# Directivity of linear sources

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 7 221 - 252 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 7.5a

In Figure 7.5a, a linear vertical source ${\displaystyle {\textit {MN}}}$ (such as a column of explosives) of length ${\displaystyle a\lambda }$, ${\displaystyle a}$ being a constant and ${\displaystyle \lambda }$ the wavelength, is activated at all points simultaneously at time ${\displaystyle t=0}$. Taking the initial waveform as ${\displaystyle Ae^{{\rm {j}}\left(\kappa r-\omega t\right)}}$, show that the effect at point ${\displaystyle {\textit {P}}}$ is

 {\displaystyle {\begin{aligned}h\left(t\right)=Aa\lambda e^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}\ {\rm {sinc}}\ \left(\pi a\sin \theta _{0}\right).\end{aligned}}} (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 ${\displaystyle {\textit {P}}}$ at different times because they must travel different distances. Denoting ${\displaystyle a\lambda /2}$ by ${\displaystyle {\textit {c}}}$, the total effect at ${\displaystyle {\textit {P}}}$ at time ${\displaystyle {\textit {t}}}$ will be

Figure 7.5a  Geometry for linear source.

{\displaystyle {\begin{aligned}h(t)=\int _{z_{0-C}}^{z_{0+C}}Ae^{{\rm {j}}\left(\kappa r-\omega t\right)}\,{\rm {d}}z=Ae^{-{\rm {j}}\omega t}\int _{z_{0-C}}^{z_{0+C}}e^{{\rm {j}}\kappa r}\,{\rm {d}}z.\end{aligned}}}

To integrate we must get a relation between ${\displaystyle {\textit {r}}}$ and ${\displaystyle {\textit {z}}}$. We assume that ${\displaystyle r_{0}\gg a\lambda }$; then

 {\displaystyle {\begin{aligned}r\approx r_{0}+\left(z-z_{0}\right)\sin \theta _{0}=r_{0}(1-\sin ^{2}\theta _{0})+z\sin \theta _{0}\\\approx r_{0}\cos ^{2}\theta _{0}+z\sin \theta _{0}.\end{aligned}}} (7.5b)

Thus,

{\displaystyle {\begin{aligned}h\left(t\right)=Ae^{-{\rm {j}}\omega t}\int _{z_{0-C}}^{z_{0}+c}e^{{\rm {j}}\kappa \left(r_{0}\cos ^{2}\theta _{0}+z\sin \theta _{0}\right)}{\rm {d}}z\\=Ae^{{\rm {j}}\left(\kappa r_{0}\cos ^{2}\theta _{0}-\omega t\right)}\int _{z_{0-C}}^{z_{0}+c}e^{{\rm {j}}\kappa z\sin \theta _{0}}{\rm {d}}z\\=Ae^{{\rm {j}}\left(\kappa r_{0}\cos ^{2}\theta _{0}-\omega t\right)}\left({\frac {e^{{\rm {j}}{\kappa }\sin \theta _{0}}}{{\rm {j}}\kappa \sin \theta _{0}}}\left|_{z_{0}-c_{0}}^{z_{0}+c_{0}}\right.\right).\end{aligned}}}

Noting that ${\displaystyle z_{0}\sin \theta _{0}=r_{0}\sin ^{2}\theta _{0}}$ and ${\displaystyle \kappa c=\pi a}$, this becomes

 {\displaystyle {\begin{aligned}h\left(t\right)=2Ace^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}\left({\frac {\sin \left(\pi a\sin \theta _{0}\right)}{\pi a\sin \theta _{0}}}\right)\\=Aa\lambda e^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}\ \mathrm {sinc} \ (\pi a\sin \theta _{0}),\end{aligned}}} (7.5c)

where sinc ${\displaystyle x=\left(\sin x\right)/x}$.

