A simple way to save a little fuel

I know that most people won't use the extreme methods of fuel consumption minimization that I've used to achieve a five tank moving average fuel efficiency of 21.09 m.p.g. in my Land Rover LR3 HSE, at least until it's a matter of taking extreme measures or not driving at all. People are repelled by the thought of (and if they're in my vehicle, the experience of) driving 55 m.p.h. on the freeway, coasting wherever possible, etc. But what about a minor adjustment that will save a little fuel?



I've previously detailed my policies on stoplights, including when I turn my engine off, coasting to minimize fuel waste when approaching red lights, whether or not it pays to speed up to attempt to make it through a green (or yellow) light, etc. Possibly, no one will adopt any of the measures I've outlined in those posts. But what about just shifting from drive to neutral? When sitting still with the vehicle in drive and brakes applied, more fuel is used than with the vehicle in neutral. I know this to be the case because I can see it on my Scan Gauge II. As I usually do, I'll run a few calculations to estimate my fuel savings from this policy, and then I'll add some more estimates to see what the effect might be on nationwide energy consumption, trade deficit, etc.



On my LR3 HSE, I typically use about 0.5 gallons/hour at idle in neutral. The absolute manifold pressure is about 4.8 p.s.i. Putting the car in gear (drive) and holding it still with the brakes causes the manifold pressure to increase to about 5.8 p.s.i. Since increasing manifold pressure results in a proportional increase in air mass flow through the engine, and hence a proportional increase in fuel consumption, we can assume that the 20.8% increase in manifold pressure results in a similar increase in rate of fuel consumption. However, we're looking at what could be saved by a driver who adopts the policy of shifting to neutral at stoplights so the appropriate way to look at the situation is that this driver will reduce his or her fuel consumption by 17.2% (1/5.8*100%).



Using estimates detailed previously (slightly modified) for stoplights hit per day, time spent per light, idling fuel consumption, etc. for the average car and driver, it looks like my "average driver" could save about 2.56 gallons/year. At current rates in Southern California, that would amount to a savings of $8.70 at the pump. I guess that would cover a single high-end caffeine product at your local Starbucks. It's probably not enough to save a homeowner headed for foreclosure though. As readers of this blog will readily infer, I certainly do it.



What about the results of nationwide application of this policy? I estimate that somewhere on the order of 333 million gallons of fuel could be saved. This is the gasoline from about 17.5 million barrels of oil. And since the other 23 gallons of oil in a barrel are not discarded when gasoline is produced at the refinery, I'll estimate that something like 8.8 million barrels could be left for other countries to purchase. This would reduce our trade deficit by just shy of $1 billion at current oil prices (about $105/barrel). Hmmm... According to the U.S. Census Bureau the January 2008 trade deficit was $58.2 billion. And here I thought I'd solved the problem.

The best speed for fuel economy

I calculated in a previous post that my "highway mileage" at 55 m.p.h. is 23.27 m.p.g. It stands to reason that there is an optimum speed for fuel efficiency based on the balance between the low efficiency at low speed due to engine friction and the increase in aerodynamic drag, proportional to the square of speed, as speed increases. Now, my Land Rover LR3 HSE is not the optimal aerodynamic shape, with a coefficient of drag of 0.41 and a frontal area of 33.9 square feet. It makes sense, and conforms with various articles I've read (see the article on HowStuffWorks.com entitled "What speed should I drive to get maximum fuel efficiency?" here for example), that the optimum speed for a box on wheels like my vehicle should have a lower optimum speed than a vehicle designed to have excellent aerodynamics.



I made some conceptual calculations that agree with the form of the equation shown in the "How Stuff Works" article linked above. There, it is stated that the power required as a function of speed is a third degree polynomial, that is, P=as^3+bs^2+cs+d where P is power required, s is speed, a,b, and c are coefficients and d is a constant specific to a given vehicle. Since power is the rate of doing work, or more importantly in this case, the rate of use of energy (burning fuel) in the appropriate units (gallons per hour for example), we can say power is proportional to gallons/mile times miles/hour. Then we can say that gallons per mile (the inverse of miles per gallon) is proportional to power divided by speed. So, substitute the polynomial above for power and divide by speed and we find that the rate of fuel consumption in gallons per mile is a second degree polynomial function of speed. Sorry for the extended math exposition!



In any case, the above leads to the following expression for fuel per unit of distance: f=ms^2+ns+p, where f is fuel consumption per distance (say, gallons per 100 miles), s is speed in miles per hour, and m and n are coefficients and p is a constant different (probably) from the previous ones. I used the same level stretch of freeway in no wind conditions that I used previously to check highway mileage (linked above) over a few weeks to check the instant miles per gallon at various speeds allowed by traffic (when I was able to maintain a set speed long enough for the display to stabilize). I then calculated the inverse in gallons per 100 miles for those speeds, plotted them in an Excel spreadsheet and made the best fit of a second degree polynomial.



From that point, it was a very simple calculus exercise to find the speed at which fuel consumption would be minimized. This turned out to be 43.4 m.p.h. I can plug this into the second degree polynomial, divide by 100, and invert the resulting number to estimate miles per gallon at that speed. The result is 27.47 m.p.g. Not too bad, but remember from the earlier post that it turns out that that stretch of freeway has a very slight downward slope in the direction I used to obtain my measurements. I have to use the method I used in that post to correct the fuel economy. To spare my patient readers the details, the correction yields a final figure of 26.18 m.p.g. at 43.4 m.p.h.



Interestingly, this speed is lower than the one previously calculated for the Jeep Grand Cherokee Limited I used to have, even though that vehicle had (according to various web sites) a higher drag coefficient. And, looking at the vehicles side by side, the Jeep looks sleeker. Even with the higher drag coefficient, the jeep feels a smaller drag force due to the smaller frontal area. But unless the figures are mistaken, the looks are deceiving as far as drag coefficient is concerned. And the estimate I had for that vehicle of optimum speed for fuel efficiency was a little over 50 m.p.h.



So, do I plan to reduce my freeway driving speed from the current 57 m.p.h. (it's 57 because I ran the above calculations based on speedometer reading, not Scan Gauge II's 2 m.p.h. lower reading since the speedometer seemed more accurate over a timed measured mile)? Probably not, since I have to be alive to continue with the experiment.