The discussion at ecomodder revolves around engine maps, gearing, etc. and of course, these are a crucial factor in the analysis. One thing I don't see, though, is the discussion of kinetic energy addition I mentioned in my first post on the subject. The fact is, if I bring my Land Rover LR3 HSE from 0 joules of kinetic energy (i.e., standing still) to about 800,000 joules (the kinetic energy at 55 m.p.h.) at a given rate of acceleration, and at half that rate, I will travel twice as far in the latter case. Since addition of kinetic energy is a matter of converting the chemical potential energy of gasoline, I've used a given amount of fuel over a longer distance. Further, I've gone a greater distance at a lower level of aerodynamic drag.
On the other hand, I've spent a greater amount of time at a speed that is less efficient. It's less efficient for two reasons: low r.p.m. and low torque is a very inefficient part of the engine map, i.e., has a high brake specific fuel consumption; and I've spent a longer time down near 0 m.p.h., where all my fuel is going to turn the engine and not overcome external forces.
Another complicating factor is "torque converter lock up." For those who don't know, a car with an automatic transmission has a torque converter between the flywheel and the transmission. This fluid coupling enables the car to stop in gear without killing the engine and serves a purpose similar to the clutch in a vehicle with manual transmission. It consists (very basically) of a case containing a fluid, a pump attached to the casing that turns with the flywheel, and a turbine that is turned by the fluid forced onto it by the pump. There are losses in the fluid coupling for various reasons, so most vehicles have a solenoid that locks the pump and turbine together when they are turning at similar speeds. This prevents the losses in the fluid coupling but is only active at speed, thus increasing the likelihood that getting to speed faster (i.e., accelerating more quickly) will save fuel.
That must be all then, right? No, of course not. There's also the ECU, or engine control unit. If the throttle is pushed to the floor, most ECU's will operate in so-called "open loop mode." Here, the computer doesn't take feedback from the oxygen sensor and estimates how much fuel to inject based on outside conditions (temperature and pressure) and throttle position. It tends to supply a very rich mixture, thus hurting fuel economy and suggesting that slow acceleration is best.
So, does one method prevail, i.e., as slowly as possible or floor it? Or is there an optimal "Goldilocks" rate (not too quick, not too slow) of acceleration? I determined to find an analytical solution. Many estimates and assumptions are necessary since I don't have an engine map for my vehicle and I don't know the parameters of the ECU. Such considerations have never stopped me before though.
I'll start with a terrific document prepared by Sierra Research for Environment Canada entitled "Alternative and Future Technologies for Reducing Greenhouse Gas Emissions from Road Vehicles." It's linked and can be downloaded as a pdf file here. This document discusses (among a huge variety of other topics) "brake specific fuel consumption" and provides a generic engine map. This map shows engine load in brake mean effective pressure as a function of engine r.p.m. and the isopleth contours show lines of equal brake specific fuel consumption. It's a composite of 1995 normally aspirated, fuel injected, two valve per cylinder engines and thus is similar but not the same as my engine, but it's the closest I can find.
Now, it's well known that losses related to heating the engine block and throttling losses as the engine turns slowly at low r.p.m. and low demand, and heating oil due to friction at high r.p.m. lead to a an island of lowest specific fuel consumption at high engine demand and mid-range r.p.m. I don't have an engine map for my Land Rover LR3 HSE so I'll use the features of the map in the Sierra Research document adjusted for the few calculated points I have for my vehicle. This, together with the calculated force required to add kinetic energy and overcome external forces, will enable me to numerically calculate (estimate) the fuel used in going from 0 to, say, 55 m.p.h. at various acceleration rates.I'll start with what I do now, which takes me from 0 to 55 m.p.h. in about 45 seconds and compare with a much brisker rate taking me to 55 m.p.h. in 10 seconds. The vehicle is rated to get to 60 m.p.h. in 8 seconds at wide open throttle. I've made a spreadsheet to analyze the energy required to add kinetic energy and to overcome external forces for each tenth of a second.
The result using slow acceleration is that I burn about 0.27 pounds of fuel in accelerating to 55 m.p.h. over a distance of 550 meters. Accelerating quickly, I burn about 0.19 pounds of fuel to get to 55 m.p.h. over 120 meters. To this I need to add the fuel used in going the additional 430 meters to get to where the slow acceleration regime took me. Using the figures from previous calculations on highway cruising fuel use, I'll use about 0.072 pounds to go 430 meters for a total of 0.262 pounds. This indicates that the quick regime is very VERY slightly more efficient. It's doubtful that my various estimates and interpolations are accurate enough to have much confidence in this result with respect to the specific numbers, but I do think it's safe to say that there isn't much difference.
Now, in gathering this data and calculating, it's clear that the poor economy is at the very low r.p.m.'s and speeds, and at r.p.m.'s above about 2600. So I believe the key is to accelerate briskly to second gear and then back off to a moderate rate to get to speed. I'll take the time to plug the numbers in for such a regime in the coming days, but I wanted to get something posted and the data gathering and calculations resulting in the conclusions above took about eight hours. So I guess the question mark in the title of this post is there for a reason, this isn't quite the final word.
8 comments:
Wow, a brilliant analysis. it really does make sense when you look at it now. Very similar fuel consumption at both levels of acceleration, I'd never guess that. I guess I can floor it now when I accelerate and know that I won't be stopping for another 500 meters or so. Once again, great article.
Take care, Ella
Thanks for the comment. An engine map for my engine and a LOT more time spent looking at various regimes would enable me to be more accurate.
I've got acceleration (and by numerical integration, velocity) data as a function of time for a few acceleration regimes, that's a start. And I have enough instrumentation on the car to get some more data points for brake specific fuel consumption at various r.p.m./brake mean effective pressure combinations. It's a matter of burning the fuel to run the experiments to get the estimates.
Maybe next time you could analyze different fuel consumption levels when going down hill and compare different gears versus neutral according to how far you get with how much fuel. Lets say the hill is 1000 meters long and then the rest of the road is straight, would it be better to glide it down and get more speed in neutral or put it in 5th and burn no fuel but get less speed and having to accelerate earlier on the straightaway. Just a thought :)
Take care, Ella
If I'm going down a hill of 1000 meters I typically have the engine off. I've actually used the gradient of a hill and the stable speed going down to confirm the external forces acting on the vehicle. I'll give your ideas some thought though, thanks.
It is, in fact, a thoughtful and well presented discussion. I'm not surprised that much conceptually as f=mv always and that will prevail once to speed. For example, if you stop your test at the disance where fast acceleration gets to speed, then you never get to the stadium to see the White Sox. No, no, that's not it. It's that acceleration becomes, obviously, less important the longer the distances between stops. You need to figure the average distance between stops (lights, pedestrians, deer, etc.) and then figure an optimal acceleration, using your beautiful calculations, for your "typical" drive(s). That would be useful. But you'd still be driving like a wuss.
To anonymous,
A thoughtful comment. Of course, in f=mv, v must be acceleration so your symbology is non-standard. Or, you may have meant f=d(mv)/dt with v as velocity.
As to its relevance, it's true that the effects are diluted by long stretches of steady state driving but I turn off my vehicle as I coast into a parking space, so "impact" is not so much of a factor as you can infer.
Finally, my level of self-confidence in my manliness is such that I can easily withstand accusations of wussiness. God knows I'm used to them.
The graph was a great reference!
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