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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 45o
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.
The Magnus effect has been shown to be the mechanism that is responsible for what is known as gravity. Levitation can be accomplished by reversing the nucleonic spin so as to reverse the effect of the waves of energy, emanating from the Earth, that creates gravity. 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 cylinder17, 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 4 In our case, the
cylinder becomes the proton with B being its diameter. The application is valid
since it is said17 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 = 4 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 [ The proton spin
angular momentum is given by [3]1/2 X h/4 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 109 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 To compare this
with gravitational force, an altered version of Newton's formula shall be used
which is [circle G]m/r2 =
F. Circle G [ EARTH WAVE FORCE Next, it has to be determined how
this amount of force is provided by the Earth waves. This is dependent upon the
radius of the proton. Though the radius of the nucleus has been experimentally
established2 it appears
that the proton radius was established by calculating a volume for the nucleons
that fitted into the volume of the nucleus according to the current theory of a
nucleus with nucleons stacked within it. At the Argonne Laboratories, Dr.
Krisch found that the proton had a hard core about three times smaller than the
proton in cross-section4.
Since the accepted radius of the proton is 1.2 X 10-13 cm. it can
be determined from this that the hard core radius is 6.92820323 X10-14
cm. Let it
be suggested that
this is the
real proton radius
and that this
proton is surrounded by an atmosphere of
etherons, due to its rotation, such as that which has been described. The energy waves
emanating from all the nucleons in the Earth [the strength of which is divided
by the Earth's radius squared] are traveling through the etheron medium, as in
space, but impeded by air molecules at the Earth's surface. Below the Earth's
surface there should be more impedance due to more obstacles creating a longer
path for the wave to follow. The velocity of light through air will be taken as
the velocity of Earth's gravitational waves, which is 2.9970254 X 1010
cm./s. The rotation of the waves as they rise from the Earth's surface [thus
moving at right angles to the forward motion] could manifest as the magnetic
field. As mentioned,
the formula for the Magnus effect is 4 Before
attempting to determine the force of the gravitational wave, its mass has to be
established that is directly involved in striking the area of the nucleon
receiving the force. This would be the cross-sectional area of the proton, for
example, thus p, which is in gms./cm.3 has to be multiplied by
the said area. Next, the penetration depth of p, which is per cm.2
X 1 cm. in depth, has to be defined as the proton diameter since that is the
length [depth] of wave that covers the proton or nucleon. The wave velocity and
the nucleon orbital velocity combined [u] is 3.154941522 X 1010 cm./s which moves
6.92820323 X 10-14 cm.
[the proton radius / the combined velocities] that it takes for the proton
radius to encounter the wave. Average t [t/2] is one half of the total t that
it takes to cover the whole diameter along the proton's y axis because the
force starts at zero and doesn't become a full force until the diameter is
covered. This makes the average force equal to that upon coverage of the
radius. The mass of the Earth wave striking the proton is p times the cross-sectional area of the proton X the proton
diameter [the depth] which equals 5.663668952 X 10-56 gms. To verify the density value, it could be derived
by using a volume of the wave covering the proton, this being To find the
average force from the momentum [mv] of the Earth wave, the formula average
force = A
final method is to go directly to the Magnus formula itself which is 4 Note: Forces have been expressed
here as dynes simply because the units commonly used are the cgs units, the
centimeter, gram and second. Recall that 1 dyne = 1 gm.cm.s-2 = 10-5
Newton. GYROSCOPIC ACTION To show that gyroscopic action
does not change the gravitational calculations of the Magnus effect, the
following analysis shall be given. A brief, but clear, explanation of
gyroscopic motion is given in Physics, a textbook for colleges, fourth edition,
by Oscar M. Stewart, 1944 [herein called Fig. 4] as follows: "In fig. 105 the wheel WW'
is rotating about the line AA as an axis. The linear velocity of a point O on
the front side of the outside edge is represented by the vector OB. Imagine
that the wheel is given an impulse which tends to move the upper end of the
axis AA toward the eye. This impulse will give to the point O a linear velocity
which will be downward [represented in Fig. 105 by the vector OC]. The
resultant motion of the point O is represented by the vector OD,
which is the vector sum of OC and OB. In order that the point O may move in the
direction OD, the axis of the wheel must shift to the position A'A', which lies
in the plane of the paper. In this case a force is applied which tends to pull
the upper end of the axis out toward the eye instead of moving in that
direction, it moves at right angles to the force." At first glance there seems to be
a similarity since there is a spinning mass, an external force and a resulting
vector force. However, they differ as to the point of application of the force
and therefore of the result. The gyroscope has a force applied to the end of
its spin axis attempting to change its axial orientation. The proton has the
force applied evenly at right angles to one side of its perimeter at 9 o'clock
thereby moving the whole particle at once. GYROSCOPIC WEIGHT REDUCTION
Reference is now made to the
Physical Review letter by Hideo Hayasaka and Sakae Takeuchi Volume 63, Number
25, 18 December, 1989. The abstract states "The weight change of each of
three spinning mechanical gyroscopes whose rotor masses are 140,175 and 176
gms. has been measured during inertial rotations -- ". "-- the right
rotations [around the vertical axis] of each gyroscope always cause weight
decreases of the order of miligrams, proportional to the frequency of rotation
at 3000 to 13000 r.p.m.
