John
The reference you have given in that Chapter 11 page 11.9 Fig.11.10 explains what I was trying to say. It gives case 1 EQUILIBRIUM FLIGHT WITH WINGS LEVEL in which the longitudinal axis of the aircraft is not in line with the relative airflow and that creates drag the side force. Case 2 is EQUILIBRIUM FLIGHT WITH ZERO SIDESLIP in which there is some bank
Case 1: 4 = 0 Figure 11.10 shows the forces and moments for Case 1, the zero bank angle case, with the left engine inoperative. The aircraft is in equilibrium with no accelerations. The pilot would note this with constant heading, ball centered, turn needle centered, rudder opposing the inoperative engine and aileron opposite the rudder to keep the wings level.
Case 2: ß = 0 Another way to balance the sideforce resulting from rudder deflection is by using the W sin if term in the sideforce equation (11.3). Figure 11.11 shows the forces and moments
for the zero sideslip case.
The aircraft is in equilibrium with some bank toward the operating engine, a constant heading and the turn needle centered. The rudder deflection is in the same direction as in the Case 1, however, less 8r is required. The ball in the turn and slip indicator will be deflected in the direction of the bank angle.
And further it states:Three important conclusions can be made from the previous discussion. First, bank angle can reduce the amount of rudder required to achieve equilibrium. Second, an increase in weight reduces the amount of bank required to reduce the sideslip to zero. Third, this configuration will have the least amount of drag. With ß = 0, no sideforce is generated, and therefore no drag due to sideforce is created