Mike Hord <mike.hord <at> gmail.com> writes:

> 
> > This is actually a reaction drive because it is based on reaction (reactive)
> > force, unless I missed something in the description.  In the regular "jet"
> > engine the reactive force is provided by expulsion of matter, in this case 
by
> > expulsion of wave, somewhat similar to photonic drive (a-la startrek etc.).
> > What they mean is the drive has no momentum because it has no moving parts.
> 
> Or, better, reaction MASS-less drive.  It uses no reaction mass to provide
> thrust.  Which is EXCELLENT, because most forms of propulsion these days
> (rocket, jet, etc.) require quite a bit of reaction mass.
> 
> > Also, they have a picture of an aircraft using this type of engine to 
provide
> > hoover capability.  I bet amount of microwave radiation it will produce will
> > wreak havoc on electronic equipment below the craft.
> 
> The end goal is to have nearly none of that microwave radiation leaking out of
> the chamber.  The energy poured into the chamber would go into lift, with
> very little (remember the goal of a Q in the billions) leaking out.
> 
> If my physics logic is right, once a craft reached altitude, the
> amount of energy
> required to keep it there would be minimal.  After all, energy is
> force integrated
> over distance.  If the distance being travelled is zero (no net motion up or
> down in the gravitational field), the only energy that would have to be placed
> into the "bottle" would be to replace leakage, and with a Q of several 
billion,
> that's not much.

Something not right here, I just cannot put a handle to it.  You still have to 
counter the force of gravity (I assume all this happens around earth).  Let's 
make an imaginary experiment: I pick a stone from ground and lift it up to the 
height of 1m.  The stone is stationary, so there is no need to spend energy to 
maintain it at that height, right?  But practical experience tells me that the 
stone will fall down.  

I also think that the engine will have to radiate  alot of energy, unless, of 
course the 3rd Newton's law is not working anymore.  

I have to go and re-read Feinman's lectures now.

Sergey Dryga



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