Ek = 1/2 m v^2

F = m a

Those two equations are all you need to show that Woodward's device would produce free energy.

We've been over this (and over it, and over it, for years). M-E thrusters (if they work) do not violate conservation of momentum or energy. They exchange momentum and energy with the rest of the observable universe.

The claim that an M-E thruster violates the entropy condition is at least plausible, though IMO it is (for a couple of reasons) on pretty thin ice as a reason to discount the whole concept. But the claim that the concept necessarily violates conservation of energy based simply on Newtonian mechanics cannot be seriously maintained; anyone capable of drawing a free-body diagram shouldn't even need to do so to understand this.

Let's take a thruster at a velocity

**v**_{1}, and a large quantity of mostly distant matter (the Far-Off Active Mass, or FOAM, the gravitational potential of which is what gives the thruster and its payload their inertia) at an average velocity

**v**_{2}. The thruster produces a thrust

**F**, which results in the FOAM experiencing some distributed force pattern that integrates to -

**F**. The thruster accelerates at an acceleration

**a**_{1} =

**F***/m*_{1}, and everything else accelerates at a mean acceleration of

**a**_{2} = -

**F***/m*_{2} (note that |

*m*_{2}| >> |

*m*_{1}|, and accordingly |

**a**_{2}| << |

**a**_{1}|). Momentum is conserved.

The rate of gain of kinetic energy of the thruster and its payload is of course

*P*_{1} =

**F**·

**v**_{1}. For the FOAM, we get

*P*_{2} = -

**F**·

**v**_{2}. This means that the input power to the thruster, assuming no extra energy from non-obvious cosmological effects, needs to be at least

*P*_{in} =

*P*_{1}+

*P*_{2}, or

*P*_{in} =

**F**·(

**v**_{1}-

**v**_{2}).

In the controversial case where the operating principle of the thruster manages to maintain an effective reaction velocity

**v**_{2} =

**v**_{1} irrespective of the value of

**v**_{1}, by somehow weighting its interaction with the rest of the observable universe, or perhaps by calling in some funky cosmological weirdness to balance the energy books, the thrust efficiency η

_{F} = |

**F**|

*/P*_{in} [N/W] is independent of

**v**_{1} and in principle unlimited. This case gives you the flywheel-type pseudo-free-energy machine, possibly violating or at least circumventing the entropy condition in the process. In the opposite case, where

**v**_{2} is essentially constant and no cosmological weirdness occurs, the maximum value of η

_{F} is the inverse of the velocity difference between the thruster and the rest of the observable universe in the axis of thrust. This case gives you the linear-brake-type pseudo-free-energy machine (which works on the same principle as a windmill), and the entropy condition is classically respected. (I don't think this second case is tenable given the form of Woodward's equations, but I haven't pored over all his papers in detail and I'm not very far into his book yet, so...) Neither of these cases results in true "free energy" in the sense of a global conservation violation, nor does any member of the family of cases that can be imagined along these lines.

BTW, the only way to get a frame-invariant kinetic energy - in other words a real energy - is to sum the values of ½

*m*_{i}**v**_{i}·

**v**_{i} in the centre-of-mass frame of reference Σ

*m*_{i}**v**_{i} / Σ

*m*_{i} =

**0**. The value ½

*m***v**·

**v** for a single object of mass

*m* and velocity

**v** is not a real energy, though it can be used for bookkeeping if you know what you're doing.

Just explain to me why I should read a ton of what will almost surely end up being a crackpot theory?

Because you insist on trying to criticize it. What you're doing is called "contempt prior to investigation". As Jim would put it, know something before posting.