
GRAVITATION
The following is part of the explanation of gravity given in Mel Winfield's book “The Science of Actuality”:
Although gravitational acceleration applies to both protons and neutrons in any object, let us examine a proton orbiting within a nucleus in an atom in an object near the Earth's surface. The behavior of a neutron is similar. Since there are waves of energy emanating from all the nucleons contained in the Earth [the average distance of which would coincide at Earth center] the proton would spin naturally with a horizontal axis in the manner of a ping pong ball on a jet of air and would maintain this attitude like a gyro as it orbited within the nucleus. Whether or not the nucleon orbits within the nucleus or just revolves with the nucleus, it still follows an orbital path. The preceding study on nuclear structure brings us to emphasize two important facts. Firstly, its orbital axis and its spin have the same orientation so consequently its orbital plane is vertical relative to the Earth's surface. Secondly, it rotates in the same direction as the atomic vortex. It could not be in any other way for these two arrangements since, in a vortex, the outer layers are orbiting faster than the inner. This would rotate all the particles in the same direction as the vortex as it is shown in Fig. 2. Energy waves coming from the Earth strike the proton on the underside and, because of its spin, are forced around it. The result is a difference of pressure between each side of the proton. This phenomenon is known as the Magnus effect from the name of the German Chemist, Heinrich Gustav Magnus who described the effect in 1852. This process is analogous to a baseball, when thrown with a spin, that forces the air to one side causing a lessening of pressure and thus a curving of its path. By replacing the ball with a proton and the air with the moving etheron field, we get the same situation, at a smaller scale, and the proton would be accelerated to the side where there is a greater velocity of the stream carried by the Earth waves. This is depicted in Fig. 2 in contrast to Fig. 1 which would be the case if the proton was not spinning. ORBITS We must consider two half orbits. The first is starting from the 9 o'clock position, through 6, and arriving at3. The top half is from 3 to 9, passing through 12. When we say acceleration, we say an increase in velocity. Since vortex density and centrifugal force hold the proton from reducing its radius, to much extent,at the 9 0'clock position, as depicted in the drawing, the total pressure is perpendicular to the orbital path and is applied against the whole vortex structure [i.e. the atom]. As the proton moves towards the 6 position, more and more of the pressure is applied to accelerate it along its orbit, the amount of which varies from 0%, initially, to 100%, finally, in accordance with curvature of its path. Since the average direction from 9 to 6 is 45^{o} from the horizontal, the accelerating pressure is 50% of the total. Fig. 3 depicts the average direction vectors. ““ represents opposition to the pressure thus decreasing velocity. “+” represents momentum in the direction ofthe pressure and thus an increase of velocity. The numbers represent clock positions. P represents Magnus effect pressure. From 6 to 3 the orbital velocity of the proton is subtracted from the stream velocity, rather than added as it was from 9 to 6, due to the fact that it is travelling with the stream coming up from the Earth rather than against it. When it travels against the stream, a unit volume of the stream passes it at a faster rate than if it was stationary, which is the equivalent of a faster stream. Therefore the two velocities are added. When it travels with the faster moving stream, its velocity is subtracted from that of the stream to obtain the relative velocity. Such is the case from 6 to 3 and from 3 to 12 which is the opposite of that from 12 to 9 and from 9 to 6. The change in the relative velocity, in turn, affects the downward velocity. The peak of acceleration was at 6 and this is the peak of the downward velocity due to maximum centrifugal force in this direction. It meets this position a little lower in the orbit, if nothing prevents it from doing so, i.e. without any resistance [such as from the Earth]. From 6 to 3 acceleration decreases to zero at 3. The orbital velocity and radius are still increased, over average, though less and less,
until at 3 they are at an average. The pressure is again perpendicular to the orbit with the full pressure being outward to the right on the proton. Pressure on other nucleons and on the vortex as a whole is in the same direction. Since all the atoms in the Earth are oriented in the same direction by the outward moving wave stream towards lesser density [hence the Earth's magnetic field], they all move to contribute to the rotation of Earth. Further explanation of this contributionto Earth rotation is given shortly. Coming back to the movement of the proton within the nucleus, it is moving against the pressure as it goes from 3, through 12, to 9 which decreases the velocity. It is the exact reverse to the bottom half and thus velocity is brought back to the low point at 9. The decrease lowers the centrifugal force. The maximum decrease in the rate of deceleration is at 12 since the proton is meeting the pressure head on and the slowing effect thereof is at a maximum. From this point the effect decreases to zero by the time it reaches 9 and the pressure is perpendicular to the orbit. At 12, the reduced centrifugal force is at a maximum, thus reducing the orbital radius equal to the increase at 6 and thereby moving