SME
Aircraft Design · Problem type
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?”
Chart
T/W(7,200) = 0.06
Level (T/W)min = 0.06 meets the curve at W/S = 7,200
This is an example. Type your own values to start a new problem.
(W/S)* (Pa)
=
q (Pa)
×
CD0
K
(T/W)min
=
2
×
×
Formulas
(W/S)*
=
q
×
√(
CD0
K
)
7,200
=
10,800
×
√(
0.02
0.045
)
(T/W)min
=
2
×
√
CD0 ×
K
0.06
=
2
×
√
0.02 × 0.045
Assumptions
•
Level cruise (G = 0) with a parabolic drag polar, CD = CD0 + KCL².
•
At (W/S)* the zero-lift and induced drag terms are equal.
•
There the airplane flies at its best L/D, so (T/W)min = 1 ÷ (L/D)max.
Step-by-step
WE KNOW
q = 10,800 Pa, CD0 = 0.02, K = 0.045
FIND
best wing loading ((W/S)*), least thrust-to-weight ((T/W)min)
Step 1 · Find best wing loading ((W/S)*)
Use (W/S)* = q × √(CD0 ÷ K)
Set
the
slope
of
T/W
to
0:
qCD0
÷
(W/S)
2
=
K
q
,
so
(W/S)
2
=
q
2
CD0
K
.
(W/S)*
=
10,800
×
√(
0.02
0.045
)
(W/S)*
=
10,800
×
√
0.4444
(W/S)* = 10,800 × 0.6666
(W/S)* = 7,200 Pa
Step 2 · Find least thrust-to-weight ((T/W)min)
Use (T/W)min = 2 × √(CD0 × K)
At
(W/S)*
each
drag
term
of
T/W
is
√
CD0K
,
so
T/W
is
twice
that.
(T/W)min
=
2
×
√
0.02 × 0.045
(T/W)min
=
2
×
√
0.0009
(T/W)min = 2 × 0.03
(T/W)min = 0.06
Check
Put the numbers back in. Both sides should match.
7,200
=
10,800
×
√(
0.02
0.045
)
✓
0.06
=
2
×
√
0.02 × 0.045
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.