Directivity of a harmonic source plus ghost

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Problem

Show that equation (6.7c) gives the directivity diagrams shown in Figure 6.9a.

Solution

The directivity is given by equation (6.7c). We take , and , 0.5, and 1.0 for the three parts of Figure 6.9a. Then equation (6.7c) gives


Substituting the three values of , we have:

The results of the calculations are shown in Tables 6.9a,b.

Figure 6.9a.  Directivity of a harmonic source at depth .

Ignoring the minus signs (which indicate phase reversals), the curves for and , shown in Figure 6.9b, conform closely to Figure 6.9a. However, we need more points to plot the -curve properly and Table 6.9b shows calculated values for intermediate points. The -curve in Figure 6.9b also conforms closely to Figure 6.9a.

Table 6.9a. Values for , , .
0 0.59 0.00 0.00
15 0.57 0.15 −0.20
30 0.52 0.44 −0.74
45 0.43 0.81 −0.97
60 0.31 1.00 −0.01
75 0.16 0.72 1.00
90 0.00 0.00 0.00
Table 6.9b. Intermediate values for .
5 −0.01 50 −0.79
10 −0.08 55 −0.46
20 −0.35 65 0.46
25 −0.54 70 0.83
35 −0.90 80 0.89
40 −0.99 85 0.52
Figure 6.9b.  Calculated directivity at source depth .

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