ON THE DEPENDENCE OF THE RATE OF ROTARY DRILLING OF BOREHOLES ON THE FEED FORCE

O. D. Alimov, L. T. Dvornikov

In work [1] we showed that between the rate of rotary drilling of boreholes and the feed force there exists the dependence

\[ V=\kappa(P_{\mathrm{п}}-P_{0}), \tag{1} \]

where \(V\) is the drilling rate, mm/min;

\(\kappa\) is the coefficient of efficiency of rotary drilling, mm/kg·min;

\(P_{\mathrm{п}}\) is the feed force on the spindle of the drilling machine, kg;

\(P_{0}\) is the feed force corresponding to the onset of volumetric failure of the rock, kg.

We determined the coefficients \(\kappa\) and \(P_{0}\) by generalizing experimental data from many researchers.

The dependence of the coefficient \(\kappa\) on the coefficient of rock strength according to the scale of Prof. M. M. Protodyakonov is shown in Fig. 1. The dependence was constructed from 11 points, each of which was obtained by processing the experimental dependence \(V=f(P_{\mathrm{п}})\), obtained by one or another researcher in drilling one or another rock. The value \(\kappa\) was determined from the dependence

\[ \kappa=\frac{V}{P_{\mathrm{п}}-P_{0}}. \]

Since the dependences \(V=f(P)\) in each particular case were not strictly linear, for different values of \(V\) \((V_{1}, V_{2}, \ldots, V_{n})\) and the corresponding \(P_{\mathrm{п}}\) \((P_{\mathrm{п}1}, P_{\mathrm{п}2}, \ldots, P_{\mathrm{п}n})\), values \((\kappa_{1}, \kappa_{2}, \ldots, \kappa_{n})\) could appear.

The most probable actual value of \(\kappa\) is the arithmetic mean \(\bar{\kappa}\) of all values \(\kappa_{1}, \kappa_{2}, \ldots, \kappa_{n}\), since all values of \(\kappa\) were obtained with the same accuracy (using the same methodology and measuring equipment).

Thus,

\[ \bar{\kappa}=\frac{\sum \kappa}{n}. \]

An objective characteristic of the accuracy in determining the arithmetic mean value is the root-mean-square error

\[ \sigma=\sqrt{\frac{\sum(\bar{\kappa}-\kappa_i)^2}{n-1}}, \tag{2} \]

where \((\bar{\kappa}-\kappa_i)\) is the mean deviation of one measurement;
\(n\) is the number of measurements.

The value \(\sigma\) determined by formula (2) is the root-mean-square error of one measurement. The root-mean-square error of the arithmetic mean is

\[ \sigma_{\kappa}=\frac{\sigma}{\sqrt{n}}. \]

Table 1

Rock: medium-grained sandstone with \(f=8\), according to A. N. Volkov [2];
optimum tool rotation speed: 270 rpm.

\[ \kappa=(1.58\pm0.02);\qquad W=\frac{0.02}{1.58}\cdot100=1.3\%. \]

The actual value of \(\kappa\) can be determined as

\[ \kappa=\bar{\kappa}\pm t\cdot\sigma_{\bar{\kappa}} \]

or

\[ \bar{\kappa}-t\sigma_{\bar{\kappa}}<\kappa<\bar{\kappa}+t\cdot\sigma_{\bar{\kappa}}; \]

here \(t\) is the confidence coefficient, any positive number.

Using the “three-sigma” rule, which consists in the assertion that the error of the approximate equality

\[ \kappa\approx\bar{\kappa} \]

does not exceed three times the root-mean-square error of the mean \(\bar{\kappa}\), one may write

\[ \bar{\kappa}-3\sigma_{\bar{\kappa}}<\kappa<\kappa+3\sigma_{\bar{\kappa}}. \]

For a comparative assessment of the accuracy of various characteristics, the so-called coefficient of variation is used; it is defined as the ratio of the root-mean-square error to the arithmetic mean value of the quantity. In the case under consideration, the coefficient of variation is

\[ W=\frac{\sigma_{\bar{\kappa}}}{\bar{\kappa}}\cdot100\%. \]

Table 2

Table 1 shows an example of processing the dependence \(V=f(P_{\text{п}})\) for the purpose of determining \(\bar{k}\), \(\sigma\), \(\sigma_{\bar{k}}\), \(W\). The results of processing eleven such dependences, with references to the sources, are given in Table 2.

As can be seen from Table 2, the coefficient of variation does not exceed 10%, which in mine-engineering calculations may be assigned to the class of “accurate” [10].

To study and describe the dependence of \(k\) on the rock-strength coefficient \(f\), we plot on the graph \(k=\varphi(f)\) the points according to Table 2 (Fig. 1). It is evident from the graph that the dependence \(k=\varphi(f)\) can be divided into two segments: \(k=\varphi_1(f)\) for \(f=2 \div 6\) and \(k=\varphi_2(f)\) for \(f \geq 6\). In both segments the dependences can be described by straight lines.

