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Nm.cum = ac.w - (:C/- f +-
a ,,
[(1-: )+2p-, (1+:,)] (3.233)
d8e CL(Xc8 - Nm.rum)
d7z == -- Cm8
(3.234)
\/2
R = g~~_j
,~7
sin ~: v
n
sl : g/~-/
V
JV~-
-6{-
I
256 PERFORMANCE, STABILiTY, DYNAMICS, AND CONTROL
N:.tum = xac,w - (:C/ )f +
x (1 - ,CC,,,.)
(d/.),um=G,(Vg),7tSeCeCc.,&~.),,._ (3.236)
Example 3.7
A trainer aircraft is initially in a steady level flight at an altitude of 5000 m
(cr = 0.6) at a forward speed'9f 100-0 m/s. The aircraft then climbs to 6000 m at
which it enters into a dive and recovers approximately in a semicircular path so
that, at the bottom of the pull- out, it is once again at 5000 m altitude.
Determine 1) the elevator setting above that required for initiallevel fligh~/an2d
2) the stick force per g, based on the following data: wing loading = isoo r
xac.w = 0.24, xc8 = 0.25, Cm f = 0.15CL, aw = 0.1, ar -0.08, t - 0-5,lr - 8 m,
VI :0.6, e - 0.35a, Ctia - - 0.003]deg, Ch8.e - - 0.006/deg, r7r = 0.9, Gi -
1.0 rad/m, Se - 2.0 m2, and Ce = 0.6 m.
So/ution. From the information grven, we find that R - 1000 m. Then,
2W
CL = p\/-2S
2 *1500
= ~225~ 0.6*1002
- 0.4
W
ALi - pgSr
1500
= ~225~0 6~9.81* 8
-. 25.4842
Cm8 = -a,Virlrr
-. -0.08 * 0.6 * 0.9 * 0.5
- -0.0216
Nv = ac.w - (-,.,)f -+ V 7 [(1- : )+-,,,]
-. 0.24 - 0.15 + 0.08 *0.6*0.9
Ot)9) [1- 0.35+ 20~25 54~8432 ]
- 0.4193
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动力机械和机身手册2(14)
