Reflector dip in terms of traveltimes squared

From SEG Wiki
Jump to navigation Jump to search
ADVERTISEMENT

Problem 4.4a

Using the dip-moveout equation (4.2b) and the result of problem 4.3, verify that

$ {\begin{aligned}\tan \xi =t/(t_{SR}^{2}-t_{0}^{2})^{1/2},\end{aligned}} $

where $ \xi ={\rm {dip}} $, $ t=t_{SM}-t_{MR} $, $ t_{SR}= $ traveltime for path $ {\rm {SR}}'{\rm {R}}={\rm {S}}'{\rm {R}}'{\rm {R}} $, $ t_{0}=2h_{c}/V= $ traveltime to receiver at $ M $ (Figure 4.4a).

Solution

Since $ t $ in problem 4.3 equals $ t_{SR} $ here, we have

$ {\begin{aligned}(Vt_{SR})^{2}=(2x\cos \xi )^{2}+4h_{c}^{2}=(2x\cos \xi )^{2}+(Vt_{0})^{2},\end{aligned}} $

$ {\begin{aligned}{\hbox{or}}\qquad \qquad \cos \xi =\left(V/2x\right)(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}} $

From equation (4.2b) we get

$ {\begin{aligned}\sin \xi =\left(V/2\right)\left(\Delta t_{d}/\Delta x\right)=\left(V/2x\right)\left(t_{SM}-t_{MR}\right),\end{aligned}} $

$ {\begin{aligned}{\hbox{or}}\qquad \qquad \tan \xi =\left(t_{SM}-t_{MR}\right)/(t_{SR}^{2}-t_{0}^{2})^{1/2}=t/(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}} $

Figure 4.4a.  Geometry for dipping bed.

Problem 4.4b

Using equation (4.2a), show that

$ {\begin{aligned}\sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}} $

Solution

Equation (4.2a) gives

$ {\begin{aligned}(Vt_{SM})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\(Vt_{MR})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\{\hbox{thus}}\qquad \qquad \sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}} $

Problem 4.4c

Under what circumstances is the result for part (b) the same as equation (4.2b) and also consistent with part (a)?

Solution

The result in part (b) can be written

$ {\begin{aligned}\sin \xi =\{[V(t_{MS}-t_{MR})]/2x\}\{[V(t_{MS}+t_{MR})]/4h_{c}\}.\end{aligned}} $

The expression in the first curly brackets is the same as the right-hand side of equation (4.2b). Hence, for the above to be the same as equation (4.2b), we must have $ V(t_{MS}+t_{MR})/4h_{c}=1 $ that is,

$ {\begin{aligned}V(t_{MS}+t_{MR})=4h_{c}=2Vt_{0},\end{aligned}} $

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\qquad\qquad t_0 = \frac{1}{2}(t_{MS} + t_{MR}). \end{align}

In part (a) we got 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/":): \tan \xi by finding 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/":): \sin \xi and 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/":): \cos\xi . The expression for 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/":): \cos \xi involves no approximation, so the only approximation is that used in the derivation of equation (4.2b) to get 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/":): \sin \xi . Thus, results in (a) and (c) are consistent provided we take 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/":): t_0 =\frac{1}{2}(t_{MS} + t_{MR}) .

Continue reading

Previous section Next section
Relationship for a dipping bed Second approximation for dip moveout
Previous chapter Next chapter
Partitioning at an interface Seismic velocity

Table of Contents (book)

Also in this chapter

External links

find literature about
Reflector dip in terms of traveltimes squared
Retrieved from "http://wiki.seg.org/w/index.php?title=Reflector_dip_in_terms_of_traveltimes_squared&oldid=121026"