Black Triangles: Triangular spacecraft
US Patent 20060145019 A1
Abstract: A spacecraft having a
triangular hull with vertical electrostatic line charges on each corner that
produce a horizontal electric field parallel to the sides of the hull. This
field, interacting with a plane wave emitted by antennas on the side of the
hull, generates a force per volume combining both lift and propulsion.
Images
Fig 1
Fig2
Fig 3
Claims(4)
1. A spacecraft comprised
of the following components:
(a) a triangular hull in the form of an equilateral
triangle;
(b) two copper plates attached on opposite vertical sides
at each of the three corners of the hull (1 a) such that a sharp
vertical edge is formed where they come together;
(c) an electrostatic generator used to charge the back
two coppercladded corners (1 b) to a high positive voltage, and
the third forward coppercladded corner to a high negative voltage;
(d) a horizontal slot antenna array mountedon the sides
of the hull; and
(e) a frequency generator, antenna and coaxial cables to
drive the antenna array (1 d).
2. To create, by claims (1
a, 1 b, 1 c), an intense vertical line
charge at the corners (1 b) and a horizontal electric field that
that is parallel to the sides of the hull (1 a);
3. To create, by claims (1
d,1 e), an electromagnetic wave with a vertically
polarized electric field traveling outward from the side of the hull (1 a);
and
4. To create, by claims (2,3),
an interaction of the electrostatic field (2) with the electromagnetic
wave (3) such that a combined spacetime curvature pressure is generated
on the hull in the upward and forward direction to produce lift and propulsion
respectively.
Fig 4
Fig 5
Fig 6
Fig 7
Description
BRIEF SUMMARY OF THE INVENTION
This invention is a spacecraft having a triangular hull
with vertical electrostatic line charges on each corner. The line charges
create a horizontal electric field that, together with a plane wave emitted by
antennas on the side of the hull, generates a force per volume providing a
unique combination of both lift and propulsion.
BACKGROUND of the INVENTION
Referring to FIG. 1,
the spacecraft has a hull in the shape of an equilateral triangle. A parabolic
antenna (E) is centrally located in the bottom of the hull. An array of
horizontal slot antennas is located along the side of the hull (A). Each back
corner (F,G) has a corner conducting plate which is charged to a positive
voltage +V. The forward corner (C) has a conducting plate charged to a negative
voltage −V. A motion control hemisphere (D) is located on the bottom surface in
each of the three corners.
Referring to FIG. 2,
two planes (A,B) intersect at the origin O at an opening angle β. Each plane
(x,y) is charged to a voltage V. The potential at point P is determined in
polar coordinates {ρφ}. The Laplace equation for the potential Φ in polar
coordinates is given by:
$\frac{1}{\rho}$ ∂ ∂ ρ ( ρ ∂ Φ ∂ ρ ) + 1 ρ 2 ∂ 2 Φ ∂ ϕ 2 = 0
Using a separation of variables solution, the potential is given as the product of two functions:
Φ(ρ,φ)=R(ρ)Ψ(φ)
which when substituted into the Laplace equation becomes:
$\frac{\rho}{R}$ ⅆ ⅆ ρ ( ρ ⅆ R ⅆ ρ ) + 1 Ψ ⅆ 2 Ψ ⅆ ϕ 2 = 0
Since the two terns are separately functions of ρ and φ respectively, each one has to be constant with the sum of the constants equal to zero:
$\frac{\rho}{R}$ ⅆ ⅆ ρ ( ρ ⅆ R ⅆ ρ ) = v 2 1 Ψ ⅆ 2 Ψ ⅆ ϕ 2 =  v 2
These two equations have solutions:
R(ρ)=aρ ^{v+bρ} ^{−v }
ψ(φ)=Acos(vφ)+Bsin(vφ)
The azimuthal angle φ is restricted to a value in the range 0≦φ≦β. The boundary condition is that the potential Φ is equal to V for any radius ρ when φ=0 and φ=β. This means that v has to be an integer value of π so that the sine function is zero:
$\mathrm{sin}$ ( v β ) = sin ( m π β β ) = sin ( m π ) = 0 m = 1 , 2 …
which in turn means that the coefficient A of the cosine term has to be zero in the solution above. Choosing b=0 makes the general solution for the potential equal to:
$\Phi $ ( ρ , ϕ ) = V + ∑ m = 1 ∞ a m ρ m π / β sin ( m πϕ / β )
which shows that when the angle is zero, the sine is zero and the potential is V. If the angle is β, then there is a multiple of π such that the sine is zero again.
$\frac{1}{\rho}$ ∂ ∂ ρ ( ρ ∂ Φ ∂ ρ ) + 1 ρ 2 ∂ 2 Φ ∂ ϕ 2 = 0
Using a separation of variables solution, the potential is given as the product of two functions:
Φ(ρ,φ)=R(ρ)Ψ(φ)
which when substituted into the Laplace equation becomes:
$\frac{\rho}{R}$ ⅆ ⅆ ρ ( ρ ⅆ R ⅆ ρ ) + 1 Ψ ⅆ 2 Ψ ⅆ ϕ 2 = 0
Since the two terns are separately functions of ρ and φ respectively, each one has to be constant with the sum of the constants equal to zero:
$\frac{\rho}{R}$ ⅆ ⅆ ρ ( ρ ⅆ R ⅆ ρ ) = v 2 1 Ψ ⅆ 2 Ψ ⅆ ϕ 2 =  v 2
These two equations have solutions:
R(ρ)=aρ ^{v+bρ} ^{−v }
ψ(φ)=Acos(vφ)+Bsin(vφ)
