Ionel Dinu, May 14, 2008:
Thank you very much indeed for your words of appreciation and for your comments.

It is apparent from these works, and I am glad that you observed it, that the notion of mass is a badly defined physical quantity, with actually no direct correspondent in the physical reality. For, as long as the inertia and the weight of an object can be explained in a hydrodynamical way, and the only parameter of the material body which enters the mathematical equations is the total atomic volume of the body (number of atoms N multiplied by the volume w of one atom), there is no need to introduce a supplementary parameter (called "mass"). So it is not that I "reject the existence of mass", it is obvious that we are better off without it and can write the equations of mechanics with "N times w" instead of "m". You will obtain the same equations of motion, the same mathematical formulas for the trajectories, etc.

I am aware of the mismatch between the sense in which I use the notions of "force" (and "pressure") and that already established and expressed in units containing kg. I took all the possible care to warn the reader about this fact and I am sure that those willing to make the shift to this new approach will know the difference.

With due respect, your proposal to use a constant is not viable to me. In this theory, the quantity of matter is measured in units of volume. The "weight" (archimedic force in the aether) of two material bodies is equal as long as their total atomic volume (volume displaced in the aether) is equal. There is really no need to use "mass" any more to express a force.

To use another term for "force" and "pressure" will also be a disaster. In common language (and historically) "force" signifies the principle which drives a body into motion or keeps it in place against another force (for example its weight). I do not think that there is a "lack of correction" in my equations and that my formulations will hamper in any way further developments.

Thierry de Mees, May 13, 2008:
Ionel Dinu's, New Fundaments for Classical Mechanics
I would like to congratulate you for your hydrodynamic approach to the gravitation and aether problem. It is an important work in order to get a better insight about interstellar trajectory possibilities, maximal interstellar speed and so on. The speed of bursts of black holes can be studied by it as well.

One problem to me is the incorrect use of the terms "force" and "pressure" compared with classic mechanics, because you reject the existence of "mass". Though, either the names, either the units of measure should be corrected. This can be done in several ways. One nice way in my opinion is to use a natural constant, but it is up to you to chose your method.

The lack of correction in your equations will make further studies almost impossible, sooner or later.

Thierry de Mees, Jan. 27, 2008:
I have read your last paper and I am very surprised by a certain number of sentences in it. You write :

"It is taught nowadays that centrifugal force is only a fictitious force that exists when we view a situation from a rotating frame of reference. If a weight is swung in circular motion on the end of a string and released, it will fly off both radially and tangentially. When asked which direction the weight will fly off in, the educated person is expected to answer ‘tangentially’ and ignore the outward radial acceleration. Only the fool is expected to draw attention to the outward radial acceleration."

This is unfortunately wrong. It is well known that the centrifugal force physically does not exist, and a former physics teacher should know that. The circular motion that is generated by gravitation, or by a rotating mass at the end of a stressed wire, is generated by a centripetal force : it is the gravitation or the stress in the wire. The centrifugal force is a non-physical force, because in the circular motion, it is just the change of direction of the inertial mass that occurs, nothing more. And any change of direction of an inertial mass gives a reaction force, felt by that inertial mass itself as long as that change of direction continues to exist. This effect is sometimes (wrongly) called centrifugal force.

Even much worser is the wrong idea that "If a weight is swung in circular motion on the end of a string and released, it will fly off both radially and tangentially. [etc...]". This is a total nonsense. Only a straight motion with constant velocity can remain after the vanishing of the centripetal force.And this motion is tangential to the orbit, just as the orbital velocity is.

An excellent website that explains how it works is :
http://www.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_2_2.html
A shorter one is :
http://www.sparknotes.com/physics/dynamics/uniformcircularmotion/section1.html
I wish you a good reading.

David Tombe, Jan. 27, 2008:
Any two objects that possess a mutual tangential velocity will have a mutual outward radial acceleration. You look again at the weight that flies off the end of a string. The distance between it and the centre of the circle that it had been orbiting will be increasing. That is an outward radial acceleration. There are no exceptions to this general rule.

I'm fully acquainted with the official teaching which tells us that centrifugal force is only fictitious. That's why I wrote the article.

