Interpretation of sonobuoy data

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

Problem 4.19a

Figure 4.19a shows a refraction profile recorded as a ship firing an air gun moved away from a sonobuoy. Identify the direct wave through the water and use its traveltime to get the source-to-sonobuoy distance (assume 1.50 km/s as the velocity in water).

Identify distinctive head-wave arrivals, determine their velocities and intercept times, and the depths of the refractors (assuming flat bedding and no velocity reversals).

Figure 4.19a.  Sonobuoy refraction profile (Courtesy Fairfield Industries.

Background

An air gun is a marine seismic energy source consisting of one or more chambers filled with air under very high pressure; very rapid release of the air into the water produces an effect similar to that of an explosion. [See Sheriff and Geldart, 1995, Section 7.4.3 for more details.] The sonobuoy is a battery-powered marine hydrophone which floats in the water and radios the seismic signals which it receives through the water to a receiving station, usually on the source boat. [See Sheriff and Geldart, 1995, Section 11.1.4 for more details.]

A strip recorder records data on a sheet of paper wrapped around a rotating cylinder. If the source is activated again before the previous record has finished, late data from the first activation may be recorded on top of the second record, an effect called paging (see Sheriff and Geldart, 1995, p. 260).

Figure 4.19b.  Measured velocities and intercept times for Figure 4.19a.

Solution

We assume that layering is planar and horizontal. The first step in the solution is to identify straight-line events that are believed to represent the direct wave and head waves. The direct wave is easily identified because it has the intercept ${\displaystyle t=0}$. Five head waves have been identified and their slopes and intercepts measured; the results are shown in Figure 4.19b and in Table 4.19a. [We realize the last integers are not significant.] The next step is to calculate the angles of incidence using equation (3.1a).

For the first and second events we have

{\displaystyle {\begin{aligned}{\frac {\sin \theta _{c1}}{V_{1}}}={\frac {1}{V_{2}}};\quad {\frac {\sin \theta _{1}}{V_{2}}}={\frac {\sin \theta _{c2}}{V_{2}}}={\frac {1}{V_{3}}}.\end{aligned}}}

Table 4.19a. Times, velocities, and calculated angles.
Event 1 2 3 4 5 6
Intercept ${\displaystyle t_{ij}}$(s) 3.70 4.90 7.80 8.60 9.00
Velocity ${\displaystyle V_{j}}$ (km/s) 1.50 3.10 4.00 7.00 9.00 13.20
${\displaystyle \theta _{1}}$ 28.9${\displaystyle ^{\circ }}$ 22.0${\displaystyle ^{\circ }}$ 12.4${\displaystyle ^{\circ }}$ 9.6${\displaystyle ^{\circ }}$ 6.5${\displaystyle ^{\circ }}$
${\displaystyle \theta _{2}}$ 50.8${\displaystyle ^{\circ }}$ 26.3${\displaystyle ^{\circ }}$ 20.1${\displaystyle ^{\circ }}$ 13.6${\displaystyle ^{\circ }}$
${\displaystyle \theta _{3}}$ 34.8${\displaystyle ^{\circ }}$ 26.4${\displaystyle ^{\circ }}$ 17.5${\displaystyle ^{\circ }}$
${\displaystyle \theta _{4}}$ 51.1${\displaystyle ^{\circ }}$ 32.0${\displaystyle ^{\circ }}$
${\displaystyle \theta _{5}}$ 43.0${\displaystyle ^{\circ }}$

We continue thus until we reach the fifth event, where we write for its angles through the various layers,

{\displaystyle {\begin{aligned}{\frac {\sin \theta _{1}}{V_{1}}}={\frac {\sin \theta _{2}}{V_{2}}}={\frac {\sin \theta _{3}}{V_{3}}}={\frac {\sin \theta _{4}}{V_{4}}}={\frac {\sin \theta _{c5}}{V_{5}}}={\frac {1}{V_{6}}}.\end{aligned}}}

The calculated angles are shown in Table 4.19a.

