MPG calculations from datalogger
 

I now have the ability to data log while I drive. There are a few parameters I can log like RPM, speed, injector PW and TPS voltage.

I tried to figure out if there is a way to calculate MPG from those pieces of data... I know the injectors are rated to flow 172ml of fuel per minute at 35PSI. There is a FPR on the vehicle that raises and lowers the pressure but I don't know if that will affect the calculation a lot or not?


For example here is an entry.

Elapsed Time (ms) = 202800
VSS = 25
RPM = 2263
INJ PW B1 (ms) = 7
TPS(Volts) = 1.2

Calculations are as follow...
injector flow rate (ml or cc per min) = 172 (given)
fuel injected per minute (us gallon) = ((172*0.000264172)*(7/100))*4 = 0.012722524
Fuel injected per hour (us gallon) = 0.012722524*60 = 0.763351411
calculated Instant MPG = 25/0.763351411 = 32.75 MPG



Is there anything I missed? The final number isn't some crazy high or low number so it seems like I could be close?

Anyone?

*crickets* *crickets*


Anyone?

the purpose of the fuel pressure regulator is to keep consistent the pressure differential at any vacuum.

ex: injectors firing at WOT will be 35PSI of guage fuel pressure with 0PSI of guage manifold pressure. injectors firing at max vacuum will be ~20.3PSI with negative ~14.7PSI manifold pressure.

it's actually easier to think about in absolute pressures, but that's how it works to keeps injector flowrate consistent across all vacuum levels.

if you had a rising-rate FPR, that's a different situation, but those are worthless hack methods of tuning fuel delivery.

I have a vacuum actuated FPR. I don't know what the different flow rates are for the injector at different pressures but I am trying to calculate MPG from the pulse width of the injectors.

here is the equation i use for my vehicles:

A*(1/3600000)*(17.4)*(3)*E*(60)*(1)

that will create fuel flow in Lb/Hr. since you have fuel flow measured in volume, not weight, your equation will be different.

A is pulse width in mSec
17.4 is injector flowrate in lb/hr
3 is the number of injections per revolution(6 cylinder engine, so 3 injectors fire per rev)
E is RPM
1 is number of times the injector is fired per 4-stroke cycle(a lot of earlier stuff fired twice per 2 revs, so those would use a 2)

then to convert that into a volume/hr(gallons/hr) value:

A/6.073

A being the previous result, 6.073 being the weight in lbs of 1 gallon of gasoline.

then simply A/B, A being MPH, B being Gallons/Hr.

How come you convert from volume to weight? 172cc/min = 16.36 lb/hr

So by using your formula...

7*(1/3600000)*(16.36)*(4)*2263*(60)*(2) = 17.28
17.28/6.073 = 2.85
25/2.85 = 8.77

I don't think I could get 9MPG if it was leaking fuel from the tank...what did I do wrong?

because i know what my injectors are rated for in lb/hr, no idea what they're rated for in volume/hr.

what i entered:

7*(1/3,600,000)*16.36*2*2263*60*1

2 because with a modern 4-cylinder, it's SFI and will fire 2 injectors per revolution(since a 4 stroke cycle takes 2 revolutions).
1 because it's SFI and only fires the injectors once per 4-stroke cycle.

makes 8.6386 Lb/Hr

/6.073

makes 1.4225 Gal/Hr

25/1.4225 = 17.5751 MPG


i don't suppose those numbers you stated are average numbers? or are they just a single frame of data? for some reason, i can never get this equation to work correctly with averages... works perfect with single-sample data though.

anyways, 17.5MPG at 25MPH with an estimated 17.5% throttle position(based off of .5 volts being idle and 4.5 volts being WOT), seems believable.

It's just a single frame of data.