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# Chapter 8 - Weights and centre of gravity - 1 Notes | EduRev

## : Chapter 8 - Weights and centre of gravity - 1 Notes | EduRev

``` Page 1

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
1
Chapter 8
Weights and centre of gravity - 2
Lecture 33
Topics
Example 8.1
Example 8.2

Example 8.1
For the sixty seater airplane considered in examples 3.1, 4.19, 5.1, 6.1, 6.2 and
6.3, obtain the following.
(i) Weights and c.g. locations of (a) wing, (b) fuselage, (c) h.tail, (d) v.tail,
(e) landing gear, (f) installed engine, (g) all else empty (h) fuel (i) payload.
(ii) Location of wing such that the c.g. of the entire airplane with MTOW is at
quarter chord of m.a.c.
(iii) Shift in the c.g. of the airplane for the following cases.
(a) Full payload but no fuel.
(b) No payload and no fuel.
(c) No payload but full fuel.
(d) Full fuel but half of the payload in the front half of passenger cabin.
(e) Full fuel but half of the payload in the rear half of passenger cabin.
Solution :
From example 3.1 the gross weight of the airplane is 21,280 kgf or 208,757 N
I)The weights and c.g. locations of various components are estimated below.
(A) Wing
From examples 5.1 and 6.1 the following data are obtained.
Wing area = 58.48 m
2

Root chord of wing = 2.636 m
Tip chord of wing = 1.318 m
Wing span = 26.49 m ; semispan = 13.245 m
Constant chord of 2.636 m upto 4.636 m from FRL
Page 2

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
1
Chapter 8
Weights and centre of gravity - 2
Lecture 33
Topics
Example 8.1
Example 8.2

Example 8.1
For the sixty seater airplane considered in examples 3.1, 4.19, 5.1, 6.1, 6.2 and
6.3, obtain the following.
(i) Weights and c.g. locations of (a) wing, (b) fuselage, (c) h.tail, (d) v.tail,
(e) landing gear, (f) installed engine, (g) all else empty (h) fuel (i) payload.
(ii) Location of wing such that the c.g. of the entire airplane with MTOW is at
quarter chord of m.a.c.
(iii) Shift in the c.g. of the airplane for the following cases.
(a) Full payload but no fuel.
(b) No payload and no fuel.
(c) No payload but full fuel.
(d) Full fuel but half of the payload in the front half of passenger cabin.
(e) Full fuel but half of the payload in the rear half of passenger cabin.
Solution :
From example 3.1 the gross weight of the airplane is 21,280 kgf or 208,757 N
I)The weights and c.g. locations of various components are estimated below.
(A) Wing
From examples 5.1 and 6.1 the following data are obtained.
Wing area = 58.48 m
2

Root chord of wing = 2.636 m
Tip chord of wing = 1.318 m
Wing span = 26.49 m ; semispan = 13.245 m
Constant chord of 2.636 m upto 4.636 m from FRL
Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
2
Fuselage width = 2.88 m
Mean aerodynamic chord = 2.295 m
Location of the leading edge of m.a.c. from the leading edge of wing = 0.237 m
Location at the a.c. of the wing from the wing root chord = 0.811 m
Referring to Fig.E5.1 and using the above data the area of the exposed wing is :
(S
exposed
)
wing
= 2
? ?
??
13.245-4.636
2.88
2.636 4.636- + 2.636+1.318
22
??
??
??
??
??
??

= 2 ? ?
2
8.425 + 17.02 = 50.89 m
Remark :
A simpler way to get the area of exposed wing in this case, where, the chord is
constant in the inboard portion of the wing, is :
58.48 – 2.88 x 2.636 = 50.89 m
2
Using Table 8.1, the estimated weight of the wing is :
50.89 x 49 = 2493.6 kgf = 24,462.3 N
Or
wing
g
W
= 24462.3 / 208757 = 11.72 %
W

From Table 8.1, the c.g. of wing is at 40% of m.a.c.
Hence, the location of the c.g. of wing from the leading edge of the root chord of
the wing is at : 0.237 + 0.4 x 2.295 = 1.155 m
(B) Horizontal tail
The horizontal tail is mounted on V.tail. Hence, its exposed area is taken roughly
equal to planform area which equals 11.11 m
2
.
