggest, but the horizontal component of the sum of gravity and the centrifugal force. And the horizontal component of this sum is zero in the case of the geocentric latitude. If one neglects the centripetal force, then "horizontal" means perpendicular to gravity, which pretty much corresponds to geocentric latitude. And yes, the difference for the Foucault pendulum is small. You claimed before that it is not, and this might be the source of your confusion here. So think about this. And by think I mean "think" and not ramble on about your ideas on this page. I don't think anyone is actually interested. If you actually have a calculation supporting your arguments (and by calculation I don't mean some pictures that come with some esoteric rambling), then please feel free to share these and we can actually have a discussion with substance. --ShanRen 17:00, 1 April 2007 (UTC)


 * Both the centrifugal term and the coriolis term are merely tools to compute the motion in an earthbound frame. It is meaningless to say that 'what counts for the Foucault pendulum is [...] the sum of gravity and the centrifugal force'. Centrifugal force is just a computational tool. --Cleonis | Talk 18:18, 1 April 2007 (UTC)


 * In choosing a calculation strategy for dealing with the angle between newtonian gravity and the plumb line, two approaches present themselves. To use a perfectly spherical planet, with the same size and the same angular velocity as the Earth, or to incorporate the oblateness of the Earth.


 * In either approach, the same angle between newtonian gravity and the plumb line is obtained: at 45 degrees latitude an angle of 0.1 of a degree.
 * If a perfectly spherical model is used, then the model predicts that newtonian gravity is everywhere perpendicular to the surface, and a plumb line that is co-moving with the rotating sphere has at all latitudes (except at the poles and the equator) an angle with respect to the local surface.
 * If the Earth's oblateness is incorporated, then the model predicts that there is at all latitudes an angle between newtonian gravity and the local surface (except at the poles and the equator), and a plumb line that is co-moving with the Earth is everywhere perpendicular to the local surface.
 * Either way, given the presence of rotation, there is in both models the same angle between newtonian gravity and the plumb line.


 * Only a model that incorporates the centripetal force will obtain the observed Foucault precession. Models that do not incorporate the centripetal force and yet obtain the Foucault precession are fudged. See http://en.wikipedia.org/wiki/User:Cleonis/Sandbox/Wheatstone-Foucault_device --Cleonis | Talk 18:18, 1 April 2007 (UTC)


 * Cleonis, the models you are talking about are not the models that I was talking about. I can't be bothered to do the calculation to check your assertion that in both of your models the angle is the same (and I doubt that you have done it), since this is irrelevant to the discussion of the pendulum. More importantly, you continue to push your assertion that the centrifugal force is central in understanding the pendulum, without providing adequate explanation. As I said above, a couple of pictures and some random assertions and formulas are not an argument. The computations in your sandbox article are not even finshed! While I appreciate your effort and some of your animations, I am tired of your bullshit. If you can't even write down the equations and solve them, how can you keep agressively making statements about what the equations are saying and how different terms in the equation affect the outcome? It's probably best if I withdraw from this discussion for a while, I just don't see how I can continue to respond to your posts in any meaningful way that moves things forward. --ShanRen 19:17, 1 April 2007 (UTC)


 * I have consistently been opposed to suggesting that there is a role for 'the centrifugal force'. All along,the problem has been that you have been atributing opinions to me that are miles away from what I actually write. Your last reaction is typical, I'm emphatically against attributing a role to "centrifugal force" and weirdly, you accuse me of the opposite; pushing for a role of "centrifugal force".
 * Both the centrifugal term and the coriolis term are merely tools to compute the motion in an earthbound frame. That has been my point all along.
 * I do need to correct myself on a particular matter. The significant feature is that there is an angle between newtonian gravity and the plumb line. Many people adopt the following erroneous reasoning: the plumb line that is co-moving with the rotating Earth is perpendicular to the local surface, ergo: the plumb line that is co-moving with the rotating Earth is not subject to any tangent-to-the surface force. To address that particular error, I emphasized the significance of the oblateness. Unfortunately, that emphasis was prone to being misunderstood. In the case of the Foucault pendulum, I now think it is better to emphasize the angle between newtonian gravity and the plumb line. --Cleonis | Talk 19:49, 1 April 2007 (UTC)

Historical Information
I commented out the new paragraph with the historical information. There are many good ideas in there, but it needs quite some work as it is. Don't have the time to do it myself right now, but will get to it eventually if nobody else does it first. :-) --ShanRen 00:20, 3 November 2007 (UTC)
 * Some of the page numbers and journal issues in the new references are not correct. Also, the references should be moved to the bottom.
 * Binet's "simple" analytic derivation is just a longer version of the dynamical argument given in this wikipedia article. Or to day it in a better way, this wikipedia article gives a simpler version of Binet's original argument.
 * "From a geometric perspective much of the explanation given above comes down to the statement that the rate of change of azimuth of a star on the horizon depends only on the observer's latitude and is the same as the rate of precession of the pendulum." -- This is not really correct. The cited article merely makes a loose argument that the rates of change are the same. The argument is made from the perspective of an inertial frame, but apart from that this has nothing to do with the geometric argument in the wikipedia article. Then the cited article relies on a special case of a more complicated formula for the rate of change of the azimuth to "derive" the rate of change of the pendulum. (I have not looked up how the formula was originally derived, so maybe the secondary citation derives the formula for the azimuth by a similar geometric argument. If that is the case this would be interesting and should be made clear.) Either way, I feel that the relation is a nice observation and should be pointed out.