The azimuthal angle φ is restricted to a value in the range 0≦φ≦β. The boundary condition is that the potential Φ is equal to V for any radius ρ when φ=0 and φ=β. This means that v has to be an integer value of π so that the sine function is zero:
$\mathrm{sin}$ ( v β ) = sin ( m π β β ) = sin ( m π ) = 0 m = 1 , 2 …
which in turn means that the coefficient A of the cosine term has to be zero in the solution above. Choosing b=0 makes the general solution for the potential equal to:
$\Phi $ ( ρ , ϕ ) = V + ∑ m = 1 ∞ a m ρ m π / β sin ( m πϕ / β )
which shows that when the angle is zero, the sine is zero and the potential is V. If the angle is β, then there is a multiple of π such that the sine is zero again.
Because the series involves positive powers of the
radius, for small enough ρ, only the first term m=1 in the series is
important. Thus around ρ=0, the potential is approximately
φ(ρ,φ)≈V+a,ρ^{π/β}sin(πφ/β)
φ(ρ,φ)≈V+a,ρ^{π/β}sin(πφ/β)
The electric field component is the negative gradient of
the potential:
${E}_{\varphi}$ ( ρ , ϕ ) =  1 ρ ∂ Φ ∂ ϕ =  π a 1 β ρ ( π / β )  1 cos ( πϕ / β )
The surface charge distribution σ at φ=0 and φ=β is equal to the electric field perpendicular to the surface times the permittivity of space ε_{0}:
$\sigma $ ( ρ ) = ɛ 0 E ϕ ( ρ , 0 ) =  ɛ 0 π a 1 β ρ π β  1
Notice that if angle of intersection β is less than π, then the equation says that there is a very small radius to a positive power which means little charge density accumulation.
${E}_{\varphi}$ ( ρ , ϕ ) =  1 ρ ∂ Φ ∂ ϕ =  π a 1 β ρ ( π / β )  1 cos ( πϕ / β )
The surface charge distribution σ at φ=0 and φ=β is equal to the electric field perpendicular to the surface times the permittivity of space ε_{0}:
$\sigma $ ( ρ ) = ɛ 0 E ϕ ( ρ , 0 ) =  ɛ 0 π a 1 β ρ π β  1
Notice that if angle of intersection β is less than π, then the equation says that there is a very small radius to a positive power which means little charge density accumulation.
Referring to FIG. 3,
the value of β, in the case of the triangular hull, is equal to 360° less 60°
for a total of 300° or:
$\beta =\frac{300}{180}$ π = 5 3 π ${\rho}^{\frac{\pi}{\frac{5}{3}}}$ π  1 = 1 ρ 2 5
which says that there is a charge density singularity to the two fifths power for small radius. Thus, the corner plates on the hull create a huge line charge density along the sharp vertical corner edge. The equation for the potential of a line charge density is given as:
$\Phi $ ( x , y ) =  λ 2 πɛ 0 Ln ( ( x  x 0 ) 2 + ( y  y 0 ) 2 )
where λ is the charge per unit length in the vertical zdirection, and x_{0 }and y_{0 }are the location of the line charge in the xyplane.
$\beta =\frac{300}{180}$ π = 5 3 π ${\rho}^{\frac{\pi}{\frac{5}{3}}}$ π  1 = 1 ρ 2 5
which says that there is a charge density singularity to the two fifths power for small radius. Thus, the corner plates on the hull create a huge line charge density along the sharp vertical corner edge. The equation for the potential of a line charge density is given as:
$\Phi $ ( x , y ) =  λ 2 πɛ 0 Ln ( ( x  x 0 ) 2 + ( y  y 0 ) 2 )
where λ is the charge per unit length in the vertical zdirection, and x_{0 }and y_{0 }are the location of the line charge in the xyplane.
Referring to FIG. 4,
the triangular hull (D) is plotted together with the potential contours (A) and
the electric field arrows (B) created by the three corner line charges. The
line charges are perpendicular to the paper. Notice that the electric field
arrows are parallel crossing the center parabolic antenna (C). The electric
field is also parallel to the sides (D) of the triangle.
Referring to FIG. 5,
along the side of the triangle (A), an array (B) of horizontal slot antennas
emit electromagnetic waves that have a vertically polarized electric E field
(C). These traveling waves interact with the electric field (D) produced by the
line charges on the corners of the triangle.
Using differential forms mathematics, this combination of
fields is represented by the Hodge star of the differential of the wedge
product of the two fields. The antenna electromagnetic field is a combination
of a traveling magnetic field B_{w}, and electric field E_{w}.
The stationary field E created by the line charges is perpendicular to the
traveling wave.
$*d$ ( E ⋀ ( B w + E w ⋀ dt ) ) ɛ c = force volume
where ε is the linear capacitance of space and c is the speed of light. Thus there is a force per volume around the hull.
$*d$ ( E ⋀ ( B w + E w ⋀ dt ) ) ɛ c = force volume
where ε is the linear capacitance of space and c is the speed of light. Thus there is a force per volume around the hull.
This combination of fields produces a spacetime curvature
as determined by Einstein's General Theory of Relativity. The traveling
electric field has an amplitude in the vertical zdirection and travels in the
xdirection
E _{w=E} _{z}cos(x−t)
