WRITERWRITERWRITERWRITER
§
AERE 261 Homework 4 Corrections
Sep 25, 2025
Given:
S
w
i
n
g
=
22.9
m
2
S_{wing}=22.9m^2
S
w
in
g
=
22.9
m
2
h
=
7000
f
t
h=7000ft
h
=
7000
f
t
μ
=
1.7
∗
10
−
5
k
g
m
∗
s
\mu=1.7*10^{-5}\frac{kg}{m*s}
μ
=
1.7
∗
1
0
−
5
m
∗
s
k
g
ρ
=
0.993
k
g
/
m
3
\rho=0.993kg/m^3
ρ
=
0.993
k
g
/
m
3
V
∞
=
89.5
m
/
s
V_\infty=89.5m/s
V
∞
=
89.5
m
/
s
M
=
0.27
M=0.27
M
=
0.27
§
1. (a)
Given:
S
=
36
f
t
2
=
3.345
m
2
S=36ft^2=3.345m^2
S
=
36
f
t
2
=
3.345
m
2
t
/
c
=
0.08
t/c=0.08
t
/
c
=
0.08
Λ
=
20
°
\Lambda=20\degree
Λ
=
20°
Q
=
1.05
Q=1.05
Q
=
1.05
x
/
c
=
0.30
x/c=0.30
x
/
c
=
0.30
c
=
3.45
f
t
=
1.052
m
c=3.45ft=1.052m
c
=
3.45
f
t
=
1.052
m
Dependencies of
C
D
,
0
C_{D,0}
C
D
,
0
:
l
=
c
=
1.052
m
l=c=1.052m
l
=
c
=
1.052
m
R
=
ρ
V
∞
l
μ
=
0.993
k
g
/
m
3
∗
89.5
m
/
s
∗
1.052
m
1.7
∗
10
−
5
k
g
m
∗
s
=
5.5
×
10
6
R=\frac{\rho V_\infty l}{\mu}=\frac{0.993kg/m^3 * 89.5m/s * 1.052m}{1.7*10^{-5}\frac{kg}{m*s}}=5.5×10^6
R
=
μ
ρ
V
∞
l
=
1.7
∗
1
0
−
5
m
∗
s
k
g
0.993
k
g
/
m
3
∗
89.5
m
/
s
∗
1.052
m
=
5.5
×
1
0
6
C
f
=
0.455
(
log
10
R
)
2.58
(
1
+
0.144
M
2
)
0.65
=
0.455
(
log
10
(
5.5
×
10
6
)
)
2.58
(
1
+
0.144
(
0.27
)
2
)
0.65
=
0.003289
C_f=\frac{0.455}{(\log_{10}R)^{2.58}(1+0.144M^2)^{0.65}}=\frac{0.455}{(\log_{10}(5.5×10^6))^{2.58}(1+0.144(0.27)^2)^{0.65}}=0.003289
C
f
=
(
lo
g
10
R
)
2.58
(
1
+
0.144
M
2
)
0.65
0.455
=
(
lo
g
10
(
5.5
×
1
0
6
)
)
2.58
(
1
+
0.144
(
0.27
)
2
)
0.65
0.455
=
0.003289
F
F
=
[
1
+
0.6
(
x
/
c
)
m
(
t
c
)
+
100
(
t
c
)
4
]
[
1.34
M
0.18
(
c
o
s
Λ
m
)
0.28
]
FF=\left[1+\frac{0.6}{(x/c)_m}\left(\frac{t}{c}\right)+100\left(\frac{t}{c}\right)^4\right]\left[1.34M^{0.18}(cos\Lambda_m)^{0.28}\right]
FF
=
[
1
+
(
x
/
c
)
m
0.6
(
c
t
)
+
100
(
c
t
)
4
]
[
1.34
M
0.18
(
cos
Λ
m
)
0.28
]
F
F
=
[
1
+
0.6
0.30
(
0.08
)
+
100
(
0.08
)
4
]
[
1.34
(
0.27
)
0.18
(
c
o
s
(
20
°
)
)
0.28
]
=
1.211
