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system to the Ox2Y2z2 system. Performing the indicated matrix multiplications,
::l:o]
O
1
0
[:,::/
1
y;
a)
::g0']
il:%]
c)
326                PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
we obtain
Then,
X2 = T12Xi
(4.22)
(4.23)
where X denotes a vector expressed in matrix form and its suffix indicates the
coordinate system in which ffie 'vector X  has its components. For example, Xi  is
a vector having its components in the OxiYizi axes system.
    Each transformation matrix A, B, or C is an orthogonal matrix. An important
property of an orthogonal matrix is that its transpose is equal to its inverse. The
transformation matrix T12, which is the product of the three orthogonal matrices,
is also orthogonal. Therefore,
T21 = (Ti2)-1 = (Ti2)t
sin0 sin 4 cos yr - sin ~ cos 4
sin Vr sin 0 sin 4 +cos yr COs 4
                    sin 4 cos 0
(4.24)
(4.25)
where superscripts -1 and ' denote the inverse and transpose of a matrix, respec-
tively.
  Transformation of vectors.  Let us consider the transformation of vectors
between various coordinate systems as follows.
  Inertial to Earth:[rxed system.  Here we are considering transformation be-
tween the Oxryizt and OxEyEzE systems. Let the OxEyEzE system coincide
with  Oxt yrz, system at t - 0. For t  > 0, ltr  - fZet, 0 - ~ - 0, where S-2e iS the
angular velocity of the Earth about Ozi or OzE axis. Then using Eq. (4.22),
XE = riExi
       cos S2et    SiD CZet   0
-. -sinS'2et COSS2et O Xi
  OO1
The reverse transformation is given by
where
Xi = T~XE
rE = (jriE)-l =
GrE)'
(4.26)
(4.27)
(4.28)
(4.29)
EQUATIONS OF MOTION AND ES11MATION OF STABILITY DERIVATIVES 327
n
Zi
Fig. 4.5    Inertial and nav:igational coordinate system,s.
   Inertial system to navigational system    Assume that the navigational system
OxeYeze is located in the northern hemisphere as shown in Fig. 4.5. Let / denote
the longitude and A the latitude of the origin of the navigational system. The
first step is to do a translation of the inertial axes system so that the origin of
the inertial axes system coincides with that of the navigational system. Then we
have to perform two Euler rotations, r/r = L and then 0 - -(90 + A,). With this.
the inertial axes system Oxiytzi coincides with the navigational system OxeYeze.
The third rotation involving 4 is not necessary. Therefore, Sb = 0. Substituting
these values of the three Euler angles in Eq. (4.22), we obtain
(4.30)
Let the vector Xi denote the position ofa particle P in the inertial frame ofreference
Oxiy,z, as shown in Fig. 4.5. We have
Xi - Xoi + Xei
(4.31)
Note that both vectors Xoi and Xet are having components in the inertial frame of
reference Ox,yizi. Then,
Xei = Xr ~ Xt,/
Xe = TreXer
    = ir/e(Xr - Xoi)
Now vector Xe has components in the OxeYeze system and Tet = ~e)'
(4.32)
(4.33)
(4.34)
                                  cos O cosy                                    cos 0 sin p                       -sin O
r12  ~     sine sin4 cos p - sinVr cos 4    sin ~ sin0 sin 4 +cos p cos4    sin ~ cos 0
                  sin O cos 4 cos p + sin p sin 4     sin ~ sin0 cos 4 - cos p sin0     cos 4 cos0
sin0 cos 4 cos p + sin ~ sin 4]
sin p sin 9 cos ~ - cos p sin 0
                  cos 4 coso                -l
      rcos O cos Vr
= l-"o-::,~
              -sin A cos L    -sin A sin/
　
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