Fig. 1

We shall use the method of least squares to describe \(k=\varphi(f)\) piecewise. According to this method, the best curve describing the dependence is the one for which the sum of the squares of the deviations of the points from it is the smallest, i.e.,

\[ \sum \varepsilon^2 = \text{minimum}. \tag{3} \]

In the range \(f=2 \div 6\) we have a series of points (Table 3).

Table 3

We describe this dependence by a straight line that passes as close as possible to all the points. We write the equation of the straight line in the form:

\[ k = B_0 + B_1 f. \]

Substituting into this equation the values of \(k\) and \(f\) from Table 3, we find the conditional equations:

\[ B_0 + 2.5 B_1 - 6.85 = \varepsilon_1, \]

\[ B_0 + 4.5 B_1 - 3.70 = \varepsilon_2, \]

\[ \begin{gathered} B_0+5B_1-3,14=\varepsilon_3,\\ B_0+6B_1-2,17=\varepsilon_4,\\ B_0+6B_1-2,06=\varepsilon_5,\\ B_0+6B_1-1,92=\varepsilon_6, \end{gathered} \]

According to condition (3),

\[ \begin{aligned} F={}&(B_0+2,5B_1-6,85)^2+(B_0+4,5B_1-3,70)^2+\\ &+(B_0+5B_1-3,14)^2+(B_0+6B_1-2,17)^2+(B_0+6B_1-2,06)^2+ \tag{4}\\ &+(B_0+6B_1-1,92)^2=\text{minimum}. \end{aligned} \]

The variables are \(B_0\) and \(B_1\). Condition (4) will be satisfied if the partial derivatives \(\dfrac{\partial F}{\partial B_0}\) and \(\dfrac{\partial F}{\partial B_1}\) are set equal to zero:

\[ \begin{aligned} \frac{\partial F}{\partial B_0} ={}&2(B_0+2,5B_1-6,85)+2(B_0+4,5B_1-3,70)+\\ &+2(B_0+5B_1-3,14)+2(B_0+6B_1-2,17)+2(B_0+6B_1-2,06)+\\ &+2(B_0+6B_1-1,92)=0. \end{aligned} \]

\[ \begin{aligned} \frac{\partial F}{\partial B_1} ={}&2(B_0+2,5B_1-6,85)\,2,5+2(B_0+4,5B_1-3,70)\,4,5+\\ &+2(B_0+5B_1-3,14)\,5+2(B_0+6B_1-2,17)\,6+2(B_0+6B_1\\ &-2,06)\,6+2(B_0+6B_1\\ &-1,92)\,6=0. \end{aligned} \]

Hence we obtain the normal equations:

\[ 6B_0+30B_1=19,84, \]

\[ 30B_0+159,5B_1=86,375. \]

We reduce the normal equations to Table 4; here \(M\) is the constant term; \(S\) is the sum of the coefficients and constant terms, determined for checking the calculations. The third row is written for checking.

Table 4

We solve the system of normal equations by means of determinants:

\[ D= \begin{vmatrix} 6 & 30\\ 30 & 159,5 \end{vmatrix} =57; \qquad D_{B_0}= \begin{vmatrix} 19,84 & 30\\ 86,375 & 159,5 \end{vmatrix} =573,23; \]

\[ D_{\alpha}= \begin{vmatrix} -16,16 & 30\\ -103,125 & 159,5 \end{vmatrix} =516,23; \qquad D_{B_1}= \begin{vmatrix} 6 & 19,84\\ 30 & 86,375 \end{vmatrix} =-76,95; \]

\[ D_{\beta}= \begin{vmatrix} 6 & -16,16\\ 30 & -103,125 \end{vmatrix} =-133,95. \]

The determinants \(D_{\alpha}\) and \(D_{\beta}\) are obtained by replacing the column \(M\) by \(S\), in order to check the solution. \(\alpha\) and \(\beta\) must be less by one than \(B_0\) and \(B_1\) [11].

\[ B_0=\frac{D_{B_0}}{D}=\frac{573{,}23}{57}=10{,}06; \qquad B_1=\frac{D_{B_1}}{D}=-\frac{76{,}95}{57}=-1{,}35; \]

\[ \alpha=\frac{D_\alpha}{D}=\frac{516{,}23}{57}=9{,}06; \qquad \beta=\frac{D_\beta}{D}=-\frac{133{,}95}{57}=-2{,}35. \]

Then \(\kappa=10{,}06-1{,}35 f\) (for \(f=2 \div 6\)). Let us determine the root-mean-square error of the equation obtained; for this purpose, using the coefficients \(B_0\) and \(B_1\), we find the errors \(\varepsilon\) and their squares (Table 5).

Table 5

The mean square deviations of the unknowns \(B_0\) and \(B_1\) of the normal equations are determined as follows [11]:

\[ \sigma_{B_0}^{2}=\frac{\sigma_0^2}{P_{B_0}} \quad \text{and} \quad \sigma_{B_1}^{2}=\frac{\sigma_0^2}{P_{B_1}}, \]

where \(\sigma_0\) is the most probable value of the root-mean-square error per unit weight;

\(P_{B_0}\) and \(P_{B_1}\) are the weights of the unknowns.