The Faraday electromagnetic tensor contains all the electric and magnetic fields in all the {x,y,z} directions. The first row and first column contain the two electric fields
${F}_{\beta}^{\alpha}=\begin{array}{c}t\\ x\\ y\\ z\end{array}$ 0 E x 0 E z cos ( x  t ) E x 0 0 0 0 0 0 0 E z cos ( x  t ) 0 0 0
The stress exerted on spacetime occurs in the xx, yy and zzdirection as calculated from the stressenergy tensor T of gravitational physics
$4$ π T μ v = F μ α F α μ  1 4 g μ v F α β F αβ
where g is the metric tensor for Cartesian space
${g}_{\mathrm{\alpha \beta}}=\begin{array}{c}t\\ x\\ y\\ z\end{array}$  1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
where the diagonal components are the coefficients of the elementary spacetime length ds squared
(ds)^{2}=−(dt)^{2}+(dx)^{2}+(dy)^{2}+(dz)^{2 }
The calculation produces three stresses T^{xx},T^{yy }and T^{zz }in their respective {x,y,z} directions.
E _{w=E} _{z}cos(x−t)
The Faraday electromagnetic tensor contains all the electric and magnetic fields in all the {x,y,z} directions. The first row and first column contain the two electric fields
${F}_{\beta}^{\alpha}=\begin{array}{c}t\\ x\\ y\\ z\end{array}$ 0 E x 0 E z cos ( x  t ) E x 0 0 0 0 0 0 0 E z cos ( x  t ) 0 0 0
The stress exerted on spacetime occurs in the xx, yy and zzdirection as calculated from the stressenergy tensor T of gravitational physics
$4$ π T μ v = F μ α F α μ  1 4 g μ v F α β F αβ
where g is the metric tensor for Cartesian space
${g}_{\mathrm{\alpha \beta}}=\begin{array}{c}t\\ x\\ y\\ z\end{array}$  1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
where the diagonal components are the coefficients of the elementary spacetime length ds squared
(ds)^{2}=−(dt)^{2}+(dx)^{2}+(dy)^{2}+(dz)^{2 }
The calculation produces three stresses T^{xx},T^{yy }and T^{zz }in their respective {x,y,z} directions.
Referring to FIG. 6,
these three stresses are plotted together as a 3D vector field animated
over time in nine frames. The graphs show that there is a lift force as
depicted by the vertical arrows as well as a force of propulsion as shown by
the interspersed horizontal arrows. With the passage of time, these vectors
exchange places with each other so that the lift becomes the propulsion and vice
versa, creating a wavy stressenergy field around the hull.
SUMMARY of the INVENTION
This invention is a spacecraft with a triangular hull
having charged flat plates on the vertical corners of the three sides. The two
rear corners are charged to a potential V. The forward corner is charged to a
potential −V. The 60° angle on the corner creates a line charge density
singularity that produces a huge horizontal electric field pointing from the
back to the front of the craft which is also parallel to the sides of the
triangle. An array of horizontal slot antennas located on the sides of the
triangular hull produce an electromagnetic wave with the electric field
polarized in the vertical direction. This combination of fields produces a
spacetime force in both the vertical and horizontal directions such that the
spacecraft receives a lift force and a force of propulsion.
A BRIEF DESCRIPTION of the DRAWINGS
FIG. 1. Perspective view of triangular
spacecraft.
FIG. 2. Drawing of the intersection of two
charged plates in order to calculate the charge density in the corner.
FIG. 3. Perspective view of the corner angle β
for the equilateral triangle.
FIG. 4. Planar 2D graph showing the
electric field produced by three line charges on the corners of the triangular
hull.
FIG. 5. Perspective view of electric field
produced by the linear charge interacting with the traveling electromagnetic
wave produced by the slot antenna.
FIG. 6. 3D vector animation of the lift
and thrust force generated by the fields.
FIG. 7. Perspective view of slot antenna.
DETAILED DESCRIPTION of the INVENTION
Referring to FIG. 7,
the antenna (A) is made out of sheet copper in which a rectangular horizontal
slot (B) has been notched out using a die press and sheet metal fixture. A
coaxial cable from the amplifier and frequency generator is attached across the
slot by soldering the outer cable (D) to one side of the slot and the inner
cable (E) to the other side of the slot. This creates the positive and negative
charges across the gap which forms the vertical electric field (F) which
radiates out perpendicularly to the copper sheet.
Although the invention has been described with reference
to specific embodiments, such as a particular antenna system, those skilled in
the art will appreciate that many modifications and variations are possible
without departing from the teachings of the invention. All such modifications
and variations are intended to be encompassed within the scope of the following
claims.
Patent
Citations
Cited Patent

