Fundamentos de Engenharia Aeronáutica - Volume único

249
φ =
(16 ⋅ h / b)
2
1 + (16 ⋅ h / b)
2
φ =
(16 ⋅ 0,35 / 2,48)
1 + (16 ⋅ 0,35 / 2,48)
2
2
φ = 0,836
A velocidade de estol é:
v
estol
= 2 ⋅ W
ρ ⋅ S ⋅ C
Lmáx
v
estol
=
2 ⋅143
1,225 ⋅ 0,9 ⋅1,65
vestol
= 12,54 m/s
A força de sustentação durante o pouso para 0,7v estol é:
L = ⋅
2
⋅ (0,7 ⋅ v
)
⋅ S ⋅
2
estol
C L
1 ρ 276
1 2
L = ⋅1,225
⋅ (0,7 ⋅12,54)
2
⋅ 0,9 ⋅ 0,
L = 11,721 N
A correspondente força de arrasto é:
1 2
L
2
D = ⋅ ρ ⋅ (0,7 ⋅ vestol
) ⋅ S ⋅ ( CD0
+ φ ⋅ K ⋅ C
2
)
1 2
2
D = ⋅1,225
⋅(0,7
⋅12,54)
2
D = 1,110 N
⋅0,9
⋅(0,022
+ 0,836 ⋅0,065⋅
0,276
Portanto, aplicando-se a Equação (4.110c) para a determinação do comprimento de
pista para o pouso da aeronave tem-se que:
)
S
L
=
g ⋅ ρ ⋅ S ⋅ C
Lmáx
2
W
⋅[ D + µ ⋅ ( W − L)]
0,7estol