Thierry de Mees, Jan. 28, 2008:
Just read http://en.wikipedia.org/wiki/Uniform_circular_motion ...

David Tombe, Jan. 27, 2008:
Centrifugal force is not confined to the limited case of circular motion. It is a much more general phenomenon. If you take an ellipse and differentiate it, you will find that there are two forces acting in the radial direction towards and away from the focus of the ellipse.

The attractive force is an inverse square law force and the repulsive force is a centrifugal force. Centrifugal force is one of a package of four forces that emerge if we consider an aether whose field velocity is A, and differentiate it with respect to time. The other three forces are collectively known as the Lorentz force. How could three of them be real and the fourth one be fictitious?

Centrifugal force in general is grad(A.v). This means that it is a convective force that arises when a particle cuts across aether flow as when the Moon orbits the Earth. It occurs everytime that two particles possess a mutual tangential speed. If we look at the particular case of the circular motion of a weight on the end of a string, there will be an outward centrifugal force given by mv^2/r. This will be balanced by an inward centripetal force also equal to mv^2/r and supplied by the tension in the string.

The combination leads to zero radial motion and so the weight moves in a circle of fixed radius. If the string is broken, the weight will fly off both radially and tangentially. If it didn't fly off radially then it would keep its same distance with respect to the pivot and it would still be moving in a circle. When it flies off, its outward radial acceleration from the pivot will be given in terms of its tangential speed relative to the pivot.

Jean de Climont, Jan. 10, 2008
(votre adresse belge me laisse penser que vous parlez aussi français, mais je n'aurais aucun problème pour vous écrire en anglais).

En analysant les sites des théories alternatives au paradigme actuel, je trouve votre intéressante contribution au problème des marées dans la page Web de Monsieur Walter Babin.

Il est vrai que les idées les plus variées ont été émises à ce sujet. Aucune ne donnait satisfaction jusqu'à l'intuition d'un astronaute (je ne me souviens plus de son nom, mais cela fait bien 20 ans). Faisant le rapprochement entre son état d'apesanteur dans sa capsule et l'état d'apesanteur de la Terre dans sa rotation autour du Soleil, il a aussitôt pensé que le problème des marées ne pouvait être qu'un problème d'attraction différentielle donc en 1/r3, du fait que l'apesanteur est relative au centre de gravité de la Terre. Les océans sont à la surface de la Terre. Côté du Soleil l'attraction est plus forte que l'accélération centrifuge d'où un renflement vers le Soleil. Aux antipodes, l'accélération centrifuge l'emporte donc il y a également renflement dans la direction opposée au Soleil. Les marées sont donc biquotidiennes.

Il faut se rappeler, à ce stade du raisonnement, que la Terre ne tourne pas en réalité autour du centre du Soleil, mais plus précisément autour du barycentre de l'ensemble Terre-Soleil. La différence est minime, il est vrai dans ce cas.

Dès lors l'évidence s'impose :la Terre tourne également autour du barycentre Terre-Lune. Ce barycentre est à peu près au 2/3 du rayon de la Terre de son centre. Dans le système Terre-Lune, le centre de la Terre est donc en état d'apesanteur dans sa rotation autour de ce barycentre. Le raisonnement est un peu différent en raison de la position du barycentre à l'intérieur de la Terre. L'attraction de la Terre est la même partout à la surface moyenne de la Terre, il ne faut donc pas en tenir compte. Du côté opposé à la Lune, la gravitation lunaire est plus faible qu'au centre de la Terre et l'accélération centrifuge plus élevée. Mais la différence est du côté de la Lune. La gravitation lunaire est plus grande, mais l'accélération centrifuge, bien que plus faible qu'au centre de la Terre, est dirigée vers l'extérieur, vers la Lune. Il y a donc également renflement. C'est un peu plus compliqué que pour le Soleil. Il y a dissymétrie entre les antipodes relatifs à la Lune, contrairement à ce qui se passe pour le Soleil. C'est le résultat auquel tout le monde parvient et que vous trouvez également. Mais il n'y a besoin d'aucun calcul pour s'en assurer.

Je vous prie d'agréer, cher monsieur, l'expression de ma considération distinguée.

