[Alternator Simulation][1]
If the alternator produces 9V peak sine waves.
\$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 \$
[![Alternator Output][2]][2]
The actual voltage seen by the battery will be line-to-line voltage across two phases minus the diode voltages.
We can determine this by looking at two points. Between \$60^{\circ}\$ and \$120^{\circ}\$, we can assume diodes associated with red and blue are conducting.
So at \$90^{\circ} \$, red and blue are conducting. Assuming 0.7V for diodes.
$$V_{Min} = V_{Red} - V_{Blue} - 2 V_{Diodes}$$
$$V_{Min} = 9 sin (90^{\circ}) - 9 sin (90^{\circ} +\ 120^{\circ}) - 2 \times 0.7V$$
$$V_{Min} = 9V - (-4.5V) - 2 \times 0.7V = 12.1V$$
At \$120^{\circ}\$, red and blue are conducting. Green is 0.
$$V_{Max} = V_{Red} - V_{Blue} - 2 V_{Diodes}$$
$$V_{Max} = 9 sin (120^{\circ}) - 9 sin (120^{\circ} +\ 120^{\circ}) - 2 \times 0.7V$$
$$V_{Max} = 7.79V - (-7.79V) - 2 \times 0.7V = 14.18V$$
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Which is \$V_{Line} = \sqrt {3}\ V_{Phase} = \sqrt {3} \times 9V = 15.59V\$, minus the 1.4V for diodes.
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So depending on the maximum and minimum you see across the battery terminal and blue dot, you should be able to work backwards to derive instantaneous waveforms for alternator. Your red and green dots.
At \$90^{\circ}\$, \$V_{Blue} = 50\% \ of \ V_{Red}\$ or at \$120^{\circ}\$, \$V_{Blue} = V_{Red}\$.
Actual voltage will also depend on speed of rotation of alternator.
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It's a 3-phase alternator.
$$P_T = \sqrt {3} V_{Line} I_{Line} cos \theta $$
$$P_T = 3 V_{Phase} I_{Phase} cos \theta $$
Turned into a dc battery charging circuit.
$$P_T = V_{Alternator} I_{Line}$$
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The Green and Blue dots should be at the same potential.
Then measure ALL 3 phase voltages at the same time. If they are all the same, then the Green and Blue dots are at the same potential.
You say they are NOT, but you provide no evidence. No measurements. No info on alternator. You say ignore this. It is not important. This (should be) a 3\$ \phi\$ alternator connected in wye with two diodes. It is fairly simple. The math is correct and the simulation gives close results. We cannot see what you see and you give us nothing to work with.
You have not been forthcoming with information, so this is my last on this. Have a nice day....
[1]: http://www.falstad.com/circuit/circuitjs.html?cct=$%201%200.000005%2010.20027730826997%2053%205%2050%0Av%20192%20224%20272%20224%200%201%2040%209%200%200%200.5%0Av%20192%20224%20160%20160%200%201%2040%209%200%202.0943951023931953%200.5%0Av%20192%20224%20160%20288%200%201%2040%209%200%20-2.0943951023931953%200.5%0Ad%20256%20256%20320%20256%201%200.7000000000000001%0Ad%20352%20256%20416%20256%201%200.7000000000000001%0Ad%20256%20288%20320%20288%201%200.7000000000000001%0Ad%20352%20288%20416%20288%201%200.7000000000000001%0Ad%20256%20320%20320%20320%201%200.7000000000000001%0Ad%20352%20320%20416%20320%201%200.7000000000000001%0Aw%20320%20320%20352%20320%200%0Aw%20320%20288%20352%20288%200%0Aw%20320%20256%20352%20256%200%0Aw%20256%20256%20240%20256%200%0Aw%20256%20320%20240%20320%200%0Aw%20256%20288%20240%20288%200%0Aw%20240%20256%20240%20288%200%0Aw%20240%20288%20240%20320%200%0Aw%20416%20256%20416%20288%200%0Aw%20416%20288%20416%20320%200%0Aw%20160%20160%20336%20160%200%0Aw%20336%20160%20336%20288%200%0Aw%20336%20288%20352%20288%200%0Aw%20160%20288%20160%20336%200%0Aw%20160%20336%20320%20336%200%0Aw%20320%20336%20320%20320%200%0Ar%20464%20256%20464%20320%200%201000%0Aw%20416%20256%20464%20256%200%0Aw%20464%20320%20464%20352%200%0Aw%20464%20352%20240%20352%200%0Aw%20240%20352%20240%20320%200%0Aw%20272%20224%20320%20224%200%0Aw%20320%20224%20320%20256%200%0Ao%2025%2064%200%204102%2020%200.05%200%204%2030%200%2023%200%2019%200%0Ao%2025%2064%200%2020738%2020%200.05%201%202%2025%203%0A
[2]: https://i.sstatic.net/MET0f.png