Lesson 2 · 35 min

Airfoil Lift, Drag and Moment

An airfoil is chosen for what it does, not how it looks. Most of what it does can be estimated before any wind tunnel: thin-airfoil theory gives the lift and pitching moment from the camber line alone. What theory cannot give, the drag and the stall, depends on the boundary layer, and so on the Reynolds number at which the airfoil works.

Learning objectives

Use thin-airfoil theory: the lift-curve slope \(2\pi\) per radian, the zero-lift angle, and the moment about the quarter chord.
Explain the aerodynamic center, and why a cambered airfoil pitches nose down.
Compute the Reynolds number and explain why airfoil data must match it.
Describe the airfoil drag polar, the laminar drag bucket, and the three kinds of stall.

The lift curve

Below the stall, an airfoil's lift coefficient grows in a straight line with the angle of attack \(\alpha\). Thin-airfoil theory, which treats the airfoil as its camber line in inviscid flow, gives the slope and the intercept:

Thin-airfoil lift curve

\[ c_l = a_0\,(\alpha - \alpha_{L0}), \qquad a_0 = 2\pi\ \text{per radian} = 0.110\ \text{per degree} \] \[ \alpha_{L0} = -\frac{1}{\pi}\int_0^{\pi}\frac{\dd y_c}{\dd x}\,(\cos\theta - 1)\,\dd\theta, \qquad x = \tfrac12(1 - \cos\theta) \]

The slope does not depend on the shape at all. Camber only moves the line: a positively cambered airfoil already lifts at zero angle, so its zero-lift angle \(\alpha_{L0}\) is negative. Real airfoils reach about 90 to 100% of \(2\pi\), because the boundary layer slightly reduces the effective camber.

The angle of attack is measured from the chord line, but a cambered airfoil does not feel the flow along its chord. It has its own zero-lift line, tilted nose-up from the chord by \(|\alpha_{L0}|\): when the flow runs along that line, the airfoil makes no lift. What the airfoil feels is the angle from its zero-lift line, \(\alpha - \alpha_{L0}\) (Figure 2.1). Lift is the force component perpendicular to the oncoming flow and drag the component along it, whatever the attitude of the airfoil.

Figure 2.1 A NACA 4412 at \(\alpha = 8^\circ\), to scale. The angle of attack is measured from the direction of the oncoming flow \(V_\infty\) to the chord line. The zero-lift line (dashed teal) passes through the trailing edge, \(|\alpha_{L0}| = 4.15^\circ\) above the chord line (dashed gray) at the nose, so this airfoil meets the flow at \(\alpha - \alpha_{L0} = 12.15^\circ\) from its zero-lift line. Lift \(L\) is perpendicular to \(V_\infty\) and drag \(D\) parallel to it, drawn with the nose-down moment \(M_{c/4}\) at the quarter chord. The drag is exaggerated: at this angle it is less than 1% of the lift.

For the NACA four-digit camber line the integral can be done once and for all. The results scale with the camber: with the maximum camber at 40% of the chord, each 1% of camber lowers \(\alpha_{L0}\) by about \(1.04^\circ\). Moving the camber aft lowers it further (Figure 2.2).

Figure 2.2 Thin-airfoil lift curves of a NACA four-digit camber line (blue) and a symmetric airfoil (gray). The zero-lift angle, the ideal lift coefficient (ring) and the lift at the chosen angle (dot) are marked. Camber shifts the line to the left without changing its slope; moving the camber aft shifts it farther and makes the moment more nose-down. The straight line ends at stall, which theory does not predict.

The aerodynamic center and the pitching moment

The lift on an airfoil acts at a point that moves with the angle of attack, the center of pressure, which is awkward for analysis. It is easier to describe the airfoil by a force and a moment at a fixed point: the aerodynamic center, the point about which the moment does not change with \(\alpha\). Thin-airfoil theory places it at the quarter chord, and real subsonic airfoils have it within a percent or two of there.

The two descriptions are the same load (Figure 2.3). Moving the lift from the center of pressure to the quarter chord adds the moment it made about that point, so \(c_{m,c/4} = -c_l\,(x_{cp}/c - 0.25)\), or

Center of pressure

\[ \frac{x_{cp}}{c} = 0.25 - \frac{c_{m,c/4}}{c_l} \]

With a nose-down \(c_{m,c/4}\), the center of pressure lies behind the quarter chord and runs aft, even off the airfoil, as the lift falls toward zero. For a symmetric airfoil it stays at the quarter chord.

