Mining Engineer V. F. Turchinskii.

Associate Professor of the Siberian Technological Institute.

Analytical and Graphical-Analytical Methods for Determining the Elements of Attitude at a Given Point on a Bed Surface

Tomsk,
1929.

Associate Professor V. F. Turchinskii

Analytical and Graphical-Analytical Methods for Determining the Attitude Elements at a Given Point on a Bed Surface.

§ 1.

Proceeding to the solution of the problem posed, we shall henceforth, in passing from the simple to the complex, take the two bounding surfaces of the bed to be parallel planes.

Let: \(H\) be some horizontal plane, \(P\) one of the two bounding planes limiting the bed (Fig. 1).

Take in the plane \(P\) an arbitrary point \(A(x_a, y_a, z_a)\). Then the position of the plane \(P\) in space, as is known, is determined by the point \(A(x_a, y_a, z_a)\) and by two distinguished straight lines—the lines of dip and strike—connected by the condition of mutual perpendicularity.

Draw through the point \(A(x_a, y_a, z_a)\) in the plane \(P\) the lines of dip and strike. In nature, in the plane \(P\), the line of dip is directly determined at the given point \(A(x_a, y_a, z_a)\) not only as a straight line passing through the point \(A\), but also as a direction: the line of dip is the line of descent of a drop of water.

Fig. 1.

By virtue of what has been said, the direction of the line of dip is naturally to be reckoned from the point \(A(x_a, y_a, z_a)\) toward an arbitrary other point on it, \(B(x_b, y_b, z_b)\), whose elevation \(z_b\) is less than the elevation \(z_a\) of point \(A\).

In solving the problem according to the principles proposed by us below, it is sufficient to give the direction of only one line of dip.

Indeed: the final aim of surveying a bed is the representation of its surfaces in the plane of the drawing. Let us construct the representation of one of the planes of the bed (Fig. 2).

We place on the plan the point \(a\)—the horizontal projection of the point \(A\). Through the point \(a\) we draw the direction of the meridian \(NS\), and then construct the direction of the projection of the line of dip, knowing its azimuth \(\alpha\). Further, knowing the dip angle \(h\) of the line of dip and the elevation of the point \(A\), it is easy, as is known, to determine the interval of the line of dip, or, what is the same thing, to determine on the projection of the latter the points corresponding to the projections of the points of intersection of the line of dip with a system of horizontal planes equally spaced from one another. Through the points thus obtained will pass the projections of the lines of strike, or horizontals—contours of the plane \(P\), perpendicular to the projection of the line of dip, since by their property they are perpendicular to the latter.

Fig. 2.

Thus we see that the problem posed—the representation of the bed plane in orthogonal projection onto the horizontal plane—has been completely solved by us, without determining the azimuth of the line of strike.

The question of the necessity of giving a definite direction specifically to the line of strike arises only in the theory of displacements in the earth’s crust, and therefore we shall omit it here.

§ 2.

Principles of the proposed analytical methods.

First principle. The first principle is based on the concept of directed quantities—the concept of the vector.

Let us take (Fig. 3) in the plane a rectangular (Cartesian) coordinate system, in which we shall take the \(X\)-axis as the principal one. Next, let us take in the same plane two points: \(A(x_a, y_a)\) and \(B(x_b, y_b)\). The points \(A\) and \(B\) will determine a straight line, which we shall take as a rectilinear vector, whose origin is point \(A\), and whose end is point \(B\).

On the drawing we shall represent this vector by a straight arrow directed from point \(A\) to point \(B\).

Fig. 3

Draw through point \(A\) a straight line parallel to the \(X\)-axis. Then the azimuth (strike, angle of position) \((AB)\) of the vector \(AB\), as is known, is determined by the angle between it and the principal direction—the \(X\)-axis (or the direction parallel to it), measured from the principal direction to the right (clockwise), or to the left (counterclockwise). Both systems of azimuths vary from \(0\) to \(2\pi\); in what follows we shall take right-hand azimuths. In the drawing the azimuth is denoted by an arc with an arrow at the end and with the index \((AB)\).

Analytically, the azimuth \((AB)\) of the vector \(AB\) is determined through the coordinates of its beginning and end by the formula:

\[ \tg(AB)=\frac{y_b-y_a}{x_b-x_a}=\frac{\Delta y_a^b}{\Delta x_a^b}, \]

or, in words: the tangent of the azimuth \((AB)\) is equal to the ratio of the increment of the coordinate \(y\) to the increment of the coordinate \(x\) from point \(A\) to point \(B\).

Let us note that in this formula the tangent of the azimuth \((AB)\) is considered as the ratio of the geometric sine to the geometric cosine. In accordance with this last remark, as is known, the following table is compiled, which is easy to remember.

Quadrant	\(\tg(AB)=\dfrac{\Gamma\sin(AB)^{*}}{\Gamma\cos(AB)}\)	Tabular angle	Sought angle (azimuth) \((AB)\)
1	\(\dfrac{+}{+}\)	\(\alpha_1\)	\(\alpha_1\)
2	\(\dfrac{+}{-}\)	\(\alpha_2\)	\(180^\circ-\alpha_2\)
3	\(\dfrac{-}{-}\)	\(\alpha_3\)	\(180^\circ+\alpha_3\)
4	\(\dfrac{-}{+}\)	\(\alpha_4\)	\(360^\circ-\alpha_4\)

\[ ^{*})\ \Gamma \text{ — geometric.} \]

The correctness of the determination of the azimuth \((AB)\) is checked by the formula:

\[ \tg [45^\circ + (AB)] = \frac{\Delta x_a^b + \Delta y_a^b}{\Delta x_a^b - \Delta y_a^b}. \]

Let us pass to the second principle. The second principle is based on the direct determination of a direction in space.

