"Harold Ensle" wrote in message
link.net...

"John Schoenfeld" wrote in message
om...
(Bilge) wrote in message
...
John Schoenfeld:
The system John S. describes works equally well
electrostaically as it does electromagnetically. The
requirement is the elements are 1/4 wavelength
apart and 90 degrees out of phase. Then one element
is repelled by the other, and one is attracted to the
other.
There was a thread a few yearss back in s.p.research
about "massless propulsion" wherein an engineer analyzed
the effect, and concluded the cause to be EM-radiation.
Anyway, if you use electrostatic plates as the elements
the system can operate at very high frequencies and
currents, and this improves it's power density.

In particular, the separation of the magnets must be
significantly
greater than the length of the wiring for this effect to occur.

The "wiring" length is adjusted so the phase relations
max the desired effect, rather like tuning an antenna.

Once again, Schoenfeld fails to follow through with the math.

Tom Davidson
Richmond, VA

Ken S. Tucker

While I'm sure this device does work theoretically, it is
constrained
by conserved of momentum which must be carried away by the radiative
emission.


Say if you want to levitate a 5000kg vehicle in earth gravity.

Levitation force = 50 000 N


Since the reactive force of this device = radiation power / c, the
net
electrical power per newton of net force is roughly 300 megawatts.



So the power to sustain such a levitating vehicle is 150 000
terrawatts, which is the daily output of around 70 000 000
medium-sized nuclear power plants.

Does that imply that the ground is supplying 150000 TW to levitate
the vehicle at ground level and keep it from falling to the center
of the earth? If not, how does the vehicle stay at ground level
according to your calculations?

Put on your tinfoil cap, the cosmic rays are bright this time of the
cycle.

You'll have to forgive Bilge. He has been so confused by relativistic
and quantum-mechanical mysticisim, that he can't do a simple classical
problem.

You are correct (as far as I can tell) to the amounts involved. This is
why it is more efficient to use mass as a propellent. The only thing
I can see (at this time) is to increase the exit velocity of the mass
to the highest possible value.

Yes, greatest efficiency is achieved for v = c, wherein
projectile mass, m, and projectile momentum. p, are related
by m p / c which perfectly characterizes a photon.

The most efficient (of power and mass-energy) propulsion
system achievable today is that of a nuclear reactor, wherein
only heat is exhausted, as infrared radiation, in a collimated
beam via parabolic reflector. Near 100% efficiency is
possible with current technology (neutrino energy is lost).
True, thrust to ship mass ratio is very low, but that doesn't
enter into efficiency. [Old Man]


It is also useful to have the source of energy directly tied into the
acceleration of the mass.

H.Ellis Ensle

"Old Man" wrote in message
...
"Harold Ensle" wrote in message
link.net...

"John Schoenfeld" wrote in message
om...
(Bilge) wrote in message
...
John Schoenfeld:
The system John S. describes works equally well
electrostaically as it does electromagnetically. The
requirement is the elements are 1/4 wavelength
apart and 90 degrees out of phase. Then one element
is repelled by the other, and one is attracted to the
other.
There was a thread a few yearss back in s.p.research
about "massless propulsion" wherein an engineer analyzed
the effect, and concluded the cause to be EM-radiation.
Anyway, if you use electrostatic plates as the elements
the system can operate at very high frequencies and
currents, and this improves it's power density.

In particular, the separation of the magnets must be
significantly
greater than the length of the wiring for this effect to occur.

The "wiring" length is adjusted so the phase relations
max the desired effect, rather like tuning an antenna.

Once again, Schoenfeld fails to follow through with the math.

Tom Davidson
Richmond, VA

Ken S. Tucker

While I'm sure this device does work theoretically, it is
constrained
by conserved of momentum which must be carried away by the
radiative
emission.


Say if you want to levitate a 5000kg vehicle in earth gravity.

Levitation force = 50 000 N


Since the reactive force of this device = radiation power / c, the
net
electrical power per newton of net force is roughly 300 megawatts.



So the power to sustain such a levitating vehicle is 150 000
terrawatts, which is the daily output of around 70 000 000
medium-sized nuclear power plants.

Does that imply that the ground is supplying 150000 TW to levitate
the vehicle at ground level and keep it from falling to the center
of the earth? If not, how does the vehicle stay at ground level
according to your calculations?

Put on your tinfoil cap, the cosmic rays are bright this time of the
cycle.

You'll have to forgive Bilge. He has been so confused by relativistic
and quantum-mechanical mysticisim, that he can't do a simple classical
problem.

You are correct (as far as I can tell) to the amounts involved. This is
why it is more efficient to use mass as a propellent. The only thing
I can see (at this time) is to increase the exit velocity of the mass
to the highest possible value.

Yes, greatest efficiency is achieved for v = c, wherein
projectile mass, m, and projectile momentum. p, are related
by m p / c which perfectly characterizes a photon.