FF=\left[1+\frac{0.6}{0.30}\left(0.08\right)+100\left(0.08\right)^4\right]\left[1.34(0.27)^{0.18}(cos(20\degree))^{0.28}\right]=1.211
FF
=
[
1
+
0.30
0.6
(
0.08
)
+
100
(
0.08
)
4
]
[
1.34
(
0.27
)
0.18
(
cos
(
20°
)
)
0.28
]
=
1.211
S
w
e
t
=
[
1.977
+
0.52
(
t
/
c
)
]
S
e
x
p
o
s
e
d
=
[
1.977
+
0.52
(
0.08
)
]
∗
3.345
m
2
=
6.752
m
2
S_{wet}=[1.977+0.52(t/c)]S_{exposed}=[1.977+0.52(0.08)]*3.345m^2=6.752m^2
S
w
e
t
=
[
1.977
+
0.52
(
t
/
c
)]
S
e
x
p
ose
d
=
[
1.977
+
0.52
(
0.08
)]
∗
3.345
m
2
=
6.752
m
2
And finally
C
D
,
0
C_{D,0}
C
D
,
0
:
C
D
,
0
=
C
f
∗
F
F
∗
Q
∗
S
w
e
t
S
w
i
n
g
=
0.003289
∗
1.211
∗
1.05
∗
6.752
m
2
22.9
m
2
=
0.001233
C_{D,0}=C_f*FF*Q*\frac{S_{wet}}{S_{wing}}=0.003289*1.211*1.05*\frac{6.752m^2}{22.9m^2}=\boxed{0.001233}
C
D
,
0
=
C
f
∗
FF
∗
Q
∗
S
w
in
g
S
w
e
t
=
0.003289
∗
1.211
∗
1.05
∗
22.9
m
2
6.752
m
2
=
0.001233
§
1. (b)
Given:
Q
=
1.3
Q=1.3
Q
=
1.3
S
w
e
t
=
0.88
m
2
S_{wet}=0.88m^2
S
w
e
t
=
0.88
m
2
Dependencies of
C
D
,
0
C_{D,0}
C
D
,
0
:
l
=
0.9
m
l=0.9m
l
=
0.9
m
R
=
ρ
V
∞
l
μ
=
0.993
k
g
/
m
3
∗
89.5
m
/
s
∗
0.9
m
1.7
∗
10
−
5
k
g
m
∗
s
=
4.705
×
10
6
R=\frac{\rho V_\infty l}{\mu}=\frac{0.993kg/m^3 * 89.5m/s * 0.9m}{1.7*10^{-5}\frac{kg}{m*s}}=4.705×10^6
R
=
μ
ρ
V
∞
l
=
1.7
∗
1
0
−
5
m
∗
s
k
g
0.993
k
g
/
m
3
∗
89.5
m
/
s
∗
0.9
m
=
4.705
×
1
0
6
C
f
=
0.455
(
log
10
R
)
2.58
(
1
+
0.144
M
2
)
0.65
=
0.455
(
log
10
(
4.705
×
10
6
)
)
2.58
(
1
+
0.144
(
0.27
)
2
)
0.65
=
0.003376
C_f=\frac{0.455}{(\log_{10}R)^{2.58}(1+0.144M^2)^{0.65}}=\frac{0.455}{(\log_{10}(4.705×10^6))^{2.58}(1+0.144(0.27)^2)^{0.65}}=0.003376
C
f
=
(
lo
g
10
R
)
2.58
(
1
+
0.144
M
2
)
0.65
0.455
=
(
lo
g
10
(
4.705
×
1
0
6
)
)
2.58
(
1
+
0.144
(
0.27
)
2
)
0.65
0.455
=
0.003376
A
m
a
x
=
π
(
0.5
∗
0.3
m
)
2
=
0.071
m
2
A_{max}=\pi(0.5*0.3m)^2=0.071m^2
A
ma
x
=
π
(
0.5
∗
0.3
m
)
2
=
0.071
m
2
f
=
l
(
4
/
π
)
A
m
a
x
=
0.9
m
(
4
/
π
)
∗
0.071
m
2
=
2.993
f=\frac{l}{\sqrt{(4/\pi)A_{max}}}=\frac{0.9m}{\sqrt{(4/\pi)*0.071m^2}}=2.993