Let us have a Cartesian coordinate system in space (see Fig. 4).

Take an arbitrary point \(A(x_a, y_a, z_a)\) and attach a direction to it. Next, through point \(A\) draw axes parallel to the axes of the given coordinate system. Then the direction is determined by the coordinates \((x_a, y_a, z_a)\) of the origin of the direction; then by the position of the vertical plane in which the direction lies, relative to the principal plane \(ZOX\), and by the azimuth \(a\) thereby determined; finally, the position of the direction in the vertical plane corresponding to azimuth \(a\) is determined by the angle \(\zeta\)—the zenith distance, measured from the \(Z\)-axis and varying from \(0\) to \(\pi\).*)

Fig. 4.

§ 3.

A complete list of the literature on the question touched upon, both in Russian and in foreign languages, is given in the work of Prof. P. Leontovsky: “Mine-Surveying Problems. Part 5. Elements of the Attitude of Beds. (Mining Geometry).” Publication of the Ekaterinoslav Higher Mining School. 1905.

The analytical solution of the question under consideration in P. Leontovsky’s work was, until recently, the last in print.

With regard to the literature of the question, I must note that in it a solution in general form is not given. The two analytical principles proposed, as we shall see below, solve the problem quite in its most general form.

The necessity of giving precisely an analytical solution to the problem of determining the attitude elements of a bed is conditioned by those practical cases in which, for a gently lying bed, three boreholes at a considerable distance from one another are specified in order to determine its position; solving the problem graphically in this case, at the scale of the plan, may entail practically significant inaccuracies.

Further, the analytical solution of the problem is most closely connected with the solution of another, no less important, problem: determining such a surface for whose points the sum of the squares of their distances from the surface of the bed is a minimum.

A solution of this problem, in the form in which it has just been stated, has not been given. There have been only attempts at solving this problem, which was called “the determination of the general strike and dip of the bed.” I shall note, incidentally, that the very approach to solving this problem is essentially incorrect (see the article by Prof. S. Yu. Dobrozhinsky, “The General Dip and Strike of a Bed”). The solution of the latter problem—the problem of “determining the general strike and dip of a bed”—I shall not present, owing to the unfinished state of its development.

*) The zenith distance \(\zeta\) and the dip angle \(h\) of the direction, as is known, are related by:

\[ \zeta + h = \frac{1}{2}\pi. \]

§ 4.

Turning now to the various cases of exploration (test-pitting, drilling, etc.), let us note that all cases, as will be shown below, in solving the problem of determining the position of the lines of dip and strike, reduce to a single one: the case of test-pitting.

Let us consider the case in which the surface of the test pit is prismatic.

As is known, in this case one determines the azimuths (strikes) and angles of dip either of the traces of the planes of the bed, or of the bedding lines of the same bed on the vertical walls of the test pit, taking the traces and bedding as directions and taking them as issuing from one point of the line of intersection of the vertical walls of the test pit.

Let \(V_1\) be one of the vertical walls of the test pit, \(V_2\) the other, adjacent to \(V_1\). \(RS\) is the vertical line of intersection of the planes \(V_1\) and \(V_2\) (Fig. 5); 1 and 2 are the traces of the plane \(P\) of the bed, issuing from point \(A\) of the edge \(RS\).

Fig. 5.

Let the point \(A\) and the traces, as directions, be referred to a Cartesian system of coordinates. Through the point \(A(x_a, y_a, z_a)\) draw the line of dip \(AF\).

The direction of the line of dip, by virtue of the basis of solution indicated by us, will be determined if we find yet another, generally speaking arbitrary, point lying on the line of dip.

Of the numerous analytical methods of solving the problem posed, found by me, I shall present, for lack of space, only some.

§ 5.

Solutions by methods of analytic geometry.

1. Forming the equation of the plane of the bed.

We shall take the point \(A\) of intersection of the traces, determined as directions, as the origin of the coordinates of three mutually perpendicular planes: the horizontal \(H\) and two vertical ones, of which one—\(ZAX\)—coincides with the plane of the meridian (Fig. 6).

The position of each of the directions in the coordinate system adopted in the observation—a spherical one—is determined, as is known, by its azimuth \(a\) and zenith distance \(\zeta\). We shall denote the azimuths and zenith distances of the directions—the first and the second—respectively by \(a_1, \zeta_1\) and \(a_2, \zeta_2\).

Let us dwell on one of the methods for forming the equation of the plane of the bed. From the origin of coordinates \(A\), as center, describe a sphere of arbitrary radius \(R\). Next, replace the directions determining the plane by three points, of which one is the origin of coordinates \((0,0,0)\), and the other two are the points of intersection of the directions with the surface of the sphere; the coordinates of the latter two points, as is known,

Fig. 6.

are determined as the projections of the corresponding vectors—\((R_1,\ a_1,\ \zeta_1)\) and \((R,\ a_2,\ \zeta_2)\)—onto the coordinate axes.