The angles in each column are the angles of incidence for the refraction generated at the ${\displaystyle j^{th}}$ interface, and the last angle in each column is the critical angle. The final step is to calculate in a step-by-step manner the thicknesses of the five beds using equation (4.18d) and the above data:

{\displaystyle {\begin{aligned}t_{i1}&=(2h_{1}\cos \theta _{c1}/V_{1}),\\h_{1}&={\frac {1}{2}}V_{1}t_{i1}/\cos \theta _{c1}=0.5\times 1.50\times 3.70/\cos \,28.9^{\circ }-3.17\,\mathrm {km} ;\\t_{i2}&=(2h_{1}\cos \theta _{1}/V_{1})+(2h_{2}\cos \theta _{c2}/V_{2}),\\h_{2}&=(V_{2}/2\cos \theta _{c2})(t_{i2}-2h_{1}\cos \theta _{1}/V_{1})\\&=(3.10/2\cos \,50.8^{\circ })(4.90-6.34\cos \,22.0^{\circ }/1.50)\\&=2.45(4.90-3.92)=2.40\,\mathrm {km} .\end{aligned}}}

Repeating the same steps, we get

{\displaystyle {\begin{aligned}h_{3}&=(V_{3}/2\cos \theta _{c3})(t_{i3}-2h_{1}\cos \theta /V_{1}-2h_{2}\cos \theta _{2}/V_{2})\\&=(4.00/2\cos \,34.8^{\circ })(7.80-2\times 3.17\cos 12.4^{\circ }/1.50\\&-2\times 2.40\cos 26.3^{\circ }/3.10)=2.44\times 2.28=5.56\,\mathrm {km} ,\\h_{4}&=(V_{4}/2\cos \theta _{c4})(t_{i4}-2h_{1}\cos \theta _{1}/V_{1}-2h_{2}\cos \theta _{2}/V_{2}-2h_{3}\cos \theta _{3}/V_{3})\\&=(7.00/2\cos \,51.1^{\circ })(8.60-2\times 3.17\cos 8.6^{\circ }/1.50\\&\quad -2\times 2.40\cos 20.1^{\circ }/3.10-2\times 5.56\cos 26.4^{\circ }/4.00)\\&=5.51\times 0.4=2.73\,\mathrm {km} ,\end{aligned}}}

{\displaystyle {\begin{aligned}h_{5}&=(V_{5}/2\cos \theta _{c5})(t_{i5}-2h_{1}\cos \theta _{1}/V_{1}-2h_{2}\cos \theta _{2}/V2\\&\quad -2h_{3}\cos \theta _{3}/V_{3}-2h_{4}\cos \theta _{4}/V_{4})\\&=(9.00/2\cos 43.0^{\circ })(9.80-2\times 3.17\cos 6.5^{\circ }/1.5\\&\quad -2\times 2.40\cos 13.6^{\circ }/3.10-2\times 5.56\cos 17.6^{\circ }/4.00\\&\quad -2\times 2.73\cos 32.0^{\circ }/7.00)\\&=6.15\times 0.74=4.86\,\mathrm {km} .\end{aligned}}}

We assumed that all beds are horizontal, except for the one with velocity ${\displaystyle V_{6}=13.2}$ km/s; the value 13.2 km/s for ${\displaystyle V_{6}}$ is unreasonably high, suggesting that the interface is dipping down toward the source. The layer between the seafloor and the first refractor is probably sediments. The 7 km/s refractor is probably the Moho.

Problem 4.19b

What is the water depth? Identify multiples and explain their probable travelpaths.

Solution

The seafloor reflection (the first arrival for short offsets in Figure 4.19a) has ${\displaystyle t_{0}=3.20}$ s. Therefore, water depth ${\displaystyle {\frac {1}{2}}\times 3.20\times 1.50=2.40}$ km. The disagreement with ${\displaystyle h_{1}}$ may indicate dip away from the receiver, that the first refractor is deeper than the seafloor, or some other discrepancy. Multiples of the seafloor reflection are observed at ${\displaystyle t_{0}=6.4,\,9.6}$, and 12.9 s. Primary reflections are probably observed only between the seafloor reflection at 3.20 s and its first multiple at 6.40 s; below 6.40 s, most of the reflection events are probably peg-leg multiples (see problem 3.8).