756 watts in a HP; calculate from there and factor in inefficiencies of conversion, etc. Also, startup will require more watts -- figure a factor of 2

The question is what the duty cycle of the pump would be.

It says "Duty continuous".
So a 1/4 horse motor would need 189 watts to run, and 378 to start up?
So, 200 watts charging from either a panel or wind generator should stay ahead of it, assuming I have wind or sun more than I run the pump?

Lose or use some watts in the inverter & charge controller too.

Here is a link that might be useful: guys inverter FAQ

Greenmile, your pump motor should be marked with an FLA, full load amps. If that is the 1.5 amp you're quoting, it takes 3 to 4 times this amount of current to start the motor. Use 4 times. Multiply this number by 230 volts to get the wattage of inverter (not converter) you would need. Don't forget to take into account the effiency of the inverter and oversize it by about 20% or more. I would say about a 2kW inverter. How much battery and how much panel would depend on how often you need to run the pump and for how long. Take 1.5 amps X 230 volts = 345 watts. A 12 volt battery would need to provide more than 28.75 Amphours to run the pump for 1 hour. Multiple batteries can be connected in parrallel to supply the required Amphours. You need enough solar panel to replentish the batteries in the amount of sunlight you receive each day. Don't forget to factor in the effiencies. By the way, it's 746 watts per HP.

One thing to consider is controls. When the wind blows hard enough to make waves on the pond, you probably do not need more aeration. Though, it would be desireable to use this windy period to charge batteries. You need to shut the motor off if the supply voltage falls too low. I'm sure that you can think of several more 'rules'. the following is a sample list of control rules:

TBD = to be determined
Time delays are needed to prevent excessive cycling and to accomodate that the wind may have short periods of quiesence followed by significant gales.

1. Run aerator if wind velocity is less than (TBD) after 15 minutes.

2. Shut off aerator if wind is greater than (TBD) for more than 5 minutes.

3. Do not turn on if battery voltage is low.

4. Turn off if battery voltage falls below (TBD).

5. Send alarm if battery voltage is less than (TBD).

6. Shut down system if battery voltage does not rise to 91% of rated open circuit voltage after (x) hours of charging with aerator off, and set alarm.

[A lead-acid cell will rise to 113% of rated voltage upon reaching full charge while being charged. 80% of this voltage is about 91% of rated voltage. A shorted cell will keep the stack from arriving at full design voltage and further attempts to charge the stack will damage good cells or vaporize the cell fluid. In a lead-acid cell, it breaks the water into hydrogen and oxygen gases.

For long cell life, do not discharge the battery below 20% of capacity.]

I'd like to add one more piece of advice to the excellent information that has been offered so far.....If you are recharging 12v batteries, use a marine or golf cart battery. These type batteries are designed to be drained and recharged many times where automobile batteries are not. They have thicker plates that do not as readily heat up, buckle and warp thus shorting cells and shortening the life of the battery as would be the case with automobile batteries which are designed to have their charge maintained.

Thanks for all of the great info. I'm searching for a book on such info. But haven't found the right one yet.
Another question; is there any reason that I couldn't use both a windmill generator, and solar panels to charge the batteries?

I would investigate DC motors to avoid the conversion loss of an inverter.

You can use both windmill and solar panel.