The nature of the Coriolis forces, centrifugal forces, turbulence and solution to the three-body problem
Abstract
According to existing concepts, the Coriolis force and centrifugal force are fictitious forces and, therefore, cannot do work. The mystical justifications for such ideas are calmly accepted by the scientific community due to “intellectual numbness” in front of the opinions of the luminaries of science. The purpose of writing this book is to substantiate the reality of these forces by showing, firstly, the physical cause of their appearance and, secondly, by showing many inexplicable natural phenomena and technical devices that become explainable based on the use of these forces. Explanation, as a rule, consists of constructing mathematical models. This second edition adds chapters that examine the three-body problem in detail and ultimately proposes a method and algorithm for solving the problem of interactions between many celestial bodies.
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S.I. Khmelnik The nature of the Coriolis forces, centrifugal forces, turbulence and solution to the three-body problem Second edition Israel 2025
0 - 2 Solomon I. Khmelnik The nature of Coriolis forces, centrifugal forces, and solution to the threebody problem Copyright © 2024 by Solomon I. Khmelnik Solomon Itskovich Khmelnik https://orcid.org/0000-0002-1493-6630 All rights reserved. No portion of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, without written permission of the author. Published by “MiC” - Mathematics in Computer Comp. BOX 15302, Bene-Ayish, Israel, 0060860 Email:[email protected] Israel 2025
0 - 3 Annotation According to existing concepts, the Coriolis force and centrifugal force are fictitious forces and, therefore, cannot do work. The mystical justifications for such ideas are calmly accepted by the scientific community due to “intellectual numbness” in front of the opinions of the luminaries of science. The purpose of writing this book is to substantiate the reality of these forces by showing, firstly, the physical cause of their appearance and, secondly, by showing many inexplicable natural phenomena and technical devices that become explainable based on the use of these forces. Explanation, as a rule, consists of constructing mathematical models. This second edition adds chapters that examine the three-body problem in detail and ultimately proposes a method and algorithm for solving the problem of interactions between many celestial bodies.
0 - 4 Table of contents Introduction / 5 Chapter 1. Four forces in mechanics / 8 Chapter 2. The nature of turbulence / 12 Chapter 3. Coriolis force and centrifugal force in electrodynamics and mechanics / 15 Chapter 4. New equations for the spinning top / 23 Chapter 5. Mathematical description of the Euler disk and experiments with it / 41 Chapter 6. Towards the substantiation of Mach's principle / 56 Chapter 7. Mass forces depending on the speed / 60 Chapter 8. Coriolis engine / 87 Chapter 9. Joke / 89 Chapter 10. Overflight anomaly / 93 Chapter 11. Coriolis force as a cause of some physical effects / 95 Chapter 12. Coriolis forces as driving forces in the swimming of marine animals / 99 Chapter 13. Rehabilitation of Rutherford's atomic model / /111 Chapter 14. On the nature of strong interactions / 117 Chapter 15. Unsupported Motion / 126 Chapter 16. Interaction of Electric Charges / 137 Chapter 17. Supplementing the Equations of Kepler's First Law and the Mercury Effect / 145 Chapter 18. On the Interaction of Celestial Bodies / 156 Chapter 19. New Statement of Three-Body Problem / 169 Chapter 20. Interaction of Two Celestial Bodies / 185 Chapter 21. Interaction of Three Celestial Bodies / 193 Chapter 22. Interaction of Many Celestial Bodies / 206 Chapter 99. References / 207
0 - 5 Introduction Currently, the prevailing idea in physics is that centrifugal forces and Coriolis forces are not real forces that do work. Mystical justifications for such ideas are calmly accepted by the scientific community due to “intellectual numbness” (in the words of M.G. Ivanov [44]) before the opinion of the luminaries of science. The book contains numerous evidence of the reality of these forces. The author is far from the first to undertake such a proof. First of all, I would like to point out Astakhov’s book [7]. Astakhov did a tremendous job of identifying numerous contradictions between physical explanations and real facts. This means that physical theory in this area cannot give reliable predictions of the development of natural processes and offer calculation and design methods acceptable for technology. “But the world did not collapse,” the reader will object. The world did not collapse only because engineers can find solutions even in the conditions of a stupid theory. We will not discuss why a bad theory can be stable. And we will not refute it in the hope that the apologists of such a theory will admit their mistakes. This is very unlikely. Let's try to propose another theory and show that it allows us to explain the previously inexplicable and solve technical problems. Chapter 1 discusses four forces: gravity, inertia, centrifugal force, and Coriolis force. Their similarities, differences, reasons for their appearance, reality, and fictitiousness are discussed. Chapter 2 shows the existence of Lorentz gravitomagnetic forces and then argues that turbulence is created by these forces. The source of these forces and additional energy that creates turbulence is the Earth's gravitational field. Chapter 3 points out that the current understanding of the nature of the Coriolis force and centrifugal force raises many puzzling questions. It is further proven that these forces can be justified asconse quence of the equations Maxwell’s or gravitomagnetism. In Chapter 4 is indicated that at present there is no complete theory of the top that answers all questions. A complete mathematical description of the top in statics using Coriolis forces is given.
0 - 6 Chapter 5 points out that there is currently no complete theory of the Euler disk. A complete mathematical description of the Euler disk in statics is given. Some experiments are described. In Chapter 6 the relationship between inertial and gravitational masses on Earth has been established. It is shown that Mach's principle and the principle equivalence of inertial and gravitational masses can be investigated experimentally. Chapter 7 assumes that in fluid dynamics, body forces are functions of velocity. Such forces are frictional forces, inertial forces, centrifugal forces, and Coriolis forces. In this case, only stationary problems for a viscous incompressible fluid are considered in some special cases, for which solutions are obtained in analytical form. Mathematical models of liquid jets, sea currents, tsunamis, dust devils, waterfalls, funnels and whirlpools are proposed. It is shown that taking into account Coriolis forces allows one to find a solution to the problems posed. Chapter 8 discusses the Coriolis force engine. Chapter 9 proposes an experiment that would be direct evidence of the reality of Coriolis forces. Chapter 10 examines the flyby anomaly and shows that it is explained by the influence of Coriolis forces. A method for calculating the flight anomaly is presented. Chapter 11 points out (without mathematical justification) some physical effects that require the Coriolis force to be explained. Chapter 12 proposes a mathematical model for the swimming of marine animals and explains the long-noted and unexplained Gray's Paradox: the actual power expended by an animal to move far exceeds its own power. Chapter 13 shows that in Rutherford's planetary model of the atom there are stationary orbits of the electron that are stable in height, and the reason for the stability and discreteness of stationary orbits is substantiated. Chapter 14 proves that nuclear forces can be substantiated as a consequence of Maxwell's equations. It is shown that the detected repulsive forces exceed the Coulomb attractive forces by a factor of ω, where ω is the rotation speed of the nucleons. It is shown that, despite their mutual attraction, rotating nucleons cannot touch. Chapter 15 analyzes a speculative construction that demonstrates unsupported motion created by the internal motion of electric charges, which can explain the flight of ball lightning.
0 - 7 Chapter 16 describes numerous variants of the interaction of electric charges. In a number of cases, these forces take the form of Coriolis forces or centrifugal forces. Chapter 17 shows that taking into account the centrifugal force allows one to substantiate the stability of elliptical orbits, and taking into account the Coriolis forces allows one to substantiate the Mercury effect. In Chapter 18, the law of universal gravitation is derived as a consequence of solving the Maxwell-Heaviside equations. Further, in accordance with these same equations, it is shown that there are flows of electro-magnetic-gravitational energy - gravitational energy even in the case when strengths do not change in time. Ultimately, it is shown that the interaction of celestial bodies occurs through the exchange of flows of gravitational energy. In this case, the force of attraction of a satellite to the Sun arises in the satellite, which itself is attracted to the Sun. In Chapter 19, it is noted that the existing solution to the threebody problem establishes the chaotic nature of this motion. It is proposed to change the mathematical formulation of this problem, which will allow us to find unambiguous solutions and stable solutions. The results of the solution of a simple three-body problem, which was found by a new method, are given. Chapter 20 discusses an algorithm for calculating the trajectories of two celestial bodies as a consequence of their interaction by centrifugal and gravitational forces. It is shown that they exchange flows of gravitational and kinetic energy. It is shown that centrifugal forces arise as a result of gravitational interaction. Chapter 21 proposes a method and algorithm for solving the problem of three bodies on a plane and in three-dimensional space. The proposed method allows calculating not only steady-state motions, but also transient processes. The results of programming some problems are considered. Chapter 22 proposes a method for calculating the motion of a set of interacting celestial bodies. The problem, considered unsolvable and proving the existence of chaos in the Universe, is solved based on strict observance of the laws of conservation of energy, momentum, and angular momentum. The possibility of solving the problem under consideration appears due to the fact that the interaction of celestial bodies is considered a result of the existence of not only gravitational forces but also centrifugal forces.
Глава 1. Four forces in mechanics 1 - 1 Глава 1. Four forces in mechanics Consider Table 1 that lists the mechanical forces acting on a separate body [29]. Table 1. The force of inertia ? Real force? Reason for appearance of force 1 ܨ Gravity force no yes distant body 2 ܨ Inertial force yes yes rectilinear motion with acceleration 3 ܨ ௭ Centrifugal force yes no body rotates 4 ܨ Coriolis force yes no body moves translationally and is in a rotating coordinate system The following several questions arise: 1. Why rectilinear motion with acceleration (see item 2) causes real force but rotation (see item 4) causes fictitious force? 2. Why and how does a noninertial (rotating) coordinate system (it does NOT interact with the body, see item 4) creates a fictitious force? 3. Why and how does the inertial coordinate system (also NOT interacting with the body, see item 2) create a real force? 4. Why is another body found for force 1 that creates this force, and for inertial forces it can be argued that such a body is absent? 5. And at the same time why its absence in item 2 is compatible with the reality of force, and its absence in items 3 and 4 requires explanations and the introduction of ideas about the existence of fictitious forces?
Chapter 3. Coriolis force and centrifugal force 3 - 1 Chapter 3. Coriolis force and centrifugal force in electrodynamics and mechanics Annotation The existing understanding of the nature of the Coriolis force and centrifugal force raises many puzzling questions. The article proves that these forces can be justified as consequence of the equations Maxwell’s for gravitomagnetism. It is further shown that the flight anomaly is a consequence of the influence of Coriolis forces and a method for calculating this influence is indicated. Table of contents 1. Introduction \ 1 2. Interaction of moving electric charges end masses\2 3. Equations Maxwell for gravitomagnetism \ 3 3.1. Coriolis force for a body that moves above the rotating Earth \ 4 3.2. Coriolis force for a body that moves and rotates above a stationary Earth \ 5 3.3. Coriolis force for a body moving and rotating above the rotating Earth \ 6 3.4. Centrifugal force \ 9 3.5. Levitation of rotating disks / 8 Conclusion \ 10 References \ 10 1. Introduction Modern ideas about the Coriolis force [1] are as follows: the Coriolis force is in no way related to any interaction of the body in question with other bodies, The Coriolis force is not a physical force and does not do work. Roughly speaking, the Coriolis force acting on a nearby body appears because another body rotates next to this body at a certain speed. Our body does not interact with this body and therefore “does not know” the magnitude of this speed, but the Coriolis force depends on this speed. The
Chapter 3. Coriolis force and centrifugal force 3 - 2 mass of this other body and the distance to it do not matter. The Coriolis force does not do work, but a freely falling body is deflected the rails of one-way railways wear unevenly, long-range artillery shells deviate from the calculated trajectory, etc. So, there is an inertial reference system rotating with the angular velocity vector, there is a non-inertial reference system that does not interact in any way with the inertial reference system, in a non-inertial frame a body of mass moves with speed , in this case, the Coriolis force is observed acting on the body perpendicular to the speed, which is determined by the formula = −2( × ). (1) This force is observed as fictitious from a non-inertial frame of reference, as, for example, in the experiment with the Foucault pendulum. But this same force is also observed from an inertial reference frame, such as, for example, the real force of coastal erosion. Observation can also be carried out from a third system, in which a non-inertial reference system rotates, in which the inertial reference system is located, as in experiment [5]. In this case, we observe how, in a non-inertial system, a fictitious force physically draws a spiral... Therefore, the Coriolis force cannot be considered fictitious and cannot be explained by the peculiarities of the observer’s perception. It is necessary to try to find a physical connection between a real rotating system and a body moving in it or near this system. These issues are discussed in more detail in [8]. It is further shown that the Coriolis force is found as a consequence of Maxwell's equations for gravitomagnetism. These forces exist in the vicinity of a gravitating body (Earth). Consequently, the Coriolis force can only arise in the vicinity of such a body and cannot exist in open space. The Coriolis force is a full-fledged force that does work. The energy expended to perform this work is supplied by the gravitating body. This proof was first given by the author in [6]. Here it is discussed in more detail. 2. Interaction of moving electric charges and masses The gravitational forces considered here exist in addition to the forces of gravity, but they have a common source of energy. Gravitational forces
Chapter 3. Coriolis force and centrifugal force 3 - 3 are completely analogous to the forces of interaction of moving electric charges. The existence of these forces is generally recognized. Moving charged bodies interact differently than stationary ones. The forces of their interaction can take the form of Coriolis forces or centrifugal forces. These issues are considered in Chapter 16. The gravitational forces of interaction of moving masses are completely analogous to the forces of interaction of moving electric charges. And when describing the gravitational forces of moving masses, we simply refer to the formulas related to the electric forces of moving charges. These formulas are given with the prefix 16, for example, (16.21). This analogy is based on the analogy between Maxwell's equations for electromagnetism and for gravitomagnetism. 3.EquationsMaxwellForgravitomagnetism In [2] the author proposes a new solution to the equations Maxwell for gravitomagnetism, which is used to build mathematical models of various natural phenomena (sand vortex, sea currents, whirlpool, funnel, water soliton, water and sand tsunami, turbulent currents, additional (nonNewtonian) interaction forces of celestial bodies). All these models use the idea of mass currents as flows of mass particles. The speed of mass particles can be very small and often their flow can be as invisible as the flow of electrons. But the existence of these phenomena and the possibility of constructing these mathematical models, similar to the mathematical models of direct current in electrodynamics [4], confirm the assumption of the existence of mass currents and the interaction of mass particles, completely analogous to the interaction of electric charges. Based on this, it can be assumed that the rotation of a body is accompanied by a mass current, similar to how the rotation of a charged body is accompanied by a convection electric current. Eichenwald [3] showed that such a current creates magnetic induction. Based on the complete analogy between Maxwell’s equations for electrodynamics and gravitomagnetism [2], it can be argued that when a body rotates, gravitomagnetic induction is created. A mass moving in a gravimagnetic field with a speed is acted upon by the Lorentz gravitomagnetic force (an analogue of the Lorentz magnetic force). Between the moving masses and , a force of interaction of moving masses (FIMM) is created (by analogy with the force of interaction of moving electric charges) – see Chapter 16. By analogy with (16.4), we write down the formula for determining FIMM: = ×(× )(). ⁄ (40)
Chapter 3. Coriolis force and centrifugal force 3 - 4 Here and below we will use the symbol μ to denote the gravitomagnetic permeability. The FIMM force is an additional non-Newtonian force of mass interaction. The Coriolis force and the centrifugal force are special cases of FIMM. Non-Newtonian forces were previously considered in [17, 18]. In [12] the authors pose, in particular, the following questions: 1. If a distant object such as a comet comes near the Sun from an almost inϐinite distance, it should fall into the Sun by virtue of gravitational force rather than making an elliptical path around the Sun. How does it know to bend around the Sun? 2. So much of matter in the Universe and the existence of Gravity in the Universe should result in the collapse of entire matter. There should be some force giving direction to the movement of the objects in the Universe. 3. To make planetary motion exist, there should be some additional force to guide the matter in the planetary motion. The authors are absolutely right in posing their questions. They find an explanation in the fact that the gravitational attraction between planets and stars is caused by external energy that our three-dimensional Universe receives from outside. Another explanation is given below. From what has been proven above, it follows that a body flying above a rotating planet is affected by the FIMM force, directed along the radius and repelling the body. It is this force that plays the role of centrifugal force in astronomy, competing with the force of attraction, which causes the body to rotate around the planet. Then the answers sound like this: 1. An asteroid approaching the Sun from afar begins to fall on it, moving past the Sun and leaning towards it. The projection of this movement onto the plane of the Sun's surface causes the appearance of centrifugal force. A situation may arise when at a certain altitude the centrifugal and gravitational forces are balanced. Then the asteroid becomes a satellite. 2. Gravity is counteracted by the Coriolis forces created by the rotation of stars. 3. These same forces create planetary motion.