We have:

\[ \begin{aligned} x_{1,2} &= R\sin \zeta_{1,2}\cos a_{1,2},\\ y_{1,2} &= R\sin \zeta_{1,2}\sin a_{1,2},\\ z_{1,2} &= R\cos \zeta_{1,2} \end{aligned} \qquad . . . . . . . . . \ (1) \]

The equation of the plane determined by the points \((0,\ 0,\ 0)\), \((x_1,\ y_1,\ z_1)\) and \((x_2,\ y_2,\ z_2)\) will be—

\[ \left| \begin{array}{cccc} x & y & z & 1\\ x_1 & y_1 & z_1 & 1\\ x_2 & y_2 & z_2 & 1\\ 0 & 0 & 0 & 1 \end{array} \right|=0, \]

or:

\[ \left| \begin{array}{cc} y_1 & z_1\\ y_2 & z_2 \end{array} \right|x - \left| \begin{array}{cc} x_1 & z_1\\ x_2 & z_2 \end{array} \right|y + \left| \begin{array}{cc} x_1 & y_1\\ x_2 & y_2 \end{array} \right|z =0 . . . . . . . \ (2) \]

According to (1), each of \(x_{1,2}\), \(y_{1,2}\), and \(z_{1,2}\) contains \(R\); therefore, after substituting them into equation (2), the latter may be divided by \(R^2\), and we obtain:

\[ \left| \begin{array}{cc} \sin \zeta_1 \sin a_1, & \cos \zeta_1\\ \sin \zeta_2 \sin a_2, & \cos \zeta_2 \end{array} \right|x - \left| \begin{array}{cc} \sin \zeta_1 \cos a_1, & \cos \zeta_1\\ \sin \zeta_2 \cos a_2, & \cos \zeta_2 \end{array} \right|y + \left| \begin{array}{cc} \sin \zeta_1 \cos a_1, & \sin \zeta_1 \sin a_1\\ \sin \zeta_2 \cos a_2, & \sin \zeta_2 \sin a_2 \end{array} \right|z =0. \]

Transforming the last equation, taking outside the determinant signs, in the first two determinants, the product \(\cos \zeta_1\cdot \cos \zeta_2\), and in the third—the product \(\sin \zeta_1\cdot \sin \zeta_2\).

We obtain:

\[ \cos \zeta_1\cdot \cos \zeta_2 \left\{ \left| \begin{array}{cc} \tg \zeta_1 \sin a_1, & 1\\ \tg \zeta_2 \sin a_2, & 1 \end{array} \right|x - \left| \begin{array}{cc} \tg \zeta_1 \cos a_1, & 1\\ \tg \zeta_2 \cos a_2, & 1 \end{array} \right|y \right\} + \sin \zeta_1\cdot \sin \zeta_2 \left| \begin{array}{cc} \cos a_1 & \sin a_1\\ \cos a_2 & \sin a_2 \end{array} \right|z =0. \]

Dividing the last equation by the product \(\cos \zeta_1 \cdot \cos \zeta_2\) and, further, expanding the determinants, we finally obtain:

\[ (\tg \zeta_1 \sin a_1 - \tg \zeta_2 \sin a_2)x - (\tg \zeta_1 \cos a_1 - \tg \zeta_2 \cos a_2)y + \tg \zeta_1 \cdot \tg \zeta_2 \sin(a_2-a_1)z = 0, \]

—the required equation of the plane determined by two traces issuing from a single point.

In what follows we shall denote the coefficient of \(x\) by \(n\):

\[ \tg \zeta_1 \cdot \sin a_1 - \tg \zeta_2 \sin a_2 = n; \]

the coefficient of \(y\) we shall denote by \((-m)\):

\[ -(\tg \zeta_1 \cos a_1 - \tg \zeta_2 \cos a_2) = -m; \]

the constant term we shall denote by \(k\):

\[ \tg \zeta_1 \tg \zeta_2 \sin(a_2-a_1)=k. \]

According to these notations, the equation of the bed plane is represented in the form:

\[ nx-my+kz=0. \]

2. Determination of the azimuth of the line of dip by the method of rotation

The equation of the plane:

\[ nx-my+kz=0. \]

Putting in the latter \(z=0\), we find the equation of the straight line of extension lying in the horizontal plane \(XAY\):

\[ nx-my=0. \]

The equation of the projection of the line of dip:

\[ mx+ny=0. \]

The angular coefficient of the projection of the line of dip is:

\[ \tg \varphi_{1,2}=-\frac{m}{n}\ldots\ldots\ldots\ldots\ldots\ldots\ldots (1) \]

Our task is to determine such an angle—the azimuth \(\varphi\)—which would correspond precisely to the direction of the projection of the line of dip (Fig. 7).

Let us proceed to analyze formula (1).

I. Let

\[ \tg \varphi > 0 \ldots\ldots\ldots\ldots\ldots\ldots\ldots (2) \]

Formula (1) may be represented in two ways:

\[ \tg \varphi=\frac{-m}{n} \quad \text{and} \quad \tg \varphi=\frac{m}{-n}. \]

1. Let us take

\[ \tg \varphi=\frac{-m}{n}; \]

the latter may be regarded as the ratio of the geometric sine to the geometric cosine.

