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Jerry Rigged Ins and Outs

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GerryS:
Jerry pointed out in the past that Jerry Rigged alternators don't fry in situations that star-rigged alternators do.  I calculated some simulations and it seems to me that this is not always true.  
Example in point:  A 9-coil stator, each coil has 75 turns of AWG#22 and each coil has .65 ohms resistance.  Each coil gives 2.0V @168RPM.  The maximum RPM this alternator sees is 600RPM in strong winds. It is charging a 12V battery.
Wired in star:  

.65Ohms x 6coils/phase x 1.3 = 5.07 ohms effective resistance

2V x 1.404 x (600RPM/168RPM) x 3coils/phase x 1.73 -1.4V = 50.64V rectified

(50.64V-12.6V) / 5.07Ohms = 7.5 Amps through AWG#22 Wire
Wired in Jerry-Rigged:

.65Ohms x 3 = 1.95Ohms/phase resistance

2V x 1.404 x (600RPM/168RPM) x 3coils/phase -1.4V = 28.68V rectified

(28.68V-12.6V) / 1.95Ohms = 8.25 Amps through AWG#22 Wire
So are my calculations wrong?  Is it ok to use the "effective 3-phase resistance" the way I did?  
What's really interesting is that if "not frying your stator" is what's important, on the graphs of Jerry-Rigged amps through a wire and 3-phase star amps through a  wire, the graphs cross where Jerry-rigged becomes a higher amps through the coil  than star-rigged.  Here is a graph of what I mean

.  In my example, if you use a coil that produces more than 1.7V at 168RPM (6V@600RPM), star has few amps running through the wires.  Using a coil that produces 1.8 volts or more, Jerry-rigged has more amps flowing through the wire.  Using a coil that produces 1.7V or less at 168RPM star-rigged has more amps going through the coils.  I created a spreadsheet with the information: http://www.otherpower.com/images/scimages/5441/Jerry_vs_3P.JPG
The point that one becomes lower amperage than the others changes depending on the # of turns of wire used in the coils.  

Ungrounded Lightning Rod:
Jerry rigged, at a given RPM, gives you the same generated voltage as delta but less current.
 - In jerry rigged the coils only conduct when they are individually putting out a higher voltage than the battery plus diode drop.  A coil will either be conducting in the one-coil path or turned off at any given moment.
 - In delta you get that current from one coil, plus half that current passing through two coils in series - 1.707 times the jerry rig heat losses in the genny for 1.5 times the charging current.  A coil will be in the one-coil or the two-coil path at any given moment (unless things are spinning so slowly that ALL the coils are off part of the time - which happens when jerry-rigged would have each coil conducting for less than a third of the cycle.)  In a situation where jerry rigged would have each coil conducting for no more than half the time the current waveforms in both the one-coil and two-coil parts of the delta's waveform will be the same as the jerry-rigged cycles (except that the two-coil waveform will be half-amplitude).  It gets more complicated when jerry rig coils would have more than 50% duty cycle but I think the results come out about the same.
 - In Y all the current goes through two coils' resistance (though you get a higher voltage so you wind things differently).
For a given amount of copper in the core, though, your power limit occurs at the same amount of heating - and heating goes with the square of the current.  Jacking up the power through the jerry rig (say, by raising RPM) to match the resistive losses in the core for delta leave jerry rig with more losses than delta.  (Delta conducts on two parts of the cycle, jerry rig on one, so delta is less uneven and the highs are less efficient than the lows.)
Multiplying jerry rig heating by a factor of 1.707 multiplies the output current by 1.307 or so, so jerry rig gets 1..307/1.5 = 87% of the output of delta for a given amount of heat.  Less than delta, but a 13% shortfall is not too shabby.
A Y/wye/star connection gets about 1/1.732 = 58% of delta's power or 67% of jerry-rigged for a given amount of heating - assuming a resistive load.  I'm not sure if the current distribution among the coils would throw it off that ratio for a charging application - but even if it did it has a long way to go to match either jerry rigged or delta.
Finally:  Delta may have additional heating from circulating currents driven by waveforms that don't add up exactly to zero around the circle.  This will reduce its advantage over these alternatives.

willib:
There is one way to find out if your stator is wound correctly or evenly .
that is to wire it in delta , and if it spins as though there is no load at all , "With no load attached " then you have done well ,

because all currents are cancelled ,
if you have resistance to turning there is a problem somewhere.

Flux:
If the phase voltage is sinusoidal the displacements are 120 deg and the phase voltages are equal then you will get no circulating currents in delta on no load.
When you connect the non linear load of a rectifier then you will get odd harmonics circulating in delta no matter how well you have built it. Star cancels the circulating harmonic currents At the expense of an oscillating star point voltage.
Flux

JW:
https://www.fieldlines.com/index.php?topic=128829.0

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