You are not wrong.... nor am I.... why?
The DC you are measuring is pulsed DC, so consists of the AC waveform, but manipulated by the diodes so that all the waveform is above the zero line. You have DC rising from zero to say 50v and back down at twice your AC frequency ( two positives per cycle now)... so it is not RMS style DC that we normally think of..... is actually a DC waveform that varies from zero to the peak of the ac wave... the meter will probably see this, and plot the rms value.... unless you have a meter that measures peak DC. ( not caring about diode loss for the moment) The DC waveform will be almost a replica of the AC wave, but all above the zero line.
To test this, get a small transformer, measure the AC.... now add diodes and rectify,.......now measure the DC, ............now put a capacitor across the output and measure again.... then tell me what happened.
When you measure say a battery, all the waveform has the potential at the terminal voltage ( straight line), with the pulsed DC waveform, the continuous DC voltage will be .7071 of the peak if the wave was a rectified sine wave..... or the RMS value. If you get all the area under the waveform, and average it ( make it a square wave) to a consistent Y axis value ( straight line), you get the same as the rms value.
Does that make sense now?.... or use the scope, and see if this is true, or your meter is smarter than mine ( cheapo things)
An analogue meter will have to display the RMS dc equivalent of the pulsed DC, as it takes work to move the needle, and the gaps between peaks will allow the needle to relax before the next impetus... so it will average the peaks for you that way.
The same is true of the current, the amp meter will show the average current, it will not show the peak currents that circulate... so if you have say a peak current of 1 amp for 1 millisecond.. followed by no amps for 9 milliseconds, then 1 amp etc etc..... you effectively have pwm... the voltage/current spikes will be terminal voltage that the switch passes to the load, but the motor or light will run/glow at 10% of it's max... and the current meter will show 10% of the peaks.... not the 1 millisecond peak.....and the power will be 10% of continuous switched on........ but all still DC.
In truth, if we use the geometry of the alternator to create an almost square wave, we limit the i^2R losses, more than a sine wave... which will have peak currents 1.4 time the square wave version. ( this will require no leakage, and very tight airgap (in the order of 4mm with windings 3mm thick) Not very practical, but Gwatpe did make one like this, and was a near square wave)
....................oztules
Edit. For those without the facilities... that transformer testing will go like this.:
For a 20v transformer, we see 20vac
When we rectify, we will drop some volts across the rectifier, depending on the diodes 1-2v so our rectified voltage will be measured by a normal multimeter as about 18vdc to 19vdc or thereabouts...
When we put an electrolytic capacitor of say 1000uf across it, the voltage will now read about 26vdc to 27vdc.
So our 20vac rectified and filtered will be say 26/18 or about 1.4 ish... or around the other way, 18/26 will be about .7........so it will work out close to the theory. The diode drops are difficult to guess at, as it depends on type, but in the ballpark of accurate enough to play with....
The startup/cut-in of the mill will be about when the DC peaks hit the battery.... then onwards and upwards (we hope)