A. Let them be such (Fig. 7) that they correspond to the azimuth of the third quadrant; then:

\[ \sin\varphi < 0 \text{ and } \cos\varphi < 0, \ldots \ldots \ldots \ldots \ldots (2_1) \]

and therefore:

\[ m > 0 \text{ and } n < 0 \ldots \ldots \ldots \ldots \ldots \ldots (2_2) \]

Let us now rotate the coordinate system about the \(Z\)-axis through the angle \(\varphi\), corresponding to the azimuth of the line of fall. Then the coordinates of points in the system \(XYZ\) will be expressed in terms of the coordinates of the same points in the new system \(X'Y'Z'\) by the formulas:

\[ \begin{aligned} x &= x' \cos\varphi - y' \sin\varphi \\ y &= x' \sin\varphi + y' \cos\varphi \end{aligned} \quad \ldots \ldots \ldots \ldots \ldots (3) \]

But in the new coordinate system \(X'Y'Z'\) the coordinates \(y'\) of points of the line of fall satisfy the equality: \(y' = 0\); in view of this, relations (3) take the form:

\[ \begin{aligned} x &= x' \cos\varphi \\ y &= x' \sin\varphi \end{aligned} \quad \ldots \ldots \ldots \ldots \ldots (4) \]

Let us determine the coordinate \(z\) of points of the line of fall. From the equation of the plane

\[ nx - my + kz = 0 \]

we obtain:

\[ z = \frac{1}{k}(-nx + my). \]

Substituting relations (4) here, we obtain:

\[ z = \frac{x'}{k}\left(-n\cos\varphi + m\sin\varphi\right) \ldots \ldots \ldots (5) \]

Let us now express the trigonometric functions entering formulas (4) and (5) in terms of the tangent. In general we have:

\[ \sin\varphi = \pm \frac{\operatorname{tg}\varphi}{\sqrt{1+\operatorname{tg}^2\varphi}} = \pm \frac{m}{\sqrt{m^2+n^2}} \]

\[ \cos\varphi = \pm \frac{1}{\sqrt{1+\operatorname{tg}^2\varphi}} = \pm \frac{n}{\sqrt{m^2+n^2}} \]

In our case, by virtue of \((2_{1,2})\), we must take:

\[ \sin\varphi = -\frac{m}{\sqrt{m^2+n^2}} \quad \text{and} \quad \cos\varphi = +\frac{n}{\sqrt{m^2+n^2}} \]

Therefore:

\[ \left. \begin{aligned} x &= \frac{n}{\sqrt{m^2+n^2}}\,x' \\[6pt] y &= -\frac{m}{\sqrt{m^2+n^2}}\,x' \\[6pt] z &= -\frac{x'}{k}\sqrt{m^2+n^2} \end{aligned} \right\} \ldots \ldots \ldots (6) \]

Expressing the coordinates \(x, y, z\) of the points of the line of dip as functions of the coordinate \(x'\) of the same points, let us proceed to analyze formula (6\(_3\)).

A\(_1\). Let us take points of the line of dip for which \(z < 0\). For these points, by virtue of our assumption (see Fig. 7), \(x' > 0\). Therefore it follows from (6\(_3\)) that \(\kappa > 0\).

The desired azimuth of the line of dip, determined by the points \((0, 0, 0)\) and \((x, y, z < 0)\), is found vectorially from the formula:

\[ \tg \varphi = \frac{y-0}{x-0} = \frac{ -\dfrac{m}{\sqrt{m^{2}+n^{2}}}\,x' }{ \dfrac{n}{\sqrt{m^{2}+n^{2}}}\,x' } = -\frac{m}{n}, \]

or \(x' > 0\).

A\(_2\). Let us now take points of the line of dip for which \(z > 0\). It is likewise easy to convince oneself that \(x' < 0\) and, correspondingly, \(\kappa > 0\).

The desired azimuth of the line of dip, determined by the points \((0, 0, 0)\) and \((x, y, z > 0)\), is found vectorially from the formula:

\[ \tg \varphi = \frac{0-y}{0-x} = \frac{ \dfrac{m}{\sqrt{m^{2}+n^{2}}}\,x' }{ -\dfrac{n}{\sqrt{m^{2}+n^{2}}}\,x' } = \]

\[ = \frac{ -\dfrac{m}{\sqrt{m^{2}+n^{2}}}\,(-x') }{ \dfrac{n}{\sqrt{m^{2}+n^{2}}}\,(-x') } = -\frac{m}{n}, \quad \text{or } (-x')>0. \]

A\(_3\). Let us further take on the line of dip two arbitrary points: for one of them—\(A(x_{z_p}, y_{z_p}, z_p)\)—let \(z_p > 0\) and, correspondingly, \(x'_n < 0\); for the other—\(B(x_{z_n}, y_{z_n}, z_n)\)—let \(z_n < 0\) and, correspondingly, \(x'_p > 0\) (where \(p\)—positiff, \(n\)—negatiff).

The desired azimuth of the line of dip, determined by the points \(A\) and \(B\), is found vectorially from the formula:

\[ \tg \varphi = \frac{y_{z_n}-y_{z_p}}{x_{z_n}-x_{z_p}} = \frac{ -\dfrac{m}{\sqrt{m^{2}+n^{2}}}\,x'_p + \dfrac{m}{\sqrt{m^{2}+n^{2}}}\,x'_n }{ \dfrac{n}{\sqrt{m^{2}+n^{2}}}\,x'_p - \dfrac{n}{\sqrt{m^{2}+n^{2}}}\,x'_n } = \]

\[ = \frac{-m(x'_p-x'_n)}{n(x'_p-x'_n)} = -\frac{m}{n}, \quad \text{or } (x'_p-x'_n)>0. \]

As we see, in all three cases one and the same formula is obtained for determining the azimuth of the line of dip: $\operatorname{tg}\varphi=\dfrac{-m}{n}$, as was to be expected.

