曝光台 注意防骗
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altitude of18,000 m, we find that the Reynolds numberis equal to 11.20x 10J. From
STATIC STABILITY AND CONTROL
291
Fig. 3.74, we get K Rt: 1.97. Substituting in the above expression for fuselage
contribution, we get (Cnp)B(W) - -0.00242]deg.
Now let us approximately evaluate vertical tail contribution. We have Av.eff =
2.4236, Au - 0.2951, and ALE., = 45 deg. From Fig. 3.14, we get t3 (CNcr)theory =
3.85/rad, where [3 : -/_ f = 1.732. -fhis gives (CNcr)theory = 2.2228/rad_
Note that p here is not sideslip angle.
Now we need to apply the sonic leading correction to the above value of CNa
using the data given in Fig. 3.15a. We have Ay = 2.5 so that AYi = Ay/cos A =
3.5360. With this, we obtain from Fig. 3.15a, CNa/(CNcr)theory = 0.825 so that
CNa - 1.8338/rad or 0.0320/deg, and ay = 0.032/deg. All the other values in the
expression for vertical tail contribution remain unchanged. Substituting, we obtain
(C,zp.v)rix = 0.00249ldeg
Then,
Cnp = (Cnp)B(W) + (Cnp.V)fix
- -0.00242 + 0.00249
= 0.00007/deg
We observe that the given aircraft has marginal static directional stability at M =
2.0. However, it should be noted that this result is based on a very crude estimation
of vertical tail contribution.
Example 3.9
An aircraft is ready for takeoff when it is detected that a crosswind of 8 m/s
is blowing across the runway. Determine the rudder angle required to maintain
a steady normal heading along the runway at unstick point using the following
data.
Wing loading ( W/S) = 2500 N/m2, span = 25 m, wing area = 70 m2, unstick
velocity -. 1.2 Vsrall, maximum lift coefficient - 1.8, lift-curve slope of t.he verti-
cal tail= 0.08/deg, (Cnp)fix = 0.012/deg, vertical tail volume ratio = 0.25, and
r7u - 0.9. Assume that 1 deg of rudder deflection changes the vertical tail incidence
by 0.4 deg.
So/ution. Assuming sea level conditions (p = 1.225 kg/m3), the stalling ve-
Iocity is given by
Vstall =
- 47.6191 m/s
t?.z:.
; 9tj
.:, -:""-.
b-: :
a- :
292 PERFORMANCE, STABtLITY, DYNAMICS, AND CONTROL
We have
Assuming k - l, we have
Vunstick - 1.2 Vsrall
- 57.1429 m/s
p = tan-l 57.18429
- 7.9696 deg
Cn8r - -kay V2r7yt2
Cn8r - -0.08 * 0.25 * 0.9 * 0.4
- -0.0072ldeg
-.p(C,,p)tjx
8r -.-
CnBr
= 13.2827 deg
The rudder should be deflected to the left by 13.2827 deg.
Example 3.10
A twin jet engine has the following data: thrust per engine = 10,000 N, span-
wise distance between the two engines = 10 m, w~~ area =50 m2, wing span =
10 m, rudder effectiveness (Cn8r) = -O.OOlldeg, and maximum permissible rud-
der deflection -.:+::20 deg. Determine the rudder deflection to maintain zero sideslip
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