The basic flight condition is steady symmetric flight, in which all the lateral variables /3, p, г, ф 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 8r 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 is a condition of nonsymmetric rectilinear translation. It is some­times used, particularly with light airplanes, to correct for cross-wind on landing ap­proaches. Glider pilots also use this maneuver to steepen the glide path, since the L/D ratio decreases due to increased drag at large /3. In this flight condition all the time derivatives in the equations of motion (except xE) and the three rotation rates p, q, r are zero. It is simplest in this case to go back to (4.9,2)-(4.9,6), from which we derive the following:

AY + mgф cos в0 = 0

AL = 0 (7.8,1)

AN = 0

We use (4.9,17) for aerodynamic forces, and for the control forces use the following as a reasonable representation:

(7.8,2)

We now add the assumption that в0 is a small angle and get the resulting equation

(7.8,3)

In this form, v is treated as an arbitrary input, and (5r, 8a, ф) as outputs. (See Exercise 7.6.) Clearly, there is an infinity of possible sideslips, since v can be chosen arbitrar­ily. Note that the other three variables are all proportional to v. We illustrate the steady sideslip with a small general aviation airplane3 of 30-ft (9.14 m) span and a gross weight of 2400 lb (10,675 N). The altitude is sea level and CL = 1.0, corre-

Table 7.2

 c, cn p -0.14 -0.0689 – 0.0917Q 0.01326 + 0.017C2 p -0.039 -0.441 -0.00109 – 0.0966Q r 0.165 -0.0144 + 0.27 CL -0.048 – 0.0238C2 0 -0.0531 0.005 0.117 0.0105 -0.0509

sponding to a speed of 112.3 fps (34.23 m/s), and the wing area is 160 ft2 (14.9 m2). The nondimensional derivatives are given in Table 7.2, from which the numerical system equation is found from (7.8,3) to be (see Exercise 7.5)

 " 280.7 0 2400 “ 4“ “ 2.991 " 755.7 -3821.9 0 = 102.93 -3663.5 359 0 _ -19.394

It is convenient to express the sideslip as an angle instead of a velocity. To do so we recall that /3 = v/u0, with u0 given above as 112.3 fps. The solution of (7.8,4) is found to be

8r//3 = .303 8Jl3 = -2.96 фф = .104

We see that a positive sideslip (to the right) of say 10° would entail left rudder of 3° and right aileron of 29.6°. Clearly the main control action is the aileron displacement, without which the airplane would, as a result of the sideslip to the right, roll to the left. The bank angle is seen to be only 1 ° to the right so the sideslip is almost flat.

We define a “truly banked” turn to be one in which (1) the vehicle angular velocity vector to is constant and vertical (see Fig. 7.22) and (2) the resultant of gravity and centrifugal force at the mass center lies in the plane of symmetry (see Fig. 7.23). This corresponds to flying the turn on the tum-and-bank indicator.[19] [20] It is quite common for turns to be made at bank angles that are too large for linearization of sin ф and cos ф to be acceptable, although all the state variables other than ф and V are small. Thus we turn to the basic nonlinear equations in Sec. 4.7 for this analysis. The large bank angle has the consequence that coupling of the lateral and longitudinal equations oc­curs, since more lift is needed to balance gravity than in level flight. Thus not only the aileron and rudder but the elevator as well must be used for turning at large ф.

 Figure 7.23 Gravity and acceleration in turn.

The body-axis angular rates are given by

(7.8,6)

We now apply the second condition for a truly-banked turn, that is, that the ball be centered in the tum-and-bank indicator. This means that the vector mg — mat, where ac is the acceleration vector of the CG, shall have no у component. But mac is the resultant external force f, so that from (4.5,6)

mg — mac = mg — f = —A

where A is the resultant aerodynamic force vector. Thus we conclude that the aerody­namic force must lie in the xz plane, and hence that Y = 0. We consider the case when there is no wind, so that

(,UE, VE, WE) = (U, V, W)

and choose the body axes so that ax = w = 0. We now use (4.7,1) with all the vari­ables constant and only и and ф not small to get:

Y = —me sin ф + mru = 0 (a)

(7.8,7)

Z = —mg cos ф — mqu (b)

When v is small, a reasonable assumption for a truly banked turn, we also have that и = V, the flight speed. It follows from (7.8,7a) that

rV

sin ф = — g

and with the value of r obtained from (7.8,6)

ojV

tan ф = —- (7.8,8)

g

The load factor nz is obtained from (7.8,7b):

Z qV

n = — —– = cos ф – I—–

mg g

With q from (7.8,6) this becomes

Vco sin ф

n = cos ф ————

g

By using (7.8,8) to eliminate Vco we get

(7.8,9)

We note from (5.1,1) that Z = —L in this case, so that n = LAV = n.. The incremen­tal lift coefficient, as compared with straight flight at the same speed and height, is

We can now write down the equations governing the control angles. From (4.7,2), to first order, L = M = N — 0, so we have the five aerodynamic conditions

Q Cm C„ Су — 0 and Д CL = (n — 1 )CW

On expanding these with the usual aerodynamic derivatives, we get

Сф + Clpp + Clrr + ClsSr + Cls8a = 0 Cm, Aa + Cmqq + CmsA8e = 0 Cnffi + Cnpp + Cnr + Cns8r + Cns8a = 0 Cyfi + Cypp + Cyr + Cys8r = 0

Я„Ла + С/С/ + CLsA8e = (n — 1 )CW

In these relations p, q, r are known from (7.8,6), that is,

(7.8,12)

When (7.8,9) is used to eliminate ф from (7.8,14), and after some routine algebra, the solution for Д<5е is found to be (see Exercise 7.6)

Except for far forward CG positions and low speeds, the angles given by (7.8,15) are moderate. The similarity of this expression to that for elevator angle per g in a pull – up (3.1,6a) should be noted. They are in fact the same in the limit n —> The eleva­

tor angle per g in a turn is therefore not very different from that in a vertical pull-up.

Finally, the lateral control angles are obtained from the solution of (7.8,13).