Determination of the Thickness of a Bed in the Case of Spatial Curvature of Exploration Wells

G. L. Kalinichenko, S. S. Sulakshin

If a well is set at the point \(M_0\) \((x_0, y_0, z_0)\) with azimuth angle \(\alpha_0\) and zenith angle \(\Theta_0\), then at the point \(M_1\) \((x_1, y_1, z_1)\), after \(l\) m (\(l\) is taken within the range \(5\text{--}10\) m), the coordinates of the point \(M_1\) \((x_1, y_1, z_1)\) will have the values (Fig. 1):

Fig. 1

\[ \begin{gathered} z_1=z_0+l\cdot \cos\theta,\\ x_1=x_0+l\cdot \sin\theta_0\cdot \cos\alpha_0,\\ y_1=y_0+l\cdot \sin\theta_0\cdot \sin\alpha_0 \end{gathered} \tag{1} \]

The derivation of these formulas is obvious.

If the angles are measured at equal intervals of well advance, then the coordinates of the end of the well in the \(i\)-th interval are found from the formulas:

\[ z_i=z_0+l\sum_{k=0}^{i-1}\cos\theta_k, \]

\[ x_i=x_0+l\sum_{k=0}^{i-1}\sin\theta_k\cos\alpha_k, \tag{2} \]

\[ y_i=y_0+l\sum_{k=0}^{i-1}\sin\theta_k\sin\alpha_k. \]

If the mineral appeared in the \(i\)-th sample, then the coordinates of the point at which the well meets the mineral are found from formulas (2). If the first sample \(i+m\) does not contain the mineral, then the coordinates of the point at which the well exits the bed are likewise found from formulas (2), except that the summation is carried out up to the index \(i+m-1\).

Having drilled 4 wells, by formulas (2) we find 4 points \((M_1, M_2, M_3, M_4)\) of intersection and 4 points \((N_1, N_2, N_3, N_4)\) at which the well exits the bed. Let us denote the coordinates of the point \(M_i\) by \(x_i, y_i, z_i\), the coordinates of the point \(N_i\) by \(x_i', y_i', z_i'\), and the coordinates of the vector \(M_iN_i\) by \(X_i, Y_i, Z_i\), where

\[ \begin{gathered} X_i=x_i'-x_i,\\ Y_i=y_i'-y_i,\\ Z_i=z_i'-z_i. \end{gathered} \tag{3} \]

From the four points at which the wells meet the bed we compute the determinant:

\[ \Delta= \begin{vmatrix} x_4-x_1 & y_4-y_1 & z_4-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1 \end{vmatrix} \tag{4} \]

If \(\Delta\), in absolute value, is close to zero, then these four points may be considered to lie in one plane. The normal vector to this plane will have coordinates:

\[ A= \begin{vmatrix} y_3-y_1 & z_3-z_1\\ y_2-y_1 & z_2-z_1 \end{vmatrix} \]

\[ B= \begin{vmatrix} z_3-z_1 & x_3-x_1\\ z_2-z_1 & x_2-x_1 \end{vmatrix} \tag{5} \]

\[ C= \begin{vmatrix} x_3-x_1 & y_3-y_1\\ x_2-x_1 & y_2-y_1 \end{vmatrix} \]

Using formulas (4) and (5), we carry out computations for the points \(N_1, N_2, N_3, N_4\). By formula (5) we find the coordinates of the normal vector of the plane bounding the bed from below. If these planes are parallel, then the thickness of the bed (its depth) is found by the formula

\[ h= \left| \frac{AX_i+BY_i+CZ_i}{\sqrt{A^2+B^2+C^2}} \right|, \tag{6} \]

where \(A, B, C\) are the coordinates of the normal vector found by formula (5), and \(X_i, Y_i, Z_i\) are the coordinates of the vector \(M_iN_i\).

Since the point where the well meets the bed and the point where the well exits the bed are relatively close to one another, the planes bounding the bed from below and above can, in practice, be considered parallel; therefore the calculation for the condition of parallelism of the lower and upper surfaces of the bed need not be carried out.

Example.
From point \(A_1\), 4 wellbores are directed; they meet the bed at points \(M_1, M_2, M_3\), and \(M_4\), and exit the beds at points \(N_1, N_2, N_3, N_4\) (Fig. 2). The coordinates of these points, in meters, were found by formulas (2):

Fig. 2

\[ \begin{aligned} M_1&\ (8.7;\quad 9.6;\quad 123.4)\\ M_2&\ (11.2;\quad 6.2;\quad 126.0),\\ M_3&\ (5.0;\quad 4.8;\quad 131.0)\\ M_4&\ (13.5;\quad 3.3;\quad 127.8) \end{aligned} \]

To compute \(\Delta\) by formula (4) and the coordinates of the normal vector of the upper plane of the bed \(A, B\), and \(C\) by formulas (5), we find:

\[ \begin{array}{lll} x_2-x_1=2.5; & y_2-y_1=-3.4; & z_2-z_1=2.6;\\ x_3-x_1=-3.7; & y_3-y_1=-4.8; & z_3-z_1=6.6;\\ x_4-x_1=4.8; & y_4-y_1=-6.3; & z_4-z_1=4.5. \end{array} \]

Substituting the values found into formula (4) and expanding it by the algebraic cofactors of the first row, we obtain

\[ \Delta = 4.8\cdot 9.96 - 6.3\cdot 25.1 + 4.5\cdot 24.6 = 47.8 - 158.1 + 110.7 = 0.4\ \text{m}^3. \]

It is easy to calculate that any point will lie on the plane drawn through the other three points within the limits of the error of the calculations (measurements).

Using formulas (5), we find the coordinates of the normal vector

\[ A=9.96;\quad B=25.1;\quad C=24.6, \]

which completely determine the direction of the bed.

Now let us find the thickness of the bed at the point of its intersection with the first borehole, which emerges from the bed at the point \(N_1(+7.3;\ 10.8;\ 127.6)\). From formulas (3) we find \(x_1, y_1, z_1\): \(x_1=-1.4;\ y_1=1.2;\ z_1=4.2\).

Now, by formula (4), we determine the thickness of the bed at this point:

\[ h=\frac{-1.4\cdot 9.96+1.2\cdot 25.1+4.2\cdot 24.6}{\sqrt{99.2+630.0+605.2}} =\frac{119.5}{36.5}=3.3\ \text{m}. \]

In the same way we can find the thickness of the bed at the points \(M_2\), \(M_3\), and \(M_4\).

REFERENCES

1. S. S. Sulakshin. Curvature of Boreholes. Gostgeoltekhizdat, 1960.
2. N. F. Frolov, E. F. Frolov. Geological Observations and Constructions in the Case of Borehole Curvature. Gostoptekhizdat, 1957.
3. I. I. Privalov. Analytical Geometry. Gostekhizdat, 1964.