The voltage rating is the maximum voltage that can be placed across the diode when reversed.  Think of a check valve for water or air that has a maximum reverse pressure rating.  If you exceed the rating the valve blows the parts out of itself backwards and is ruined.

The amperage rating is how much current can you flow through in the forward direction before you burn it up.    Generally the amperage rating has to do with how well the diode can dissipate the heat that's generated by the minor inefficiency of the device.

Diodes "watts" aren't figured by that exact Volts times Amps like the output of a genny---diodes only have to dissapate/eat their voltage drop X the amps through (I x voltage drop + a teensy bit of R thrown in).
And Schottkey's are low voltage drop across types.

You should be able to use that diode.  RP already explained most of it.

With a DC motor as generator the diode is blocking the battery current from driving backward into the genny and running it as a motor (and the mill as a fan).  Wind is unlikely to ever drive the blades BACKWARD at any appreciable speed (and thus generate a significant reverse voltage.)  So the voltage rating only needs to be as large as the battery voltage (including the boost from attachment of other chargers or pulse desulfators), plus a safety margin.  If you're using a 12 or 24V system, 40V should be  fine.

You'll have to consider the current at lower voltage loads and in wind gusts before furling.  But an 80 amp rating and a good heatsink sounds reasonable for a genny rated at 20 amps.  (Be sure your furling takes it out of the wind before it fries itself or the diode.)

The diode in your case is used to insure that when the Generator is OFF there is not a PATH to discharge the Battery bank.

You have a 12 volts battery bank, so the DIODE in your case needs to have a reverse breakdown good to cover the 12 volts, You can use a 25 volts or higher up diode.

You problem is the current capability of the diode, that should be at least 2 times the peak current to reduce the voltage drop

You have 80 amps 45 Volts = that is the solution for You -- use two in parallel with a good heat sink.

Your concept of:

>If my genny is putting out 800 watts can I just use a diode that is capable of 1600 watts? [ 40 amps x 40 volts ] Or tell me diodes don't work like that

is erroneous, the diode is not carrying 1600 watts, it is carrying the current ( 40 amps) and the voltage drop, that for a Schottky may run around 0.5 to 0.6 volts for a total of 0.5 * X 40 amps = 20 to 24 watts

Your heat sink should be capable of dissipating around 40 to 50 watts

I was not happy with the one that I did find, but it's working.

Does anyone know of any good suppliers ? maybe a couple of URLs ?

Thanks