If the linear source is replaced by a concentrated source of equal strength at the center ${\displaystyle {\textit {O}}}$, the effect at ${\displaystyle {\textit {P}}}$ would be ${\displaystyle Aa\lambda e^{{\rm {j}}\left(\kappa r_{0}-\omega t\right)}}$. Dividing the right-hand side of equation (7.5c) by this quantity, we get for the array response ${\displaystyle {\textit {F}}}$

 {\displaystyle {\begin{aligned}F={\rm {sinc}}\ \left(\pi a\sin \theta _{0}\right).\end{aligned}}} (7.5d)

## Problem 7.5b

An explosion initiated at the top of the explosive column ${\displaystyle {\textit {MN}}}$ in Figure 7.5a travels down the column with velocity ${\displaystyle V_{e}}$. Show that the array response is

 {\displaystyle {\begin{aligned}F=-{\rm {sinc}}[\pi a(\sin \theta _{0}-V_{r}/V_{e})],\end{aligned}}} (7.5e)

where ${\displaystyle V_{r}}$ 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 ${\displaystyle {\textit {MN}}}$ exploded at ${\displaystyle t=0}$. We now consider the case where the explosion starts at point ${\displaystyle {\textit {M}}}$ and travels down the column with velocity ${\displaystyle V_{e}}$, that is, the explosion starts at ${\displaystyle z=z_{0}+c}$ at ${\displaystyle t=0}$, where ${\displaystyle c=a\lambda _{e}/2}$. Writing ${\displaystyle \kappa _{e}=\omega /V_{e}}$, ${\displaystyle \kappa _{r}=\omega /V_{r}}$, ${\displaystyle \gamma =V_{e}/V_{r}}$ the wave generated by the element dz arrives at ${\displaystyle {\textit {P}}}$ with the phase ${\displaystyle \omega [\left(z_{0}+c-z\right)/V_{e}+(r/V_{r})]}$ Using equation (7.5b) the phase becomes

{\displaystyle {\begin{aligned}\omega \left\{\left(z_{0}+c\right)/V_{e}+\left(r_{0}\cos ^{2}\theta _{0}\right)/V_{r}+z\left[\left(\sin \theta _{0}\right)/V_{r}-1/V_{e}\right]\right\}\\=\kappa _{e}[z_{0}+c+\gamma r_{0}\cos ^{2}\theta _{0}+z(\gamma \sin \theta _{0}-1)].\end{aligned}}}

Assuming a harmonic wave function ${\displaystyle \psi _{P}}$, we can write

{\displaystyle {\begin{aligned}\psi _{P}=AK\int _{z_{0}+c}^{z_{0-C}}e^{{\rm {j}}mz}{\rm {d}}z,\end{aligned}}}

where ${\displaystyle A={}}$ amplitude, ${\displaystyle K=e^{{\rm {j}}\kappa _{e}\left[\left(Z_{0}+c\right)+\gamma r_{0}\cos ^{2}\theta _{0}\right]\}}}$, ${\displaystyle m=\kappa _{e}\left(\gamma \sin \theta _{0}-1\right)}$. Integrating, we obtain

{\displaystyle {\begin{aligned}\psi _{P}=\left(AK/{\rm {j}}m\right)e^{{\rm {j}}mz}\left|_{z_{0}+c}^{z_{0-C}}\right.\\=\left(AK/{\rm {j}}m\right)e^{({\rm {j}}mz_{0})}\left[e^{-{\rm {j}}mc}-e^{{\rm {j}}mc}\right]=-\left(2AK/m\right)e^{{\rm {j}}mz_{0}}\sin \left(mc\right)\\=-\left(2AKc\right)e^{{\rm {j}}mz_{0}}\ {\rm {sinc}}\left(mc\right)\\=-2Ac\exp \left\{{\rm {j}}\kappa _{e}\left[\left(r_{0}+c\right)+\gamma r_{0}\cos ^{2}\theta _{0}+r_{0}\left(\gamma \sin \theta _{0}-1\right)\right]\right\}{\rm {sinc}}(mc)\\=-2Ac\exp \left\{{\rm {j}}\kappa _{e}\left(\gamma r_{0}+c\right)\right\}{\rm {sinc}}(mc).\end{aligned}}}

If we locate the same amount of explosive at ${\displaystyle z_{0}}$ and explode it at ${\displaystyle t=0}$, we get at ${\displaystyle {\textit {P}}}$

