Things Everybody Knows...that aren't so

Imagine that we have two prospective airplane designs, with the same weight W, wingspan b and airspeed v. One has a wing that has twice the chord, hence twice the area, of the other.

The equation for the induced drag coefficient is typically presented in textbooks this way:

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c_di = c_l²/(pi)AR e, where

c_l is the lift coefficient of the wing, pi is the Greek letter symbolizing the ratio of a circle’s circumference to its diameter, AR is the aspect ratio and e is the span efficiency factor.

The lift coefficient c_l is the lift force exerted by the wing, which must equal the airplane’s weight W, divided by the dynamic pressure and the wing area A,

c_l = W/0.5 (rho) v² A, where rho is the mass density of the air and v is the airspeed.

Taking the area of the smaller wing to be A, that of the larger wing is 2A. Therefore the lift coefficient of the large wing will be half that of the smaller wing.

Aspect ratio AR is the ratio of the square of the wingspan to its area

AR = b²/A

so the aspect ratio of the larger wing is half that of the smaller wing.

[We’ll disregard the span efficiency factor e and simply assume that its value is always 1.]

The induced drag coefficients of the two wings are, for the smaller wing, cl2/(pi) AR , and for the larger wing (0.5 c_l)²/(pi) 0.5 AR or or one-half the induced drag coefficient of the smaller wing with the higher aspect ratio. If we accept this as the final word, reducing aspect ratio should give better induced drag.

Now if I and many others had bothered to carry my calculations this far, it would have raised suspicion. Unfortunately, I hadn’t – a friend had to enlighten me.

You see, when we convert the coefficient to an actual force, here is what happens:

D_i = c_di 0.5 (rho) v² A

Plugging in values to this formula makes it clear that doubling wing area cuts the induced drag coefficient in half, while the area of the wing doubles; one-half times two equals 1, so at equal span, changing aspect ratio neither increases nor decreases induced drag.

Using the formula for the actual drag force directly would have saved us time and confusion.

Di = W²/0.5 (pi)(rho)v² b²

Here it’s obvious that if span is constant, the induced drag can’t change. Wing area and aspect ratio don’t appear in the equation.

Note that there is nothing wrong with the equation for the coefficient - it is physically and mathematically valid - but many students (and not a few teachers) have been misled in interpreting it. They see AR in the denominator, and that’s it. If they had calculated it through for our two cases, they would have noticed the paradox of lower AR giving a lower induced drag coefficient and would have become curious. I’m sorry to admit that I didn’t do that calculation until a friend asserted the irrelevance of aspect ratio.

It is span, more precisely span loading W/b, that governs induced drag: a lower span loading results in lower induced drag.

It would be better if, for textbooks, the authors should set aside their preference for dimensionless coefficients and concentrate on dimensional numbers.

My thanks to Hernan A. Posnansky, who set me straight about this.

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