the entire nucleus downward. Whatever the proton position in the orbit may be, the stagnation point is always on the left of the proton because it is always rotating in the same direction. Consequently, the direction of the pressure is mainly directed continuously to the right. The difference in centrifugal force between the top and bottom of the proton orbit [ together with lower starting positions in each case] drives the proton downward. Naturally, the same effect applies to all the other protons and neutrons in the nucleus. They each fall so that the whole atom, and any mass of which it is a part, falls at the same rate. A more massive body than the Earth would emit more forceful waves and the gravitational push would be more important. This is what gravity is on Earth and, by extension, universal gravitation between bodies in space. This is a pressure phenomenon from a purely mechanical origin which has long eluded physicists by not having been united with the other fundamental forces of nature. GRAVITATIONAL FORCE
According to this description we can easily imagine why the gravitational force is so weak since this is the consequence of a velocity differential relative to the ongoing wave. Then we will no longer be surprised to notice that it is 10^{40} times the strong force [vortex forces] in strength. Its range is long, contrary to this last force which directs its force inwards, because these are radiant energy waves propagating outward with a force proportional to the inverse square law. They move through space at the same velocity as that of light according to the medium density and in the case of the Earth, for example, they do not move as fast because they are impeded by the numerous atoms and their fields within the Earth. In fact, the carrier of the force wave is the etheron or more precisely a collection of etherons. This is what has been called a gravitational wave which has never been detected and this is not without reason! It would be difficult to detect movement of etherons in the Aether except possibly by temperature as has been the 3K particle, or kaytron as it is called herein, which is called background radiation.
Gravitational force has been determined to be 10^{40} less than the binding energy for medium elements which is about 8 MeV [8 million electron volts] per nucleon [proton and neutron]. This is because most of the energy of gravitation lies in the nucleons of the atom. A nucleon has the same structure and method of producing a field of waves as does the atom except that instead of nucleons in orbit, as in the atom, the orbiting particles are the basic etherons. The volume of a proton is 10^{15} times that of an atom. The energy of the orbiting particles in the nucleon is also of this ratio. Since etherons contain equivalent energy to one another, less etherons by volume means the same reduction in energy. However, a wave is produced within a nucleon by a revolution of the orbiting etherons of an inner vortex in the nucleon, hence the number of waves, or the strength of the field of waves, emanating from a nucleon is dependent upon the velocity of these orbiting etherons.
Alan D. Krisch, in an article called the Spin of the Proton, Scientific American, May 1979, states that "in Newtonian Mechanics, the kinetic energy of a moving particle is proportional to the square of its velocity". Thus the strength of the field is reduced again since the velocity is proportional the square root of the nucleon energy. If we multiply [10^{15}]^{2} by the strong force of the atom, we bring the field strength of the proton down to 10^{30}. However, the force of the waves at the surface of the proton, which have been emitted from the inner vortex at the center of the proton, also depend upon its radius, which is 10^{5} times that of the atom, because the gravitational effect is proportionate to the square of the radius. Thus the total gravitational effect of the nucleon is [10^{5}]^{2} X 10^{30} which equals 10^{40} times the strong force of the atom. MAGNUS EFFECT
For the quantitative approach to demonstrate the Magnus effect on a proton, we must refer to the Fluids Mechanic field. According to the analysis of ideal fluid flow around a cylinder^{17}, The formula for the lift force at right angles to the free streaming fluid where the two stagnation points come together at the bottom of the cylinder, with the two streamlines making an angle of 60 degrees with the tangent to the cylinder, is FL = pBUT. p = density of the undisturbed stream, U = steady flow of uniform velocity of the undisturbed stream, B = length of the cylinder, T = the circulation and equals 4RU. R = the radius of the cylinder [proton]. The maximum peripheral velocity of the flow is 4U at the top. In our case, the cylinder becomes the proton with B being its diameter. The application is valid since it is said^{17} that it applies not only to the circular cylinder but to a cylinder of any shape including the lifting vane or air foil. Also it has been applied to a baseball which is spherical. There may be a slight difference in the case of a solid sphere since, in that case, there are different angles of deflection of the approaching wave. In the case of a proton, there are smaller particles that make up the vortex of which it is composed. It is structured in the same manner as is the sun. The waves strike these smaller particles head on without deflection, regardless of the curvature of the vortex. They strike them as if they were on a flat surface so therefore the cross sectional area of the sphere is receiving the effects of the force. This is the area of a circle which is used in the next calculations.
