NUMERICAL EXAMPLE

The rudder and aileron angles in a steady truly-banked turn are calculated by way of example for the same general aviation airplane as was used above for the sideslip. The altitude is sea level, the speed is 125 fps (38.1 m/s) and the stability and control derivatives are as in Table 7.2. The solution of (7.8,13) for climb angles between — 10° and +10° shows that the sideslip angle /3 remains less than 1.5° and the rudder and aileron angles are as shown in Figs. 7.24 and 7.25. The value of CL varies over the range of bank angles used from 0.8 to 1.6, so several of the stability derivatives are significantly affected. It is seen that the aileron angle is always positive for a right turn—that is, the right aileron is down (stick to the left), and that the rudder is usually negative (right rudder) although its sign may reverse in a steep climb. The strong ef­fect of the climb angle derives from the fact that the roll rate p is proportional to в and changes sign with it. Thus the terms Clpp and СПрр in the moment equations change sign between climbing and descending and affect the control angles required to produce zero moment.

Finally, it may be remarked that the control angles obtained would have been substantially different had it been stipulated that /3, not Cy, should be zero in the turn. It would not then be possible, however, to satisfy the requirement that the ball be cen­tered in the tum-and-back indicator.