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

Lets use my old VX as an example.

At idle fuel consumption is about .2 gallons per hour.

At 55 MPH fuel consumption is 1 gallon per hour (close approximate).

What would fuel consumption be at 2000 RPM with no load?

Compare this to the fuel consumption at 55 MPH=1 GPH.

I woudl appreciate it if someone with a scan guage could do this calculation, especially the fuel consumption at 3 times the idle speed, compared to the idle speed fuel consumption.

This would answer a question that has perplexed me for some time, which is whether fuel consumption is a linear or exponential function of engine speed?

It may be that in reality it is a combination of both.

regards
gary

It has been said that internal engine friction is an exponential function of RPM, so I would assume that no-load fuel usage is not linear (even if it does not follow RPM perfectly).

Maybe we can extrapolate something from this: https://metrompg.com/posts/rpm-mpg.htm

https://metrompg.com/posts/photos/chart-rpm-mpg-z.gif

Using a model of MPG = 1/rpm (x 120,000 to normalize the numbers into the same range), we get the following:
That's a 97% correlation, according to Excel.

Unfortunately, there are too many variables involved there to help us with the no-load RPM question. For example, aerodynamic drag...

No, the test is run with the same speed each time, eliminating aero drag. It is, however, loaded and not unloaded, but loaded is where most of us drive anyway.

Oh....in that case, it's probably pretty good, the only non-engine variable would be drag in the transmission which is probably minimal.

:eek: Wow.
Using that same ratio, on my car, to calculate the value of a transmission swap.

My car currently runs 2750 rpm at 60 mph. Changing to a CX or VX transmission would drop that to 2200 rpm at 60 mph. Using this calculation, it would raise my steady-state mpg at 60 from 43 to 54 mpg!

Actually, does it work? Does that chart need to be GPH instead of MPG? Also, there is another variable...pumping loss from throttle position.

Maybe I'll try in neutral with my meter and see what I can learn.

The greatest variable in engine efficiency is how much of the available atmospheric pressure is being introduced into the combustion chamber.

Best efficiency will always be at moderate RPM with the lowest percentage of throttle position that allows each cylinder filling event to be as close to 100% of atmospheric pressure as possible.

The reason is:

An engine is basically a lever. It compresses a fuel air mixture, then ignites the mixture, which expands and applies pressure to the top of the piston.
Your best leverage is when you have the maximum possible pressure differential between compression and combustion (cylinder filling event).

The longest lever if you catch my drift.

Under higher loads you can get maximum atmospheric induction at relatively low throttle positions.

Using a vacuum guage, if you apply just enough throttle to get the lowest vacuum reading on your guage (while under some load) you have found the most efficient "sweet spot" of the engine. This will be true at relaitively low RPM, where the losses due to friction and reciprocation are at their lowest amounts.

A BSFC map clearly demonstrates this phonemon as well as illustrating that WOT is not the most efficient, when you consider throttle position. Throttle position can be as low as 30% up to 80% depending on the load applied.

This is why I always tried to stay in higher gears in my VX. You could easily check by adding more throttle, with no significant increase in power.

There may be specific examples where this does not apply perfectly in every situation, but as a general rule it's hard to beat.

In most cases, to achieve higher mileages, you have to make adjustments that are directly related to you specific driving environment. The best example in my case is the timing of the traffic lights on my normal route, which numbered over 50 in a 32 mile round trip.

If I accelerated too slowly I would catch almost every light changing yellow. This situation required me to accelerate more aggressively to avoid getting caught by the numerous lights.

I would still be interested if anyone could do a quick test with their scan guage to get an idea of what the unloaded fuel consumption of their engine would be at 2000 RPM compared to idle speed.

It would also be nice to see what horsepower was required to maintain both idle and 2000 RPM speeds using the same engine as the example.

My engine design eliminates almost 100% of reciprocation losses. It would be nice to be able to quantify the percentage of those losses compared to the cumulative losses of the same example engine, but the calculations include so many offsetting variables that precision may be impossible. This would allow basic efficiency calculations for a design that has never been built.

regards
gary

Of course I didn't think this. But when I am driving it says it is only like 2 hp. This just isn't correct.

Ok, if it's reading 2hp while steady-cruising, that's definitely got to be incorrect. At what speed did you see that reading?