The weight reduction
occurs for both normal and reverse attitudes. -- [here reverse attitude of a
gyroscope means merely upside down attitude, or orientation]. On the other
hand, the left rotations of each gyroscope yield zero weight change for all
frequencies of rotation and both attitudes. --The experimental results cannot
be explained by the usual theories." The result of
this experiment can be accounted for by the explanation of gravity contained
herein. The attitude of the gyroscope would not make any difference because the
constituents, within its atoms, of neutrons and protons [nucleons] would always
spin with a horizontal axis, due to the Earth's vertically rising waves, just
as does a ping pong ball on a jet of air. In this attitude the nucleon is
naturally forced towards the source of waves, i.e. the Earth, by the Magnus pressure
as described. If the Magnus pressure was from any other direction, this is the
direction of any object's motion. Ordinarily,
nucleons in matter will spin with a horizontal axis, as previously mentioned,
as the waves from the Earth move upward and drive etherons against them. In the
Earth's field they are generally spin aligned, end to end, creating the Earth's
magnetic lines of force. In
the case of a spinning rotor, as in this experiment, this horizontal attitude
is not changed by other forces. The nucleons still have a horizontal axis. With
respect to the rotor, the nucleon always spins in the same direction but as the
rotor rotates 180o the axis rotates 180o to the observer
above or with respect to the Earth's field without [i.e. its orientation in
space]. The torque on the nucleon's axis by the Earth's field and its own
gyroscopic resistance to reorientation in space are both offset on the opposite
side of the rotor. nucleons in
rotor The
nucleons are drawn in the diagram as wheels on an axle. The dot is the point of
a rotation arrow coming up out of the paper and the plus sign is its tail
entering the paper. The rotor is in a horizontal plane with a view from above.
When the gyroscopes were spun clockwise, as viewed from above, the researchers
found no change in their weight. When they were spun counter-clockwise they
appeared to lose weight. As the rotation of the rotor, as shown, carries the
nucleon with the dot end into the etheron field, etherons are forced over the
nucleon, by its spin, causing a pressure beneath that lifts. In the opposite
direction, with the plus sign end leading, the rotor etherons are carried below
which would cause a downward pressure.
In the direction shown, the etherons are moving up and over on the dot side
which causes an upward force at the bottom that works against the acceleration
of the nucleon towards the Earth. When the force is
with the nucleonic orbit, in the same direction as spin, as shown, the lift is
manifested and this subtracts from
the downward force
exerted by the
Earth wave thus causing a weight loss. When the force
is in the opposite direction to the nucleonic orbiting, as in opposite rotor
rotation, the force cannot accelerate the nucleon. Thus, in
this direction of the rotor rotation, there is no change of weight of the
nucleons or of the rotor which is composed of them. Due to this concept that explains
the result of the said experiment in detail, a weight loss occurs when the
rotation of the rotor does not cause an opposition to the orbital rotation of
the nucleon and not in the opposite direction. The Physical
Review letter states "a rotor is composed of various materials and
domains". Due to this, one can only approximate the values of the average
nucleon in the rotor including its radius of gyration. To calculate the latter,
the use of its actual radius is probably more accurate than the calculated
equivalent radius. Variations in the materials composing the silicon steel
rotor can be reduced by using the aluminum rotor which is said to have had
"nearly" the same weight loss even though it has one gram less mass.