Let us also note that, in this case, the condition shown above must be satisfied: $k>0$.

1. Let us now turn to the next supposition, made at the outset: $\operatorname{tg}\varphi=\dfrac{m}{-n}$. In accordance with the drawing, and consequently with the condition: $\sin\varphi<0$ and $\cos\varphi<0$, we must put:

\[ m<0 \text{ and } n>0. \ldots \ldots \ldots \ldots \ldots \ldots (7). \]

In this case, by reasoning similarly to the preceding case, we find that:

\[ \sin\varphi=+\frac{m}{\sqrt{m^2+n^2}} \quad \text{and} \quad \cos\varphi=-\frac{n}{\sqrt{m^2+n^2}} . \]

Therefore (4) and (5) take the form:

\[ \left. \begin{aligned} x&=-\frac{n}{\sqrt{m^2+n^2}}\,x'\\ y&=\frac{m}{\sqrt{m^2+n^2}}\,x'\\ z&=\frac{x'}{k}\sqrt{m^2+n^2} \end{aligned} \right\} \ldots \ldots \ldots \ldots \ldots \ldots (8) \]

The analysis of the formulas is analogous to the preceding case.

$A_1$. Let $z<0$; then, according to the drawing, $x'>0$ and, according to (8), $k<0$.

The sought azimuth of the line of dip, determined by the points $(0,0,0)$ and $(x,y,z<0)$, is found vectorially from the formula:

\[ \operatorname{tg}\varphi = \frac{y-0}{x-0} = \frac{\dfrac{m}{\sqrt{m^2+n^2}}\,x'} {-\dfrac{n}{\sqrt{m^2+n^2}}\,x'} = \frac{m}{-n} \]

for $x'>0$.

$A_2$. Let $z>0$; then, according to the drawing, $x'<0$, and from (8) it follows that $k<0$.

The sought azimuth of the line of dip, determined by the points $(0,0,0)$ and $(x,y,z>0)$, is found vectorially from the formula:

\[ \operatorname{tg}\varphi = \frac{0-y}{0-x} = \frac{ -\dfrac{m}{\sqrt{m^2+n^2}}\,x' }{ \dfrac{n}{\sqrt{m^2+n^2}}\,x' } = \frac{ \dfrac{m}{\sqrt{m^2+n^2}}\,(-x') }{ -\dfrac{n}{\sqrt{m^2+n^2}}\,(-x') } = \frac{m}{-n} \]

for $(-x')>0$.

\(A_3\). Let us take, on the line of dip, two points \(P\) and \(Q\); for \(P\) let \(z_p > 0\) and, correspondingly, \(x'_n < 0\); for \(Q\) let \(z_n < 0\) and, correspondingly, \(x'_p > 0\) [where \(p\)—positif, \(n\)—negatif].

We find the desired azimuth of the line of dip, determined by the points \(P\) and \(Q\), from the formula:

\[ \operatorname{tg}\varphi = \frac{y_{z_n}-y_{z_p}}{x'_{z_n}-x'_{z_p}} = \frac{ \dfrac{m}{\sqrt{m^2+n^2}}x'_p + \dfrac{m}{\sqrt{m^2+n^2}}x'_n }{ -\dfrac{n}{\sqrt{m^2+n^2}}x'_p + \dfrac{n}{\sqrt{m^2+n^2}}x'_n } = \]

\[ = \frac{m(x'_p-x'_n)}{-n(x'_p-x'_n)} = \frac{m}{-n}, \]

for \((x'_p-x'_n)>0\).

As we see, in all three cases one and the same formula is obtained for determining the azimuth of the line of dip: \(\operatorname{tg}\varphi=\dfrac{m}{-n}\), as was to be expected.

Let us note further that, in this case, as was shown above, the condition must be fulfilled: \(\kappa<0\).

Thus we arrive at the conclusion that the formula \(\operatorname{tg}\varphi=\dfrac{-m}{n}\) was obtained by taking \(\kappa>0\); while the formula \(\operatorname{tg}\varphi=\dfrac{m}{-n}\) was obtained by taking \(\kappa<0\). Both cases correspond to one and the same tangent: \(\operatorname{tg}\varphi=-\dfrac{m}{n}\).

We shall come to the same conclusion by assuming the azimuth of the line of dip to correspond to other quadrants. It is clear that the sign of the quantity \(\kappa\) determines the order of notation of the directions—traces.

Indeed: \(\kappa=\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)\).

The product \(\operatorname{tg}\zeta_1\cdot\operatorname{tg}\zeta_2\) does not change when the order of notation of the directions—traces is changed.

Let the order of the directions \(a_1\) and \(a_2\) be chosen so that \(\kappa>0\); this choice is evidently always possible.

Now change the order of notation of the azimuths; then \(\kappa<0\).

Further, when the order of notation of the azimuths is changed, the quantities \(m\) and \(n\) change their signs.

Indeed, this follows directly from the formulas:

\[ m=\operatorname{tg}\zeta_1\cos a_1-\operatorname{tg}\zeta_2\cos a_2, \]

\[ n=\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2 \]

by mutual interchange of the minuend and the subtrahend.