The most efficient (of power and mass-energy) propulsion
system achievable today is that of a nuclear reactor, wherein
only heat is exhausted, as infrared radiation, in a collimated
beam via parabolic reflector. Near 100% efficiency is
possible with current technology (neutrino energy is lost).
True, thrust to ship mass ratio is very low, but that doesn't
enter into efficiency. [Old Man]

Interesting..............but I think you missed my point. It would be
more useful for the ship to use the nuclear reactor to accelerate
a mass. For example, imagine a person in space and he throws a rock.
He will be propelled in the opposite direction. Now how much energy
did he expell? He expelled E=mc^2 (where m is the mass of the rock)
plus the kinetic energy of the rock. If one uses photons, the photons
are being created and thus the ship has to create the equivalent
of E=mc^2.... a tremendous cost of energy in contrast to the original
case were the E=mc^2 was freely supplied by the rock.

H.Ellis Ensle

Harold Ensle wrote:

plus the kinetic energy of the rock. If one uses photons, the photons
are being created and thus the ship has to create the equivalent
of E=mc^2.... a tremendous cost of energy in contrast to the original
case were the E=mc^2 was freely supplied by the rock.

Tossing rocks does not convert their mass into energy.

Bob Kolker

"Robert J. Kolker" wrote in message
news:fMNac.150721$1p.1937182@attbi_s54...


Harold Ensle wrote:

plus the kinetic energy of the rock. If one uses photons, the photons
are being created and thus the ship has to create the equivalent
of E=mc^2.... a tremendous cost of energy in contrast to the original
case were the E=mc^2 was freely supplied by the rock.

Tossing rocks does not convert their mass into energy.

I know........you apparently missed the point.

H.Ellis Ensle

"Harold Ensle" wrote in message
link.net...

"Robert J. Kolker" wrote in message
news:fMNac.150721$1p.1937182@attbi_s54...


Harold Ensle wrote:

plus the kinetic energy of the rock. If one uses photons, the photons
are being created and thus the ship has to create the equivalent
of E=mc^2.... a tremendous cost of energy in contrast to the original
case were the E=mc^2 was freely supplied by the rock.

Tossing rocks does not convert their mass into energy.

I know........you apparently missed the point.

H.Ellis Ensle

Kolker has the right of it. You have thrown away mc^2 + mv^2 / 2
of energy for p = mv worth of momentum, whereas directed heat
radiation from a nuclear reactor yields p = mc from mc^2 worth of
energy. The ratio of efficiencies is v / c. The two techniques
approach equivalency as v = c, but current techniques for particle
acceleration are extremely inefficient, and equivalency is not attained
unless m p / c, as in gamma = 1000. This is most easily satisfied
by ultra-relativistic electrons, but this requires a 100% efficient, mile-
long, linear electron accelerator. No such beast exists. [Old Man]

"Old Man" wrote in message
...
"Harold Ensle" wrote in message
link.net...

"Robert J. Kolker" wrote in message
news:fMNac.150721$1p.1937182@attbi_s54...


Harold Ensle wrote:

plus the kinetic energy of the rock. If one uses photons, the
photons
are being created and thus the ship has to create the equivalent
of E=mc^2.... a tremendous cost of energy in contrast to the
original
case were the E=mc^2 was freely supplied by the rock.

Tossing rocks does not convert their mass into energy.

I know........you apparently missed the point.

H.Ellis Ensle

Kolker has the right of it.

What are you talking about? He misread my post, so he can't
possibly be correct.

You have thrown away mc^2 + mv^2 / 2
of energy for p = mv worth of momentum, whereas directed heat
radiation from a nuclear reactor yields p = mc

What's 'm' here? The momentum from radiation is p=E/c where
E is the energy of the photon. Or given the fields
p=Integral(all space) eExB dV.

[...]

H.Ellis Ensle

Yes, indeed!

............ ...However, as along any case, we cannot be without to
remark, therefore, if x and y are commuted to one an other, it would be as
it should be a clarified along that matter, what would as should follows:

( x, y )sq/n = ( - 1 )sq/n ( y I x )sq/n = ( x - y )sq/n

As in this case, we can arrive to an equation, which would be as should be
as follows:

p ( x , y ) = R ( x - y ) = r ( x - y )
E

As in all cases we can be allow to attain what would follows:

p ( x , y ) = p ( y , x )

P ( x , º ) = R ( x ) = r ( x )
E


............ ...Finally, therefore, this is what it would be as it
should what is all about, a definitely as a matter a
fact!!!!!!!!!!!!!!!.......... ...


--
Ahmed Ouahi, Architect
Best Regards!



"Harold Ensle" wrote in message
hlink.net...

"Old Man" wrote in message
...
"Harold Ensle" wrote in message
link.net...

"Robert J. Kolker" wrote in message
news:fMNac.150721$1p.1937182@attbi_s54...


Harold Ensle wrote:

plus the kinetic energy of the rock. If one uses photons, the
photons
are being created and thus the ship has to create the equivalent
of E=mc^2.... a tremendous cost of energy in contrast to the
original
case were the E=mc^2 was freely supplied by the rock.

Tossing rocks does not convert their mass into energy.

I know........you apparently missed the point.

H.Ellis Ensle

Kolker has the right of it.

What are you talking about? He misread my post, so he can't
possibly be correct.

You have thrown away mc^2 + mv^2 / 2
of energy for p = mv worth of momentum, whereas directed heat
radiation from a nuclear reactor yields p = mc

What's 'm' here? The momentum from radiation is p=E/c where
E is the energy of the photon. Or given the fields
p=Integral(all space) eExB dV.

[...]

H.Ellis Ensle