mechano

MULTIMOPLECULES

A very much important value is that of the multimolecule. Since a long time ago I am aware of this; for obtain it had to consider several things. For all people that had read my work, it is not necessary to explain what is a multimolecule. In my investigation work of an electric current had to meditate in many facts. In first place concluded that a model of electric current as that of the free electrons was with too many deficiencies; I have been commenting many times in other theories, here will not repeat them: With respect to the resistance of the current, the practical experience for us is very much valuable and the first useful experience we can take for structure our model. As the current electrons do not get free so easy as is considered in the theory of the free electrons, we had to take in account that they moved along a chain of atoms, that we called multimolecule. This means that in all the conductors the current electrons move at the same velocity on their conductor wires, along the multimolecules; some with less resistance (good conductors), and other with more resistance. Our first problem was to determine how move the current electrons in every one of the atoms of the multimolecule. Based in many other phenomena, this solution was relative easy, as could be understand in the numerical problems. Other problem was to determine how many atoms has a multimolecule. Really this is not an easy problem. Since the beginning we got the idea that each current electron jumps from an atom to the forward one of the multimolecule in some way, similar as an electron of a molecule. How many jumps could make in this way, in order to move along the multimolecule? To answer the fore question a logic and scientific way, is not easy for us; so we studied it considering the behavior of some conductors: For this we choose the mercury atom, that practically is the atom most resistant of all conductors, and with its liquid structure there is less risk that its resistance can be altered by irregular distribution of its atoms. Some kind of atoms, as iron are more resistant to the flow of electrons, but they could be magnetized, etc., etc. Depending of the length and the cross section of a mercury conductor, can give to it any one resistance. By first instance this does not affect us. Here will be accepted the value given in a conventional way; and from this value will determine an electric intensity that will be a balanced one; that is, that will give: Wr = Wi (see the fore problem). If the resistance R is reduced to its half, its intensity I is duplicated, and the energies Wr and Wi get incremented in the same magnitude than I. In few words, the maximum variation could have the mercury conductor could not affect to the other conductors that are better ones. For instance, in the fore numerical problems we had: RHg = 1,362.3 ohms; RCu = 24.1747. We can give equal values to both conductors if: LHg = 142,204 cm.; LCu´ = 1,342.3 LHg / 24.1747 = 59.8774 LHg = 8,514,807 cm.

For instance, we have for Hg: L = 100 m: A = 1 cm2; I = 120 Amps.;
 = 0.0000968; i = 10.4 eV.

RHg =  L / A = 9.58 x 10—5 x 10,000 / 1 = 0.958 ohms.
ik´ = 10.4 x 1.6 x 10—19 = 1.664 x 10—18 joule / atom
I = 120 Amps.; Ie = 120 x 6.3 x 1018 = 7.56 x 1020 electrons / Sec
Wr = R I2 = 0.958 x 1202 = 13,705 joule / Sec.
Wi = ik´Ie = 1.664 x 10—18 x 7.56 x 1020 = 1,260 joule / Sec.
Wr / Wi = 13,795 / 1,260 = 10.95
If: I´ = 120 / 10.95 = 10.96 Amps. = balanced intensity.
Wr´ = 0.958 x 10.962 = 115 joule.
Ie´ = 10.96 x 6.3 x 1018 = 6.9048 x 1019 electrons / Sec.
Wi´ = 1.664 x 10—18 x 6.9048 x 1019 = 115 joules.

For a copper conductor we have: A = 1 cm2.; I = 120 Amps.; i = 7.7 eV.;
 = 0.000001725.
We can have equal resistance than the Hg conductor if it is of a length:
LCu = Hg LHg / Cu = 0.0000958 x 100 / 0.000001725 = 5,553.62 m.
RCu´ = Cu LCu / A = 0.000001725 x 555,362 /1.0 = 0.958 ohm.

