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 ${\displaystyle C}$ is vertically above the well geophone at ${\displaystyle C^{\prime },A,B,D}$, and ${\displaystyle E}$ are equidistant from ${\displaystyle W}$ in the cardinal directions, and the traveltimes from sources ${\displaystyle D}$ and ${\displaystyle E}$ to ${\displaystyle C^{\prime }}$ are equal.

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

Solution

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

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

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

Subtracting, we find

{\displaystyle {\begin{aligned}V^{2}(t_{BC^{\prime }}^{2}-t_{AC^{\prime }}^{2})=4xy,\end{aligned}}}

 {\displaystyle {\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

 {\displaystyle {\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 ${\displaystyle C^{\prime }C}$.

Figure 4.23a.  Mapping a crooked borehole (from Barton, 1929). (i) Plan view; (ii) vertical section.

Problem 4.23b

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

Solution

From equation (4.23a),

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

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

Solution

{\displaystyle {\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 ${\displaystyle t_{B}}$ fixed, we get

 {\displaystyle {\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

{\displaystyle {\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

{\displaystyle {\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}}}

Figure 4.23b.  Snell’s law raypaths.

For ${\displaystyle \Delta t_{AC^{\prime }}=1}$ ms, ${\displaystyle CW=-1.6}$ m.

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

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

{\displaystyle {\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

{\displaystyle {\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, ${\displaystyle y}$ varies only 5%, mainly because we subtract the squares of traveltimes, thus partially canceling errors. However, the change in ${\displaystyle h}$ is more than 20%.