Any aluminum alloy should have a greater proportion of aluminum than that
of iron in the silicon steel rotor. No
mention is made of the composition of the rotor shaft. The mass of the
average nucleon in the aluminum atom can be calculated from the atomic weight
of 26.9815 u. When the mass of the electrons [13 X .00055u =.00715 u] is
subtracted, the mass of the nucleus = 26.97435u. Multiplied by 1.66057 X 10-24 gms., the mass = 4.479279638 X 10-23
gms. When divided by 27 nucleons, the average mass of each nucleon is
1.658992459 X 10-24 gms. R, the radius of gyration, for
the aluminum atom and the radius at which the average nucleon would be located,
equals the sq. rt. of r2 X 2/5. r varies from 1.4 X 10-13
for the light elements to 1.2 10-13 for the heavy elements times the
mass number to the power of one third. 1.3 X 10-13 X 271/3
will be used. Thus R = 2.466576575 X 10-13 cm. The
angular momentum of the atomic nucleus would be the sum of that of the 27
nucleons. Protons have an angular momentum of 9.13300574 X 10-28
gm.cm2/s and neutrons have 6.255438586 X 10-28 gm.cm.2/s.
13 protons and 14 neutrons total 2.063052148 X 10-26 gm.cm.2/s.
Angular momentum = mvr. From this v of the average nucleon is found to be
angular momentum divided by atomic mass [4.479279638 X 10-23]
divided by R = 1.867271322 X 109 cm./s. The only concern
is the nucleon's horizontal movement into the etheron field where it meets
etherons to force them over it by its rotation. If the atom was stationary, the
nucleon would not meet new etherons. Thus the rotation of the rotor is
essential to a reduction of weight. The speed of the rotor is so low, in
comparison, it adds or subtracts very little velocity to or from the nucleon.
It is mainly to bring the nucleon into the field and to orient the spin axis at
90o to the flow of etherons. Waves from the horizontal motion do not
interfere with waves from the Earth because past a critical angle waves pass
through each other without distortion. Two forces can also act in different
directions at once, such as those of the sun and moon when acting on the Earth. Only
the forward side of the nucleon can encounter the etherons. The lower quadrant
is occupied by the Earth stream so this leaves only the upper quadrant. The
nucleon motion is vertical at the equatorial position and 100% horizontal at
the polar position. The average horizontal motion, between 0% and 100% is 50%
of this distance which is 1/8 of the circumference. If the nucleon is covering,
in horizontal distance, only 1/8 of the circumference each time that the nucleon
goes around the orbit, the equivalent velocity is 1/8 of 1.867271322 X 109 cm./s which equals
2.334089153 X 108 cm./s. The rotor radius
is 2.9 cm. 1/2 r2 = R2 so R = 2.050609665 cm. The
circumference X 13000 r.p.m. = 2791.611447 cm./s. If this is subtracted from
the above velocity, since etherons are moving in the same direction, the
relative velocity is 2.334061237 X 108 cm./s. If the Magnus formula,
F = 4 The Earth's
gravitational force on the average aluminum nucleon equals F = To calculate the
percentage of weight loss there is, the Magnus force X 100 divided by Earth's
gravitational force = 9.134030291 X 10-21 dynes = .00539%. This is
nearly equal to the experimental result of .00678% of weight loss for the
silicon steel rotor as it was said in the said Physical Review letter. It is
nearly equal despite a possible slight error due to the constituents of the
rotor, as mentioned, or the average Earth's gravity due to location of the
experiment. However, at Japan it is hardly noticeable. There is a change in weight
loss in proportion to the change in rotor speed. The ratio of low to high speed
= 3000/13000 = .23. The ratio of low to high weight loss is 2.6/11.9 = 2.18.
The reason for this is that with increased speed, the number of etherons
encountered by the nucleon increases for the same distance across the nucleon.
