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Sample text

Do not confuse Ny here with the acceleration potential of Section XI. ) *(x (ý, 71), t (ý, 1)) =Vp (ý, St3(X, (56) ) (57) t) =/ (11) We ignere the role of the y and z coordinates. It will be seen that they have no influence on the final solution. By the chain rule, we have from equation (56). (58) ýt = Vz~t + Vnri, (59) + ~xwnrix Substituting equations (57), (58) and (59) into equation (53), we obtain P= (',+ (60) W1,1%) + U(Mk+ n+l) Rearranging the terms gives P = V4(t (61) k + w(i, + Ur1 1 ) We are free to choose the relationship between the (i, 'l) coordinates and the (x, y) coordinates.

This is the fundamental solution for the potential which arises due to a source moving along the x axis with constant velocity of -UV. In the next section, we transform the coordinates to a moving frame. 51 SECTION VIII The Elementary Solution to the Aerodynamic Potential Equation Our objective of the past three sections has been to derive elementary solutions to the aerodynamic potential equation (42) which may be used to model the flow over wings and bodies. in Section V, we recognized that the aerodynamic potential equation is related to the acoustic potential equation by a simple Gaussian transformation.

According to the sifting principle of equation (120), we evaluate the integrand at E) = 0 only. This gives the result C(xO,YO, Zo, t) = [ ! (12 r (132) We now evaluate equation (132) for the potential which arises from a moving source. This is achieved in equation (145) using equations (135) and (143). From equation (130), obtain the following quadratic in Tfor E = 0. __T+ ( 2 +Y2+Z) 0 -[0 0 (133) where p2 = 1 - M2 and M = U/a is the Mach number. From this quadratic equation, we expect two solutions for T.

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