Figure 2.3 Two ways to describe the same load on an airfoil, from thin-airfoil theory. Top: the lift alone, at the center of pressure, which moves with the angle of attack. Bottom: the lift at the quarter chord (the aerodynamic center) with a moment \(M_{c/4}\) that stays the same at every angle. Lower the angle of attack: on a cambered airfoil the center of pressure runs aft while the moment does not change. Arrow lengths are proportional to \(c_l\).

Moment about the quarter chord (thin-airfoil theory)

\[ c_{m,c/4} = \frac{\pi}{4}\left(A_2 - A_1\right), \qquad A_n = \frac{2}{\pi}\int_0^{\pi}\frac{\dd y_c}{\dd x}\cos n\theta\,\dd\theta \]

Zero for a symmetric airfoil; negative (nose down) for positive camber, and more negative the farther aft the camber. Every cambered wing therefore needs a tail, or a canard, to balance it: Lesson 8 and Module 11.

One more number comes out of the same integrals: the ideal lift coefficient, \(c_{l,i} = \pi A_1\), at which the flow meets the leading edge smoothly with no suction peak there. Near it the airfoil has its lowest drag and its gentlest pressure distribution, and it is the natural design point (Lesson 3).

Example 2.1 — The NACA 2412

Thin-airfoil theory gives the NACA 2412 a zero-lift angle of \(-2.08^\circ\), a quarter-chord moment of \(-0.053\) and an ideal lift coefficient of 0.256. Find its lift coefficient at \(\alpha = 4^\circ\), and its pitching moment per metre of span about the quarter chord at that angle, for a chord of \(1.5\ \text{m}\) at \(50\ \text{m/s}\) at sea level.

Show solution

\[ c_l = 2\pi\,\frac{4 - (-2.08)}{57.30} = 2\pi(0.1061) = 0.667 \] \[ q = \tfrac12(1.225)(50^2) = 1531\ \text{Pa} \] \[ M' = q\,c^2\,c_{m,c/4} = 1531(1.5^2)(-0.053) = -183\ \text{N m per metre} \]

The moment is the same at every angle below the stall: that is what makes the quarter chord the aerodynamic center. Measured lift curves of the NACA 2412 give a zero-lift angle close to \(-2^\circ\) and a slightly lower slope than \(2\pi\), as expected.

Reynolds number

Lift below the stall depends little on viscosity. Drag and the stall depend on it entirely, through the boundary layer, and its behavior is set by the Reynolds number, the ratio of inertial to viscous forces based on the chord:

Reynolds number

\[ Re = \frac{\rho V c}{\mu} \]

Sea level: \(\rho = 1.225\ \text{kg/m}^3\), \(\mu = 1.789 \times 10^{-5}\ \text{Pa s}\). Above 11 km: \(T = 216.65\ \text{K}\), \(\mu = 1.422 \times 10^{-5}\ \text{Pa s}\). Viscosity depends on temperature only.

The boundary layer starts laminar at the nose: thin and smooth, with little friction, but easily pushed off the surface by a rising pressure. Somewhere along the chord it turns turbulent: thicker and with more friction, but its fuller velocity profile carries more momentum close to the wall, so it follows the surface much farther against a rising pressure. Where even the turbulent layer runs out of momentum, it separates: the flow near the wall reverses, and the outer flow leaves the surface (Figure 2.4).

Figure 2.4 The boundary layer on the upper surface of an airfoil at a high angle of attack, schematically, with its thickness greatly exaggerated: laminar from the nose (violet), turbulent after transition (blue), and separated near the trailing edge (orange), with a velocity profile from each region below. The laminar profile is thin; the turbulent one is fuller near the wall, which makes more friction but resists separation; in the separated region the flow near the wall runs backward.

Wind-tunnel models fly at a few hundred thousand, light aircraft at 3 to 10 million, airliners at 20 to 50 million. A higher Reynolds number gives a boundary layer that is thinner relative to the chord and turns turbulent sooner: a lower friction coefficient once the flow is mostly turbulent and, up to a point, a higher maximum lift. Airfoil data measured at the wrong Reynolds number can mislead by more than the difference between two airfoils, so always compare airfoils at the Reynolds number of the design.

Example 2.2 — The business jet's Reynolds numbers

The business jet's mean chord is about \(2.0\ \text{m}\) (Lesson 6). Find the Reynolds number in cruise at Mach 0.80 and \(12\ \text{km}\), and on approach at \(220\ \text{km/h}\) at sea level.

