# Locating the bottom of a borehole

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 4.23a

Barton (1929) discusses shooting into a geophone placed in a borehole (Figure 4.23a) to determine where the bottom is located. Point $C$ is vertically above the well geophone at $C^{\prime },A,B,D$ , and $E$ are equidistant from $W$ in the cardinal directions, and the traveltimes from sources $D$ and $E$ to $C^{\prime }$ are equal.

Assuming straight-line travelpaths at the velocity $V$ , derive expressions for $C^{\prime }C$ and $CW$ in Figure 4.23a(ii) in terms of the traveltimes to $C^{\prime }$ from $A$ and $B,\,t_{A}$ and $t_{B}$ .

### Solution

Since $EW=DW$ , $t_{DC^{\prime }}=t_{EC^{\prime }}$ , and $AB\perp DE$ , point $C^{\prime }$ must be in the vertical plane through $AWB$ and $C$ must lie on the straight line $AWB$ . Then, letting $h=C^{\prime }C,y=CW,x=AW=BW=DW=EW$ , we have

 {\begin{aligned}(Vt_{AC^{\prime }})^{2}=(x-y)^{2}+h^{2},\end{aligned}} (4.23a)

 {\begin{aligned}(Vt_{BC^{\prime }})^{2}=(x+y)^{2}+h^{2}.\end{aligned}} (4.23b)

Subtracting, we find

{\begin{aligned}V^{2}(t_{BC^{\prime }}^{2}-t_{AC^{\prime }}^{2})=4xy,\end{aligned}} {\begin{aligned}{\hbox{hence}}\qquad \qquad CW=y=\left(V^{2}/4x\right)\left(t_{BC^{\prime }-t_{AC^{\prime }}^{2}}^{2}\right).\end{aligned}} (4.23c)

Adding equations (4.23a) and (4.23b) gives

 {\begin{aligned}&V^{2}\left(t_{AC^{\prime }}^{2}+t_{BC^{\prime }}^{2}\right)=2(x^{2}+y^{2}+h^{2}),\\{\hbox{and}}\qquad C^{\prime }C=h&=\left[(V^{2}/2)\left(t_{AC^{\prime }}^{2}+t_{BC^{\prime }}^{2}\right)-(x^{2}+y^{2})\right]^{1/2}.\end{aligned}} (4.23d)

Since all quantities on the right are known, we can find $C^{\prime }C$ . Figure 4.23a.  Mapping a crooked borehole (from Barton, 1929). (i) Plan view; (ii) vertical section.

## Problem 4.23b

What are the values of $t_{AC^{\prime }}$ and $t_{BC^{\prime }}$ for $V=2500$ m/s, $AW=BW=C^{\prime }C=1000$ m, $CW=200$ m?

### Solution

From equation (4.23a),

{\begin{aligned}(Vt_{AC^{\prime }})^{2}=(x-y)^{2}+h^{2},\end{aligned}} {\begin{aligned}{\hbox{so}}\qquad \qquad t_{AC^{\prime }}=[(1.00-0.20)^{2}+1.00^{2}]^{1/2}/2.50=0.512\,\mathrm {s} ,\\(Vt_{BC^{\prime }})^{2}=(x+y)^{2}+h^{2}=1.20^{2}+1.00^{2},t_{BC^{\prime }}=0.625\,\mathrm {s} .\end{aligned}} ## Problem 4.23c

How sensitive is the method, that is, what are $\Delta (CC^{\prime }$ /$\Delta t_{AC^{\prime }}$ and $\Delta (CW)/\Delta t_{AC^{\prime }}$ ? For the specific situation in part (b), how much change is there in $CW$ and $C^{\prime }C$ per millisecond error in $t_{AC^{\prime }}$ ?

### Solution

{\begin{aligned}\Delta (CC^{\prime })/\Delta t_{AC^{\prime }}\approx \mathrm {d} h/\mathrm {d} t_{AC^{\prime }},\quad \Delta (CW)/\Delta t_{AC^{\prime }}\approx \mathrm {d} y/\mathrm {d} t_{AC^{\prime }}.\end{aligned}} From equation (4.23c), assuming $t_{B}$ fixed, we get

 {\begin{aligned}\mathrm {d} y/\mathrm {d} t_{AC^{\prime }}=-V^{2}t_{AC}/2x.\end{aligned}} (4.23e)

Differentiating equation (4.23a) and using equation (4.23e) gives

{\begin{aligned}V^{2}t_{AC^{\prime }}&=(x-y)(-\mathrm {d} y/\mathrm {d} t_{AC^{\prime }})+h\,\mathrm {d} h/\mathrm {d} t_{AC},\\\mathrm {d} h/\mathrm {d} t_{AC^{\prime }}&=(1/h)[V^{2}t_{AC^{\prime }}-(x-y)V^{2}t_{AC^{\prime }}/2x]=(V^{2}t_{AC^{\prime }}/h)[1-(x-y)/2x]\\&=(V^{2}t_{AC^{\prime }}/2h)(1+y/x).\end{aligned}} Using values from part (b), we obtain

{\begin{aligned}\mathrm {d} y/\mathrm {d} t_{AC^{\prime }}=-2.50^{2}\times 0.512/(2\times 1.00]=-1.60\,\mathrm {km/s} =-1.6\,\mathrm {m/ms} .\end{aligned}} For $\Delta t_{AC^{\prime }}=1$ ms, $CW=-1.6$ m.

Also, $\mathrm {d} h/\mathrm {d} t_{AC^{\prime }}=[2.50^{2}\times 0.512/(2\times 1.00)](1+0.20/1.00)=1.92\,\mathrm {km/s}$ .

For $\Delta t_{AC^{\prime }}=1$ ms, $\Delta h=1.9$ m.

## Problem 4.23d

Assume a velocity of 1500 m/s for the first 500 m and 3500 m/s below 500 m. What are the traveltimes now and how would these be interpreted if straight raypaths are assumed?

### Solution

By trial and error we find that the angles should be as shown in Figure 4.23b. Then

{\begin{aligned}t_{AC^{\prime }}=500/(1500\cos 22.5^{\circ })+500/(3500\cos 51.2^{\circ })=0.582\,\mathrm {s} ,\\t_{BC^{\prime }}=500/(1500\cos 22.5^{\circ })+500/(3500\cos 63.2^{\circ })=0.678\,\mathrm {s} .\end{aligned}} Interpreting these results as in part (a), we get

{\begin{aligned}{\overline {V}}&=(1/2)(1500+3500)=2500\,\mathrm {m/s} ,\\y&=({\overline {V}}^{2}/4x)\left(t_{BC_{\prime }}^{2}-t_{AC^{\prime }}^{2}\right)\\&=(2500^{2}/4\times 1.00)(0.678^{2}-0.582^{2})=189\,\mathrm {m} ,\\h^{2}&=({\overline {V}}^{2}/2)(t_{AC^{\prime }}^{2}+t_{BC^{\prime }}^{2})-(x^{2}+y^{2})\\&=(2500^{2}/2)(0.582^{2}+0.678^{2})\\&\quad -(1.00^{2}+0.19^{2}),\\h&=1210\,\mathrm {m} \end{aligned}} Thus, $y$ varies only 5%, mainly because we subtract the squares of traveltimes, thus partially canceling errors. However, the change in $h$ is more than 20%.