Re: Electrogravity From Accelerated B Fields

[Frederick Sparber]

Oops.

 

Make that 20KW input per tonne lift and 10KW/tonne for kinetic energy

since the voltage across the impedance of the coax line is 1/2 the input pulse voltage

and the other 1/2 is across the gas discharge heat energy dissipating "radiator"

for a theoretical maximum mechanical  efficiency.

Also  note that the maximum PRF is limited by the "speed limit" of the coax line:

 

Line impedance Zo = (L/C)^1/2   and Time T =(LC)^1/2.

 

 

"The 10KW/tonne to just hover (same as a Helicopter or Harrier fighter) at the earth's surface,

plus the 1/2 mv^2  = 500 watts/tonne x v^2 kinetic energy to impart velocity to the craft shows up

as the heat energy rejected by the gas discharge radiators from the PRF and amplitude of

the charge-generating pulses moving down the coax line generating the Hypocharge."

 

----- Original Message -----

From: Frederick Sparber

To: vortex-l

Sent: 4/1/2006 11:34:15 AM

Subject: Re: Electrogravity From Accelerated B Fields

The hidden Hypocharge (*Q) in this, sheds some light on how

mechanical work is derived from running pulses down a coax and

burning it up in a discharge impedance:

>

> running a pulse around a coaxial "wire-in-a-tube" device
> to get a combination  1/R^2 Anti-Gravity and Force Field, was that
> although the Charge Q = capacitance x pulse voltage V  "walking around
> the loop"  charging-discharging the L-C increments, setting up a primary B field that
> was shielded and would cancel any AG-FF effect , was that the capacitance charge displacement

> current Id = C*dV/dt sets up a secondary dB/dt  field which in turn sets up a Secondary  
> E field that doesn't "see" the "ordinary" Electromagnetic "shielding" Effects
> and therefore only "sees" the 1/R^2 gravity field properties of matter.

>

> If the coax is terminated with a 75 ohm resistor (a light bulb or gas discharge tube)

> the line will be non-reflective, but the input pulse needs to be twice the amplitude.

>

>

The "E Field" or (*q ) = 1.6e-19/gamma = 4.78e-38 coulomb, literally develops a net charge

(*Q ) on the craft that repels the approximately 13 coulomb (*Q' ) Gravitational Hypocharge at

the earth's surface.

 

F = 9e9 x Q x Q'/R^2   Newtons

 

This is analgous to work done in electrical devices by manipulating charge.

 

There is No Free Lunch. Sorry Jones.  :-)

 

The 10KW/tonne to just hover (same as a Helicopter or Harrier fighter) at the earth's surface,

plus the 1/2 mv^2  = 500 watts/tonne x v^2 kinetic energy to impart velocity to the craft shows up

as the heat energy rejected by the gas discharge radiators from the PRF and amplitude of

the charge-generating pulses moving down the coax line generating the Hypocharge.

 

An on-board gyro will suffice for lateral maneuvering.

 

Fred