直升机飞行力学复习题答案

1、Chapter OneA helicopter of central articulated rotor makes a level flight with cruse speed. In this flight condition, the pitching attitude angle is : - -20, longitudinal cyclic pitching angle is Bi 70, rotor longitudinal flapping angle is a1s = -30. Assuming the tilted angle of rotor shaft is :. =

2、00, please determining the following angles: Helicopter climb angle1二 Fuselage attack angle ;二 Rotor attack angle : s =Rotor flapping due to forward speed a10 =中心较式旋翼直升机以巡航速度前飞。俯仰角 -2。，纵向周 期变距7 ,纵向挥舞角-3 ,旋翼轴前倾角0 平飞，爬升角0机身迎角-2 桨盘平面迎角-2 吹风挥舞4Chapter Two1. For the main/tail rotor configuration helicopt

3、er, the pilot applies which stick or rudder to control what kind of surfaces and corresponding aerodynamic forces?2. Why the gradient of control stick forces can t be too large or small?3. Co-axis, tandem and tilted-rotor helicopters have no tail rotor. How to change the direction in hover for these

4、 helicopters?1. 操纵一一气动面一一响应P13表2-1前推/后拉杆纵向周期变距，桨盘前倾/后倒前飞/后飞，俯仰左推/右推杆操操横向周期变距，桨盘侧倒操操侧飞，滚转油门/总距杆操操改变总距操操改变垂向速度脚蹬操操改变尾桨总距操操改变航向2. 为什么杆力梯度不能太大也不能太小P16太大时大操纵较吃力，太小了不易感觉当前位移量。杆力梯度适中有利于精确操纵。3. 共轴、纵列、倾转旋翼机如何悬停转弯？共轴操操上下旋翼总距差动纵列操操前后横向周期变距一个向左一个向右倾转旋翼操操一侧后倒一侧前倒Chapter Three1. The rotation direction of Black ha

5、wk helicopter rotor is anticlockwise from the top view. Please answer the following questions:How the rotor flaps when the helicopter comes up gust in hover?How to adjust the controller to keep the hover condition?How the blade changes its feathering angle following the pilot s adjustment?2. During

6、left and right turn flights with constant speed and altitude, is there any different for pilot controls?3. Is there flapping motion for hinge-less rotor?1 . 黑鹰直升机旋翼右旋悬停时遇到阵风旋翼如何挥舞，此时如何操纵前方来的阵风会导致旋翼后倒，右倒。应向前、向左推杆。前推杆导致90桨距变小，270桨距变大左推杆导致180桨距变小， 0桨距变大2 .不改变高度和速度时，左转弯和右转弯有区别没？以右旋直升机为例，左转时需要增加尾桨总距，为了平衡

7、尾桨拉力的增加需要左推杆以增加侧向力，这却导致了旋翼拉力的降低，进而需要提总距，又导致向前拉力的增加，因而需要后拉杆。总之，右旋直升机前飞左转弯的操纵为，踩脚蹬左压杆提总距后拉杆。同理， 右旋直升机右转弯的操纵为：松脚蹬减小左压杆减总距减小前推杆3 .无铰式旋翼有挥舞吗？没有挥舞铰但是通过桨叶根部的柔性段或者桨榖柔性件的弹性变形实现挥舞运动。Chapter FourWhich items in the linear equations of helicopter motion reflect the longitudinal and lateral cross-coupling effect?

8、直升机线化运动方程中哪些项体现了横纵向的耦合P53操纵输入系数矩阵C 中， FX 对横向周期变距A1 、 FZ 对纵向周期变距B1 的导数均不为0，即横向周期变距对后向力有影响，而纵向周期变距对侧向力也有影响。Chapter Five1. What is the mission of helicopter trim calculation?2. In order to keep the center of helicopter mass in the specified range it is necessary to check the helicopter loadings before

9、flight. Why the center of helicopter mass must be in the specified range?3. The tail rotor of some helicopters is lower than main rotor. What is the advantagesand disadvantageswhen the tail rotor is lower than main rotor?4. Bell-412 helicopter releases left and right buoys for making an urgent ladin

10、g on the water surface. Compare the differences of pilot controls and helicopter attitude before and after releasing buoys.配平的任务是什么根据平衡条件确定直升机稳定飞行所需的操纵量和直升机的姿态角。为什么重心在特定的范围内重心太前抑制直升机后飞及悬停低速性能，而太靠后则影响大速度前飞性能以及稳定性，因此重心需要限制在一定的范围内。尾桨低于旋翼的好处及坏处P70好处：尾梁扭转载荷不大，不需要加固垂尾以安装尾桨。总之可以减小结构重量。Chapter Six坏处：前飞时尾桨桨盘

