" We need to look at odd, obscure effects that physics has overlooked."
Not physics but conventional physicians and engineers !
It is a step up/down (= lever) game with the "parametric" functions time and distance !
Pulse with modulation;PWM and the "duty cycle" https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=1&ND=3&adjacent=true&locale=en_EP&FT=D&date=20121011&CC=US&NR=2012256422A1&KC=A1
 Power from Ph.1 stator winding is effectively the root-mean-square value of its stator current multiplied by its rms stator voltage=(Ph.1 peak current)*(Ph.1 peak voltage)*(0.5). Likewise, Ph.2 power=(Ph.2 peak current)*(Ph.2 peak voltage)*(0.5). Since Ph.1 and Ph.2 are essentially equal in magnitude and time displaced by 90° phase relative to each other, the Ph.1 plus Ph.2 power sum is (peak current)*(peak voltage) of either Phase 1 or Phase 2.
Inasmuch as peak current and voltage of Ph.1 and Ph.2 are equal to each other, and each is sinusoidal with 90° relative phase, and sin<2>(A)+cos<2>(A)=1, and the sum of power from Ph.1 and Ph.2 equals power fed to the DC power bus IDC*VDC, then, for either phase: (peak stator current)*(peak stator voltage)=(IDC,*VDC).
 The above equation explains why controlling peak stator current so it is proportional to speed squared, when multiplied by peak stator voltage, which is proportional to speed, results in output power (IDC*VDC) proportional to the third power of speed. Coupling this generator to a wind turbine having blade pitch control or other means to limit its speed maximizes energy yield from the most prevalent winds, when power usually is most needed. It also protects the turbine from mechanical stress incurred by turbines that do not have said speed-limiting features, while still providing controlled DC electric power at levels its loads can accept.
 Generator power and efficiency with wind turbine drive is computed below, for a representative example of the present invention, at maximum shaft speed, mid-speed, and minimum usable speed, using a few simplifying approximations. Shaft speed, power, and the other variables in the computations herebelow are exemplary, and not intended as limiting the present invention in any way. This will help explain FIG. 1 and FIG. 2 configuration operation, distinctions and improvements over widely used prior art generators:
 Let maximum speed equal 1000 revolutions per minute (rpm), mid-speed equal 500 rpm, and minimum speed equal 100 rpm. Also, let maximum stator current Imax=10 amperes, and nominal VDC=100 volts. Further, let Q1-Q4 power MOSFET ON resistance Rdson=0.01 ohm, inductor L1-L4 series pair winding resistance RL=0.1 ohm. Also, stator winding resistance Rs=0.15 ohm, stator voltage Vmax=100 volts at 1000 rpm, and fly-back (free-wheeling) diode D1-D8 forward drop Vf=1-volt at 10 amp. These parameters are consistent with a test prototype, according to the present invention, developed to generate power from wind turbines.
 At 1000 rpm, Vmax=100 volts, so PWM duty-cycle (Ton)/(Ton+Toff) is essentially zero. Therefore, losses=Imax<2>(RL+Rs)+2 VfImax=(10 amp)<2>(0.25 ohm)+(2 volt)(10 amp), amounting to 45 watts loss. Output power=(Imax)*(Vmax)=(Imax)*(VDC)=(10 amp)(100 volts)=1000 watts. So, generator efficiency at maximum speed and maximum power is about 95% for this example of generator and integrated electronics parameters.
 At 500 rpm, Imax=(10 amp)/(4)=2.5 amps; and Vmax=(100 volts)*(0.5)=50 volts. So PWM duty-cycle=1⁄2. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/2=(2.5 amp)(50 volt)=125 watts. Losses to maintain inductor current=Imax<2>(RL+Rs+Ron)=(2.5 amp)<2>(0.26 ohm)=1.6 watts. Fly-back diode losses=2 Vf*Imax/2=(0.6 v)(2.5 amp)=1.5 watts. So total losses=3.1 watts. Therefore, mid-speed generator efficiency is about 97%.
 At 100 rpm, Imax=(10 amp)/(100)=0.1 amp; and Vmax=(100 volts)/(10)=10-volts. So PWM duty-cycle= 9/10. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/10=(0.1 amp)(10 v)=1 watt. Losses to maintain stator and inductor current=Imax<2>(RL+Rs+2*Rdson)=(0.1 amp)<2>(0.27 ohm)=0.0027-watt. Fly-back diode losses=(2*Vf)*(Imax)/10=(0.6 v)(0.1 a)/5=0.012 watts. So total losses=0.015-watt. Thus, generator efficiency at low speed is about 98%.
rotorical,static/MEG/linear ,kinetical/mechanical ?
Motor : B/EMF ? F= BIL F= Compensation ?https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=5&ND=3&adjacent=true&locale=en_EP&FT=D&date=19790215&CC=DE&NR=2733719A1&KC=A1
partial translation by google:
In a conventional generator the field strength B, the length of his head with L and the current flowing through is denoted I, the result is the size of the counter force F, which is exerted on the conductor and thus to the shaft to F = BLI.
If the field strength of the electrical device 2B, the length of its conductor L is and only one current durchfiiesst, corresponds to the thickness of the half of the current flowing in the generator current, is the force exerted on the armature force F '= 2B x L x 1 = LI.
2 This shows that the force exerted on the armature of the electrical device and hence on the driving shaft power is equal to the counter force, which is exerted on the shaft from the conventional generator. The force is generated to act in a direction such that it compensates the reaction force. A stronger current flow provides the required torque to the shaft. Therefore, the shaft and thus the armature are moved on and generates electric power.In order to flow current through these half the electrical device, half the generator voltage is required. So with is 1/4 of the power generated is lost to engegenzuwirken the deceleration force. The remaining 3/4 of the energy generated or slightly less of them are for outdoor use are available.
If the field strength of the electrical device is a multiple of the field strength of the generator, for example, 5-fold, it is necessary that 1/5 of the current generated flows in the electrical device to generate the compensating force. 1/5 of the generated voltage is required for the current flow. Thus, 1/25 or 4% of the energy produced is consumed in the electrical device, while 96% or less of them are for outdoor use are available. In this way it is not necessary that any electrical energy is introduced from an external source in the generator according to the invention for generating electrical energy.
read also Kanarev "Pulse Power" explaination http://gsjournal.net/Science-Journals/Communications-Mechanics%20/%20Electrodynamics/Download/1180https://www.google.pt/search?q=ohmsche+gesetz&client=opera&hs=8XG&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwirk-bShqXOAhWIIcAKHYUED0wQsAQIPQ&biw=1366&bih=640