The voltage you see at the output terminals of the alternator is the generated voltage minus the resistive voltage drop - which is current times series resistance.

The current produces a magnetic field in the coils that causes a drag between the rotor and the stator.  This drag is directly proportional to the current.

If the alternator is hooked to a resistive load the current (and its resulting drag) will go up in proportion to the generated voltage (and the RPM).  The voltage at the output terminals is a fixed fraction of the generated voltage.  Everything rises in step.  Power delivered is current times voltage (and power pulled from the wind is drag time RPM).  These go up with the square of the RPM.

However if the alternator is hooked to a battery-charging load something very different happens.

First, up to the "cutin" RPM the generated voltage is below the battery voltage (plus the diode drops).  No current flows, there's no drag on the turbine shaft (except for things like bearing friction), and the voltage at the alternator terminals is the open circuit voltage.  The mill starts easily and spins up with the wind speed.

But once you reach cutin the current starts flowing.  The battery is a pretty good approximation of a fixed voltage.  So the voltage driving the current is the EXCESS of the generated voltage over the battery voltage (plus the diode drops, which are also approximately a constant), and the resistance this is driving current through is not the load resistance, but just the resistance of the coils and transmission lines.  So current climbs very steeply with the EXCESS of the RPM above the cutin RPM.  And so does drag.

Power is still current times voltage.  But the power delivered to the battery is the current times the BATTERY voltage, while the power dissipated in the wiring is the current times the DIFFERENCE voltage (generated - (battery plus diodes)).  You see the battery+diode voltage at the rectifier input, that plus the transmission line's portion of the voltage drop at the alternator terminals.  But most of your resistive drop is proabably inside the coils (unless you used really wimpy transmission wiring).

Just as the current climbs steeply with above-cutin RPM, so does the drag.  And that means so does the "slip" of the turbine versus the wind.  If the alternator, transmission wiring, battery internal resistance, etc. were all superconductors, the RPM would track the wind speed up to the cutin RPM, then never exceed that RPM no matter how hard the wind blows.  Because there's nontrivial resistance the RPM continues to rise, but more slowly, once the RPM is above cutin.  You get a straight line with a kink in it:  Rising steeply with windspeed up to cutin, rising more slowly afterward.

Now the amount of power available from the wind goes up with the CUBE of the wind speed.  (Energy in a particular mass of air goes up with the square of the speed, but the amount of mass goes up with the speed as well.  m * v^2 * v = m v^3)  To extract that power optimally you'd want to keep the angle-of-attack of the blades constant.  So the RPM (and generated voltage) would go up with the wind speed and the drag (and current drawn) with its square.  Resistive loads have the current going with the first power of RPM leading to collected power going with the square - a not-too-bad approximation for a while, until the cube function gets away from it at higher RPMs - by which time you probably want to furl anyhow.  For battery-charging loads the kink of the line goes the right way, but it's a single segment linear approximation.

In both cases the failure to track the available power means the angle-of-attack of the blade versus the apparent wind - and thus the efficiency of the blades - changes with RPM.  In the case of the resistive load the current, and the drag, rises too slowly.  But that lets the RPM climb more rapidly than the wind speed, giving a bit of compensation.  (It also puts a stronger load on the turbine relative to the available torque at startup than during normal operation and a weaker one at high wind speeds, leading to both starting difficulties and runaways if furling is absent.)

In the battery-charging case startup is easy because there's no load up to cutin.  But then the load starts rising steeply.  You want to "match" the load to the blades so the slope is about tangent to the ideal cube function in the common operating region.  Get it too high and you'll stall the blades at low wind speeds.  The airflow detaches, you lose torque, and you never get up to proper RPM at the typical operating range wind speeds.  Too low and you don't get the current you could have gotten.  Within the "ok" range you can balance it to get that stall when the wind speed gets excessive (though this means you're not getting all the power you could have below stall).  But that's a critical tuning to hit.  Easier to tune it to pull the best power during operation at the cost of being subject to runaway at high winds, then prevent runaway with a separate furling mechanism.

That will show itself as current 'I' equal to the open-circuit voltage (minus that of the battery) divided by the total resistance as given above.

