I'm not looking to run it from the generators. More or less a semi-hybrid. Where the gas engine is only to expand the range. (Slow the rate of discharge)
On Tue, Apr 1, 2008 at 6:49 PM, Jeffrey Blamey <[EMAIL PROTECTED]> wrote: > If you figure watts are watts, take the number of watts required by > the drive motor to maintain a speed, let's say 45 mph - 50A @ 72 V = > 3600 watts, so I would then look at the power plant paired with the > generator on a commercially available unit, you may well find a higher > horsepower engine than you are proposing, then again maybe not. > > Jeff > > On Tue, Apr 1, 2008 at 2:53 PM, john fisher <[EMAIL PROTECTED]> wrote: > > focussing on the connecting the genset to the battery pack: > > > > jeff said > > > If it is a normal AC genset you have the AC (120/240) to DC > conversion > > > losses through your charger. Ideally if you find a DC genset that can > > > output near 85VDC, or modify a genset to output directly to 85VDC > you > > > can minimize your losses. > > > > > > Jeff > > > > so in looking at DC motors to use as a generator. > > Is it reasonable to work backwards from the motor input to figure out > the likely generator output? For instance, if I > > see a DC motor that makes 3hp at say 3000 rpm ( pulling numbers out of > the air here) and requires about 3Kw to do it, is > > it reasonable to assume a 4hp ( allowing for losses) ICE driving the > same motor at 3000 rpm will produce about 3Kw, more > > or less? > > > > John > > > > > >