f
=
(
4/
π
)
A
ma
x
l
=
(
4/
π
)
∗
0.071
m
2
0.9
m
=
2.993
F
F
=
1
+
0.35
f
=
1
+
0.35
2.993
=
1.117
FF=1+\frac{0.35}{f}=1+\frac{0.35}{2.993}=1.117
FF
=
1
+
f
0.35
=
1
+
2.993
0.35
=
1.117
And finally
C
D
,
0
C_{D,0}
C
D
,
0
:
C
D
,
0
=
C
f
∗
F
F
∗
Q
∗
S
w
e
t
S
w
i
n
g
=
0.003376
∗
1.117
∗
1.3
∗
0.88
m
2
22.9
m
2
=
1.884
×
10
−
4
C_{D,0}=C_f*FF*Q*\frac{S_{wet}}{S_{wing}}=0.003376*1.117*1.3*\frac{0.88m^2}{22.9m^2}=\boxed{1.884×10^{-4}}
C
D
,
0
=
C
f
∗
FF
∗
Q
∗
S
w
in
g
S
w
e
t
=
0.003376
∗
1.117
∗
1.3
∗
22.9
m
2
0.88
m
2
=
1.884
×
1
0
−
4
§
1. (c)
I will be doing calculations for just a single landing gear and will introduce the coefficient fo
3
3
3
in the build up later summation.
Given:
diameter
=
14
i
n
=
0.356
m
\text{diameter}=14in=0.356m
diameter
=
14
in
=
0.356
m
width
=
7
i
n
=
0.178
m
\text{width}=7in=0.178m
width
=
7
in
=
0.178
m
C
D
,
w
h
e
e
l
=
0.25
C_{D,wheel}=0.25
C
D
,
w
h
ee
l
=
0.25
Q
w
h
e
e
l
=
1.2
Q_{wheel}=1.2
Q
w
h
ee
l
=
1.2
Dependency of
C
D
,
0
C_{D,0}
C
D
,
0
:
S
f
r
o
n
t
a
l
=
diameter
∗
width
=
0.356
m
∗
0.178
m
=
0.0634
m
2
S_{frontal}=\text{diameter}*\text{width}=0.356m*0.178m=0.0634m^2
S
f
ro
n
t
a
l
=
diameter
∗
width
=
0.356
m
∗
0.178
m
=
0.0634
m
2
And finally
C
D
,
0
C_{D,0}
C
D
,
0
:
C
D
,
0
=
C
D
∗
S
f
r
o
n
t
a
l
S
w
i
n
g
∗
Q
=
0.25
∗
0.0634
m
2
22.9
m
2
∗
1.2
=
8.306
×
10
−
4
C_{D,0}=C_D*\frac{S_{frontal}}{S_{wing}}*Q=0.25*\frac{0.0634m^2}{22.9m^2}*1.2=\boxed{8.306×10^{-4}}
C
D
,
0
=
C
D
∗
S
w
in
g
S
f
ro
n
t
a
l
∗
Q
=
0.25
∗
22.9
m
2
0.0634
m
2
∗
1.2
=
8.306
×
1
0
−
4
§
1. (d)
Given:
C
D
,
0
,
vertical tail
=
0.0006
C_{D,0,\text{vertical tail}}=0.0006
C
D
,
0
,
vertical tail
=
0.0006
C
D
,
0
,
fuselage
=
0.0047
C_{D,0,\text{fuselage}}=0.0047
C
D
,
0
,
fuselage
=
0.0047
C
D
,
0
,
wing
=
0.0076
C_{D,0,\text{wing}}=0.0076
C
D
,
0
,
wing
=
0.0076
The final
C
D
,
0
:
C_{D,0}:
C
D
,
0
:
C
D
,
0
=
C
D
,
0
,
