SME
Aircraft Design
Wing loading and thrust-to-weight (constraint diagram)
Use this for “Cruise at 230 m/s where ρ = 0.4 kg/m³, CD0 = 0.02, K = 0.045. What T/W does W/S = 5000 Pa need?”
Chart
T/W(5,000) ≈ 0.06359
This is an example. Type your own values to start a new problem.
T/W
=
q (Pa)
×
W/S (Pa)
+
×
W/S (Pa)
q (Pa)
+
G
ρ
Air density
example
V
Flight speed
example
Formulas
q
=
0.5
×
ρ
×
V
2
10,580
=
0.5
×
0.4
×
230
2
T/W
=
q
×
CD0
W/S
+
K
×
W/S
q
+
G
0.06359
=
10,580
×
0.02
5,000
+
0.045
×
5,000
10,580
+
0
Assumptions
•
A parabolic drag polar, CD = CD0 + KCL².
•
Lift equals weight: L = W cos γ ≈ W for a shallow climb.
•
T/W and W/S at this flight condition, not at takeoff.
•
The climb gradient G = sin γ; G = 0 in level cruise.
Step-by-step
WE KNOW
ρ = 0.4 kg/m³, V = 230 m/s, CD0 = 0.02, K = 0.045, G = 0, W/S = 5,000 Pa
FIND
dynamic pressure (q), thrust-to-weight (T/W)
Step 1 · Find dynamic pressure (q)
Use q = 0.5 × ρ × V²
Dynamic pressure: half the density times the speed squared.
q
=
0.5
×
0.4
×
230
2
q
=
0.5
×
0.4
×
52,900
q
=
0.2
×
52,900
q
=
10,580
Pa
Step 2 · Find thrust-to-weight (T/W)
Use T/W = q × CD0 ÷ (W/S) + K × (W/S) ÷ q + G
Thrust balances drag plus the climb share of weight; divide both by W.
T/W
=
10,580
×
0.02
5,000
+
0.045
×
5,000
10,580
+
0
T/W
=
211.6
5,000
+
225
10,580
T/W = 0.04232 + 0.02127
T/W = 0.06359
Check
Put the numbers back in. Both sides should match.
10,580
=
0.5
×
0.4
×
230
2
✓
0.06359
=
10,580
×
0.02
5,000
+
0.045
×
5,000
10,580
+
0
✓
Topic check
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More problem types
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Use this for “The approach needs a stall speed of 60 m/s with CLmax = 2.2 at sea level. What is the largest wing loading?”
Wing loading for the least thrust
Use this for “In cruise q = 10,800 Pa, CD0 = 0.02 and K = 0.045. Which wing loading needs the least T/W, and what is that T/W?”