The formula as applied to the proton is F = 4Pr^{2}U^{2}. Since, in our model, the proton is descending on the left side of its vertical orbit, counterclockwise, while rotating counterclockwise [an observer on the opposite side would see the opposite], the Earth wave stream rising vertically strikes the proton and turns into it like a water wave striking the shore an angle and then progressively turning parallel to it. As each minute portion of the wave, in turn, meets the shore, its direction of motion turns at 90^{o} to the shore line which is why the water molecules move a short distance, in this direction, on to the sands. Thus the full dynamic [velocity] force of the wave is transmitted to the proton which, by its rotation, carries it around to its right side. The stagnation point is now at the nine o'clock position with the force difference, known as lift force, being applied horizontally to the right. There is no force pushing back from the right because the wave stream has sufficient momentum so that centrifugal force holds it away from that side. Since the proton is essentially traveling at 45^{o} to the horizontal by moving from the nine o'clock position of the orbit to the six o'clock position, this force of the Magnus effect is divided into the horizontal and vertical directions. Thus the full force drives the proton or neutron.
The Magnus force accelerates the proton from zero acceleration at the nine o'clock position to its maximum downward acceleration at the six o'clock position of the orbit. The average change in velocity [V] thus takes place over 1/8 of the circumference of the orbit. The radius of a nucleus is given as 1.2 X 10^{13} X [mass no.]^{1/3} cm. If iron is used as an example, the mass no. is 56 and the radius is then 4.591034839 X 10^{13} cm. Since there are 56 nucleons in an iron nucleus that are orbiting, an average radius of orbit will be taken which is the radius from the axis to where all of the mass would be located if it was at a point or on a circumference line. The formula for this radius is [2/5 r^{2}]^{1/2} = R. R would equal 2.903625382 X 10^{13} cm. which is the orbital radius of the average nucleon in the Fe atomic nucleus. The circumference of this radius would be 1.824401634 X 10^{12} cm.
The proton spin angular momentum is given by [3]^{1/2} X h/4 which is 9.13300574 X 10^{28} gm. cm.^{2}/s which = Iw. The neutron spin angular momentum is .684926602 times this which is 6.255438586 X 10^{28} gm.cm.^{2}/s. The 26 protons and 30 neutrons added together = 4.251213068 X 10^{26} gm.cm.^{2}/s. They are added together because it has been shown herein that the neutron is a proton with a third vortex revolving in the opposite direction around it with an electron between. At least it is opposite relative to the vortexes within it as would a slower orbiting one which would also stop the angular motion of any waves emanating from the inner structure. The proton, making up most of the mass of the neutron, will revolve in the same direction as the bare proton and therefore all angular momentum should be added together with the third vortex of the neutron subtracting [by counteracting] almost one third of the angular momentum from its proton. This agrees with accepted data. Now, the mass of the Fe atom, derived from its atomic weight, 55.847u  26 electrons X .00055u [which is .0143u] = 55.8327u. Multiplied by 1.66057 X 10^{24} gms. per u = a total of 9.271410664 X 10^{23} gms. which is the mass of the nucleus. Angular momentum of the atom = mvr so if 4.251213068 X 10^{26} is divided by the mass of the nucleus and by R, the average velocity of 1.579161219 X 10^{9} cm./s for the orbit around the nucleus is obtained. The mass of the average nucleon = 9.271410664 X 10^{23} gms. / 56 = 1.655609047 X 10^{24} gms. The nucleon period in the atom = R X 2 / 1.579161219 X 10^{9} cm./s = 1.155297896 X 10^{21 }s. The full velocity increase from zero to maximum takes place over 1/4 of the orbit thus the full time is the period divided by 4 which = 2.88824474 X 10^{22}. To verify that the average velocity [] increase is one half of this, F = ma can be used where a = g. The value of will be shown to be 982.9514643 in section II, Cosmology. g = v/t or t = v/g. During the interval of the full time, the Magnus force drives the nucleon downward through a distance of 1/2at^{2} which equals .5 X 982.9514643 cm./s^{2} X [2.88824474 X 10^{22}]^{2}s = 4.099869758 X 10^{41}cm. = 4.099869758 X 10^{41}^{ }cm. in 2.88824474 X 10^{22} s = 1.419502198 X 10^{19} cm./s. t for this average velocity = /g = 1.419502198 X 10^{19} cm./s /982.9514643 cm./s^{2} =1.44412237 X 10^{22} s. The average force = 1.655609047 X 10^{24} gms. X 1.419502198 X 10^{19} cm./s /1.44412237 X 10^{22} s = 1.627383337 X 10^{21} dynes.