Thus the density p increases which
creates greater Magnus force. A greater Magnus force due to the density
increase by rotation would most likely increase the percentage to that of the
experiment. . The more
neutrons relative to protons there are in an atom, the less angular momentum,
and thus velocity, there is for the average nucleon. Consequently, the slower
the element will fall as shown by the dropping experiment. If one calculates
the velocities of the average nucleon within the atom at the radius of
gyration, one finds that the velocity is greater in the silicon atom, for
instance, that has 14 protons and 14 neutrons than it is in the iron atom that
has 30 neutrons and 26 protons. This is because that when the Earth wave meets
the nucleon moving downwards the two velocities are added to get the relative
velocity. Thus the silicon nucleon has a faster stream moving on one side of
its surface than that of the iron nucleon and this produces a greater force
which, in turn, causes a greater gravitational acceleration and the silicon
will fall faster. In the case of
aluminum, the nucleonic speed around the average orbit was calculated to be
1.867271322 X 109 cm./s. For iron, it was calculated to be 1.579161219
X 109 cm./s.The flow is faster on one side of the Al nucleon, when
added to the Earth flow, than on one side of the Fe nucleon. The faster flow
produces greater Magnus force and thus the Al falls faster than Fe. Anomalous Gravitational
Acceleration To the Sun by Spacecraft In a similar manner to that of
the previous experiment a solution is now offered, utilizing the Magnus
formula, to the following problem: According to John D. Anderson et
al in the Physical Review letters, 5 October 1998, Volume 81, Number 14, data
from Pioneer 10, Pioneer 11, Galileo and Ulysses indicated an apparent
anomalous, constant, acceleration towards the sun acting on the spacecraft,
with magnitudes of 8.09 X 10-8 cm./s2 [Ap] and 8.56 X 10-8
cm./s2 for Pioneers 10-11 respectively and 12 X 10-8
cm./s2 for Ulysses. The Pioneers were launched in 1972 and 1973.
Their orbit determination program analysis of unmodeled accelerations began in
profile with distance. The conclusion was that Ulysses was
subjected to an
unmodeled acceleration towards the sun of 12 X 10-8 cm./s2. It is said that no
plausible explanation was found so far. It was said that
no magnitude variation of Ap with distance was found with the Pioneers but
there was a difference between 10 and 11, which were structurally the same, and
with Ulysses, though the difference was small. Since they were moving through
different areas of space, a difference in density of medium particles could
account for this small discrepancy. It has been
stated herein that a medium of etherons could account for the transmission of
electro-magnetic energy waves with a greater density causing a greater
velocity. It has also been shown that such etherons [kaytrons] could be the
"dark matter" that provides the 90% of the mass required by our
galaxy now deemed to be missing and that such density causes the average
velocity of light, in our galaxy, to be six times that measured on our Earth.
The density of etherons would be at a maximum between stars as the stars [suns]
and planets drive etherons away from them by the emission of energy waves. As
energy waves move outward they decrease in energy according to the inverse
square law and thus etheron density will gradually increase with distance. The
etherons moving outward would be replaced by others entering the center of the
vortex of etherons orbiting the sun or planet en masse just as our atmosphere
moves at one with our planet as it rotates. The etherons enter the vortex
center just as the electrons etc. from the sun enter our atmosphere at the
poles. Also, just like our atmosphere,
it will contain currents and slightly different velocities over different
areas. The solar wind
would be one big influence in this respect as it would partially interact with
the Aether [the collective name]. This explanation involves only the orbital
velocity of the spacecraft and of the etheron field. The solar wind, which is
nearly radial would have some orbital component which would coincide with that
of the etherons. The radial velocity is perpendicular to the orbit and so does
not enter into the following calculations. Any Magnus effect, laterally, from
this radial velocity would be minimal because it is too slow to affect the
energy waves from the sun travelling at light speed in the same direction. The orbital
velocity of the etherons minus that of a spacecraft is the relative velocity
that would be the only other influence on radial acceleration besides the sun's
waves. First, the orbital component of the velocity of the Aether, which would
be similar to this component of the solar wind, will be determined. The orbital
velocity at the sun's surface shall be the escape velocity since the particles
have to leave the sun before orbiting around it. Escape
velocity from Earth is usually given as [2GM/r]1/2. [GM/r]1/2
is the velocity of of an object when there is a balance between gravitational
force and centrifugal force. GM is This
escape velocity should not be confused with the explanation for a stable orbit
of a planet, given later. The planets were born in an orbit where centrifugal
force was greater than gravitational force. When they were freed they moved out
to the balanced position. Momentum would carry them past where the forces were
reversed which drew them back. Oscillation would take place until they settled
in. In
this case, particles leave the balanced orbit , no doubt due to the said
pressure and emission by the sunspots and continued pressure by the faster
light waves is maintained. The extra velocity given to a spacecraft from Earth
ensures that it escapes Earth's gravity until the sun's gravity becomes
stronger and takes over. Next,
the velocity at the craft's orbit [rb]
needs to be determined. There is a relationship in the ratios of the orbital
velocity and the orbital radius evident in the movements of the planets. This
could be expressed as r is proportional
to 1/v2 or 1/v2 [orbit b] / 1/v2 [orbit a] = rb/ra. The following ratios are an
example: Planet r [ Earth =
1] 1/v2 [Earth = 1] Earth 1 1
Mars 1.52647478 1.52647479 Jupiter 5.201003829 5.201003831 Saturn 9.538003703 9.538003705 [The planetary data was taken
from calculations in Section II with v = circumference divided by period]. PIONEER 7 The sun's radius is 7 X 1010 cm. but before using the
