That's not a simple question to answer. You'd need to know exactly how much of a car's energy is used up overcoming drag as opposed to overcoming the various other losses in the system. The best you could say is that the losses due to drag would also be halved.
Thank you,
According to that article, if we reduce nearly half the drag, we will gain around 2.5 miles per gallon extra. I was hoping to get higher mileage advantage that that :)
Again, there are many variables in the equation. Looking at speed, weight, and efficiency of the engine, are all important.
The reduction of drag reduces the load on the engine. By cutting resistance in half, you get closer to the unloaded efficiency of the motor.
This might be a greater improvement for smaller engines than for larger engines designed to handle heavy loads.
Is there a particular speed you are concerned about? Type of vehicle? Size of engine?
I'm thinking about Tesla Model S. Let's say if we reduce the drag to 0.06 from 0.24, How far the car could run on single charge. Currently they say it runs 200 miles or so. I know they won't be able to reduce the drag so low... but this is just imaginary situation I'm thinking of.
Say you had a fixed journey of say 10,000 m to travel in a given time, i.e. at a given speed, then with a fixed set of vehicle variables (drag coefficient, rolling resistance coefficient, vehicle mass) you could figure the energy required (Power required * time) to complete the journey, then change the drag coefficient to any other value, recalculate and compare.
the simplest approach to enhance a vehicle's mileage is to make it littler and lighter and give it a littler motor. In any case, we need 400-hp sports autos and seven-traveler SUVs and 5,000-pound-limit tow vehicles — and we need great gas mileage, as well.