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
Fig. 3.13c.
From Figs. 3.13a and 3.13b, we obtain ao/(ao)ttieory = 0.887 and (ao)cheory =
6.58 so that ao - 5.8365/rad. Using Fig. 3.35, we obtain (a8)a/(a8)C, = 1.083.
Using the data of Fig. 3.36, we obtain Kb - 0.55. From Fig. 3.37b, we obtain
Cl8/(Cl8)trieory = 0.817 and, from Fig. 3.37a, (CLa)aleory = 3.77]rad so that C/8 =
3.08lrad. Then, subsrituting in Eq. (3.98), we obtain r - 0.3143.
Example 3.6
Estimate the hinge-moment coeffcients for a horizontal tail fitted with an eleva-
tor and having the following data: A = 4.0, Ac/4 = 45 deg, AHL - 45 deg, ct,/c f =
0.09, cf /c = 0.16, Reynolds number Ri - 6.1 x l06, airfoil section NACA
65A006, rk - 0.25, r7o = 0.65, and tcl2cf - 0.08.
So/ution. The first step is to calculate the balance ratio as given by
BR -.
-. 0.0412
The next step is to evaluate various parameters appearing in Eq. (3.121).
From Fig. 3.46, we note that Chabal/C;ra, = 1. Using Fig. 3.13c, we obtain
4TE -. 7 deg for the given 65A series airfoil with 6% thickness ratio so that
tan 4-rE]2 = 0.0616. Then, from Fig. 3.13a, we obtain ao/(ao)meofy - 0.887 and,
from Fig. 3.13b, (ao)theory = 6.58/rad so that ao = 5.8365/rad. With this and
using cflc = 0.16 from Fig. 3.47a, we obtain, C;a/(Chtr)theory = 0.65. Using
Fig. 3.47b for t]c -. 0.06, we get (Cha)theory = -0.39 so that cLa = -0.2535lrad
and Chaw - -0.2535]rad.
We have c,lc f = 0.09 (streamwise) or cZ]c; = 0.09]cos 45 deg = 0.1272 (nor-
mal to quarter chordline). Similarly, with c f ]c = 0.16, we get c}-lc' = 0.226.
We have i7i = 0.15 and r7o = 0.65. Using these values and Fig. 3.48b, we get
(Kct)", = 1.45 and (Ka)0o = 2.4. Then, using Eq. (3.125), we obtain Ka - 0.6188.
Furthermore, using Fig. 3.48c, we get B2 - 0.92.
From Fig. 3.48a, for A = 4.0, we get ACha,/aoB2Ka cos Ac/4 ~ 0.012. With
all these values, we get ACha - 0.0296/rad. Substitution in Eq. (3.121) gives
Chtr = -0.1028/rad or -0.0018/deg.
Now let us calculate Ch8.e. From F,g. 3.49, we firicl (Ch8.e)bal/CL8.e = 1. Ffom
Fig. 3.50a, using c f /c = 0.16 and aol(ao)rriemy = 0.887, we find cL8.e/cch8.e)theory
=0.90. Using t/c = 0.06 and cf /c = 0.16 from Fig. 3.50b, we find (Cha.e)a}eory
= -0.83 so that C28.e = -0.747lrad and Ch8.ew -. -0.747.
248 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Using Fig. 3.37b, we obtain Cl8/(CL3)theory = 0.817 and, from Fig. 3.37a,
(Cl8)theory = 3.77 so that CLb - 3.08]rad. We have cra = -C18/ao. Then, with
ao -. 5.8365, we get a8 - -0.5277. Using cj{ld = 0.226 and A - 4.0 from
Fig. 3.51a, we get ACh8, = 0.0156(CL8 B2Ka cos Ac/4 cos AHL).
We have T71 - 0.25 and T,o - 0.65. With these values and from Fig. 3.51b.
we obtain (K6),,, = 1.30 and (K8),ro = 2.25. Then, using Eq. (3.130), we obtain
K8 = 0.4688. Further substitution grves AChB.e - 0.01036/rad.
Finally, substitutingin Eq. (3.126), we find Cha.e -. -0.3457/rad or -0.0060/deg.
3.4 Stability in Maneuvering Flights
The class of flight paths when the load factor exceeds unity is called maneuvers.
The load factor n is defined as the ratio oflift to weight.
L
n = W (3.171)
Therefore, during a maneuver, the lift exceeds the weight and the airplane struc-
turc is subjected to higher stresses than those encountered in steady level flight
with aload factor ofunity.ln general, during a maneuver, the airplane experiences
accelerations, which makes it necessary for us to consider inertia forces in the
analysis. We will study such motions later in the text. However, by introducing
some simplifications and ignoring inertia forces, we can extend th~ methods of
static stability and controlto study simple maneuvers like the prdl-up from a dive in
a vertical plane and the coordinated turn in a horizontal plane. Even though not ex-
act, this approach gives us an idea about thelevel of stability, control requirements,
and the stick force gradients that can be expected in such maneuvers.
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