These are some of my thoughts and notes to give myself some understanding of the topic of how to design an economical car, how to estimate their drag coefficients, and some of the design considerations. I think you'll find it interesting. If you are bored, skip the math and look at the graphs. If you are mathematically inclined, check over my figures.

Have a look at the following

diagram.

Designing an automobile is an exercise in juggling various priorities. Note that it is possible to design a shape that gets a drag coefficient far below modern cars - the streamlined body, teardrop or icecream cone. Most cars have a Cd of around 0.35. The streamlined body has 11% the drag of a typical automobile, and at highway speeds, at least 60% of the work of an automobile is fighting this drag.

There are numerous ergonomic reasons why the teardrop shape is not optimal ergonomically. It's harder to manufacture, consumers aren't used to it, handling will not be optimal because the wheels will be required to be shorter and narrower than most cars, the trunk space will be reduced, the rear seats will have less room, etc.

However, this is a site about saving fuel. Mother nature has provided us with the teardrop shape; we have to work around that. One thing to note is that with a rake of greater than 11 degrees for the rear cone part of the icecream cone shape, there is just as much drag if not more than if it was just severed with a 90 degree bend. (Such as done with the VW 1L car).

We know that drag is primarily a function of the largest area of turbulence. Minimize turbulence completely and you are left with the a Cd of near zero - 0.04. If you have a streamlined shape up until the end (as in the above car), how big the area of that end of the car will determine largely what Cd the car has.

As such, I thought it would be useful to postulate a formula for roughly working out drag coefficients in such cars and our own modifications, relative to vehicle length so that we can get a rough idea of the optimization process and why the cars are built this way.

The first thing to do is realize that we can approximate the car as a choice between two extremes. The first of those is the half sphere. The second is the teardrop, the streamlined body, which is approximately a half sphere and a cone with 11 degree taper. (Obviously this is an approximation, but quite close.)

So, if we say that the drag should be approximately the same as the final area where we cut off the icecream cone shape, and we know that the drag has to lie between two extremes, 0.42 for the half sphere and 0.04 for the streamlined body, we can propose a formula.

Cd = 0.04 + (Af/Ahs) * 0.38

Where

Cd = drag coefficient

Af = final area at rear of car

Ahs = area of half circle (or cross sectional area of car)

Note that when Af = Ahs, we get a Cd of 0.42, which is the same as the hemisphere. And when Af = 0, we get a Cd of 0.04, as we would expect from the full streamlined body.

Next we want to get Cd in terms of l, or length from the half circle part of the car to the rear, as indicated in the above diagram. This will enable us to graph Cd versus l, give us an indication why car manufacturers make the tradeoffs they do, and give us an indication for our own car modifications.

First of all, we know that Af corresponds to the following formula for area of a circle:

Af = pi * r2 ^2

Where r2 is the radius of the final portion of the car.

If r1 = r-r2,

And since tan (theta) = opp/adj

tan theta = r1/l

therefore r1 = l * tan (theta)

therefore r2 = r - r1 = r - l * tan (theta).

Now, Af = pi * r2 ^2

= pi * (r- l * tan (theta))^2

So, Cd = 0.04 + (Af / Ahs) *0.38

= 0.04 + [(pi * (r- l * tan (theta))^2)/Ahs] * 0.38

But since Ahs = pi * r^2,

Cd = 0.04 + [(pi * (r- l * tan (theta))^2)/(pi * r^2)] * 0.38

**Cd = 0.04 + [(r- l * tan (theta))^2 / (r^2)] * 0.38**
So now we have a formula for drag in terms of length and radius at the widest point. Note that the above formula can be simplified in terms of (l/r) ratio. Also note that the formula will only apply with theta <=11 degrees.

If we adopt theta = 11 degrees, l/r = 5.14 gives us the full streamlined body, and l/r = 0 gives the half circle. We can now graph the result to give us a handy curve by which we know when it is appropriate to "cut off" our streamlined body.

Here are the same figures, but plotted against with the y-axis as logarithmic scale to give a better indication of where the diminishing returns are - around l/r = 4.

So, now we have a much better grasp on how far to extend the tail of our car in order to achieve a particular coefficient of drag, or how much we sacrifice by cutting that tail short. And note that the constants in the formula will vary from car to car, depending on how streamlined the front is, etc, but it's still good to get an overall picture of how the "boattail" thing works and the design considerations involved.

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