I found this site with formulas to calculate for the parasite drag. But I wondered if these formulas are to calculate parasite drag or the parasite drag coefficient? So do you have to multiply the calculated value with the square of the velocity, as in the formula for the general drag force?

I also wonder if there is a difference between the wetted area of an airfoil and the normal planform area?

  • $\begingroup$ Can you add which formula you mean in your question? $\endgroup$
    – user2821
    Commented Dec 21, 2014 at 9:11
  • $\begingroup$ I mean this formula: f = kc_fS_wet $\endgroup$ Commented Dec 21, 2014 at 11:03

1 Answer 1


The philosophy employed on the Stanford site is to convert the product of drag coefficient and corresponding surface area to a "drag area", which is technically the area of a body with a drag coefficient of 1. See page 1-8 in Sighard Hoerner's book "Fluid Dynamic Drag" for the definition and tons of helpful real-world data on drag.

To convert this into a drag force, it still needs to be multiplied with the dynamic pressure, which is the product of the square of the velocity, the air density and the factor 0.5. No further inclusion of the surface area is needed, as this is already factored into the drag area values of each drag component.

Using the drag coefficient lets you compare the relative dragginess of drag components (friction drag, pressure drag, ...). If you want to compare the dragginess of aircraft components (wing, fuselage, tail surfaces, ...), using their drag area adds a factor for their size, making it easier to compare the relative influence of the drag of different aircraft components on total drag.

To answer your second question: Planform area is the wing area you see in the top view of an aircraft, and the wetted area is the actual area which is exposed to airflow. If you start with the planform area, you need first to subtract the part of the wing which is inside the fuselage. Next, you double the result to account for both sides, top and bottom, and then you divide this sum by the cosine of the dihedral angle and the cosine of the incidence angle to arrive at your wetted area.

  • $\begingroup$ So if i get it right, the draf coefficient already has a factor of relative dragginess in it, like S_object/S_total. So the drag area is proportional to S_object^2/S_total? $\endgroup$ Commented Nov 14, 2014 at 23:22
  • $\begingroup$ @SimonRavelingien: The drag coefficient itself is dimensionless, so it is independent of scale, area or speed. The drag area is no dimensionless coefficient any more, but depends on the size of the corresponding part $S_{object}$. By multiplying it with dynamic pressure, you get the drag force of the part. Add all parts up, and you have the total drag. Only when you want to get one drag coefficient for the whole airplane, you need to scale the contributions of single parts with their relative area $\frac{S_{object}}{S_{total}}$. A factor like $\frac{S^2_{object}}{S_{total}}$ is never needed. $\endgroup$ Commented Nov 15, 2014 at 16:02

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