I believe that regardless of the Cd number for the Mercedes, the Model S's design/form is still more "aerodynamic". Here's a point that I think needs to be made, and I'll support my argument below.
Cd is an incomplete measure of aerodynamic efficiency. CdA is the correct value, and I think that's what Doug was getting at.
Review for the non-engineers/science folks: The equation for drag force for high Reynolds number situations (such as a car passing through air) is F = 1/2*ρ*v^2*Cd*A, where F is the drag force, ρ is the density of the fluid, v is the velocity of body relative to the fluid, Cd is the drag coefficient, and A is some reference area--often times the frontal area of the vehicle is used.
What contributes to drag? For cars, primarily form drag (based on the shape of the body), and skin friction (due to the frictional drag of the air on the surface of the vehicle).
Form drag is reduced by having an aerodynamic shape.
Skin friction drag is reduced by reducing the surface area of the vehicle, among other things. That is, a smaller car (more importantly shorter) car will have less skin friction than a similar longer car.
Next, recognize that the drag coefficient is experimentally determined. Essentially, you can directly measure F, ρ, v, and A--you basically just plug in those values and solve for Cd.
Now here's the important part: What is used for A? Do you use frontal area? Area of the surface of the vehicle? A mix of the two?
NASA engineers often need to specify what they used for the "A" reference area when reporting the Cd value.
If you use surface area of the vehicle, then longer cars (the Model S in this case) have an advantage in the published Cd value. Why? Because long/bigger cars have more surface area. This makes the "A" reference area larger, and when you solve for Cd the Cd value becomes smaller.
If you use the frontal area, then the longer/larger car (Model S) has a disadvantage.
Consider two cars with the same frontal area and the same shape (form drag)--one long, and one short. In this case, the longer car will have more skin friction, because there is more skin surface exposed to the fluid. This will result in higher drag force--and since the reference "A" value between the two cars is the same, the longer car will appear to have a higher published Cd value.
This is why publishing Cd alone is misleading. Cd*A is what really ought to be published.
Take a look at the Model S and the Mercedes. The Model S *looks* more aerodynamic, right? I'm willing to bet that it *is*, at least with regard to the design/shape of the vehicle. When you look at a car and state that it looks aerodynamic, you're thinking about form drag. Now look how much bigger the Model S is than the Mercedes. The Model S is much longer. This means more skin friction, higher drag force, and therefore a higher published Cd value, since the frontal area is typically used for the "A" value.
So to summarize: The Model S design is likely more aerodynamic than the Mercedes, but because the Model S is a larger/longer car, its published Cd value is higher.
In other words:
Model S wins the form drag battle.
Mercedes wins the skin friction drag battle, because it's a shorter/smaller car. Less surface area.
Based on how the math works out, and how the Cd is calculated, the Mercedes has the lower Cd value. But make no mistake. If the Mercedes and the Model S were the same size, I have no doubt that the Model S would win hands down.
Make sense?
- - - Updated - - -
I should add--to reiterate what I, Doug, and others have mentioned--that the true measure of the aerodynamics of a vehicle is not the Cd....but the drag force developed by the car.
Auto reviewers are wrong when they state that the car with the lowest Cd is the most aerodynamic. That is just plain wrong. Wrong wrong wrong. It completely disregards the size of the vehicle.
The most aerodynamic is the one with the lowest "F" or drag force.
Since ρ and v are completely independent of the car, the Cd*A is the true measure of the car's aerodynamic efficiency.
So it really ought to be:
F=1/2*ρ*v^2*e, where e is the "correct" drag coefficient.
Therefore, e = 2*F/(ρ*v^2). Higher drag forces yield a higher "correct" drag coefficient. "A" shouldn't even enter the picture. A is arbitrary and irrelevant when you're comparing two completely different vehicles.
I think some car reviewers have figured this out. Others have clearly not.
Is the Model S more aerodynamic? We don't have enough information to know. After all, it is longer and wider--just a bigger car overall. However, I'm sure that if you look at drag force relative to the size of the vehicle, the Model S will have the Mercedes beat. In other words, Tesla probably does better with the design (form drag) given the size constraints of the vehicle.
), the *yawn* was not for the OP, but for the car. Next time you're not sure, just ask me.