To compare this with gravitational force, an altered version of Newton's formula shall be used which is [circle G]m/r^{2} = F. Circle G [] will be defined in PART II, Cosmology, and is equal to 3.989644343 X 10^{20} cm.^{3}/s^{2}. The mass is that of the average iron nucleon which is 1.655609047 X 10^{24} gms. and r = the average radius of the Earth, as determined in Part II, of 637090389.9 cm. The result is 1.627383338 X 10^{21} dynes. This is an exact correlation. Note: The top and bottom half of the orbit around the atom can be treated separately because they have separate time intervals and are identical in result even though the effect at the top half is the result of deceleration rather than acceleration. DROPPING TESTS Now the dropping tests conducted by this author in 1981 and 1983 will be examined. Atoms of different elements, because of the different forces from different neutronproton combinations, would have slightly different acceleration. The amount of element does not matter because each atom in the element is forced downward at the same rate as its neighbor. It is only different elements that would show different accelerations. If such could be shown to be the case, it would be demonstrated by experiment that gravity is due to nucleonic force. Different elements should fall at different rates according to the neutron proton composition per equivalent mass. This premise was proven by experiment in the years above stated. In the dropping analysis following , only the experimental drop ratio is given since it would not matter if the timer's base time was accurate, slow or fast. All the drops would be timed by the same base time, thus their ratios to one another would be accurate. The calculated gravitational force will be given in terms of proton force. The gravitational effect of a neutron is 84.24633% of that of a proton as explained herein. In the example of the iron atom, 30 neutrons are equivalent to 25.273899 protons. If we add 26 protons, we have the equivalent of the force of 51.273899 protons.
Each element's drop rate is calculated by the amount of nucleonic force per unit mass. It is determined for that element by its neutronproton ratio, which establishes the force, which is then divided by the mass of the neutrons and protons [or atomic weight in mass units] to determine the force per unit mass. It should be explained why the force is determined per unit mass. If forces exerted on different substances to create accelerations are to be compared, their masses must be normalized or in other words made equal. Whatever the substances, whether atom, molecule, etc., the masses of each kind can be in terms of one amu or one proton mass, whichever is the easiest to calculate, or any other mass as long as they are equal. Whatever calculation is done to the mass of a substance, such as dividing by an equal amount to bring it to unity or multiplying by a factor so as to bring it equal to another mass, the same must be done to the force associated with it because the force varies with the mass. Only then can the forces of both substances be compared to see which is the greater to drive the substance faster towards a larger body of matter. When then compared with the known acceleration of another substance, the acceleration of the first substance, can, by ratio, be determined.
