I am a DIY mechanic who is looking to build a vehicle to maximize fuel efficiency at highway speed. I have been working on cars all my life, and I am a scientist, so I prefer to plan and calculate rather than just throwing things together. Each engine has it’s own fuel consumption curve, and I want to ask any of the actual automotive engineers out there how one might approximate fuel economy losses due to air resistance. It is very straightforward to calculate the theoretical maximum fuel economy for a given speed using the specific fuel consumption and power output at a given RPM, and then plugging that RPM into gear ratios to arrive at a certain speed and fuel consumption. However, that only reflects the engine and drivetrain (which are weightless in the model) traveling at speed with zero air resistance. Obviously that is not a realistic model, and so I am interested in calculating adjusted fuel economy assuming I know the drag coefficient and frontal cross-sectional area of the vehicle. Does anyone have an idea of how to do that (in a general sense, I can do the math myself with my own numbers)? Thanks in advance for your help!

“It is very straightforward to calculate the theoretical maximum fuel economy for a given speed using the specific fuel consumption and power output at a given RPM”

You need calculate the power required to propell the car at highway speed. More specifically, the power requirement to overcome all the resistances (rolling resistance, wind resistance, etc.).

Power requirement = Resistance X Velocity

Once you know the total power requirement, you need to compensate for drive train efficiency. The exact efficiency is best determined through experimentation on the specific vehicle in question. But for estimation purposes assume it’s 80% - 85%

Also, be aware that there are many other factors that can affect fuel economy. So calculations are really nothing more than an educated guess.

Right, so if I figure that the total resistance for a very small vehicle comes out to be 5 horsepower at 75 mph, then the engine must produce at least 5 hp to travel at that speed. However, that doesn’t explain (at least I don’t think it does) how air resistance impacts fuel economy. The assumption in my calculation is that the engine produces a certain, fixed amount of power at a given rpm, and that is a constant. I believe (again, I could be mistaken) that specific fuel consumption is performed under a no-load situation. If the engine produces a given amount of power using a certain amount of fuel, how does that consumption translate to a load environment? Perhaps that is the question I ought to have asked. The only relationship I have been able to draw from series of drag equations is that at some point the vehicle will become drag-limited (the engine can only produce a certain amount of horsepower, and at some speed, that horsepower will equal the power from the air resistance, which means that the car cannot go faster than that speed unless the engine’s horsepower is increased). After determining that yes, apparently a 10 HP engine CAN go 75 mph, I now need to decide how fuel economy is related to that. Any ideas? Perhaps my logic is unsound, I will certainly allow that that could be possible. Thanks again.

I prefer to plan and calculate rather than just throwing things together.

Yea, but there is a very good reason they use wind tunnel test to get those air resistance numbers. It is just too complex and variable to compute. It is much cheaper and more accurate to measure.

You might research the drag calculations developed by the WW-2 aircraft designers…Drag increases with the square of speed or perhaps the cube of speed, something like that… Once you reach 60 mph, the drag starts to consume some serious power. At 100 mph, any thought of fuel mileage is gone as air drag, at this point, requires SERIOUS horsepower to overcome. These ponies must be fed…

The frontal area of automobiles could be reduced, giving a worthwhile reduction in drag, but the practicality of the vehicle is also reduced as it’s shape becomes more and more radical…

The force needed to overcome air drag increases with the square of the speed.

The power needed to overcome air drag increases with the cube of the speed.

If you doubt that it takes an eight fold increase in power to double the speed of a vehicle, compare the top speed of a Suzuki Hayabusa motorcycle (178 horsepower, 191 mph in the models that were built before top speed was electronically limited to 187 mph by an agreement amonst the Japanese manufacturers) to the horsepower of a motorcycle that can only go half that fast. (1970 Honda CB350, about 25 horsepower realistically unless you belive the advertising hype of that era, and about 95 mph, unless you actually believe the hilariously optimistic speedometers found on most bikes of that era.

Incoming wall of text…

I appreciate everyone’s help and advice very much. I already have researched this area extensively, and calculating power and force, etc, is trivial. The problem I am having is that I can’t find any information correlating the work an engine does with its fuel consumption while under load. If one assumes for a moment that there is no air resistance, then the fuel consumption (in gallons) is without doubt solely related to engine RPM (although fuel injector geometry and such things make the relationship extraordinarily complex). Many engines have published figures on this sort of thing. Without taking air resistance into account, then the engine will consume a certain amount of fuel while outputting a certain horsepower at a given RPM. If you then use a transmission to convert engine RPM to wheel RPM, you can get significantly more wheel revolutions than engine revolutions. In this case, the fuel consumption is constant, so as speed increases, one discovers that gas mileage INCREASES without limit. SInce there is no work being done by the engine, the fuel consumption is solely related to the consumption at whatever engine RPM relates to the speed. The problem is that this calculation breaks down in the real world, or else everyone would drive as fast as possible to get the best mileage. I already mentioned how drag power would eventually become the limiting factor to top speed (and thus fuel economy), however, I am having trouble relating increases in air resistance with increases in fuel consumption FOR THE SAME RPM. Consider that for a given car, at 3000 rpm in third gear, you might be traveling at 30 mph. At 3000 rpm in fifth gear, you might be traveling at 75 mph. Obviously increasing the distance you cover per revolution would increase your fuel economy (that’s why every conventional, non-hybrid car gets better gas mileage on the highway than in the city, although the city has stops and starts to consider as well) However, air resistance must necessarily increase your fuel consumption for the engine to turn over at that RPM while fighting that resistance. I think that this issue comes down to the following question:

