Since I'm the one that produced those numbers I better explain:
The table as published is missing a whole set of (more) numbers, and the efficiency you see in there is strictly that of the alternator, based on resistive losses (and yes, I'm fully aware that ignores many other losses, such as friction and eddy currents). The objective was to keep an eye on those losses and avoid letting out the magic smoke. Maybe that is a non-issue, if the alternator simply won't deliver the power requested by the inverter, I just wanted to be on the safe side. Keep in mind that the MPPT info that I derived is just a 'best effort' based on very, very little information to go on. I don't have a power curve for the alternator, there is no power curve for the turbine, I even don't have a solid number for the unloaded AC voltage per RPM (it's derived somewhat circuitously from a DC number). The calculations also ignore a great many things and assume nice sine waves, good power factors (which is not true for a bridge rectifier) etc. So, the margin of error in all the numbers is large. Lucky for me an MPPT curve can be off by quite a bit without much effect on energy production or behavior of the turbine.
To take the entry for 8 m/s:
I start with the energy in the wind for that wind speed, derated for a reasonable rotor efficiency, which works out to 3011 Watt.
The optimal TSR for these blades and the angle they are mounted at is (I was told) 7. In this case I use a TSR of 9 to get a reasonable alternator efficiency and not generate too much heat in there. So, for a TSR of 9 the rotor RPM at 8 m/s works out to 250 RPM.
At 250 RPM I'm calculating an unloaded AC voltage of 288 Volt (3-phase).
For the rotor to not speed up, nor slow down, the alternator needs to load it up to the same power as the wind is generating, that 3011 Watt, which calculates to a current of 6.0 Amp per phase (that power than splits between losses and what is delivered to the inverter).
The internal resistance of the alternator measured at 16.2 Ohm phase-to-phase, it's a wye configuration, so that's 8.1 Ohm per phase leg. With that, the voltage drop can be calculated and I get a loaded AC voltage of 203 Volt.
With the current and internal resistance the resistive losses in the alternator can be calculated, which works out to 885 Watt.
Now that I have the losses, the available power at the output is the wind input minus losses, which is 2126 Watt. The alternator efficiency is output divided by wind input, or 2126 / 3011 = 71% (this is an approximation, ignoring a few more things).
From the loaded AC voltage the loaded DC voltage follows, coming out at 260 Volt DC.
The power left after losses is taken, together with the inverter efficiency, to generate the power coming out of the inverter, into the grid, which works out to 1980 Watt. That forms the MPPT pair: 260V DC and 1980 Watt.
All the other points are calculated similarly. The reason the TSR increases from its ideal of 7 for the lower wind speeds to 11 at the higher wind speeds is to keep the alternator losses down. As you can see from the table it gets up to 1500 Watt at 10 m/s. That is a lot of heat! My question to Jarrod was if the alternator can handle this without starting to smoke (there's lots of air cooling at that wind speed, but still).
SparWeb, the power in the wind would show a cube curve (though I derate for decreased efficiency at higher wind speeds, where drag increases since the blades are truly whirling around at those RPMs). The MPPT curve shows anything but a cube in this case because the TSR is not kept constant. So, the MPPT curve does attempt to truly track the power in the wind, but it's not a cube. The MPPT curve does not affect furling, that is based on the mechanical/aerodynamic properties of the blades vs. tail, since it is the moment between them that causes furling, and it should furl at just about the same wind speed with or without a load (it better does, since there is no guarantee that the grid is always "on", so there may not be a load at times). The Aurora inverters do not clamp voltage. All the inverter does is present a load, based on the voltage. So, for each DC voltage on its input it will present a resistance of XX Ohm, based on the MPPT curve. This load tops out at the highest point of the MPPT curve, or 4200 Watt in that table, with the actual resistance that the inverter presents decreasing as the RPM/voltage of the alternator increases (higher voltage means higher resistance to keep that load at 4200 Watt).
Hopefully this explains things a bit!
-RoB-