WRITERWRITERWRITERWRITER
§
AERE 261 Homework 9 Corrections
Sep 25, 2025
§
1.
General givens:
m
=
1680
k
g
m=1680kg
m
=
1680
k
g
S
r
e
f
=
22.9
m
2
S_{ref}=22.9m^2
S
re
f
=
22.9
m
2
ρ
=
1.225
k
g
/
m
3
\rho=1.225kg/m^3
ρ
=
1.225
k
g
/
m
3
C
L
,
m
a
x
=
1.37
C_{L,max}=1.37
C
L
,
ma
x
=
1.37
C
L
,
r
o
l
l
i
n
g
=
0.6
C_{L,rolling}=0.6
C
L
,
ro
ll
in
g
=
0.6
K
=
0.056
K=0.056
K
=
0.056
C
D
,
0
=
0.017
C_{D,0}=0.017
C
D
,
0
=
0.017
Takeoff givens:
μ
r
o
l
l
i
n
g
=
0.05
\mu_{rolling}=0.05
μ
ro
ll
in
g
=
0.05
h
obstacle,takeoff
=
50
f
t
h_\text{obstacle,takeoff}=50ft
h
obstacle,takeoff
=
50
f
t
η
p
r
=
0.9
\eta_{pr}=0.9
η
p
r
=
0.9
P
s
h
a
f
t
=
180000
W
P_{shaft}=180000W
P
s
ha
f
t
=
180000
W
climb angle
=
3
°
\text{climb angle}=3\degree
climb angle
=
3°
n
liftoff
=
1.05
n_\text{liftoff}=1.05
n
liftoff
=
1.05
V
L
O
=
1.2
V
s
t
a
l
l
V_{LO}=1.2V_{stall}
V
L
O
=
1.2
V
s
t
a
ll
Landing givens:
n
f
l
a
r
e
=
1.08
n_{flare}=1.08
n
f
l
a
re
=
1.08
V
f
l
a
r
e
=
V
T
D
=
1.3
V
s
t
a
l
l
V_{flare}=V_{TD}=1.3V_{stall}
V
f
l
a
re
=
V
T
D
=
1.3
V
s
t
a
ll
θ
a
=
3
°
\theta_{a}=3\degree
θ
a
=
3°
h
obstacle,landing
=
50
f
t
h_\text{obstacle,landing}=50ft
h
obstacle,landing
=
50
f
t
μ
b
r
a
k
i
n
g
=
0.5
\mu_{braking}=0.5
μ
b
r
akin
g
=
0.5
§
Takeoff ground roll
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
V
s
t
a
l
l
=
2
W
ρ
S
C
L
,
m
a
x
=
2
∗
16480
N
1.225
k
g
/
m
3
∗
22.9
m
2
∗
1.37
=
29.29
m
/
s
V_{stall}=\sqrt{\frac{2 W}{\rho S C_{L,max}}}=\sqrt{\frac{2 * 16480N}{1.225kg/m^3 * 22.9m^2 * 1.37}}=29.29m/s
V
s
t
a
ll
=
ρS
C
L
,
ma
x
2
W
=
1.225
k
g
/
m
3
∗
22.9
m
2
∗
1.37
2
∗
16480
N
=
29.29
m
/
s
V
L
O
=
1.2
V
s
t
a
l
l
=
1.2
∗
29.29
m
/
s
=
35.15
m
/
s
V_{LO}=1.2V_{stall}=1.2 * 29.29m/s=35.15m/s
V
L
O
=
1.2
V
s
t
a
ll
=
1.2
∗
29.29
m
/
s
=
35.15
m
/
s
P
a
=
η
p
r
P
s
h
a
f
t