Thierry de Mees, Jan. 10, 2008:
Oui, je suis belge et également bilingue néerlandophone-francophone. Merci pour votre intérêt au site web de Walter Babin et à mon article. J'ignorais qu'un astronaute aurait été le pionnier à la découverte du système des marées. On n'est jamais assez âgé pour apprendre... Pour en revenir à votre hypothèse, en effet, l'explication des marées parait facile à développer par un raisonnement ayant rapport aux forces centrifuges et les décalages des objets célestes par rapport à leur barycentre. Mais nous sommes aveuglés par les expériences de notre enfance avec le seau d'eau qu'on fait tourner très vite à tour de bras, et par l'idée que l'eau de la Terre serait un peu comme de l'eau dans un bol ou un verre.

Malheureusement, le raisonnement que vous émettez n'est pas correct.

Qu'est l'accélération centrifuge?

A chaque endroit d'une orbite circulaire, l'accélération centrifuge est donnée par a=v²/r et la force de gravitation par g=Gm/r². Mais l'accélération centrifuge n'est qu'une force fictive, inexistante! Ce qui se passe vraiment est qu'un objet ayant une certaine vitesse rectiligne et une accélération gravitationnelle, perpendiculairement sur cette vitesse, décrit une courbe circulaire, qui peut être trouvée par la simple trigonométrie. L'accélération centrifuge n'existe pas. Elle a été introduite artificiellement pour parfaire à la loi action=réaction.

Ceci dit, l'attraction du Soleil est toujours identique à l'accélération centrifuge, je dirais presque par définition, par la loi trigonométrique, ou, si vous préférez, par la loi action=réaction, et ceci, pour chaque particule indépendante ou chaque objet indépendant.

En fonction de l'accélération centrifuge, la seule chose qui se passe est que l'eau, se trouvant du côté Soleil se trouve dans une orbite (une courbure trigonométrique) plus serrée (rayon plus court et vitesse tangentielle plus grande) et ira donc plus vite dans son orbite que le centre de la Terre. L'eau du côté opposé ira plus lentement que la vitesse d'orbite de la Terre. Mais il s'agit là de vitesses tangentielles qui n'ont rien à voir avec des renflements. Ce sont des déplacements des eaux dans le sens des "latitudes" de la Terre. Ces mouvements définissent tous les grands courants du monde, et même rend impossible l'arrêt du "Courant Indien", "The Main Conveyor Belt" qui nous apporte des courants chauds en Europe. Pourtant un certain ex-vice président des états unis semble y croire.

Toutefois, un phénomène est crée. Puisque ces changements de vitesse tangentielles le long des latitudes ont lieu deux fois par jour, une fois accéléré, deux fois neutre et une fois ralentie, dans le cas d'une obstruction par une côte terrestre, on peut avoir des 'courants de marée', qui ne sont pas les marées strictement parlant. Cet addition d'effets peut provoquer des hauteurs de marée significativement plus importantes ou moins importantes à tel ou tel endroit. Ces effets devraient surtout se sentir en latitude équatoriale côté Soleil, soit coté opposé au Soleil. Mais ce ne sont pas des renflements. L'accélération centrifuge n'a strictement rien à voir avec des renflements dont on parle traditionnellement, avec un renflement des deux côtés de la Terre, simultanément.

En ce qui concerne la Lune, l'erreur de raisonnement est encore plus clair. La Lune tourne autour de la Terre et non pas l'inverse, donc, l'accélération centrifuge ne se trouve pas côté Terre mais côté Lune, où il n'y a pas d'eau. Dans le cas contraire, la Terre qui tourne autour du barycentre Terre-Lune, l'effet est inexistant comme expliqué plus haut, hormis la présence d'un courant équatorial. En tout cas, je le répète encore une fois, l'action de l'accélération centrifuge se limite aux différences de courbures des orbites de l'eau côté Soleil et côté l'opposé du Soleil, et le seul résultat est le mouvement équatorial des mers contre le sens de rotation (spin) de la Terre. Les effets secondaires de ces flux d'eaux sont la déviation en cercles vers l'hémisphère nord et sud, le long des côtes terrestres.

