Directivity of linear sources

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Problem 7.5a

In Figure 7.5a, a linear vertical source (such as a column of explosives) of length , being a constant and the wavelength, is activated at all points simultaneously at time . Taking the initial waveform as , show that the effect at point is


(7.5a)

What is the array response?

Background

A distributed source can be thought of as an array. The array response is the ratio of the output of an array to the output when all of the elements are concentrated at the midpoint of the array.

Detonating cord is an explosive cord with a constant velocity of detonation; it is used to connect two charges in order to delay detonation of the second charge. Detonation of the first charge initiates detonation in the cord, which in turn detonates the second charge. By varying the length of cord, detonation of the second charge can be delayed a desired amount.

Solution

Although the explosive is exploded instantaneously, energy from different parts of the column arrive at at different times because they must travel different distances. Denoting by , the total effect at at time will be

Figure 7.5a  Geometry for linear source.

To integrate we must get a relation between and . We assume that ; then


(7.5b)

Thus,

Noting that and , this becomes


(7.5c)

where sinc .

If the linear source is replaced by a concentrated source of equal strength at the center , the effect at would be . Dividing the right-hand side of equation (7.5c) by this quantity, we get for the array response


(7.5d)

Problem 7.5b

An explosion initiated at the top of the explosive column in Figure 7.5a travels down the column with velocity . Show that the array response is


(7.5e)

where is the velocity in the rocks. Under what circumstances does equation (7.5e) reduce to equation (7.5d)?

Solution

In part (a), the entire column exploded at . We now consider the case where the explosion starts at point and travels down the column with velocity , that is, the explosion starts at at , where . Writing , , the wave generated by the element dz arrives at with the phase Using equation (7.5b) the phase becomes

Assuming a harmonic wave function , we can write

where amplitude, , . Integrating, we obtain

If we locate the same amount of explosive at and explode it at , we get at

hence the array response is

Omitting the first factor, which is independent of and and hence is merely a scale factor, we have

But

so


(7.5f)

[The minus sign for sinc occurs here and not in equation (7.5d) because the direction of integration is opposite to that assumed in deriving equation (7.5d)]. For an instantaneous explosion the result is given by equation (7.5d), namely

Equation (7.5c) reduces to equation (7.5d) whenever . For , i.e., for rays traveling almost vertically downward, the required condition is that . For most explosives, km/s, so should be no more than about 1.5 km/s, the velocity of water, for the two equations to give nearly the same result.

Problem 7.5c

Calculate the array response for a column 10 m long, given that m, km/s, km/s, and , , , .

Solution

thus

Table 7.5a. Array response as a function of direction.
0.99
30 0.094 0.094 1.00
60 0.38 0.37 0.97
90 0.49 0.47 0.98

Differences in directivity are negligible when the charge length is much smaller than the wavelength.

Problem 7.5d

If the column in part (c) is replaced by six charges, each 60 cm long and equally spaced to give a total length of 10 m, the charges being connected by spirals of detonating cord with detonation velocity 6.2 km/s, what length of detonating cord must be used between adjacent charges to achieve maximum directivity downward?

Solution

Let be the length of detonating cord between successive charges; maximum directivity downward is achieved when the traveltime through the explosive column is the same as that in the adjacent rocks. In part (c) we were given km/s, km/s, so , hence m.

Problem 7.5e

What are the relative amplitudes (approximately) of the waves generated by the explosives in part (d) at angles , , , and when m?

Solution

An approximate solution can be obtained by assuming that the average velocity is equal to ; this means that the traveltime down through the 10 m column of explosives is the same as that for a wave in the adjacent 10 m of rock. In this case, , and

Table 7.5b. Relative amplitudes.
–0.79 –0.71 0.90
30 –0.39 –0.38 0.97
60 –0.11 –0.11 1.00
90 0.00 0.00 1.00

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