How is the induced drag calculated for a wing with elliptical planform ? Is this wing shape the most efficient ?

  • $\begingroup$ @mins The most induced drag-efficient according to Lifting Line Theory, but other shapes can be more efficient overall when wing weight and other things are considered $\endgroup$
    – costrom
    Jan 20, 2016 at 19:35

2 Answers 2


Induced drag is caused by the downward deflection of the air streaming around the wing. The resulting aerodynamic force is tilted backwards by half the deflection angle, and the air flows off the wing with an added vertical speed component, producing downwash. Increasing the downwash angle means increasing both lift and the backward tilt, so the induced drag goes up with the square of the lift produced.

If you want to minimize induced drag for a given lift, this quadratic dependence means the optimum is reached when the downwash angle is constant over span.

How is the induced drag calculated for a wing with elliptical planform?

The elliptical, untwisted wing has the same angle of attack and the same lift coefficient over span, and produces the desired constant downwash angle. To simplify things, let's assume the wing is just acting on the air with the density $\rho$ flowing with the speed $v$ through a circle with a diameter equal to the span $b$ of the wing. If we just look at this stream tube, the mass flow is $$\frac{dm}{dt} = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v$$

Lift $L$ is then the impulse change which is caused by the wing. With the downward air speed $v_z$ imparted by the wing, lift is: $$L = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v\cdot v_z = S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho$$

$S$ is the wing area and $c_L$ the overall lift coefficient. If we now solve for the vertical air speed, we get $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$ with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. Now we can divide the vertical speed by the air speed to calculate the angle by which the air has been deflected by the wing. Let's call it $\alpha_w$: $$\alpha_w = arctan\left(\frac{v_z}{v}\right) = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$

The deflection happens gradually along the wing chord, so the mean local flow angle along the chord is just $\alpha_w / 2$. Lift acts perpendicularly to this local flow, thus is tilted backwards by $\alpha_w / 2$. In coefficients, lift is $c_L$, and the backwards component is $\alpha_w / 2 \cdot c_L$. Let's call this component $c_{Di}$: $$c_{Di} = arctan \left(\frac{c_L}{\pi\cdot AR}\right)\cdot c_L$$

For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the reaction force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR}$$

If the circulation over span has an elliptic distribution, the local change in circulation times the local amount of circulation is constant, and the induced drag $c_{Di}$ is at it's minimum. If this would be different, a higher local $v_z$ causes a quadratic increase in local induced drag, so the whole wing will create it's lift less efficiently.

Is this wing shape the most efficient ?

Only if you ask an aerodynamicist would the answer be yes. An elliptic wing will give you the best ratio of lift to drag, which clearly is one way to express efficiency.

In reality, the wing has to lift itself plus a payload, but only lifting the payload should be considered when formulating efficiency. Therefore, pure lift/drag optimization is too narrow. What should count is the best ratio of lift minus wing weight relative to drag. R. T. Jones wrote a NACA Technical Note back in 1950 in which he looked at this problem analytically. Wing weight goes up when much lift is created near the tips, because this lift will cause a disproportional root bending moment, and the wing spar which has to carry this bending moment is a significant part of the wing structure. Therefore, reducing lift at the tips and adding more lift at the root will create a lighter wing for a modest drag increase, resulting in an overall optimum for an almost triangular lift distribution. When compared to an elliptical wing planform, the total wing span of such an optimized wing is bigger for the same overall drag, but this wing will weigh less.

Spanwise loading comparison

Spanwise loading comparison for wings of the same fixed lift, from NACA Technical Note 2249.

But this is too easy. Scaling laws must be considered in addition. You know that elephants have much more massive legs relative to their body size than antelopes (or even ants, for an even more drastic comparison), since body mass scales with the cube of linear dimension while structural strength scales only with the square of linear dimension. This means that wing spar weight will be proportionally higher for larger aircraft.

As a consequence, insects have more elliptic wings than albatrosses, and model aircraft have optimum wings which are much more elliptic than the optimum wing of an airliner. The optimum shifts from an elliptic load distribution at very small scales to an almost triangular distribution at large scales.

