Calculating phase voltage
Completely revised answer...

Alternator Simulation

If the alternator produces 9V peak sine waves.  Your Red Dot to Green Dot voltage will vary from -9V to 9V.

\$V_{Red} = 9 sin (\omega t) V \$

\$V_{Green} = 9 sin (\omega t\ -\ 120^{\circ}) V \$

\$V_{Blue} = 9 sin (\omega t\ +\ 120^{\circ}) V \$

$$V_{Min} =  9 sin (90^{\circ}) - 9 sin (90^{\circ} +\ 120^{\circ}) - 2 \times 0.7V = 12.1V$$

$$V_{Max} =  9 sin (120^{\circ}) - 9 sin (120^{\circ} +\ 120^{\circ}) - 2 \times 0.7V = 14.19V$$



The actual voltage seen by the battery will be line-to-line voltage across two phases minus the diode voltages.

\$V_{Line} = \sqrt {3}\ V_{Phase} = \sqrt {3} \times 9V = 15.59V\$.

\$V_{Bat} = V_{Line} - V_{diodes} = 15.59V - 1.4V = 14.19V\$.

\$I_{Bat} = \frac {V_{Bat}} {R} = \frac {14.19V} {100\Omega} = 141.9mA\$.

Maximum Power at Resistor:
$$ P = \frac {V_{Bat}^2}{R} = \frac {(14.19V)^2}{100\Omega} = 2.01W $$
Three-phase Power:
$$ P = \sqrt {3}\ V_{Bat}\ I_{Line}\ cos \theta = \sqrt {3} \times 14.19V \times 141.9mA \times 1 = 3.487W$$
The Three-phase Power calculation is wrong because of the action of the diodes. 
 Only two sets are on at the same time as shown by the drawing.  This also means there will be current flowing in the neutral.

Single-phase Power with Line voltages:
$$ P = V_{Bat}\ I_{Line}\ cos \theta = 14.19V \times 141.9mA \times 1 = 2.01W$$
This agrees with the load calculation.

Single-phase Power with Phase voltages:
$$ P = \sqrt {3}\ V_{Phase}\ I_{Phase}\ cos \theta = \sqrt {3} \times (9V - 0.7V) \times 141.9mA \times 1 = 2.04W$$
Again, this agrees.  Because of the \$\sqrt {3} \$, I'd go with line voltages.

The action of the diodes also means the Green and Blue dots will NOT be at the same potential.  Image shows +Bat varying from 8.35V to 3.86V, while -Bat varies from -3.86V to -8.35V.  This brings us back to the \$V_{Min}\$ calculation.   



$$V_{Min} = 8.35V - (-3.86V) = 12.21V $$

$$V_{Min} =  9 sin (90^{\circ}) - 9 sin (90^{\circ} +\ 120^{\circ}) - 2 \times 0.7V = 12.1V$$

Your Green Dot will be at 0V and the Blue Dot to Green Dot voltage will vary from -3.86V to -8.35V.

Minimum Power:
$$ P = \frac {V_{Bat}^2}{R} = \frac {(12.1V)^2}{100\Omega} = 1.46W $$

Numbers should scale to real world.



To calculate phase voltage:


Measure \$V_{Bat_{Min}}\$.
Calculate \$V_{Phase} = (V_{Bat_{Min}}\ +\ 1.4V) \times \frac {2}{3}\$.


To calculate power (Max or Min):


Measure \$V_{Bat_{Min}}\$.
Measure \$I_{Bat_{Min}}\$.
\$P = V_{Bat_{Min}} \ I_{Bat_{Min}}\$
Repeat using Max quantities.