# Cross-dip

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.10a

Two intersecting spreads have bearings N10${\displaystyle ^{\circ }}$E and N140${\displaystyle ^{\circ }}$E. If the first spread shows an event at ${\displaystyle t_{0}=1.760}$ s with dip moveout of 56 ms/km down to the south and the same event on the second spread has a dip moveout of 32 ms/km down to the northwest. Assume average velocity of 3.00 km/s.

1. Find the true dip, depth, and strike.
2. What are the values if the dip on the second spread is southeast?

### Solution

We first give numerical solutions, then graphical solutions.

i) We take the ${\displaystyle x}$-axis in the N40${\displaystyle ^{\circ }}$W direction, the ${\displaystyle y}$-axis in the N130${\displaystyle ^{\circ }}$W direction, and the ${\displaystyle y'}$-axis in the N170${\displaystyle ^{\circ }}$W direction (see Figure 4.10a). We now use equation (4.2j) to calculate ${\displaystyle (\Delta t/\Delta y)}$ for ${\displaystyle \alpha =130^{\circ }}$.

{\displaystyle {\begin{aligned}56=32\cos 130^{\circ }+(\Delta t_{y}/\Delta y)\cos 50^{\circ },\end{aligned}}}

Figure 4.10a.  Dips northwest and southwest.

hence, ${\displaystyle (\Delta t_{y}/\Delta _{y})=100}$ ms/km; total dip moveout${\displaystyle {}=(32^{2}+100^{2})^{1/2}=105}$ ms/km, dip ${\displaystyle \xi =\sin ^{-1}(3.00\times 0.105/2)=9.1^{\circ }}$,

{\displaystyle {\begin{aligned}\mathrm {strike} \ \Xi &=\tan ^{-1}(32/100)\\&=17.7^{\circ }\ \mathrm {relative\ to} \ x-\mathrm {axis} \\&=\mathrm {N} (40^{\circ }-17.7^{\circ })\mathrm {W} =\mathrm {N} 22.3^{\circ }\mathrm {W} ,\end{aligned}}}

depth ${\displaystyle h={\frac {1}{2}}\times 3.00\times 1.76=2.64}$ km (normal to bed).

ii) We take the ${\displaystyle x}$-axis in the S10${\displaystyle ^{\circ }}$W direction (see Figure 4.10b) where ${\displaystyle \alpha =50^{\circ }}$, so

{\displaystyle {\begin{aligned}32=56\cos 50^{\circ }+(\Delta t_{y}/\Delta y)\sin 40^{\circ },(\Delta t_{y}/\Delta y)=-5.2\ \mathrm {ms/km} ,\end{aligned}}}

The positive ${\displaystyle y}$-axis is toward S80${\displaystyle ^{\circ }}$E, so the minus sign means that the ${\displaystyle y}$-component of dip is in the N80${\displaystyle ^{\circ }}$W direction.

{\displaystyle {\begin{aligned}{\hbox{Total dip moveout}}&=(56^{2}+5.2^{2})^{1/2}\\&=56.2\ \mathrm {ms/km} \end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \mathrm {dip} \ \xi =\sin ^{-1}\left({\frac {1}{2}}\times 0.056\times 3.00\right)=4.8^{\circ },\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{and}}\qquad \qquad \mathrm {strkie} \ \Xi =\tan ^{-1}[56/(-5.2)]=-84.7^{\circ }.\end{aligned}}}

Figure 4.10b.  Dips southeast and southwest.

The strike is measured relative to the negative direction of the ${\displaystyle x}$-axis as shown in Figure 4.2d where both dip components are positive; when the ${\displaystyle y}$-component of dip is negative, as it is here, the strike line goes from ${\displaystyle M}$ to a point on the negative ${\displaystyle y}$-axis. Referring to Figure 4.10b, we see that the strike line is rotated counter-clockwise ${\displaystyle 84.7^{\circ }}$ from the negative direction. Graphical solutions for (i) and (ii) are shown in Figure 4.10c.