Naturally, we must now turn to the determination of the order of notation of the directions—traces.

3. Determination of the order of notation of the directions—traces.

The case corresponding to \(\kappa>0\).

\[ \kappa=\operatorname{tg}\zeta_1\cdot\operatorname{tg}\zeta_2\sin(a_2-a_1). \]

In proceeding to the investigation, let us note that:

1) the angle \(A\), enclosed between the walls of the pit, is always less than \(\pi\) \((A<\pi)\);

2) the azimuths of the directions—traces are taken as positive (counting them clockwise), and

3) the zenith distances of the directions—traces satisfy the inequality:

\[ \pi > z_{1,2} > 0. \]

Therefore: \(2\pi > a_{1,2} > 0\), and, as a consequence, \(\left|a_2-a_1\right|^{*)} < 2\pi\).

From the inequality written above, \(\kappa > 0\), it follows that two cases are possible.

I-st:

\[ \tan z_1 \tan z_2 > 0 \quad \text{and simultaneously} \quad \sin(a_2-a_1)>0 \ldots\ldots (1) \]

II-nd:

\[ \tan z_1 \tan z_2 < 0 \quad \text{and simultaneously} \quad \sin(a_2-a_1)<0 \ldots\ldots (2) \]

Let us consider the 1st case. By virtue of inequality (1), it splits further into the following cases:

\[ \begin{aligned} \text{A)}\quad &\tan z_1>0;\quad \frac{\pi}{2}>z_1>0\\ &\tan z_2>0;\quad \frac{\pi}{2}>z_2>0\\[4pt] \text{B)}\quad &\tan z_1<0;\quad \frac{\pi}{2}<z_1<\pi\\ &\tan z_2<0;\quad \frac{\pi}{2}<z_2<\pi, \end{aligned} \qquad \ldots\ldots (3) \]

for which

\[ \sin(a_2-a_1)>0 \ldots\ldots (3_1) \]

The inequality \(\sin(a_2-a_1)>0\) entails the possible inequalities:

\[ a_2-a_1>0 \ldots\ldots (4) \quad \text{and} \quad a_2-a_1<0 \ldots\ldots (5) \]

From inequalities \((3_1)\) and (4) it follows that the angle \((a_2-a_1)\) satisfies the inequality:

\[ \pi>a_2-a_1>0 \ldots\ldots (6) \]

From inequalities \((3_1)\) and (5) it follows that the angle \((a_2-a_1)\) satisfies the inequality:

\[ -2\pi<a_2-a_1<-\pi \ldots\ldots (7) \]

Of the two azimuths—\(a_2\) and \(a_1\)—let us denote the larger by \(a_{\max}\), and the smaller by \(a_{\min}\). It is clear that \((a_{\max}-a_{\min})=\left|a_2-a_1\right|\); therefore inequalities (6) and (7) give, respectively:

\[ \pi>\left|a_2-a_1\right|>0 \]

\[ 2\pi>\left|a_2-a_1\right|>\pi, \]

or:

\[ \pi>a_{\max}-a_{\min}>0 \ldots\ldots (8) \]

\[ 2\pi>a_{\max}-a_{\min}>\pi \ldots\ldots (9) \]

Comparing (8) and (6), we obtain: \(a_{\max}\equiv a_2\).

Comparing (9) and (7), we obtain: \(a_{\max}\equiv a_1\);

or, in words: if the directions—traces are such that their zenith dis-

\[ ^{*)}\ \text{absolute value.} \]

distances are equal, and the azimuths satisfy inequality (8) or (9), then \(a_{\max}\) must necessarily be denoted, respectively, by \(a_2\) or \(a_1\).

Let us consider case II, in which \(\operatorname{tg}\zeta_1 \operatorname{tg}\zeta_2 < 0\), and simultaneously

\[ \sin(a_2-a_1)<0 \qquad \ldots \qquad (10) \]

This case further splits into two cases:

\[ \left. \begin{array}{ll} \mathrm{C}) & \operatorname{tg}\zeta_1>0;\quad \dfrac{\pi}{2}>\zeta_1>0,\\[6pt] & \operatorname{tg}\zeta_2<0;\quad \dfrac{\pi}{2}<\zeta_2<\pi,\\[10pt] \mathrm{D}) & \operatorname{tg}\zeta_1<0;\quad \dfrac{\pi}{2}<\zeta_1<\pi,\\[6pt] & \operatorname{tg}\zeta_2>0;\quad \dfrac{\pi}{2}>\zeta_2>0, \end{array} \right\} \qquad \ldots \qquad (10) \]

for which

\[ \sin(a_2-a_1)<0 \qquad \ldots \qquad (11) \]

The inequality \(\sin(a_2-a_1)<0\) entails the possible inequalities:

\[ a_2-a_1>0 \qquad \ldots \qquad (12) \]

and

\[ a_2-a_1<0 \qquad \ldots \qquad (13) \]

From inequalities (11) and (12) it follows that the angle \((a_2-a_1)\) satisfies the inequality:

\[ 2\pi>a_2-a_1>\pi \qquad \ldots \qquad (14) \]

From inequalities (11) and (13), however, it follows that the angle \((a_2-a_1)\) satisfies the inequality:

\[ -\pi<a_2-a_1<0 \qquad \ldots \qquad (15) \]

Of the two azimuths \(a_2\) and \(a_1\), let us denote the larger by \(a_{\max}\), the smaller by \(a_{\min}\).

