I have found a lot of information to calculate drag coefficients and I know that total drag can be calculated by $$ D = \frac{1}{2} \rho u^2 S C_d $$ However I was wondering, if I substitute the zero lift drag coefficient into this equation is it correct to assume the value produced for the drag force would be the parasitic drag whilst the remainder of the total drag is made up of lift induced drag?

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    $\begingroup$ Related: how is the induced drag calculated for a wing with elliptical planform $\endgroup$ – Peter Kämpf Dec 7 '20 at 19:16
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    $\begingroup$ Just a reminder, the above formula is not total drag, you might know this already, what you are calculating is drag produced in response to the generation of lift with respect to an airfoil, remember there are many other forms of drag, for example profile drag, parasitic drag etc. Total drag will be the sum of all of them. $\endgroup$ – Raffles Dec 7 '20 at 21:40

Induced drag is the portion of the lift force that opposes motion. Induced drag exists even when the drag coefficient CD in your equation is zero.

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Lift is perpendicular to the air flow relative to the wing. Imagine a wing moving right to left across your screen. The air is moving left to right relative to the wing. If there's no vertical induced airflow then the lift is vertical and there's no induced drag. However, lift deflects the air down, so typically the air relative to this wing is not moving exactly left to right, but is tilted down slightly. Hence, the lift, which is perpendicular to that airflow, is actually tilted back (right) slightly. That means a (typically small) portion of the lift is "pushing this wing back," opposing its right-to-left motion.

If you know trigonometry, induced drag is lift times the sine of the inflow angle, $ L\sin \phi $. The inflow angle is the angle of the airflow relative to the horizontal in this picture.

There's a picture of exactly this at the bottom of this page in the induced power section. (Induced power is the power required to overcome induced drag.)

  • $\begingroup$ If lift is the component perpendicular to the air flow, as you correctly state in the second paragraph, then it cannot oppose motion. Also the question is about wing, so you shouldn't mix in factors only relevant for propellers—in which case you should also be talking about induced power, not drag. $\endgroup$ – Jan Hudec Dec 10 '20 at 23:23
  • $\begingroup$ Lift can oppose motion. This is exactly the case provided in the example. Motion is right to left, but net airflow is NOT left to right because you must include induced airflow, which "tilts it down" as shown in the picture. The reason it's called induced drag is because it's associated with this induced airflow. $\endgroup$ – Mat Dec 12 '20 at 0:10
  • $\begingroup$ No! Lift is defined as the component perpendicular to relative wind, and relative wind is defined as the free stream velocity, that is reciprocal of the aircraft motion in the air. The force due to circulation is tilted back, and this is the induced drag. But the definitions are such that this isn't what is called lift. Lift is one component of this force and induced drag is the other. $\endgroup$ – Jan Hudec Dec 12 '20 at 21:50
  • $\begingroup$ Please check wikipedia or any standard book or any aerodynamic software. It's well defined - lift is perpendicular to the flow. The flow includes more than motion, e.g. wind, induced velocity, sometimes interference from other objects. Another way to think of this is that it wouldn't make sense otherwise - a wing cannot discriminate flow due to motion from other flow sources. $\endgroup$ – Mat Dec 13 '20 at 22:50
  • $\begingroup$ Wikipedia says: “Lift is the component of this force that is perpendicular to the oncoming flow direction.” (emphasis mine). So it does not include induced velocity and other things. And of course there is no wind in the reference frame of air mass. $\endgroup$ – Jan Hudec Dec 13 '20 at 23:02

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