Solving for a and 6 gives: 



a = D""'" (k - nuo^) BS^ ; 6 = - D~^u)(c + BS) BS^ 



where 



D = (k3_ - mi/)^ + ^ (c + BS)^ 



The phase angle of v relative to 6 is thus tan (6/a) or - tan 

 to(c + BS)/(k - nuo ) . This "phase shift enables BS Q to supply the energy 

 lost from the foil through the v term. It may be worthwhile to verify this 



compensation. 



•2 

 The rate of energy loss is (c + BS) v ; the rate of energy supply is 



p. .2 222 2 22 



BS 6v. Now V = 0) C (3 sin tot +a cos ut - 2a6 sin wt cos tot). The time 



2 2 



average, over a cycle, of sin wt or cos cat is 1/2, whereas sin tot cos tot 



averages zero. Hence, v*^ averages (1/2) to C (a + 6 ), but a + 3 = 



_-i 2 2 



D (BS ) . The average rate of energy loss is, therefore: 



I (c-i^ + BS) (O^C^ D~^ (BS^)^ 



On the other hand, 



2 2 2 



0v = - (oBC sin tot + (oaC sm tot cos ii)t 



2 • 

 Th\js the rate of energy gain, or BS Sv, averages: 



- \ toBC^BS^ = I to^C^ D"^ (c^ + BS)(BS^)^ 



This equals the average rate of energy loss. 



In conclusion, to have a typical flutter situation with a positive 

 critical flutter speed, it is clearly necessary in the arrangement under 



79 



