SME
Aerodynamics · Problem type
Drag polar and maximum L/D
Use this for “A wing has CD0 = 0.025, AR = 8 and e = 0.8. Find CD at CL = 0.5 and the best L/D.”
Chart
CD(0.5) ≈ 0.03743
CD(CL) = tangent(CL)
The curves cross at CL ≈ 0.709.
This is an example. Type your own values to start a new problem.
K
=
1
π
×
×
CD
=
CD0
+
K
×
CL
2
(L/D)ₘₐₓ
Maximum lift-to-drag ratio
calculated
CLb
CL at best L/D
calculated
Formulas
K
=
1
π ×
e
× AR
0.04974
=
1
π × 0.8 × 8
CD
=
CD0
+
K
×
CL
2
0.03743
=
0.025
+
0.04974
×
0.5
2
(L/D)ₘₐₓ
=
1
2 ×
√
CD0 ×
K
14.18
=
1
2 ×
√
0.025 × 0.04974
CLb
=
√(
CD0
K
)
0.709
=
√(
0.025
0.04974
)
Assumptions
•
Parabolic drag polar: zero-lift drag CD0 plus induced drag that grows with CL squared.
•
Oswald efficiency e (about 0.7 to 0.9) measures how close the lift spread is to elliptic.
•
Best L/D sits where a line from the origin just touches the polar; there induced drag equals CD0.
Step-by-step
WE KNOW
CD0 = 0.025, AR = 8, e = 0.8, CL = 0.5
FIND
induced drag factor (K), drag coefficient (CD), maximum lift-to-drag ratio ((L/D)ₘₐₓ), CL at best L/D (CLb)
Step 1 · Find induced drag factor (K)
Use K = 1 ÷ (π × e × AR)
One
over
π
times
e
times
the
aspect
ratio.
K
=
1
π × 0.8 × 8
K
=
1
6.4 × π
K
=
1
20.11
K
=
0.04974
Step 2 · Find drag coefficient (CD)
Use CD = CD0 + K × CL²
Zero-lift drag plus the induced part KCL².
CD
=
0.025
+
0.04974
×
0.5
2
CD = 0.025 + 0.04974 × 0.25
CD = 0.025 + 0.01243
CD = 0.03743
Step 3 · Find maximum lift-to-drag ratio ((L/D)ₘₐₓ)
Use (L/D)ₘₐₓ = 1 ÷ (2 × √(CD0 × K))
One
over
twice
the
root
of
CD0
times
K
.
(L/D)ₘₐₓ
=
1
2 ×
√
0.025 × 0.04974
(L/D)ₘₐₓ
=
1
2 ×
√
0.0012435
(L/D)ₘₐₓ
=
1
2 × 0.035263
(L/D)ₘₐₓ
=
1
0.070526
(L/D)ₘₐₓ = 14.18
Step 4 · Find CL at best L/D (CLb)
Use CLb = √(CD0 ÷ K)
Induced drag KCL² equals CD0 here.
CLb
=
√(
0.025
0.04974
)
CLb
=
√
0.5026
CLb = 0.709
Check
Put the numbers back in. Both sides should match.
0.04974
=
1
π × 0.8 × 8
✓
0.03743
=
0.025
+
0.04974
×
0.5
2
✓
14.18
=
1
2 ×
√
0.025 × 0.04974
✓
0.709
=
√(
0.025
0.04974
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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