You may end up, depending on the relative voltages, resistances, etc, losing/dissipating a lot of that I*V power in the windings and wiring rather than getting it into the battery.

Rgds

If I am off base here, then someone more knowledgeable set me straight

IMHO, You cannot directly connect the battery bank to the generator that can produce 100+ DC volts.
The diff between the generator voltage and battery bank would be the system loss or 50% in the coils of the generator.
In a direct connection, whatever current is going thru the battery is going thru the coils.
So an electronic means is required that translates the variable generator voltage to a more fixed voltage range that allows the battery to effectively charge ...but the generator is still running at >100v ..has to!  volts is proportional to rpm ...period

When you have a resistive load, then if the generator is 1.2 ohms  and the load say is 12 ohms, then you have effectively a voltage divider,
so the actual voltage produced is say 100v across the 12 ohms = 8.33 amps ...
BUT, that same 8.33 amps goes thru the 1.2ohm generator coil for a voltage "drop" of  8.33 x 1.2 = 10 volts
In effect the generator produced 110v, but only 100 was usable ...10 % heated the atmosphere

the total energy produced by the turbine is 833 watts in load (measured) + 83.3 watts in generator = 916.3watts ( per leg) which is what the wind is producing as mechanical power after friction, etc is accounted for.

In building an axial flux, the lower the generator coils resistance ie heavier wire, and the bigger the load resistor still able to run at high Cp, the less energy is wasted up the tower. That is why I like the idea of a variable PWM or a means to switch in discrete loads at certain WS ( reads: voltage)

When your fancy unit feeds the grid, I have no idea how the inverter determines how much load to put on the generator.
Where does it tell you the amps at 220v feeding the house circuits, that you reported earlier in a diff thread at 3 ->4 amps??

but that is at the use end , not the generating end

There is a term for the "potential" of the turbine, and it's called "electromotive force".  The phrase is used for the theoretical voltage that can be obtained, but it is only measurable when no current flows.  It is measured in volts, which causes many people to avoid talking about it.  The EMF is equal to Open-Circuit Volts.  When the circuit is open, not current can flow, and you can measure EMF pretty directly.  (There is a caveat on that statement, but there's no point in bringing it up, now).

Once a current starts to flow in the circuit, you start to see voltage drops.  Current multiplied by the impedance of the complete system equals the voltage drop.  If you knew the current, the impedance (resistance) of all components, and the net voltage, you can estimate the EMF.  Your measured voltage is equal to EMF minus all the voltage drops.

In the battery-charging situation, the battery takes control of the measured voltage, but it doesn't regulate the current.  So your measured voltage stays constant, but the EMF can change as the windmill turns faster/slower.  Equilibrium arrives with the current going up and down in proportion.  If you were to check the mathematics, a slow windmill, that has not achieved "cut-in" would have a negative current.  This isn't a false conclusion: the rectifiers that make the AC into DC are preventing reverse current.

In the resistive load, nothing regulates the voltage or the current.  All you have are the sum of the resistances in load and generator, and the EMF "pushing" the electrons.

The voltage created inside the coils of the alternator, once it starts to flow, (i.e. creating current, and moving through a load) generates a magnetic field of it's own that is directly opposite the field generated by the permanent magnets, and is proportional to the current flowing.  

If you find a value of load resistance for which its voltage exactly matches the battery voltage, you can easily find the current flowing through the resistor, which would be the same as the current into the battery, when that is connected in place of the resistor.  From that, you can calculate the power into the battery.  If you know the open circuit voltage you can find the alternator resistance, as well.  You only need to do it for one value of RPM.

In the case of a non linear alternator, such as indicated by your numbers it is more complicated.  Firstly, you need to find the open circuit voltage versus RPM.  That's the easy part.  Then you would likely make a series of plots of voltage (and current) for the alternator as a function of RPM.  The reason for this is that the RPM is likely to vary as the load resistance is changed.  You can then interpolate between the curves to find the value of load resistance that gives a voltage to match the battery's.  Or you could simply connect the battery and measure its voltage and current as a function of RPM.  Well, of course, battery voltage voltage varies considerably, so maybe doing those plots might make sense after all.