Locating the bottom of a borehole

Problems in Exploration Seismology and their Solutions

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 4.23a

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

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

Figure 4.23b.  Snell’s law raypaths.

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%.

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Table of Contents (book)

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

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