Replying to myself ;)
John Berry wrote:
<snip>
I do have a concern with your diagram though, if you go from wire or coax to plates the current branch out to fill the plates.
The problem is that the currents branching out should create a force in the opposite direction this becomes more clear if you substitute the capacitor plates with wires.
But what can be done is to use the same metal you use with the plates with the same width for the whole construction (except the the toroid and wires leading up to it, doesn't matter that they are thinner as the currents must both converge and diverge going in and out.
This would only be a problem with non ampere type forces at 90 degrees to the current.
Actually at this moment, I'm not quite sure why two hoop coils in attraction at an angle to each other shouldn't create a unidirectional force if the forces are at right angles, or indeed why two permanent magnets in attraction also at an angle wouldn't create the same effect.
Though I can't believe this is the case, either I'm missing something or Ampere was correct, which I think he was. (or both)
O ^ O
\ /
\ N S /
S \ / N
\ /
O O
Hoop coil diagram.
Horace Heffner wrote:
The following design of a resonant Coaxial Capacitor Thrustor indicates a
thrust of about 1 kilogram force per kilowatt is feasible.
DIAGRAM OF RESONANT OPEN ENDED THRUSTER
Toroid o========o::: | | ---------- -----o-xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Capacitor plates | B1 P1 | | B2 P2 ----------------o-xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Note -
B1 and B2 are the bends with partially unopposed self-forces
P1 and P2 are power supply points
It is assumed the toroid coil can be taken off vertically without inducing a net lateral force.
Fig. 1 - DIAGRAM OF RESONANT OPEN ENDED THRUSTER
For coaxial versions see Figs. 2 and 3 below.
TOROIDAL CONDUCTOR
Assume the conductor is made of tubing about 0.5 cm diameter. Small radius
of the torus is 4 cm. Inner radius of torus is 15 cm. Major radius Mr is
thus 19 cm. and outer radius is 23 cm. Total of N = 45 turns Coil area A
is about 50 cm^2. Coil conductor length is about 11.3 m. Inductance is
approximated by:
L = u N^2 A (1/Mr) (1.26x10^-6 H)
L = (1) (45^2) (50) (1/(19)) (1.26x10^-6 H)
L = 6.71 mH
CAPACITOR
Plate size is 23 cm x 50 cm, giving an area of 1150 cm^2. Plate separation
is 0.5 cm, of which 0.12 cm is 20 kV insulation with dielectric constant Ke
= 8 and the rest is dielectric constant Ke = 1. Capacitance is given by:
C = Ke (A/d) (8.85x1010^-12 F)
For the insulating layer:
Ci = 8 (1150/0.12) (8.85x10^-12 F)
Ci = 6.79x10^-7 F
For the air layer:
Ca = 1 (1150/0.38) (8.85x10^-12 F)
Ca = 2.68x10^-8 F
Total capacitance:
1/Ct = 1/Ci + 1/Ca
1/Ct = 1.473x10^6 + 3.73x10^7 (1/F)
Ct = 2.58x10^-8 F
ALTERNATE DESIGN FOR CAPACITOR - COAXIAL
A design with similar values can be obtained by making the capacitor
coaxial In that case the plate fed by P1 would be the outer sheath and P2
the inner conductor. This has the advantage of minimizing external waste
AF radiation. Except for concerns about resistance, the coax can be made
as long as necessary to accommodate the desired capacitance.
To approximate an equivalent coaxial capacitor to the above flat plate
design we can use an inner diameter of 7 cm and an outer diameter of about
7.5 cm, and length still of about 50 cm. The thrust comes from the fact
the trailing end of capacitor is open, which is clear with the flat plate
capacitor design, but not so clear with the coaxial design. Proof of that
the two are equivalent may require running a high resolution FEA model.
However, it appears that both are equivalent, and in any case that this
point is comparatively moot in that the principle issue involved, namely
waste audio frequency radiation, is a comparatively small issue compared to
the issue of whether the basic concept works.
RESONANT FREQUENCY
The resonant frequency f0 is given by:
f0 = 1/(2 Pi (L*C)^0.5)
f0 = 1/(2* 3.1415 * 1/(1.73118x10^-10)^0.5)
f0 = 12,096 Hz
REACTANCE AND IMPEDANCE
Capacitive reactance Xc is given by:
Xc = 1/(2 Pi f0 C)
= 1/(2 * 3.1415 * 12,096 * 2.58x10^-8) ohms
Xc = 510 ohms
Inductive reactance Xl is given by:
Xl = 2 Pi f0 L
Xl = 2 * 3.1415 * 12,096 * 6.71x10^-3 ohms
Xl = 510 ohms (check)
The impedance Z = (R^2 + (Xl-Xc)^2)^0.5 is thus equal to resistance R.
