Carl White wrote:93143 wrote:A flywheel with radius R, with Mach-effect thrusters of thrust efficiency E positioned tangentially on the edge. Power (P) is torque (T) times angular velocity (w), so given a thruster power p, the output power of an ideal generator hooked to the flywheel is P = Tw = (pE)Rw, and if ERw > 1, well...
I don't really understand this argument. If you use the motion of the flywheel to generate power, the wheel has to decelerate. You can't take any more power out of the apparatus than the thruster is putting in, except transiently (e.g. as it slows down until a balance is reached, after a period of acceleration during which power wasn't generated). So what you really seem to be saying is that if the thruster is producing more power than it is using, it's overunity. Which is true by definition, but I don't see why you need to bring a flywheel into it.
You've confused momentum and energy again. Force and power are two different things. A Mach-effect thruster uses power. It outputs force. Therefore it is not sensible to say that an M-E thruster puts power into a flywheel; what it is doing is applying a torque. And the power input to the flywheel by that torque (which is equivalent to the power output of the generator, if inefficiencies are ignored) is proportional not only to the torque, but also to the rotational speed.
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In more detail:
A Mach-effect thruster is supposed to be able to generate thrust when power is supplied. The amount of thrust generated is related solely to the power input and the thrust efficiency of the thruster in newtons per watt. If it doesn't work this way, then either it will be extremely finicky or even downright unusable as a space drive, or Lorentz invariance is wrong. Frankly I don't see why the former case should be true, given the extremely high operating frequencies of these devices - even spinning around on a big flywheel at 3600 rpm or something shouldn't be substantially different from stationary operation.
Thus, a Mach-effect thruster can be used to generate a certain level of torque on a flywheel for a given input power, completely independent of the rotational speed of the flywheel.
It is possible in principle to apply a counter-torque to the flywheel in such a way as to harvest energy while maintaining a steady-state rotational speed against the torque supplied by the thruster - a generator is a good example. And there is no real reason this has to happen at any
specific rotational speed.
It can be shown (in fact I did in the quoted post) that
in the steady-state case, if the product of the flywheel's radius and angular velocity is high enough, the output power is sufficient to power the Mach-effect thruster and then some.
As for why I used a flywheel - basically it's because, given working Mach-effect technology at a good thrust efficiency, this setup is an eminently practical steady-state power generation scheme, and it is thus very easy to understand the energy balance. Linear acceleration can illustrate the point too, but it tends to cause people to confuse themselves...
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Numbers: A common ballpark figure for thrust efficiency used by M-E proponents, as an example of a well-engineered future space drive they believe is possible, is 1 N/W. Slap that on a flywheel with a 1 m radius, and stabilize it at 1 rad/s. The (ideal) output power is P = p(1)(1)(1), exactly equal to the input power.
Now let it spin at 2 rad/s. The torque is the same, but the speed is doubled; the output power from the (ideal) generator is now twice what the M-E thruster consumes.
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This is not a violation of conservation if the Mach-effect thruster is allowed to interact with distant matter, since momentum and energy exchange can then occur with reaction mass that is not immediately apparent to an observer of the assembly. The assembly is then essentially harvesting energy from the rest of the causally-connected universe. It is not a perpetual motion machine of the first kind, but it is doing a d@mn good imitation of one...