vertical tail
+
C
D
,
0
,
fuselage
+
C
D
,
0
,
wing
+
C
D
,
0
,
horizontal tail
+
C
D
,
0
,
weather radar
+
3
C
D
,
0
,
landing gear
C_{D,0}=C_{D,0,\text{vertical tail}}+C_{D,0,\text{fuselage}}+C_{D,0,\text{wing}}+C_{D,0,\text{horizontal tail}}+C_{D,0,\text{weather radar}}+3C_{D,0,\text{landing gear}}
C
D
,
0
=
C
D
,
0
,
vertical tail
+
C
D
,
0
,
fuselage
+
C
D
,
0
,
wing
+
C
D
,
0
,
horizontal tail
+
C
D
,
0
,
weather radar
+
3
C
D
,
0
,
landing gear
C
D
,
0
=
0.0006
+
0.0047
+
0.0076
+
0.001233
+
1.884
×
10
−
4
+
3
∗
8.306
×
10
−
4
=
0.0168
C_{D,0}=0.0006+0.0047+0.0076+0.001233+1.884×10^{-4}+3*8.306×10^{-4}=\boxed{0.0168}
C
D
,
0
=
0.0006
+
0.0047
+
0.0076
+
0.001233
+
1.884
×
1
0
−
4
+
3
∗
8.306
×
1
0
−
4
=
0.0168
§
2.
Given:
m
=
1680
k
g
m=1680kg
m
=
1680
k
g
V
∞
=
89.5
m
/
s
V_\infty=89.5m/s
V
∞
=
89.5
m
/
s
h
=
7000
f
t
h=7000ft
h
=
7000
f
t
K
=
0.056
K=0.056
K
=
0.056
Thrust required in steady level flight:
q
∞
=
1
2
ρ
∞
V
∞
2
=
1
2
∗
0.993
k
g
/
m
3
∗
(
89.5
m
/
s
)
2
=
3977
P
a
q_\infty=\frac{1}{2}\rho_\infty V_\infty^2=\frac{1}{2}*0.993kg/m^3 * (89.5m/s)^2=3977Pa
q
∞
=
2
1
ρ
∞
V
∞
2
=
2
1
∗
0.993
k
g
/
m
3
∗
(
89.5
m
/
s
)
2
=
3977
P
a
g
=
9.81
m
/
s
2
g=9.81m/s^2
g
=
9.81
m
/
s
2
W
=
m
g
=
1680
k
g
∗
9.81
m
/
s
2
=
16480
N
W=mg=1680kg*9.81m/s^2=16480N
W
=
m
g
=
1680
k
g
∗
9.81
m
/
s
2
=
16480
N
T
R
=
q
∞
S
C
D
,
0
+
2
K
W
2
q
∞
S
=
3977
P
a
∗
22.9
m
2
∗
0.0168
+
2
∗
0.056
∗
(
16480
N
)
2
3977
P
a
∗
22.9
m
2
=
1864
N
T_R=q_\infty S C_{D,0}+\frac{2KW^2}{q_\infty S}=3977Pa * 22.9m^2 * 0.0168+\frac{2 * 0.056 * (16480N)^2}{3977Pa * 22.9m^2}=\boxed{1864N}
T
R
=
q
∞
S
C
D
,
0
+
q
∞
S
2
K
W
2
=
3977
P
a
∗
22.9
m
2
∗
0.0168
+
3977
P
a
∗
22.9
m
2
2
∗
0.056
∗
(
16480
N
)
2
=
1864
N
IMPORTANT
P
R
=
1864
N
∗
89.5
m
/
s
=
167
k
W
P_R=1864N*89.5m/s=\boxed{167kW}
P
R
=
1864
N
∗
89.5
m
/
s
=
167
kW
It's a little off due to rounding errors along the way (I retained more precision.)