SG Fuel cut off value
 

I just got my SGII in today and am setting it up. I will be putting it in my HX. What is the value for the fuel cut off screen?

I'd like to jump in and ask the same question but for a 2004 Subaru WRX STi. I left this setting as it was.

The fuel cutoff setting isn't in most of the manuals, only the absolute most recent edition. According to the manual you take your idle TPS reading, and add 4 to it, and that's your cutoff value. The default of 25 works for most vehicles without any further changes though.

^ That's a good place to start. Then, do the following and adjust as needed.

Watch the Loop gauge and the MPG gauge. You want to see 9999 mpg any time it's showing Open Loop, under DFCO (engine braking).

Be aware that the HX has lean-burn, and the Scangauge can't properly read this. It assumes a standard air/fuel ratio all the time.

Thanks for the reply. I figured it out after playing with the SG last night for about an hour. The only thing I hate about the SG so far is that the horse power reading is wrong. You can't tell me when I am idling that the engine is only producing .9 horse power. Unless I have it improperly installed or misunderstand this gauge, I think it is useless.

I can believe .9 HP for idling. How many horsepower do you think it takes to turn a 4 cylinder engine at 800rpm?

Any guesses at how to do the math? Or, maybe we could compare to starter motors?

yea did you think you were using all 140+ hp all the time? lol i hope not!

at idle there's virtually no load on the engine hence why in a manual tranny car if you dont press the gas some when trying to take of from a dead start its gonna stall(generally speaking).

An idling engine is producing nothing, just sitting there wasting energy.

13% of the fuel consumed by vehicles in the US is wasted idling.

"Idling", the word itself means of no consequence.

Now when you try to identify the cost of idling, it can get very complicated.

Alternator

Water Pump

Power Steering Pump

Sucking air through a restriction

Trying to produce power while only utilizing about 25% of atmospheric pressure

Oil pump drive losses

Friction of moving parts, including viscosity friction

The physical act of accelerating each reciprocating part from 0 speed at TDC to max speed at 90 degrees ATDC, then decelerating those same reciprocating parts to 0 speed again at BDC, only to reaccelerate them again to 270 degrees then stopping again at TDC. Two times this whole cycle for a single combustion pulse of 180 degrees out of 720 total degrees.

The physics calculations necessary to calculate the percentage of force attributable to reciprocation versus the part attributable to rotation, of the connecting rod, is so complicated even the most powerful computer would have trouble doing the calculations. This transition occurs over the complete mass of the connecting rod, with the most rotation at the bottom and the most reciprocation at the top.

Imagine the difference between coasting in neutral and coasting in gear tp understand the total losses in reciprocating engines.

Now imagine the difference if you removed the spark plugs, which would eliminate almost all "pumping losses", if you consider pumping losses as the actual moving of the air through the engine.

Now imagine the difference if you removed the cylinder head and all valve train related components, friction, and the corresponding losses. Even in this scenario you would still be losing your inertia moving the reciprocating components up and down in the engine, violating Newtons laws of inertia 8 times for each revolution of each reciprocating assembly.

I am actually surprized it only takes .9 horsepower to keep an engine "idling". Thats the equivilant work it would take to raise over 475 pounds of weight 1 foot every second, or 60 feet per minute.

regards
gary

Now the question?

Does that energy requirement increase in a linear or exponential fashion as engine speed increases?

Flywheels are exponential, twice the speed requires 4 times the energy.

Is this also true of reciprocating engines, especially knowing a lot of their energy losses are in rotating parts?

Or are reciprocation losses linear and ratational losses exponential?

If both are exponential, then the .9 HP required to idle would double with every 100% increase in engine speed, above idle.

regards
gary

Hmm...that right there is something I can imagine.

Well, when you put it that way...drat, totally inconclusive. :D