Chapter 3. Coriolis force and centrifugal force 3 - 5 3.1. Coriolis force for a body that moves above the rotating Earth Consider row 3 in the table. 1. Based on the above, let us rewrite formula (16.21), obtained above for the interaction of electric charges, as applied to the interaction of mass charges: = ( × ), (41) where ≈ −45, , are mass and speed of a moving body, is surface density of masses, as elements of mass current, - angular velocity of rotation of the plane on which these elements are evenly distributed. Comparing formulas (1, 24) we find that −2 = , (42) whence it follows that the mass density = − ≈ 0.044 = 4.4 ∙ 10 . (43) “Excuse me,” the attentive reader will be surprised. “You reject the Coriolis theory and at the same time use his formula?” I can only join in his surprise and be even more surprised that such different methods of reasoning led to the same formulaic result! Nevertheless, the conclusion of mass density (26) can be accepted only if there is a confident experimental verification of the value of the coefficient “2” in formula (1) of the Coriolis force. The method used to derive the Coriolis force proves the reality and not the fictitiousness of this force and reveals the source of power for this force - the Earth's gravitational field. 3.2. Coriolis force for a body that moves and rotates above a stationary Earth Consider row 1 in the table. 1. This case assumes that the Earth is stationary. In practice, this means that the speed of rotation of the Earth is negligible compared to the speed of the body. But to describe it, the same formula (16.21) is used, which was used above in section 3.1, i.e. the same formula (16.24) and the same constant . This means that the body is acted upon by the Coriolis force, which depends only on the parameters of the movement of the body itself, but the Coriolis force appears only due to the existence of the Earth next to the body! Next we will show that in deep Space this Coriolis force is absent! The source of power for this force is the Earth's gravitational field.
Chapter 3. Coriolis force and centrifugal force 3 - 6 3.3. Coriolis force for a body moving and rotating above the rotating Earth Ermolin in article [10] examined the influence of Coriolis forces on the planets of the Solar system. «The article considers the effect of the Coriolis acceleration forces, inertia forces and the gravity forces of on the planets of the Solar system, as well as all existing objects of the Universe. It explains also the effect of the Coriolis acceleration forces on the formation of the ocean tides, with the greatest force acting in the Earth’s equatorial zone. The authors have established the effect of the Coriolis acceleration forces on the position of the rotary axes and planes of the orbits of any space objects, including planets in the Solar system, with the simultaneous rotation of these objects around their own axes and around the center of rotation, which is typical of all the planetary systems of the Universe.» This serves as a clear proof of the reality of the Coriolis forces. The author does not note this, because in an edited official publication it is better to about it and not mention it. The author did everything according to the requirements of official mechanics, which found a way out of the impasse "Coriolis force is fictitious, but manifests itself in reality." Everything is simple, dear reader! Acceleration is proportional to force. Therefore, we can talk about Coriolis acceleration instead of Coriolis force, and the question of the reality or fictitiousness of acceleration is simply not raised. In Chapter 17, the Mercury effect is explained by the existence of the Coriolis force. 3.4. Centrifugal force Consider row 2 in the table. 1. It is shown that the rotation of a charged plane under a moving charge can be replaced by the rotation of a charge above the charged plane. Then in formula (16.21) is the vector of the angular velocity of the rotating charge, the linear speed of this charge = . (57) In applying the formula(23)to the interaction of mass charges we obtain: = . (58) This formula is different from the formula for centrifugal force С= , (59) only by coefficient. By analogy with the previous one, here we find the mass density = − ≈ 0.022 = 2.2 ∙ 10 . (60) Thus, the nature of the centrifugal force is the same as the nature of the Coriolis force, and the source of power for this force is the Earth's
Chapter 3. Coriolis force and centrifugal force 3 - 7 gravitational field. The difference between formulas (60) and (16.26) raises doubts about the correctness of the coefficient “2” in formula (1). In Chapter 17, the stability of elliptical orbits is explained by the existence of centrifugal force. 3.5. Levitation of rotating disks Consider row 4 in the table. 1. It is shown that a rotating body interacts with the rotating Earth. This manifests itself as levitation of rotating disks. This phenomenon is discussed in detail in [11]. Conclusion Coriolis forces and centrifugal forces are real physical forces. Their existence is explained by the interaction of mass currents existing in moving bodies and the rotating Earth. The interaction of mass currents is similar to the interaction of electric currents. The indicated forces exist in the vicinity of any rotating celestial bodies, and this vicinity is a sphere, the radius of which is equal to the diameter of the body. It follows from this that these forces must be taken into account in any calculations of mechanics, hydrodynamics, aerodynamics, and astronautics. These forces must be especially taken into account when designing any rotating structures in spacecraft intended for flights in the vicinity of massive celestial bodies. References 1. Coriolis force, Wikipedia,https://ru.wikipedia.org/wiki/Coriolis_Force 2. Khmelnik S.I. Gravitomagnetism: natural phenomena, experiments, mathematical models. 5th edition, 2020, ISBN 978-1365-62636-4. Printed in USA, Lulu Inc., ID 20262327, http://doi.org/10.5281/zenodo.140366 3. A. Eichenwald. Electricity, M.L. 1933, paragraph 282,http://lib.izdatelstwo.com/Papers2/Eyhenvald.djvu 4. Khmelnik S.I. Consistent solution of Maxwell's equations.18th edition, 2020, ISBN 978-1-329-96074-9. Printed in USA, Lulu Inc., ID 18555552, http://doi.org/10.5281/zenodo.3783458 5. A ball rolling on a rotating platform NRNU MEPhI, https://www.youtube.com/watch?reload=9&v=LkrmALM8TsA 6. Khmelnik S.I. Coriolis force and centrifugal force in electrodynamics and mechanics. Papers of Independent Authors,
Chapter 3. Coriolis force and centrifugal force 3 - 8 ISSN 2225-6717, 2020, 48(2), 65–73, https://doi.org/10.5281/zenodo.3900260 7. https://en.wikipedia.org/wiki/Flyby_anomaly 8. Khmelnik S.I. Four forces in mechanics. Papers of Independent Authors, ISSN 2225-6717, 2022, 54, 174–178, https://doi.org/10.5281/zenodo.7004069 9. Zilberman G.E. Electricity and magnetism, Moscow, ed. "Science", 1970. 10. Ermolin V.B. Coriolis acceleration and its effect on space objects. SCIENCE WITHOUT BORDERS, No. 2 (7), 2017, https://cyberleninka.ru/article/n/koriolisovo-uskorenie-i-ego-vliyaniena-kosmicheskie-obekty 11. Coriolis force, https://drive.google.com/file/d/1na73RPBOa3wTc3ltd_Zlv0u6f3 FP1QIv/view?usp=sharing 12. Pushpak N Bhandari and Nandan M Bhandari. Fundamental Forces are not Fundamental as our 3-D Universe is Driven by an External Energy Source. International Journal of Physics Research and Applications , https://journals.indexcopernicus.com/search/article?articleId=3908705 13. Законы орбитального движения планет, http://www.solarclimate.com/sc/zodv.htm 14. К.В. Бычков, А.С. Нифанов, И.М. Сараева. Задачи по теме «Динамика материальной точки» для студентов астрономического отделения. МГУ им. М.В. Ломоносова. 15. Хмельник С.И. Прав был упрямый Хэвисайд! Доклады независимых авторов, ISSN 2225-6717, 2024, 62, 101–106, https://doi.org/10.5281/zenodo.11102511 16. https://ru.wikipedia.org/wiki/Законы_Кеплера 17. Хмельник С.И., Хмельник М.И. Дополнительные силы взаимодействия небесных тел. Доклады независимых авторов, ISSN 2225-6717, 2012, 21, 57–64, https://doi.org/10.5281/zenodo.3551381 18. Хмельник С.И., Хмельник С.И. Еще о дополнительных силах (неньютоновских) взаимодействия небесных тел. Доклады независимых авторов, ISSN 2225-6717, 2013, 24, 149–159, https://doi.org/10.5281/zenodo.3551391
Chapter 4. New equations for the spinning top 4 - 1 Chapter 4. New equations for the spinning top Annotation It is pointed out that at present there is no complete theory of the top that answers all questions. A complete mathematical description of the top is given, incl. equations of top dynamics at any speeds. New well-known experiments that currently do not have any explanation are considered in detail. The resulting equations use the fact, disputed by today's science, that the Coriolis force and the centrifugal force are real forces that do work. The coincidence of the calculation results and experiments is proof of this fact. Contents 1. Introduction 2. State equations 3. Forces acting on the spinning top 4. Examples 5. Dynamics 6. Conclusions Appendix 1 References 1. Introduction The question of why the top does not fall is constantly raised despite the fact that there is a well-founded theory of how the top works. This question is not new. It was asked in 1890 by prof. John Perry [10, p. 93]. He wrote “... in a spinning top, obviously, only with rotation does life and stability appear, or, in other words, only then do forces act that oppose the earth's gravity, which tends to overturn the spinning top. Where do these forces come from and how are they explained? ” The questioner intuitively feels that the initial push cannot give the energy that is needed for a long and vigorous rotation. The questioner intuitively feels that there must be a real force that keeps the top from falling. But the theory explains why it spins, and the unspoken sounds like answer is "doesn't fall because it spins." But maybe our intuition is deceiving us and the spinning top actually has enough energy? This issue first of all discussed below.
Chapter 4. New equations for the spinning top 4 - 2 2. State equations On fig. 1 shows a spinning top in its simplest form. The spinning top has rotation of the top around its own vertical axis with an angular speed , rotation of a top inclined at an angle ∝ to the plane around its own axis with an angular speed , precessionon the circumference of a top inclined at an angle ∝ to the plane, with an angular speed . In Table 1 of Appendix 1 lists the parameters of the state of the top at the initial moment 1 and at the moment 2, when the top is in a position at which the angle ∝< 2 ⁄. At moment 1, there is only rotation around the vertical axis. At moment 2, a precession additionally appears. Let us write down for moment 2 the equations of the laws of conservation of momentum and energy , which do not depend on how and by what forces the top passed into this state: +=, (2) +=. (3) Substituting the equations from Table 1 into equations (1, 2), we get: +=, (4) + = . (5) where , - angular speeds of rotation of the top at the moment 1 and 2 around the axis, which is the rod, - angular speed of precession around the vertical axis, ,, - moments of inertia during rotation with speeds ,,. ,, - angular momentum during rotation with speeds ,,. Moments of inertia == (6) The moment of inertia changes depending on the angle ∝ - see Fig.2. For ∝=0 this moment can be taken equal to =ℎ. For ∝= this moment is = by the Steiner theorem. So,
Chapter 4. New equations for the spinning top 4 - 9 Example 3. In this example, the functions listed above are found at h=0.8; R=0.2; =60, m=1 - see fig. 4. The dotted vertical in the third window highlights the point where ∝о= 2. At this point, the function (∝)=0 and its derivative (∝)∝ <0. Therefore, at this point the top is in a stable position. Fig. 4. The practical implementation of this case, i.e. when the barbell of top is in a vertical position, and the disk is horizontal, can be found on the Internet - see fig. 5. It can be seen that the top in this case hangs motionless in the air, i.e. the vertical force acting on the top is zero. In our example, you can also see that the vertical force acting on the top is zero - see the F2x force graph in the first window. 0 0.5 1 1.5 2 -1000 -500 0 500 1000 1500 F2x(g), F2y(b), F3(r) 0 0.5 1 1.5 2 alfa (Volchok99.m) -1000 -800 -600 -400 -200 0 M(r) T(b) 0 0.5 1 1.5 2 -50 0 50 100 150 om1(g) om2(b) om3(r) 0 0.5 1 1.5 2 alfa (mode=5) -1000 -500 0 500 1000 FSx(g) FSy(b)
Chapter 4. New equations for the spinning top 4 - 10 Fig. 5. Example 4. In this example, the functions listed above are found at h=0.6; R=0.15; oml=6.9 - see fig. 6. The dotted vertical in the third window highlights the point where ∝о=0. At this point, the function (∝)=0 and its derivative (∝)∝ <0. Therefore, at this point the top is in a stable position. Fig. 6. 0 0.5 1 1.5 -10 -5 0 5 10 15 F2x(g), F2y(b), F3(r) 0 0.5 1 1.5 alfa (Volchok99.m) -2 -1 0 1 M(r) 0.1*T(b) 0 0.5 1 1.5 -10 -5 0 5 10 om1(g) om2(b) om3(r) 0 0.5 1 1.5 alfa (mode=7) -10 -8 -6 -4 -2 0 FSx(g) FSy(b)
Chapter 4. New equations for the spinning top 4 - 11 The practical implementation of this case, i.e. when the barbell of top is in a horizontal position, and the disk is vertical, is considered in many publications - see, for example, fig. 7 from [3], fig. 8 from [4, 9]. It can be seen that the top in this case rotates on a horizontal rod. Measurements in [4] show that its top weight is zero. There are no explanations. In our example, you can also see that the vertical force acting on the top is zero - see the F2x force graph in the first window. Thus, both practically and theoretically it is shown that in this position the top is weightless. Such an experiment is also considered in [8] with reference to [9]. The article begins by stating that for such a device "the detailed mechanics of which are still an enigma". The author of this article has developed three new Euler equations that are much longer than those found in textbooks. The resulting nonlinear equation is modeled in the MATLAB system to obtain and visualize a numerical solution. Under certain conditions, providing small oscillations of the gyroscope axis (maximum oscillation of eight degrees in the angle of inclination) near the horizontal plane through the fulcrum, linearization is performed, which is successfully compared with the above-mentioned nonlinear numerical solution. The author argues that the numerical solution under certain conditions «is crucial to the debate about whether such an engine may produce a net thrust, or not. A relevant paradox is resolved». The question of where the source of forces is is up in the air, as is the device in question. In our example, you can also see that the vertical force acting on the top is zero - see the F2x force graph in the first window. Thus, both practically and theoretically it is shown that in this position the top is weightless. Fig. 7.
