I am a new aviation student and I was reading about induced drag the other day. I know that it is produced as a result of the tip vortices and that the greater the aspect ratio of an airplane the less the induced drag force. But when it came to the equation of the force, it is equal to:

$D_i = \frac{1}{2}\rho V^2 S \frac{C_L^2}{\pi AR \epsilon}$

If we substitute aspect ratio $AR$ with span/chord $\frac{b}{c}$ and plan area $S$ as $b\cdot c$, the span term will be cancelled and the induced drag will be affected by the chord length only.

It kind of contradicts the effect of aspect ratio on the induced drag force, doesn't it?


Aspect ratio $AR$ can be written as $\frac{b}{c}$ which is equal to $\frac{b^2}{S}$.

Before we start substituting, note that $C_L$ also depends on the wing surface area $S$.

$L = \frac{1}{2}\rho V^2 C_L S$


$C_L = \frac{L }{\frac{1}{2} \rho V^2 S}$

Substituting all this in the induced drag formula yields:

$D_i = \frac{1}{2}\rho V^2 S \frac{C_L^2}{\pi AR \epsilon} = \frac{L^2}{\frac{1}{2}\rho V^2 S \pi AR \epsilon} = \frac{L^2}{\frac{1}{2}\rho V^2 \pi b^2 \epsilon}$

This shows that the induced drag is proportional to the inverse of the square of the wingspan.


Let's work this through...

The coefficient of induced drag is inversely proportional to the aspect ratio.

$C_{di} = \frac{C_L^2}{\pi AR e}$

NASA Page on induced drag coefficient

The overall coefficient of drag is the form/skin drag plus the induced.

$C_D = C_{d0} + C_{di}$

NASA Page on drag formula

The actual force of drag assuming $C_{d0} = 0$ is

$D = \frac{1}{2}\rho V^2 S \frac{C_L^2}{\pi AR e}$

Can be reduced to

$D = \frac{1}{2}\rho V^2 c^2 \frac{C_L^2}{\pi e}$

Where $c$ is the mean chord.

As DeltaLima pointed out in their answer, you could then substitute $C_L$ using the lift equation to show it is inversely proportional to the span squared.

Both equations are correct so what we can take away is:

  • As the chord is increased induced drag increases
  • As the span is increased induced drag decreases

In other words, the induced drag is inversely proportional to the aspect ratio.

  • $\begingroup$ May I disagree with your first conclusion? If lift stays constant, increasing chord will decrease lift coefficient, so induced drag will not be affected. Only the induced drag coefficient is inversely proportional to aspect ratio. Induced drag is inversely proportional to span loading. $\endgroup$ Mar 1 '17 at 23:05
  • $\begingroup$ @PeterKämpf I thought it was the other way around, the coefficient was constant and the lift changes? $\endgroup$ Mar 2 '17 at 6:35
  • $\begingroup$ If lift changes, drag has to change, too. The goal should be to not change lift, so you can see what the parameter change does to drag. $\endgroup$ Mar 2 '17 at 12:31
  • $\begingroup$ @PeterKämpf but to me that's the point of the question, the does the drag change? Unless I'm mistaken (entirely possible), if we kept the aerofoil shape the same and just increased the size (chord length), the lift would change not the coefficient of lift. $\endgroup$ Mar 2 '17 at 13:58
  • 1
    $\begingroup$ Yes, you are not mistaken. In the context of your question the first conclusion is correct, even trivially so. But it plants a wrong idea about induced drag into the mind - just what prompted the question in the first place. If you look at a real airplane and play with the parameters, a chord increase will reduce angle of attack and result in constant induced drag. From there it becomes easier to see that span loading determines induced drag, not chord. $\endgroup$ Mar 2 '17 at 17:56

As we know that induced drag is due to vortex generated due to pressure diffrence at chord.So if the chord is longer more induced drag will be produced.From mathematical equations we can say that coefficient of induced drag depend on A.R while total induced drag doesn't depend on A.R

  • $\begingroup$ Could you include those mathematical equations in your answer? $\endgroup$
    – Bianfable
    Nov 11 '19 at 10:33

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