Angle of attack, lift coefficient, drag coefficient: Question bank

0. Symbols, axes, and pictures you need first

Before the traps, three quick self-contained anchors so no symbol below is a mystery.

Definition The lift-curve symbols

Relative wind $V_{\infty}$ : the velocity of the air far upstream, i.e. the direction the air appears to come from in the wing's own frame. In level flight it is horizontal and points backward along the flight path. This is the reference direction for everything — see the figure below.
Chord line: the straight line from the leading edge to the trailing edge of the airfoil.
Angle of attack $α$ : the angle from the relative wind up to the chord line. Positive $α$ = leading edge tilted up into the wind.
Lift $L$ : force component perpendicular to the relative wind. Positive lift points to the airfoil's upper (suction) side. Drag $D$ : component parallel to the relative wind, pointing downstream.
Zero-lift angle $α_{L = 0}$ : the specific $α$ at which $C_{L} = 0$ . For a symmetric section $α_{L = 0} = 0$ ; for a normally-cambered wing $α_{L = 0} < 0$ (it lifts even at $α = 0$ ).
Lift-curve slope $a_{0} = \frac{d C _{L}}{d α}$ : how fast $C_{L}$ climbs per unit angle. It carries units — either per radian ( $2 π \approx 6.28/ rad$ from thin-airfoil theory) or per degree ( $\approx 0.11/ deg$ ). Always state which; they differ by the factor $π /180 \approx 1/57.3$ .

Figure — Angle of attack, lift coefficient, drag coefficient

True or false — justify

A flat plate at zero angle of attack produces zero lift.

True for a symmetric shape. A symmetric section (or flat plate) has

α_{L = 0} = 0

, so

C_{L} = a_{0} (α - 0) = 0

α = 0

. A cambered airfoil has

α_{L = 0} < 0

, so it makes lift even at

α = 0

Doubling the airspeed doubles the lift at fixed angle of attack.

False. Lift scales with

q = \frac{1}{2} ρ V^{2}

, so doubling

V

multiplies lift by

2^{2} = 4

, not

2

C_{L}

itself does not change with speed (until Mach effects appear).

The lift coefficient

C_{L}

has units of newtons per square metre.

False.

C_{L} = L / (\frac{1}{2} ρ V^{2} S)

is force ÷ (pressure × area) = force ÷ force = dimensionless. That dimensionlessness is exactly why a wind-tunnel model predicts the full-scale wing.

Lift always acts vertically upward relative to the ground.

False. By the sign convention above, lift is perpendicular to the relative wind, not to gravity. In a climb, dive, or bank the relative wind is tilted, so lift tilts with it — that is how banked turns work.

At the stall angle, lift is at its maximum.

True.

C_{L}

rises with

α

up to

α_{stall}

where it peaks (

C_{L, m a x}

); past that angle it drops. So the stall angle marks maximum lift, not zero lift — see Boundary Layer Separation & Stall.

Induced-drag force can be made small by simply flying faster.

True as a limit, but read it carefully. To hold a fixed weight

W

C_{L}

must satisfy

W = C_{L} q S

, so

C_{L} \propto 1/ V^{2}

; then the induced-drag force

D_{i} = C_{D, i} q S \propto C_{L}^{2} \cdot q \propto (1/ V^{2})^{2} \cdot V^{2} = 1/ V^{2} \to 0

V \to \infty

. What can never be zero at nonzero lift is the induced-drag coefficient

C_{D, i} = C_{L}^{2} / (π e AR)

— don't conflate the coefficient with the force (see Induced Drag & Wingtip Vortices).

Increasing aspect ratio reduces induced drag at a fixed lift coefficient.

True. Induced drag

= C_{L}^{2} / (π e AR)

, so a larger

AR

(longer, slimmer wing) shrinks it — hence gliders have very long wings.

Profile drag

C_{D, 0}

is exactly constant for all angles of attack.

False (approximately true only at low $α$ ).

C_{D, 0}

is roughly constant in the linear region, but near stall the separated flow enlarges the wake and both form drag and total

C_{D}

climb steeply.

The angle of attack equals the pitch angle of the aircraft.

False. Pitch is measured against the horizon; angle of attack is measured against the relative wind. A plane descending nose-level still has a positive

α

because the wind comes from slightly below.

Spot the error

"Since force is

ρ V^{2} S

, we can write

L = C_{L} ρ V^{2} S

The missing

\frac{1}{2}

is the error. Lift uses the dynamic pressure

q = \frac{1}{2} ρ V^{2}

, so

L = C_{L} \frac{1}{2} ρ V^{2} S

C_{L}

is calibrated with the

\frac{1}{2}

; dropping it makes every coefficient wrong by a factor of two.

"A wing with

α_{L = 0} = - 2^{\circ}

makes no lift at

α = 0^{\circ}

Wrong. Using a per-degree slope

a_{0} \approx 0.11/ deg

so angles stay in degrees,

C_{L} = a_{0} (α - α_{L = 0}) = 0.11 \times (0 - (- 2)) = 0.11 \times 2 \approx 0.22 > 0

. (If you insist on the per-radian

a_{0} = 2 π

, first convert

2^{\circ} = 0.0349

rad, giving

2 π \times 0.0349 \approx 0.22

— same answer.) Zero lift happens at

- 2^{\circ}

, so at

α = 0

this cambered wing is already lifting.

"Thin-airfoil theory gives a lift-curve slope of

2 π

per degree."

The

2 π

is per radian, not per degree. Converting:

2 π / rad \times (π /180) rad/deg \approx 0.11

per degree. Mixing the units inflates the slope by a factor of about 57.

"Faster planes have a higher lift coefficient."