On a power curve of an engine, if you run the thing at a certain RPM, does it always produce the amount of horsepower indicated on the curve (I suspect not), or is that a measure of the upper limit of power the engine can produce at that RPM? If the former is true, then further correction for fuel economy must be put in place to adjust for air resistance. However, if the latter is true, and the engine would only consume enough fuel to generate the necessary horsepower to achieve the desired speed, then I would have all the information I needed to approximate the answer.

When I do figure out how this is done, I will happily publish my results on an excel spreadsheet on google docs for all of you to play with. I greatly appreciate everyone’s assistance, and I’m sure if we all put our heads together we can figure it out [or maybe there is an automotive engineer waiting in the wings to answer it for us ; )]

Also, in response to Joseph E Meehan, I agree that the issue is very complex, and I have no doubt that they use wind tunnel testing to aid them, however, I am positive that no giant corporation would waste tons of money on a guess-and-check approach to car design. I suspect they use windtunnel testing to determine the drag area to plug into their equations, but that they must have equations to approximate the end result of their designs. After all, much of the trouble for car manufacturers is (going to be) meeting ever increasing fuel efficiency standards, and they must certainly have some sort of method of approximating the efficiency of a design when considering it for production. I suspect that with today’s computer technology (or even computers from the early 90’s), they could come up with some sort of workable model. I have heard it said that excel is the most powerful program the average person will ever use in their life. Having seen what it can do, I don’t doubt this assertion, and I will try my utmost to make it work so that everyone can benefit (in case anyone else is enough of an idiot to try building his own car from scratch)

Thanks again.

One last post and I’ll shut up. Upon re-reading some of the comments, perhaps I can enlighten you all with respect to what I am trying to accomplish. I can readily lay my hands on an old motorcycle, some transmission parts, and a 10 HP diesel engine. I want to build a three wheeled vehicle for commuting to work. Unfortunately, the nature of my work requires that I travel off-site a bit, and compensation for travel time is not an encouraging number. I only need the thing to be one or two seats, and tandem at that, and small, lightweight, and very aerodynamic. I’m not overly concerned with performance, although some consideration must be made for simple highway safety (it better accelerate fast enough not to get creamed from behind), and my driving motivation is to affect absurdly high gas mileage. Given the minimal needs for the design, I don’t doubt that such a thing is possible, and has indeed been done by several car manufacturers as prototype concept cars. Additionally, I don’t need to worry about automotive safety standards since I will be registering the vehicle as a motorcycle (that was the FIRST thing I researched), but I will obviously try my hardest not to make the thing a rolling death-trap. With that in mind, I hope that my goal is a little clearer. After all, I need to fuel my '88 Porsche 928 that is sitting in the driveway all alone ; )

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"On a power curve of an engine, if you run the thing at a certain RPM, does it always produce the amount of horsepower indicated on the curve (I suspect not), or is that a measure of the upper limit of power the engine can produce at that RPM? "
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You are right to suspect not. The typical horsepower curve is based on WOT (Wide Open Throttle). There is a different curve for every possible throttle position from just off idle to WOT. Thats a lot of curves.

Driving is a very dynamic environment. In addition to the air resistance due to the speed of the vehicle, there is the air resistance due to wind, which is often varying with time. Additionally the terrain affects the rolling resistance (drag due to other things, like tire resistance, momentum and internal resistances). For that reason, the driver is always adjusting the throttle position to find the curve to maintain the desired speed., or change the speed as needed.

BTW, the wind tunnel measurements mentioned earlier are done with small clay models. Real cars and trucks have higher coefficients of drag than the models, usually about 50% higher. There are formula’s using the c.d. (coefficient of drag) of a vehicle to create a fuel economy model but most of use just use them as a relative comparison of one vehicle to another. That is the vehicle with the lower c.d. will get better fuel economy and go faster, all other things being equal (which they never are).

Most of the aerodynamic drag formulas were developed by the Wright Brothers. For a couple of bicycle mechanics, they were pretty smart.

Excellent! So would you say then that one could approximate fuel consumption by estimating the drag power on the car (hp) and then multiplying that power by the specific fuel consumption (gal/hp-h) of whatever RPM produces the desired speed? That would give gal/h, which you could then divide speed by, and net out miles/gal (mi/h/gal/h = mi/gal). Or perhaps are you suggesting that the specific fuel consumption curve only applies to WOT? I would suspect that it must be listed as *specific* fuel consumption because that roughly correlates to the fuel consumption over the given range of throttle positions for that RPM (which would make it an actually meaningful number rather than a curiosity of WOT). Thanks so much for your help!