Hence, weight of h.tail = 11.11 x 27 = 300 kgf = 2943 N = 1.41 % of W
g

To arrive at the location of the c.g. of the horizontal tail, the following are noted
from example 6.2.
Mean aerodynamic chord of h.tail = 1.52 m
The distance between the leading edge of the root and the leading edge of m.a.c.
is 1.711 x tan 11.24 = 0.34 m
From Table 8.1, the c.g. of the h.tail is at 40% of its m.a.c. Consequently, the

Page 3

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
1
Chapter 8
Weights and centre of gravity - 2
Lecture 33
Topics
Example 8.1
Example 8.2

Example 8.1
For the sixty seater airplane considered in examples 3.1, 4.19, 5.1, 6.1, 6.2 and
6.3, obtain the following.
(i) Weights and c.g. locations of (a) wing, (b) fuselage, (c) h.tail, (d) v.tail,
(e) landing gear, (f) installed engine, (g) all else empty (h) fuel (i) payload.
(ii) Location of wing such that the c.g. of the entire airplane with MTOW is at
quarter chord of m.a.c.
(iii) Shift in the c.g. of the airplane for the following cases.
(a) Full payload but no fuel.
(b) No payload and no fuel.
(c) No payload but full fuel.
(d) Full fuel but half of the payload in the front half of passenger cabin.
(e) Full fuel but half of the payload in the rear half of passenger cabin.
Solution :
From example 3.1 the gross weight of the airplane is 21,280 kgf or 208,757 N
I)The weights and c.g. locations of various components are estimated below.
(A) Wing
From examples 5.1 and 6.1 the following data are obtained.
Wing area = 58.48 m
2

Root chord of wing = 2.636 m
Tip chord of wing = 1.318 m
Wing span = 26.49 m ; semispan = 13.245 m
Constant chord of 2.636 m upto 4.636 m from FRL
Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
2
Fuselage width = 2.88 m
Mean aerodynamic chord = 2.295 m
Location of the leading edge of m.a.c. from the leading edge of wing = 0.237 m
Location at the a.c. of the wing from the wing root chord = 0.811 m
Referring to Fig.E5.1 and using the above data the area of the exposed wing is :
(S
exposed
)
wing
= 2
? ?
??
13.245-4.636
2.88
2.636 4.636- + 2.636+1.318
22
??
??
??
??
??
??

= 2 ? ?
2
8.425 + 17.02 = 50.89 m
Remark :
A simpler way to get the area of exposed wing in this case, where, the chord is
constant in the inboard portion of the wing, is :
58.48 – 2.88 x 2.636 = 50.89 m
2
Using Table 8.1, the estimated weight of the wing is :
50.89 x 49 = 2493.6 kgf = 24,462.3 N
Or
wing
g
W
= 24462.3 / 208757 = 11.72 %
W

From Table 8.1, the c.g. of wing is at 40% of m.a.c.
Hence, the location of the c.g. of wing from the leading edge of the root chord of
the wing is at : 0.237 + 0.4 x 2.295 = 1.155 m
(B) Horizontal tail
The horizontal tail is mounted on V.tail. Hence, its exposed area is taken roughly
equal to planform area which equals 11.11 m
2
.
Hence, weight of h.tail = 11.11 x 27 = 300 kgf = 2943 N = 1.41 % of W
g

To arrive at the location of the c.g. of the horizontal tail, the following are noted
from example 6.2.
Mean aerodynamic chord of h.tail = 1.52 m
The distance between the leading edge of the root and the leading edge of m.a.c.
is 1.711 x tan 11.24 = 0.34 m
From Table 8.1, the c.g. of the h.tail is at 40% of its m.a.c. Consequently, the

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
3
distance between the leading edge of the root chord of h.tail and the c.g. of h.tail
is :
0.34 + 0.4 x 1.52 = 0.95 m
(C) Vertical tail
The contribution of dorsal fin to the weight of v.tail is ignored at this stage of
preliminary design.
Area of v.tail = 12.92 m
2
Hence, weight of v.tail = 12.92 x 27 = 348.8 kgf = 3422 N = 1.64 % of W
g

To arrive at the location of the c.g. of the v.tail, the following are noted from
example 6.2.
