曝光台 注意防骗
网曝天猫店富美金盛家居专营店坑蒙拐骗欺诈消费者
velocity and spin radius.
7,,5-2 Balance of Moments
For balance of moments, the sum of all the moments must be zero, regardless
of the axes system chosen* Suppose we consider the airplane motion with respect
to the spin axis; then the centrifugal forces acting on all the components of the
airplane acting radially outward from the spin axis do not produce any moment
about the spin axis. Therefore, the net aerodynamic moment about the spin axis
must also be zero. This means that the resultant aerodynamic force must also pass
through the spin axis. In other words, to determine the equilibrium spin modes,
we simply have to find the combination of angle of attack, sideslip, and spin rate
at which the resultant aerodynamic force passes through the spin axis. However,
the scenario is not this simple. Usually, tre aerodynamic data does not contain
the magnitude, direction, and point of action of thejresultant aerodynamic force.
Instead, we usually have-the aerodynamic data in the form of lift, drag, side force,
pitching, rolling, and yawing moment coefficients in the stability axes system.
Using this information, we have to compute the balance of pitching, rolling, and
yawing moments about the body axes system to determine the equilibrium spin
modes.
For steady-state spin, p = q - r - 0. With this, moment Eqs. (7.77-7.79) take
the following form:
/ = qr(lz - /y)
M = rp(lx - Iz)
N = pq(ly - /x)
(7.110)
(7.111)
(7.112)
Here, L, M, and N denote the net external rolling, pitching, and yawing moments
acting on the airplane duning the steady-state spin. Because we have ignored the
power effccts,the only external moments acting on the airplane are the aerodynamic
moments. The right-hand side of Eqs. (7.110-7.112) represent the moments due
to inertia cross coupling effects. Substituting for p, q, and r from Eqs. (7.91-7.93)
656 PERFORMANCE, STABfLITY, DYNAMICS, AND CONTROL
in Eqs. (7.110-7.112), take the following form:
Q2 .
L = - 2 sin2asiriX(Iz - Iy)
M = S22 sin2a cosX(lx - Iz)
g22 -2
N = - 2 COS2 tX sin2X(ly - Ix)
The above Eqs. (7.113-7.115) can be written as
L+Li - O
M + M, - 0
N+Nt -0
(7.113)
(7.114)
(7.115)
(7.116)
(7.117)
(7.118)
where Lr is the inertia-rolling moment, Mi is the inertia-pitching moment, and N,
is the inertia-yawing moment as given by
Li = qr(ly - Iz)
~2
= 22 sin2asinX(Iz -Iy)
. Mt -rp(lz - Ix)
Q2
中国航空网 www.aero.cn
航空翻译 www.aviation.cn
本文链接地址：
动力机械和机身手册3(114)