For copper we have:  = 0.000001725; i = 7.7 eV.
Determination of: nm = quantity of atoms that has a multimolecule.
In the Hg conductor we have: r = 1.13763 x 10—8 cm.; 2 r = 2.2752 x 10—8 cm.
A = 1.0 cm2 = 1.0 / /2 r)2 = 1.0 / (2.27526 x 10—8)2 = 1.931692245 x 1015 atoms.
Ie = 6.3 x 1018 I´ = 6.3 x 1018 x 10.96 = 6.905 x 1019 = quantity of current electrons that flow in one second.
6.905 x 1019 tj = 6.905 x 1019 x 1.12 x 10—11 = 7.733 x 108 electrons / tj.
1.9317 x 1015 = quantity of multimolecules there are in a volume: A nm of atoms.
-- - - - - - -- - - - - - - -- - - - -
Originally, when I made the model of the polygonal orbits, I considered that the magnetron of :  = 9.2754 x 10—21 was an energy value; afterward, I saw that could work as an energy if it was affected by a term: Q, that is expressed in (gr. / erg). The formula, expressed in e,m, units of the magnetron of Bohr, has an energy would be: M = e h / 4  m. Here the term (e / m) only gives a mathematical proportion, not a physical one… I did not pay much attention to this problem because I was working in other problems of physics; but now that I am studding the flux of the atomic particles have find many interpretations in the problem of the magnetron of Bohr…. Because this, I decided to search other energy that could explain the perpendicular deflection of the orbit of the electron. With this modification will vary the time of jump (tj) and the time of multimolecule (tm): But this will not affect our theories, only will be some quantitative variations in some problems, that in the future, and without precipitation will be corrected.
-- - - - -- - - - - - - - - -- -
Considering that a current electron moves along a multimolecule in a time: tm = nm tj, we have:
1.9317 x 1015 = 7.733 x 108 nm;
nm = 1.9317 x 1015 / 7.733 x 108 = 2.498 x 106  2.5 x 106 atoms = number of atoms that has a multimolecule. Also is very much important to point out that the fore number is equal to the quantity of atoms there are in the cross section of the conductor for each active atom in any given moment. -

Monterrey, México; September / 2003 Manuel de Hoyos Robles - --)









TWO HOLES EXPERIMENT

In the XIX century the undulated theory of light was imposed because with waves could explain many phenomena, that they could not do with a corpuscular theory. In the XX century, also happened the same thing; and also, afterward were found some phenomena they could not be explained by an undulated theory; so they considered that matter (corpuscles) could be transformed in energy (waves). The partisan of the undulated theory considered that the experiment of the two holes proved that light is produced by waves. They used two instruments, one that produces waves, and a slug gun. They had two parallel walls; one with two holes, and in front of it other in which they measure the intensity of the waves and of the slug particles. In the opposite side are the instruments that produce the two mentioned effects. When only one hole is opened in the middle , and the experiment is made with both instruments, the maximum intensity is appreciated similar in the testing wall, as could be seen with a symmetric curve with one maximum values. When both holes are opened

















THE RETURN DISTANCE (NO GROWING OF ENTR0PY)

Since the epoch of Newton, Coulomb, etc. the fields of forces have got a fundamental importance ; nevertheless if it is considered the infinite time of existence of the Universe, in this time all the fields of the Universe would be exhausted because the bodies that produce them would not be able to continue producing the microatomic particles that form such fields, but this is not so, because the particles that produce such fields, when they are expelled, recede from such bodies at light velocity, move during millions of years and after that time they return to the place were they were emitted, to this process we call return distance in such way that the fields never get exhausted ; to these fields we call mechanized fields for to differentiate of the structured ones that were accepted since the Newton’s epoch, and in which there were not considered as of corpuscular nature. With the mechanized fields of mechanophysics these problems are solved. Mechanized fields are formed by ultramicroscopic particles emitted at light velocity by the bodies that form such fields in all directions, when these field particles reach another body they act in this one, making the body to acts in accordance with the kind of field is considered, and as there is reciprocity between causes and effects, the second body affects to the first one in a similar way. As can be appreciated the mechanized fields by far are more simple than the structured ones. The first step of structure of the mechanized field has been given.