Generally, the denser the element the less force per unit mass there is pushing down since the neutron excess usually increases with density or atomic weight. However, there are exceptions if two elements have the same neutron excess such as beryllium and aluminum which each have one neutron excess. When the force is calculated for each, in units of one proton force [the amount of force exerted on one proton] with a neutron force being calculated as a fraction of a proton force, it is found that the aluminum has more push downward, instead of less, and so will drop faster than the beryllium. In the case of the horizontally suspended ring composed of one half of each of these metals, designed and tested in an experiment by University of Washington physicist Paul Boynton^{18}, The aluminum should, as it did, turn towards the mountain in front of which it was hung. This is contrary to Fischbach's idea^{13} that all elements with greater atomic mass would be forced away by some fifth force emanating from the Earth. Boynton's experiment did validate Fischbach's observation that different elements exhibited different accelerations but disputed his explanation of it. This author's experiment consisted of dropping Al, Fe, Cu and Pb through two laser beams, one below the other, spaced four feet apart. The distance covered enabled a slight difference in accelerations to more greatly influence the difference in velocities and thus create large differences in the time dropped. Each element dropped any number of times within its own distinct time range zone with no more than .145% variation. The release of the weights was consistent, the air pressure on the tapered 1/4 X 1 1/4 inch knife edge bottoms was consistent and the lengths of the weights were adjusted to eliminate air turbulence. The bottoms of the weights were suspended .025 inches above the top laser beam by a fine wire from the weight to a horizontal pin, the end of which was inserted through a loop in the end of the wire. Though the pin was made to move vertically by means of a screw thread, all wires were made to accommodate the different lengths of weights so that the vertical adjustment did not have to be used. The periphery light from the laser light, striking the bottom of the weight, cast a shadow line on a paper beside the sensor which coincided with a line on the paper. The height of each weight was thus checked and was accurately the same within thousands of an inch. The pin was drawn back by an electromagnet that was well away from the weights, causing the release of the weight which fell through a three inch diameter tube [to minimize air movements] cutting both beams one above and one below the tube. The breaking of the first beam started the timer and the second stopped it. The drop time differences were in the area of the thousandth and ten thousandths of a second but because the timer showed numbers to the ten millionth of a second the averages of the drop times for each element became a very accurate value. The calculated and the experimental drop time ratios are given in table 1. The Cu mass number is 69.09% of element #63 and 30.91% of #65. After subtracting the number of protons from the mass number, the average number of neutrons in Cu is 34.6182. It will be seen that the experimental and the calculated drop time ratios, in round numbers to the third decimal place, are exactly equal. < The force ratio for each of the four metals in terms of proton force and the comparison with experimental drop time ratio is calculated as follows: Wt. Neutrons X .8424633 + protons = Total P.F. / Atomic Wt. Ratio [unit mass] Al 14 11.7944862 13 24.7944862 .9189950408 1.001013655Fe 30 25.273899 26 51.273899 .9180644405 1. Cu
34.6132 29.16456301 29 58.16456301 .9154007398 .9970985689
Pb 126 106.1503758 82 188.1503758 .9081054868 .989152228 TABLE 1_{a} Wt. Calculated Experimental drop time ratio Contin. Ratio Inverse [Average of 20 drops each] Fe = 1 Al .999 .999Fe 1. 1. Cu 1.0029 1.0029 Pb 1.010 1.01
_{}
It was previously mentioned that it is known that the angular momentum of the neutron is 68.4926602 % of that of the proton. It was shown that if the neutron is really a proton with a counterrotating vortex around it, the vortex would have an angular momentum of half of the remaining 31.5073398% which is negating the other half. The portion of the neutron's angular momentum in the direction of the proton spin is 68.4926602 plus one half of the remaining 31.5073398 or 84.24633% of that of the proton. Angular momentum of the proton within the neutron has been reduced, by the outer vortex, by 15.7536699% and the counterrotation of the outer vortex reduces it further by an equal amount relative to the environment. Thus, angular momentum inside the vortex is still 84.2463301%. This percentage was used in determining the proton force of neutrons in the foregoing table. By dividing by the atomic mass to bring all elements to unit mass, it is leaving a ratio of angular speeds since speed times mass equals momentum. It is nucleonic angular speeds that match the ratio of dropping times, and it is these that bring the changes in the Magnus force effect. The outer neutron vortex has only approximately 15% of the neutron angular momentum and is therefore only a small deterrent to etherons, moved by waves of energy from the Earth, reaching the inner proton in an object above the Earth. Here the angular speed of the neutron's inner proton together with the angular speed of the free proton determine the accelerating force from the Magnus effect to create the gravitational acceleration. The results obtained from the dropping experiments substantiate the concept of gravitation being due to the Magnus effect as well as verify that different elements fall at different rates due to their different neutron proton ratios. The denser an atom, the greater is the neutron excess over protons. This is usually a rule. In these cases, the proton force and the neutron equivalent per unit mass becomes less and less with increase in density. Less force means slower spin, less frequency and longer wavelength in atomic emission. Higher frequency is thus related to greater energy. This is proven when high frequency power is supplied to fluorescent lights and more light is created. Conversely, power is saved for the same light level. Transmutation of elements from one to another has taken place in our experiments by a change in the number of electrons around the nucleus and hence the energy level of the atom and its frequency. Nucleons [protons and neutrons] thus can change their energy level which would result in a different spin rate and a different gravitational acceleration. Different gravitational accelerations were evident in the drop tests.
The following photographs [taken with this author's camera] are the first experiments in levitation achieved by using a special kind of a moving electric field. The method and results substantiate the theory herein:
Copyright © 19772015 Mel Winfield. All rights reserved.
