“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

Sunday, March 09, 2008

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.


maastrichtian said...

You are WAY above my head with your calculations, but it seems odd that there is no discussion of gearing. Does your analysis include your 6-spd ATM? At 40-mph, with any elevation change I'm thinking your truck is going to shift, and isn't that throwing your numbers out of whack?
Thanks for the always thoughtful analyses and for the diversion from work as well.

King of the Road said...

Yes, those calculations are run based on numbers gathered on (nearly) level ground, with the transmission using whatever gear the engine map assigns in cruise. I'm in the process of a lengthy experiment and calculations regarding acceleration that will (hopefully) take the engine map, torque converter lock up, etc. into account. It's a very very tough and complex problem and, like you, I have a living to earn!