Show solution

At 12 km, \(\rho = 0.3108\ \text{kg/m}^3\), \(a = 295.1\ \text{m/s}\), so \(V = 0.80(295.1) = 236.1\ \text{m/s}\):

\[ Re = \frac{0.3108(236.1)(2.0)}{1.422 \times 10^{-5}} = 10.3 \times 10^6 \] \[ Re = \frac{1.225(61.11)(2.0)}{1.789 \times 10^{-5}} = 8.37 \times 10^6 \quad \text{(approach)} \]

The cruise speed is nearly four times the approach speed, but the air at 12 km is a quarter as dense: the two Reynolds numbers are close. Airfoil data near 10 million suit both.

Drag and the drag polar of an airfoil

Airfoil drag is skin friction plus pressure (form) drag from the thickening boundary layer. Plotted against \(c_l\), it is nearly flat over a range of lift coefficients and rises at both ends as the flow begins to separate. For a smooth airfoil at a Reynolds number near 10 million the minimum is about 0.005 to 0.006: about a third of the business jet's whole-aircraft \(\CDz\) of 0.0165 (Module 3), which also counts the fuselage, tails and nacelles.

A laminar-flow (six-series) airfoil adds a drag bucket: within its design range of \(c_l\) the boundary layer stays laminar farther aft, and the drag drops by a third or more. The bucket is narrow, needs a smooth surface, and shrinks as the Reynolds number rises. A designer who picks a six-series airfoil should check that the cruise lift coefficient sits in the bucket, and that the bucket survives the real surface (Figure 2.5).

Figure 2.5 Drag polars of a four-digit airfoil (blue) and a six-series laminar-flow airfoil with a design lift coefficient of 0.2 (violet), schematically, with values typical of airfoils 12 to 15% thick near \(Re = 9 \times 10^6\). The six-series airfoil has a drag bucket from \(c_l \approx 0\) to 0.4 and slightly more drag outside it. Switch to a rough surface: the laminar flow is lost, the bucket disappears, and both airfoils have about the same, higher drag.

Stall

At high angle of attack the boundary layer cannot follow the upper surface to the trailing edge, and the lift stops rising. How it stops depends mostly on the thickness and the leading-edge radius:

Trailing-edge stall (thick airfoils, about 14% and more): separation starts at the trailing edge and creeps forward as \(\alpha\) increases. The lift curve rounds over gently. Pilots get warning, and recovery is easy.
Leading-edge stall (moderate thickness, about 9 to 13%): a small separation bubble near the nose suddenly bursts and the flow leaves the whole upper surface at once. The lift drops abruptly from a high maximum.
Thin-airfoil stall (thin airfoils with a sharp nose, below about 9%): a long bubble forms at the leading edge and grows aft with \(\alpha\). The stall is gentle but the maximum lift is low.

Figure 2.6 The three kinds of stall, schematically: the flow over a thick, a moderate and a thin airfoil (separated flow orange), and their lift curves. Raise the angle of attack. On the thick airfoil the separation creeps forward from the trailing edge; on the moderate one a small bubble near the nose bursts all at once; on the thin one a long bubble grows aft until it reaches the trailing edge. The highest maximum lift comes from airfoils of about 12 to 15%, where trailing-edge and leading-edge stall meet; thinner airfoils lose maximum lift, thicker ones gain drag. Flaps, slats and the planform (Lesson 6) change the picture for the whole wing.

For a first choice this is enough: a gentle trailing-edge stall suits a trainer or a light aircraft, a high maximum lift suits a short-field design, and a thin airfoil suits a fast one that will rely on flaps and slats for its low-speed lift.

Check your understanding

Key takeaways

\(c_l = 2\pi(\alpha - \alpha_{L0})\) per radian; camber makes \(\alpha_{L0}\) negative, about \(-1^\circ\) per 1% of camber at 40% of the chord.
The aerodynamic center is at the quarter chord; cambered airfoils have a nose-down \(c_{m,c/4}\) that the tail must balance.
\(Re = \rho Vc/\mu\): compare airfoils only at the design's Reynolds number.
Drag is nearly flat in \(c_l\) near the design point; laminar airfoils add a narrow bucket.
Thick airfoils stall from the trailing edge (gently), moderate ones from the leading edge (abruptly), thin ones with a long bubble (gently, at low \(c_{l,\max}\)).
Next, Lesson 3 chooses the business jet's design lift coefficient and thickness.