Les renflements, décrites par le développement en série de Legendre, sont des forces perpendiculaires sur ces dernières vitesses, et forment les marées par déplacement des eaux dans le sens "longitudes + latitudes" de la Terre, et ce, des deux côtes (face et opposé) au Soleil et à la Lune. Les "vrais renflements classiques", c'est ça.

Egalement, les barycentres n'ont aucune influence sur les marées, puisque la distance entre les objets célestes restent identiques, pour chaque système de paires. La rotation de la Terre par rapport au barycentre du système Terre-Lune n'a aucune influence sur les marées pour la même raison. Uniquement l'interférence entre les deux systèmes Terre-Lune et Soleil-Terre pourrait donner une influence qui par-contre me parait assez négligeable. Cela mériterait néanmoins un calcul de contrôle.

J'espère avoir levé le voile plus encore que de par mes calculs, pour clarifier certaines fausses idées et paradoxes autour du phénomène des marées. I faut approcher cette étude de façon extrêmement rigoureuse, en essayant de comprendre chaque petit pas du point de vue physique, jusque dans les moindre détails.

J'avoue que moi-même, ces finesses ne me viennent pas tout de suite, mais après longue réflexion sur les formulations mathématiques obtenues. Et au départ, la vue des choses doit être à ce point lucides qu'un début de développement mathématique puisse se faire. Bien-sûr, je ne reste pas à l'abri d'erreurs. Il ne serait utile de savoir de votre part, à quels endroits de mon exposé de mon article, les choses devenaient trop compliquées, confuses, ou contradictoires. Vous pourriez certainement m'aider à les améliorer. Merci d'avance.

Jean de Climont, Jan. 11, 2008
Cher monsieur,
Vous ne pouvez imaginer le plaisir que j’ai eu à vous lire. Il est si difficile de faire comprendre que l’accélération centrifuge n’a aucune existence physique que j’ai depuis longtemps renoncé. Il est très rare de rencontrer des personnes, même dans les milieux scientifiques qui comprennent cela. On reconnaît que la force centrifuge n’existe pas. C’est ce qui m’a été enseigné dans ma jeunesse. Mais quant à l’accélération centrifuge elle reste pour beaucoup une réalité. Cela vient, je crois, de la conception newtonienne d’un espace mathématique absolument vide qui existerait physiquement comme contenant des choses.

Je suis un spécialiste de la mécanique des fluides. Dans un fluide, il y a des pressions et des vitesses mais nulle part d’accélération centrifuge. C’est précisément ce que vous dites : un artifice mathématique. Newton a accusé injustement Descartes de n’avoir pas donné de moyen d’équilibrer la gravité afin que les astres restent sur leur orbite. C’est une profonde injustice car Descartes à consacrer justement une très grande partie de son ouvrage à ce problème. Mais à l’immense différence de Newton, il raisonnait sur un éther fluide qui aurait été la cause de la gravitation. Il ne pensait qu’en termes de vitesse et de pression, notion déjà assez claire pour lui suite aux expériences de Torricelli. Il a attribué la gravitation à l’effet gyrocyclone (comme on peut l’observer dans une tasse de thé), mais en restant purement qualitatif, ce qui est évidemment le problème de sa théorie.

Il est bien évident que la proposition de mon courrier n’a de sens physique que dans un fluide ou dans un éther cartésien. Il faut la reformuler entièrement en termes de vitesse et de pression à l’exclusion de toute accélération y compris d’ailleurs de toute notion d’attraction qui sous-tend également une accélération.

Thierry de Mees, Jan. 11, 2008:
Cher Monsieur de Climont,
Je suis heureux d'apprendre que vous êtes spécialisé dans la mécanique des fluides. C'est un domaine passionnant que, hélas, je connais très mal. L'éther cosmique est en effet une réalité indéniable, ne fût-ce parce qu'elle contient les constantes de la perméabilité, la permittivité, la constante de gravitation, la constante de gyrotation (hypothétique) et donc aussi, la vitesse de la lumière. Un vide absolu ne pourrait contenir ces constantes, avec leurs dimensions physiques. Et peut-être l'éther peut-il être vu et décrit comme un fluide, proche à ce que nous savons des fluides terrestres. On peut dire qu'Einstein a agilement évité le problème de l'éther, mais n'a pas réfuté son existence. Les adeptes d'Einstein en ont profité pour descendre définitivement les adeptes de Poincaré en suggérant que si la théorie de relativité n'avait pas besoin d'éther, l'éther n'existerait pas non plus. Et comme la plupart des moutons suivent le chef du troupeau, jusqu'à aujourd'hui...