Mean aerodynamic chord of v.tail = 2.60 m
The distance between leading edge of the root chord of v.tail and leading edge of
m.a.c. = 2.366 tan 30
o
= 1.366 m
From Table 8.1 the c.g. of v.tail is at 40% of its m.a.c. Consequently, the distance
between the root chord of v.tail and c.g. of v.tail is :
1.366 + 0.4 x 2.6 = 2.41 m
(D) Engine
As mentioned in example 6.3, the weight of each engine is 450 kgf.
Hence, from Table 8.1, the installed weight of the two engines is :
2 x 1.3 x 450 = 1170 kgf = 11478 N = 5.5 % of W
g

As regards the location of c.g. of the engine, Ref.1.24, chapter 11, mentions that
for gas turbine engines the location of c.g. from the engine inlet is between 30 to
45% of engine length.
In the present case the engine length is 2.13 m. From Fig.6.7a it is assumed that
the engine inlet is just behind the boss of the propeller. From example 6.3 the
boss of the propeller is at 1.58 m from the leading edge of the wing. Taking c.g.
of engine to be at 40 % of engine length, the location of engine c.g. from the
leading edge of the wing is :
-1.58 + 0.4 x 2.13 = - 0.728 m i.e. 0.728 m ahead of the leading edge of the root
chord of the wing.

Page 4

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
1
Chapter 8
Weights and centre of gravity - 2
Lecture 33
Topics
Example 8.1
Example 8.2

Example 8.1
For the sixty seater airplane considered in examples 3.1, 4.19, 5.1, 6.1, 6.2 and
6.3, obtain the following.
(i) Weights and c.g. locations of (a) wing, (b) fuselage, (c) h.tail, (d) v.tail,
(e) landing gear, (f) installed engine, (g) all else empty (h) fuel (i) payload.
(ii) Location of wing such that the c.g. of the entire airplane with MTOW is at
quarter chord of m.a.c.
(iii) Shift in the c.g. of the airplane for the following cases.
(a) Full payload but no fuel.
(b) No payload and no fuel.
(c) No payload but full fuel.
(d) Full fuel but half of the payload in the front half of passenger cabin.
(e) Full fuel but half of the payload in the rear half of passenger cabin.
Solution :
From example 3.1 the gross weight of the airplane is 21,280 kgf or 208,757 N
I)The weights and c.g. locations of various components are estimated below.
(A) Wing
From examples 5.1 and 6.1 the following data are obtained.
Wing area = 58.48 m
2

Root chord of wing = 2.636 m
Tip chord of wing = 1.318 m
Wing span = 26.49 m ; semispan = 13.245 m
Constant chord of 2.636 m upto 4.636 m from FRL
Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
2
Fuselage width = 2.88 m
Mean aerodynamic chord = 2.295 m
Location of the leading edge of m.a.c. from the leading edge of wing = 0.237 m
Location at the a.c. of the wing from the wing root chord = 0.811 m
Referring to Fig.E5.1 and using the above data the area of the exposed wing is :
(S
exposed
)
wing
= 2
? ?
??
13.245-4.636
2.88
2.636 4.636- + 2.636+1.318
22
??
??
??
??
??
??

= 2 ? ?
2
8.425 + 17.02 = 50.89 m
Remark :
A simpler way to get the area of exposed wing in this case, where, the chord is
constant in the inboard portion of the wing, is :
58.48 – 2.88 x 2.636 = 50.89 m
2
Using Table 8.1, the estimated weight of the wing is :
50.89 x 49 = 2493.6 kgf = 24,462.3 N
Or
wing
g
W
= 24462.3 / 208757 = 11.72 %
W

From Table 8.1, the c.g. of wing is at 40% of m.a.c.
Hence, the location of the c.g. of wing from the leading edge of the root chord of
the wing is at : 0.237 + 0.4 x 2.295 = 1.155 m
(B) Horizontal tail
The horizontal tail is mounted on V.tail. Hence, its exposed area is taken roughly
equal to planform area which equals 11.11 m
2
.
Hence, weight of h.tail = 11.11 x 27 = 300 kgf = 2943 N = 1.41 % of W
g

To arrive at the location of the c.g. of the horizontal tail, the following are noted
from example 6.2.