Figure 4.10c.  Graphical solutions to cross-dip problem.

## Problem 4.10b

Calculate the position of the reflecting point (migrated position) for each spread in (i) as if the cross information had not been available and each had been assumed to indicate total moveout; compare with the result of part (a). Would the errors be more or less serious if the calculations were made for the usual situation where the velocity increases with depth?

### Solution

We find the coordinates of the migrated reflecting points assuming the velocity is constant at the value of the average velocity. We take the ${\displaystyle x}$-, ${\displaystyle y}$-, and ${\displaystyle z}$-axes positive to the south, west, and downward, respectively, the source being at the origin.

In (i) the spread along the ${\displaystyle x}$-axis with bearing N10${\displaystyle ^{\circ }}$E has dip moveout down to the south and west (Figure 4.10b), hence

{\displaystyle {\begin{aligned}\xi =\mathrm {dip} =\sin ^{-1}(3.0\times 0.056/2)=4.8^{\circ }.\\h={\hbox{depth normal to bed}}\approx {\hbox{vertical depth}}={\frac {1}{2}}\times 3.00\times 1.760=2.64\ \mathrm {km} .\end{aligned}}}

Since the dip is mainly south and west, the reflection point is shifted north and east along the spread direction, the distance ${\displaystyle 2640\sin 4.8^{\circ }=221}$ m. Resolving this along the ${\displaystyle x}$- and ${\displaystyle y}$-axes, we get ${\displaystyle x=-221\cos 10^{\circ }=-218}$ m and ${\displaystyle y=-221\sin 10^{\circ }=-38}$ m. The vertical depth is ${\displaystyle 2640\times \cos 4.8^{\circ }=2630}$ m, giving coordinates (${\displaystyle -218,-38,2630}$).

For the spread bearing S140${\displaystyle ^{\circ }}$E (Figure 4.10b),

{\displaystyle {\begin{aligned}\xi =\sin ^{-1}(3.0\times 0.032/2)=2.8^{\circ }\ \mathrm {S} 40^{\circ }\mathrm {E} .\end{aligned}}}

The reflecting point is ${\displaystyle 2640\times \sin 2.8^{\circ }=129}$ m north and west of the source. Thus, ${\displaystyle x=-129\cos 40^{\circ }=-99}$ m, ${\displaystyle y=129\sin 40^{\circ }=83}$ m, ${\displaystyle z=2640\cos 2.8^{\circ }=2640}$ m, so the coordinates are ${\displaystyle (-99,83,2640)}$.

Taking into account cross-dip, the total dip is ${\displaystyle 9.1^{\circ }}$ down to the south and west. The horizontal displacement of the reflecting point is ${\displaystyle 2640\sin 9.1^{\circ }=418}$ m in the direction ${\displaystyle \mathrm {N} (22^{\circ }+90^{\circ })\mathrm {E} =\mathrm {N} 112^{\circ }\mathrm {E} (\mathrm {since} \Xi =\mathrm {N} 22^{\circ }\mathrm {W} )}$. Thus, ${\displaystyle x=-418\cos(180^{\circ }-112^{\circ })=-157}$ m, ${\displaystyle y=-418\sin(180^{\circ }-112^{\circ })=-388}$ m, ${\displaystyle z=2640\cos 9.1^{\circ }=2610}$ m. The coordinates are now ${\displaystyle (-157,-388,2610)}$. The change in ${\displaystyle z}$ is small ${\displaystyle (30\ \mathrm {m} \approx 1\%)}$ but the ${\displaystyle x}$- and ${\displaystyle y}$-coordinates vary considerably, both percentagewise and in absolute values.

The errors become more serious when the velocity increases with depth because these calculations are based on the average velocity ${\displaystyle {\bar {V}}}$ rather than the interval velocity ${\displaystyle V_{i}}$, which is usually greater than ${\displaystyle {\bar {V}}}$.