In the infinite time of existence of the Universe all the bodies producer of mechanized fields would be exhausted of their field fluids they emit, unless we consider other property of such fluids that we call: return distance, this return distance can vary in considerable magnitudes of time and distance in accordance with many facts; when a cycle of return distance is effectuated the fluid that original was emitted return toward the emitter body and is “recuperated”. Not all the corpuscles or particles emitted by a body return to the place they were emitted, because are absorbed by other celestial bodies. This, and other celestial bodies also emit particles that are absorbed by the first mentioned body. The second step of the mechanized fields has been made, that is, of no exhausted fields and with this solution is obtained other solution of a problem that have been of preponderant importance since the Boltzmann’s epoch, that is, the growing of entropy. It has been considered that the energy moves always from a higher level toward a lower one. In some way this is a simple criterion that with the mechanophysics is seen in a more subtle way, in accordance with this all particles (entropy particles) of the Universe always had have a determined energy that remain in them in the infinite time in the past and in the future, this energy acts in such particles in accordance with many facts in any determined process and in accordance with one of the fore mentioned facts the particle lose energy, but in accordance with other not evident facts the particle win energy (law of conservation of energy) in such way that all the particles of the Universe have not lost nor win energy in the infinite time of their existence in the Universe; forward it is going to see a subtle case, in this (returning distance) there is not the problem given by the growing of entropy, because the energy (in form of mechanized fields) that is emitted away in all directions, because the returning effect return to the opposite way (maybe in millions years).

There is a fundamental difference between the structured fields and the mechanized ones. It is considered that the first ones have an innate existence and that a body is affected by them in proportion to the magnitude of the physical body properties. The mechanized fields can not have an existence if there not exist the interacting bodies, this means that this bodies are the producer of such fields, sc the theory of the mechanized fields is very much simplified. How many mechanized fields we can have? We can answer the fore question if we observe all kinds of radiation that can emits a body in different circumstances: a gravity field, an electromagnetic field,, a light field, etc. With the mechanized fields we return the acceptance of the Newton’s law of gravity attraction. That was considered was not true because they fail in the precession of the perihelion of Mercury and in the deflection of a light ray moving tangential and near of a heavy celestial body as the Sun? I am not in accordance with the fore trace because in it was not considered that the gravity velocity was incremented (aberrated) by the velocity of the affected or receptive bodies: (Mercury planet and the light corpuscles mentioned); the effect that produces the fore increment of receiving gravity velocity is proportional to the square of such increment. The structured fields depart from the amplitude of the Universe affecting to the interacting bodies: the mechanized fields depart from the interacting bodies affecting the space of the Universe in which they interact.

The problem of the red shift can be explained with a numerical example and at the same time will be given a logical explanation of what has been interpreted as the expansion of the Universe given by modern physics. It has been observed that while farther recede a celestial body, its light radiation (gamma,.....radio radiation) move toward the red frequency, this is considering the constancy of light velocity, about this model (expansion Universe) there are some contradictory points we are not going to mention because the lack of space. In mechanophysics are considered the light radiation formed by material corpuscles (corpuscular nature). A light ray is formed by a straight line of light corpuscles spaced one from each other at equidistant we call: corpuscular distance (wave length in the undulate theory of light). In our model of mechanophysics the frequency of radiation is given by the corpuscular distance, and for distant celestial bodies diminishes not because the corpuscular distance grows, but because the light velocity diminishes due to the action of the remanent gravity (as we will see forward) that acts at equal intervals of time we call: braking time in each corpuscle of the ray of light At the present time the advance of the technology permit us to measure the velocity of light affected by the mentioned braking effect, who does this must not be influenced with the idea that light velocity is a constant value..

With respect to the action of remanent gravity in the light radiation’s we can make more than one model with a structure criterion. Of them and with a fundamentalist criterion we will choose one on which is considered that the electrons that emit the light are not affected by gravity, only their “corresponding” proton in the nucleus of the atom. Each corpuscle of a ray of light emitted, has a quantity of remanent gravity (nugravity) enough to acts at each time of brake during all the returning distance trajectory in such a way that the ray of light emitted by the electron in a given point of space, return to the same point after millions years, as could be appreciated by the numerical example will be given forward. In other themes, forward will be seen that particles with inherent energy can multiply this effect, with multiple interchanges of this energy. Here the kinetic energy produced by the remanent gravity is double than the kinetic energy with which the light corpuscle was emitted; this in accordance with the entropy particles of mechanophysics that consider that all the particles of the Universe have a fixed energy that is not lost nor win in any process we consider In accordance with our model the corpuscles of light are integrated with the nugravity in the surface of the “corresponding” proton with the quantity of nugravity in each corpuscle, enough to act producing the braking effect during all the returning time, with a force can be determined with the law of Newton acting in a corpuscle placed in the surface of the proton, by the proton exclusively. If the rays of the light are produced by atomic particles that are in a heavy celestial body, the remanent gravity in each light corpuscle will grow, so is required to sum the remanent gravity of the proton (nugravity) plus the remanent gravity produced by the heavy celestial body (astrogravity) all these in such way that the return distance trajectory will be smaller than that produced by a body (in vacuum space) emitted with a velocity little bad smaller than that of escape velocity as would be in the case of a celestial body with a mass that approximates to that of a black hole.