above equation, the craft's orbital radius has to be determined. The most
accurate data of such radius would come from Pioneer 7 which was launched to
study the sun's radiation. When a craft is launched it follows the arc of a
parabola but after a million miles the sun's influence takes over and its orbit
becomes a heliocentric ellipse. It will be assumed that the gravitational
anomaly calculated for Pioneer 10 will also apply to pioneer 7. It
is given that the perigee [P] of Pioneer 7 is 1.010 AU which, when multiplied
by 1.4959787 X 1013, =
1.510938487 X 1013 cm.
The apogee [A] was 1.125 AU which, = 1.682976038 X 1013 cm. The period was 34810560s. [P+A] /2 = the
semi-major axis = 1.596957263 X 1013
cm. = a. a - perigee [P] = 8.6018776 X 1011
cm. which is the base [y] of a right triangle with the side x which is the
semi-major axis [b] of the ellipse. r = hypotenuse of triangle which = a. y/r = sine e
= [a2- b2]1/2 divided by a = .0538642077 Circumference of an ellipse, C, =
2 Going
back to the equation; 1/vb2 divided by 1/va2 = rb
/ ra, vb/va = 228. 1/vb2
= 228 X 1/[4.78 X 107]2. [1/va2 was previously determined to be 4.78 X 107]. 1/vb2 = 9.978816898 X 10-14 cm.2/s2. Vb = the velocity in the orbit of Pioneer 7 = 3.165632 X
106 cm./s which is the
orbital velocity of the etherons encountered by Pioneer 7. The
velocity of Pioneer 7, in the same direction, is 2 The
orbital velocity of the solar wind and etherons would not be the same as that
of Earth, as suggested by two University and Astronomy professors, because the
Earth is in a relatively stable orbit. If one uses the said equation with
Earth's orbital velocity, it is found that at the sun's surface the velocity
would have to be that of a stable orbit which is [ If
b represents Earth velocity then, using the sun surface escape velocity, 1/vb2 divided by [1/
4.8 X 107]2 = 1.4959787 X 1013
cm.2/s2 divided by 7 X 1010 cm.2 and vb
= 3.283428416 X 106
cm./s. This is a little faster than the Earth velocity which is determined by
Section II data to be 2.986631488 X 106
cm./s. The
etherons in Pioneer's orbit strike its nucleons from behind and sweep by on the
sunward side. This would cause the nucleon to align its axis vertically and
rotate clockwise as viewed from above. This would mean that the etherons
striking the nucleon from behind would be rotated to the sunward side thus
adding to the acceleration towards the sun due to the Magnus effect. If Pioneer 7
showed an anomalous acceleration of 8.09 X 10-8 cm./s the same as Pioneer 10 and if its components
from iron to plastic had nucleons of the average mass of aluminum, which is
1.658992459 X 10-24 gms.
each, the force involved [ma] = 1.342124899 X 10-31 dynes on the average nucleon. By the Magnus formula
this force = 4 It
is expected that etheron density would be greater near the spacecraft than near
the Earth's surface. Gravitational force [ The
calculation is as follows: 1.342124899 X 10-31 / 4 p = 2.73417673 X10-17 gms/cm3. This is close enough to
the 2.71 X 10-17 gms/cm3 as determined for the
density of etherons on Earth to verify the accuracy and the reason for the
anomaly and to once again show that the Magnus effect formula can be effective
in calculating gravitational 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 neutron-proton 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 neutron-proton 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 Boynton18, The aluminum should, as
it did, turn towards the mountain in front of which it was hung. This is
contrary to Fischbach's idea13
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] Fe 30
25.273899 26 51.273899 .9180644405 1. Pb 126
106.1503758 82 188.1503758 .9081054868
.989152228 TABLE 1a Wt.
