Math Equation help for MPG average History table

Found this on TunerPro forum from TGO forum. This is the equation used in 93V8S10 $OD, $OE, etc files

Quote:

---

found an old post at TGO with instructions and a formula for calculating
instant MPG:

For the MPG conversation equation, use:


X/(C*0.0000212*Y*Z)

Where:

X = Vehicle Speed
Y =
Engine Speed
Z = Base Pulse Width
C = Injector constant (lbs/hr)

---

Have no idea what 0.0000212 is?

Then Robert said a better formula is

Quote:

---

i use a different equation, one that works a bit more accurately for me...


A*(1/3600000)*(16.7)*(3)*E*(60)*(2) < What is 60? No mention below? I'm guessing 1/3600000 is 1 rpm?
where 16.7 is your injector
flowrate in lb/hr, the 3 represents half of the cylinders in the engine, and the
2 represents how many injector pulses per 4 stroke cycle(double fire, SFI or
single fire mode would need a 1). A is BPW in mSec, and E is RPM.

and
that will create a value showing injector flowrate in lb/hr.

A/6.073 < Is 6.073 the equation from above??

A is the value that was just created, and will
now create a injector flowrate in gallons/hour value.

A/B

A is MPH, B is gal/hr of fuel flow, then you'll have a MPG value.

---

Trying to figure out the more complex version from Robert, my questions are in Red above. This is for a MPG history table. For my LT1 it would be A*(1/3600000)*(24.87)*(4)*E*(60)*(1) if 60 is correct? And since it's SFI 1 is correct. With BPW of right and left divided by 2(R+L)/2

First one is easy...

Quote:

---

Originally Posted by EagleMark

Found this on TunerPro forum from TGO forum. This is the equation used in 93V8S10 $OD, $OE, etc files
Have no idea what 0.0000212 is?

Then Robert said a better formula is
Trying to figure out the more complex version from Robert, my questions are in Red above. This is for a MPG history table. For my LT1 it would be A*(1/3600000)*(24.87)*(4)*E*(60)*(1) if 60 is correct? And since it's SFI 1 is correct. With BPW of right and left divided by 2(R+L)/2

First one is easy...

---

I think it should be "A=(1/3600000)*(24.87)*(4)*E*(60)*(1)". when you input the RPM, then this should output your LB/HR

It looks like 1/3600000 is the number of msec in one hour, and 60 is there to convert the RPM(minute) into hours. because everything has to be in the same term to put it all together.

then once you have LB/HR, you divide by 6.073 which is the density of gasoline, and that gives you GAL/HR.

Looks like you are on the right track.

Thanks for the help Greg! The answer to what 60 and 6.073 are makes the calculation correct in my head as well!

Not my work though, I dislike taking credit, just trying to apply someone elses work to my car and another guy with same car started the project.

I don't know who did origanal TGO conversion? But the more complicated and supposed to be more accurate was from Robart Saar.

I think it might have been Fast305.