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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
at t - 0. Therefore,
Xo, = [ R:e ]
EQUATIONS OF MOTION AND ESTfMATION OF.STABILITY DERIVATIVES 349
and L - }. = 0. Then, using Eq. (4.32),
00
Tte= 0 : ']
-1 0
Furthermore,
Xe = Tte(Xt _ Xoi)
or
xt + 36,250
X;e:l = I-i : '] [R'6+2,50050] _ [-] , '] [R:']
Ze 0
= [-32gOj250]
Example 4.2
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