An emitted corpuscle of light will return to the point it was emitted due to the return effect produced by the nugravity, the force produced by nugravity, in accordance with our model is:

Fp = G M m / r2 = 6.673 x 10-- 11 x 1.6725 x 10-- 27 x 1.47236 x 10-- 50 /(1.7142951 x 10-- 14)2 =
5.5915227 x 10-- 60 Kg m / Sec2
= 6.673 x 10-- 11 m3 / Kg Sec2. = constant of universal gravitation.
M = mass of proton (Kg)
m = mass of a corpuscle (Kg)
r = radius of the proton (m)
0.5 m c2 = 0.5 x 1.47236 x 10-- 50 (3 x 108)2 = 6.62562 x 10-- 34 joule = 1 quantum of energy

As the positive particle of the binary system is affected directly by the remanent gravity (nugravity), we suspect that the electrons are not affected directly by the gravity. The time required for the corpuscle affected by the remanent gravity to return due to the corresponding force F would be:
2 tp = 2 m c / Fp = 2 x 1.47236 x 10-- 50 x 3 x 108 / 5.5915227 x 10-- 60 = 1.57992 x 1018 Sec.
In accordance with the theories of expanding Universe, they consider that at a distance equal to:
4 x 1010 to 6 x 1010 light years there are galaxies that recede to a velocity equal to that of light
5 x 1010 x 31,557,600 = 1.57788 x 1018 Sec. = 2 tp
31,557,600 Sec. = 1 year.

About the return distance in a black hole it could be from a limited distance to practically zero distance, so the light corpuscles do not escape from it. In accordance with the concept of continuous action, if an electron emits (n) corpuscles (as those of light) they get light velocity with a kinetic energy equal to n h (n quantum) and the electron gets a kinetic energy equal to the fore one, but in the opposite way (action = reaction). In accordance with this and with mechanophysics explained the theory of the polygonal orbits of the orbital electrons around the nucleus of the atom. In each vertex of the polygonal orbit of an orbital electron are acting (n) light corpuscles that deflect the trajectory of the electron in order to follow its corresponding polygonal orbit. If no spectrums are seen of the acting light corpuscles;: How can be explained the action of them? If we consider that in each vertex of the polygonal orbit the return distance of the corpuscles is equal to the diameter of the electron, when they move from the interior face to the exterior one of the electron, they do not escape from this last face, but begin their return trajectory, from zero velocity to c velocity, toward the interior face of the electron and from here they escape toward the interior face of the positive particle (see binary systems, theme : Hydrogenated Model of Atom).

Monterrey, Mexico, August 12, 1996













HYDROGENIZED MODEL OF ATOM

In the fore theme of Return Distance was simplified in a great deal the concept of fields, because they were considered in their structure, not as something characteristic of all the space of the Universe, but only characteristic of a limited space of the mentioned Universe, corresponding to the interacting bodies (or particles), in such limited space.

Now that we are working in the electrotherapy have had all kind of problems, from that produced by impertinent persons, to technical ones, because this we developed a very much satisfactory model of electric current, but this was not enough, due some persons are too much skeptic. We are not skeptics, but too much exigent with our work, so we began to work trying to make a model of atom that fulfill a series of required conditions, with much more success than that we expected. In doing this I remember a good friend of mine, that since I began to work with mechanophysics, he was very much cooperative with me, encouraging and stimulating my work, all this in spite he was supporter of modern physics, I refer to Dr. Rodolfo Castillo Bahena, he was director and professor of the Physics Department in the Instituto Tecnológico y de Estudios Superiores de Monterrey: his indications, commentaries, and critics were very much useful, I remember in one occasion he commented me that in order that the mechanophysics could be accepted, it would be necessary to give a model of atom with certain characteristics. Because then I had many ideas in my head, I considered this as a simple conversation, but now I consider it as a prophesy (now also have many ideas, but try to be more organized).

