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In the abo've calculations, we have assumed CDu = CLu - 0.
                                          Example 4.12
    For the generic wing-body configuration shown in Fig. 4.38, estimate the sta-
bility derivatives CLq, Cmq, CUt, and Cmd fQr a flight Mach number of 0.10 at sea
level and based on the following data.
Fig.4.38   Generic wing-body ofExample 4.12.
0-85m
EQUATIONS OF MOTION AND ESTIMATION OF STABILI-fY DER"VATIVES 429
                          Table 4.1  Fuselage
                                                                         geometrical data
x.
m
bf,
m
o
0.25
0.50
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
9.5
10,0
0.05
0.4
0.50
0.60
0.70
0.8
0.825
0.85
0.85
0.825
0.80
0.65
0.5
0.3
    Exposed wing area Se = 31.0 m2, total (theoretical) wing area S = 35.94 m2,
exposed aspect ratio Ae -3.22, theoretical aspect ratio A = 3.27, exposed span
be = 10 m, leading-edge sweep ALE = 45 deg, exposed root chord Cre  -. 5.6  m,
tip chord cr  = 0.6 m, total (tip-to-t.ip) span b:10.85 m, maximum fuselage cross-
sectional area Snmax  = 0.5675 m2, fuselage length l.f  = 10 m, maximum fuselage
width b f,max  = 0.85 m, fuselage apparent mass coefficient k2 - ki - 0.95, and
distance of center of gravity from fuselage nose Xcg =5.0 m. Assume that the  u
cross section of the fuselage is circular and the variation of the width/diameter
along the fuselage axis is shown in Table  4.1.
  Solution. Wehave
            (CLq)WB = [KW(B) + KB,W,] (SS   )(C,q)e + (Clq)B (S  S~f )
From Chapter 3, we have
     KWB=0.1714(bb )2+0.8326(bfb )+0.9974
     KBW=0.7810(bb )2+1.1976(b )+0.0088
With b f,max - 0.85 m and b -.10.85 m, we obtain K}w:1.0636 and KBW  -.
0.1074.
   We have
                                                   (CLq)e = (~ + 2€) (CLtr)e
;\.
 : /;
 : :t
.VI
.
'a
 t tA/
~..
!t
ct
~
n.
.N
~
r:
/?
':ri
 . :N
 .'
  .
..
ir
 ..
430              PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
where
 g-I
       ct
x = (Xac)e - Xcg,le
The exposed taper ratio A,t  - 0.6]5.6  -. 0.1071. The data on Xac givenin Fig.   3.19
is for A, - O and 0.2. Using these two sources of data, we find that the mean value
of (Xac/Cr)e = 0.575 for Ae -. 0.1071. Then, we get (Xoc)e - 0.575 *5.6 -- 3.22.
We have x  = (Xac)e - (Xc8)lc = 3.22 - 2.0 -  1.22. The mean aerodynamic chord
is given by   .
                          -= (23 )(:A~+A )
Substituting Ae  = 0.1071 and Cre = 5.6 m, we get Ce  - 4.1058 m and g  : 0.2971.
Similarly, with theoretical root chord Cr - 6,025 and theoretical taper ratio A, =
0.09958, we get the theoretical mean aerodynamic chord c - 4.0529 m.
       The subso:ruc lift-curve slope of a straight tapered wing is given by the following
expression:
CLat -
27r A
2+
where 8  = ^~~ and the midchord sweep angle Ad2 is given by the following
expression:
tfulAc/2=tariA.E-( b )
With ALE  - 45 deg, Cre = 5.6 m,cr  - 0.6 m forthe exposed wing, we get tan Ac/2  =
　
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