The quantity

\[ \sigma_0^2=\frac{\sum \varepsilon^2}{n-m}; \]

here \(\varepsilon\) are the deviations (see Table 5),

\(n\) is the number of conditional equations,

\(m\) is the number of unknowns,

i.e.

\[ \sigma_0^2=\frac{0{,}1885}{6-2}=0{,}047; \qquad \sigma_0=0{,}217. \]

We determine the weights of the unknowns by means of determinants:

\[ P_{B_0}=\frac{D}{D_{11}}, \qquad P_{B_1}=\frac{D}{D_{22}}, \]

where \(D_{11}\) and \(D_{22}\) are the cofactors of the first and second elements of the main diagonal of the determinant of the system of normal equations.

Thus,

\[ P_{B_0}=\frac{57}{159{,}5}=0{,}358; \qquad P_{B_1}=\frac{57}{6}=9{,}5; \]

then the mean square deviations of the unknowns are:

\[ \sigma_{B_0}^{2}=\frac{0{,}047}{0{,}358}=0{,}131; \qquad \sigma_{B_1}^{2}=\frac{0{,}047}{9{,}5}=0{,}0045; \]

\[ \sigma_{B_0}=0{,}36; \qquad \sigma_{B_1}=0{,}067; \]

or

\[ B_0=10{,}06 \pm 0{,}36; \qquad B_1=-1{,}35 \pm 0{,}067. \]

Rounding, \(B_0=10{,}0 \pm 0{,}4;\qquad B_1=-1{,}35 \pm 0{,}07.\)

Thus, the equation of the straight line \(\kappa=\varphi_1(f)\) in the range \(f=2 \div 6\) will have the form:

\[ \kappa=10-1{,}35 f. \tag{5} \]

In the range \(f \geqslant 6\) for \(f\) and \(\kappa\) we have (Table 6).
We carry out analogous calculations and obtain

Table 6

\[ 6B'_0+58.5B'_1=7.42, \]

\[ 58.5B'_0+633.2B'_1=60.35. \]

Solving the equations by means of determinants, we obtain:

\[ D=376.95;\quad D_{B'_0}=1167.869;\quad D_{B'_1}=-71.97; \]

\[ D_\alpha=790.919;\quad D_\beta=-448.92, \]

whence

\[ B'_0=3.098;\qquad \alpha=2.098, \]

\[ B'_1=-0.1909;\qquad \beta=-1.1909. \]

Thus,

\[ \kappa=3.098-0.1909f. \]

We compile Table 7.

Table 7

From Table 7

\[ \sigma_0^2=\frac{0.0284}{4}=0.0071,\quad \sigma_0=0.084. \]

Weights of the unknowns:

\[ P_{B'_0}=\frac{376.95}{633.20}=0.595; \]

\[ P_{B'_1}=\frac{376.95}{6}=62.825. \]

Mean-square deviations of the unknowns

\[ \sigma_{B'_0}^{2}=\frac{0.0071}{0.595}=0.119;\qquad \sigma_{B'_1}^{2}=\frac{0.0071}{62.825}=0.00011. \]

\[ \sigma_{B'_0}=0.11;\qquad \sigma_{B'_1}=0.01. \]

Then

\[ B'_0=3.098\pm0.11;\qquad B'_1=-0.1909\pm0.01. \]

Taking \(\sigma_{B'_0}\) and \(\sigma_{B'_1}\) into account, we write

\[ B'_0=3.1\pm0.1;\qquad B'_1=-0.19\pm0.01. \]

The equation \(\kappa=\varphi_2(f)\) for \(f \geqslant 6\) will take the form:

\[ 3.1-0.19f=\kappa. \tag{6} \]

We have recommended determining the minimum necessary feed force corresponding to the onset of volumetric destruction of rocks in rotary drilling by the formula

\[ P_0=L\cdot f; \]

here \(L\) is a proportionality coefficient. For cutters of \(42\ \text{mm}\), \(L\) may be taken equal to 35 [1].

Table 8 presents the results of processing the experimental data \(P_0\) for various rocks.

Table 8

\[ L = 30.2 \pm 0.8; \qquad W = 3\% \]

Thus, on the basis of the available experimental data, the coefficient relating \(P_0\) and \(f\) lies in the range \(27.8 < L < 32.6\), from which it is evident that the coefficient 35 adopted by us in [1] agrees sufficiently well with the experimental data.

Conclusion

The calculation of rotary drilling regimes for boreholes may be carried out by the formula

\[ V = \kappa (P_{\mathrm{p}} - P_0). \]

The drilling efficiency coefficient is

\[ \kappa = 3.1 - 0.19f \quad \text{(for rocks with } f > 6\text{)} \]

and

\[ \kappa = 10 - 1.35f \quad \text{(for rocks with } f = 2\text{--}6\text{)}. \]

The value

\[ P_0 = 35f. \]

Literature

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