Supercharger info

Dan Zorde dzorde at soanar.com.au
Wed Aug 26 08:29:06 GMT 1998


Sounds logical, but how do you divide a voltage output by a frequency signal ? 
 Do you turn the rpm into a voltage first ?

Dan	dzorde at soanar.com.au

On Wednesday, 12 August, 1998 1:39 PM, Walter Petermann 
[SMTP:corsaro at brokersys.com] wrote:
> Mike,
> I think what he's saying is this:
>
> Say the MAF output is read as kg/min of air then
>
> kg/min (air)*14.7= kg/min (fuel)
>
> but what's needed is kg/rev of fuel to get injector on time. Deviding
> kg/min (fuel) by rpm gives kg/rev (fuel). Or, MAF reading needs to
> be devided by rpm.
>
>  Walter
>
> Mike Morrin wrote:
> >
> > >Chris Morriss wrote:
> > >>  Imagine that the engine is
> > >> loaded such that it requires the same amount of fuel per revolution at
> > >> 6000 rpm as it does at 2000 rpm.  The MAF at 6000 rpm is going to be
> > >> roughly 3 times as much as at 2000,
> >
> > Are you saying that the air/fuel ratio is varying in proportion to RPM?
> > smells fishy to me...
> >
> > regards,
> >
> > Mike Morrin




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