# Interpreting marine refraction data

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 11 415 - 468 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 11.17

How many distinct separate head waves are indicated in Figure 11.17a, and what are their apparent velocities? Calculate the depths and velocities of the respective refractors, assuming (i) no dip, (ii) ${\displaystyle 5^{\circ }}$ dip to the right, (iii) ${\displaystyle 5^{\circ }}$ dip to the left.

Figure 11.16a.  Wave travel in a salt dome (from Barton, 1929). (i) Vertical section; (ii) plan view.

### Solution

We take ${\displaystyle S}$ at ${\displaystyle x=0}$ as the source and the geophone at ${\displaystyle x=12}$ km as ${\displaystyle R}$. Four wavetrains are observed (Figure 11.17b). The slowest (which one has to view at grazing incidence to pick) has an apparent velocity of about 1.40 km/s; the arrival time at ${\displaystyle R}$ is off scale. This is the direct water wave, but its velocity is low for water, perhaps because of interference with a dispersive channel wave, loss of early cycles, or because the distance (calculated from water-borne direct waves seen at a floating sonobuoy) is not exact. Because of the uncertainty in the traveltime, we shall use the value 1.50 km/s for the water velocity.

The next event (L) has an apparent velocity of about 1.70 km/s and an arrival time of 6.84 s; the intercept time is difficult to determine because the event seems to be curved. It is probably a head wave at or near the sea floor.

The next fastest refraction (M) has an apparent velocity of about 2.19 km/s between ${\displaystyle x=1}$ and 4 km and an apparent velocity of 2.67 km/s for ${\displaystyle x>4}$ km; this change in velocity may be due to a change in dip or it may indicate the presence of two separate events. We shall consider only the velocity 2.67 km/s, the travel time being 5.38 s, and the intercept time 0.95 s.

Refraction N has a linear alignment and an apparent velocity of 5.45 km/s, traveltime of 3.97 s, and an intercept of 1.77 s; it is probably from a basaltic basement surface.

The measured data are:

{\displaystyle {\begin{aligned}\mathbf {L:} &\quad V_{\rm {L}}=1.70\ {\rm {km/s}},\quad t_{R{\rm {L}}}=6.84\ {\rm {s}},\quad t_{iL}=?;\\\mathbf {M:} &\quad V_{\rm {M}}=2.67\ {\rm {km/s}},\quad t_{R{\rm {M}}}=5.38\ {\rm {s}},\quad t_{iM}=0.95\ {\rm {s}};\\\mathbf {N:} &\quad V_{\rm {N}}=5.45\ {\rm {km/s}},\quad t_{R{\rm {N}}}=3.97\ {\rm {s}},\quad t_{iN}=1.77\ {\rm {s}}.\\\end{aligned}}}

If L is a head wave, the intercept time should be ${\displaystyle (6.84-12.0/1.70)=0.22}$ s. Thus we conclude that L is not a head wave from a planar refractor; it may be a dispersive water wave or part of a reflection, but we have insufficient data to identify it. Disregarding L, we are left with only the water layer and the two refractors M and N.

Figure 11.17a.  Marine refraction profile (from Ingham, 1975, 130).

Figure 11.17b.  Interpreted marine refraction profile (from Ingham, 1975,130).

(i) Assuming no dip:

Refraction M

We have the following data: ${\displaystyle V_{\rm {I}}=1.50}$ km/s, ${\displaystyle V_{\rm {M}}=2.67}$ km/s, ${\displaystyle t_{i{\rm {M}}}=0.95}$ s.

Then,

{\displaystyle {\begin{aligned}\sin \theta _{c}/1.50&=1/2.67,\quad \theta _{c}=34.2^{\circ };\\h_{\rm {M}}&=V_{1}t_{i{\rm {M}}}/2\cos \theta _{c}=1.50\times 0.95/2\times \cos 34.2^{\circ }=0.82\ {\rm {km}}.\end{aligned}}}

Refraction N:

{\displaystyle {\begin{aligned}\sin \theta _{1}/1.50&=\sin \theta _{c}/2.67=1/5.45;\quad \theta _{1}=16.0^{\circ },\quad \theta _{c}=29.3^{\circ };\\1.77&=2\times 0.82\cos 16^{\circ }/1.50+2h_{2}\cos 29.3^{\circ }/2.67;\\h_{2}&=\left(1.77-1.10\right)/0.65=0.67/0.65=1.03\ {\rm {km}};\\z_{\rm {N}}&=0.83+1.03=1.86\ {\rm {km}}.\end{aligned}}}

(ii) Assuming ${\displaystyle 5^{\circ }}$ dip to the right

In this case, the profile is in the downdip direction, the arrival times, velocities, and intercept times are unchanged except that the velocities are now apparent velocities and the intercept times give slant depths normal to the beds. Note that the water depth increases to the right because horizon M dips to the right.

Refractor M

We have: ${\displaystyle V_{1}=1.5}$ km/s, ${\displaystyle V_{d{\rm {M}}}=2.67}$ km/s, ${\displaystyle \xi =5^{\circ }}$, ${\displaystyle t_{i{\rm {M}}}=0.95}$ s. Using equation (4.24d), we write

{\displaystyle {\begin{aligned}&\sin \left(\theta _{c}+5^{\circ }\right)=V_{1}/V_{d{\rm {M}}}=1.50/2.67;\\&\left(\theta _{c}+5^{\circ }\right)=34.2^{\circ },\quad \theta _{c}=29.2^{\circ };\\&V_{\rm {M}}=V_{1}/\sin \theta _{c}=1.50/\sin 29.2^{\circ }=3.07\ {\rm {km/s}};\\&h_{\rm {M}}=V_{1}t_{i{\rm {M}}}/2\cos \theta _{c}=1.50\times 0.95/2\cos 29.2^{\circ }=0.82\ {\rm {km}}\end{aligned}}}

(this is the slant water depth at ${\displaystyle S}$).

