# Refractions and refraction multiples

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.16a

Determine the traveltime curve for the refraction ${\mathit {SMNPQR}}$ and the refraction multiple ${\mathit {SMNTUWPQR}}$ in Figure 6.16a.

### Solution

We assume that the velocities are known to three significant figures. Then, using equation (3.1a),

{\begin{aligned}\theta _{2}={\rm {sin}}^{-1}(2.00/4.20)=28.4^{0};\theta _{1}={\rm {sin}}^{-1}(2.80/4.20)=41.8^{\circ }\end{aligned}} The traveltimes can be obtained either graphically or by calculation. Calculating, we get for the refraction traveltimes $t_{R}$ {\begin{aligned}t_{R}=2SM/2.80+2MN/2.00+NP/4.20\\=2\times 0.75/2.80{\rm {\;cos\;}}41.8^{\circ }+2\times 3.25/(2.00{\rm {\;cos\;}}28.4^{\circ })\\+(x-2\times 0.75{\rm {\;tan\;}}41.8^{\circ }-2\times 3.25{\rm {\;tan\;}}28.3^{\circ })/4.20\\=x/4.20+2\times 0.75{\rm {\;cos\;}}41.8^{\circ }/2.80+2\times 3.25{\rm {\;cos\;}}28.4^{\circ }/2.00\\=x/4.20+3.26.\end{aligned}} The critical distance (see equation (6.15a) is

{\begin{aligned}x^{'}=2(0.75{\rm {\;tan\;}}41.8^{\circ }+3.25{\rm {\;tan\;}}28.4^{\circ })=4.86\ {\rm {km}}.\end{aligned}} The traveltime curve for SMNTUWPQR is parallel to that for SMNPQR and displaced toward longer time by the amount $\Delta t$ where

{\begin{aligned}\Delta t=2TU/2.00-TW/4.20\\=2\times 3.25/(2.00{\rm {\;cos\;}}28.4^{\circ })-2\times 3.25({\rm {\;tan\;}}28.4^{\circ })/4.20=2.86\ {\rm {s}}.\end{aligned}} The critical distance for SMNTUWPQR is increased to

{\begin{aligned}x^{'}=2\times 0.75{\rm {\;tan\;}}41.8^{\circ }+4\times 3.25{\rm {\;tan\;}}28.4^{\circ }=8.37\ {\rm {km}}.\end{aligned}} The traveltime curves are plotted as curves (a) in Figure 6.16b. Figure 6.16b.  Traveltime curves. Letters denote curves for respective parts (a), (b), (c).

## Problem 6.16b

Determine the traveltime curves when both refractor and reflector dip $8^{\circ }$ down to the left, the depths shown in Figure 6.16a now being the slant distances from $S$ to the interfaces.

### Solution

A combined graphical and calculated solution probably provides the easiest solution although Adachi’s method (see problem 11.5) could be used to give greater precision if the data accuracy warranted. A large-scale graph was used to achieve better accuracy; Figure 6.16c is a reduced-scale replica. The traveltime curves are shown in Figure 6.16b labeled (b).

The critical distance for the refraction is $SR_{1}=4.38$ km, and

{\begin{aligned}t_{R_{1}}=(1.00+0.18)/2.80+2\times 3.71/2.00=4.13\ {\rm {s}}.\end{aligned}} The $V_{2}$ -layer outcrops at $R_{2}=5.39$ km,

{\begin{aligned}t_{R_{2}}=1.00/2.80+2\times 3.71/2.00+1.12/4.20=4.33\ {\rm {s}}.\end{aligned}}  Figure 6.16c.  Geometry and raypaths for dip $8^{\circ }$ to the left.

The headwave has a different slope to the right of $R_{2}$ . To plot the curve in this zone, we use point $R_{4}$ at the offset $x=8.79\ {\rm {km}}$ . Then,

{\begin{aligned}t_{R_{4}}=1.00/2.80+(3.71+3.21)/2.00+4.71/4.20=4.94\ {\rm {s}},\end{aligned}} and the headwave curve is a straight line joining the traveltimes at the points $R_{2}$ and $R_{4}$ .

