曝光臺(tái) 注意防騙 網(wǎng)曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費(fèi)者

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
at 100 mts in level fiight at sea level with one engine completely out.
Solution.    Yawing moment due to asymmetric thrustis given by
Nr - Thrust *distance
        ~  10,000 * 5.0
      - 50,000 Nm
The yawing moment due to rudder defiection:
For equilibrium,
Nr = ;:p V2SbCn2Sr8r
Nr + Nr -0
so that
STATIC STABILITY AND CONTROL
Fig. 3.92    De\relopme/t ofsideslip due to bank.
8r - pV~;VC
                             -2 * 50,000
-   .225~0~2~50~~0~0 00 )
= 16.32 deg
3.6 LateraIStability
       Lateral stability is the inherent capability ofthe airplane to counter a disturbance
in bank. In level flight, both wings are in a horizontal plane and the bank angle is
zero. However, because of some disturbance,if the airplane banks but very slowly
so that the roll rate is negligibly small, then there is no aerodynamic mechanism
to generate a restoring rolling moment unless sideslip develops. Therefore, the
airplane is neutrally stable with respect to a disturbance in banikv without sideslip.
Fortunately, once banked, the airplane develops a sideslip in the direction of the
bank because ofa spanwise component ofthe weight W sin ~ as shown in Fig. 3.92.
As a result of this sideslip, if a restoring rolling moment is induced, then the
airplane is said to be Iaterally stable. Once the wings are back in level condition, the
disturbances in bank angle and sideslip are eliminated, and the airplane returns to
its original, steady level flight condition. On the other hand, if the induced rolling
moment causes the bank angle to increase further, generating more and more
sideslip, the aircraft is said to be laterally unstable. If the induced rolling moment
　
中國航空網(wǎng) m.k6050.com
航空翻譯 www.aviation.cn
本文鏈接地址：PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL2(74)