Of course it does. The engine will never put out as much energy as it
consumes, and electrolysis will never produce as much fuel as its power
source, the engine, burns to produce the energy needed.
What sort of engine is it he is using? If it is an internal combustion
engine, you can expect a peak efficency of 46%. With an electrolysis
system with a maximum efficiency of 99%. Assume an engine which puts
half its output toward propelling the car, and half which produces
electricity for electrolysis. SO for every 100 units of hydrogen
consumed by the engine, it produces enough electricity to produce 23
units of hydrogen fuel. If the system starts off with 200 units of
hydrogen, and uses 100 per hour, the after one hour of operation there
will be 123 units of hydrogen in the tank. One hour later, there is 46
units of hydrogen left. 46 minutes later there are 17 units of hydrogen
left,and 15 minutes later there are only 6 units left, and so on.
SO, essentially, such a system requires 7 hours worth of electrolysis to
produce a little over 3 hours of automotive propulsion, for a cumulative
efficiency of of just over 45%. Now, this is merely the engines
efficiency, which does not count the losses due to powertrain
efficiency, rolling and aerodynamic drag, etc.
-- TANSTAAFL!!! Michael Lorrey ------------------------------------------------------------ mailto:retroman@tpk.net Inventor of the Lorrey Drive Agent Lorrey@ThePentagon.com Silo_1013@ThePentagon.com http://www.tpk.net/~retroman/Mikey's Animatronic Factory My Own Nuclear Espionage Agency (MONEA) MIKEYMAS(tm): The New Internet Holiday Transhumans of New Hampshire (>HNH) ------------------------------------------------------------ #!/usr/local/bin/perl-0777---export-a-crypto-system-sig-RC4-3-lines-PERL @k=unpack('C*',pack('H*',shift));for(@t=@s=0..255){$y=($k[$_%@k]+$s[$x=$_ ]+$y)%256;&S}$x=$y=0;for(unpack('C*',<>)){$x++;$y=($s[$x%=256]+$y)%256; &S;print pack(C,$_^=$s[($s[$x]+$s[$y])%256])}sub S{@s[$x,$y]=@s[$y,$x]}