Hello SamoaPower,
Astounding, nope. Interesting maybe!
Considering your past posts your going to enjoy this.
You painted a very good story, for a moment I though I screwed up. Nope!
For the answer we need to take a look at a bigger picture.
Your example used 13v so assuming 10' system. 36% eff.
Vb=13v, Vd=2.4V (two diode drops as in a bridge) and Ra=1 ohm.
For low currents the diode drops well be less, keeping fixed to make it easier to simulate.
Cut in is 5mph, available wind power is 18watts, want 1 amp charge current. Setting this up to establish the voltage needed at cut in is 16.4v = 13v + 2.4v + Vra (1 * 1).
For a given alternator speed, we have a certain open circuit voltage, Voc.
All other voltages are a ratio of this voltage based on the RPM.
For a given wind speed there is a limited available power out, if loaded greater then the available power out the RPM will drop lowering the forcing voltage resulting in matching the load to available power. The power will remain based on the wind speed not the RPM, the blades are being stalled a little, so the output power will be based on the profile of the blade. Here we will assume is within range of the blade and the efficiency has not changed.
This explains why with the generator shorted it spins slowly, also why in very high speed winds it can turn enough to burn out or at least spin faster. Efficiency at this condition sucks, we are out of the power profile of the blade, second reason it spins slowly.
Example: 6mph Pwind = 31 watts, 19.68Voc, 1.58a for 31 w.
V = 13 + 2.4 + 1*1.58 = 16.98v or 2.7v mis-match causing I to exceed 1.58a. But not goto 2.7a
RPM will lower until 17.2v is reached and I = 1.8 a for a 31 watt. Pb = 23.4w
Short diodes. Vd = 0. For 5.1v mis-match. I = 2.05amps. 0.25a different. Pb = 26.65w
Had to use a quadratic solver to get the numbers above.
Equation: RI^2 + 15.4I = 31 for the first, and RI^2 + 13I = 31 for second.
Example: 15mph, Pwind = 485 watts, 49.2Voc at 9.87a
With diodes: I = 15.6amps, 31.0v Pb = 202w. Pheat = 283 w
Short diodes, I = 16.46amps, 29.46v Pb = 213w. Pheat = 272 w
Pheat is the power as heat which is the diode and stator. With the short diodes the stator heat goes up by 283 - 2.4*15.6 = 283 - 37.44 = 245.56 or 26.44 watts.
Most systems furl in the early 20's mph, let's pick 22mph for the next example:
22mph Pwind = 1531watts, 72 Vos, 21.22amps.
With diodes: I = 32.2a, 47.6Vos, Pb = 418w
Short diodes: I = 33.2, 46.2Vos, Pb = 431w, 13w gain. 0.8%
3.25watts extra for your trouble for the first 10% gain, I knew the benefit near cut-in was the best, goes down hill from there. 11 watts extra for the second 2% gain.
I did over simplify it with my statement, I don't think your doing this for 3watts gain at cut in.
Some people have put series heaters in line lets try 2 ohms. Reports were current went down a few amps, 15mph I = 10.4amps. Pb = 135w, or 68w less. Current dropped more than I was expecting.
2*10.4^2 = 216w in the heater, stator heat is 108w or half. If the heat is as usable as charging the battery then the system is much more efficient.
I'm very glad we had this talk, I have a much better understanding. This took hours to write. What do you think?
Have fun,
Scott.