With the structure of the atoms it is possible to make a great simplification, and we believe that with this it is possible to make great progress in this field. Till now it has been considered the nucleus of an atom formed by a determined quantity of protons and neutrons, making a unitary structure, affecting in this way to all the orbital electrons, and viceversa. With this criterion have been given different models of nuclei affected with many complexities and deficiencies. Here will be given a model by far more simple that will simplify its investigation and understanding; in this model will be considered that each orbital electron is only affected in a direct way by its corresponding proton and neutron; to this structure we will call: binary system. The intervention of a binary system in the other ones of the atoms will be minimum in all the normal conditions and only could be appreciated in an evident way in phenomena of radioactivity or half life of atoms; in which is manifested the interference of a binary system with another one. With respect to the spectrums produced by the excited atoms, they could be explained by indirect or secondary interaction of a binary system with other ones.

The medullar point of a binary system is that the electron and the positive particle work in a synchronized way emitting jn an interior way their propeller fluids every time of vertex, but this is because other effects also are synchronized; a car can move because has a motor, but its movement is not arbitrary, but in accordance with a driver that controls it, next are going to see some facts that contribute in the control of the behavior of the binary systems. In order that a binary system is not affected by other (s) one and vice-versa, as was say, it will be considered that in a general way, either the electrons, as the protons of each binary system, have a determined side or face were their propeller fluids are emitted, in the corresponding side are received and absorbed the microcorpuscles of the electric fields that excite the atomic particles (their propeller fluids); the amplitude of this side permits a small tolerance given by a small solid angle, in such way that can exist variations (in the orbital electrons) in the spectral lines (s, p, d, f). If the atomic particles (electron or proton) receive the microcorpuscles of the electric field in a different side than the fore mentioned, the microcorpuscles are not absorbed, not exciting the atomic particle. Another property of the microcorpuscles of the electric fields, in our model of binary system is that they have a return distance equal to the distance that there is between the orbital electron and the nuclear proton (or positive particle), of the binary system. Continue with the required conditions, will be considered that the electric fields formed by emitted microcorpuscles, are not formed in an immediate way, but in a progressive one, from a null (maybe small) intensity to its maximum value; the variation of intensity of the electric fields emitted by the orbital electron and the positive particle coincide with the angular rotation or spin, that is manifested in each vertex of their polygonal orbits, here the intensity is null (or small) when the deflection start, and gets the maximum and final value when the deflection finish, When the charged (-) particle has turned an angle of 360o. In accordance with the binary system, as will be appreciated with the numerical example will be given later.. Let see a binary system formed with an orbital electron as negative particle, and a proton as a positive one. The orbital electron is excited in each vertex (with a deflection angle = ) of its orbit.

The positive particle (proton) is excited in each vertex by a fields. Although the fields act fast in the vertices of the orbits, and with the same magnitude in the positive particles as in the negative particles, this does not mean that the positive particle get equal quantity of charge than the negative one, If the positive particle got its charge in one instant. In the binary system the orbital electron and the corresponding proton, always have their corresponding side of absorbing microcorpuscle of the electric fields in front one from each other, this because in this way are oriented the particles in the binary system. A binary system in the hydrogen atom is formed by one orbital electron and one nuclear proton; in a deuteron atom the binary system is formed by one orbital electron and one positive particle formed with one proton + one neutron; in the triton atom by one orbital electron and a positive particle; formed by one proton and two neutrons. From the point of view of the mass of the positive particle, there are three types of binary systems, as just have seen. From the point of view of the orbital electrons moving in any one of the 7 shells of the atom, obviously there are 7 types of binary systems. In the fore cases the negative particle (orbital electron) has the same mass. In the way can be supposed how is formed the positive particle can be imagined one sphere (the proton) two or three spheres are fusion, forming one bigger sphere,

In previous investigation works in mechanophysics, and based in some experimental data, and considering that all the elementary particles have the same density, we obtained that the radius of the electron is: re = 1.4 x 10-- 13 cm and of the proton: rp = 1.7 x 10-- 12 cm. About the nucleus of an atom, there are many objections, for instance, if we consider the nucleus of hydrogen atom as a proton or as a nushell (see forward), will be a great difference of density. In the value of: r7 = 1.1376 x 10-- 8 cm. = radius of the 7 orbit = radius of the atom, we can confide more because the Avogadro’s number, also in the mass of electron: me = 9.1091 x 10-- 28 gr. and the mass of proton: mp = 1.6725 x 10-- 24 gr., because the fields effects. Some one could object that can not confide in the binary systems, but if with two complete different ways can arrive to the same results, the fore objection is demerit.