Chapter 4. New equations for the spinning top 4 - 12 Fig. 8. 5. Dynamics Above, we considered a sequence of static states that differ in the value of the angle ∝. In the simplest case, we can assume that the top stays in position with given ∝ for a time inversely proportional to the overturning moment M. Under this assumption, we can calculate the duration of the top in position ∝ by the formula (∝)=1/abs(∝). (1) The duration of the fall of the top from the position ∝= to the given position ∝ is determined by the formula: Т(∝)=∫(∝)∙∝ ∝ / . (2) The stable position ∝, as mentioned, is determined by the condition of the form (∝)=0 and (∝)∝ <0. The top remains in this position until the friction against the air reduces its kinetic energy. Example 1 Consider example 4.2 and fig. 4.2. Under the conditions of this example, we construct the functions t (∝), Т(∝), (∝) - see fig. 1. The left border of the graphs corresponds to the steady position of the top.
Chapter 4. New equations for the spinning top 4 - 13 Fig. 1. Let us now consider the above equations, taking into account the equation of the dynamics of the rotational motion of a rigid body around a fixed axis, which has the following form: о=о , (1) where о is the moment of forces acting on the body, о is moment of inertia of a rotating body, is angular velocity of body rotation. The top performs two rotations simultaneously and its equivalent moment of inertia and the equivalent rotation speed are connected with the moments of inertia and the speeds of the terms of the rotations by the equation of the momentum conservation law (2.4). Therefore, in our case (1) takes the form =1 , (2) where M is the overturning moment of the top, considered above. Then = . (3) Appendix 1 shows that 1 1.1 1.2 1.3 1.4 1.5 1.6 0 0.5 1 t 1 1.1 1.2 1.3 1.4 1.5 1.6 0 50 100 150 T 1 1.1 1.2 1.3 1.4 1.5 1.6 alfa (Volchok99.m) -0.5 0 0.5 M
Chapter 4. New equations for the spinning top 4 - 14 = . (4) Hence, = . (5) This means that the speed , given at the initial moment, changes depending on (5) from the overturning moment of forces. With a large mass of the top, acceleration (5) can be considered zero and the above method for calculating the top can be used. Taking into account (5), the calculation of the dynamics of the top should be performed according to the following algorithm: 1. At the initial moment ∝= ,=,=0. 2. The system of equations (2.4, 2.5) is solved, as shown above, and thus the speeds , are determined. 3. The moment M is calculated as shown above. 4. The condition of the form =0 and ∝ <0 is checked. The fulfillment of this condition means that the top has passed into a steady state and the calculation is terminated. 5. The new value ∝=∝+∙∝ is calculated. 6. is calculated using (5) with the obtained value of M. 7. The new value =+ is calculated. 8. Go to step 2 Consider in equation (2.5) the left term of the form . It corresponds to the total kinetic energy of the top. Since (as follows from the algorithm) the speed changes, the kinetic energy of the top also changes - it can decrease and increase. The power delivered by the overturning moment in a given position of the top depends on ∝ and is defined as (∝)=(∝)∙1(∝). (6) 6. Conclusions An algorithm for calculating the dynamics of a top is proposed. It can be used at any top rotation speed. It allows us to explain known but still unexplained experiments. He explains, in particular, the increase in weight and energy of the spinning top. The algorithm is based on the use of the equation of dynamics of rotational motion of a rigid body and the laws of conservation of
Chapter 4. New equations for the spinning top 4 - 15 momentum and energy. In addition, it uses the notion that Coriolis forces and centrifugal forces are real forces. Such an action of the Coriolis forces and centrifugal forces is possible only if they can do work, i.e. are real powers. This proves the reality of these forces. On the other hand, a mathematical proof of this fact is given in [1]. It shows that these forces can be justified as a consequence of Maxwell's equations for gravitomagnetism, and the energy source for these forces is the Earth's gravitational field. But even in the absence of such evidence, there are many doubts about the assertion that these forces are fictitious [2]. Another proof of the reality of these forces is the explanation of many astronomical facts found in [11] using the Coriolis forces. So, the article gives a mathematical description of the top, which is still missing, which uses the known facts of mechanics and the notion of the Coriolis force, which is NOT accepted in mechanics, as a real force. The author is repeatedly pointed out the MEPhI experiment [5], in which the fictitiousness of the Coriolis force is convincingly proved - see fig. 1. Consider this proof. But first of all, I want to note that I am not criticizing the lecturer, nothing personal. The experimenter is one of the best teachers in Russia at one of the best physics institutes in Russia. Fig. 1. The disk rotates with an angular speed ω. The ball is pushed out by the experimenter from the central hole along the radius of the disk and moves along the radius under the action of the inertial force with a linear speed . In this case, the Coriolis force с, perpendicular to the radius , acts on it. As a result, the ball describes a spiral, moving in the direction opposite to the rotation of the disk (therefore, we cannot suspect that the
Chapter 4. New equations for the spinning top 4 - 16 disk is pulling it with the force of friction). The force that pulls the ball tangentially is the Coriolis force: с=−2×, (1) The question is, where did this power come from? Further, the experimenter argues that this force appeared because the ball moves in a coordinate system associated with a rotating disk, and the angular speed of the disk enters formula (1). The experimenter is one of the best teachers in Russia at one of the best physics institutes in Russia. Modern Physics speaks through him. We can suggest a modification of the experiment. Let there be a thin plane above the disk and let the ball lie on this plane. At the same time, we completely exclude the mechanical influence of the disk. Only the coordinate system of the disk remains. So, we push the ball and bring the rotating disk to the plane. In this case, the Coriolis force appears, moving the ball. No fraud! No wonder, because there really is no power. The question of where this power came from is superfluous. Such is nature, says physics. Physics is fine with that. But how can a physicist accept such an explanation?! It would be more honest to admit that physics has no explanation and it must be sought. Or, following the example of Mach, to assume the influence of celestial bodies. But in this experiment, the explanation is much simpler. The ball, lying in the central hole, rotates together with the disk with an angular speed ω and continues to rotate after the central impact of the experimenter. The speed in formula (1) is the speed of the ball, not the disk. It is the lack of a clear answer to the question “where does the power come from?” and led to the emergence of such a theory: it was necessary to find the answer so that the students respected the teachers! It would be possible not to build hypotheses about the nature of this force (as Newton did with the force of inertia). But the times at that time were, apparently, not the same. And off we go. The force was not recognized as real, but fictitious and incapable of doing work. The following physicists had to show miracles of ingenuity in order to find the coordinate system due to which the Coriolis force appeared, and the source of energy that works for it. This issue is considered in detail by Astakhov in [2].
Chapter 4. New equations for the spinning top 4 - 17 Appendix 1 Table 1 Angular speed Moment of inertia Angular momentum Kinetic energy = 1 2 = = 1 2 = 1 4 = 1 8 2 1 2 = = = 1 2 = 1 4 = 1 8 2 2 2 = , где = ℎ − ℎ − ∝ - см. (1.8) = = 1 2 = 1 4 = = 1 4 Here we determine the parameters of the state of the spinning top at moments 1 and 2. In Table. 1 shows the basic formulas, where the following notation is accepted ∝ is the angle of inclination of the spinning top to the rolling plane, is spinning top mass, is the acceleration of gravity, is gravity, is the spinning top radius, ℎ is the height of the spinning top - see the segment OB in fig. 1, is angular speed of spinning top rotation around the vertical diameter at moment 1, is angular speed of spinning top rotation around the vertical diameter at moment 2, is angular speed of precession at moment 2, is linear speed of precession, is angular momentum, is moment of inertia, is energy.
Chapter 4. New equations for the spinning top 4 - 18 References 1. Khmelnik S.I. Four forces in mechanics. https://www.academia.edu/80854442/, https://doi.org/10.5281/zenodo.7004069 2. Астахов А.А. Физика. Порядок вещей или осознание знаний, 2006, https://www.litres.ru/a-a-astahov/fizika-dvizheniyaalternativnaya-teoreticheskaya-mehanika-ili-osoznanieznaniya/chitat-onlayn/ 3. The experiment concerning wheel versus gravity at the University of Sydney: Anti-Gravity Wheel? https://www.youtube.com/watch?v=GeyDf4ooPdo 4. И. Белецкий. ГИРОСКОП ТЕРЯЕТ ВЕС ? https://www.youtube.com/watch?v=FwrlRpC8BDA 5. Шарик, катящийся по вращающейся платформе, НИЯУ МИФИ, https://www.youtube.com/watch?reload=9&v=LkrmALM8TsA 6. Летающий спинер, https://www.youtube.com/watch?v=rDDfKVjjG2g 7. Хмельник С.И. Новые уравнения для волчка. Доклады независимых авторов, ISSN 2225-6717, 2020, 50(1), 65–74. https://doi.org/10.5281/zenodo.4047902 8. Christopher G. Provatidis. Forced precession of a gyroscope and its application to Laithwaite's engine. European Journal of Physics, Volume 42, Number 3, https://iopscience.iop.org/article/10.1088/1361-6404/abce88/meta 9. Professor Eric Laithwaite gives a demonstration of a large gyro wheel, https://www.youtube.com/watch?v=JRPC7a_AcQo 10. John Perry. Spinning Tops. The "Operatives' Lecture" of the British Association Meeting at Leeds, 6th September, 1890, https://www.gutenberg.org/ebooks/34268 11. Ермолин В.Б. КОРИОЛИСОВО УСКОРЕНИЕ И ЕГО ВЛИЯНИЕ НА КОСМИЧЕСКИЕ ОБЪЕКТЫ. НАУКА БЕЗ ГРАНИЦ, № 2 (7), 2017
Chapter 5. Mathematical description of the Euler disk 5 - 7 The given equations allow us to find all the parameters of the disk for a given and R. In the table. 1 shows some examples in the SI system. The following shows graphs of functions , , and specific Coriolis force ⁄ at =1000 and =0.25 - see Fig. 2. It can be seen that there is an angle ∝≈1.1 where the Coriolis force takes on the greatest value. At this point the disk remains in a stable state for a long time - see fig. 3 from [23]. Another experiment is considered in Section 5, where it is shown that the considered forces act on each element of the disk. Fig. 3. 5. Chain ring The experiment shown in Fig. 1 is known from the Internet. 4. “Chain circle” - the CPU spins up on a cylinder at high speed and collides with this cylinder. In this case, the CPU continues to move like an Euler disk, maintaining its shape. Here forces A and B act on each link in the chain. From the previous it follows (see Fig. 4) that a force acts in the direction of vector A, and a forc e acts in the direction of vector B. In addition, each chain link is subject to centrifugal force caused by the rotation of the CPU around its own axis . This force adds to the force and ensures that the CPU maintains its shape. Thus, =+ and =. Both of these forces keep the CPU from falling. This means that the forces considered act on each element of the Euler disk, and in this case, on the links of the chain.