Mean aerodynamic chord of h.tail = 1.52 m
The distance between the leading edge of the root and the leading edge of m.a.c.
is 1.711 x tan 11.24 = 0.34 m
From Table 8.1, the c.g. of the h.tail is at 40% of its m.a.c. Consequently, the

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
3
distance between the leading edge of the root chord of h.tail and the c.g. of h.tail
is :
0.34 + 0.4 x 1.52 = 0.95 m
(C) Vertical tail
The contribution of dorsal fin to the weight of v.tail is ignored at this stage of
preliminary design.
Area of v.tail = 12.92 m
2
Hence, weight of v.tail = 12.92 x 27 = 348.8 kgf = 3422 N = 1.64 % of W
g

To arrive at the location of the c.g. of the v.tail, the following are noted from
example 6.2.
Mean aerodynamic chord of v.tail = 2.60 m
The distance between leading edge of the root chord of v.tail and leading edge of
m.a.c. = 2.366 tan 30
o
= 1.366 m
From Table 8.1 the c.g. of v.tail is at 40% of its m.a.c. Consequently, the distance
between the root chord of v.tail and c.g. of v.tail is :
1.366 + 0.4 x 2.6 = 2.41 m
(D) Engine
As mentioned in example 6.3, the weight of each engine is 450 kgf.
Hence, from Table 8.1, the installed weight of the two engines is :
2 x 1.3 x 450 = 1170 kgf = 11478 N = 5.5 % of W
g

As regards the location of c.g. of the engine, Ref.1.24, chapter 11, mentions that
for gas turbine engines the location of c.g. from the engine inlet is between 30 to
45% of engine length.
In the present case the engine length is 2.13 m. From Fig.6.7a it is assumed that
the engine inlet is just behind the boss of the propeller. From example 6.3 the
boss of the propeller is at 1.58 m from the leading edge of the wing. Taking c.g.
of engine to be at 40 % of engine length, the location of engine c.g. from the
leading edge of the wing is :
-1.58 + 0.4 x 2.13 = - 0.728 m i.e. 0.728 m ahead of the leading edge of the root
chord of the wing.

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
4

(E) Landing gear
From Table 8.1 the weight of the nose wheel plus the main landing gear is 4.3 %
of W
g
i.e. 0.043 x 208757 = 8977 N
Out of this total weight, the nose wheel and main wheel account for 15 % and
85 % respectively. Hence, nose wheel weighs 0.15 x 8977 = 1347 N and the
main wheels weigh 0.85 x 8977 = 7630 N.
As regards the locations of the c.g.’s of nose wheel and main wheels, it is
recalled that the nose wheel and main wheels share respectively 10 % and 90 %
of the airplane weight. From example 6.3, the wheel base is chosen as 9.78 m.
Hence, the c.g. of the nose wheel is 0.9 x  9.78 = 8.802 m ahead of the c.g. of
the airplane. The c.g. of the main wheels, as a group, is :
0.1 x 9.78 = 0.978 m behind the c.g. of the airplane.
(F) Fuselage and systems
Wetted area of fuselage is obtained as follows. Various dimensions are obtained
from Fig.6.8c
(a) Nose portion
Length of nose = 0.7 m
Diameter at the end of nose = 1.64 m
Hence, average diameter = 1.64/2 = 0.82 m
Consequently,wetted area or surface area of nose ?
2
p ×0.82×0.7 =1.80m
(b) Cockpit portion
Length of cockpit = 2.54 m
Diameter at the end of cockpit = 2.88 m
Hence, wetted area of cockpit portion ?
2
1.64+2.88
p× ×2.54 = 18.03 m
2
??
??
??

(c) Midfuselage portion
Length of midfuselage = 12.83 m
Diameter of midfuselage = 2.88 m
Hence, wetted area of midfuselage = p x 2.88 x 12.83 = 116.08 m
2

(d) Tail cone
Page 5

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
1
Chapter 8
Weights and centre of gravity - 2
Lecture 33
Topics
Example 8.1
Example 8.2

Example 8.1
For the sixty seater airplane considered in examples 3.1, 4.19, 5.1, 6.1, 6.2 and
6.3, obtain the following.
