Speo,
You, along with Ghurd, seem to have a bit of a problem understanding the concept of a wind turbine's power output.
The rating of a given power at a certain windspeed is simply an upper limit imposed by the capability of the machine and the prudence of the user. It does not mean that, at the rated windspeed, it will always be delivering that rated power.
The power delivered under those conditions will primarily be determined by the load, whether batteries, resistors or a combination determined by a controller such as a C40.
The C40, in Diversion Mode, shunts a certain amount (via PWM) of the WT available current to the dump load resistor to maintain the battery terminal voltage for the charge algorithm phase it happens to be in at the time. This is true whether in absorption or float phase. Thus, the actual power delivered by the WT depends on these loads.
You may notice that I tend to talk in terms of current rather than power. It clarifies the issue.
Ghurd: "However if the windmill is making 100W and the resistor is dumping 47W, then 53W will still be going into the battery."
This is incorrect. The WT will only deliver as much power as the loads demand. If the battery is far from full, all of the current available will continue to charge the battery with nothing going to the dump load, or if it's close to full, it will divert some of the current to the dump load while continuing to charge the battery according to its needs, or if the battery is full, it will divert all of the available current (less a little) to the dump load.
Your assumption that with that resistor, the turbine will speed up and will not be protected against higher windspeeds, is correct. Not recommended.
Ghurd: "The windmill will still be charging the battery. Possibly at up to 100W.
The resistor will be draining the battery, but only draining the battery at 47W.
The other 53W will be overcharging the battery."
This is also incorrect. It wouldn't be much of a controller if it allowed this to happen. The dump load resistor does not "drain the battery". The controller only diverts enough current from the WT to maintain a given battery voltage according to its algorithm and does not allow battery overcharging.
As to your basic question, the Xantrex C40 manual says that a good rule of thumb is to size the dump load to 80% of the controller rating, i.e., 32 amps. At 14.4V, that means 0.45 ohms. This would also allow for a larger mill expansion than you now have. The power rating should be at least twice your current WT power. Personally, I would chose 1000W to allow for future expansion and a 100% safety factor.
By the way, your formula for parallel resistors is incorrect. It's R total = the recipical of the sum of the recipicals, i.e., Rt = 1/((1/R1)+(1/R2)+(1/R3)...+(1/Rn)). For two resistors it simplifies to Rt = (R1*R2)/(R1+R2).