Triangular spacecraft US20060145019A1 Patent Files

                                       




DOWNLOD FILE HERE -->>United States US 20060145019A1(12) Patent Application Publication 

Conceptual

A shuttle having a triangular body with vertical electrostatic line charges on each corner that deliver an even electric field parallel to the sides of the frame. This field, connecting with a plane wave produced by radio wires in favor of the frame, creates a power for each volume consolidating both lift and drive.

Portrayal



BRIEF Synopsis OF THE Creation

[0001]

This innovation is a rocket having a triangular structure with vertical electrostatic line charges on each corner. The line charges make an even electric field that, together with a plane wave discharged by recieving wires in favor of the frame, produces a power for each volume giving a one of a kind mix of both lift and drive.

Foundation OF THE Innovation

[0002]

Alluding to FIG. 1, the rocket has a structure in the state of a symmetrical triangle. An illustrative radio wire (E) is halfway situated in the base of the body. A variety of flat space recieving wires is situated at the edge of the body (A). Each back corner (F,G) has a corner leading plate which is charged to a positive voltage +V. The forward corner (C) has a directing plate charged to a negative voltage −V. A movement control half of the globe (D) is situated on the base surface in every one of the three corners.


[0003]

Alluding to FIG. 2, two planes (A,B) meet at the starting point O at an opening edge β. Each plane (x,y) is charged to a voltage V. The potential at point P is resolved in polar directions {ρφ}. The Laplace condition for the potential Φ in polar directions is given by:

1 ρ ⁢ ∂ ρ ⁢ ( ρ ⁢ ∂ Φ ∂ ρ ) + 1 ρ 2 ⁢ ∂ 2 ⁢ Φ ∂ ϕ 2 = 0

Utilizing a partition of factors arrangement, the potential is given as the result of two capacities:

Φ(ρ,φ)=R(ρ)Ψ(φ)

which when substituted into the Laplace condition progresses toward becoming:

ρ R ⁢ ⅆ ρ ⁢ ( ρ ⁢ ⅆ R ⅆ ρ ) + 1 Ψ ⁢ ⅆ 2 ⁢ Ψ ⅆ ϕ 2 = 0

Since the two terns are independently elements of ρ and φ separately, every one must be steady with the aggregate of the constants equivalent to zero:

ρ R ⁢ ⅆ ρ ⁢ ( ρ ⁢ ⅆ R ⅆ ρ ) = v 2 ⁢   ⁢ 1 Ψ ⁢ ⅆ 2 ⁢ Ψ ⅆ ϕ 2 = - v 2

These two conditions have arrangements:

R(ρ)=aρ v+bρ −v

ψ(φ)=Acos(vφ)+Bsin(vφ)

The azimuthal edge φ is limited to an incentive in the range 0≦φ≦β. The limit condition is that the potential Φ is equivalent to V for any range ρ when φ=0 and φ=β. This implies v must be a whole number estimation of π with the goal that the sine work is zero:

sin ⁡ ( v ⁢   ⁢ β ) = sin ⁡ ( m ⁢   ⁢ π β ⁢ β ) = sin ⁡ ( m ⁢   ⁢ π ) = 0 ⁢   ⁢ m = 1 , 2 ⁢   ⁢ …

which thusly implies that the coefficient An of the cosine term must be zero in the arrangement above. Picking b=0 makes the general answer for the potential equivalent to:

Φ ⁡ ( ρ , ϕ ) = V + ∑ m = 1 ∞ ⁢ a m ⁢ ρ m ⁢   ⁢ π/β ⁢ sin ⁡ ( m ⁢   ⁢ πϕ/β )

which demonstrates that when the edge is zero, the sine is zero and the potential is V. On the off chance that the edge is β, there is a numerous of π with the end goal that the sine is zero once more.

[0004]


Since the arrangement includes positive forces of the span, for little enough ρ, just the principal term m=1 in the arrangement is vital. Consequently around ρ=0, the potential is roughly


φ(ρ,φ)≈V+a,ρπ/βsin(πφ/β)

[0005]

The electric field segment is the negative angle of the potential:

E ϕ ⁡ ( ρ , ϕ ) = - 1 ρ ⁢ ∂ Φ ∂ ϕ = - π ⁢   ⁢ a 1 β ⁢ ρ ( π/β ) - 1 ⁢ cos ⁡ ( πϕ/β )

The surface charge dissemination σ at φ=0 and φ=β is equivalent to the electric field opposite to the surface occasions the permittivity of room ε0:

σ ⁡ ( ρ ) = ɛ 0 ⁢ E ϕ ⁡ ( ρ , 0 ) = - ɛ 0 ⁢ π ⁢   ⁢ a 1 β ⁢ ρ π β - 1

Notice that in the event that edge of crossing point β is not exactly π, the condition says that there is a little sweep to a positive power which implies little charge thickness amassing.

[0006]

Alluding to FIG. 3, the estimation of β, on account of the trian

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