On Wed, Mar 9, 2011 at 12:35 AM, Kirk Wallace
<kwall...@wallacecompany.com>wrote:

> Let me try to provide more details on my understanding of the phase
> timing of DIY converters. Attached is a schematic of a common rotary
> converter. The source is here:
> http://www.metalwebnews.com/howto/ph-conv/ph-conv.html
> http://www.metalwebnews.com/howto/ph-conv/fig1.html
>
> I used this to make my converter and is the only design I have some
> understanding of. As the schematic shows L1 and L2 go straight to the
> output and are unaffected. Since L1 and L2 are single phase, L1 is a
> mirror of L2, therefore 180 degrees out of 360 apart.
>
>
But, but, you are committing a tautology. If you reference L1 to L2 and vice
versa, of course L1 is a mirror of L2. In your scheme there is no 'ground'
or reference point, so you only have one phase coming in, and the motor
generates the second (and third) phase L3, shifted with respect to L1/L2.
One way to see it comes from the trigonometric identity  holding that sum of
three sines shifted by 0, 120 and 240 degrees is zero:
sin(x)+sin(x+2*%pi/3)+sin(x+4*%pi/3)=0

Let's say we take some arbitrary reference point, so that L1(t) is the
voltage on the first leg. Assume L2(t)-L1(t) = V sin(2 pi f t). If the
converter generates L3(t)=L1(t)-V sin(2 pi f t + 2/3 pi), then L1-L3 is V
sin(2 pi f t + 2/3 pi), i.e. shifted by 120 degrees, and L3-L2 is -V sin(2
pi f t + 2/3pi) - V sin(2 pi f t)  which reduces to V sin(x+4*%pi/3) i.e.
the same AC shifted 240 degrees. Note that it doesn't matter what the
reference point is, because its potential L1(t) drops out --- all that
matters are voltage differences between the legs of the circuit, not the
potentials of the legs themselves.
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