Filing date

Publication date

Applicant

Title

Sep
2, 1969

Jul
11, 1972

Fuchs
Harry B

Method
and means for creating artificial gravity in spacecraft


Sep
30, 1991

Dec
14, 1993

Shearing
Ernest J

Protective
enclosure apparatus for magnetic propulsion space vehicle


Dec
27, 2002

Dec
13, 2005

Peter
Grandics

Method
and apparatus for converting electrostatic potential energy


May
9, 2002

Nov
13, 2003

St.
Clair John Quincy

Electric
dipole moment propulsion system


May
9, 2002

Nov
13, 2003

St.
Clair John Quincy

Rotating
electrostatic propulsion system


Aug
4, 2004

Feb
23, 2006

St
Clair John Q

Electric
dipole spacecraft

*
Cited by examiner
Classifications
U.S.
Classification


International
Classification


Cooperative
Classification


European
Classification

B64G1/40Z

Publication
number

US20060145019
A1

Publication
type

Application

Application
number

US
11/017,093

Publication
date

Jul 6, 2006

Filing date

Dec 20,
2004

Priority
date

Dec 20,
2004

Inventors


Original
Assignee


Export
Citation


Patent
Citations (6),
Classifications
(4)


External Links: USPTO,
USPTO
Assignment, Espacenet

And SEE Black Triangles: Cornerstone of the US Space Fleet @ https://nexusilluminati.blogspot.com.au/2014/06/blacktrianglescornerstoneofusa.html
Richard D Hall: Black Triangle Craft
(27:04)
Richard D Hall The Secret Space Fleet
(1:03:04)
For more information about electromagnetic propulsion see http://nexusilluminati.blogspot.com/search/label/electromagnetic%20propulsion
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The triangle is the correct geometrical design, and there's no need to confuse people with a bunch of mathematical BS!
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