by Zoltan Losonc - courtesy
Zoltan Losonchttp://feprinciples.true.ws/liftvac.htmDocument source :
The
purpose of the following examination is to find out
whether the asymmetrical electric forces could result in
a reaction-less total force upon the lifter (as in the
case of Forlov’s
T-capacitor) in perfect vacuum, and explain the
measured values at least partially. If there is such
reaction-less force, then what is its magnitude? If there
is no such force from this effect in vacuum (or
negligible), then is there any possibility to obtain a
resultant reaction-less force form electrostatic forces
in the presence of air.
For the following examination we assume that the lifter consisting of a thin corona wire and a conducting plate with rounded edge facing the wire is much longer than its greatest cross sectional dimension, and it is in perfect vacuum with no other objects or external E-fields nearby. This assumption is needed to simplify the problem to a degree when it can be easily solved. The cross-section of the lifter in x-y plain is shown in fig 1.
Since
the length of the lifter is usually much greater then its
cross-sectional dimensions, its charge distribution and
E-field will differ very little from the case of an
infinitely long lifter (except in the proximity of the
ends). The analysis of an infinitely long (in both z
directions) lifter is easier because we can neglect the
influence of the ends, but the results and conclusions
will be also valid for real lifters of finite length with
good approximation. After deriving the x component of the
resultant force per unit length, the total force upon a
real lifter can be calculated by multiplying this value
with its length. For the calculations we use MKS system
of measures.
If there is electric charge on the surface of the bodies then electric field lines will originate and terminate in these charges. Since the electric field lines are also force lines (unlike the magnetic field lines) a force will attempt to pull these charges out of the surface in the direction of the E-field. We know further that the vector of E-field is always perpendicular to the surface of a conductor in electrostatics. The magnitude of this force per unit surface (or the electrostatic pressure) can be calculated with the formula: ( To calculate the forces acting on the lifter, first we have to find the charge densities in each point on the surface of the conductors. Then dividing the whole surface into small segments we calculate the elementary forces on each of these segments. Since these forces are not parallel with the x-axis but perpendicular to the surface element, we have to calculate their x-components in order to get the thrust in x direction as shown on fig 2.
In lack of a better alternative I have written a program for this special case to solve the problem. A variant of the collocation BEM method have been used to find the charge distribution, and then calculated the forces on each of the 4 parts of the lifter. The cross section of the examined geometry for this program is shown in fig 3.
By
changing the parameters shown on the drawing we can
calculate the charge distribution and forces for
different dimensions and voltages in search for the
optimal proportions for the highest resultant force (if
any). If the examined case does not have cylindrical edge
in part 4, then we can take
Before accepting the results generated by the computer
as a valid solution, we have to find some way to develop
trust in the accuracy of the program. There are several
possibilities for this verification. Perhaps the easiest
and most convincing is to calculate the specific capacity
between two infinitely long cylinders from the obtained
charge distribution, and to compare it with the
analytically derived value. For the sake of this
comparison the following input parameters create a
geometry, which represents a very good approximation of
two parallel infinitely long conducting cylinders with
identical radius
After giving the initial input parameters, the program
slightly modifies them to insure integer number of
segments on each part, so the calculations are made
actually with the normalized parameters. The small
relative dimensions of To verify the accuracy of the charge distribution
computed by the program, we calculate the capacity of the
cylinders from these charge densities. This is done by a
separate subroutine, which multiplies the local charge
density in each segment with the length of the segment For the above input parameters the program has given the following charge distribution and capacity:
The next step is to calculate the capacity of these cylinders with the same parameters using the analytically derived exact formula that can be found in the electromagnetic textbooks: If we
substitute the given (normalized) parameters into this
formula the
Now let us see the charge distribution and calculated forces in perfect vacuum for a typical lifter (lifter 1) built by JL Naudin. The input parameters for this case are:
The following diagram shows the plotted coordinates of each segment’s center above the x-axis:
Since the cross-section of the lifter is symmetrical to the x-axis it is enough to calculate the charge distribution and forces only for the upper part. The lower part will have the same values mirrored below the x-axis, and the total force on the bodies will be 2-times the calculated forces for the upper half. Therefore only the upper part is visible in the diagram. The charge distribution of this setup is shown on the following diagram:
The numbers along x-axis represent the segment numbers
in which the charge densities are calculated. Since the
diameter of the wire and the thickness of the plate are
much smaller than the width of the plate, even by
dividing the upper surface into Using the above described method (based on electrostatic pressure) the program calculates the forces only upon the parts 1, 2 and 4, since the surface of the plate (part 3) is parallel with the x-axis and the vectors of E-field on the surface are parallel with the y-axis. This yields no force component in x direction and owing to the symmetry, the y components will cancel each other. The calculated forces on these parts are:
Although the resultant To solve this dilemma the program has been extended
with a subroutine, which calculates the forces using the
Coulomb’s law and gives a more accurate final
result. The third column shows the forces calculated this
way. According to this method the program chooses a
segment, and calculates the x-component of the E-field
intensity created by the charges from all other segments
at its center. Then multiplying it with the charge per
unit length (in z direction) present in the chosen
segment an elementary force is obtained. Calculating and
summing up all these elementary forces, we get the total
force (per unit length) in x direction. The resultant By increasing the number of segments the accuracy should also increase. If we take n = 3000 (6000 segments covering the whole geometry) instead of n = 1000, we get the following results:
In this case 24 segments cover the wire and 12 the
edge of the plate which gives higher accuracy. Comparing
this table with the table 4. we can see that The
order of magnitude of the
Mr. Stein has proved in his paper “
According to Coulomb's law the force The explanation for this is that the charges on the surface of a conductor are not in the form of an infinitely thin layer of charge, but the layer has a finite thickness. Therefore they build a thin layer of space charge below the surface as shown in the following figure: *Fig. 5.*
The E-field intensity beneath the surface (around the
charges) is not constant but it changes linearly from
zero to
The fundamental rule of the electrostatics says that
no tangential electric force component can exist on the
conductor’s surface, since then the charges would
move in the direction of the force (which is not a static
case) until the equilibrium is attained. That means,
since the program calculated a small negative tangential
E-field and force |