Fig. 1. Air Flow About a Flat Normal Plate. Pressure Zone at Front and .#. Turbulent Zone at Rear (H). Arrows Show Direction O OW.
Air Pressure on Normal Flat Plates. When a flat plate or "plane" is held at right angles or "normal" to an air stream, it obstructs the flow and a force is produced that tends to move it with the stream. The stream divides,as shown in Fig. 1 and passes all around the edges of the plate (P-R), the stream reuniting at a point (M) far in the rear. Assuming the air flow from left to right, as in the figure, it will be noted that the rear of the plate at (H) is under a slight vacuum, and that it is filled with a complicated whirling mass of air. The general trend of the eddy paths are indicated by the arrows. At the front where the air current first strikes the plate there is a considerable pressure due to the impact of the air particles. In the figure, pressure above the atmospheric is indicated by *, while the vacuous space at the rear is indicated by fine dots. As the pressure in front, and the vacuum in the rear, both tend to move the surface to the right in the direction of the air stream, the total force tending to move the plate will be the difference of pressure on the front and rear faces multiplied by the area of the plate. Thus if F is the force due to the impact pressure at the front, and G is the force due to the vacuum at the rear, then the total resistance (D) or "Drag" is the sum of the two forces.
Contrary to the common opinion, the vacuous part of the drag is by far the greater, say in the neighborhood of from 60 to 75 per cent of the total. When a body experiences pressure due to the breaking up of an air stream, as in the present case, the pressure is said to be due to "turbulence," and the body is said to produce "turbulent flow." This is to distinguish the forces due to impact and suction, from the forces due to the frictional drag produced by the air stream rubbing over the surface.
Forces due to turbulent flow do not vary directly as the velocity of the air past the plate, but at a much higher rate. If the velocity is doubled, the plate not only meets with twice the volume of air, but it also meets it twice as fast. The total effect is four times as great as in the first place. The forces due to turbulent flow therefore vary as the square of the velocity, and the pressure increases very rapidly with a small increase in the velocity. The force exerted on a plate also increases directly with the area, and to a lesser extent the drag is also affected by the shape and proportions. Expressed as a formula, the total resistance (D) becomes: D = KAV², where K = co-efficient of resistance determined by experiment, A = area of plate in square feet, and V= velocity in miles per hour. The value of K takes the shape and proportion of the plate into consideration, and also the air density.
Example. If the area of a flat plate is 6 square feet, the co-efficient K = 0.003, and the velocity is 60 miles per hour, what is the drag of the plate in pounds? Solution. D = KAV² = 0.003 × 6 × (60 X 60) = 64.80 pounds drag. For a square flat plate, the co-efficient K can be taken as 0.003.
Aspect Ratio. The aspect ratio of a plate is the ratio of the length to the width. Thus, with an aspect ratio of 2.0, we understand that the plate is twice as long as it is wide. The ratio of the length to the width has a very considerable influence of the resistance or drag, this increasing as the ratio is made greater. If the resistance of a square plate is taken as 1.00, the resistance of a plate with an aspect ratio of 20 will be about 1.34 times as great. The following table will give the effects of aspect ratio on the resistance of a flat plane.
| Aspect Ratio. Length/Width | Resistance K as a Multiple of a Square Plate. |
| 1.00 (square) | 1.00 |
| 1.50 | 1.04 |
| 2.00 | 1.05 |
| 3.00 | 1.07 |
| 4.00 | 1.08 |
| 5.00 | 1.09 |
| 6.00 | 1.10 |
| 7.00 | 1.12 |
| 9.00 | 1.14 |
| 10.00 | 1.15 |
| 15.00 | 1.26 |
| 20.00 | 1.34 |
| 30.00 | 1.40 |
To convert the values of a square plate into a flat plate of given aspect ratio, multiply the resistance of the square plate by the factor under the "K" heading. For example: The resistance of a certain square plate is 20 pounds, find the resistance of a plate of the same area, but with an aspect ratio of 15. Solution. The factor for a ratio of 15 will be found to be 1.26, hence the resistance of the required plate will be 20 × 1.26=25.2 pounds.
Fig. 2. Air Flow About a Streamline Body Showing an Almost Complete Absence of Turbulence Except at the Extreme Rear Edge. Resistance Is Principally Due to Skin Friction.