Chapter 4. Mathematical description of the Euler disk 4 - 8 Omega_3 B A Omega_2 Fig. 4. 6. Disk rolling in a circle A disk rolling in a circle is one of the states of the Euler disk. Let's take a closer look at it. Let's denote: - disk radius, – disk thickness, - density of the disk material, - mass of the ring. Disk weight = , (1) and the moment of inertia of such a disk =. , (2) From (1, 2) we find: =. . (3) Using Steiner's theorem, we find the moment of inertia of a ring rotating around a point lying on the outer radius, =+, (4) where A is the distance between the point - the center of rotation with a known moment of inertia and point – the center of rotation with the desired moment of inertia .Obviously, in our case
Chapter 5. Mathematical description of the Euler disk 5 - 9 =. (5) Combining (4, 5), we get: =. . (6) Thus, the ring rotates around a point on the outer radius and the radius of this rotation is equal to. The linear velocity of the center of gravity of the ring and the angular velocity of rotation of the ring during such rotation are related by an equation of the form =2 . (7) Then the ring is subject to centrifugal force =/ . (8) Let us assume that the ring rotates continuously at this speed without friction. If the ring moves horizontally, then a lifting force (8) is constantly acting on it, which keeps it from falling. 7. Wheel set Let's consider a further modification of this experiment. Let there be a wheel pair - two rings rotating on a common axis along a horizontal plane on a rail circle, and the plane is connected to the common axis so that they can only rise together. In this case, the lifting force (6.8) of the ring becomes the lifting force of the entire structure. Instead of a rail circle, a gear train can be used. Let's also consider the power of the engine rotating this ring. If the traction force is such that the structure moves vertically at speed , then the engine power =. (1) According to the law of conservation of energy, this power is equal to the power expended by the engine to rotate the pair with angular velocity , i.e. =, (2) where is the torque applied to the axis of the pair. From (6.8, 1, 2) we find: ==/. (3) Let us also consider the angular momentum of the rotating pair: =. (4) In our case, the angular momentum turns into a vertical impulse of the structure, i.e. =, (5) where is the mass of the entire structure. Hence, =. (6)
Chapter 4. Mathematical description of the Euler disk 4 - 10 Equations (3, 6) combine all design and dynamic parameters of the structure. Here the law of conservation of momentum is satisfied. But it is calculated on the assumption that centrifugal force does work. Thus, this conclusion is proof that centrifugal force is a consequence of the law of conservation of momentum and centrifugal force does work. Finally, the lifting force of the structure can be found directly using the Coriolis force formula: =. (7) The speed of rotation of a pair in a horizontal circle with a radius о is equal to о=/(2о) (8) or, taking into account (12), о= /о. (9) From (7, 6.7) we find: = . (10) Structure weight =. (11) Then = ≈ . (12) 8. A round island floats on a round lake Let's consider the case when ∝=0 we assume that the disk lies on a surface that does not create a friction force. In this case, you cannot use the formula p2.2) due to division by zero. But the Coriolis force, determined by (p2.1), exists: =−2×=−2. (1) Centrifugal force also exists, but its action is limited by the rigid circle along which the disk rolls, remaining in a horizontal position. Thus, in this position, the disk makes two rotations with speeds and and rolls with a linear speed determined by(p1.3), =−cos(∝)=(−). (2) around the circumference, pressing against it with centrifugal force. At the same time, from(1.18b) we have: =() (). (3) The disk must rotate forever under the influence of the Coriolis force, which compensates for the frictional force. Nature performed the corresponding experiment: in Argentina, a round island floats on its own
Chapter 5. Mathematical description of the Euler disk 5 - 11 in a round lake. Despite the ideal proportions, people did not create them - see fig. 1 [70]. Rice. 1. Appendix 1 Here we will determine the parameters of the state of the disk at moments 1 and 2. In Table. 1 shows the basic formulas, where the following notations are adopted ∝- angle of inclination of the disk to the rolling plane, - disk mass, – acceleration of gravity, - gravity, - disk radius, ℎheight of the center of the disk - see fig. 1, - angular speed of rotation of the disk around the vertical diameter at moment 1, - angular speed of rotation of the disk around the vertical diameter at moment 2, - angular velocity of partial rotation at moment 2, - linear precession speed, - moment of inertia, - angular momentum, – energy.
Chapter 4. Mathematical description of the Euler disk 4 - 12 From Fig. 1 follows: ℎ=tg(∝). (p1.0) From the table 1 follows: = =1 1.5 ⁄, (p1.1) The linear speed of precession is the speed of movement of m. B at radius =AB, rotating at angular speed (see Fig. 1): =−cos(∝). (p1.3) Table 1. Angular velocity Moment of inertia Momentu m Kinetic energy Rotating the disk aroundown vertical axis = 1 2 = = 1 2 = 1 4 Rotation of a disk inclined at an angle to the plane around its own axis ∝ = 1 2 = = 1 2 = 1 4 Precessional movement of a top inclined at an angle to the plane.∝ = + (theorem Steiner) = ∙ , Where = 1 . 5 = = 1 2 = 1 2 = = 1 4 = = 1 4 Appendix 2 The fall of the disk is counteracted by Coriolis =−2×, (p2.1) where is the linear precession speed (p1.3), and centrifugal force =− −cos(∝) (p2.2) In steady state, force and gravity are balanced by force – see fig. 1. Therefore from ( p.2.1, p.2.2, p.1.3) we find: =− (∝)=3 2−cos(∝)/sin(∝), (p2.4) =−cos(∝)= −cos(∝)∙ctg(∝) (p2.5)
Chapter 5. Mathematical description of the Euler disk 5 - 13 or = −cos(∝)∙ctg(∝). (p2.6) If this equality is not satisfied, then the disk falls under the action of a force determined from (p2.5): =− −cos(∝)∙ctg(∝) (p2.6a) It can be seen that a solution to this equation exists only for ≥. (p2.6b) Consequently, for the Euler disk the condition (p2.6b). When tilted at an angle, the disk moves vertically by and, therefore, loses potential energy∝(1−sin∝) =(1−sin∝). (p2.7) Combining this formula with (p2.6), we find: =3 2 −cos(∝)ctg(∝)(1−sin∝). (p2.8) Next, we combine this formula with the formula for from table. 1. Then we find: = 3 2 −cos(∝)ctg(∝)(1−sin∝). or = (p2.9) Where =2 −cos(∝)ctg(∝)(1−sin∝). (p2.10) Appendix 3 Here we will take a closer look at the Coriolis force determined by (p2.1,p1.3): =−2×=2×−cos(∝) or =2 −cos(∝), (p3.1) Further from (p.3.1, 17,18) let's find =2 −cos(∝) (−) (+)(3−) 2(+) or =1 2, (p3.3) where = −cos(∝)(−)(3−)(+).(p3.4)
Chapter 4. Mathematical description of the Euler disk 4 - 14 Appendix 4 From (p2.6) And(18) we get: =1 2 −cos(∝)∙ctg(∝)() (). (p4.1) Let's denote: (∝)= −cos(∝)∙ctg(∝)2(+) (3−)2. (p4.2) Then we get: =(∝)1 2. (p4.3) The function (∝) has a maximum at some ∝ - see fig. 4. Let us denote: =(∝о). (p4.4) From (p4.3) and (p4.4) we find: = (p4.5) or = (p4.6) So, there is a minimum initial speed of rotation of the top at which it can tilt and assume a stable equilibrium position in an inclined position. Accordingly, there is a maximum rate of precession at which the Coriolis force keeps the disk from falling. Before reaching this speed, the disk gradually falls under the influence of a force determined from (p2.5): =− −cos(∝)∙ctg(∝) (p4.7)
Chapter 5. Mathematical description of the Euler disk 5 - 15
Chapter 6. To substantiate Mach's principle 6 - 1 Chapter 6. To substantiate Mach's principle Annotation The relationship between the inert and gravitational masses on Earth has been established. It is shown that Mach's principle and the principle of equivalence of inert and gravitational masses can be investigated experimentally. There is a well-known Mach principle, which states that the inert properties of each physical body are determined by all other physical bodies in the Universe and depend on their location [1]. The first idea to test this principle is to temporarily remove all bodies in the universe. But the current amount of technology is not enough to implement this idea. And therefore (and until the "salutary" idea of unknowability has not finally taken possession of the minds, as happened, for example, in quantum mechanics), we will try to understand how the Earth could create inertia (all the other bodies of the Universe without us, someone pushed to sufficient distance). How could the Earth create an inertial force acting on an accelerating body? Maybe, about the same as a large stationary electric charge on another accelerating charge ... After all, there between electrodynamics and gravitomagnetism, which are described by the same Maxwell equations, are many analogies - see for example [2]. Consider an electric charge q moving relative to the charge Q with acceleration . The strength of the electric field E, created by the charge q at the center of the charge Q, is determined by the Larmor formula [3]. In the case when the speed of the charge is much less than the speed of light c, this formula takes the form [4]: ≈×[×] , (1) where R is the distance between charges, and is the unit vector of the vector directed from charge q to charge Q - see Fig. 1. If the acceleration vector is directed perpendicular to the vector , then the vector о=×[×], (2)
Chapter 7. Mass forces depending on speed 7-4 along the radius, along the circle and along the central axis are observed in it. Velocity corresponds to angular velocity = ⁄. (8) In the presence of angular velocity , centrifugal and Coriolis forces arise =2+, ( 9) =2, (10) directed along the radius and along the central axis, respectively. There is also gravity =. (10a) The hydrodynamic equation (1.1) in this case has the form ⋅Δ+2+=0, (11) ⋅Δ+=0, (12) ⋅Δ+2−=0, (13) where is the unknown function and Δ=Δ Δ Δ= ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ + + + + + + + + + ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , (14) Coriolis forces bring momentum and energy to the system. Let us write down the equation of the law of conservation of momentum for such motion: (++)=−++1 2∙, (15) where on the left is written the impulse added by the acting forces during the time , and on the right is the change in the momentum of the ring during the same time. Obviously, = . (16) From (9, 10, 15, 16) we get: (2+2+)=−++1 2 (17) or (2+2++)=− + +1 2 . (18) Thus, the presence of angular velocity , i.e. the rotation of the vortex follows from the law of conservation of momentum.,
Chapter 7. Mass forces depending on speed 7-5 We will look for a solution to the system of equations (11, 12, 13) in the form: =()+()∙exp(∝+), (19) =(), (20) =()+()∙exp(∝+), (21) where :∝, are some constants, and () are unknown functions of the argument . Then equations (11, 12, 13) after reduction by common factors will take the form + +∝ ++(2+)=0, (22) + +∝ ++()=0, (23) + +∝ ++2=0, (24) 2 −=0, (25) 2 +=0, (25a) where =1/. (26) From (8, 25) we find: = (2), ⁄ (27) =−0.5. (27a) From (23) it follows that =0. (28) From (28, 18) we find: (2+2)=− + −. (29) From (29, 19, 21) we find: 2(+)=−(+)− (30) or =−0.5(+ (+)⁄ ). (31) From (31, 20, 8) we find: =−(+ (+)⁄ ), (32) From (22, 24, 8, 32) two equations follow (we neglected the second term in formula (9); this is justified by numerical simulation) + +∝ ++2=0, (33) + +∝ ++2=0. (34) One solution to this system of equations could be:
Chapter 7. Mass forces depending on speed 7-6 =, (35) + +∝ ++2=0. (36) So, the solution comes down to solving equations (35, 36, 32), from which we find: + +∝ ++(+ 2 ⁄ )=0, (37) =−(+ (2)⁄ ). (38) Let us also determine the energy flux density along the jet: = (39) or =2=2exp(), (40) It follows that when >0 the energy flow increases as the jet moves, which is a consequence of the work of Coriolis forces, which constantly add energy to the jet. The power transferred by exp()≈ 1 the entire jet is equal to =∫2∙2∙ =4∫∙ . (41) Finally, we find the pressure inside the jet using (5, 6): =− 2+ . (42) This pressure turns the jet into a solid and holds the liquid within the volume of the jet. Example 1. Let us consider the conditions of one process from[27]. "Water compressed by one ofmain components of the system -high pressure pump(4000 bar), passes through a water nozzle, forming a jet with a diameter of 0.35 mm, entering the mixing chamber. In the mixing chamber, water is mixed withabrasive(granitesand) and then it passes through a second, carbide or diamond nozzle with an internal diameter of 1 mm. From this nozzle a stream of water with abrasive comes out at a speed of about 1000 m/sec and hits the surface of the material being cut.”
Chapter 7. Mass forces depending on speed 7-7 Fig. 2. In Fig. 2 shows the calculation resultsin the SI system with: =0.001, ==1000. Shown as a function . , ,=,,=, of radius The total power transferred by the entire jet is =1.6∙10 Wt. Finally, we find the pressure inside the jet using (6): =10N/m2. Example 2. Example 1 was calculated for water with=1000,=0.0009. For waterWithabrasivewe will assume that the density has increased to =5000, the coefficient of internal friction remained equal = 0.0009. In this case we find: =8.5∙10 Wt,=5∙10N/m2. bz2 bz1 bz omega FzC Ff
Chapter 7. Mass forces depending on speed 7-8 Example 3. For comparison, consider a stream of air with =1.3,=1.5 ∙ 10. In this case we find: =2.1∙10 Wt,=1.3∙10N/m2. Example 4. Let us finally consider a stream of air with an abrasive, taking for it=5000,=1.5∙10. In this case we find: =3.6∙10 Wt,=5∙10N/m2. Thus, it is shown that the jet acquires energy as it spreads and this energy is supplied to it by Coriolis forces. 3. Sea currents and tsunamis Sea currents, water and sand tsunamis can be represented as a limited or closed strip, moving at a constant speed and at the same time maintaining a cross-sectional shape throughout its entire length. These phenomena are striking in their grandeur, organization and demonstration of the existence in their volume of an inexhaustible source of colossal energy. What is the internal structure of such a strip and how does its engine work? Knowing the answers to these questions, we can further speculate about the cause of such phenomena. But first of all, you need to find answers to the first questions, which, in principle, you can try to do without knowing the conditions for their occurrence. First of all, there is a desire to identify such phenomena with an electromagnetic wave - also infinite, preserving its shape, moving and transferring energy. But in a vacuum, such a wave does not waste energy along the path of its propagation and, therefore, does not need an internal source of energy. There are always internal losses in a fluid flow and therefore such a source is necessary. We will immediately begin with the statement that the driving forces in the phenomena under consideration are Coriolis forces. The author has already considered mathematical models of currents and tsunamis based on the equations of gravitomagnetism [2]. Below we propose another model based on the use of Coriolis forces. It is possible that these models can be combined, but no such attempt is made here and no comparison is made between the old and new models. Accepted ideas about the causes of ocean currents do not agree well with the existence of closed flow trajectories and the stability of the configuration and cross-sectional shape.Typically, gradient currents,
Chapter 7. Mass forces depending on speed 7-9 wind-induced currents, and tidal currents are distinguished.These factors can cause the emergence of currents, but cannot maintain the existence (for centuries) of a closed flow trajectory (since oppositely directed sections of this trajectory must be subject to oppositely directed influences). However, flows, as a rule, are closed (as can be seen in Fig. 1 from [33]). Also, to explain these phenomena, they usually point to differences in the composition and properties of the jet water and the surrounding waters. It is more natural (in our opinion) to assume that these differences are a consequence of the isolation of the jet, and not the cause of this isolation. Fig. 1. Five major oceanic cycles To explain the reasons for the existence of a tsunami, they usually point to the initial shock from an earthquake, and to predict the behavior of a tsunami (which is extremely important for practice), statistics of past tsunamis and maps of the seabed are used. Dozens of institutes and hundreds of scientists are doing this [34]. But here we are interested in the general patterns of tsunami movement. Let's look at Fig. 2 from [2]. It seems unconvincing to think of the initial shock asthe reason for the long movement of this colossus. It seems that this “device” has its own engine inside, and the resistance of the environment is only a catalyst, a force that presses on the gas pedal. So, there must be internal forces that ensure the movement, stability of the configuration and cross-sectional shape of the indicated flows. The jet is an absolutely closed system and, consequently, is described by an equation of the form (1.1).In our case, the Coriolis forces are unknown, because depend on speeds. Therefore, we have to perform
Chapter 7. Mass forces depending on speed 7-10 the calculation under some additional assumption. We will assume that the boundaries of the jet (its radius) are known and the longitudinal velocity at the jet boundary is known. Fig. 2. We will consider a mathematical model of the jet in a system of rectangular coordinates ,, and we will assume that an ideal tsunami has the form of a flat wall with axes ,, directed respectively horizontally along the wall, vertically and along the thickness. In equation (1.1), the Lagrangian in rectangular coordinates is determined by a formula of the form Δ=Δ Δ Δ= ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ + + + + + + ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ . (8) We define the mass Coriolis forces created by the rotation of the Earth with angular velocity о as
Chapter 7. Mass forces depending on speed 7-11 =2о, ( 9) =2о. (10) For middle latitudes о≈10. We define the mass force of gravity as =−. (11) Let us write equation (1.1) taking into account (8-11): ⋅+2о − 2о=0. (12) In this case, our problem takes the form of a system of three equations with three unknowns ,,: ⋅Δ+2о=0, (13) ⋅Δ−=0, (14) ⋅Δ+2о=0. (15) Next, we will look for a solution in the following form: =ex, (16) =, (17) =ex, (18) ex=exp(++), (19) where ,,, are some constants. Substituting (16-18) into (8), after differentiation we obtain: Δ Δ Δ=ex+ex+ex ex+ex+ex (20) or Δ Δ Δ=ex ex (21) Substituting .(16-17, 21) into (13-15), we get ex+2exо=0, (22) −=0, (23) ex+2siо=0 , (24) where = . (25) One of the solutions to the system of equations (22, 24) may be: =, (26) =2о. (27) From (23) we obtain:
Chapter 7. Mass forces depending on speed 7-12 =. (28) From (9, 10, 16, 18, 26_28) we get: ==оex, (29) ==(о)ex, (30) Let us also determine the densities of energy flows circulating in a rectangular jet along the coordinate axes: = , = . (31) From (29-31) we get: ==(2о)ex. (32) Thus, we have defined an analytical description of a jet with a rectangular cross-section. Specific values of all parameters can be determined with known statistics of measurements of such jets (ocean currents and tsunamis). The author does not have such information and would be glad to cooperate in any form. Note that at v the magnitude of the exponent increases with increasing . Hence and from (32) it follows that the energy flow increases as the jet moves. This is a consequence of the work of Coriolis forces, which constantly add energy to the jet. This is why sea currents do not die out, and tsunamis accelerate and expand. Finally, we find the pressure inside the jet using (6): =− ++ex. (33) This pressure turns the jet into an almost solid body and holds the liquid within the volume of the jet. 4. Dust devil and top A dust devil is widely known, which is an almost vertical column of dust - see fig. 1. Such a vortex has a vertical axis of rotation, a height of several tens of meters, a diameter of several meters, a speed inside the vortex of about 10 m/sec, and a lifetime of several tens of seconds [35]. There are also phenomena similar to it - air, ash, water vortices - see fig. 2 from [36]. This vortex has a diameter of about 0.1 m. The causes of sand vortices are considered to be various atmospheric phenomena (wind, heating of the atmosphere). However, the very existence of a sand whirlwind - the preservation of shape and movement - is difficult to explain by the same reasons. In addition, such vortices exist and move on Mars, where there is no atmosphere [35]. Therefore, when explaining such vortices, the main questions are about the source of energy and the reasons for stability in such an unusual form.
