@Hayden Watson , I don't know where to begin. You've written two long, authoritative-sounding posts that, in my opinion, are just conceptually and practically incorrect. Rather than attempt a fisking of them, which would be time consuming and perhaps pointless, please allow me to make a couple of observations that are contrary to your presentation. Before I do some let me say that I am fine with being shown I'm wrong, and learning something. My academic experience with alternators, generators, and motors was my undergraduate "power" lab, a 1 year lab while I was in school for electrical engineering. But that was 42 years ago, and I, unfortunately, discarded my textbooks and notebooks several years ago, and, of course, my recollection could be foggy. My practical knowledge is only with alternators on boats in the last 20 or so years.
Specifications
I have never seen an alternator specification that specifies alternator operating characteristics in terms of Watts. It has always been, for a given nominal voltage output of 12 or 24 or 48VDC, a maximum current output at a particular RPM and ambient temperature. For example, a 12V Bosch rated at 122A at 6,000 RPM and 80º C.
Example specification:
Bosch Motorsport | Alternator B5
Operation
In that it's the magnetic flux of the field (carried on the rotor) that induces the EMF in the stator, it is the field current that controls the alternator output, not the field voltage. The simplest, internal regulators adjust field current to maintain the set output voltage, which, for example is 14.4V.
Output
The alternator doesn't "know" its ratings. I t must be integrated into a properly designed system in order to not exceed its ratings. System considerations are environment (temp), desired output, and load characteristic. So, if you put a bigger load on an alternator, by, for example, attaching it to a bank that has a lower impedance than perviously, like switching from FLA to LFP, the output current will increase. Ohm's Law, in the simplest analysis: I = V/R. Note that Volts and Amps go up proportionately, not as you stated:
The amps are dependent on a variable of Volts so as volts to up, amps go down and vise-versa.
That statement is incorrect. If you increase the voltage, the current will increase, not decrease.
Note that batteries are more complex than just a simple resistance, but in the first order analysis the analogy is useful.
Note also that in your analysis you are using a Watt rating as an independent variable, and saying that the current output is a function of a fixed power rating and the output voltage. This is also incorrect. The power output is a product of the current output and output voltage, the latter two being controlled by the regulator (assuming sufficient rotor speed). So, put 0.1Ω across the output of that Bosch B5 and spin it at 6k RPM in 80ºC and it will put out 144A all day. Put 0.05Ω across the output and it will put our 288A for a short period of time (assuming the regulator can put out sufficient field current), until it burns out - probably wreck the stator.
And that's all there is to it.