# Refraction interpretation by stripping

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 11.6a

Solve problem 11.5 by stripping off the shallow layer.

### Background

Stripping is a method of interpreting refraction data by removing the effect of upper layers, the removal being accomplished by reducing the traveltimes and distances so that in effect the source and geophones are located on the interface at the base of the “stripped” layer. Stripping can be accomplished by calculation or graphically, or by a combination.

### Solution

We wish to compare our results with those of problem 11.5, so we use the same measurements, namely ${\displaystyle V_{1}=2.02}$ km/s and

{\displaystyle {\begin{aligned}V_{d2}=3.73\ {\rm {km/s}},\quad V_{u2}=4.51\ {\rm {km/s}},\quad t_{u1}=0.92\ {\rm {s}},\quad t_{d1}=0.46\ {\rm {s}};\\V_{d3}=4.29\ {\rm {km/s}},\quad V_{u3}=5.81\ {\rm {km/s}},\quad t_{u2}=1.28\ {\rm {s}},\quad t_{d2}=0.66\ {\rm {s}}.\end{aligned}}}

(To avoid triple subscripts, we denote intercept times at downdip and updip source locations by ${\displaystyle d}$ and ${\displaystyle u}$.)

We start by using equations (4.24f) to get ${\displaystyle V_{2}}$:

{\displaystyle {\begin{aligned}1/V_{2}=\left(1/V_{d2}+1/V_{u2}\right)/2,\quad V_{2}=4.08\ {\rm {km/s}}.\end{aligned}}}

Equations (4.24b,d) can be written

{\displaystyle {\begin{aligned}\sin \left(\theta _{c1}+\xi _{2}\right)=V_{1}/V_{d2},\qquad \sin(\theta _{c1}-\xi _{2})=V_{1}/V_{u2};\\{\hbox{so}}\quad \quad \sin(\theta _{c1}+\xi _{2})=2.02/3.73,\qquad (\theta _{c1}+\xi _{2})=32.8^{\circ };\\\sin(\theta _{c1}-\xi _{2})=2.02/4.51,\qquad (\theta _{c1}-\xi _{2})=26.6^{\circ };\end{aligned}}}

hence ${\displaystyle \theta _{c1}=29.7^{\circ }}$, ${\displaystyle \xi _{2}=3.1^{\circ }}$. These are the same as those in problem 11.5.

Next we calculate the distances perpendicular to the first refractor at ${\displaystyle A}$ and ${\displaystyle B}$ (Figure 11.6a). We use equation (4.24b) to get ${\displaystyle h_{d}}$ and ${\displaystyle h_{u}}$:

{\displaystyle {\begin{aligned}h_{d}=\left(V_{1}t_{d1}\right)/2\cos \theta _{c1}=2.02\times 0.46/2\times \cos 29.7^{\circ }=0.53\ {\rm {km}};\\h_{u}=\left(V_{1}t_{u1}\right)/2\cos \theta _{c1}=2.02\times 0.92/2\times \cos 29.7^{\circ }=1.07\ {\rm {km}}.\end{aligned}}}

These results are identical with those in problem 11.5. We verify the dip using these depths:

{\displaystyle {\begin{aligned}\xi _{2}=\tan ^{-1}\left[\left(1.07-0.53\right)/10.0\right]=3.1^{\circ }.\end{aligned}}}

The first step in stripping is to plot the shallow refractor; we do this by swinging arcs with centers ${\displaystyle A}$ and ${\displaystyle B}$ and radii 1.07 and 0.53 km, the refractor being tangent to the two arcs. To get the “stripped” time values, we subtract the times down to and up from the first refractor, i.e., traveltimes along ${\displaystyle AA'}$ and ${\displaystyle BB'}$ for sources ${\displaystyle A}$ and ${\displaystyle B}$. Although maximum accuracy would be achieved by stripping times for all geophones, the curves for the shallow refraction are so nearly linear that we calculate the stripped times only for each source and one intermediate point on each profile (${\displaystyle Q}$ and ${\displaystyle N}$). We obtain the required distances by measuring the paths in Figure 11.6a. Calculation of the stripped times is given below. Path lengths: ${\displaystyle AA'\approx 1.30}$, ${\displaystyle MN\approx 0.83}$, ${\displaystyle BB'\approx 0.65}$, ${\displaystyle PQ\approx 1.08}$ km.

Figure 11.6a.  Stripping for refraction interpretation. The numbers below the zero-time line are distances from ${\displaystyle A'}$.