Chapter 7. Mass forces depending on speed 7-13 The author has already considered the mathematical dust devil in [2]. In contrast, below is the proposed model, which explicitly uses Coriolis forces as internal forces that ensure the stability of the vertical jet. It is possible that the two models could be combined, but no such attempt is made here and no comparison is made between the old and new models. Next, the question is considered: “How does a dust devil work?”, and the question “How does it arise?” remains unanswered. Rice. 1. Fig. 2. In Fig. 3 shows the top in its simplest form. In this case, the top with the mass has the form of a cylinder with a radius and height ℎ, and is inclined to the horizontal plane at an angle ∝. The top has
Chapter 7. Mass forces depending on speed 7-20 ==() ⁄⁄ . (4) The unknown functions (),() are the solution to equations (2.322.34). Let's rewrite these equations: =−0.5(+ (+)⁄ ), (5) =, (6) + +∝ ++2=0, (7) + +∝ ++2=0. (8) From the experiments discussed in the introduction, it can be assumed that in the jet the central region rotates in one direction, and the outer region in the other, and, therefore, there is a certain radius where the function, ()=0, ( 9) From (5, 9) we find: ()+ ()+() ⁄=0 (10) or ()=− () ⁄−(). (11) From (7, 8, 9) it follows that when = an equation of the form + +∝ +=0, (12) where = or =. We will look for a solution in the form =cos() And=cos(), (13) where ,, are some coefficients. Then equation (13) takes the form: −+ −+∝ =0. (14) This equation is easily solved by the numerical method and its solution is =. If the value of the constant is determined, then the value of the constant is determined by (11):. cos()=− cos() ⁄−cos(). (15) or =− cos() ⁄−. (16) Example 1. Let =20,∝=1.4,=2.5. In the MATLAB system polynomial (14) take the form: . = [-1 0.0156 -1 0.005];
Chapter 7. Mass forces depending on speed 7-21 = roots(p)=0.05; This solution is the only real solution to this polynomial. Let = −4.5. From (16) we find: =−5.34. With the found parameters, the functions (),(),() are determined which are shown in Fig. 2.. It can be seen that at a constant direction of the longitudinal and radial velocities, the direction of rotation of the jet changes to the opposite at a certain radius. Section 2 defines mass forces (gravity forces, centrifugal forces, Coriolis forces) - see (2.9, 2.10): =2+, (17) =2−. (18) In Fig. 3 shows graphs of forces and . Let us also determine the energy flows flowing along the radius and along the jet: =, (20) =. (21) These functions are shown in Fig. 4. It is clear that radial energy flows are directed in the opposite direction and do NOT leave the volume of the jet, due to which its integrity is preserved, the longitudinal energy flow in the central part of the jet is directed upward; It is these currents that carry the trout up the falls. It is important to note that the flow of water in the central part of the jet is unidirectional with the flow in the outer part of the jet! These facts are possible only if there is rotation of the jet around the vertical axis, i.e.∝≠0. The resulting functions allow you to calculate the power consumed by the jet and the water flow in the jet:, =2∫ −∫ , (23) =2∫ −∫ , (24) Note on sections 2 and 5. In these sections, essentially the same mathematical problem is solved. But section 5 uses the fact that equation (14) can have a solution for . Section 2 does not verify the existence of such a solution, i.e. by
Chapter 7. Mass forces depending on speed 7-22 default it is assumed that such a solution exists for .=0<< > 0 0.005 0.01 0.015 -4 -2 0 bz 0 0.005 0.01 0.015 0 2 4 6 br 0 0.005 0.01 0.015 Fig. 2. (Vodopad.m) 0 0.2 0.4 omega 0 0.005 0.01 0.015 -10000 -8000 -6000 -4000 -2000 0 Fz 0 0.005 0.01 0.015 Fig. 3. (Vodopad.m) -4000 -3000 -2000 -1000 0 1000 Fr
Chapter 7. Mass forces depending on speed 7-23 6. Funnels and whirlpools First we will look at a funnel filled with water. The question of the direction of rotation of water in a funnel and its connection with the direction of the Coriolis force created by the rotation of the Earth is very often discussed. The period of this rotation depends, as is known, on the period of the Earth’s daily rotation and the latitude of the funnel’s location: =sin()⁄ . (1) When = we have, for example, =24∙ ≈10. (2) Angular frequency of such rotation о≈10. Such a speed of rotation of water in the funnel could not be noticeable. Therefore, the real rotation speed is not related to the rotation of the Earth. Therefore, when describing a funnel mathematically, it can be considered as a structure that is motionless relative to the Earth. We will consider the jet arising in the funnel by analogy with the jet discussed in Section 2. For the convenience of the reader, we will rewrite the basic formulas from section 2. In the jet, the velocities ,, of mass particles directed along the radius, along the circle and Sz Sr
Chapter 7. Mass forces depending on speed 7-24 along the central axis are observed. Velocity corresponds to angular velocity = ⁄. (3) In a whirlpool, the rotation of water is a precession that occurs when water moves towards the center of the whirlpool, which follows from the law of conservation of momentum and is proven in the same way as in section 2. In the presence of angular velocity , centrifugal and Coriolis forces and gravity arise =2+, (4) =2, (5) =. (6) Verin in article [32] performed a comprehensive analysis of the funnel, waterfall and other natural phenomena. But he did not analyze the influence of Coriolis forces and the reasons for the appearance of rotation. Therefore, radial and vertical velocities remained unconsidered. The following discussion concerns only the calculation of these velocities. The hydrodynamic equation (1.1) taking into account the Coriolis forces has the form ⋅Δ+2+=0, (7) ⋅Δ+=0, (8) ⋅Δ+2−=0, (9) Where is the unknown function and Δ=Δ Δ Δ= ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ + + + + + + + + + ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , (10) Coriolis forces bring momentum and energy to the system. The equation for the law of conservation of momentum takes the form: (2+2++)=− + +1 2 . (11) Thus, the presence of angular velocity , i.e. the rotation of water in the funnel follows from the law of conservation of momentum., We will look for a solution to the system of equations (7, 8, 9) in the form: =()+()∙exp(∝+), (12)
Chapter 7. Mass forces depending on speed 7-25 =(), (13) =()+()∙exp(∝+), (14) Where ∝, are some constants, and are unknown functions () of the argument . Then equations (7, 8, 9, 11) will take the form: = (2), ⁄ (15) =−0.5, (16) + +∝ ++2=0, (17) + +∝ ++2=0. (18) =−0.5(+ (+)⁄ ). (19) Looking at photographs of waterfalls, you can see that there is a radius at which in the area of smaller radius ≤ there are no radial flows, i.e.=0.From this and from (17) it follows that the root of the equation + +∝ +=0. (20) Let us denote the function - the solution of this equation as , and the value of this function at = as . Thus, when = we have: ()=0,()=. (21) The left side of equation (18) coincides with equation (20) and, therefore, this part, as a function, also vanishes when = and the value of this part ()=0, i.e. ()= ()=0. (22) From (5, 19) we find that ()=() =0, (23) those. The inner surface of the whirlpool hole does not rotate. Thus,the whirlpool, as a system of equations (17-19), can be calculated in exactly the same way as the outer part of the jet in section 11.5. But this calculation applies only to the surface of the whirlpool. Verin in article [39] showed that the shape of an ideal funnel is determined by an equation of the form: ℎ=1− , (24) where is the undisturbed water level, ℎ is the height of the funnel at the level of the radius , is the internal radius of the funnel - see Fig. 1. In this case, the linear speed of rotation does not depend on the depth. This means that it is enough to solve the equations for the surface
Chapter 7. Mass forces depending on speed 7-26 with coordinates (ℎ, ), and then use the found functions ,, for any ℎ. Verin writes: What happens under the surface of the water funnel? In order to answer this question, we don’t even need to make any additional calculations (this is one of the advantages of an ideal “portrait” of a phenomenon!). Indeed, knowing the shape of the envelope line of the funnel (7), we can determine imaginary surfaces of equal pressure under water. The fact is that these surfaces of equal pressure completely repeat the shape of the surface of the funnel itself, since the centrifugal force has a horizontal direction and is perpendicular to the force of gravity, as a result of which the change in vertical pressure is determined only by the value of the water column (Fig. 2). This figure shows the envelope lines of surfaces of the same pressures, differing by the same value ΔР. In fact, it is the same curve, drawn several times with the same vertical offset. Fig. 1. Shape of the envelope of an ideal water funnel h
Chapter 7. Mass forces depending on speed 7-27 Fig. 2. Surfaces of equal pressure follow the shape of the funnel. When comparing the mathematical models of a waterfall and a whirlpool, the question remains unanswered: why is there a central part in a waterfall jet with a flow of water and an oppositely directed flow of energy, while in a whirlpool the central part is empty?
Chapter 8.Coriolis engine 8-1 Chapter 8. Coriolis engine Let's look again at Figures 7 and 8 in Chapter 3 - see fig. 1. If everything is so, then why not make an aircraft in the form shown in Fig. 2? There, the motor M rotates the disks D through the gearbox R in opposite directions with angular velocity around their common axis and rotates the disks D themselves around a vertical axis with angular velocity. In this case, disks D move at linear speed = (1) and each of them is acted upon by the Coriolis force = −2( × ). (2) These forces add up to the lifting force of the device. Fig. 1. R M D D Fig. 2.
Chapter 8.Coriolis engine 8-2 This idea is not new and is embodied in a patent [45]. The proposed implementation turned out to be very complex and, apparently, that is why the patent did not find a buyer. It is interesting to note that the reviewers were not confused by the author's honest reference to the use of a "fictitious" Coriolis force. The author points out that “the proposed method makes it possible to create a prime mover capable of moving vehicles or spacecraft over planets and in the observable Universe without the use of rocket or propeller thrust.” However, in Chapter 10 it was proven that the Coriolis force acts only in the immediate vicinity of the planet, at a height above the surface not exceeding the radius of the planet.
