Diff between theory and practice

[M HILL]

>     (vol eff @ 6000)* 3000 * disp > (vol eff @4800) * 2400 *disp
> 
> therefore, if  (vol eff @ 6000) is 80% or greater than (vol eff @ 4800), he will need 
> more fuel at 6000 RPM.  According to the numbers provided, he used 170 units at 6000 RPM 
> and 155 at 4800 RPM.  This means (assuming constant A/F), the engine flowed 9.6%  (which 
> is (170/155) - 1)more air.  If volumetric efficiency was constant between the two 
> speeds, the engine would have flowed 25% (which is (6000/4800) - 1)).
> 
> Given the above info., VE at 6000/VE at 4800 = (4800*170)/(6000*155) = 0.877 which means the 
> volumetric efficiency dropped 13% at 6000 RPM compared to 4800 RPM.
> 
> This is assuming engine temp. and everything else was constant.

This is not the case if the fuel is measured on a per cycle basis as 
in measuring the fuel pulse width.  When measuring pulse width engine 
speed has nothing to do with the amount of fuel.  Your calculations 
assume the fuel measurement is per unit time, not per cycle.  If the 
AFR is held constant then the peak fuel should be at peak torque as 
here the volumetric efficiency is highest and so most air is used per 
cycle, therefore most fuel per cycle.

Martin