C_{L}

depends on

α

(and secondarily Mach/Reynolds), not directly on

V

. Faster flight raises

q

, so lift rises even though

C_{L}

is unchanged at fixed

α

"The Kutta–Joukowski result

L^{'} = ρ V Γ

works because pressure pushes up on the wing."

The subtle slip:

L^{'} = ρ V Γ

is a statement about circulation

Γ

set by the Kutta-Joukowski Theorem and the Kutta condition, not a direct pressure balance. Pressure is the mechanism; circulation is the bookkeeping that predicts the total force.

"Because

L / D = C_{L} / C_{D}

, flying at the highest possible

C_{L}

gives the best glide."

Wrong — high

C_{L}

also blows up induced drag (

\propto C_{L}^{2}

), so

C_{D}

grows faster than

C_{L}

eventually. The best

L / D

sits at an intermediate

C_{L}

where profile and induced drag are balanced, not at

C_{L, m a x}

Why questions

Why do we bother defining a dimensionless coefficient instead of just tabulating force versus speed?

Because force changes with every speed, density, and size, but

C_{L} (α)

is the shape's DNA — measure it once on a small model and predict lift at any

V

ρ

, and

S

via

L = C_{L} q S

Why does lift die suddenly at stall rather than fading gradually?

The boundary layer stays attached and orderly until the adverse pressure gradient becomes too steep; then it separates over a large region almost at once, collapsing the low-pressure suction on the upper surface — a threshold effect, per Boundary Layer Separation & Stall.

Why is the

\frac{1}{2}

q = \frac{1}{2} ρ V^{2}

the same one that appears in Bernoulli's Equation?

Because

q

is the kinetic-energy density of the flow, and Bernoulli's pressure balance

p + \frac{1}{2} ρ V^{2} = const

trades exactly that KE term against static pressure — the aerodynamic coefficients inherit it directly.

Why does making lift force a wing to make induced drag?

Generating lift sheds wingtip vortices that tilt the local flow downward (downwash) by an induced angle

α_{i} \propto C_{L}

; this rotates the lift vector backward, and its backward component

D_{i} \approx L α_{i} \propto C_{L}^{2}

is induced drag — you cannot get the vertical force without the tilt (see Induced Drag & Wingtip Vortices).

Why does thin-airfoil theory give a universal slope

2 π

(per radian) independent of the shape?

The slope comes from the linearised circulation of a flat vortex sheet plus the Kutta condition, which are geometry-independent to first order. Camber and thickness shift

α_{L = 0}

and change

C_{L, m a x}

but leave the leading-order slope near

2 π

per radian.

Why do the coefficients still depend on Reynolds and Mach number if they are "dimensionless"?

Dimensionless means free of units, not free of all physics. Reynolds Number governs boundary-layer behaviour and Mach Number governs compressibility — both reshape the pressure field, so

C_{L}

and

C_{D}

shift with them even at fixed

α

Why can a symmetric airfoil never lift at zero angle of attack, while a cambered one can?

Symmetry forces the upper and lower flow to be mirror images at

α = 0

, so pressures cancel and

C_{L} = 0

. Camber breaks that mirror, giving net upward suction even when the chord is aligned with the wind (

α_{L = 0} < 0

Edge cases

What is

C_{L}

exactly at the zero-lift angle

α = α_{L = 0}

Exactly zero:

C_{L} = a_{0} (α_{L = 0} - α_{L = 0}) = 0

, independent of whether

a_{0}

is in per-radian or per-degree. This is the reference point of the whole lift curve — the effective angle is measured from here, not from the chord.

What happens to lift and drag at

V = 0

(aircraft at rest)?

Both vanish because

q = \frac{1}{2} ρ (0)^{2} = 0

, so

L = C_{L} q S = 0

and likewise

D = 0

. No relative wind means no aerodynamic force regardless of angle of attack — which is why takeoff needs a ground run.

At a negative angle of attack (below

α_{L = 0}

), what direction does lift point?

Lift becomes negative by the sign convention (points toward the "belly"/lower surface), because

C_{L} = a_{0} (α - α_{L = 0}) < 0

once

α < α_{L = 0}

. Inverted flight and downforce wings exploit exactly this sign.

If a wing had infinite aspect ratio, what would happen to induced drag?

It would go to zero:

C_{L}^{2} / (π e AR) \to 0

AR \to \infty

. This is the idealised two-dimensional airfoil with no tips, hence no tip vortices — the limiting case thin-airfoil theory assumes.

In the deep-stall region well past the stall angle, is

C_{L}

zero?

No —

C_{L}

drops but stays positive (a flat plate at high angle still deflects some air); it just falls far below

C_{L, m a x}

while

C_{D}

rises sharply, wrecking the

L / D

ratio.

As Mach number approaches 1, what does the constant-slope thin-airfoil result require?

A compressibility correction. The Prandtl–Glauert factor multiplies the slope, so

a_{0}

is no longer a fixed

2 π

per radian — the low-speed formula only holds while the flow is effectively incompressible, per Compressible Flow.

At exactly the stall angle, is the flow fully separated?

Not fully — separation is beginning to dominate at the peak of

C_{L}

. Full separation happens past the stall angle; right at the peak, attached and separated flow are in a losing balance that tips over as

α

increases further.

Connections

Angle of attack, lift coefficient, drag coefficient — the parent this bank tests.
Bernoulli's Equation — the $\frac{1}{2}$ shared with dynamic pressure.
Kutta-Joukowski Theorem — behind the $2 π α$ slope.
Boundary Layer Separation & Stall — the stall traps.
Induced Drag & Wingtip Vortices — the $C_{L}^{2}$ traps.
Reynolds Number & Mach Number — why dimensionless still means shape-dependent.
Compressible Flow — the high-Mach edge case.