Chapter 11.Coriolis force as a cause of some physical effects 11-1 Chapter 11. Coriolis force as a cause of some physical effects 1. Lead To shorten the text, we will call the movement of a body rotating around its own axis with angular velocity and moving along this axis with speed VV-motion or -movement or, simply, VVD. In Chapter 6 it is proven that during explosive motion in the vicinity of a rotating planet, the mass currents of this body and the planet interact in such a way that the Coriolis force acts on this body. If we accept this statement, then it becomes obvious that to explain the effects associated with VVD, it is necessary to take into account the Coriolis force. Below we will indicate several such effects without mathematical justification.ω(ω,) 2. Bullet derivation It is obvious that the Coriolis force acts on the rotating bullet and projectile. This does not negate the fact that they are also affected by the same forces that are used in the modern explanation and calculations of dervation. 3. Magnus Effect The effect is that when a flow of liquid or gas flows around a rotating body, a force directed towardsperpendiculardirectionflow or direction of speed of a body in a calm fluid. In this case we see a typical BB movement. To explain this effect, a circulation of velocity around the body is introduced. However, such a circulation does not exist, as Prandtl showed 100 years ago in [40]. Thus, to explain and calculate the Magnus effect, the Coriolis force must be taken into account. 4. Levitation of rotating disks Can the Coriolis force be a lifting force? In Chapter 2, the centripetal force and the Coriolis force were defined as, respectively, С= , (1) = −( × ). (2)
Chapter 11.Coriolis force as a cause of some physical effects 11-2 (in the last formula we discarded the coefficient “2”, as was justified in Chapter 2). Let's denote: - first escape velocity directly above, – the corresponding angular velocity, - radius of the Earth, - the speed at which the Coriolis force is equal to the force of gravity, оangular velocity of the Earth. For mid latitudes о≈ 10. (3) Obviously, = . (4) From (1, 2) it follows that the corresponding forces are defined as С = = , (5) = о. (6) But both of these forces are equal to the weight of the body, i.e. С = . (7) From here we find that о = (8) or = о ⁄. (9) A body must fly at such a speed for it to be held by the Coriolis force. At the same time, it will, of course, fly into space. Let us consider a ring that is motionless relative to the Earth and rotating with angular velocity . There will be no centripetal force acting on it. Let the ring have a radius , thickness ℎ and rim width . The Coriolis lift force acting on the ring is = о, (10) where is the mass of the ring = 2ℎ, (11) where is the density of the ring. Now consider a disk with radius . His weight = ∫ . (12) and the Coriolis lift force acting on it is =∫ = о ∫ . (13) The disk will levitate if these forces are equal: ∫ = ∫ (14) or
Chapter 11.Coriolis force as a cause of some physical effects 11-3 ∫ =∫, (15) Where = о . (16) Taking into account (11), we obtain: ∫ =∫ . (17) After integration we get: = (20) Taking into account (16), we obtain: = о (21) or, taking into account (3), ≈∙. ∙ ≈. ∙ 10. (22) It can be seen that the angular velocity of rotation of a disk levitating under the influence of the Coriolis force is practically impossible. 4. Sports There are many sports games with balls and balls (table tennis, golf, tennis, football, baseball, volleyball, basketball, handball, billiards). In these games, the ball or ball often moves along a trajectory that is difficult to explain. These cases are also of scientific interest - see, for example, [41]. Obviously, the motion of a spinning ball or sphere is motion, which can be calculated taking into account Coriolis forces. Let's look at an example from[50] – see fig. 1. A real physicist, sitting on a carousel, throws a ball along a radius towards the center. And the ball miraculously turns to the side! This, according to the physicist, is a consequence of the Coriolis force, which has no reason for its appearance and does not consume energy: it arose due to the fact that a carousel was spinning nearby. But let us remember the law of conservation of angular momentum. The ball, while on a rotating carousel, has angular momentum (as part of the angular momentum of the carousel). Having flown out of the carousel, the ball did not lose this angular momentum - it turned into the angular momentum of the ball’s own rotation. A process occurred that was the opposite of the process of converting the angular momentum of the top's own rotation into the angular momentum of the top's precession. So, the ball, flying out of the hands of a real physicist, flies and rotates. Thisa typical motion in which a true Coriolis force acts on the ball.
Chapter 11.Coriolis force as a cause of some physical effects 11-4 Fig. 1
Chapter 12.Coriolis force as driving forces in the swimming of marine animals 12-1 Chapter 12. Coriolis force as driving forces in the swimming of marine animals Annotation The method of swimming of marine animals - fish, mammals, snakes, as well as flying snakes - is considered. It is shown that the main method of swimming of marine animals consists of transverse oscillation of parts of their body. It is proved that in this case a longitudinal current arises along the body, carrying the body of the animal in a direction perpendicular to these vibrations. The mass forces in such a flow are the Coriolis forces. The energy supplied by these forces far exceeds the energy expended by animals to create their own transverse vibrations. This energy allows animals to swim almost continuously and quickly, overcoming the resistance of water. The source of energy is the Earth's gravitational field. Table of contents 1. Introduction 2. Mathematical model of the “fish” swimming method 3. Mathematical model of the “dolphin” swimming method 4. Water snakes 5. Flying kites 6. Conclusion Appendix 1 Appendix 2 1. Introduction The whale moves at a speed of 30 km/hour,thanks to the wavelike movements of the tail in the vertical plane[60, 61]. For a prairie resident, such a movement seems inexplicable, because... he did not see the animals rushing across the prairies,thanks to the wave-like movements of the tail in the vertical plane.Other marine inhabitants have smaller tails and therefore, “in addition to their tails,” they wriggle their whole body, but in a plane perpendicular to the direction of
Chapter 12.Coriolis force as driving forces in the swimming of marine animals 12-2 movement - dolphins wriggle in a vertical plane, and fish in a horizontal one. “The frequency of fish body bending, or undulation, is quite high. It basically determines the relative speed of the fish. The more often a fish bends its body, the greater the speed it develops.”[68]. In essence, the idea is the same: you need to move something perpendicular to the direction of movement and then you will float forward! Let us add that man has long ago (but much later than the whale) also mastered the method of moving forward with the vertical movement of an oar - see the mentions of Venetian gondoliers and others in [62]. And Appendix 2 tells how a child instinctively learns to swim, apparently in the same way. Water works wonders!This was first proven in independent studies conducted in 1909 and 1912 by Knoller[59]and Betz[59]. They discovered that an inclined wing creates an effective angle of attack, which results in thrust and lift. This phenomenon was called the effectKnoller-Betz. This effect underlies many inventions in water and air transport [5]. An attempt was made in [63] to create a mathematical model of this effect. It cannot be said that it ended in success. But toIn addition to explaining the mechanism by which force appears, it is necessary (and more important) to find a source of energy that helps marine inhabitants move tirelessly, overcoming the resistance of water. A shark moves continuously all its life, and the power exerted by a whale is many times greater than the power of a submarine of comparable size and speed - see Appendix 1. Detection of an energy source is also necessary (naturally) for technical design. Let us also note that whales, sharks, submarines and airships are placed in conditions much more severe than those created by the authorseffectKnoller-Betz. Experimenters created vertical vibrations of an object (wing) by an external force, but animals and technology must create these vibrations with their own forces. IN[68]we read: “In physiology, the so-called “Gray's paradox” is known. Back in the 30s, the American researcher J. Gray, when comparing the calculated energy costs required for a dolphin to move in water, with the actual energy costs of the animal, discovered a paradoxical phenomenon. For movementAt a speed of 40 km/h, a 180 cm long dolphin needs to develop a power of at least 2.5 liters. With. In reality, a dolphin is not capable of developing even 0.5 liters of power. With. A similar paradox has been found in fish.... It should be recognized that the ability of fish to reduce hydrodynamic resistance still remains largely a mysterious natural phenomenon" Appendix 1 provides a calculation of the power of a whale in comparison with the power of a submarine of similar size and speed. It
Chapter 12.Coriolis force as driving forces in the swimming of marine animals 12-3 has been shown that the power required to move the whale is several times greater than the power generated by the whale itself. Summarizing the above, it can be assumed thata floating object must create a vertical force (like dolphins) or a horizontal force (like fish) for the water to create a longitudinal force. 2. Mathematical model of the “fish” swimming method We will solve the problem posed as a problem of hydrodynamics with mass forces depending on speed - similar to the problems considered in [65]. The closest approach to this problem is the calculation of sea currents and tsunamis - see section 3 in [65]. We will consider a stream of water in which a parallelepipedshaped object moves along its axis length and square section with side (very rough fish model). We will consider the mathematical model of this jet in a system of rectangular coordinates ,,, where is the coordinate axis along which the axis of the parallelepiped is located, is the vertical axis, and is the horizontal axis.We will assume that this object creates a wave in its body that propagates along an axis and floats along this axis at a speed. In this case, the object carries away water, which moves at the same speed . This follows from the boundary layer effect Prandtl: The viscosity of a fluid causes the boundary layer of fluid to stick to the solid boundary and move with it. In [66, Chapter 5] it is shown that for absolutely closed systems the equations of hydrodynamics take the form ⋅+⋅=0, (1) where is the coefficient of internal friction, is the fluid density, is the mass force, is the Lagrangian, which in rectangular coordinates is determined by a formula of the form Δ=Δ Δ Δ= ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ + + + + + + ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ . (2) Since water moves at a certain speed , mass Coriolis forces arise, created by the rotation of the Earth at an angular velocity о. They are defined as =2о, (3)
Chapter 12.Coriolis force as driving forces in the swimming of marine animals 12-4 =2о. (4) For middle latitudes о≈10. We define the mass force of gravity as =−+, (5) where the force is not defined. Let's write equation (1) taking into account (2-5): ⋅+2о −+ 2о=0. (6) In this case, our problem takes the form of a system of three equations with three unknowns ,, and an unknown force : ⋅Δ+2о=0, (7) ⋅Δ−+=0, (8) ⋅Δ+2о=0. (9) First, we will look for a solution in a form that will allow us to explain the “fish” way of swimming (using horizontal oscillations of the body). To do this, we will look for a solution in the following form: =si, (10) =si, (11) =si, (12) si=sin(++), (13) where ,,, are some constants. Substituting (10-13) into (6), after differentiation we obtain: Δ Δ Δ=si∙γ si∙γ si∙γ, (15) where γ=++. (16) Substituting .(10-12, 15) into (7-9), we get si∙γ+2оsi=0, (17) si∙γ−+=0, (18) si∙γ+2оsi=0, (19) where = . (20) One of the solutions to the system of equations (17, 19) may be: =, (21) =2оγ ⁄, (22) From (3, 4, 10, 12, 21, 22) we get: ==(2оγ ⁄)si, (24)
Chapter 12.Coriolis force as driving forces in the swimming of marine animals 12-5 ==(2о)si γ ⁄. (25) We will assume that the object, by its own means, creates a horizontal force . Let us also determine the densities of energy flows circulating in the jet along the coordinate axes:. =+ , = . (26) From (24-26) we get: =(2о)si ⁄, (27) =+, (28) where = = γоsi. (28) We do not consider vertical forces and energy flows, since fish resist gravity with the help of a swim bladder. "Fish" way of swimming(with the help of horizontal oscillations of the body is that a floating object must produce a horizontal force in order for the water to produce a longitudinal force. In other words, in parallelepiped horizontal force is created, acting on water, and in water are being created Coriolis forces ,, current on parallelepiped. These forces act in a certain area with a length parallelepiped and square cross-section with side . In this area extends horizontal energy flows with density , and longitudinal flow with density . This means that power is transmitted horizontally = (29) from the object, power is transmitted horizontally from the water =, (29a) and along parallelepiped and power is transferred from water = (30) where is the surface area of one side parallelepiped, a is cross section of the indicated area. Hence, =, (31) ≈ . (32) Let's find a relationship = = . (33) The indicated power flows depend on the corresponding forces and on the frequency. In [58] also it is indicated what exactly frequency mainly determines the speed of fish.
Chapter 12.Coriolis force as driving forces in the swimming of marine animals 12-6 To estimate the frequency of water vibrations, we will assume that the speed of water waves arising around a moving fish is equal to the speed of the fish. Then the speed we found is the speed of a sinusoidal wave (13) with amplitude =2оγ ⁄ . (34) On the other hand, with the speed of wave motion is determined by a formula of the form [71] =. (35) In our case, the amplitude of the wave function is the speed of its movement, therefore, = , (35) where the circular frequency of this wave is determined by its frequency : =2. (36) We will assume that the main role is played by transverse vibrations of the fish’s body, creating a longitudinal wave in the fish’s body along the axis . We will assume that the water wave oscillates only along this same axis. Then, (see (13, 16, 34)): =2о ⁄ . (37) γ=, (38) si=sin()=sin( )., (39) where is the wavelength = . (40) Let's find the frequency of this wave from (35-37): = = =о . (41) Let us find at the known speed from (37, 40): =2о (42) or =о =.о (43) or =о =.о. . (44) Ivanov in [58, table. 5.5]indicates that the undulation of fish is 10≤≤300, and the ratio is 2≤ ≤3.3
Chapter 13. Rehabilitation of Rutherford's atomic model 13 - 1 Chapter 13. Rehabilitation of Rutherford's atomic model Annotation It is shown that in the planetary model Rutherford atom there are stationary electron orbits that are stable in height, the reason for the stability and discreteness of stationary orbits is substantiated. They write[1], that “Rutherford’s planetary model of the atom corresponds to modern ideas about the structure of the atom with refinements... The disadvantage of the planetary model was the impossibility of explaining the stability of atoms with it. Since electrons move ..., then according to the laws of classical electrodynamics they should radiate electromagnetic waves, while losing to radiation kinetic energy orbital motion and, as a result, “fall” onto the core. Calculations performed using classical electrodynamics methods show that electrons should “fall” onto the nucleus in a time of the order of 10−11 s.” These refinements, as is known, led to the creation of quantum mechanics. However, let us consider how destructive the radiation processes are for the conservation of the kinetic energy of the electron. Consider an electron with mass , which rotates around its own axis with angular velocity and moves along a certain trajectories around the nucleus with linear speed see fig. 1. The electron is subject to the force of gravity (the Coulomb force of attraction of a negatively charged electron to a positively charged nucleus) =, (1) where is the acceleration of free fall of an electron onto the nucleus; Coriolis force =−2× (2) and centrifugal force = , (3) where ,2 are the instantaneous radius and angular velocity of rotation at a given point of the specified trajectory.
Chapter 13. Rehabilitation of Rutherford's atomic model 13 - 2 Omega F2 F1 F4 F3 Fig. 1. The electron loses its kinetic energy due to radiation. The power lost by an electron can be defined as the power of a certain force of resistance to the movement of the electron, i.e. =. (4) Let a vector have three projections - a projection onto the radius, a projection onto the tangent to the circle and a projection onto the vertical. Let us also consider the projection onto the plane of the circle, the projection onto the vertical plane tangent to the circle, and the projection onto the vertical plane passing through the radius. The movement of the body will be stable if three conditions are met: =−21 (5) =−21 (6) =−21, (7) where the projections of the Coriolis force are indicated on the right. From (4)taking into account (7) we find: =−2. (8) Formulas (1, 5) can be combined. Then we get: =−2 (9) Formulas (3, 6) can be combined. Then we get: =−2 (10)
Chapter 13. Rehabilitation of Rutherford's atomic model 13 - 3 From (8, 9, 10) we find: =− , (11) =− , (12) =− . (13) In these formulas, formula (7) is not obvious. It indicates the force of resistance to the movement of the electron acting along the velocity , and the projection of the Coriolis force to the same speed. By definition (2), such a projection is equal to zero. The explanation is given in the appendix and is based on the fact that the real trajectory of the electron is spiral line. It is shown there that at a certain “height” above the nucleus there can be an electron orbit that maintains its height, and the stability and discreteness of such stationary orbits. Thus, if for an electron we know ,,2,R, then from (11, 12, 13) the projections and the velocity vector itself can be found. And vice versa, if the vector is known, then from (11, 12, 13) the velocity and orbital radius R can be found and acceleration . The latter value for a particular nucleus determines the “altitude” of the orbit. Hence, an electron with a certain speed has a uniquely defined orbit. The energy of rotation of an electron around its own axis with a speed and its angular momentum does not change. Energy of electron rotation in a circle =0.5=0.5 (14) does not change, since the speed and radius of the orbit do not change. The conservation of these values occurs despite the fact that the electron is constantly losing its energy, because the electron energy is constantly restored by the Coriolis force. Such an action of the Coriolis force is possible only if it can do work, i.e. this power is real power. A mathematical proof of the reality of this force is given in Chapter 3. There it is shown that the Coriolis force can be justified as con sequence of the equations Maxwell, and the source of energy for this force is the electric core. Rutherford's model of the atom must be returned to physics.
