SME
Aerodynamics
Airfoil theory
Use this for “A cambered airfoil with zero-lift angle −2° flies at 4°. Find cₗ.”
Diagram
cₗ = 2π(α − α₀) with α₀ the zero-lift angle: 2π × 0.1047 rad = 0.658.
Drawn with 1.926% camber at 40% of the chord, which thin-airfoil theory gives α₀ = −2°.
This is an example. Type your own values to start a new problem.
Δα (rad)
=
(
α (°)
−
(
αL0 (°)
)
)
×
π
180
cₗ
=
2π
×
Δα (rad)
Formulas
Δα
=
(
α
−
αL0)
×
π
180
0.1047
=
(4
−
(−2))
×
π
180
cₗ = 2π × Δα
0.658 = 2π × 0.1047
Assumptions
•
Thin airfoil, small angles, no stall: the theory ignores separation past about 12–15°.
•
A symmetric airfoil has αL0 = 0; camber makes αL0 negative.
•
The lift-curve slope a₀ is 2π per radian (0.1097 per degree).
Step-by-step
WE KNOW
α = 4°, αL0 = −2°
FIND
angle from zero lift (Δα), lift coefficient (cₗ)
Step 1 · Find angle from zero lift (Δα)
Use Δα = (α − αL0) × π ÷ 180
Measure
α
from
the
zero-lift
angle,
then
turn
the
degrees
into
radians.
Δα
=
(4
−
(−2))
×
π
180
Δα
=
6
×
π
180
Δα
=
18.85
180
Δα = 0.1047 rad
Step 2 · Find lift coefficient (cₗ)
Use cₗ = 2π × Δα
Thin-airfoil theory: cₗ grows by 2π for each radian past zero lift.
cₗ = 2π × 0.1047
cₗ = 0.658
Check
Put the numbers back in. Both sides should match.
0.1047
=
(4
−
(−2))
×
π
180
✓
0.658 = 2π × 0.1047 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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