“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, August 10, 2008

Specifics of a high mileage car

In my previous post, I discussed what, outside of the engine and driveline, could be modified to increase fuel mileage. What are the specifics of such a car? Since the laws of physics are unchanging as far as is known and reasonably well known at the macro scale at which cars travel down roads, certain conclusions can be drawn. Let's start with the obvious: fuel is burned to overcome forces acting on the car to take it down the road. So there are two fundamental approaches to high gas mileage, i.e.: put more of the energy in a given amount of fuel to work; and reduce the forces acting on the vehicle.

I'll save maximizing the utilization of energy available in the fuel for another post. Here, I'd like to see what it would take to make a car that gets, say, 75 m.p.g. with currently achievable engine and drive line efficiency by reducing the forces acting on the car. I'll look at achieving this fuel mileage at 55 m.p.h. As I've previously mentioned, force times speed is power, and power is the rate of doing work or, equivalently, using energy.

So, we should be able to say that force times speed equals energy (fuel) divided by time, if the appropriate adjustments are made for units. Or, rearranging, force equals energy divided by speed multiplied by time. And, as would be expected, this simplifies to energy divided by distance. So if I assume 125 million joules/gallon, 25% drive line efficiency, and that I use that gallon in 75 miles, I can determine that the maximum combined force of aerodynamic drag and rolling resistance that I can overcome is about 260 Nt (Newtons). For the SI challenged reader, this is 59.6 pounds.

Referring to my previous post, at a fixed speed the only variables available to control are mass, rolling resistance, frontal area, and drag coefficient. Let's assume that tandem seating isn't a saleable option at this point. What can we do? Well, let's start with vehicle weight. In this article, it's estimated that about 40% of the weight of an average car could be eliminated through replacing steel with carbon fiber. Let's use a conservative estimate of 25%. Then, in this article it's stated that the lowest coefficient of rolling resistance on tires currently available is 0.0062, the highest checked was 0.0152. Let's assume that we can utilize tires with a coefficient of 0.008.

Let's get started. We'll take a small four seat sedan, something like a Toyota Yaris. This vehicle has a curb weight of 2293 pounds, a drag coefficient of 0.29 and a frontal area of (as best I could find) 2.282 meters squared. Let's predict the highway m.p.g. at a steady 55 m.p.h. using, from the previous post, the equation for joules/meter (which is another measure for the inverse of miles per gallon, using the appropriate unit conversions and efficiencies). We'll assume two 170 pound adults to make total weight 2633 pounds. Finally, I'll assume a coefficient of rolling resistance of 0.0115. Running through the calculations, we find that about 355 Newtons are required. To apply this force over a mile, assuming 25% efficiency in the engine, we'd use 0.01828 gallons, or a fuel efficiency of 51.8 m.p.g. Not bad, we're a good part of the way there.

But the car is rated at 36 m.p.g., what gives? Well certainly the EPA tests are more demanding than a steady 55 m.p.h. on level ground. Beyond that, it could be that the new tires with fresh tread have a higher coefficient of rolling resistance. Or, it could be that the engine is able to deliver significantly less than 25% of the energy available in the fuel. If we assume a rolling resistance coefficient of 0.0130 and 20% efficiency, the figure is 36.7 m.p.g. This seems close, and is typical of the types of iterative calculations that are necessary. I'm going to stay in the middle, since I should calculate better than the EPA mileage, due to the rigors of their test. I've verified this in my own LR3. I'm going to assume that the Yaris has a rolling resistance coefficient of 0.0122 and is able to deliver 22% of the energy in the fuel it burns to the wheels. This yields 43.0 m.p.g. Close enough.

Now, what do we get if we reduce the weight by 25%, use tires with a coefficient of rolling resistance of 0.009, and a coefficient of drag of 0.24? Running the numbers, we get 60.8 m.p.g. This is not good, let's see what the maximum credible reductions of coefficient can give us. Using 0.0062 and 0.16 for the coefficients of rolling resistance and drag respectively, we get 90.3 m.p.g. Thus, we conclude that a small car like the Yaris, with the maximally achievable modifications for efficiency, can exceed the target 75 m.p.g. But remember that we've replaced most of the steel with carbon fiber, taken every conceivable measure to reduce drag, and installed tires that are exceptionally efficient and may not wear well, handle well, or be very comfortable. And the fact of the matter is that I very much doubt if a vehicle can be brought to market with a 0.16 coefficient of drag. Let's see what we get with 0.22 and call it good. After all, tires with rolling resistance coefficient of 0.0062 currently exist according to the above-cited article. The answer is 71.5 m.p.g., slightly below the target.

So we conclude that it can be done but the price, both economic and in terms of comfort, is quite high. Clearly, attention to the engine is warranted, as is consideration of drive train modifications. A hybrid engine, combined with pulse and glide driving techniques, could greatly increase efficiency of fuel utilization but it would increase the weight. There is just no free lunch. Tandem seating anyone?

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