I'll bite... here's how I'd went about calculating my injector requirements... (straight from EEC Analyzer documentation)
((Power x BSFC) x (1 + Safety Margin))/Number of Injectors = pounds/hour
((375 x .50) x (1 + .20))/8 = 28.13lb/hr
I'm using 375FWHP (300RWHP + 25%), a BSFC of .50 (.45-.50 for N/A, .65-.68 for F/I) and a safety margin of 20%. My calculations err on the side of being over-injected, which as Grady said, is way better than being under-injected.
Run the math yourself, see what you come up with.
Wes
I have a Q or two for you since you gave a reply
We have fwhp and rwhp to be concerned about ... correct
If the inj formula is based upon fwhp
and
Since rwhp is a lesser value than fwhp based upon a % of drive train loss
When talking about a 300rwhp combo suffers a 15% dt loss .........
We are talking about a fwhp value of 353 ... 353x.85=300.05
and
We are not accurate if we figure it like this
15% of 300rwhp is 45 and then add that to 300 for a working fwhp value of 345
Not too much difference ... I know ... but ... still ... it would be more if dealing
with a forced combo
Then there is the 80% duty cycle which is considered safe by most.
Then there is the fact that Ford runs fuel pressure at a lesser value (39lbs) than
what you see given in those formulas which ... IIRC ... is around 43.5lbs
I wanna say someone said at the lower pressure that Ford runs at, you are not
getting the full rating of the size of injector you have chosen
anyway ... stuff to think about
The formula I have seen used goes like this:
fwhp TIMES bfsc DIVIDED BY # of inj's used TIMES inj duty cycle
For a NA 300rwhp stick trans combo ...............
I would use 353fwhp X .85 (drive train loss) = 300.5
I would use a bsfc of .50
I would use a inj duty cycle of .80
the formula would then be ........
353x.50=176.5
then
8x.80=6.4
then
176.5/6.4=27.58 lb injector size
Very close to what you came up with Wes
Now as I said above .....
I used 350rwhp as a value for my peace of mind
and
an inj duty cycle of 80%
which worked out to a final inj value size of 32.19 lbs
If one was using an auto trans ... how much more loss would one
have to figure in the formula
I can't believe nobody but Wes has anything to say about this issue
considering I see this issue being asked about over and over here
and on other sites as well
Grady