Entre parenthèses, la théorie que je développe ne tient pas directement compte de l'éther non plus, sans pour autant la nier.
Je vous remercie de notre intéressant entretien et je vous souhaite beaucoup de succès dans votre impressionnante quête sur les théories et les propositions alternatives. C'est un travail titanesque ! Aussi, je ne manquerai pas, si vous le permettez, de faire appel à votre expertise concernant la dynamique des fluides durant mes recherches.
Cordialement,

David Tombe, Nov. 10, 2007:
The textbook explanation of a fictitious Coriolis force is correct. But that is not the effect that explains the cyclonic patterns in the weather.

We need to consider a Coriolis force that is caused by the actual rotation of space itself. That can arise from an alignment of fine-grain rotating vortices ie. the magnetic field.

Thierry de Mees, Nov. 10, 2007:
Oh ! then that is the reason of the confusion : the textbook is wrong. I found a marvelous explanation of the "Coriolis Force" on Wikipedia instead.

David Tombe, Nov. 9, 2007:
What I am saying is that the Coriolis force in the atmosphere is indeed very real. I am also saying that it cannot be explained using the standard textbook approach. The applied maths textbooks teach Coriolis force to be a fictitious effect that occurs when a linear motion is observed from a rotating frame of reference.

That textbook approach cannot explain the very real cyclonic effects in the atmosphere.

I have made the Coriolis force real by involving the idea that space itself rotates. It rotates within the fine-grain vortices of the magnetic field.

Thierry de Mees, Nov. 9, 2007:
Thank you for your email. So, I suppose you agree with my analysis of the Foucault pendulum. The Coriolis problem still remains, and I think I discovered a confusion in your explanation.

Your explanation of what you call the "fictious Coriolis effect" is not the Coriolis effect at all. What you explain is only the difference between the 'absolute' spherical path (for the absolute observer away from the Earth's surface), and the 'relative' spherical path (for the relative observer on the Earth itself). That is of course not what I was talking about.

There is no fictious Coriolis effect. The Coriolis deflection is indeed real. If the wind tries to follow a 'straight' path over the sphere from North to South (or South to North), while connected to the Earth's rotation, it will be deflected by the Coriolis Force. Thus, it is definitely applicable to the atmosphere.

Only the word "force" in Coriolis Force is fictious, because it is not a real physical force, but the Coriolis effect can be described as if it were a force.

I hope this will clarify some more about the Coriolis Force and about the Coriolis effects on the atmosphere.

P.S. I correct my earlier e-mail: the Foucault pendulum rotates to the right at the North pole and at the left at the South pole. But this doesn't change anything to the essence of my explanation.

David Tombe, Nov. 8, 2007:
Let's concentrate on your first sentence below.
The fictitious Coriolis effect that you are talking about would indeed apply to a North -South movement of wind. But only if that movement was not already in any way connected with the rotation of the Earth. It would then appear to be deflected relative to the surface of the Earth due to the Earth's rotation, but it would perform a straight North-South line in space.

This is not what happens in real life. The Coriolis effect is real. The cyclones can be seen from space. And the fictitious Coriolis effect does not apply to the atmosphere because it is already rotating as a body with the Earth. That rotation cancels any artificial tangential effects.

Thierry de Mees, Nov. 7, 2007:
You write :"Besides that, it would not apply to something like the atmosphere which is already partaking in the Earth's rotation." This is indeed incorrect. At the contrary, the N-> S or S -> N velocity must be partaking in the Earth's rotation in order to invoke the Coriolis Force.

