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~g. 6.17    Physical motion of the airplane during a Dutch roll.
a2,= },  C"p + ~8bCCBb )
(6.217)
 1 CnBbi(mi-biCy)+C,.)  (6.218)
an= jl ( C.,  C,Bb
For free response, A8 r  - A8r = 0-
    To have an understanding of the physical parameters that have a major effect
on the damping ratio and frequency of                                  lation,let us introduce
some add      al simplifications. Usu'~_f ybe8B- 3wk:ro~l.;tsa, a~tmall so tlaat we ,an
 assum  od~'t:/to.  _ 0. }/Vith this,it ca                                        characteristic equation
 correspon          to the system given by Eq. (6.214) is given by
                                                  A2 + BA+C = 0                                   (6.219)
where
B=-(C+bjC )
(6.220)
C = (   11  ) [C,pC.rbi + Cnp(JTIi - bi Cyr)]             (6.221)
AIRPLANE RESPONSE AND CLOSED-LOOP CONTROL         575
Comparing Eq. (6.219) with the standard second-order Eq. (6.34), we obtain
COn ==
<=-(2 )(~p+b]C )
(6.222)
(6.223)
Usually, C,tr < 0 (side force due to sideslip), Cnp > O (static directional stability
parameter), Cyr > O (side force due to yaw rate), and Cnr < 0 (damping in yaw).
Generally, Cyr iS small so that mi   >  b] C).r. The majorcontribution to the frequency
term comes from Cnp, and that for damping ratio comes from Cnr.
  Spiral mode.  For the general aviation airplane, we have observed that,
during the slow convergence corresponding to the small negative real root, the
sideslip varies very slowly so that Ap t- O. Hence, the side force equation can
be ignored. Furthermore, the roll rate is practically zero during this slow spi-
ral motion so that tlhe net rolling moment must be zero. Furthermore, we ignore
the contribution due to the product of inertia term Ixzi. With these simplifying
assumptions, the rolling moment and yawing moment Eqs. (6.169) and (6.170)
reduce to
Ctt*Ap + Ctrbir + Ct8a A8a + CL8r A8r ~ 0
Izir = Cnp Ap + Cwrbi + Cn8o A8a + Cn8r A8r
For free response, A8a - A8r - 0.
   Rearranging Eqs. (6.224) and (6.225), we get
.    (-CnpClr + CnrCip)b,r
r --
                                     Iz1 Cip
(6.224)
(6.225)
(6.226)
We can obtain an analytical solution to this equation. Let r =roeA,p' so that
r = ro AspeA.Pr. Then,
ro\speX,Pr - (-Cnp Clr + CnrqO)biroe.,Pr                   (6.227)
                                          Iz I Clp
Asp
bi (Cip Cnr - Ctr Cnp)
         Izl CW
(6.228)
Usually, CLp  < 0 (stable dihedral effec0 and C"  < 0 (darnping in yaw) so that t_he
term CipCnr > O. Also, C,,p > 0 (directionally stable) and Clr > 0
due to positive yaw) so that ClrCnp  > 0. Therefore, when lcipcnrl
Asp < O, and the spiral mode will be stable. Therefore, for spiral stabi
positive rol
" }ClrCn/31
ity, we mus
have Cip Cn,   >  Ctr Cnp. In other words, the airplane should have a relatively higher
　
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