  • $\begingroup$ Why in computing the lift from downwash did you assume that the wing influenced uniformly just a circle of air of diameter equal to the span ? I would think that the downwash effect of the wing would be variable from near to far away. $\endgroup$ Jan 21, 2016 at 21:58
  • $\begingroup$ @EarlQuandt: Yes, in reality the influence of the wing fades the more you move away from it. But this is a simplification which you will also find in propeller theory, and it gives the correct result, because what you neglect in the far field you compensate by assuming a constant influence over the near field. $\endgroup$ Jan 21, 2016 at 23:27
  • $\begingroup$ I am uneasy about calculating induced drag from a model of the wake that is not really there. I thought the wake field of the elliptic wing before roll-up was more like a flat sheet. $\endgroup$ Jan 24, 2016 at 16:16
  • $\begingroup$ @EarlQuandt: Yes, there is a flat vortex sheet in potential flow. But there is no such sheet in reality, only parcels of air. An untwisted elliptic wing will have all parcels of air flowing off the trailing edge moving downwards at the same speed, and also those above and below. The movement will extend up- and downwards more where wing chord is higher. In the center of the elliptic wing a bigger air mass is affected per unit of time than at the tips. This is nicely reflected in my simplification. Your uneasiness comes from the simplifications and misinterpretations of potential flow theory. $\endgroup$ Jan 25, 2016 at 0:23
  • 1
    $\begingroup$ Mr Kämpf is trying to refute real world wind tunnel tests with unsupported personal philosophy. What he refers to as "scaling" is Reynold number and the statement that " almost triangular distribution at large scales" is false; look at the Antonov An-225 (worlds largest), C5 Galaxy, or Spruce Goose. Any similarity to a triangle is simply structural limitations and fuel requirements of a high aspect ratio which is always the most efficient wing shape for subsonic $\endgroup$
    – jwzumwalt
    Dec 12, 2017 at 21:16

For a wing with elliptical span loading, the induced drag can be directly calculated from the lift coefficient. The induced drag coefficient $C_{D_{i}}$ can be calculated as,

$C_{D_{i}} = \frac{C_{L}^{2}}{\pi A}$

where $C_{L}$ is the lift coefficient and $A$ is the aspect ratio.

Elliptical loading produces the minimum induced drag according to the lifting line theory, when only the span and lift are considered. If other considerations come into play (like wing bending moment), the most efficient shape varies.

As for why the best distribution is elliptic, the equations can be readily derived from lifting line theory; basically this is because the downwash is constant along the span. A good way of reasoning why this is so is given in The Minimum Induced Drag of Airfoils by Max Munk, NACA Report No. 121.

If the distribution is the best one, the drag cannot be decreased or increased by transferring one lifting element from its old position to some new position. Now, the share of one element in the drag is composed of two parts. It takes a share in producing a downwash in neighborhood of other lifting elements and, in consequence, a change in their drag. It has itself a drag, being situated in the downwash produced by the other elements.

... In case of the lifting straight line, the two downwashes, each produced by one element in the neighborhood of the other, are equal. For this reason, the two drags of the two elements each produced by the other are equal too, and hence the two parts of the entire drag of the wings due to one element. ... hence, the entire drag due to one element is unchanged when the element is transferred from one situation to a new one of the same downwash, and the distribution is best only if the downwash is constant over the whole wing.

For this reason, when only the span and lift are considered, the elliptical loading gives the minimum induced drag as the downwash is constant over the wing. When the constraints are modified, other distributions and wing shapes become more efficient. For example, from On the Minimum Induced Drag of Wings by A. H. Bowers:

$\diamond$ Prandtl/Munk (1914)


Constrained only by span and lift

Downwash: $y = c$

$\diamond$ Prandtl/Horten/Jones (1932)

Bell shaped

Constrained by lift and bending moment

Downwash: $y = bx + c$

$\diamond$ Klein/Viswanathan (1975)

Modified bell shape

Constrained by lift, moment and shear (minimum structure)

Downwash: $y = ax + bx + c^{2}$

  • $\begingroup$ I don't understand the formula proposed by Klein/Viswanathan, I assume x is spanwise coordinate, so why is there ax, and bx? Also, what is the benefit of squaring a constant? I suspect there is a mistake, because If we reduce the formula we get the linear equation of Horten. $\endgroup$
    – ROIMaison
    Apr 29 at 7:40
  • $\begingroup$ I checked the source you indicated, and it's indeed an error in the conclusion slides. See slide 16 for the right formula: “Hence the required downwash-distribution is parabolic. y = ax^2 + bx + c" $\endgroup$
    – ROIMaison
    Apr 29 at 7:58

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