{\displaystyle {\begin{aligned}\psi _{P}^{*}=2Ace^{-{\rm {j}}\kappa _{r}r_{0}}=2Ace^{-{\rm {j}}\kappa _{e}\gamma r_{0}},\end{aligned}}}

hence the array response is

{\displaystyle {\begin{aligned}F=\psi _{P}/\psi _{P}^{*}=-e^{{\rm {j}}\kappa _{e}c}{\rm {sinc}}(mc).\end{aligned}}}

Omitting the first factor, which is independent of ${\displaystyle \theta _{0}}$ and ${\displaystyle r_{0}}$ and hence is merely a scale factor, we have

{\displaystyle {\begin{aligned}F=-{\rm {sinc}}[c\kappa _{e}(\gamma \sin \theta _{0}-1)].\end{aligned}}}

But

{\displaystyle {\begin{aligned}ca=\left(a\lambda _{e}/2\right)\left(2\pi /\lambda _{e}\right)=\pi a,\end{aligned}}}

so

 {\displaystyle {\begin{aligned}F=-{\rm {sinc}}[(\pi a(\sin \theta _{0}-V_{r}/V_{e})].\end{aligned}}} (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

{\displaystyle {\begin{aligned}F={\rm {sinc}}\left(\pi a\sin \theta _{0}\right).\end{aligned}}}

Equation (7.5c) reduces to equation (7.5d) whenever ${\displaystyle \left(V_{r}/V_{e}\right)\ll \sin \theta _{0}}$. For ${\displaystyle \theta _{0}\approx 90^{\circ }}$, i.e., for rays traveling almost vertically downward, the required condition is that ${\displaystyle V_{r}\ll V_{e}}$. For most explosives, ${\displaystyle V_{e}\approx 6{-}7}$ km/s, so ${\displaystyle V_{r}}$ 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 ${\displaystyle {\textit {F}}}$ for a column 10 m long, given that ${\displaystyle \lambda _{r}=40}$ m, ${\displaystyle V_{e}=5.5}$ km/s, ${\displaystyle V_{r}=2.1}$ km/s, and ${\displaystyle \theta _{0}=0^{\circ }}$, ${\displaystyle 30^{\circ }}$, ${\displaystyle 60^{\circ }}$, ${\displaystyle 90^{\circ }}$.

### Solution

{\displaystyle {\begin{aligned}a=10/40=0.25,\;\;V_{r}/V_{e}=2.1/5.5=0.38;\end{aligned}}}

thus

{\displaystyle {\begin{aligned}F={\rm {sinc}}\left[\left(\pi /4\right)\left(\sin \theta _{0}-0.38\right)\right]={\rm {sinc}}\,x.\end{aligned}}}

Table 7.5a. Array response as a function of direction.
${\displaystyle \theta _{0}}$ ${\displaystyle x}$ ${\displaystyle \sin x}$ ${\displaystyle F}$
${\displaystyle 0^{\circ }}$ ${\displaystyle -0.30}$ ${\displaystyle -0.29}$ 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 ${\displaystyle {\textit {L}}}$ 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 ${\displaystyle V_{e}=5.5}$ km/s, ${\displaystyle V_{r}=2.1}$ km/s, so ${\displaystyle 10/21=6\times 0.60/5.5+5L/6.2}$, hence ${\displaystyle L=5.1}$ m.

## Problem 7.5e

What are the relative amplitudes (approximately) of the waves generated by the explosives in part (d) at angles ${\displaystyle \theta _{0}=0^{\circ }}$, ${\displaystyle 30^{\circ }}$, ${\displaystyle 60^{\circ }}$, and ${\displaystyle 90^{\circ }}$ when ${\displaystyle \lambda =40}$ m?

### Solution

An approximate solution can be obtained by assuming that the average velocity ${\displaystyle V_{e}}$ is equal to ${\displaystyle V_{r}}$; 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, ${\displaystyle a=10/40=1/4}$, ${\displaystyle V_{r}/V_{e}=1.0}$ and

{\displaystyle {\begin{aligned}F={\rm {sinc}}\left[\left(\pi /4\right)\left(\sin \theta _{0}-1.0\right)\right]={\rm {sinc}}\ y.\end{aligned}}}

Table 7.5b. Relative amplitudes.
${\displaystyle \theta _{0}}$ ${\displaystyle y}$ ${\displaystyle \sin }$ ${\displaystyle |F|}$
${\displaystyle 0^{\circ }}$ –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