It is clear that

\[ a_{\max}-a_{\min}=\left|\,a_2-a_1\,\right|; \]

therefore inequalities (14) and (15) give, respectively:

\[ 2\pi>\left|\,a_2-a_1\,\right|>\pi \]

\[ \pi>\left|\,a_2-a_1\,\right|>0, \]

or:

\[ 2\pi>a_{\max}-a_{\min}>\pi \qquad \ldots \qquad (16) \]

\[ \pi>a_{\max}-a_{\min}>0 \qquad \ldots \qquad (17) \]

Comparing (16) and (14), we obtain: \(a_{\max}=a_2\).

Comparing (17) and (15), we obtain: \(a_{\max}=a_1\),

or, in words: if the directions—traces—are such that their zenith distances are unequal, and the azimuths satisfy inequalities (16) or (17), then \(a_{\max}\) must necessarily be denoted, respectively, by \(a_2\) and \(a_1\).

4. Geometric interpretation of conditions (6) and (7) of the first case, in which \(\zeta_1\) and \(\zeta_2\) are identical, and of conditions (14) and (15) of the second case, in which \(\zeta_1\) and \(\zeta_2\) are not identical.

Case I (\(\zeta_1\) and \(\zeta_2\) are identical).

Conditions: \(\pi > a_2-a_1 > 0\) and \(-2\pi < a_2-a_1 < -\pi\) correspond to those cases in which the point \(N\) of the principal direction \(NS\) (meridian) lies respectively outside and inside the angle corresponding to the shaft (see drawings 8 and 9).

Both drawings indicate, in accordance with what has been said, that the position of direction 2 is obtained from direction 1 by rotating the latter, within the angle corresponding to the shaft, to the right.

Drawing 8.

Drawing 9.

Case II (\(\zeta_1\) and \(\zeta_2\) are not identical).

Conditions: \(-\pi < a_2-a_1 < 0\) and \(2\pi > a_2-a_1 > \pi\) correspond to those cases in which the point \(N\) of the principal direction \(NS\) (meridian) lies respectively outside and inside the angle corresponding to the shaft (drawings 10 and 11).

Drawing 10.

Drawing 11.

Both drawings indicate, in accordance with what has been said, that the position of direction 2 is obtained from direction 1 by rotating the latter outside the angle corresponding to the shaft, to the right.

Conclusion of the analysis. If, in the notebook (journal), a drawing has been made (a horizontal projection) of directions—traces with marks determining their position in space, then, in order to designate the order of the directions—traces, it is sufficient, in the case of identical \(\zeta_1\) and \(\zeta_2\), to draw, inside the angle corresponding to the shaft, and, in the case of non-identical \(\zeta_1\) and \(\zeta_2\), outside the angle corresponding to the shaft, an arcuate arrow to the right; the direction adjacent to the beginning of the arrow must be designated by the number 1. The last sentence may be replaced by the following: the older direction corresponding to the drawn arcuate arrow must be designated by the number 2.

The determination of the order of designations of directions—traces for which the condition \(k < 0\) is fulfilled is left to the reader to consider.

In what follows, unless otherwise stated, we shall take, for simplicity, \(k > 0\).

Thus, we have chosen (for \(k > 0\)) the order of designation of directions—traces; it remains now to turn to the calculation of the azimuth \(\varphi\) by the formula:

\[ \tg \varphi = \frac{-m}{n} = \frac{\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1} {\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2}, \]

taking into account that the numerator of the fraction is the geometric sine, and the denominator the geometric cosine, and, in accordance with this, za-

by marking, using the table that was given above in setting forth the first principle for solving the problem.

We shall regard as known the method of determining, from logarithm tables, the smallest angle \(\varphi_{\min}\) by the formula given; further, in order to avoid the use of Gaussian logarithms, we shall give formulas convenient for ordinary logarithming.

§ 6.

Derivation of logarithmic formulas for determining the azimuth \(\varphi\) of the line of dip.

I.

We have:

\[ \tg \varphi = \frac{\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1} {\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2} \ldots\ldots\ldots\ldots\ldots\ldots\ldots (1) \]

a) Let:

\[ \tg \zeta_2 \cos a_2 > 0. \]

Divide the numerator and denominator of formula (1) by the product:

\[ \tg \zeta_2 \cos a_2. \]

Then:

\[ \tg \varphi = \frac{ 1-\dfrac{\tg \zeta_1\cdot \cos a_1}{\tg \zeta_2\cdot \cos a_2} }{ \dfrac{\tg \zeta_1 \sin a_1}{\tg \zeta_2 \cos a_2}-\tg a_2 } \ldots\ldots\ldots\ldots\ldots\ldots\ldots (1) \]

We equate to \(\tg u\) the minuend of the denominator of expression (1)

\[ \frac{\tg \zeta_1 \sin a_1}{\tg \zeta_2\cdot \cos a_2}=\tg u, \ldots\ldots\ldots\ldots\ldots\ldots\ldots (1_1) \]

whence

\[ \frac{\tg \zeta_1}{\tg \zeta_2} = \frac{\cos a_2}{\sin a_1}\tg u \]

and, consequently, the subtrahend of the numerator of expression (1) assumes the form:

\[ \frac{\tg \zeta_1 \cos a_1}{\tg \zeta_2 \cos a_2} = \tg u\cdot \ctg a_1 \ldots\ldots\ldots\ldots\ldots\ldots\ldots (1_2) \]

Therefore expression (1) assumes the form:

\[ \tg \varphi = \frac{1-\tg u\,\ctg a_1}{\tg u-\tg a_2} \ldots\ldots\ldots\ldots\ldots\ldots\ldots (2) \]

We transform the numerator and denominator of expression (2).