Calculated
Experimental drop time ratio Contin. Ratio Inverse [Average of 20 drops each] Fe = 1 Fe 1. 1. Cu 1.0029 1.0029 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 counter-rotating 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
counter-rotation 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:
Additions >>
Copyright © 1977-2008 Mel Winfield. All rights reserved.
GRAVITATION
The key piece of equipment is an electron generator, in our case a Van der Graaf generator, that fires electrons at a target that we intend to levitate. It is said to produce a static charge which, by definition, is static or stationary. However, as the charge builds up on the sphere, electrons are emitted due to mutual repulsion. These can be detected at a distance from the sphere. Once emitted, these electrons are independent of the sphere. When they strike the target, their energy is absorbed by the nucleons near the surface of the target. The nucleonic spin axis is reversed due to the direction from which the electrons come [similar to a ping pong ball spinning on a jet of air]. The Magnus effect is now reversed and causes the target to move towards the electron source direction, which is the generator sphere. If this sphere is stationary will stop at its closest approach. If the sphere was attached to the target, The motion of the target would move the sphere which, if it continues firing electrons, would continue to move the target from the force of the targets own nucleons in the direction of the sphere. If the sphere is above or even partially higher than the target, the target will levitate.
If the generator were the only piece of equipment the electrons would only penetrate the skin of the target and would not be able to reach over 50 per cent of the nucleons in the target that is necessary to have the target free of the Earth's gravitational effect. However, it would lose weight. This is shown in a video where the target is on a balance scale. Both sides of the scale are affected but with one farther away than the other there would be more electrons affecting the closer one due to the inverse square law. The difference between the two levitates the closer one towards the sphere showing a loss of weight. This takes place consistantly every time the generator is turned on.
To accomplish full levitation, the next key piece of equipment is a special coil that emits a pure electrical field without a magnetic component. This field is composed of spiraling energy waves emitted from the coil. These waves add energy to drive the electrons from the Van der Graaf With much greater force which, when it is great enough, will cause the energy to reach over 50 per cent of the targets nucleons, past the electron barriers around each atom and thus cause full levitation. Of course, frequency also plays an important part and a special unit has been built to control this. The electrons must strike the nucleon at a particular point in their orbit around the nucleus so as to accelerate it rather than retard it. Thus the frequency of the electron pulses should equal the period of the atomic vortex. Levitation occurs when the electron wave phase is adjusted so as to strike the nucleon in its position where acceleration can take place. The frequency can also be set on the harmonics of the main one. For example, the electron pulse could strike the nucleon every second time around. However, the closer it is to the prime frequency the faster and more powerful is the thrust.
In the future, especially when electrons are replaced by finer particles, such as xrays, the worlds energy requirements will be met at minimal cost. This author has already designed an electrical generator powered by nucleonic energy that will change the world due to minimal cost as long as leasing fees , especially in the third world countries, are kept much below standard energy supplies.
RU. R = the radius of the cylinder [proton]. The
maximum peripheral velocity of the flow is 4U at the top. 
Pr2U2. Since, in our model, the
proton is descending on the left side of its vertical orbit, counter-clockwise,
while rotating counter-clockwise [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 90o 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 45o 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.
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 r2]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.
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 counter-acting] almost one third of the angular
momentum from its proton. This agrees with accepted data.
/
1.579161219 X 109 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 1/2at2 which equals .5 X
982.9514643 cm./s2 X
[2.88824474 X 10-22]2s = 4.099869758 X 10-41cm. 
=
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./s2
=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.
] will
be defined in PART II, Cosmology, and is equal to 3.989644343 X 1020 cm.3/s2.
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.
pr2u2. If this is
to be equal to the force of 1.627383338 X 10-21 dynes [determined
from the nucleon's angular momentum and g] that has been shown to be
gravitational force, pr2
has to equal 1.301060145 X 10-43 gm.cm.2 per cm.2.
The density p would then equal
2.710541969 X 10-17 gms./cm.3. r = nucleon radius. In
this formula, u is the combined velocity of the wave and of the proton moving
against it. The latter was derived from its angular momentum. 
r2 X d of the proton which =
2.089496868 X 10-39 cm.3. p = m/vol. =2.710541969 X 10-17
gms./cm3 which is the
same result.
mv/t
is used. t = the stopping time by the proton while the semi-diameter of its
cross-section encounters this portion of the wave. t = 2.195984674 X 1024s.
