曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

Because 0 is supposed to be in the range -7tl2 < 0 < lr]2, the angle 0 is
uniquely determined by Eq. (4.217). Furthermore, we have from Eqs. (4.203) and
Eq. (4.206),
so that
C33 = COS 0 COS 4 = eg - e? - e~ + e32
C32 = c.os 0 siri~ = 2(e2e3 + eoei)
4 =,os-" (,C, o)
(4.218)
(4.219)
(4.220)
=COS-I 9-ei-e+e),lsgn[2(e2e3+eOel)] (4.221)
    .jC --eOe2):
                          ! - eg- - e
yr=COS-I ~F4-C,:  :--),lsgn[2(e,e2+eOe3)]   c4.222)
                                             'i e3 - eoe
As said before, we will not encounter the singularity at 0 -. 7rl2 if we obtain
the Euler angles using the method of quaternions. We can summarize all three
approaches to calculate the Euler angles as follows:
   1) Euler angle rates: r/o, 00, 4o     body rates p, q, r + Euler rates I4(, 0,
4       p(t), O(t), ~(t).
    2)  Direction cosine matrices (DCM): yro, 0o, qbo, DCM at t = 0 ~ body rates
 p, q, r -* DCM updates equations + DCM at time t -* yr (t), O(t), ~(t).
    3)  Quaternions:  po, 00, ~0 ~ body rates p, q, r      quaternions eo, ei, e2, and
 e3 + DCM ~ rly(t), O(t), ~(t).
Example 4.1
  With respect to an Earth-centered inertial frame, a vehicle has the following
velocity components:
ui(t) = uo, + axit
   vi(t) = ayit
       Wi = O
Assuming Uor = 100 ft/s , ax,  = 25 ft/S2, ay, = 50 ft/S2, determine the position
and veloatjt with reference to the Earth-fixed XEyEZE and navigationafxeYeze
systems. Assume that, at t - 0, Lhe vehicle is located on the equator with / - 0.
where the sgn(-) function has the following properties: if the argument is positive,
then the sgn(-) function returns the value of +1. If it is negative, then sgn(-)
function returns the value of -1. Equation (4.221) uniquely determines the sign
of ~. Similarly, using Eqs. (4.198) and (4.199), we get
348             PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
    Solution.    At t - O, we have xt = Re, yi - Zi -. 0. Here, Re - 2.0973364 x
l07 ft is the radius of the Earth at the equator. For t  > 0, we have
                                  Xr(t) = Xor + [," .tdt
          =Re+uoit,+ant2 '
                                           yt(t) = ay,t2
With ti - 50 s, uo, - 100 ft/s, al/ -25 f1/S2, and ayt =50 ft/S2, we get xt  -
Re +36,250 -. 2.1009614 x 10J ft and y, - 62,500 ft. Note that z, - O.
    The angular velocity of the Earth about the Oz, or OzE axis is 1 rev/day or
With ti -. 50 s, we have
From Eq. (4.27), we have
g2e -
 27r
24~36 O
-. 0.7272 x  10-4 rad/s
cos g2etl - 0.999993
sin f2eti - 0.003636
sin <2etl
,osCZet {l[X:.]
  O
   2.100984
                                                             _1388.61,~5 0' ]
                               0
The altitude is given by
                                he =       -- Re
Substituting for XE, yE, and Re, we obtain he = 36,000.0 ft.
      The origin of the navigational system OxeYeze lies directly beneath the vehicle
　
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