You also write : "The large scale effect that you are talking about applies to something like a Foucault pendulum at the poles, and to a lesser extent as the Foucault pendulum descends to lower latitudes." This is neither correct. The Foucault pendulum at one Earth's pole is a simple pendulum that is fixed at an 'absolute' point. It just shows the Earth's rotation (360° angular rotation each day). At the equator, the Foucault pendulum is fixed at a 'relative' point that moves along an 'absolute' equatorial circle around the Earth. At that location, the Foucault pendulum doesn't show any angular variation. Between the poles and the equator you get all kinds of mixes of both behaviors.

Your above sentence would mean that cyclones would preferably occur at the poles, if comparable with the Foucault effect, which is not true.

Besides that, the Foucault pendulum always rotates to the left (relative motion), on all locations on the Earth (with exclusion of the equator where it doesn't rotate), while the cyclones show inversed rotation directions in each hemisphere, just as the Coriolis Force does. The Coriolis Force occurs preferably neither at the poles, neither at the equator, but between-in, just as it happens with the cyclones.

As a conclusion, your explained magnetic influence might be acceptable, but the arguments against using the Coriolis Force and comparing the large effect of cyclones with the Foucault pendulum are definitely wrong.

David Tombe, Nov. 5, 2007:
I am not actually denying the Coriolis force. I studied the Coriolis force in depth in an advanced applied mathematics course at university and I know all about it. (see section VI of my paper The Coriolis Force in Maxwell's Equations I am devolving the source of the Coriolis force in atmospheric cyclonic behaviour to the fine-grain aether vortices in the magnetic field. The large scale effect that you are referring to is fictitious. It's effects would not be viewable from space. Besides that, it would not apply to something like the atmosphere which is already partaking in the Earth's rotation. The large scale effect that you are talking about applies to something like a Foucault pendulum at the poles, and to a lesser extent as the Foucault pendulum descends to lower latitutes. The Coriolis maths is fine for explaining cyclonic behaviour but I think that there will need to be some re-assessing of the coefficients.

Thierry de Mees, Nov. 4, 2007:
When reading your paper of cyclones, http://wbabin.net/science/tombe22.pdf and as far as I interpreted your paper correctly, I was surprised by the fact that you refute the major influence of the general Coriolis force of the Earth in the origination of cyclones. At the other hand, your proposal of the Earth's magnetic influence is very acceptable, and rather explains (in my opinion) the local rotations of small typhoons for example, in dry and electrically charged air.

The Coriolis force on the Earth occurs when a path is described, at a certain velocity, from the North to the South or vice versa, while following the Earth's rotation, upon the spherical earth, or in the spherical air surrounding the Earth.

The mathematical deduction is very real and correct. When you set a step from N to S, the Earth's rotation velocity increases tangentially, and results in a retardation of the new longitudinal position versus the original longitudinal position. This effect can be interpreted and expressed as a force : the Coriolis force. Thus, winds from N -> S and S -> N , or components following these paths, generate the Coriolis force and thus rotation.

There is an amplifying effect as well, once there is a cyclone that is born, due to the fact that the circular motion of the cyclone is constituted of parts that go from North to the South and vice versa. We can also expect an increase of the cyclone force over warm seas due to the lifting effect (with N -> S and S -> N components as well) and due to the injection of (heavier) water vapour.

So, all this really means very much to me, physically, and the magnitudes of these forces should not be underestimated.

MATHIS - DE MEES

Thierry de Mees, Sep. 3, 2007:
I am surprised of your answer. The maths do work, as I proved in my last email. But people publish wrong things because they assemble data here and there without knowing what they are doing. So, gravitation of course explains the tides perfectly.

This closes the chapter of tides. Concerning your E/M theory, I hope I will find some time to read it and let you know my opinion.