POWER AND CURRENT
If we assume a coil resistance of 1 ohm per 1000 m, or 0.001 ohms per
meter, we have a coil resistance of 11.3 m * 0.001 ohms per meter = 0.0113
ohms. Assume the power supplied Ps = 1500 W. If It is the resonance tank
current then the heat Ph dissipated in the coil is given by:
Ps = Ph = 1500 W = It^2*(0.0113 ohms)
So:
It = 364 A
We have a very high Q coil, because:
Q = Xc/R = 510/0.0113
Q = 1858
This gives Is the supply current:
Is = It / Q = 636/1858
Is = 0.196 A
And the supply voltage:
Vs = Ps / Is = 1500/ 0.196
Vs = 7653 V
The apparent power in the tank circuit is:
Pt = It * Vt = 7653 * 364 W
Pt = 2.79 MW
MAXIMUM RADIATION FORCE
Note that the above numbers all ignore radiation. Additional power must be
supplied to account for any radiation. However, note that, at 1500 W power
supplied, that the *force* from any radiation can be ignored. There is
2.94x10^9 photonic watts per kg-f of thrust, so the 1500 watts could only
produce 5.1x10^-7 kg-f thrust.
SELF-FORCE OF THE CLOSED END COAXIAL CAPACITOR
Estimating the cap force for a capped coax:
Fc = 1.61x10^-3 N
Fc = 0.165 g-f
Though small, a force of about 1/10 gram is sufficiently large to attempt
detection, and to be very useful. Also, this rough design is only to
provide sample calculations as a starting point for an actual experiment
design. Such a device, or one of approximately the same dimensions can be
fairly readily constructed from ordinary copper materials and a custom
power supply.
At 2.94x10^9 photonic watts per kg-f of thrust, the 10^-4 kg-f would
require 294 kw of broadcast power. This number is way above the 1.5 kw
power supplied, but interestingly way below the Pt = 2.79 MW apparent power
of the resonant circuit.
It is also notable that by using a larger inductance and capacitance that
the frequency could be dropped further and thus Q improved by reducing the
skin effect, and also by using a solid conductor for construction of the
toroidal inductor. The estimated 0.5 cm wire size corresponds to 4 gage
copper, which is 0.2533 ohms per 1000 feet, or 0.831 ohms per 1000 m.
Using 3 gage copper, a 10 percent increase in wire diameter, gives only
0.659 ohms per 1000 m, thus improving R by 51 percent and net force by 230
percent, to 0.378 g-f. However, resistance of the capacitor has been
ignored, due to its small length. Achieving low resonant frequency
involves use of a much longer capacitor, thus capacitor resistance will
play a significant role. Increasing capacitance by reducing the capacitor
gap width is fruitless because net thrust is roughly proportional to the
gap width. The thrust is essentially provided by the fact that some
(angular) percent of a closed current loop is open.
CRYOGENIC ALUMINUM
Aluminum has a conductivity of 0.377x10^6 ohm^-1 cm^-1, while copper is
0.596 x10^6 ohm^-1 cm^-1, thus copper provides 58 percent better
conductivity, but 250 percent better thrust per watt. Aluminum has
advantages over copper for propulsion due to a 70 percent savings in
weight. At 21 deg. kelvin high purity aluminum resistance drops to less
than 1/500 to 1/1000 times that of room temperature, thus providing an over
100,000 improvement in thrust/watt over copper. At this amplification the
0.1 g-f thrust becomes 10 kg-f. This amply repays the cost of 1500 W of
refrigeration, which in space can hopefully be provided at a power cost of
less than 10 kW. If so, a thrust/power ratio of about 1 kg-f/kW is
achieved.
SUPERCONDUCTING RESONATOR
The lower resonant frequency achieved by use of much larger capacitors and
inductors should permit use of superconductors, with a much larger net
thrust feasible, and lower refrigeration cost. The major impediment to the
use of superconductors may be the radiation from SC surfaces exposed to
high voltage gradients.
COAXIAL DRIVE CONFIGURATIONS FOR SHIP
Figs 2 nd 3 show coaxial direves as they might be implimented in a large ship.
------------ | | | --------------------------------- |===T-------------------------------- | --------------------------------- | | ------------
Note - T represents power, oscillator, and toroidal coil unit
Fig. 2 - Open ended coaxial drive with bulged power housing
------------ | | | --------------------------------- |===T-------------------------------- | | --------------------------------- | | ------------
Note - T represents power, oscillator, and toroidal coil unit
Fig. 3 - Closed ended coaxial drive with bulged power housing
The toroidal coil and osciallor (denoted "T") can be enclosed in a bulge at
the end of the coaxial coil. External power could be supplied by leads
inside a smaller "head" coaxial lead, denoted "===", or the power unit,
probably a nuclear reactor, could be enclosed within the bulged power
housing itself.
It is notable that, provided the central conductor terminates well before
the end of the coaxial sheath, an end cap on the sheath should have no
effect other than to capture or reflect any radiation. As shown above,
this can only have negligible effect on thrust.
Regards,
Horace Heffner