The force is continuous as the proton encounters the next portion of the wave
and so on. Average force = 5.663668952 X 10-56 gms. X 3.154941522 X
1010 cm./s divided by t which equals 8.136916689 X 10-22
dynes. Multiplied by 2 [since this calculation is for coverage of one half of
the nucleon in average time by a portion of the wave whereas the full area of
the nucleon is covered at all times due to the expanse of the Earth field] the
full force = 1.627383338 X 10-21 dynes. This is an exact correlation
to the 1.627383338 X 10-21 dynes of the known gravitational force
and it shows that the Magnus effect is the true cause of gravity.
pr2u2 = F. pr2 = 1.301060145 X 10-43
gm.cm.2/cm.3 and u = the combined velocity of 3.154941522
X 1010 cm./s. By this formula the force of the Magnus effect =
1.627383338 X 1021 dynes.
pr2u2, is used
where p = 2.710541969 X 10-17
gms./cm.3, r = proton radius = 6.92820323 X 10-14 cm. and
u = 2.334061237 X 108 cm./s then F = 8.907005689 X 10-26
dynes. The density p of the Earth
wave can be used for the Magnus formula for aluminum as well as for any other
element because it only relates the nucleon to the Earth stream.
m/r2
= 1.72279986 X 10-21 dynes. This would make Al fall faster than Fe
as will be shown in the dropping tests. F for iron was only 1.627383338 X 10-21
dynes.
s,
calculated in Section II as 1.33440816 X 1026
cm.3/s2. If 1.2 is used, instead
of 2, as the lowest escape velocity from the sun, the result is 4.782841 X 107 cm./s. The formula,
[2GM/r]1/2, is given for
the escape velocity for Earth with the 2 inserted to make sure that it doesn't
settle back into a balanced orbit. The sun is much more massive than the Earth
and thus radiates much stronger nucleonic waves as well as creating strong
light pressure. This extra pressure would extend well beyond the sun's surface
and would tend to ensure continuous escape. Thus the reduced escape velocity
from the sun, given herein, is logically within reason.
= [a-p]/a = 5.38641690 X 10-2. y = a-p. y/b = tan 
= .053942479. Therefore 8.6018776 X 1011/ .053942479 = b =
semi-major axis = 1.594638912 X 1013.
The sum of two focal radii = 2a, therefore, when equal, one radius = a =
hypotenuse.
a [1-e2/4 - 3e4/64 -45e6/2304].
C= 1.002669639 X 1014 cm.
C/2
= r =
1.595798293 X 1013 cm.
[which is the average radius].
r
divided by the period which is 2.880360554 X 106 cm./s. The relative velocity is, when this is
subtracted from the etheron orbital velocity, 2.85271 X 105 cm./s. This is the velocity at which the etherons
strike the nucleons of Pioneer 7 from behind [since they are moving faster than
the craft].
s/rs]1/2.
All particles of the solar wind would have to have a greater velocity to escape
and measurement of its velocity in Earth's orbit shows that they are still in
the escape mode and therefore must move a little faster than the Earth.
pr2u2 as previously expressed.
The only uncertain value is that of the density p which was previously determined to be 2.710541969 X 10-17 gms. per cm.3 here on Earth. r is the
radius of the nucleon which is 6.92820323 X 10-14 cm. as previously determined. U is the
aforementioned velocity of 2.85271 X 105
cm./s. p shall now be calculated to
see how close it is to the Earth calculated value when the anomalous
acceleration value is used. It should be fairly close because light speed varies
in different densities but this would become apparent in areas not too far from
Earth if there were any extensive errors in distance due to this. Earth waves
drive away etherons, lowering the density, but so do the sun's waves outside of
the protective magnetic envelope around Earth. This envelope is also the reason
that planets would not exhibit the anomalous acceleration that shows up with
the spacecraft.
m /r2] on the craft [63049 gms.]
was 187 times stronger on the Earth than that of the sun in space. This would
mean that waves from the Earth would have more force to drive away etherons
than the sun would in space, thus allowing a greater density in the solar area.
The amount of greater density would depend upon the interaction of the large
particles of the solar wind with the etherons and kaytrons.
[6.92820323
X 10-14]2.[2.85271 X 105]2 = pAl 14 11.7944862
13 24.7944862 .9189950408 1.001013655
Cu
34.6132 29.16456301 29 58.16456301 .9154007398 .9970985689
TABLE 1bAl .999 .999
Pb 1.010 1.01
This page last modified May/2008.
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