Miles Mathis, Sept. 3, 2007:
Thierry, congratulations on finding the error on your own. It is pretty clear once you do the math yourself. I never said that tides were caused by the Sun. You must have just read the first paragraph of my paper and assumed that is what I was arguing. I accept the current data which tells us that Sun tides are approximately 46% of Moon tides. And I accept all other data that you have mentioned, such as that the swells follow the Moon and so on. All this is clear by the simple fact that the tides are on a monthly cycle. There is no arguing with that, of course. But since these equations don't work, and since the barycenter math doesn't work either, I think we have to consider that the current mathematical description of gravity is incomplete, and that tidal theory is incomplete also. What my papers show is that the E/M field is exactly the right size to fill this hole. In short, the E/M field causes tides, not the gravitational field. For tides we require a force at a distance, and gravity simply cannot provide that. Gravity only provides an acceleration, not a force. But the E/M field supplies us with a mechanical force, mediated in a direct manner by subparticles (photons). THe mechanisim is there, and I have shown that the math is also there. My math proves that E/M fits the hole and gravity does not. The E/M field also gives us two orthogonal forces simultaneously (electric and magnetic) and this explains perturbations. Gravity was never able to explain perturbations mechanically, since the gravitational field cannot supply any forces at a tangent to the orbit. The magnetic field can. I recommend you read my paper on Celestial Mechanics. This shows where the problems are. The paper on tides addresses the tidal problem only. My two new papers on tides show the solution. My paper on the ellipse also shows an interesting solution to the orbital problem. All these papers are on my personal site, www.geocities.com/mileswmathis. You may still be reading me at Walter Babin's site, which doesn't have all my papers.

Thierry de Mees, Sep. 2, 2007:
When recalculating the whole chapter "17.4 Theory of Ocean Tides" http://oceanworld.tamu.edu , (see former emails), I come to the following. Equations (17.5) until (17.8) are correct. The conclusion below (17.8): "The second term produces a constant force parallel to OA." is of course wrong, since the second term contains a "cos (phi)". It is indeed the second term which will have influence on flows, as I told before, and I conclude even that the tides are generated by this second term. The contribution of (17.9) is insignificant and equations (17.10) until (17.14) are totally wrong. Let's calculate it over, this time correctly.

We have to bear in mind that phi and r are constants in this calculation. Only r1 and R vary. The approach to come to (17.10) is wrong. We do not look for variations of phi but we are looking what displacement the potential will give (the work). Thus, the second term of (17.8) brings us to the force, if we know which force is needed over an infinitesimal distance d r1 to obtain the given infinitesimal change of potential (only the second term).

I follow now the annexed handwritten page.

The correct force, generated by the second term of (17.8) is found by considering the change of potential when displacement occurs over an infinitesimal distance d r1. The horizontal part of this force, this is the force at the sea's surface of the spherical Earth, is given by the cos (beta) component. The vertical component (towards the bottom of the sea) does not interest us, because it will not influence tides much, since the sea-layer is very small compared to the Earth's radius.

I wrote down the geometric relationships as well, on the right side of the page. When simplifying, by putting r1 equal to R in equation (3), I come to the inversed cubic relationship in R.

I conclude that the deduction shown on the "oceanworld.tamu.edu" website is wrong. On the other hand, when using my deduction, the tides are generated by the Moon and not by the Sun, as expected. I guess that the numerous wrong deductions on the websites concerning tides are essentially due to the copy-paste facilities we are beneficing of these days, and, you are right, of the lack of criticism of scientists.

I thank you very much, dear Miles, to having opened my eyes once again: few people are using their brains...

Miles Mathis, Sept. 1, 2007:
Yes, as I said before, you have to actually analyze the equations. Don't take their word for it. I have closely analyzed all the math in my papers, and it doesn't work. It looks good from a distance, but close up it is a fudge.

Thierry de Mees, Aug. 31, 2007:
Concerning the tides-theory, I admit that it would appear "normal" that the inverse square gravitation formulas would be used and that the Sun would then rule the tides and not the Moon.

Observation however shows that the Moon rules the tides (at least, the quasi-resonance complies). So, I as well would look closely at the phenomenon of the water hydraulics in relation to the Sun and the Moon, which certainly are not identical to simple rigid bodies' dynamics.

