LATERAL STEADY STATES
The basic flight condition is steady symmetric flight, in which all the lateral variables /3, p, r, <f> are identically zero. Unlike the elevator and the throttle, the lateral controls, the aileron and rudder are not used individually to produce changes in the steady state. This is because the steady-state values of /3, p, г, ф that result from a constant 8a or 6T are not generally of interest as a useful flight condition. There are two lateral steady states that are of interest, however, each of which requires the joint application of aileron and rudder. These are the steady sideslip, in which the flight path is rectilinear, and the steady turn, in which the angular velocity vector is vertical. We look into these below before proceeding to the study of dynamic response to the lateral controls.
THE STEADY SIDESLIP
The steady sideslip is a condition of nonsymmetric rectilinear translation. It is sometimes used, particularly with light airplanes, to correct for crosswind on landing approaches. Glider pilots also use this maneuver to steepen the glide path, since the LjD ratio decreases due to increased drag at large /3. In this flight condition D = d/dr = 0, and p = r = yE = 0. Thus, with reference to (5.13,17), the only nonzero state variables are /3, ф, and ip. For the control terms we use the following, which is a good representation for conventional airplanes:
With (10.4,1), and the special conditions for steady sideslip, (5.13,17) reduces
QypP + ^w„ cos УеФ + CvsT dr = 0
clfp + cltrdr + cliasa = 0
CnpP + Спіт dr + CnSa ^0 = 0 /3 + f cos ye— 0
The fourth equation is the only one containing p, and may be dropped from consideration. The three preceding ones contain the four variables /5, ф, дт, да. Hence an infinite set of solutions exists, in which any one of the four may be selected arbitrarily. If we choose ф to be arbitrary the equations can be solved for the corresponding /3, dr, da (provided of course that its matrix is not singular). Thus
As an example, consider the jet transport used previously, at CWe — 1.0, ye = 0, with the /3 derivatives as in See. 9.6. In addition to these we need the control derivatives, for which we use
СПт = .067 G4r = .003 СЩг = -.040
G4a = -.065 <7^ = .005
It is evident from (10.4,3) that jj, dr, and da are all proportional to ф, hence the ratios of the angles are constant. The numerical result is:
^ = .0558; ^ = 1.675; ^ = -1.800
I» (8 0
so that for a sideslip of 10°, the other angles are ф = .56°, dr = 16.75°, da = —18.00°. As expected, a slip to the right requires left rudder and right aileron. The control angles are seen to be large; powerful controls are needed to sideslip at large angles. When the matrix A is singular, it only indicates that ф is zero in the sideslip. In that case the equations can be rearranged to put ф on the l. h.s. and /3 on the r. h.s., in which case the new matrix is very unlikely to be singular.