Diff between theory and practice

[jon hanson]

 [snip]
>The engine IS most efficient at peak torque, but that is not why he needs
more fuel at 
>higher RPMs.  If he needed more FUEL PER HORSEPOWER produced, then you are
right.  But 
>what he is indicating is that he is maintaining constant air fuel ratio
(constant CO), 
>and is using more fuel at 6000 RPM.  This simply means he is flowing more
air which 
>means that:
>
>	(vol eff @ 6000)* 3000 * disp > (vol eff @4800) * 2400 *disp
>
> [snip]

Thanks to all who replied.
The fuel no I talked about is a no between 0 and 255 which this particular
system translates into an injector open time in ms. the larger the no the 
more fuel per injection event.

The above explanation still confuses me as it is based around the rate at which
fuel is being used. As I understand it (and this not neccesarily correct)
at Peak VE each intake stroke draws in the greatest mass of air into the
cylinder
compared to any other rpm which is off peak torque rpm. To maintain the
desired A/F
ratio (in my case 3.5% CO) I then need the greatest mass of fuel per
injection event.
Note i see the whole thing very much as a per
induction/compression/power/exhaust
event and the particular revs of the engine dont come into it.
Rpm comes into it when you are interested in the rate at which you are using
fuel.
ie the more induction/power strokes you make per second the more fuel per
second you use.
But a fuel no of 150 represents an injector open time in ms for each single
injection.
my figures suggest that each induction stroke at 6000 rpm is drawing in a
geater mass of
air than each induction stroke at 4800 rpm and I know the torque peak is
quoted as 4800rpm
I live at 5,500 ft altitude could this affect the breathing characteristics
of the engine?
If I'm being thick please dont hesitate to bash some sense into my head.

Regards
Jon Hanson
Johannesburg
South Africa