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A few of the consequences of these assumptions are now discussed. For our linearized boundary value problem, the wing and therefore the trailing edge are in the z = 0 plane. In time, the mathematical representation of the wing slices out a plane region as it moves forward through the air at velocity U. We will treat this planar region as the trailing wake. (We could have a non-planar wake. ) It is assummed that there is no jump in the non-zero velocity component, w = •z across the wake. Therefore, 4, is symmetric with respect to z in its first approximation (eg.

43 SECTION VII The Moving Source In this section, we obtain the elementary formula (145) for the moving source. In the next section, we transform this solution to the moving frame of reference. Then one may demonstrate that this formula does indeed satisfy the aerodynamic potential equation (42). The following explanation is a review of the derivation provided by Garrick 1. We superimpose a train of stationary sources, all in a line. These sources are pulsed in a sequence, thus giving the same effect as a single source moving at a constant speed.

Given the linear aerodynamic potential equation 10it =ui' (163) and the fundamental source solution of equation (152). Os1= ikf(t - T) (164) we show that the elementary solution d= (165) (•s() is also a solution. We substitute equation (165) into equation (163) to obtain pa+ + -0 (166) Next, the order of differentiation is changed %)x + (0$), + (0) 2 0 (167) The term in the square brackets is known to be zero from equation (163). Equation (167) reduces to a o =2o] 62 (168) The Source Doublet Thus, we have shown that equation (165) is also a solution to the aerodynamic potential equation.

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A COMPILL4TIBN OF THE MATHEMATICS LEADING TO THE DOUBLET LKI TICE METHOD


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