Wing Sizing: Part 1

How much wing do I need? I’ve read a lot of textbooks and other sources and the thing I’ve learned is that you can make this as simple or complex as you like. Nearest I can tell, all of the methods out there are approximations of the reality that is actually taking place. Naturally some are better than others (and some are just junk).

Tip losses

Regardless of the planform shape and twist (washout) and airfoils selected (maybe even multiple ones at different places on the same wing), the resulting lift distribution tends to be elliptical-ish, like we see on the F-18 above.

C_L vs. c_l : Looking at a Hershey bar

Looking at Peery’s figure 9.10 above, the lift (per unit span) is greatest at the center. He expressed it in terms of circulation () but expressing it in lb/ft of span is probably a little easier to grasp. If we nondimensionalize the loading by dividing by the wing chord at that point and the dynamic pressure, we get the section lift coefficient, . (Notice it is a little “c,” denoting airfoil section lift coefficient.) The section lift coefficient for various airfoils can be looked up in books like Theory of Wing Sections or Riblett’s GA Airfoils.

If we’re thinking of a rectangular planform (Hershey bar) with an aspect ratio of 8 (8 times as much span as chord), the lift distribution for half of the wing looks like this, according to the method of calculation found in Theory of Wing Sections or NACA Technical Report 572. Because it has a consistent chord across its span, a graph of the section lift coefficient has the same shape.

Even though the maximum of this particular airfoil is 1.55, the average is about 88% of that, or 1.37. Therefore we say that the span efficiency (not to be confused with Oswald’s span efficiency) is 0.88 and wing lift coefficient (big “C”) is 1.37. Note that this is a fairly high aspect ratio; for lower aspect ratios, the span efficiency and would be less. John Roncz, for example, assumes all lift distribution is elliptical, and therefore span efficiency is . This is probably not a bad assumption for determining because it leaves a little room for losses due to fuselage interference, imperfect surfaces, gaps for the ailerons and flaps, etc.

Elliptical planforms and span efficiency

Instead of a Hershey bar, let’s look at an elliptical planform and its lift and lift coefficient distribution. Now we have ideal elliptical lift distribution. But what is going on with the distribution?

Remember we said earlier that the lift coefficient at any point along the span is the lift distribution divided by (which is constant along the span) and the chord at that point along the span. Since we are dividing elliptical lift distribution by elliptical chord distribution, the graph is a flat line.

Consequently, in theory anyway, the span efficiency is 100%. But if we decide to add twist–aerodynamic or geometric–to improve stall characteristics (altering lift distribution from its ideal elliptical shape), the distribution will be affected. Additionally, we still have the losses we mentioned earlier, so we’ll assume span efficiency takes a 10% hit. This is what’s found in Raymer’s eq. (12.21) in Aircraft Design: A Conceptual Approach and also Gudmundsson’s eq. (9-72) in General Aviation Aircraft Design. Is that the right number? Short of CFD (VLM or panel method), which I don’t have access to, it seems reasonable to me. We’ll design the wing under this assumption and then we can compare the result to some aircraft with known performance to check if we are in the ballpark.

First take on wing size

Earlier we looked at the lift equation:

$$\scriptsize L=qSC_L $$

In sizing the wing, we’re concerned with the wing’s ability to generate enough lift at stall speed, when is smallest.

$$ \scriptsize q = \tiny \frac{1}{2} \scriptsize \rho V_{stall}^{2} $$

$$ \scriptsize q = \tiny \frac{1}{2} \scriptsize \left( 0.002377 \text{ slugs/ft}^{3} \right) \> \left(58 \text{ mi/hr} \cdot \frac{5,280\text{ ft}}{\text{mi}} \cdot \frac{\text{hr}}{3,600 \text{ sec}} \right)^{2} = 8.60 \text{ lb/ft}^{2}$$

It’s interesting that this is only about one tenth of the dynamic pressure at a cruise speed of 200 mph! Our wing has to deal with a lot of different conditions.

Earlier we determined that wing loading would be 17.2 lb/ft² (assuming a wing lift coefficient of 2.0) and that gross weight is 1,924 lb, which we’ll increase by 8% to account for some down force on the tail.

$$\scriptsize L=1.08 \cdot 1,924 \text{ lb} = 2,080 \text{ lb} $$

Actually, it’s 2,077.92, but three significant figures will do just fine. So now if we rearrange the lift equation to solve for S, it appears we have everything we need to find the wing area.

$$\scriptsize S = \frac{L}{qC_L} $$

$$ \scriptsize S = \frac{2,080 \text{ lb}}{8.60 \text{ lb/ft}^{2} \cdot 2.0} = 121 \text{ft}^{2}$$

But we’ve made a big assumption here, which is that we’ll actually get a wing lift coefficient of 2.0. Next, we’ll see if we can get to that figure with flaps.