At the first place, what does the Sun and the Moon do to the hydraulics of the oceans? They attract the oceans facing them, almost in line with their distances rS and rM. The same happens with the oceans which are at the back side of the Earth in relation to the sun and the moon. These oceans will however not move at all, since their gravitation point in line with the distances rS and rM, and since water is not compressible. This forms most of the first term of the equation (17.8) in http://oceanworld.tamu.edu/resources/ocng_textbook/chapter17/chapter17_04.htm The only places where gravitation can act, is more near the poles. It can create motion of the oceans as far as the oceans and the land create a kind of valleys wherein the flow can occur. Up to now, these motions must be ruled by the Sun and not by the Moon. The question is now: are these dynamic motions ruling tides? Or is it only limited to some flows? Is it possible to get these flows coming and going back 2 times a day? Remember: tides go essentially up and down and are not huge flows. So, I would conclude that the Sun's gravitation (which overrules the Moon's gravitation) is not necessarily the origin of tides, but only of huge flows. This forms most of the second term of the equation (17.8) in http://oceanworld.tamu.edu/resources/ocng_textbook/chapter17/chapter17_04.htm Honestly, I do not see any wrong maths up to now, and I see a physical meaning of these terms as well. The theory of tides has more terms in (17.8). The third term would explain the tides themselves. I did not analyse these terms yet, but I will do it when I find some time, because I like finding the physical meaning of things.

Miles Mathis, Aug.30, 2007:
I went to your link. All that math is fake. It is the same math I took apart at Wiki, and that Wiki took down in embarrassment. The math looks pretty, I admit, but it is full of holes. Look at the equation with r^3 in the denominator. The guy uses similar equations for sun and moon, but I show in my paper that the two equations can't be the same like that. If you pull the equation apart he is using the wrong radius in the moon equation. I show it all in great detail in my first paper http://geocities.com/mileswmathis/tide.html

You obviously didn't read that paper, you just assumed I couldn't be right. I recommend actually reading it. I am right. The math on the internet is garbage. ANother problem is the barycenter, which your guy also mentions. He mentions it in passing, as just the first term of his expansion, but he is hiding a big problem. If you do the barycenter part of the math in full, it turns out that the tides created by the earth orbiting the barycenter would swamp all the tides from the Moon and Sun, making them invisible. I do all the math and show this.

I know that it is easier just to accept pretty math when it seems to work, but I really think people need to dig a little deeper than that. And scolding me for publishing papers when you haven't even studied them is kind of strange, don't you think?

Try again to read the papers. It is worth your time.

Thierry de Mees, Aug. 6, 2007:
I thank you for your answer, and I will not go into discussion about it, since the motivation of your paper is not a classical approach as I thought it was. Indeed, I start from the point of view that only gravitation should generate such an effect, and I respect that you have another point of view. However, the theory I referred to seems to be extremely efficient, even if the mathematical deduction is quite unexpected and spectacular. But it is strictly done according some classic physical and mathematical rules.

Miles Mathis, Aug. 8, 2007:
I will look at your link. I cannot accept that it works until I do all the math. But I am highly skeptical, since I do not think the gravitational field causes tides. There is no mechanism capable of doing it,and I require a mechanism.

Thierry de Mees, Aug. 6, 2007:
I have read your papers on the tides (Miles Mathis - The Solution to Tides). The theories one find on the internet are indeed not always representing the actualized and accepted theories. Wikipedia or some university websites are certainly incorrect to some (perhaps many) points. And I agree that it is not always easy to know if these websites represent the general accepted theories.

Actually, the classical subject of Tides is fully understood, but sometimes incorrectly published. In http://oceanworld.tamu.edu/resources/ocng_textbook/chapter17/chapter17_04.htm is clearly explained how the -correct- theory works and how this complies with reality.

Thus, since the actual theory works perfectly, is it convenient to discuss the websites' wrong theories in papers on the WBabin website? Or shouldn't these issues be discussed in the forums of Wikipedia and in the University's forums directly?

In my opinion it is better, even if it is interesting to play with new ideas and with new theories, that only papers with a new insight, disproving other -generally accepted- theories should be published, even on an amateur Physics Website. Wrong websites should be discussed directly with the authors of the wrong websites.

It happened to me twice to freak on an unacceptable article, from Nasa. The first time I had contacted the responsible editor directly. I was right and Nasa removed the article immediately; the second time I had published a furious paper on WBabin's website as well. But, thanks to a website's colleague researcher, I found out that I was wrong and I had to remove the paper and apologise humbly to Nasa.

This has certainly been profitable for the overall quality of the scientific value of the internet, and particularly of the General Science Journal.

Wishing you a lot of success in your further research.