You are assuming the engine can still produce enough torque that low to still keep you at 60 efficiently.

You'll most definitely find yourself downshifting to accelerate and using more throttle than you used to. In town you might feel you can rev lower but freeway is a whole new ball game. I always wanted a 5th gear for very low rpm in city-only driving and on the freeway it just went back up into 4th for a 'normal' highway rpm. I obviously won't be doing that now but it was planned.

When I was watching it I was going 60 mph for 2 hp. I know I know...this isn't right, but that is what it is reading. I don't really care that it doesn't work. I just wanted to let people know that the xgauge isn't great, but if someone can correct me I would love to hear from them.

My personal number is MUCH less. I am anti-idle.

I see the same thing, maybe it doesn't like Honda's? I have an 2002 si, and its never read over 100, with my foot all the way to the floor. :eek:

How high was your RPM when your foot was on the floor?

No-Load fuel usage
 

Ok, so I spent some gas on research this morning. We wanted to know how much gas is used at various RPM while not loaded -- in this case, using the normal meaning of "load", not the OBDII "load".

I tried various RPM in neutral in my 2002 GMC 5.3l V8, measuring with my fuel injector duty cycle meter. I repeated it multiple times and got some steady values, some that varied from one test to the next.
RPM Duty cycle
525 1.1 (Idle)
600 1.3 (Idle)
1000 1.7
1200 1.6 (WTF?)
1500 2.5 - 3.0
2000 4.0 - 4.4
2500 5.5
3000 6.9
4000 8.2 - 9.2

There's no redline marked but it's never shifted higher than 5500 and I hate all the ruckus involved with high RPM, so even getting up to 4000 rpm was quite distasteful for me...I wasn't going to bring it up any higher.

Looks nearly linear to me, except for that one anomoly at 1200 rpm. I added in a linear best-fit model, with a 99% correlation between the two. (using excel's correl() function). A linear fit matches with the results from MetroMPG's test above. The only difference is, this is duty cycle and not mpg, so the linear fit is inverse of Metro's.
Attachment 1536

I wonder if there's some sort of intake or exhaust resonance at 1200 rpm. That might explain the dip.

Thanks a lot HC and Pale, I was very curious as to what the potential for my original engine design would be.

Linear means my old VX would have used 80% of the fuel at 2000 RPM in maintaining engine speed.

I know increasing the load would increase efficiency, but a significant majority of that 80% is just keeping the engine running at that RPM.

This corresponds with some members who manage to achieve 100% over EPA ratings through hypermiling, because they are reducing (to the greatest extent possible) unnecessary engine running.

In the era of the rotary aircraft engine, prior to 1918, a direct comparison of the 1913 Mercedes grand prix engine and the Gnome racing aircraft engine reveals an interesting comparison.

Mercedes
22 liters@ 2300 RPM= 200 HP
Gnome
11 liters@ 1300 RPM= 160HP

Twice the engine displacement at 40% greater RPM only produced 25% more horsepower!!!!!!!!!!

That is the best direct comparison to the two engines I have ever found.

The question is:

With modern improvements in design and technology, could the basic rotary design offer similar improvements in power compared to a conventional reciprocating engine. With such an obvious difference in the direct comparison, it has been my belief (for almost 10 years) that a current roatry design, especially one that could transform itself into a flywheel, would offer a dramatic improvement in efficiency over any reciprocating engine that will ever be made.

Also you must consider that the air cooled rotary would have lost significant power in the aero drag of cooling itself. My best guess would be 20 HP.

Now we are talking about a theoretical situation (emphasis on theoretical) where you could build a modern rotary engine (not a wankel, but similar to the original aircraft rotary) that would approach 60% efficiency, cpmpared to modern passenger car diesels at 41%. This would also utilize Homogenous Charge Compression Ignition, and not need any emissions controls.

If this assumpition is correct and my in wheel drive produces the simple powertrain with efficiencies equivalent to my VX, we would be talking about a real 150-200 MPG 5 passenger car.

I hope I live long enough to see it happen.

regards
gary

I did it today, and I was in 5th at 4000 rpms, and it was at 80 something.
I will try on the way home at a higher RPM, like 6500 in 3 or something