It seems that you can get all kinds of different info on the internet (or even these forums) on the best shift points for fuel economy. Thought I'd start a thread specifically for addressing this. Does driving at the lowest RPM possible increase fuel economy? Or should we aim to shift closer to the peak of the torque curve? Or at some other "sweet spot"? Perhaps those who've experimented with a ScanGauge can chime in. . .

 

Starting out I wouldn't shift into 5th until I was past 45mph. That got me about 30-32mpg babying it a bit. I now shift to 5th for all speeds over 35mph and I'm getting 32-35mpg, depending on how much highway driving I do (I usually avoid the highway if possible). My 1-2-3 shifts are usually at no lower than 3,000 rpm, after which I pop into 4th and 5th below 3,000 rpm. Mind you, I have a GD3, though the ideal shift points should be about the same.

 

Shift point for good FE is 2000rpm-2500rpms.

 

I usually shift around 2000-2100 before and was getting 45+mpg. Now I don't care much because the gas prices is not that bad compare to $4/gal

 

it isnt about the RPM. its about the throttle position

 

how does the scan gauge work how does it know your MPG how does it calculate it? were is it wired to?

 

I don't know how the scan gauge is wired. The scan gauge works by telling you how much mpg you are getting with your present driving style, it can make you change your driving style so you can get better mpg,

 

I wanna know how were it wired to!!! lol damn

 

It's not about one or the other, RPMs or throttle position. It's about both, and the best shift points depend on how quickly you're accelerating. For instance: if you upshift early but are still trying to accelerate fairly quickly you're working against yourself. If you're accelerating fairly aggressively the best shift points would be straddling our max torque @ 3400 rpm, no?

 

it plugs directly into your obd2 port. under the driver's side dash area. it reads the computer and displays everything about what your car is doing

 

1900 to 2100 every gear

 

its that easy to install?? some said here after 24hrs or installation i was like what!! so it just plugs directly into your OBD2 port nice!!

 

I'm actually trying a couple tanks of that right now. I'm sick of my 31-32 mpg results on these past 2500 rpm tanks. It's really annoying crossing half tank at 135 miles.

Assuming a stock GD airbox, the peak torque is at or around 4500 rpm on the L15A1. Lowest BSFC should be at an rpm lower than that but not by much, with a high but less than WOT throttle opening (enough to nearly equalize manifold pressure, but not enough to raise dynamic compression to the point fuel richening is required).

Argument. Sorry, gonna give you some redneck physics:

-lowest fuel consumption per mile is in cruising conditions (steady state only, P&G does not apply), so an increased portion of a trip in this state will not decrease mileage. Let's concentrate on acceleration.

Acceleration is change in velocity, thus with mass of car, a certain fixed change in amount of kinetic energy. Energy (joules) is the integral of horsepower (joules/second), so for the constant change in energy E, horsepower being H, and S being the elapsed time, the equation E[car] = H*S must be satisfied. Also written layman as Joules = (Joules/second)[engine] * seconds elapsed. (a Joule is a unit of energy.)

Notice if you cut horsepower in half, the time must be doubled to reach the same speed. (let's forget the e=mv^2 for the moment, this is the overall delta v).

The exact same equation would apply to fuel as well. Assume gasoline has x amount of energy in a given unit of volume, so the rate of fuel consumption is also the rate of joules in the gas per second, J/s. The energy released can be described as E[fuel] = J/s[fuel] * seconds. That is, the E[fuel] is the total amount of energy released by the gas.

Since in any given acceleration example, the total kinetic energy given to the car is the same, and the number of seconds is the same for both equations, we can put stuff together.

The amount we're looking for is the waste heat, the energy released by the fuel we never get to use. For that, just a simple equation: E[car] / E[fuel] = R, a ratio, and the least fuel is consumed during an acceleration to speed with the highest R (not the acceleration itself, that's a change in velocity; this is for the complete effort to get to speed, regardless how long it takes). Let's make an equation from that by substituting what we know about these variables, and make seconds a constant since the time is the same in both due to both equations describing the same situation.

What we have so far:
E[car] = H * S
E[fuel] = fuel rate * S
E[car] / E[fuel] = R

So:
(H*S) / (fuel rate *S) = R

Guess what, seconds cancel out. S/S = 1.

H / fuel rate = R.

Makes sense, less energy that's wasted in heat, the more power you'll get from the same amount of fuel.

Now let's flip that, and call 1/R = r.

r = fuel rate / H, or r = (J/s)[fuel] / (J/s)[engine].

r = J[fuel] / J/s/s[engine], and since fuel has a constant energy density:

r = fuel amount per Horsepower per second, the equation for Brake Specific Fuel Consumption.

So why do all that math to find it? Look back to find what we were dealing with-- the total acceleration effort, beginning to end. Regardless of duration, the car will use the least fuel overall if accelerated at the point of lowest BSFC, even if the instantaneous fuel use is high.

Now read this excellent article:
famous article from autospeed

From there, you should be able to take a reasonable guess at shift points. I'm currently trying out 4500 from first and 4000 from there on out, with a relatively high load (but not WOT).

Hope that helps

 

i was driving tonight. in fifth gear on a flat road, accelerating VERY slow, like 1mph per 10 seconds, and going from 23mph to 39mph, i was getting between 45-55mpg.