We can continue without impediment, so for a heavy atom as gold: Au, for example: Z = 79; A = 197, we can imagine a structure formed by 79 binary systems, of them 39 have the positive particle formed with one proton + one neutron; and 39 formed with one proton + two neutrons. in the 7 shell (?) 2 with one proton: 39 x 3 + 39 x 2 + 2 x 1 = 197.

In accordance with the characteristics of a binary system, the orbital electron and the positive particle spin around the center of the nucleus of atom in a synchronized way, the orbital electron at a distance equal to the radius of the shell; the positive particle to the fore distance divided by 1836 / 4 = 459; . 459 me = m1+ / 4. The positive particle could be triple mass = 1 proton + 2 neutrons, double mass = 1proton + 1 neutron; single mass = 1 proton. Although we are ignorant of chemistry knowledge, also we believe in the future could be determined different chemical affinities in function of the mass of the positive particles of the binary systems of the seven shell . Both particles (+ -) describe their corresponding polygonal orbit of a similar shape, around the center of the nucleus, but of course of different radius of gyration, so that it is considered that the orbital electrons move in shells, the positive particles will move in çnushells (forming the nucleus of the atom). The size of the nushells, by far are smaller that their corresponding shells, also the size of the orbital electron compared with that of the positive particles, in the first or most interior nushell hardly is space for two moving positive particles, so it can be explained why in the first shell only can spin two orbital electrons. In Fig.(1), second orbit, with O is represented the center of the nucleus of the atom, that also is the center of gyration of the binary system; PN is the positive particle; E the orbital electron; rN the radius of gyration of the positive particle; Rn the radius of gyration of the negative particle (orbital electron)

When the orbital electron moves in the: n = 7 shell (most external shell);
Rn = R7 = 1.1376 x 10—8 cm = radius of the atom.
The radius of gyration of the positive particle:
rN = r7 = 4 Rn / 1,836 = 4 R7 / 1,836 = 4 x 1.1376 x 10—8 / 1,836 = 2.484 x 10—11 cm = radius of the nucleus (nushell)

For: n = 1 orbit :r1 = 2.484 x 10—11 / 7 = 3.5486 x 10-- 12 cm.(*about two radiuses of proton, (see Fig. 4)

The electron E moves in a polygonal orbit, for correlated reasons the proton PN moves in a
synchronized way in another polygonal orbit, smaller than that of E. For the orbital electron to deflect in a vertex of its orbit (7) it is required an energy to produce the velocity of deflection v_l_7 (see Fig.2)
v_ = 2 x 309 Sin 6o 43 = 69.1 Km./ Sec. (7a orbit); 6º.43 x 28 = 180o

The energy of deflection of the electron is: E7 = 0.5 me v2 = 0.5 x 9.1091 x 10-- 28 (6.91 x 106 )2 = 2.175 x 10-- 14 erg.

In a vertex of the 7 polygonal orbit is formed a field affecting the orbital electron of the binary system and other similar one affecting the positive particle (this in accordance with a model).

Considering the length of each side of the polygonal orbit: d = 2.55 x 10-- 9 cm, and also considering in Fig.(2) there is a proportion between velocities and distances, is obtained the deflection distance:
d_l_7 = 69.1 d / 308 = 69.1 x 2.55 x 10-- 9 / 308.6 = 5.73 x 10-- 10 cm.

Next is going to see a binary system corresponding to the 2 orbit. From Fig.(3), corresponding to two sides of the second orbit, is obtained the deflection velocity: v = 826.58 Km./ Sec. The deflected distance: d -= 1.96 x 10-- 9 cm. = 2.55 x 10-- 9 x 2 Sin 22o.5

The energy of deflection is:
E2 = 0.5 x 9.1091 x 10-- 28 (8.2658 x 107)2 = 3.112 x 10-- 12 erg.

In Fig.(4) is represented the orbit of a positive particle in the first nushell, from this can be deduced that in the first shell can move as much as two binary systems. Next are given some relations in which are harmonized geometrical and kinetic conditions.
v_l_n = 2 vn Sin(45o / n) . . . . ( c) Monterrey, México, September 10, 1996