Okay I'll take a break from the thread.
The first question that comes to my mind is about the analog voltage and current meters. I have limited experience with analog meters. I am wondering if they are AC or DC meters, you can't tell from the markings. I suspect that either type of meter might act like a peak detector for the voltage/current waveforms, and that would exaggerate the output power measurement.
The motor is set up with nine stator coils and eight rotor magnets. Two of the stator coils are drive coils and seven are pick-up coils.
For a minute, let's assume that you have nine pick-up coils. In that case as the rotor turns you have "poly-phase" full-wave rectification (akin to mains power three-phase rectification). At any given time one of the pick-up coils is generating the most EMF and drives the output. Which pick-up coil drives the output fires off in a round-robin pattern like the firing of the cylinders in a piston engine.
Getting back to the real setup, there is a major difference. There are seven pick-up coils and not nine. That means that two out of nine "cylinders" "don't fire" as the generator runs. It's like the output voltage waveform (and current waveform for a resistive load) has two "missing teeth."
Another issue that would have to be checked on the scope, is how much of a "dead band" is there between voltage peaks. When you look at the geometry of the setup and pretend for a second that the rotor and stator have nine pick up coils and nine magnets (i.e.; eliminate the poly-phase rectification) then you can see that there is a significant "dead band" between magnet passes where the AC voltage output would be very low.
I am suspecting that the analog meters can't really detect the "missing teeth" and the "dead bands" in the waveforms and are therefore acting like peak detectors instead of giving you the true average voltage.
Let's assume for the sake of argument that my hypothesis is correct. Keeping it really simple let's try to make a derating factor that we can apply to the voltage and current measurements.
The "missing teeth" derating factor = (7/9) = 0.78
For the "dead band" derating factor, let's make a conservative guess of 0.85.
Therefore the total derating factor = 0.78 * 0.85 = 0.66
Let's look at the numbers:
The setup is not fully completed but these are the results so far:
in 12.50v * 0.94a = 11.75W
out 11.8v * 1.8a = 21.24W
11.8 V * 0.66 = 7.79 V
1.8 A * 0.66 = 1.12 A
7.79 V * 1.12 A = 8.72 Watts
This is the output power from the generator after applying a derating factor to the voltage and current measurements from the analog meters.
There is an easy way to make an accurate generator power output measurement that Romerouk could do if he has a true-RMS digital multimeter.
We are going to assume that Romero can determine the resistance of his light bulb filament when illuminated with a fair degree of accuracy. This might require doing a separate simple experiment which we won't discuss now.
Assuming this is true then using a true-RMS multimeter all that he has to do is calculate Vrms-squared/Filament_resistance to make a very accurate generator output power measurement.
Even easier, drop the light bulb all together and use a 10-watt power resistor that has the same approximate resistance of the light bulb filament.