=
0.9
∗
180000
W
=
162000
W
P_a=\eta_{pr} P_{shaft}=0.9 * 180000W=162000W
P
a
=
η
p
r
P
s
ha
f
t
=
0.9
∗
180000
W
=
162000
W
P
a
=
T
V
L
O
P_a=TV_{LO}
P
a
=
T
V
L
O
IMPORTANT
T
=
P
a
V
L
O
=
162000
W
35.15
m
/
s
=
4609
N
T=\frac{P_a}{V_{LO}}=\frac{162000W}{35.15m/s}=4609N
T
=
V
L
O
P
a
=
35.15
m
/
s
162000
W
=
4609
N
D
=
1
2
ρ
(
0.7
V
L
O
)
2
S
(
C
D
,
0
+
K
C
L
,
r
o
l
l
i
n
g
2
)
D=\frac{1}{2} \rho (0.7 V_{LO})^2 S (C_{D,0} + K C_{L,rolling}^2)
D
=
2
1
ρ
(
0.7
V
L
O
)
2
S
(
C
D
,
0
+
K
C
L
,
ro
ll
in
g
2
)
D
=
1
2
∗
1.225
k
g
/
m
3
∗
(
0.7
∗
35.15
m
/
s
)
2
∗
22.9
m
2
∗
(
0.017
+
0.056
∗
(
0.6
)
2
)
=
315.5
N
D=\frac{1}{2} * 1.225kg/m^3 * (0.7 * 35.15m/s)^2 * 22.9m^2 * (0.017 + 0.056 * (0.6)^2)=315.5N
D
=
2
1
∗
1.225
k
g
/
m
3
∗
(
0.7
∗
35.15
m
/
s
)
2
∗
22.9
m
2
∗
(
0.017
+
0.056
∗
(
0.6
)
2
)
=
315.5
N
L
=
1
2
ρ
(
0.7
V
L
O
)
2
S
C
L
,
r
o
l
l
i
n
g
=
1
2
1.225
k
g
/
m
3
(
0.7
∗
35.15
m
/
s
)
2
∗
22.9
m
2
∗
0.6
=
5095
N
L=\frac{1}{2} \rho (0.7 V_{LO})^2 S C_{L,rolling}=\frac{1}{2} 1.225kg/m^3 (0.7 * 35.15m/s)^2 * 22.9m^2 * 0.6=5095N
L
=
2
1
ρ
(
0.7
V
L
O
)
2
S
C
L
,
ro
ll
in
g
=
2
1
1.225
k
g
/
m
3
(
0.7
∗
35.15
m
/
s
)
2
∗
22.9
m
2
∗
0.6
=
5095
N
IMPORTANT
s
g
=
1.44
W
2
g
ρ
S
C
L
,
m
a
x
(
T
−
D
−
μ
r
(
W
−
L
)
)
s_g=\frac{1.44 W^2}{g \rho S C_{L,max} (T - D - \mu_r (W - L))}
s
g
=
g
ρS
C
L
,
ma
x
(
T
−
D
−
μ
r
(
W
−
L
))
1.44
W
2
s
g
=
1.44
∗
(
16480
N
)
2
9.81
m
/
s
2
∗
1.225
k
g
/
m
3
∗
22.9
m
2
∗
1.37
∗
(
4609
N
−
315.5
N
−
0.05
∗
(
16480
N
−
5095
N
)
)
s_g=\frac{1.44 * (16480N)^2}{9.81m/s^2 * 1.225kg/m^3 * 22.9m^2 * 1.37 * (4609N - 315.5N - 0.05 * (16480N - 5095N))}
s
g
=
9.81
m
/
s
2
∗
1.225
k
g
/
m
3
∗
22.9
m
2
∗
1.37
∗
(
4609
N
−
315.5
N
−
0.05
∗
(
16480
N
−
5095
N
))
1.44
∗
(
16480
N
)
2
s
g
=
278.5
m
s_g=\boxed{278.5m}
s
g
=
278.5
m
§
Takeoff transition