{\displaystyle {\begin{aligned}{\hbox{Time along path:}}\quad \quad \ AA'\approx 0.64,\ MN\approx 0.41,\ BB'\approx 0.32,\ PQ\approx 0.53\ {\rm {s}}\\{\hbox{Time along path:}}\quad \quad \ AA'MN=2.50\ {\rm {s}}\\{\hbox{Stripped time for}}\quad \quad \ AA'MN\approx 2.50-\left(0.64+0.41\right)\approx 1.45\ {\rm {s}}\ ({\rm {point}}\ E)\\{\hbox{Time along path}}\quad \quad \ AA'B'B=3.00\ {\rm {s}}\\{\hbox{Stripped time for}}\quad \quad \ AA'B'B\approx 3.00-\left(0.64+0.32\right)\approx 2.04\ {\rm {s}}\ ({\rm {points}}\ F,H)\\{\hbox{Time along path}}\quad \quad \ BB'PQ=2.30\ {\rm {s}}\\{\hbox{Stripped time for}}\quad \quad \ BB'PQ\approx 2.30-\left(0.32+0.53\right)\approx 1.45\ {\rm {s}}\ ({\rm {point}}\ G)\end{aligned}}}

Stripping off the first refractor in effect moves sources ${\displaystyle A}$ and ${\displaystyle B}$ down to ${\displaystyle A'}$ and ${\displaystyle B'}$ and geophones at ${\displaystyle Q}$ and ${\displaystyle N}$ down to ${\displaystyle P}$ and ${\displaystyle M}$, so the stripped times are plotted above these shifted points, the new traveltimes curves being ${\displaystyle EF}$ and ${\displaystyle GH}$. Measurements on these stripped curves give the following:

{\displaystyle {\begin{aligned}V_{d3}=4.65\ {\rm {km/s}},\quad V_{u3}=5.13\ {\rm {km/s}},\quad t_{d3}=0.19\ {\rm {s}},\quad t_{u3}=0.30\ {\rm {s}}.\end{aligned}}}

Table 11.6a. Comparison of results of Adachi’s and stripping method.
Item Problem 11.5 Problem 11.6 Difference
${\displaystyle V_{3}({\hbox{km/s}})}$ 4.92 4.88 0.8%
${\displaystyle \theta _{c2}}$ ${\displaystyle 56.0^{\circ }}$ ${\displaystyle 56.7^{\circ }}$ 1%
${\displaystyle \xi _{3}}$ ${\displaystyle 5.8^{\circ }}$ ${\displaystyle 5.7^{\circ }}$ 2%
${\displaystyle h_{2}^{*}({\hbox{km}})}$ 1.32 ${\displaystyle 1.20^{*}}$ 10%
 *Vertical depth measured at source A.

We now get

{\displaystyle {\begin{aligned}V_{3}&=[\left(1/4.65+1/5.13\right)/2]^{-1}=4.88\ {\rm {km/s}},\\\theta _{c2}&=\sin ^{-1}\left(4.08/4.88\right)=56.7^{\circ },\\h_{u2}&=V_{2}t_{u3}/2\cos \theta _{c2}=4.08\times 0.30/2\cos 56.7^{\circ }=1.11\ {\rm {km}},\\h_{d2}&=4.08\times 0.19/2\cos 56.7^{\circ }=0.71\ {\rm {km}},\\\xi _{3}&=\tan ^{-1}\left[\left(h_{u2}-h_{d2}\right)/A'B'\right]=\tan ^{-1}\left[\left(1.11-0.71\right)/8.90\right]=2.6^{\circ }.\end{aligned}}}

This dip is relative to ${\displaystyle A'B'}$, so the total dip is ${\displaystyle \xi _{2}+\xi _{3}=\left(3.1^{\circ }+2.6^{\circ }\right)=5.7^{\circ }}$

## Problem 11.6b

Compare the solutions by stripping with those using Adachi’s method (problem 11.5).

### Solution

To compare depths, we measured vertical depths below A. Results for the first layer are the same for both methods, those for the next layer are given in Table 11.6a.

## Problem 11.6c

### Solution

• Easy to understand
• Straightforward in application
• Can be used with beds of different dips if the strike is the same
• As rapid as other methods when done graphically
• Can be used to interpret irregular or curved surfaces

• Very sensitive to velocity errors
• Like most methods, assumes the same strike for all refractors
• Difficult to apply when dips are steep
Figure 11.7a.  Refractors with the same strike but different dips.