Refractor ${\displaystyle N}$

We could use Adachi’s method (problem 11.5), but the given data are not in a form suitable for this method. Instead we shall strip off (problem 11.6) the surface layer after which we have two parallel horizons, and therefore can use equation (4.18a) to solve for horizon N.

The slant depth of horizon M at the source ${\displaystyle S}$ is 0.82 km. At ${\displaystyle R}$ the slant depth is

${\displaystyle h_{\rm {M}}=0.82+12.0\times \sin 5^{\circ }=1.87\ {\rm {km}}.}$

Thus to locate horizon M we swing arcs at ${\displaystyle S}$ and ${\displaystyle R}$ with radii 0.82 and 1.87 km, respectively, then draw horizon M tangent to the arcs (see Figure 11.17c).

Figure 11.17c.  Construction for stripping off the water layer.

To solve for N, we need the total time that the refraction from N spends in the water layer. For this we need the angle of approach to the surface of head wave N. Equation (4.2d) states that the angle of approach ${\displaystyle \alpha }$ is given by

{\displaystyle {\begin{aligned}\sin \alpha =V_{1}\left(\Delta t/\Delta x\right)=V_{1}/V_{d{\rm {N}}}=1.50/5.45,\quad \alpha =16.0^{\circ }.\end{aligned}}}

Figure 4.2c shows that ${\displaystyle \alpha }$ is relative to the vertical, so if the ray emerges from horizon M at the angle of refraction ${\displaystyle \theta }$, ${\displaystyle \theta +5^{\circ }=16.0^{\circ }}$, whereas the angle at which the ray left source ${\displaystyle S}$ is ${\displaystyle \theta -5^{\circ }=\theta ^{\circ }}$. Therefore we draw a ray from ${\displaystyle S}$ down to M at the angle ${\displaystyle 6^{\circ }}$ to the vertical and another down from ${\displaystyle R}$ to M at the angle ${\displaystyle 16^{\circ }}$ to the vertical (see rays ${\displaystyle SS'}$ and ${\displaystyle RR'}$ in Figure 11.17c). The total length (${\displaystyle {\rm {SS}}'+{\rm {RR}}'}$) is (${\displaystyle 0.80+1.80}$) km; dividing by 1.50 gives 1.73 s to be subtracted from ${\displaystyle t_{N}}$, leaving ${\displaystyle t_{N'}=3.97-1.73=2.24}$ s for the traveltime relative to M. Also the distance ${\displaystyle S'R'=x'=11.5}$ km. The intercept times for both M and N are based on the normals to the beds; since the beds are parallel, we can subtract ${\displaystyle t_{iM}}$ from ${\displaystyle t_{in}}$ [see part (i)] to get ${\displaystyle t_{N'}}$, the intercept time of N for the virtual source ${\displaystyle S'}$. This gives ${\displaystyle t_{iN'}=\left(1.77-0.95\right)=0.82}$ s.

We must correct the apparent velocity ${\displaystyle V_{d{\rm {N}}}=5.45}$ km/s. We write this as ${\displaystyle 1/5.45=\Delta t/\Delta x=2.20/12.0}$ s/km. The event N in Figure 11.17b is linear from ${\displaystyle x=12}$ to beyond ${\displaystyle x=6}$km, so we write ${\displaystyle 1/5.45=0.110/6}$ s/km. The numerator is the difference between ${\displaystyle t_{N}}$ at ${\displaystyle x=12}$ and 6 km, and Figure 11.17c shows that the correction is ${\displaystyle RT=0.50/1.50=0.33}$ s. Thus ${\displaystyle \Delta t'=1.10-0.33=0.77}$ s. The correction to ${\displaystyle \Delta x}$ is negligible, so we get ${\displaystyle V_{N}\approx 6.00/0.77=7.79}$ km/s.

We can now get the depth of N below M. We have

{\displaystyle {\begin{aligned}\theta _{c}=\sin ^{-1}\left(V_{\rm {M}}-V_{\rm {N}}\right)=3.07/7.79=23.3^{\circ }.\end{aligned}}}

Because horizons M and N are parallel, we find the intercept time of N relative to M by subtracting the intercept times in part (i), so

{\displaystyle {\begin{aligned}t_{iN^{'}}=1.77-0.95=0.82\ {\rm {s}}.\end{aligned}}}

Thus

{\displaystyle {\begin{aligned}h_{N^{'}}=3.07\times 0.82/2\cos 23.3^{\circ }=1.37\ {\rm {km}}.\end{aligned}}}

Then, since the depth of M at ${\displaystyle S}$ is 0.82 km,

{\displaystyle {\begin{aligned}z_{N}=0.82+1.37=2.19\ {\hbox{km}}\ {\hbox{(slant depth at S')}}.\end{aligned}}}

(iii) Assuming ${\displaystyle 5^{\circ }}$ dip to the left

Because of the shallow depth of M we check to see where it outcrops. The horizon M passes through the point ${\displaystyle \left(x,\;z\right)=\left(0,0.82\cos 5^{\circ }\right)=(0,0.817)}$, (see part (ii)) with slope ${\displaystyle \tan 5^{\circ }}$, Thus it will outcrop at ${\displaystyle x=0.817/\tan 5^{\circ }=9.34}$ km. But Figure 11.17c shows that event M exists at offset of 12 km, hence the assumption of ${\displaystyle 5^{\circ }}$ dip to the left is not consistent with the given data.