Scientific
Teaching, Analysis and Research
Findings on
"Levitation in Earth's E-field" Technology
The Only Thing Left to
do is the Math
Since we know the field strength of the "earth
capacitor" the math calculations are simple and straight
forward. The measurment of capacitance is universal for both
metric and english calculations. Dielectric break down voltages
are normally expressed as volts per mil, mil being an english
unit for .001 inch. I will use english units for the majority of
the calculations. Please forgive me for this.
First and formost I believe in using standard "off the
shelf" devices when ever possible. (I don't propose to
reinvent the wheel.) A 20,000 volt DC power supply is easily
available from just about any (working) small black and white
television set. There are also a number of web sites devoted to
the "how to" on building hv machines such as the Wimshurst
machine, the Van deGraff
generator, and the like. Obtaining or making a high voltage power
supply is not an insurmountable task. But please be very, very
careful when working with voltages of this magnitude. With the described capacitor a wrong
move could mean instant death.
I will start with the assumption of a 20 kv dc power supply.
At a 1 coulomb ( A coulomb is just the unit of electrical
charge.) charge on the capacitor we would need a 50 microfarad
capacitor. (That sounds like it might be achievable.)
Capitance =charge/voltage
farads = 1 coulomb / 20,000 volts
farads = 0.00005 = 50 microfarads
To build this capacitor requires conducting plates and a
dielectric material. (Dielectric is just the insulating material
between the capacitor plates.) The plates can be constructed out
of a foil material (aluminum?). The dielectric material should be
chosen to have the highest combination factor of break down
voltage and dielectric constant. (A number of dielectric
materials and some of thier properties are listed in a table
farther on at this web site.) Common polyethelene sheeting seems
a reasonable choice at first. It has a break down voltage of 1200
volts per mil and a dielectric constant of 2.3 and it is widely
available (hardware store). But when I ran the calculations the 3
plate capacitor had to be nearly 88 feet in diameter. Not
acceptable. (Use 3 layers of 6 mil sheets ( 18 mils) to meet to
minimum 20 kv break down voltage.) There is a very good web site
devoted to capacitor construction. Visit Morning
Star II Resources and see for yourself.
The formula for constructing a parellel plate capacitor is:
C = (.224 * K *A / d) * (N - 1)
C = capitance in pico farads
K = the dielectric constant
A = the surface area of one plate in square inches
d = the seperation between the plates in inches
N = the total number of plates
50,000,000 pico farads = (.224 * 2.3 * A / .018 inches) * (3
plates - 1)
50,000,000 = 28.6 * A * (2)
50,000,000 = 57.2 *A
874125.8 = A (in square inches)
SQR (874125.8 / pi) = 527.5 inches for the radius (assuming a
circular capacitor)
527.5 / 12 = 43.9 to convert inches to feet
43.9 * 2 = 87.9 feet (double the radius to get the diameter)
Next choice for a dielectric is Mylar®polyester
film (DuPont brand name). Mylar® is not as easy to get
as polyethelene but is available through DuPont's
retail suppliers. Mylar® has a breakdown voltage
rating of 7500 volts per mil and a dielectric constant of 3.2
(Use a single 3 mil sheet to meet the minimum 20kv break down
voltage.) The calculated diameter is:
C = (.224 * K *A / d) * (N - 1)
C = capitance in pico farads
K = the dielectric constant
A = the surface area of one plate in square inches
d = the seperation between the plates in inches
N = the total number of plates
50,000,000 picofarads =(.224 * 3.2 * A / .003)*(3 - 1)
50,000,000 =(238.9 *A)*(2)
50,000,000 =(477.8 * A)
104646.3 = A (square inches)
SQR(104646.3 / pi) = 182.5 inch radius for the circular
capacitor
That is just over 31 feet diameter for the capacitor.
It would seem the size of these capacitors are unworkable
without going to a larger number of plates. If more plates were
added then the diameter would decrease proportionally. But all of
the interior plates would be shielded from the electric field and
not contribute to the resultant force.
Copyright © 1999 C.
Brauda. All Rights reserved.