We have two types of reflected refractions: a typical path for the first type is $SMNP_{1}U_{2}WP_{4}R_{4}$ , the reflection occurring at the shallow dipping interface, The second type, $SMNP_{5}R_{6}P_{7}R_{7}$ , involves reflection at the surface. The first type exists between $R_{3}$ and $R_{4}$ , and the curve is parallel to the head-wave curve to the right of $R_{3}$ . The second type exists to the right of $R_{4}$ and the curve is parallel to the other reflected refraction. To plot the reflected-refraction curves, we need one point on each curve and then use the refraction-curve slope to the right of $R_{3}$ . For the first type, we find the coordinates of $R_{5}$ :

{\begin{aligned}x=6.80\ {\hbox{km}},\\t_{R_{5}}=1.00/2.80+(2\times 3.71+3.32+3.12)/2.00+1.12/4.20=7.55\ {\rm {s}}.\end{aligned}} For the second type we find coordinates of $R_{7}$ :

{\begin{aligned}x=9.15\ {\rm {km}},\\t_{\rm {R_{7}}}=1.00/2.80+(3.71+3.30+2.97+2.78)/2.00+3.85/4.20=7.65\ {\rm {s}}.\end{aligned}} ## Problem 6.16c

What happens when the reflector dips $3^{\circ }$ to the left and the refractor $5^{\circ }$ to the left? Figure 6.16d.  Geometry and raypaths for $3^{\circ }$ and $5^{\circ }$ dips to the left.

### Solution

A summary of the detailed graphical solution is as follows. The traveltime curves are shown in Figure 6.16b labeled (c). The various angles in Figure 6.16d are

{\begin{aligned}\theta _{c}=28.4^{\circ }\;,\;\theta _{2}=(28.4^{\circ }+5^{\circ }-3^{\circ })\;=30.4^{\circ },\\\theta _{1}={\rm {sin}}^{-1}[(2.80/2.00){\rm {\;sin\;}}30.4^{\circ }]=45.1^{\circ },\\\alpha _{1}={\rm {angle\ of\ approach}}\ =45.1^{\circ }+3^{\circ }=48.1^{\circ },\\\theta _{2}^{'}=(28.4^{\circ }-5^{\circ }+3^{\circ })\;=26.4^{\circ },\\\theta _{1}^{'}={\rm {sin}}^{-1}[(2.80/2.00){\rm {\;sin\;}}38.5^{\circ }]=38.5^{\circ },\\\alpha _{1^{'}}=38.5^{\circ }-3^{\circ }\;=35.5^{\circ }\end{aligned}} The refraction curve is a straight line though $R_{5}$ and $R_{7}$ :

{\begin{aligned}R_{5}:x^{'}=4.64\ {\rm {km}}={\rm {critical\ distance}},\\t_{R_{5}}=(1.09+0.44)/2.80+(3.65+3.50)/2.00=4.20\ {\rm {km}};\\R_{7}:x=8.15\ {\rm {km}},\\t_{R_{7}}=(1.09+0.44)/2.80+(3.64+3.35)/2.00+3.80/4.20\\=4.95\ {\rm {s}}.\end{aligned}} The incident angle at $W$ is $\theta _{3}=24.4^{\circ }$ , which is less than the critical angle, so that no refraction will be generated there, only a reflection. However, the refraction that starts at $N$ will give rise to upgoing rays which will be reflected, giving a reflected refraction, such as the ray that ends at $R_{8}$ . The traveltime curve is parallel to the refraction curve and exists beyond $R_{6}$ whose coordinates are

{\begin{aligned}x=7.79\ {\rm {km}},\\t_{R_{6}}=(1.09+0.48)/2.80+(3.64+3.50+3.42+3.36)/2.00=7.52\ {\rm {s}}.\end{aligned}} 