Lesson 3 · 40 min

Choosing the Airfoil

An airfoil is chosen for a design point. The wing will spend most of its life in cruise, at one lift coefficient, one Mach number and one Reynolds number, and the airfoil should be at its best there. Two numbers come out of that design point: the lift coefficient the airfoil should be designed for, and the thickness the cruise Mach number allows.

Learning objectives

The design lift coefficient

In level cruise the lift equals the weight, so the wing's lift coefficient follows from the wing loading and the dynamic pressure. At high speed the dynamic pressure is most easily written with the Mach number and the static pressure:

Cruise lift coefficient

\[ C_L = \frac{W/S}{q}, \qquad q = \tfrac12\rho V^2 = \tfrac12\gamma p M^2 \]

\(W/S\) in N/m² (weight, not mass), \(\gamma = 1.4\). The weight falls as fuel burns, so \(C_L\) falls through the cruise unless the aircraft climbs.

Which \(C_L\) should cruise be at? Module 1 found the best \(L/D\) where the induced drag equals the zero-lift drag, and Module 3 (Lesson 4) showed that a jet's best range is a little faster, at a lower lift coefficient:

Lift coefficients for best range

\[ C_L^{*} = \sqrt{\frac{\CDz}{K}} \ \ \text{at}\ \LDmax, \qquad C_{L,\text{range}} = \frac{C_L^{*}}{\sqrt3}\ \ \text{(jets)} \]

\(K = 1/(\pi A e)\). A propeller aircraft's best range is at \(C_L^{*}\) itself.

The airfoil should then be designed to work best near that lift coefficient: choose its design (ideal) lift coefficient close to the wing's cruise \(C_L\). The sections do not all carry the same lift (Lesson 6), and on a swept wing they work normal to the sweep at a higher local lift coefficient, so the section design \(c_l\) is often set a little above the wing's \(C_L\).

Example 3.1 — The business jet's cruise lift coefficient

From Module 3: \(\Wo = 6943\ \text{kg}\), \(S = 28\ \text{m}^2\), \(A = 8\), \(e = 0.75\), \(\CDz = 0.0165\), and cruise starts at \(0.970(0.985) = 0.9555\,\Wo\) after takeoff and climb. Find \(C_L\) at the start of cruise at Mach 0.80 and \(12\ \text{km}\) (\(p = 19\,330\ \text{Pa}\)), compare it with the best-range value, and find the altitude at which they would match.

Show solution
\[ \frac{W}{S} = \frac{0.9555(6943)(9.81)}{28} = 2324\ \text{N/m}^2 \] \[ q = \tfrac12(1.4)(19\,330)(0.80^2) = 8660\ \text{Pa} \] \[ C_L = \frac{2324}{8660} = 0.268 \] \[ K = \frac{1}{\pi(8)(0.75)} = 0.05305 \] \[ C_L^{*} = \sqrt{\frac{0.0165}{0.05305}} = 0.558 \] \[ C_{L,\text{range}} = \frac{0.558}{\sqrt3} = 0.322 \]

At 12 km the jet cruises below its best-range lift coefficient: the air is too dense for it. Matching \(C_L = 0.322\) needs \(q = 2324/0.322 = 7218\ \text{Pa}\), so \(p = 7218/(0.7 \times 0.64) = 16\,110\ \text{Pa}\). Above 11 km, \(p = 22\,632\,e^{-(h - 11\,000)/6341.6}\) with \(h\) in m:

\[ h = 11\,000 + 6341.6\ln\frac{22\,632}{16\,110} = 13\,160\ \text{m} \]

About 13 km: business jets do cruise there, above the airliners. Its airfoil should be designed for a lift coefficient of about 0.3. As fuel burns, the best-range altitude rises; a cruise-climb (or step climbs) keeps \(C_L\) near the design value.

Figure 3.1 The business jet's cruise lift coefficient against altitude (blue), and its best-range value (green). Move the weight from the start to the end of cruise, and the Mach number. The design lift coefficient of the airfoil should sit near the green line at the altitudes the aircraft will actually cruise.

What thickness buys, and what it costs

Thickness ratio is the airfoil decision with the largest effect on the rest of the design:

Wings are usually thickest at the root, where the bending moment and the fuel volume are largest, and thinner at the tip: 14% at the root and 10% at the tip is a common choice for a subsonic transport.

Thickness and the Mach number

The critical Mach number is the flight Mach number at which the local flow somewhere on the airfoil first reaches Mach 1. A little above it a shock forms and the drag starts to rise; the drag-divergence Mach number \(\MDD\) is where the rise becomes steep (by a common definition, where \(\dd\CD/\dd M = 0.1\)). A subsonic jet cruises just below \(\MDD\).

A simple estimate of \(\MDD\) is Korn's equation, extended to swept wings by simple sweep theory: the airfoil feels only the flow normal to the sweep (Lesson 5).

Korn equation for a swept wing

\[ \MDD = \frac{\kappa_A}{\cos\Lambda} - \frac{t/c}{\cos^2\Lambda} - \frac{c_l}{10\cos^3\Lambda} \]

\(\kappa_A = 0.87\) for conventional (six-series) airfoils, 0.95 for supercritical ones; \(\Lambda\) is the sweep (use the mid-chord or quarter-chord line), \(t/c\) and \(c_l\) are measured along the flight direction. Thicker airfoils and higher lift lower \(\MDD\); sweep and supercritical design raise it.

Solved for the thickness, it gives the thickest airfoil a cruise Mach number allows:

Thickness allowed by a target \(\MDD\)

\[ \frac{t}{c} = \kappa_A\cos\Lambda - \frac{c_l}{10\cos\Lambda} - \MDD\cos^2\Lambda \]

Example 3.2 — How thick can the business jet's wing be?

The business jet cruises at Mach 0.80 with \(c_l = 0.30\). Its wing is swept \(25^\circ\). To leave a margin, require \(\MDD = 0.82\). Find the thickest conventional and supercritical airfoils, and \(\MDD\) for the 12% airfoil of Module 3's sketch.

Show solution

\(\cos 25^\circ = 0.9063\), \(\cos^2 25^\circ = 0.8214\):

\[ \frac{t}{c} = 0.87(0.9063) - \frac{0.30}{10(0.9063)} - 0.82(0.8214) = 0.0818 \ \ \text{(conventional)} \] \[ \frac{t}{c} = 0.95(0.9063) - 0.0331 - 0.6735 = 0.154 \ \ \text{(supercritical)} \] \[ \MDD = \frac{0.95}{0.9063} - \frac{0.12}{0.8214} - \frac{0.30}{10(0.7444)} = 1.048 - 0.146 - 0.040 = 0.862 \]

With a conventional airfoil the wing could be only 8% thick; a 12% conventional airfoil would reach drag divergence at Mach 0.774, below the cruise speed. A supercritical airfoil allows almost twice the thickness, and the 12% wing of the sketch has a comfortable margin. That difference, a lighter wing holding more fuel, is why every modern jet uses supercritical sections.

Figure 3.2 Drag-divergence Mach number against thickness ratio from the Korn equation, for conventional (gray) and supercritical (blue) airfoils. The dashed line is the required \(\MDD\); where each line crosses it is the thickest airfoil allowed. Increase the sweep to see how it buys thickness, and the lift coefficient to see what it costs.

Check your understanding

Key takeaways