You have to distinguish the open battery voltage, the open DC voltage coming from the rectifier of the wind turbine and the loaded voltage. Lets first assume that no battery charge controller is connected and that the DC output of the rectifier isn't connected to the batteries.
If you measure the open battery voltage of a 48 V lead acid battery, it is about 48 V for a 10 % full battery and about 50.4 V for a 90 % full battery. If you measure the open voltage at the DC output of the rectifier, it depends on the rotational speed of the generator and so it depends on the wind speed. If it is a nominal 48 V wind turbine, the open voltage can easily become a factor three higher than 48 V so about 150 V at high wind speeds. How high depends on the safety system of the wind turbine. If the DC output of the rectifier is coupled to the battery, a current starts flowing as soon as the open DC voltage of the rectifier is higher than the open battery voltage. The current increases about proportional with the voltage difference but the current also depends very much on the generator size. So during charging there is only one voltage and that is the loaded voltage.
It has to be prevented that a full battery is over charged. This is normally done by a battery charge controller which consists out of a voltage controller and a dump load. This device limits the maximum charging voltage up to 2.3 V per cell and so up to 24 * 2.3 = 55.2 V for a 48 V battery. So if this voltage is reached, a part of the generated current goes into the battery and a part goes into the dump load. However, if the battery charge controller is adjusted at for instance 2.1 V per cell so at 24 * 2.1 = 50.4 V, the dump load starts extracting energy much to early.
I would do the following. Disconnect the battery charge controller and disconnect the rectifier from the battery. Measure the open DC voltage at the rectifier. You should measure an open voltage of 48 V already at a wind speed of about 3 m/s. Don't do this at high wind speeds because the you may blow your rectifier diodes if the voltage becomes very high. Next connect the DC output of the rectifier to the battery and measure voltage and current. You should see that the loaded voltage increases a little at increasing current. If you measure almost no current, it means that the generator is very small and may be too small for a rotor with a diameter of 2 m. But if the combination of rotor, rectifier and battery works nicely, you know that your problem comes from the battery charge controller.
Next, couple the battery charge controller to the battery. If this controller works nicely, the dump load of it should only take a current if the loaded voltage has reached the 55.2 V level. If it starts taking energy at a lower voltage, the adjusted maximum voltage is too low. At my website:
www.kdwindturbines.nl there is a free manual at the bottom of the list with KD-reports of a 27.6 V, 200 W battery charge controller in which this procedure is explained in detail. The maximum charging voltage of its voltage controller can be adjusted but it might be that this isn't possible for your controller.
It might also be that you think that you have a problem but that there is no problem at all. You write that you are using four 12 V, 220 Ah batteries. These are very large batteries and for a small current, the required loaded voltage is only a little higher than the open voltage. If the generator of the wind turbine is too small, it can supply only a low current of some Amps at moderate wind speeds. The charging voltage of a very large battery will hardly increase for this low current. If the batteries are charged at the same time by a big solar pannel which can supply large currents, the real charging voltage will mainly be determined by the big solar panel and so a small wind turbine will have only a very little influence on the charging voltage.