I take a single day off, and look what happens...
The question we've been arguing about is whether or not the work done by an M-E thruster necessarily equals the energy input required to run it. You attempt to answer this question by calculating the work done by an M-E thruster in the laboratory reference frame, make the unstated and unsupported assumption that this equals the energy input required to run it, and then triumphantly declare that there is no difference between the value you've calculated and the value you've calculated.
Listen closely, because I'm tired of explaining this:
The reaction mass used by a Mach-effect thruster is not in the laboratory frame of reference. It therefore makes no sense at all to calculate the thruster's power requirements based on its velocity in the laboratory frame of reference.
What mental block is preventing you from understanding this?
An M/E thruster will operate on the same idea, but the 'road' is beyond the scope of our research. It will still have a limited efficiency ('grip') with which it can exert a force on that 'road', and thus generate thrust.
That is EXACTLY IT. Now, all you need to do is realize that the effective average relative velocity between the thruster and its 'road'
is what determines the minimum power required to generate a certain amount of thrust (ie: the maximum possible "thrust efficiency"). It has nothing to do with how fast the thruster appears to be moving relative to the building it's in.
It makes no difference what frame of reference the power supply is in, either, because electrical power transmission doesn't care about piddly little velocities like these, so the power leaving the power supply in the lab frame will be basically the same as the power arriving at the thruster in the thruster's frame. And once the thruster has been supplied with electrical power, its mode of operation has nothing further to do with the lab frame or anything in it. It's not interacting with the floor or the atmosphere; it's interacting with the gravinertial field, which (if it is in fact the cause of inertia, and if M-E thrusters work we can assume that it is) has been observed to behave the same independent of velocity state.
In essence, an M/E thruster that causes some object to get into an equilibrium state will, theoretically, require zero power. For example, if you have an airship that uses M/E thrusters to counteract its weight and stay afloat at a constant altitude, it would, in theory, require no power to stay put because the change in potential energy (as a result of changing altitude) is zero.
Right, just like a helicopter can hover without expending any energy. Wait...
The reason conventional 'engines' (e.g. jets or props) require continuous power to keep an aircraft aloft in hover is because they're pushing air to work against gravity. Since air is notoriously inefficient at that and has a tendency to disperse, you will have to keep pushing to maintain a constant upward force.
A helicopter needs to expend energy to hover because the reaction mass the helicopter is pushing on is a fluid, and is thus unable to sustain a static stress. So far your handwaving is more or less on the mark. But the real reason, resulting from the above fact, is this: the reaction mass is moving relative to the helicopter in the axis of thrust, so finite power is required to exert a force on it. (You can "hover" by walking up a down escalator, but this does require power even in the ideal case, because the relative velocity between you and the escalator is nonzero, so you need to do work in the course of exerting a force. Conversely, you can 'windmill' a helicopter in a sufficiently strong updraft.)
So what is the M-E thruster pushing on and how fast is it going?
gravity is used to generate power because it moves water down a drop. It's not the gravity that does the work, but the drop
Wrong. It's gravity that's doing the work, because the object it's exerting a force on is moving in the axis of the force vector. It's that simple - in the laboratory frame of reference
. The amount of work done by a force is frame-dependent
. (Don't get too scared of this; it always ends up working out...)
That's where the 1N/W figure comes from. It's the maximum in a static thrust situation, e.g. when there is equilibrium, the thing requires 1W for each Newton it has to push. This doesn't mean it's necessarily the same for a dynamic (moving) situation as well, since that has not been tested.
Here's the misunderstanding.
The solution (which should be obvious by now) is that according to Lorentz invariance, a particular Mach-effect thruster fed a certain quantity of power should produce the same thrust no matter what inertial reference frame it's in. There is no 'preferred' reference frame. So the idea of "static" vs. "dynamic" thrust, which is derived from aeronautical engine testing, has no basis here.
In other words, if inertia is the same regardless of velocity (and it is), M-E thrust efficiency should be too, because the field it's interacting with behaves the same.
If this were not true, you would then have to explain why the preferred velocity state of the M-E thruster just happens to follow the testing lab around the Earth as it rotates and orbits the Sun, galactic centre, etc., not to mention why the inertial effect it's supposedly taking advantage of shows no such preference...
Oh, and both you and Carl miscalculated the rocket-on-flywheel problem. His is worse, but they're both pretty bad. (For one thing, his heat of combustion is off by a factor of 1000, and you didn't correct him, choosing instead to neglect a factor of 9 in a follow-on calculation...) Has either of you ever heard of specific impulse?
At 62.8 m/s tangential velocity, almost all of the energy expended by the rockets will end up in the ~4 km/s exhaust stream, not the motion of the generator.
And that should be a clue to where the 'extra' energy comes from in my M-E flywheel example...
I just thought of a great example. Take a rocket engine, in a vacuum. At full throttle, it uses the same amount of energy per second to produce the same thrust, completely independent of how fast it's moving or how much mass it's pushing. You can calculate a frame-independent thrust efficiency value for it, even though the amount of work it does NEGLECTING THE KINETIC ENERGY OF THE PROPELLANT is frame-dependent.
The above is exactly true of a Mach-effect thruster.