# Triangular spacecraft US20060145019A1 Patent Files

DOWNLOD FILE HERE --&gt;&gt;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



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



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.



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.



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)



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.



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