R
=
1.44
V
s
t
a
l
l
2
g
(
n
−
1
)
=
1.44
∗
(
29.29
m
/
s
)
2
9.81
m
/
s
2
∗
(
1.05
−
1
)
=
2519
m
R=\frac{1.44V_{stall}^2}{g(n-1)}=\frac{1.44 * (29.29m/s)^2}{9.81m/s^2 * (1.05 - 1)}=2519m
R
=
g
(
n
−
1
)
1.44
V
s
t
a
ll
2
=
9.81
m
/
s
2
∗
(
1.05
−
1
)
1.44
∗
(
29.29
m
/
s
)
2
=
2519
m
s
t
r
=
R
sin
θ
=
2519
m
sin
3
°
=
131.8
m
s_{tr}=R \sin \theta=2519m \sin 3\degree=\boxed{131.8m}
s
t
r
=
R
sin
θ
=
2519
m
sin
3°
=
131.8
m
§
Takeoff air
h
=
50
f
t
=
15.24
m
h=50ft=15.24m
h
=
50
f
t
=
15.24
m
h
a
=
h
−
R
+
R
cos
θ
h_a=h - R + R \cos \theta
h
a
=
h
−
R
+
R
cos
θ
h
a
=
15.24
m
−
2519
m
+
2519
m
cos
3
°
=
11.788
m
h_a=15.24m - 2519m + 2519m \cos 3\degree=11.788m
h
a
=
15.24
m
−
2519
m
+
2519
m
cos
3°
=
11.788
m
s
a
=
h
a
tan
θ
=
11.788
m
tan
3
°
=
224.9
m
s_a=\frac{h_a}{\tan \theta}=\frac{11.788m}{\tan 3\degree}=\boxed{224.9m}
s
a
=
tan
θ
h
a
=
tan
3°
11.788
m
=
224.9
m
IMPORTANT
§
Takeoff total
s
takeoff
=
s
g
+
s
t
r
+
s
a
=
278.5
m
+
131.8
m
+
224.9
m
=
635.2
m
s_\text{takeoff}=s_g+s_{tr}+s_a=278.5m+131.8m+224.9m=\boxed{635.2m}
s
takeoff
=
s
g
+
s
t
r
+
s
a
=
278.5
m
+
131.8
m
+
224.9
m
=
635.2
m
§
Landing approach
h
f
=
R
−
R
cos
θ
a
=
2519
m
−
2519
m
cos
3
°
=
3.452
m
h_f=R-R \cos \theta_a=2519m - 2519m \cos 3\degree=3.452m
h
f
=
R
−
R
cos
θ
a
=
2519
m
−
2519
m
cos
3°
=
3.452
m
h
=
50
f
t
=
15.24
m
h=50ft=15.24m
h
=
50
f
t
=
15.24
m
h
f
+
h
a
=
h
h_f+h_a=h
h
f
+
h
a
=
h
h
a
=
h
−
h
f
=
15.24
m
−
3.452
m
=
11.788
m
h_a=h-h_f=15.24m-3.452m=11.788m
h
a
=
h
−
h
f
=
15.24
m
−
3.452
m
=
11.788
m
s
a
=
h
a
tan
θ
a
=
11.788
m
tan
3
°
=
224.9
m
s_a=\frac{h_a}{\tan \theta_a}=\frac{11.788m}{\tan 3\degree}=\boxed{224.9m}
s
a
=
tan
θ
a
h
a
=
tan
3°
11.788
m
=
224.9
m
§
Landing flare
IMPORTANT
R
R
R
must've been recalculated with
n
=
1.08
n=1.08
n
=
1.08
which was not performed.