(i) Weights and c.g. locations of (a) wing, (b) fuselage, (c) h.tail, (d) v.tail,
(e) landing gear, (f) installed engine, (g) all else empty (h) fuel (i) payload.
(ii) Location of wing such that the c.g. of the entire airplane with MTOW is at
quarter chord of m.a.c.
(iii) Shift in the c.g. of the airplane for the following cases.
(a) Full payload but no fuel.
(b) No payload and no fuel.
(c) No payload but full fuel.
(d) Full fuel but half of the payload in the front half of passenger cabin.
(e) Full fuel but half of the payload in the rear half of passenger cabin.
Solution :
From example 3.1 the gross weight of the airplane is 21,280 kgf or 208,757 N
I)The weights and c.g. locations of various components are estimated below.
(A) Wing
From examples 5.1 and 6.1 the following data are obtained.
Wing area = 58.48 m
2

Root chord of wing = 2.636 m
Tip chord of wing = 1.318 m
Wing span = 26.49 m ; semispan = 13.245 m
Constant chord of 2.636 m upto 4.636 m from FRL
Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
2
Fuselage width = 2.88 m
Mean aerodynamic chord = 2.295 m
Location of the leading edge of m.a.c. from the leading edge of wing = 0.237 m
Location at the a.c. of the wing from the wing root chord = 0.811 m
Referring to Fig.E5.1 and using the above data the area of the exposed wing is :
(S
exposed
)
wing
= 2
? ?
??
13.245-4.636
2.88
2.636 4.636- + 2.636+1.318
22
??
??
??
??
??
??

= 2 ? ?
2
8.425 + 17.02 = 50.89 m
Remark :
A simpler way to get the area of exposed wing in this case, where, the chord is
constant in the inboard portion of the wing, is :
58.48 – 2.88 x 2.636 = 50.89 m
2
Using Table 8.1, the estimated weight of the wing is :
50.89 x 49 = 2493.6 kgf = 24,462.3 N
Or
wing
g
W
= 24462.3 / 208757 = 11.72 %
W

From Table 8.1, the c.g. of wing is at 40% of m.a.c.
Hence, the location of the c.g. of wing from the leading edge of the root chord of
the wing is at : 0.237 + 0.4 x 2.295 = 1.155 m
(B) Horizontal tail
The horizontal tail is mounted on V.tail. Hence, its exposed area is taken roughly
equal to planform area which equals 11.11 m
2
.
Hence, weight of h.tail = 11.11 x 27 = 300 kgf = 2943 N = 1.41 % of W
g

To arrive at the location of the c.g. of the horizontal tail, the following are noted
from example 6.2.
Mean aerodynamic chord of h.tail = 1.52 m
The distance between the leading edge of the root and the leading edge of m.a.c.
is 1.711 x tan 11.24 = 0.34 m
From Table 8.1, the c.g. of the h.tail is at 40% of its m.a.c. Consequently, the

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
3
distance between the leading edge of the root chord of h.tail and the c.g. of h.tail
is :
0.34 + 0.4 x 1.52 = 0.95 m
(C) Vertical tail
The contribution of dorsal fin to the weight of v.tail is ignored at this stage of
preliminary design.
Area of v.tail = 12.92 m
2
Hence, weight of v.tail = 12.92 x 27 = 348.8 kgf = 3422 N = 1.64 % of W
g

To arrive at the location of the c.g. of the v.tail, the following are noted from
example 6.2.
Mean aerodynamic chord of v.tail = 2.60 m
The distance between leading edge of the root chord of v.tail and leading edge of
m.a.c. = 2.366 tan 30
o
= 1.366 m
From Table 8.1 the c.g. of v.tail is at 40% of its m.a.c. Consequently, the distance
between the root chord of v.tail and c.g. of v.tail is :
1.366 + 0.4 x 2.6 = 2.41 m
(D) Engine
As mentioned in example 6.3, the weight of each engine is 450 kgf.
Hence, from Table 8.1, the installed weight of the two engines is :
2 x 1.3 x 450 = 1170 kgf = 11478 N = 5.5 % of W
g

As regards the location of c.g. of the engine, Ref.1.24, chapter 11, mentions that
for gas turbine engines the location of c.g. from the engine inlet is between 30 to
45% of engine length.