11、处于旋翼下洗流中，直升机需侧倾一个角度以平衡尾桨、旋翼高度差所引起的滚转力矩。比较有无浮筒时驾驶员的操纵增加浮筒，相当于在前飞时增加一个作用于重心以下的后向力，相比于无浮筒时驾驶员的操纵，需要增加纵向周期变距、提高总距、减小横向周期变距、增加尾桨总距。1. The feathering angle of horizontal tail is negative. Analysis the speed stability and pitch damping if the feathering angle of horizontal tail is positive.2. Analysis the

12、effect of flapping hinge offset on the speed stability, attack angle stability, diheadral effect and yaw stability.3. What is the different of longitudinal hover oscillation from lateral hover oscillation?4. Why co-axis helicopter always uses two vertical tails?分析速度稳定性及俯仰阻尼，当平尾正装的时候稳定前飞时，前方来流速度的增量将导

13、致平尾上产生一个额外的低头力矩，因而是速度不稳定的。当直升机以俯仰角速度绕Z 轴转动时，比如抬头，将使得平尾的迎角增加，产生额外的低头力矩，阻碍直升机抬头，因而是阻尼力矩。分析挥舞铰偏置量对速度稳定性、迎角稳定性、上反效应及航向稳定性的影响。当飞行速度增加时，来流左右不对称性增加导致桨尖平面后倒，所引起的桨榖力矩使得直升机抬头，因而挥舞铰偏置量对速度的变化是起静稳定作用的。当机身迎角增加时，前行桨叶升力增加的多些，后行桨叶升力增加的少些，导致桨尖平面后倒，所附加的桨裁力矩也使 得直升机抬头，因而挥舞较偏置量是迎角静不稳定的。当直升机向右倾斜时，拉力分量导致直升机向右移动，进而 导致旋翼前后气流

14、不对称引起桨尖平面左倒，所附加的桨裁 力矩也使得机身左倾，因而是静稳定的。纵向悬停振荡模态与横向悬停振荡模态的区别纵向悬停振荡：该模态类似于单摆运动，表现为速度与俯仰 角两个运动变量之间的耦合，但由于旋翼是迎角静不稳定 的，该模态往往是振荡发散的。横向悬停振荡：与纵向悬停振荡类似，但由于尾桨垂尾的存 在，该模态除了滚转角与侧向速度两个变量的耦合，还附加 了航向的摆动；如果尾桨垂尾提供的侧向阻尼足够大，该模 态将是振荡收敛的。为什么共轴双旋翼需要两个垂尾因为没有尾桨，共轴双旋翼需要更大面积的垂尾来提供航向 稳定性、横向运动阻尼、航向运动阻尼。从减轻结构重量的 角度上讲，需要面积相同时，双垂尾比单

15、垂尾更有利。Chapter Seven1. What is the physical meaning of helicopter control bandwidth?2. Analysis the effects of following factors on the damping, control sensitivity and time constant of helicopter vertical motion.Increasing rotor solidityIncreasing rotor rotational speed3. There is no oscillation for

16、 the response of helicopter to collective pitch control, while the responses of helicopter to stick or rudder are oscillation. Why?带宽的物理含义直升机对于低频操纵有较好的跟随性，操纵响应幅值与操纵量成正比，相位之后也较小；在进行高频操纵时，直升机跟 随性变差，滞后变大且响应幅度减小。因而直升机存在一个 操纵频率的适应范围，称为带宽。分析下列因素对直升机垂直运动的阻尼、操纵灵敏度及时间常数的影响增加实度增加桨尖速度仅仅考虑旋翼的作用Fy 4?，1R2 R2 1(vy) 2322Fy 7 J DR2-：R2H ，6_V1_2Fy =D JR 二R ra.4d. Vy VmFyy Vy Fy，：7dt. ：Vy(S)Fy 7 k=v 二17(s) ms -Fyy Ts 1其中 k=-Fy*/F：y =|cR, T=mF：y当旋翼实度增加时，运动阻尼增加，操纵灵敏度不变，时间常数减小。当桨尖速度增加时，运动阻尼增加，操纵灵敏度增加，时间 常数减小。直升机对总距控制的