Bob Swinney wrote:
....
Assume the motor is 80% efficient and that it is exactly as capable of being
a generator as it was a motor; i.e., also 80%. Take the input power rating
for the motor X 80%. This is the input power rating of that motor ran as a
generator. ...

no no no, this is the _output_ power rating used as a generator!!

So the input to get the rated output would be 100/80 = 125% of rating.
A 20% loss in the windings gives 100% of rating as output.

Bob