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Izl = 2~pU2Sb (6.172)
Ixz] = pU S- (6.173)
However, Eqs. (6.169) and (6.170) are not in the standard state-space form because
t3, p, and r terms appear on their right hand sides. To get around this problem, we
566 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
proceed as follows. From Eqs. (4.394) and (4.396), we have
. ACt = plxi - Ixzlr
ACn = r Izi - Ixzl p
(6.174)
(6.175)
Multiply Eq. (6.174) by Izi and Eq. (6.175) by Ixzl, add the two resulting equations,
and simplify to obtain
P = I;i ACt + 4zi ACn
r = I;z] ACt +I:iACn
where
Ix I
I;1 = Il z~2 1
Izl
I;l = Il z~2z~
Ixzl
I:zl = Ixl z~x2z~
Then,
(6.176)
(6.177)
(6.17 8)
(6.179)
(6.180)
p = (Cip I;] + CnPCzi) Ap + (CiBbi I;i + Cnpbi Ixzi) AB
+ (Cipbi Izl + Cnpbi Ixzi) p + (Ctrbi Izl + Cnrbi Ixzi) r
+. (Ctl.l;l + Cn8"lj:zl) A8a + (Ct8,1;1 + Cnai,lrzi) A8r (6.181)
r = (C,.p Ixl + CiPI:zi) Ap + (C Bbi I;i + CtBbi I;zi) Ap
+ (Cnpbi I;rl + Cipbi Ixzi) P + (Cnrbic + Ctrbi Uzi) r
+ (Cn8" Ct + Cthl;zi) A8a + (Cnt.lxl + Cl8,l:czl) A8r (6.182)
Now we have to substitute for p from Eq. (6.168) in Eqs, (6.181) and (6.182)
and do some simplification. With this, Eqs. (6.168), (6,181), and (6.182) can be
expressedin the standard state-space form as follows: .
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