In the present case the engine length is 2.13 m. From Fig.6.7a it is assumed that
the engine inlet is just behind the boss of the propeller. From example 6.3 the
boss of the propeller is at 1.58 m from the leading edge of the wing. Taking c.g.
of engine to be at 40 % of engine length, the location of engine c.g. from the
leading edge of the wing is :
-1.58 + 0.4 x 2.13 = - 0.728 m i.e. 0.728 m ahead of the leading edge of the root
chord of the wing.

Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
4

(E) Landing gear
From Table 8.1 the weight of the nose wheel plus the main landing gear is 4.3 %
of W
g
i.e. 0.043 x 208757 = 8977 N
Out of this total weight, the nose wheel and main wheel account for 15 % and
85 % respectively. Hence, nose wheel weighs 0.15 x 8977 = 1347 N and the
main wheels weigh 0.85 x 8977 = 7630 N.
As regards the locations of the c.g.’s of nose wheel and main wheels, it is
recalled that the nose wheel and main wheels share respectively 10 % and 90 %
of the airplane weight. From example 6.3, the wheel base is chosen as 9.78 m.
Hence, the c.g. of the nose wheel is 0.9 x  9.78 = 8.802 m ahead of the c.g. of
the airplane. The c.g. of the main wheels, as a group, is :
0.1 x 9.78 = 0.978 m behind the c.g. of the airplane.
(F) Fuselage and systems
Wetted area of fuselage is obtained as follows. Various dimensions are obtained
from Fig.6.8c
(a) Nose portion
Length of nose = 0.7 m
Diameter at the end of nose = 1.64 m
Hence, average diameter = 1.64/2 = 0.82 m
Consequently,wetted area or surface area of nose ?
2
p ×0.82×0.7 =1.80m
(b) Cockpit portion
Length of cockpit = 2.54 m
Diameter at the end of cockpit = 2.88 m
Hence, wetted area of cockpit portion ?
2
1.64+2.88
p× ×2.54 = 18.03 m
2
??
??
??

(c) Midfuselage portion
Length of midfuselage = 12.83 m
Diameter of midfuselage = 2.88 m
Hence, wetted area of midfuselage = p x 2.88 x 12.83 = 116.08 m
2

(d) Tail cone
Airplane design(Aerodynamic)   Prof. E.G. Tulapurkara
Chapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras
5
Average diameter ? 2.88 / 2 = 1.44 m
Length of tail cone = 9 m
Hence, wetted area of tail cone  ?
2
p×1.44×9 = 40.72 m
Consequently, wetted area of fuselage is :
1.80 + 18.03 + 116.08 + 40.72 = 176.63 m
2

Using value in Table 8.1, the weight of fuselage is :
176.63 x 24 = 4239.1 kgf = 41585.84 N or 19.92 % of W
g

Remarks:
(i) An expression for quicker,but approximate, estimation of fuselage wetted area
is :
??
f
0.75×l × Perimeter of the cross section with maximum area
In the present case the approximate estimate would be:
0.75 x p x 2.88 x 25.07 = 170.12 m
2

This value is 96.3 % of the value obtained earlier.
(ii) The weight of the fuselage as percentage or gross weight (W
f
/W
g
) appears
high. However, as noted below Table 8.1, the quantity (W
f
/W
g
) includes weights
of fuselage, furnishing, consumables etc. Reference 1.19, chapter 8 presents
information on weights of various items, as percentage of W
g
, for different types
of airplanes. For the turboprop airplane the data indicates that the weight of
fuselage structure, furnishings and consumables would respectively be 9 to11 %,
4 to 6 % and 1.5 to 2 % of gross weight. These would add upto 15 to 17 % of
gross weight. Further as noted in Remark (i), after the subsection 8.2.2, the
weight of fuselage, along with furnishing etc., and the systems is taken as one
unit. Accordingly,
landinggear fuselage system empty wing engine
h.tail V.tail
gg ggg g g
W W+W W W W
WW
=- + + + +
WW WWWW W
??
??
??
??

From example 3.1, (W
empty
/W
g
) is 0.559.
Hence,
fuselage system
g
W+W
W
=  0.559 – (0.117 + 0.0141 + 0.0164 + 0.055 + 0.043) = 0.3135
```
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