HA, HA, 1 and 1/3 HP! Good luck with that jj. Maybe if you can make that 1 and 1/3 HP at 12000 rpm. I think you will find that to turn that alternator fast enough with a B&S to make full output you will need about 5 HP. I am talking about driving the alternator with a lawn mower type engine, right?
In all seriousness, you can run your 60AMP 24V alternator (talking MV here) on less than 15HP you also know that heat, drive losses, the fan, etc. take more HP.
I think the problem with your HP calculation is the alternator's driven speed. You are talking HP at the alternator at the optimum rpm with no losses considered. I'm talking HP at the drive engine, in the real world.
A car alternator is full load at 6000 rpm. That 6000 rpm is crank rpm so you can figure out the alternator rpm based on the pulley diameters. A typical automotive alternator is designed for 12000 rpm. That is a 2:1 drive ratio typically. Some will do more but I don't remember specifics.
The Leece-Neville or Prestolite on the Deuce is going to be driven at a different ratio because a diesel is not going to turn 6000 rpm. I haven't checked to see the difference in crank and alternator pulleys on the multi-fuel. My guess is the alternator is spun at least 2:1 and maybe 3:1 or more of the crank speed.
In a nut shell, the alternator has to be driven at the optimum rpm for full output. If the alternator has to be over driven because the drive motor or engine can't turn that fast. then you need more engine or motor HP because of drive losses.
Well, we got pretty far from what the OP was asking.
TWELVE HORSEPOWER! FIFTEEN HORSEPOWER!! My physics book says watts equals volts times amps. So my 66amp alternator at 14.8volts makes less than 1000watts. ONE horsepower is 746watts, so, ignoreing thermodynamics, i should get the full output of the alernator from a 1-1/3HP motor. Yeah, i know it won't work out quite that way, but i surely do not need twelve or fifteen horsepower to drive a measly 66amps!