I can't tell you exactly what you need to know since I am new at this stuff but maybe reading a very long post about my learning experience might help.
When building my first wind turbine which was a HAWT (3 blade), I built a 1/2 scale model (0.52m diameter) to figure out what size axial flux alternator I would need. I didn't know how to calculate the size of the alternator I would need for the full scale turbine and I also didn't know how much power the turbine could put out. Based on what I had read on this forum and the web, I did know that if I knew how much power the turbine could output at different wind speeds then that would be how much power the alternator could put into the battery if there were no losses (100% efficient).
I built the 1/2 scale version and also built a small pony brake attached to the back of the blade hub with a spring scale to measure the force exerted at a specific distance from the centre of rotation. With the pony (prony) brake I could extract a fixed amount of mechanical power to see how the turbine would respond and be able to see what the maximum mechanical power output was at specific wind speeds. I used a large variable speed fan as the wind source. I measured turbine rpm and spring force at different fan speed settings and then calculated the power output based on the formulas obtained from wikipedia. I measured wind speed using a home made anemometer. You can find pictures of the anemometer in my files along with some of my turbine testing.
Now that I had the turbine mechanical power output, I could build an alternator that could convert the mechanical power available into electrical power, or could I? In a perfect world the alternator would convert 100% of the mechanical power available into electrical power. In my imperfect world I would get much less than 100% conversion. Not knowing how to calculate the losses and not willing to take the time to learn how to calculate, I took the time to build several small scale alternators. Not sure if I saved any time doing it that way.
No, I still can't do the calculations that well but one thing I did learn is that an alternator in a perfect world could do: mechanical power in = electrical power out. Another way of saying it is I learned that in our world the alternator can not extract more power than the wind turbine can put out at a specific wind speed. If the alternator tries to extract more power than what the turbine can provide then the turbine just slows down and you end up with less power output from the alternator (since it slows down). So building a more powerful alternator than what the turbine can handle has no advantage. You must match the alternator output (including losses) to the turbine output and this is far from simple (flux has some great info on this in his posts (matching the load)).
In the end the methodology I used to determine the alternator specifications was to use the guide lines posted on this forum, Hugh piggott's books and trial and error.
Hugh's books are a very worthy investment if you want all the info in one place.
After reading all that you still don't have any formula's but hopefully you can see that if your turbine puts out 32 watts as measured with a pony brake then your alternator could only output 32 watts to the battery if it was 100% efficient.
Now for some formulas. I am not an electrical or mechanical engineer so there may be errors here.
Coil wire in the alternator stator has resistance and the power lost as heat in the wires when current flows is P = I * I * R. P is power in watts, I is current (amps) and R is resistance (ohms). Can you see that if current flow doubles then the power lost as heat will be four times as great in the alternator? So as current flow increases in the coils of the alternator you start loosing power in the alternator as heat. If 50% of the turbine power is lost in heating the coils of the alternator, you only have 50% of the turbine power available to get to the battery, assuming no other losses. Ok, so make the coils as low a resistance as possible so that we don't lose power to heat and all of it makes it to the battery. Well, that leads us into the next formula and problem.
Determining how much current will flow into the battery can be roughly estimated by I = (Alternator voltage(DC) - 1.4(diode drop) - Battery voltage)/ R. R is the total resistance of two phases of the star wired alternator including line resistance from alternator to battery. If R is really really small then we can get a huge amount of current flowing into the battery without losing it to heat, right? The power into the battery would be P = I * Battery voltage. But remember, we can only extract so much power from the turbine at each wind speed. As soon as we start extracting power from the turbine, it starts to slow down. We can continue to extract more power but the turbine continues to slow down.
Now the problem. The alternator voltage is directly proportional to its rpm. So if alternator rpm drops, its voltage output drops and therefore so does the current. If rpm goes up, current output from the alternator goes up and power into the battery goes up. It now becomes a balancing act. The rpm settles out when mechanical power available from wind turbine = power into battery + alternator copper losses(coils heating) + bridge rectifier losses + cable line losses + alternator bearing friction losses + alternator windage(spinning rotor, air friction) losses + many other losses I failed to mention. In order to "Match the Load"(Flux TM) we need to control how much power we extract from the turbine at each wind speed so that we can maximize the amount extracted with out stalling the turbine and limiting power output.
I'm not going to get into stalling in to much detail but lets just say that we need to keep the blades spinning just above a minimum rpm for each wind speed to extract maximum power. If we slow down the blades to much then the turbine rpm will increase very little when the wind speed increases and therefore alternator power output will increase very little.
Others have already given you the formula for calculating turns per phase based on magnets, rpm etc to reach cut-in. After all that reading the punch line is I can't tell you exactly what you want to know. But I hope you can see from what you have read so far that you can make a "guestimate".
Alternator open circuit voltage is directly proportional to its rpm. This is because the induced voltage is proportional to the number of turns in the coils and the rate of magnetic field change. The rate of Magnetic field change is determined by rpm, the size and strength of magnets, air gap, flux path etc. Ok, so there are a lot of variables, way to many to start tinkering with by trial and error to find the optimum solution for matching the alternator to the turbine. Some people have done this and have posted this info within this forum and others have published designs in inexpensive books (OtherPower(The Dans), Hugh Piggott). Also posted in this forum are rules of thumb guide lines if you don't want to read the books but its not easy to find all of them.
One key thing I found that helped me determine coil winding count, rotor diameter, number of magnets and size, etc, was the alternators Volts/rpm and R. Volts/RPM is the ratio of the alternator's open volts(DC) to alternator rpm. If this ratio is too large then the alternator will stall the turbine since the voltage rises too quickly and therefore power taken from the turbine rises faster than what the turbine can provide. If that ratio is a little bit too large for the turbine then you can add line resistance to decrease the current/power if its too late to change magnets, rotor diameter, winding count or air gap. Its a balancing act because you do want the Volts/rpm high enough to get the proper cut-in and prevent over speeding in high winds by having the blades go into stall. If the ratio is too small for the turbine then you don't extract all the turbine has to offer at the low power end of the curve and in heavy winds the turbine will over-speed and destruct before it furls. R is the total resistance that the alternator sees as mentioned before. If R is too low then its similar to having Volts/rpm too high. If R is too high then its similar to having Volts/RPM too low.
How do you use the volt/rpm ratio and R? I set-up a spread sheet using the formula's already given to plot the power curve of the alternator which was compared to the power curve of the blades I was going to use. You can not get them to match exactly but you can make the generator work ok with in a portion of the wind turbines power curve. I assumed that the power curve of the blades was 25% of the theoretical wind power curve for the blade diameter. Once I had Volts/rpm and R it was a matter of balancing cost of best magnets you can buy, magnet wire cost, cost of steel for rotors etc which meant another spread sheet.
In the end its all about compromises between: cost, efficiency, time, safety.
Having formulas to determine your final design help a lot but I think building a small scale working model will go a long way in helping you understand what you need to do for the final design. You are building a wind turbine that has an unknown power curve and therefore you can not design the matching alternator for it. The key to the success of you VAWT turbine will be determining usable power that you can extract at different wind speeds. Once you have the curve for usable mechanical power from the wind turbine then you can design the alternator that will convert that usable mechanical power into electrical power.