θ
f
=
θ
a
=
3
°
\theta_f=\theta_a=3\degree
θ
f
=
θ
a
=
3°
s
f
=
R
sin
θ
f
=
2519
m
sin
3
°
=
131.8
m
s_f=R \sin \theta_f=2519m \sin 3\degree=\boxed{131.8m}
s
f
=
R
sin
θ
f
=
2519
m
sin
3°
=
131.8
m
§
Landing ground roll
V
T
D
=
1.3
V
s
t
a
l
l
=
1.3
∗
29.29
m
/
s
=
38.077
m
/
s
V_{TD}=1.3V_{stall}=1.3 * 29.29m/s=38.077m/s
V
T
D
=
1.3
V
s
t
a
ll
=
1.3
∗
29.29
m
/
s
=
38.077
m
/
s
D
=
1
2
ρ
(
0.7
V
T
D
)
2
S
(
C
D
,
0
+
K
C
L
2
)
D=\frac{1}{2} \rho (0.7 V_{TD})^2 S (C_{D,0} + K C_L^2)
D
=
2
1
ρ
(
0.7
V
T
D
)
2
S
(
C
D
,
0
+
K
C
L
2
)
D
=
1
2
∗
1.225
k
g
/
m
3
∗
(
0.7
∗
38.077
m
/
s
)
2
∗
22.9
m
2
∗
(
0.017
+
0.056
∗
(
0.6
)
2
)
=
370.3
N
D=\frac{1}{2} * 1.225kg/m^3 * (0.7 * 38.077m/s)^2 * 22.9m^2 * (0.017 + 0.056 * (0.6)^2)=370.3N
D
=
2
1
∗
1.225
k
g
/
m
3
∗
(
0.7
∗
38.077
m
/
s
)
2
∗
22.9
m
2
∗
(
0.017
+
0.056
∗
(
0.6
)
2
)
=
370.3
N
L
=
1
2
ρ
(
0.7
V
T
D
)
2
S
C
L
=
1
2
∗
1.225
k
g
/
m
3
∗
(
0.7
∗
38.077
m
/
s
)
2
∗
22.9
m
2
∗
0.6
=
5979
N
L=\frac{1}{2} \rho (0.7 V_{TD})^2 S C_{L}=\frac{1}{2} * 1.225kg/m^3 * (0.7 * 38.077m/s)^2 * 22.9m^2 * 0.6=5979N
L
=
2
1
ρ
(
0.7
V
T
D
)
2
S
C
L
=
2
1
∗
1.225
k
g
/
m
3
∗
(
0.7
∗
38.077
m
/
s
)
2
∗
22.9
m
2
∗
0.6
=
5979
N
s
g
=
1.69
W
2
g
ρ
S
C
L
,
m
a
x
(
D
+
μ
r
(
W
−
L
)
)
s_g=\frac{1.69 W^2}{g \rho S C_{L,max} (D + \mu_r (W - L))}
s
g
=
g
ρS
C
L
,
ma
x
(
D
+
μ
r
(
W
−
L
))
1.69
W
2
IMPORTANT
s
g
=
1.69
∗
(
16480
N
)
2
9.81
m
/
s
2
∗
1.225
k
g
/
m
3
∗
22.9
m
2
∗
1.37
(
370.3
N
+
0.05
∗
(
16480
N
−
5979
N
)
)
=
1360
m
s_g=\frac{1.69 * (16480N)^2}{9.81m/s^2 * 1.225kg/m^3 * 22.9m^2 * 1.37 (370.3N + 0.05 * (16480N - 5979N))}=\boxed{1360m}
s
g
=
9.81
m
/
s
2
∗
1.225
k
g
/
m
3
∗
22.9
m
2
∗
1.37
(
370.3
N
+
0.05
∗
(
16480
N
−
5979
N
))
1.69
∗
(
16480
N
)
2
=
1360
m
§
Landing total
s
landing
=
s
a
+
s
f
+
s
g
=
224.9
m
+
131.8
m
+
1360
m
=
1716.7
m
s_\text{landing}=s_a+s_f+s_g=224.9m + 131.8m + 1360m=\boxed{1716.7m}
s
landing
=
s
a
+
s
f
+
s
g
=
224.9
m
+
131.8
m
+
1360
m
=
1716.7
m