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   (Cmq)B=2(C;na,)B(l-x:') :(V., )]
                                                                 - Xml -
Here,
                VB
Xml=~f Xcl=lf VBI=SB,maxlf
VB .= l," SB(X) dx
         1
x. = V[, [,'f SB(X)X dx
PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Furthermore,
(C;na,)B = (C,nu)B ( S.,f )
                    2(k2 -                    ~l( ) (Xm - X) ck
(Cma)B = ~k~ [,xo ~Sd~
Here, Xm - Xcg - 5.0 m. From Table 4.1, we find that xo = 6.0 m. Performing
the reqtrired caglculations, we get VB = 4.2762 m3, xc = 5.1510 m, Xml  ~ 0.5,
Xcl - 0.515, VB1 - 0.75351, (Crrta)B - 0.9105/rad, (C:na)B = 0.6860, and
(Cmq)B - -1.2820/rad. With these, we get (Cmq)vnl - -1.4355/rad.
   Now let us calculate the acceleration derivatives CLd and Cmct for the given
wing-body configuration. We have   .
(CLa)WB = [KW(B)+ KB,W,](SS  )(C  ).+ (C d)BS S~f 7,ad
where
                                (CU)e = 1.5 (-,, )e (CLa)e + 3CL(g)]rad
   C~(g) =  (- 22-) (0.0013 r4 _ 0.0122.c3 + 0.0317 r2 + 0.0186 c - 0.0004)
                                  p =,W-
      r-pA
Substituting we obtain r = 3.22, CL(g) = -0.6322, and (CLtr)e = 0.60278/rad.
    rfhe body contribution is given by
                             (CLa,)B = 2(C2a)B(S.tf )
Furthermore, using Eqs. (4.533) and (4.534), we obtain (CL&)B -.2.8634/rad.With
these values, we obtain (CLcr)lW = 0.7287/rad.
Finally,
(Cmde)WB = [KWcB)+KB,W)](SS  )(Cmdr)e+ (Cma)BS St:, /,ad
where
                          (Cmdt)e = (Cmi,)e + (-: )(C dl)e
                            (C:nct.)e = -(;7;   (-,, ):(C,.)e + gC, o(9)
    Cmo(9)  =  ( 22   ) (0.0008 r4 _ 0.0075 r3 + 0.0185 r2 + 0.0128 t - 0.0003)
EQUATIONS OF MOTfON AND ESTIMATION OF STABILITY DERIVATIVES 433
Substituting, we obtain Cmo(g):0.3591, (C:_na.)e=-0.8094, and (Cmd,)e =
-0,51189/rad. Furthermore, we have
     (Cma.)B=2(C;na)B1~.., V,,]~)
We have (C;ntr)B = 0.6860. Substitution grves (C,m)B - - 0.06117/rad./7Vith these
values, we get (Cnui)VW~ - -0.5095/rad.
Example 4.13
   An aircraft wing has the following characteristics: leading-edge sweep = 0,
aspect ratio = 6, taper ratio -. 1.0, sectional Iift-curve slope = O.1/deg, zero-
lift sectional drag coefficient CDO - 0.022, and increase in sectional profile drag
coefficient per deg angle of attack CDaJ  - O.OOl/deg.
     Assuming thatit is a midwing configuration, determine Ctp, Cnp, CLr, and Cnr
using strip theory for ct = 5 deg and M = 0.20. Compare your'results with those
obtained using Datcom methods.
    So,ution.   The strip theory gives
                                 (Ctp)W = -g(ao + CDJ)
We have ao - 0.1 * 57.3 -. 5.73/rad and
                             CD,I = CDO,I + CDa,ICL
                                                          -. 0.022 + 0.001* 5 - 0.027
　
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