My daughter is a mechanical engineer for Firestone Industrial Products. As such she is involved in vehicle design, among other things. I put the following question to her:
Is there a magic speed number where a car will get the best cost for a particular trip.
i.e - suppose all things are equal and two cars travel the same 500 mile route with the same conditions and equipment.
The first car goes an imaginary 100 MPH and it therefore take about 5 hours.
The second car drives it at 50 MPH and will take about 10 hours.
Now, the second car gets better gas mileage because of the slower speed, but is using the engine twice as long as the first car. So, the first car, while not getting good mileage gets there quicker and only runs its engine half the time the second car does. (make sense I hope)
There has to be a way to figure a magic number for a particular car going a set distance as to what its speed should be to achieve the best/least amount of gas used for the trip.
And, her response was:
The first car going 100 MPH will chew through a TON more fuel than the one traveling at 50 MPH to cover the same distance.
The crux of the question is how to minimize the amount of energy (fuel) needed to push a given mass (the car) a given distance (500 miles). While the first car does use the engine for half the amount of time, that is offset by several factors. The first one, and probably the biggest issue, is that it takes a lot more energy (fuel) to push a car at 100 MPH for 500 miles. The wind resistance is, in part, a factor of the drag coefficient of the car, multiplied by the square of the velocity. So as velocity increases, the drag increase is substantial. The net result is that you have to mash the accelerator a lot more to keep the engine bubbling along at whatever RPM is required to hold a particular speed. So, with all else being equal, a car running at 100 MPH is spinning twice as many RPM in the engine as one running at 50 MPH (for arguments sake, let's say 4000 RPM at 100 vs 2000 RPM at 50). In a frictionless world, one without any wind resistance, the cars' fuel consumption would only be variable by the cars BSFC (Brake specific fuel consumption, more on that later). However, with the wind resistance, the car's engine needs to produce a lot more horsepower (and torque) at 4000 RPM to keep pushing the car through the air. For instance, if it takes only 50 HP to overcome the wind at 50 MPH, it may take 150 HP to keep the car steady at 100 MPH.
To expand a bit further, a car running at 50 MPH into a headwind of 10 MPH will use more fuel than a car at 50 MPH with no headwind. The engine is running at the same speed, but the engine needs more fuel to overcome the wind resistance, so you're giving the car more gas.
As far as ideal conditions, it's very difficult to itemize. When an engine is dynoe'd, you can get the BSFC (Brake specific fuel consumption). This basically gives you the engines fuel economy for every engine speed (RPM) and torque output. If you run the engine at peak efficiency, where the BSFC is optimized, then theoretically you're getting the best fuel economy. This is generally optimized at around 35% of maximum RPM, and 75% of the peak torque for unmodified passenger cars. In terms of what that means with the throttle, it suggest "bogging" the engine for city driving (if you have a manual). Offering more throttle at lower RPM will raise the torque output, but keeping the engine RPM at less than 35% of redline should keep the fuel consumption more moderate.
At highway speeds, it means about the same. Keep the engine RPM below 35% of redline RPM. Exceeding that means the engine is getting a greater thirst for fuel for every increment above that. Coupled with the fact that the wind resistance is also increasing as engine speed increases, you get a double-whammy and an overall much thirstier car.
(As an aside, I'm not sure exactly how the BSFC is calculated or measured. I believe that the engine's volumetric efficiency has a little to do with this, as well. The volumetric efficiency is how much air/fuel each cylinder is actually taking in on each engine revolution. in a 3.0 liter v-6, each cylinder has a volume of 0.5 liters. However, on the intake stroke, the cylinder cannot take in that much air/fuel. Turbulence in the intake charge, temperature and pressure variations, intake charge velocity all take part in the volumetric efficiency. Lower volumetric efficiency generally means less power produced for a given engine revolution. Volumetric efficiency is generally highest (best) near the engine's peak torque RPM. Volumetric efficiency is basically a measure of how efficiently an engine is using the available engine displacement at any given time.
So, the ideal realm to operate a car's engine, very generally speaking, would be to have the engine operating at or near the optimized BSFC and torque. However, if this peak requires running the car at a speed where the wind resistance begins to offset any efficiency gains in the engine, then you're still a net negative on fuel economy for the car. Let's say the best BSFC is around 2200 RPM. Ideally, you'd want to operate the engine here. But, if this ratchets the speed to a point that wind resistance is out gaining efficiency increases, you'd be better off running the engine at a less optimal 2,000 RPM.
So, to find the "optimum" for any car would be very difficult. We'd need a BSFC map for the engine, the overall gearing (to know what vehicle speed a given engine RPM will produce), know the drag coefficient, and probably a litany of other items I'm overlooking.
But if it was me, I'd try to keep the engine at or below 35% of redline RPM (2100 RPM for a 6000 RPM redline) to get the best bang for my fuel buck.