Chapter 13. Rehabilitation of Rutherford's atomic model 13 - 4 Application In Chapter 3 it is shown that an electron in a vacuum moves along a spiral trajectory, which is a solution to the system of Maxwell’s equations [2]. In cylindrical coordinates ,,, the stream functions have the form: =cos(+), (1) =sin(+), (2) =sin(+), (3) where is the charge of the electron, , are some constants, and are the constants , , of the speed of motion of the electron along the coordinates. At the same time. In Fig. Figure 2 shows three helical lines described by functions (10, 11) of the current with projections and . In Fig. 1 shows as an example: a thick line at =2,=0.8, a middle line at =0.5,=2 and a thin line at =2,=1.6. Rice. 2 Next we will consider functions of coordinate velocities form = =cos(), (4) in parametric form with parameter =+. (5) Thus, the electron moves at a speed =,,, (6) and =cos(), (7) =sin(), (8) =sin(). (9) Let's move on to Cartesian coordinates and then consider the vector
Chapter 13. Rehabilitation of Rutherford's atomic model 13 - 5 =,,, (10) where =cos()−sin(), (11) =sin()+cos(). (12) or, taking into account (8, 9), =cos()cos()−sin()sin(), (13) =cos()sin()+sin()cos(). (14) It is important to note for further purposes that functions (13, 14) are periodic functions of the coordinate , which follows from (5). If an electron rotates around its own axis with angular velocity , then the Coriolis force acts on it =−2×, (15) where is the mass of the electron. At the same time =,,, (16) =,,. (17) From (10-12) we find: ==− =− =−. (18) It can be seen that the Coriolis force has a projection onto the axis , i.e. on the trajectory of the electron. The force is determined through functions (13, 14) and is therefore periodic. Let us assume that the electron rotates around the nucleus and denote the length of the electron trajectory on one closed turn as . Let's consider the average value of the force at length and denote it as . Due to the periodicity of the force , the quantity can take on values of different signs. Along with this force, a force acts on the electron resistance to electron movement, which is proportional to the length . Let us consider the total force acting on the electron along the trajectory: =−. (19) At the same time 1. if =0, then the length of the trajectory does not change. 2. if >0, then the electron accelerates, gradually moves away and its trajectory lengthens, 3. if <0, then the electron slows down, gradually falls onto the nucleus and its trajectory shortens.
Chapter 13. Rehabilitation of Rutherford's atomic model 13 - 6 We can also consider the derivative at a point with the value , where =0. At <0 the trajectory is stable. Indeed, as the height decreases, the length decreases, but at the same time it increases and the electron accelerates, and at the same time it rises. Thus, there are trajectories that are stable in height - stationary orbits. These orbits are located discretely in height. The stability of stationary ones is ensured by the Coriolis force acting on the emitting electron. References 1. Planetary model of the atom, https://ru.wikipedia.org/wiki/Planetary_model_of_atom. 2. Khmelnik S.I. Coriolis force and centrifugal force in electrodynamics and mechanics. Papers of independent authors, ISSN 2225-6717, No. 48, p. 67, 2020, https://zenodo.org/record/3900260. 3. Khmelnik S.I. Equations of motion of a single charge in vacuum, Papers by independent authors, ISSN 2225-6717, no.50, page 32, 2020, https://zenodo.org/record/4164972 4. Khmelnik S.I. Rehabilitation of the Rutherford model of the atom, Papers of independent authors, ISSN 2225-6717, no.49, 2020, https://zenodo.org/record/4024950
Chapter 14. On the nature of strong interactions 14 - 1 Chapter 14. On the nature of strong interactions Annotation The existing concept of the nature of nuclear forces has a number of disadvantages. The article proves that these forces can be substantiated as a consequence of Maxwell's equations. In this case, it is assumed that the nucleons rotate around their own axis with a certain angular velocity . It is shown that the detected repulsive forces exceed the Coulomb forces of attraction by a factor of . It is shown that, in spite of mutual attraction, rotating nucleons cannot touch. Contents 1. Introduction \ 2. Interaction of moving electric charges \ 3. Interaction of rotating electrically charged bodies \ 4. Maxwell's equations for gravitomagnetism 5. Some properties of nucleons \ Application \ References \ 1. Introduction They write: "The need to introduce the concept of strong interactions arose in the 1930s, when it became clear that neither the phenomenon of gravitational nor the phenomenon of electromagnetic interaction could answer the question of what binds nucleons in nuclei." Strong interactions between nucleons today are described in a very complex theory, containing many assumptions and representing an "eclectic picture: next to mathematically rigorous calculations, semi-quantitative approaches based on quantum mechanical intuition are adjacent, which, however, perfectly describe experimental data." Below, we again attempt to describe strong interactions between nucleons as a variant of gravitational interactions. The proposed theory can be easily tested, since (as just said) there are excellently described
Chapter 14. On the nature of strong interactions 14 - 2 experimental data - there will be no need for long discussions and arguments. 2. Interaction of moving electric charges In Chapter 16 discusses the interaction of moving charges. In particular, Section 7 of this chapter discusses the interaction of two moving charges. It is shown that the total force of attraction between charges and , located at a distance r and moving with velocities and , respectively, is equal to = , (1) where =×(×)−×(×) (2) Here we neglect mechanical forces, considering massless charges. Here we also neglect the forces of interaction of static charges. However, the energy of this pair of electric charges is the energy of the electric field of these charges. The appearance of the second charge next to the first does not change the energy of the first and second charges. Their mutual movement cannot change their overall energy. Thus, the considered movement occurs without energy consumption. This process can be compared with the movement of an electromagnetic wave: there is movement, but the energy of the wave is conserved; there is a flow of energy, but there is no change in the energy of the wave. In our case, there is also a movement of charges (and the associated movement of energy), but there is no change in the total energy. 3. Interaction of rotating electrically charged bodies In Chapter 16 discusses the interaction of moving charges. In particular, Section 8 of this chapter discusses the interaction of two rotating charges. Let the charges and q_2 be in the bodies and , respectively, which rotate around their axis with angular velocities and , respectively. Then the vectors of linear velocities =×, (3) =×, (4) In fig. 2 shows these bodies, charges and and vectors and of the positions of these charges. It is shown that the force of their interaction is equal to:
Chapter 14. On the nature of strong interactions 14 - 3 = 4(, ), (5) where (, )=∙ (6) is a certain function of known , , . Lq2 q2 q1 r a1 a2 o2 o1 Fig. 2. 4. Maxwell's equations for gravitomagnetism In [3], the author proposed a new solution of Maxwell's equations for gravitomagnetism, which is used to construct mathematical models of various natural phenomena (sand vortex, sea currents, whirlpool, funnel, water soliton, water and sand tsunami, turbulent flows, additional (nonNewtonian) interaction forces celestial bodies). All these models use the concept of mass currents as flows of mass particles. The speed of mass particles can be very low and often their flow can be invisible as well as the flow of electrons. But the existence of these phenomena and the possibility of constructing these mathematical models, similar to mathematical models of direct current in electrodynamics [4], confirm the assumption of the existence of mass currents and the interaction of mass particles, completely analogous to the interaction of electric charges. Based on this, it can be assumed that any movement of a body is accompanied by a mass current, similarly to how the rotation of a charged body is accompanied by a convection electric current. Eichenwald [5] showed that such a current creates magnetic induction. Based on the complete analogy between Maxwell's equations for electrodynamics and gravitomagnetism [3], it can be argued that when a body moves, gravitomagnetic induction is created. A mass m moving in a gravimagnetic
Chapter 14. On the nature of strong interactions 14 - 4 field with a speed v is acted upon by the Lorentz gravitomagnetic force (analogue of the Lorentz magnetic force). In electrodynamics, the magnetic induction is defined through the magnetic strength as =, where is the absolute magnetic permeability of the medium, and for vacuum ≈10.. In electrodynamics, gravitomagnetic induction is defined through gravitomagnetic strength as = , (18) where ≈10 is the gravitational constant, ≈10 is the coefficient of the gravitational permeability of the vacuum determined experimentally. For what follows, note that ≈10 , (19) Thus, the formulas used above are applicable in gravitomagnetism when the coefficient is replaced by the coefficient . At the same time, we can use the entire mathematical apparatus (including designations) described above in application to electrodynamics to describe the interaction of bodies with mass, and by charges we mean elementary masses - EM. 5. Some properties of nucleons. We use the above to describe the interaction of nucleons. In this case, we will assume that the nucleon is a ball with a uniform distribution of mass over the volume, has an angular velocity of rotation around its own axis, and this axis retains its orientation when the nucleon moves. We can apply several previous conclusions directly to nucleons. We use the above to describe the interaction of nucleons. In this case, we will assume that the nucleon is a ball with a uniform distribution of mass over the volume, has an angular velocity of rotation around its own axis, and this axis retains its orientation when the nucleon moves. We can apply several previous conclusions directly to nucleons. Nucleons contain many uniformly distributed EMs. By integrating functions (5, 6) over the volumes of two nucleons, one can find the function of attraction of these nucleons. If , is EM, then the force of attraction of two nucleons with volumes is determined by (5, 6) as =∫ = 4∫(, ) . (21) Total nucleon mass =∫ . (22) If the nucleon body is divided into elements containing EM , then
Chapter 15. Unsupported movement without violating physical laws 15-2 q1 q2 yx z v1 v2 Fig. 1. Fig. 2. -2 -1 0 1 2 -0.2 -0.1 0 0.1 0.2 -1.5 -1 -0.5 0 0.5 1 1.5 -->x, <--y, |z
Chapter 15. Unsupported movement without violating physical laws 15-3 Fig. 3. Fig. 4. -1 -0.5 0 0.5 -0.05 0 0.05 0.1 0.15 -0.5 0 0.5 1 -->x, <--y, |z -3 -2 -1 0 1 0 0.2 0.4 0.6 0.8 0 0.5 1 1.5 2 -->x, <--y, |z
Chapter 15. Unsupported movement without violating physical laws 15-4 The force acting on this structure as a whole can be found using formula (8). Fig. 2 shows a spatial graph of the change in this force during the time of one charge rotation (thick line) and the projections of this graph onto coordinate planes (thin lines). From here on the projections are designated by lines as follows: green – xz, blue – xy, red – yz; the directions of the axes are indicated under the figure. With a known force and given zero initial values, the speed and trajectory of the structure are found for the same period – see Fig. 3 and Fig. 4, respectively. During this period, the structure moves a certain distance Rmax=2.8. Fig. 5 shows the trajectory structure in two periods when it is displaced a certain distance Rmax=5.6. Fig. 5. 4. The Second Experiment In the design shown in Fig. 1, one charge was located on each circle. Now let us consider a design in which several charges are located on each circle, but they are all concentrated in one semicircle and distributed evenly over the semicircle – see Fig. 5a. Here, too, the force acting on this design as a whole can be found using formula (8). It turns out that the vector of this force lies on the plane with any number of charges > 1. The velocity vector and trajectory also lie on the plane . Fig. 6 shows, as an -6 -4 -2 0 2 0 0.5 1 0 1 2 3 4 -->x, <--y, |z
Chapter 15. Unsupported movement without violating physical laws 15-5 example, the trajectory of the structure for one period for the case when the structure contains 5 charges on each circle. y z Fig. 5a. Fig. 6. -0.5 0 0.5 1 1.5 -2 -1.5 -1 -0.5 0 x 10-4 0 0.5 1 1.5 -->x, <--y, |z
Chapter 15. Unsupported movement without violating physical laws 15-6 Fig. 7. Fig. 7 shows for the same case the graphs of the change in force (window F) and speed (window V) during the time of one revolution of the charges and the trajectory of the structure (window T) in coordinates . In this and the following figures it is assumed that the axis is directed horizontally, and the axis - vertically. In Fig. 7 it is evident that in one period the structure shifts a certain distance Rmax=2. Fig. 8 shows the same graphs for the same design for two periods. It is clear that the design is shifted by a distance Rmax=4. -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 F -0.2 0 0.2 0.4 0.6 -0.2 0 0.2 0.4 0.6 V -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 0.5 1 1.5 T
Chapter 15. Unsupported movement without violating physical laws 15-7 Fig. 8. Fig. 9 and Fig. 10 show the same graphs for two periods for designs containing 15 and 25 charges, respectively. For all designs, the value of one charge is taken to be equal to = 1/. It is evident that under this condition the graphs of the change in force and speed do not depend on the number of charges, and the trajectories are practically independent of the number of charges. Thus, such a design, with an increase in the number of charges, "tends" to a design with an infinite number of charges. In it, the linear density of charge distribution along the length charged semicircle is equal to = . As for the implementation of such a design, the charges in it should touch, but not merge into a solid strip, since the charge distribution density function along the strip is uneven (charges accumulate at the edges of the strip). Charges in such a design can be constantly restored from a constant voltage source through brush contacts. -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 F -0.2 0 0.2 0.4 0.6 -0.2 0 0.2 0.4 0.6 V -0.5 0 0.5 1 1.5 2 2.5 3 0 1 2 3 T
Chapter 15. Unsupported movement without violating physical laws 15-8 Fig. 9. Fig. 10. -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 F -0.5 0 0.5 1 -0.4 -0.2 0 0.2 0.4 0.6 V -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 T -2 -1 0 1 2 -1 -0.5 0 0.5 1 F -0.5 0 0.5 1 -0.4 -0.2 0 0.2 0.4 V -1012345 0 0.5 1 1.5 2 T
Chapter 15. Unsupported movement without violating physical laws 15-9 In conclusion, let us consider the calculation results for the same conditions that were used for the calculation in Fig. 9, but for 20 periods – see Fig. 12. In this figure, the red vector on the velocity hodograph depicts the average velocity ≈ 0.32 movements of the structure. Over 20 periods, the structure moved a distance ≈ 40. Fig. 12. 5. Movement Parameters Let's take a closer look at some of the characteristics of such a movement. In this case, we will not consider the energy required to rotate the structure at a constant speed. On the kinetic power , consumed by the structure for the movement of the structure as a whole, the average speed of movement and displacement designs influence speed of the structure as a whole = (1,2,3), driving force = (1,2,3), developed by the design, number of revolutions , rotation frequency or circular rotation frequency = 2, radius of the structure , linear velocity of charges = , total charge , quantitycharges, each of which has a magnitude ⁄, mass of the structure . -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 F -0.5 0 0.5 1 -0.4 -0.2 0 0.2 0.4 0.6 V -5 0 5 10 15 20 25 30 35 40 0 5 10 15 T
