Resolution of cross-dip
|
| |
| Series | Geophysical References Series |
|---|---|
| Title | Problems in Exploration Seismology and their Solutions |
| Author | Lloyd P. Geldart and Robert E. Sheriff |
| Chapter | 4 |
| Pages | 79 - 140 |
| DOI | http://dx.doi.org/10.1190/1.9781560801733 |
| ISBN | ISBN 9781560801153 |
| Store | SEG Online Store |
Problem 4.9a
Sources $ B $ and $ C $ are, respectively, 600 m north and 500 m east of source $ A $. Traveltimes to zero-offset geophones at $ A $, $ B $, and $ C $ for a certain reflection are 1.750, 1.825, and 1.796 s. What are the dip and strike of the horizon, the average velocity being 3.25 km/s?
Solution
The dip moveouts are $ 0.075/0.600=0.125 $ s/km to the north and $ 0.046/0.500=0.092 $ s/km to the east. We take the $ x $-axis east-west (see Figure 4.9a). If we apply equation (4.2b) as in problem 4.2c, we get
$ {\begin{aligned}\sin \xi _{y}&=(3.25/2)(0.075/0.60)=0.203\\&=m;\qquad \xi _{y}=11.7^{\circ };\\\sin \xi _{x}&=(3.25/2)(0.046/0.50)=0.150\\&=\ell ;\qquad \xi =8.6^{\circ }.\end{aligned}} $

Using equations (4.2g,h), we find that
$ {\begin{aligned}\sin \xi &=(\ell ^{2}+m^{2})^{1/2}=(0.150^{2}+0.203^{3})^{1/2}=0.252,\quad \xi =14.6^{\circ };\\\mathrm {strike} \ \Xi &=\tan ^{-1}(\ell /m)=\tan ^{-1}(0.150/0.203)\\&=36.4^{\circ }\ {\hbox{with respect to the}}\ x{\hbox{-axis}}{}={\hbox{N}}53.6^{\circ }{\hbox{W}}.\end{aligned}} $
Problem 4.9b
What are the changes in dip and strike if line AC has the bearing N80Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): ^{\circ} E?
Solution
We use equation (4.2j) and Figure 4.2e to obtain the dip moveout along the east-west Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): x -axis when the measured moveout along the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): x' -axis with bearing N80Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): ^{\circ} E is 0.092 s/km. Equation (4.2j) gives
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} \frac{3.25}{2}\left(\frac{0.046}{0.500}\right) &= 0.150 \\ &= \ell \cos 10^\circ + 0.203 \sin 10^{\circ}; \end{align}

thus,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} \ell = 0.150/\cos 10^{\circ} - 0.203\tan 10^\circ = 0.117. \end{align}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} \hbox{So} \qquad\qquad \sin\xi = (0.117^2 + 0.203^2)^{1/2} = 0.234,\quad \xi = 13.6^\circ,\\ \tan \Xi = 0.117/0.203 = 0.576, \end{align}
and the strike angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \Xi = 30.0^{\circ} , that is, N60.0Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): ^{\circ} W. Thus, the $ 10^{\circ } $ change in line direction changes the dip about Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): 1.0^{\circ} and the strike about Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): 6^{\circ} .
Problem 4.9c
Solve part (b) graphically.
Solution
The graphical construction, shown reduced in Figure 4.9c, yields dip moveout of 0.145 s/km and strike of N61Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): ^{\circ} W.
The dip moveout 0.145 gives a dip 13.6Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): ^{\circ} . Thus the graphical solution gives almost exactly the same values as in part (b).

Continue reading
| Previous section | Next section |
|---|---|
| Effect of migration on plotted reflector locations | Cross-dip |
| Previous chapter | Next chapter |
| Partitioning at an interface | Seismic velocity |
Also in this chapter
- Accuracy of normal-moveout calculations
- Dip, cross-dip, and angle of approach
- Relationship for a dipping bed
- Reflector dip in terms of traveltimes squared
- Second approximation for dip moveout
- Calculation of reflector depths and dips
- Plotting raypaths for primary and multiple reflections
- Effect of migration on plotted reflector locations
- Resolution of cross-dip
- Cross-dip
- Variation of reflection point with offset
- Functional fits for velocity-depth data
- Relation between average and rms velocities
- Vertical depth calculations using velocity functions
- Depth and dip calculations using velocity functions
- Weathering corrections and dip/depth calculations
- Using a velocity function linear with depth
- Head waves (refractions) and effect of hidden layer
- Interpretation of sonobuoy data
- Diving waves
- Linear increase in velocity above a refractor
- Time-distance curves for various situations
- Locating the bottom of a borehole
- Two-layer refraction problem