"Also, If this can be geared 3:1 and get 4X rated voltage.  That would be around 600V  three phase.  Is that correct? "

no not really ... I had a 24v DC PM motor that was rated at 194rpm ..we used a vbelt at 2:1 and got it up to 100v, so the turbine was spinning around 400rpm, motor shaft at 800rpm ...you have to know how fast you expect the turbine to spin, then figure the pulley ratio to match. You can't get out more than 35% of the power in the wind.

If you don't charge batteries, you can heat at 100% efficiency using a calrod heating element (cheap) in a domestic hot water tank like DaveB does ie preheat your water when the wind blows

Charging 12v battery is the worst option you can choose, and 48 volt charging is commonplace. Remember that battery charging is inherently 70% efficient.
Simple ohm's law dictates how much voltage loss you get in the wire that brings the juice to your home. I can run 230v/3phase at 4amps (3KW) 250 feet to the house with #12 wire

V=IxR
P=VxI=V^2/R=I^2xR

if the voltage is 5x, then the amps is 1/5x ...resistance in the wire stays the same once you buy it
so, either you choose very thick expensive wire for low voltage/high current, or very thin inexpensive wire for high voltage/low current

If you look at my calculator, you can get the rpm for any combo of turbine diameter built, TSR (assume=6) and WS ...then you pick the pulley ratio to get the voltage you want out

From that rpm, you now know what voltage you will get BUT, you have to look to see how much power is in the wind and multiply by the efficiency Cp= 0.35
Once you know max power you get when the rotor furls and some WS, you can pick out a resistance element that matches it ...you can't pick a low ohms load and put 600v thru it ...that would be 10KW and the wind only has 1 KW ...can't be done ...might as well short the leads

Another factor is the servo winding resistance.
For a 12 volt system, it is possible that the voltage drop at the servo = voltage across the load ...so 50% of power generated is lost to atmosphere up the tower.
If I run my 220v/3phase into a 3phase bridge to get me 1.4x DC voltage, then 230v becomes 322 volts DC ..if for my 10 footer, I furl at 25mph, then the max power I expect is 1.5KW, so I'd need a 69 ohm load   R=V^2/P = 322x322/1500=69 ohms

However, if I furl at 20mph, the max load will be 1KW and I can use a 103ohm load
Since my servo has (3) 3ohm windings, the most I lose up the tower is < 10% or so.