Why is a half-streamlined body less aerodynamic?

Question

As seen here, a streamlined half body is less aerodynamic than a fully streamlined body. The half streamlined one has less surface area (kind of) and less frontal area. Why is its drag more?

Answer 1

The drag coefficients are normalized by a reference area. For something like this, it is going to be the frontal area.

So, if the streamlined body and half-body had the same frontal area, the drag on the half body would be larger. That isn't particularly surprising.

This figure is misleading because all the objects are not drawn to equal frontal areas.

It is hard to know what differences there are in the half-body shape -- for example, does the aft end close out in a 2D or 3D manner? I.e. is it half of an axissymmetric body, or more like a 2D airfoil shape with the nose rounded over.

Edits:

Here is a quick model of a streamlined body and matching half-body I threw together.

These two bodies have identical frontal area. Therefore, their drag coefficient ratios should equal their drag ratios.

First, to achieve the same frontal area, the linear dimensions of the half body were scaled up by $\sqrt{2}$ (i.e. 1.414). So it is 41% longer and 41% wider than the full body.

The wetted area divided by the reference area is the same for both bodies (the flat area against the wall is not counted against the half body).

$S_{wet}/S_{ref}$ = 10.689

So, these two bodies have identical wetted area.

If we assume the wall extends far from the body, it will act like a symmetry plane -- and the inviscid portions of the flow (or the entire low if it was inviscid) would be identical. I.e. we expect the pressure distributions around the two bodies to be similar.

This leaves us with differences in Reynolds number. At the same flow condition, the half body will have a Reynolds number 41% higher than the full body -- which could place it in a dramatically different place.

Of course, from the original drawing, it is possible the half body looks like this:

Notice the 'flattened' tail. You can see it from the top view, but the side view looks identical.

Answer 2

The numbers given in the table are coefficients, not force values. They have to be referenced to a given frontal cross-section area.

The wrong way to interpret the table: "If I take a streamlined body, and then I cut it in half and plop it down against the floor of a wind tunnel, then the drag will have increased by a factor of .09/.04 = 2.25."

The right way to interpret the table: "If I take a streamlined body, and then I cut it in half and plop it down against the floor of a wind tunnel, and then I magically expand the size (volume) of the body so that its new cross-sectional frontal area is exactly the same as before I cut it in half, then the drag will have increased by a factor of .09/.04 = 2.25."

Here's an observation: using an ellipsoid (for simplicity) rather than a streamlined body, when we go through this procedure, the actual exposed surface area remains exactly the same, not counting the surface area in contact with the floor.

For example, an ellipsoid with a major axis of 3 units, and with minor axes of 1 unit each, has a surface area of 30.83 units, and a frontal cross-sectional area of pi.

If we want to enlarge the ellipsoid so that (before slicing in half) it has double the frontal cross-sectional area, the minor axes will be enlarged to each be the (square root of 2) units long, and the major axis will be three times that length. The surface area is now exactly doubled, to 61.66 units, so after cutting it in half, we are back to a surface area of 30.83 units, not counting the surface area in contact with the floor.¹

However, we've increased the enclosed volume by a factor of 1.414, or the square root of 2.

These relationships remain true regardless of the ratio between the length of the major and minor axes of the ellipsoid. For example, if we increase the initial length of the major axis from 3 to 5, while leaving the minor axes at length = 1, then after we expand the ellipsoid to double its cross sectional frontal area, and then slice it in half, we'll still end up with the same surface area we started with (not counting the flat side), and we'll still end up with 1.414 times the enclosed volume that we started with.

The surface areas and volumes noted here were calculated by this on-line calculator, which also states the formulae involved.

Since we increase the enclosed volume when we expand the streamlined body to double its cross-sectional frontal area and then slice the object in half, it's not surprising that we'll find that the drag coefficient has increased. There's more "bulk" there to deflect the streamlines.

Footnotes:

1. Another related answer has noted that this same relationship-- total surface area remaining constant, not counting the area in contact with the floor-- also applies when we apply the same manipulation to an actual streamlined body, rather than to an ellipsoid.

Answer 3

In terms of drag coefficient, it is important to have "clean", undisturbed air flowing over a surface to minimize drag.

The half airfoil shape near the ground has a higher coefficient of drag because airflow near the ground is more chaotic than airflow further away.

Through no fault of the shape itself, the measured drag coefficient will be higher near the ground.

Move the same shape away from the ground, maintaining 0 angle of attack, it's drag coefficient will be lower, especially if the bottom leading edge is rounded.

One can see why engineers of the 1940s became enamored with "laminar flow" to reduce drag for long distance aircraft. Turbulence disrupting the boundary layer can dramatically increase drag.

Scale can also affect drag coefficient because, again, of greater distance from the ground. The 1938 Volkswagen Beetle has a drag coefficient of 0.48. The more modern Volkswagen XL1 has a drag coefficient of 0.19, and the Ecorunner 8 has 0.045.

Finally we can see the number values in the question chart refer to coefficient of drag, seen in the drag formula as:

Drag = Cdrag × Area × Air density × Velocity$^2$

To get a comparison of Drag for given shapes, the frontal areas must be equal.

Answer 4

This is a typical example where a picture is worth a thousand words... or is it?

This picture is a potpourri of several experiments, some of them performed and recorded in very old NACA reports. The coefficients for the circular shapes have been extracted for example from this NACA report dated 1937.

The coefficients in that picture are valid for fully turbulent flow, at Reynolds numbers from $10^4$ to $10^6$ and, as usual for subsonic aerodynamic coefficients, they have been adimensionalised via the "frontal area". This explains the difference between the "streamlined body" and the "streamlined half-body": the latter possesses half of the frontal area and therefore its coefficient would be twice as much as the full body, even if they had the same drag.

Even if it's not specified, the coefficient of the cube has been obtained from a cube lying on the wind tunnel's wall.