The resistance will always change. This is a factor of the SOC of the bank, so dont plan on having one simple answer. You need to understand the resistance curve.

Battery resistance is meaningless in this case. If you short a large 48v battery bank the potential current is in the thousands of amps. The actual battery resistance is in the milliohm region and not relevant to the charging requirements.

There is probably a minimum  safe ( prefer to call it practical ) battery capacity as you would probably need something in the region of 100Ah to safely absorb your charging current in the bulk charge region without damaging the battery.

"If the rectified voltage is 1.4 x the AC this means it's possible I could see 140 VDC ? "

No this is not going to happen, that is probably where you are struggling. If it is a partly charged 48v bank then you are going to see something a bit over 48v and it is not going to rise ( except very slowly as the battery charges).

Just above cut in your current will try to rise far more rapidly than with the heating load and if you had an engine driving this then you would have serious currents to contend with. With wind, you will just drive your prop into stall and you will likely get less current than with the better matched heating load. Unless you have some serious line resistance or you deliberately add some I doubt that you will get the speed up to anywhere near the 300 rpm. If you cut in at 105rpm I would guess that you will hold hard in stall and be surprised indeed if you can get it up to 200 rpm.

Whatever you do you will need to limit the alternator ac line current to the same value as your heating scheme to keep the stator safe.

Your load voltage will be the charging battery volts and the load power will be this voltage x the dc charging current, if you are trying to think in terms of an equivalent load resistance you will get seriously confused as it is a non linear load. If you try to base things on Current squared x true battery internal resistance your figure will be in watts or fractions of and the result will be meaningless.

This is an energy balance situation, you are balancing the mechanical input of the prop against the electrical battery charging load. You don't have unlimited power from the prop and the power you have will be a function of the wind speed and the operating tsr. Your tsr will drop like a stone above cut in unless the alternator is seriously inefficient.

Not going to try again, I can't explain it any other way.

"Battery resistance is meaningless in this case. If you short a large 48v battery bank the potential current is in the thousands of amps. The actual battery resistance is in the milliohm region and not relevant to the charging requirements."

  How is the battery resistance not relevant ? If say someone measures 30 A charging doesn't ohms law state that there must be a resistance and voltage to measure current ? And, won't this battery resistance change as the state of charge changes ?

  "If the rectified voltage is 1.4 x the AC this means it's possible I could see 140 VDC ? "

"No this is not going to happen, that is probably where you are struggling. If it is a partly charged 48v bank then you are going to see something a bit over 48v and it is not going to rise ( except very slowly as the battery charges)".

  Yes, I understand this also (if charging) I just wondered if I had the 1.4 x the AC voltage correct for figuring the (open) DC voltage. If so then I could expect to see approx. 140 open volts DC from 100 VAC. Correct ?

   "This is an energy balance situation, you are balancing the mechanical input of the prop against the electrical battery charging load. You don't have unlimited power from the prop and the power you have will be a function of the wind speed and the operating tsr. Your tsr will drop like a stone above cut in unless the alternator is seriously inefficient."

  This makes perfect sense to me also, only though if in your first sentence ... battery charging LOAD is a "relevant" resistance.

  I am hoping to understand battery charging, IE: current, resistance, charge rate, amp hour rating etc. with regard to a constantly changing DC charge voltage such as with a wind generator.

  I am seriously trying to understand this and you are correct that if the resistance component is "meaningless" then the other calculated relationships I am trying to figure will be impossible to figure and therefore will be "meaningless" also.

  Going back a few postings under controls maybe someone can answer my following question : Let's say I have a constant wind for as long as I want and this produces a constant rectified 70 VDC voltage. I have an ammeter properly connected to the charging circuit to a 48 VDC battery bank that is at 50% fully charged. What will I read at the ammeter and what components of the circuit are responsible for this reading ? As charging time passes, what will I notice for  the ammeter reading ? I'm not looking for exact figures just : what happens as the battery bank charges ? When during this process does the resistance of the stator + line resistance become a concern for stator heat and or burnout ?

The battery resistance is almost 0.00 ohms.
'Nothing' compared to the ohm or 2, or 10 ohms, of the rest of the circuit.
Not enough to bother figuring in any calculations.

Maybe consider the battery the same as a perfect 1Giga Watt 48V Zener diode.  Or 56V if you like.  Same idea.

seems to me that trying to get to a definitive answer is much like holding water in a sieve or shooting at a moving target. (a very irratically moving target)

and i am not at all convinced that ohm's law is the last word on figuring where the power goes.

considering ohm's law is most useful in purely dc circuits, and somewhat less useful in ac circuits,, the question then becomes

is an AC generator/alternator charging a battery via rectifiers more like a purely dc circuit?   more like an ac circuit?     something in the middle?   or worse ... something in the middle that moves around acting more like a dc circuit at one point in the charging process and then more like an ac circuit at another point?

and is this relationship linear?

seems like a pretty good case for more effort in development of controlled rectification  or some form of switch mode conversion scheme.

there is just so many friggin variables, i can't see how anyone is going to do all th math needed to get a load match between generator and battery that suits a broad spectrum of windspeeds, and the broad spectrum of power that goes into charging a battery.

sincerely,, what am i missing here?

"Battery resistance" is one of those things that you never worry about because you never see it manifest itself in your design.  It has no effect on anything.

Then one day you hear that batteries have internal resistance.

To add a few more variables, how about reactance, inductance, and AC ripple passing the diodes.

It sounds like you have a package unit from plans or a kit. The maths will have been worked out, and I think you said you had 700 AH of battery power, that should sink any surplus power from the PMG and the cable run will heat up a bit if it is pushed.

Interesting reading though and it shows there is a place for a very simplyfied manual for people starting to be interested in RE.  All the sites I have looked at assume pre knowledge.  I am sure if the knowledge was readily available it would make RE less of a black art followed by geeks and strange people.  It would also stop some of the companies advertising turbines that defy the laws of physics.