SME
Aerodynamics
Finite wings
Use this for “A wing of aspect ratio 8 and span efficiency 0.9 flies at CL = 0.6. Find CDi and αᵢ.”
Diagram
Drawn
rectangular
with
span
chord
=
AR
=
8.
Induced
drag:
0.6
2
π × 0.9 × 8
=
0.01592.
The vortices push the air behind the wing down; a longer wing pushes it less.
This is an example. Type your own values to start a new problem.
CDi
=
CL
2
π
×
×
αi,rad
Induced angle in radians
worked out (not typed) · rad
αᵢ
Induced angle
calculated · °
Formulas
CDi
=
CL
2
π ×
e
× AR
0.01592
=
0.6
2
π × 0.9 × 8
αi,rad
=
CL
π × AR
0.02387
=
0.6
π × 8
αᵢ
=
αi,rad
×
180
π
1.368
=
0.02387
×
180
π
Assumptions
•
Lifting-line theory: the trailing tip vortices push the air down, tilting the wind by the induced angle αᵢ.
•
The induced drag coefficient CDi is the drag that downwash adds, with no friction at all.
•
Span efficiency e = 1 only for an elliptic lift distribution; αᵢ = CL ÷ (πAR) is the elliptic value.
•
Long wings (large aspect ratio AR) cut induced drag.
Step-by-step
WE KNOW
CL = 0.6, AR = 8, e = 0.9
FIND
induced drag coefficient (CDi), induced angle in radians (αi,rad), induced angle (αᵢ)
Step 1 · Find induced drag coefficient (CDi)
Use CDi = CL² ÷ (π × e × AR)
Square
CL,
then
divide
by
π,
e
and
AR.
CDi
=
0.6
2
π × 0.9 × 8
CDi
=
0.6
2
7.2 × π
CDi
=
0.6
2
22.619
CDi
=
0.36
22.619
CDi = 0.01592
Step 2 · Find induced angle in radians (αi,rad)
Use αi,rad = CL ÷ (π × AR)
Elliptic wing: divide CL by π times AR for the angle in radians.
αi,rad
=
0.6
π × 8
αi,rad = 0.02387 rad
Step 3 · Find induced angle (αᵢ)
Use αᵢ = αi,rad × 180 ÷ π
Turn the radians into degrees.
αᵢ
=
0.02387
×
180
π
αᵢ
=
4.2966
π
αᵢ = 1.368°
Check
Put the numbers back in. Both sides should match.
0.01592
=
0.6
2
π × 0.9 × 8
✓
0.02387
=
0.6
π × 8
✓
1.368
=
0.02387
×
180
π
✓
Topic check
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