Chapter 15. Unsupported movement without violating physical laws 15-10 It can be shown that when > 4 quantitycharges does not affect the parameters of movement and = (,), (1) =(,,,), (2) = (,,,,). (3) Fig. 13 shows graphs of changes in instantaneous values of motion parameters when = 5, = 5, = 1,= 1,= 1. Here - trajectory of movement, 1,3 - coordinates , depending on time, - hodograph of the total velocity and the vector of the average velocity - hodograph of force 1,3 - projections of force,depending on time, - instantaneous power depending on time, - average power, 1,3 - velocity projections,depending on time, - speed amplitude Fig. 13. -0.05 0 0.05 0.1 0.15 0 0.1 0.2 T 0 1 2 3 4 -0.2 0 0.2 x1,x3 0 1 2 3 4 -1 0 1 f1,f3 0 1 2 3 4 -0.05 0 0.05 P 0 2 4 -0.1 0 0.1 v1,v3 0 2 4 0 0.05 0.1 vm -0.1 0 0.1 -0.1 0 0.1 V -1 0 1 -1 0 1 F
Chapter 15. Unsupported movement without violating physical laws 15-11 6. Resistance to Movement There is always a force acting on the structure resistance to motion - friction or payload. Usually such a force is proportional to the instantaneous velocity , i.e. ≈ ⋅, (4) where - a known coefficient. In this case, the instantaneous power of resistance to movement =(⋅)= ⋅, (5) Fig. 14. Fig. 14 shows graphs of changes in instantaneous values of motion parameters when = −0.75 and = 5, = 5, = 1,= 1,= 1. In the "P" window, the horizontal line is the power graph (5). As you can see, the trajectory gradually turns into circular movements of the entire structure "on the spot", the instantaneous amplitude of the velocity tends to some constant value (since the movement turns into rotation "on the spot"), Thus, the considered structure performs unsupported movement even in the presence of resistance. The power of the structure's engine is spent on rotating the charges and overcoming resistance. -0.02 0 0.02 0.04 0.06 0 0.02 0.04 T 0 1 2 3 4 -0.05 0 0.05 x1,x3 0 1 2 3 4 -1 0 1 f1,f3 0 1 2 3 4 -0.05 0 0.05 P 0 2 4 -0.1 0 0.1 v1,v3 0 2 4 0 0.05 0.1 vm -0.1 0 0.1 -0.1 0 0.1 V -1 0 1 -1 0 1 F
Chapter 16. Interaction of moving electric charges 16 - 7 where is the absolute electrical permittivity. Therefore, = =4. (43) We will call this value the effectiveness of the Lorentz forces The total force of interaction of two charges (4, 40) = +, (44) From here we find: = , (45) where =×(×)−×(×) (46) The sign "-" appeared because the radii in formulas (4, 40) have opposite signs. Here we neglect mechanical forces, considering massless charges. Here we also neglect the forces of interaction of static charges. However, the energy of this pair of electric charges is the energy of the electric field of these charges. The appearance of a second charge next to the first does not change the energy of the first and second charges. Their mutual movement cannot change their total energy. Thus, the movement in question occurs without energy expenditure. This process can be compared to the movement of an electromagnetic wave: there is movement, but the wave energy is preserved; there is an energy flow, but there is no change in the wave energy. In our case, there is also movement of charges (and the associated movement of energy), but there is no change in the total energy. In the general case ≠, i.e. Newton's third law is not observed - unbalanced forces arise that act on charges and and bend the trajectories of these charges. If charges and do not leave some common structure during the process of movement, then force (45) acts on it. This force can move the structure. The energy for the displacement is the energy for the internal motion of charges. It can be assumed that such forces provide the flight of ball lightning. An example of such a structure is considered in Chapter 15 8. Interaction of rotating electrically charged bodies Let charges and be in bodies and , respectively, which rotate around their axis with angular velocities ) and , respectively. Then the linear velocity vectors =×, (47) =×. (48)
Chapter 16. Interaction of moving electric charges 16 - 8 Lq2 q2 q1 r a1 a2 o2 o1 Рис. 2. Fig. 2 shows these bodies, charges and and vectors and of the positions of these charges. We have: +−−=0 (49) or =+−. (50) Thus, from (40, 47, 48) it follows that =(, ), (51) where (, )=∙ (52) is function determined by (45, 46, 47, 48) for given , , . References 1. Зильберман Г.Е. Электричество и магнетизм, Москва, изд. "Наука", 1970.
Chapter 17 17 - 1 Chapter 17. New Derivation of Equations of Satellite Motion in an Elliptical Orbit and the Mercury Effect Contents 1. Centrifugal force acts on the satellite 2. Stability of an elliptical orbit 3. The Coriolis force acts on the satellite Appendix 1 Appendix 2 References Abstract It is proved that the equation of an elliptical orbit obtained taking into account the action of gravity can also be obtained taking into account the action of two forces - gravity and centrifugal force. It is shown that taking into account the centrifugal force allows us to justify the stability of elliptical orbits. This property of the new solution allows us to assert the reality of centrifugal force. It is further shown that the satellite is affected by the Coriolis force, which creates a rotating moment of the satellite around the Sun. It is this moment that explains the Mercury effect. This fact allows us to assert the reality of the Coriolis force. 1. Centrifugal force acts on the satellite The natural satellite revolves in an elliptical orbit around the Sun. This first law of Kepler follows analytically from Newton's laws of universal gravitation and inertial motion [1]. There is also centrifugal force (CFF). However, it is not used in the equations of celestial mechanics, since it is believed that the Central Library Systemthis is one of the fictitious forces of inertia. The author has already proven that the CFF is a real force, the source of energy for which is the Earth's gravitational field [3]. If we accept this statement, then A)Kepler's first law should be analytically derived taking into account the existence of the CFF.
Chapter 17 17 - 2 The force of attraction between two bodies acts on both bodies. Each body is acted upon by a force arising in this body and pushing it towards the second body. When calculating the trajectory of the first body, one cannot assume that the force of gravity comes from the second body - on the contrary, it comes from this body, but in a coordinate system with the center in the second body, it acquires a negative sign. Thus, B) it cannot be assumed that the first body is in the area of action of central forces of gravity of the second body. The known equations of motion of a satellite in an elliptical orbit in polar coordinates with the center of coordinates in the Sun have the form [1]: = +, (1) =(×), (2) =−/, (3) where ,- polar coordinates of the radius, -specific central force, - the mass of the Sun, – gravitational constant. It is assumed that the value =(×)=const, (4) since it represents the specific angular momentum that is conserved in the central force field. This means that the quantity can be calculated once (when determining the initial conditions). The equations taking into account the existence of the centralized system have the form: =+, (5) =(×), (2) =−/, (3) = sin(), (6) = + , (6a) where is specific centrifugal force, is the angle between the vectors and and the value (2) is not constant. Fig. 1 shows an elliptical trajectory and vectors ,,. Figure 1a shows an illustration of formula (6a), where the projections of the vector [4] are shown:
Chapter 17 17 - 3 = , = . (6b) Fig. 1а. The force of attraction is distributed together with the flow of energy. The flow of energy can only be distributed from the source of energy. Therefore, the energy of gravity of a satellite to the Sun is generated by the attracted body, the satellite (and not the Sun). In this case, the specific angular momentum does not remain constant in our case (×)=∙∙sin(), (7) moreover =−, (8) where is vector inclination angle – see Fig. 1. V r Z Vs Fig. 1.
Chapter 17 17 - 4 From (6, 7) we find (×)=√. (9) Then from (2, 9, 6) we find: =√, (10) or = (11) or, taking into account (6), =∙() . (11а) From (11) it follows that =. (12) This means that equation (1) can be written as (5). Note that the angular velocity ω= , (12а) From (12a, 12, 9) it follows that the centrifugal force creates the angular momentum. In other words, it is the centrifugal force that creates the rotation of the satellite around the Sun. From (6a, 12) we find: Из (6а, 12) находим: = + (13) or = +. (14) Thus, the proposed change in the equation of motion of the satellite is as follows 1) expression (4) cannot be considered a constant, 2) equation (1) should be replaced by equation (5), 3) equation (6a) should be added. The relative difference between the found solution and the known one is for the orbit of Mercury and for the speed of motion along the orbit <10. (15) So, a new solution to the well-known problems has been found equations of motion of a satellite in an elliptical orbit. First of all, we note that in the well-known decision the assertion was made that the specific angular momentum is a constant (does not depend on the coordinates), since it is assumed that only one force acts. In the obtained solution, the specific angular momentum is NOT a constant, since two forces act,
Chapter 17 17 - 5 created by different bodies. The following statements were used in proving the proposed solution: 1) centrifugal force is a real force acting in celestial mechanics; 2) the source of energy for the force of attraction is a body that seeks to approach another body, giving it its energy - this second body is the receiver of the gravitational energy of the first body; 3) each body in an attractive pair generates two forces: centrifugal force acting on the opposite body, pushing the bodies apart; the force of attraction acting on itself, bringing bodies together; 4) in this case, each body is subject to two oppositely directed forces coming from different sources, which ensures the stability of the orbit$ 5) the energy constant in one body does not remain constant – see Appendix 1; this means that when bodies interact, energy is transferred between bodies: forces are transferred in the flow of gravitational energy (as in electrodynamics). Summarizing the above, we write down the complete system of equations for the satellite's trajectory =+, (5) =∙∙sin(), (2) =√, (2) =−/, (3) = sin(), (6) = +, (6a) =+, (8) =∙exp(), (21) =∙exp(). (22) For illustration, Fig. 2 show functions obtained as a solution to the system of equations (5, 2, 3, 6) for the orbit of Mercury: 1) ellipse =(), 2) function sin(), 3) function (7) ()=(×), 4) energy constant ℎ() .
Chapter 17 17 - 6 It is important to note that the angular momentum and the "energy constant" are not constants. Thus, the found and elliptical solutions differ very slightly. We can confidently neglect this difference. But the above statements make it possible to solve the problem of the interaction of many celestial bodies in a completely different way. At the same time, chaos in these solutions disappears, chaos that some people present as an achievement of physics. Certainty returns to physics. Fig. 2. 2. Stability of an elliptical orbit The satellite experiences random disturbances. We will not consider disturbances caused by the influence of celestial bodies on each other. But even a single satellite experiences disturbances that constantly exist. Indeed, Kepler's law is valid for point or spherical objects. But the planet is a shapeless lump, not a perfect sphere. Even if there are no external disturbances, there must be disturbances caused by the rotation of the planet around its own axis. It is obvious that a satellite that accidentally deviates from the ideal trajectory towards the Sun will be deviated even more by the increasing force of gravity. It is obvious that a satellite that elips: y=f(x) v=f(x) L=f(teta) h=f(teta)
Chapter 17 17 - 7 accidentally exceeds its speed will not be slowed down by anything and will move to another orbit. It has been proven (see, for example, [2]) that if these disturbances do not exceed a certain limit, then the planet oscillates around a constant orbit. This regularity has been verified by numerous computational experiments, and theorems describing these regularities have been obtained. But why is the next random perturbation on average zero (as assumed in the mentioned theorems)? But why are random disturbances limited and do not lead to a change in orbit for centuries? But why doesn't the planet continue to rotate along the other elliptical path that the random disturbance pushed it onto? It can be argued that the elliptical orbit is unstable. Fig. 3. To estimate the stability of the orbit using the new equation, we will consider how the forces and change when a random effect changes the current radius by the value Δ. Fig. 3 shows the graphs of the dependence of these forces on Δ. Point Δ=0 corresponds to condition (5), which we show for clarity in Fig. 3 we will depict as ||≈||. This condition is fulfilled for a point located on an ideal orbit. For further reasoning, we recall that the forces , depend on according to formulas (6, 3), respectively. When Δ>0 the force || decreases faster than the force || , and the resulting force ||−||<0 is directed toward the center of the ellipse, and when Δ<0 the force || decreases slower than the force
Chapter 17 17 - 8 || , and the resulting force ||−||>0 is directed away from the center of the ellipse. Thus, when the radius of the satellite changes externally in one direction or another from its position in an ideal orbit, a force appears directed along the radius in the opposite direction - the satellite returns to an ideal orbit. This means that the orbit calculated by the new equation is stable. Thus, the observed stability of the orbits indicates that the satellites are moving by gravity and centrifugal force. This, in turn, proves the reality of centrifugal force. 3. The Coriolis force acts on the satellite It is known that there is slow precession of the perihelion of the orbTы Mercury, openUrbain Le Verrierin 1859, and the less noticeable precessions of the orbits of Mars, Venus, and Earth, which were discovered by Simon Newcombeom back in the century before last. It can be assumed that similar orbital shifts exist for all satellites. There is no explanation for such shifts within the framework of classical mechanics. In addition to the above, a planet that rotates around its own axis and moves around the Sun is also affected by the specific Coriolis force: =−2(×), (31) where is the vector of the angular velocity of the satellite's rotation. Let us first consider the case when the vectors and are perpendicular. In this case, the vector lies in the plane of the orbit and =−2. (32) The force creates an acceleration directed along the plane of the orbit and perpendicular to the trajectory. This case applies to the motion of Mercury =2∙10, since its axis of rotation is directed almost perpendicular to the plane of the orbit. In addition, Mercury has , which greatly exceeds the value of this value for all planets except Venus. The rotational moment of a satellite around the Sun, created by the Coriolis force, is equal to =sin() (33) - see Fig. 4. In this case =− −, (34) =+ −. (35) Hence, =sin− −2. (36)