Numerator:

\[ 1-\tg u\,\ctg a_1 = \frac{1}{\tg a_1}(\tg a_1-\tg u) = \]

\[ = \frac{\sin a_1\cos u-\cos a_1\sin u}{\cos a_1\cos u}\cdot\frac{1}{\tg a_1} = \]

\[ = \frac{1}{\tg a_1}\cdot \frac{\sin(a_1-u)}{\cos a_1\cos u} = \frac{\sin(a_1-u)}{\sin a_1\cos u} \ldots\ldots\ldots\ldots\ldots\ldots\ldots (1_3) \]

Denominator:

\[ \tg u-\tg a_2 = \frac{\sin u\cos a_2-\cos u\sin a_2}{\cos u\cdot \cos a_2} = \]

\[ = \frac{\sin(u-a_2)}{\cos u\cdot \cos a_2} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (1_4) \]

And finally we have:

\[ \tg\varphi= \frac{\dfrac{\sin(a_1-u)}{\sin a_1\cos u}} {\dfrac{\sin(u-a_2)}{\cos a_2\cos u}} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (3) \]

Since we determine \(\varphi\) vectorially, and \(\cos u\) enters both the numerator and the denominator of the expression \(\tg\varphi\), it is natural to set two assumptions.

\[ 1)\ \cos u>0. \]

Then:

\[ \tg\varphi= \frac{\dfrac{\sin(a_1-u)}{\sin a_1}} {\dfrac{\sin(u-a_2)}{\cos a_2}} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (II) \]

\[ 2)\ \cos u<0. \]

Then:

\[ (-\cos u)>0. \]

Transforming (3), we obtain:

\[ \tg\varphi= \frac{\dfrac{\sin(a_1-u)}{-\sin a_1(-\cos u)}} {\dfrac{\sin(u-a_2)}{-\cos a_2(-\cos u)}} \]

and, finally:

\[ \tg\varphi= \frac{-\dfrac{\sin(a_1-u)}{\sin a_1}} {-\dfrac{\sin(u-a_2)}{\cos a_2}} \ldots\ldots\ldots\ldots\ldots\ldots\ldots (III) \]

b) Let now \(\tg \zeta_2\cos a_2<0\).

Divide the numerator and denominator of (I) by:

\[ (-\tg \zeta_2\cdot \cos a_2)>0. \]

A priori we see that in this case the transformation will be entirely analogous to the preceding transformation, and the formulas will be obtained from the preceding ones, changing only the signs of the numerator and denominator to the opposite ones.

Therefore, if:

\[ 1)\ \cos u>0,\ \text{then} \]

\[ \tg\varphi= \frac{\dfrac{\sin(u-a_1)}{\sin a_1}} {\dfrac{\sin(a_2-u)}{\cos a_2}} \ldots\ldots\ldots\ldots\ldots\ldots (IV) \]

2) if \(\cos u < 0\), then

\[ \operatorname{tg}\varphi = \frac{ -\dfrac{\sin (u-a_1)}{\sin a_1} }{ -\dfrac{\sin (a_2-u)}{\cos a_2} } \ldots\ldots\ldots\ldots\ldots (V) \]

As a result:

I. If \(\operatorname{tg}\zeta_2 \cdot \cos a_2 > 0\), then

\[ \operatorname{tg}\varphi = \frac{ \pm \dfrac{\sin (a_1-u)}{\sin a_1} }{ \pm \dfrac{\sin (u-a_2)}{\cos a_2} } \ldots\ldots\ldots\ldots\ldots (P) \]

The sign \((+)\) of the numerator and denominator corresponds to \(\cos u > 0\).
The sign \((-)\) of the numerator and denominator corresponds to \(\cos u < 0\).

II. If \(\operatorname{tg}\zeta_2 \cos a_2 < 0\), then

\[ \operatorname{tg}\varphi = \frac{ \pm \dfrac{\sin (u-a_1)}{\sin a_1} }{ \pm \dfrac{\sin (a_2-u)}{\cos a_2} } \ldots\ldots\ldots\ldots\ldots (Q) \]

The sign \((+)\) of the numerator and denominator corresponds to \(\cos u > 0\).
The sign \((-)\) of the numerator and denominator corresponds to \(\cos u < 0\).

In computations we first use simply any one of the formulas:

\[ \operatorname{tg}\varphi = \frac{\cos a_2 \sin (a_1-u)} {\sin a_1 \sin (u-a_2)} \ldots\ldots\ldots\ldots\ldots (A) \]

or

\[ \operatorname{tg}\varphi = \frac{\cos a_2 \sin (u-a_1)} {\sin a_1 \sin (a_2-u)}, \ldots\ldots\ldots\ldots\ldots (B) \]

from which we obtain the value \(\varphi_0\) of the angle \(\varphi\) that is smallest in absolute magnitude.

The azimuth \(\varphi\) of the line of dip is then obtained vectorially, using the inequalities and formulas indicated under the signs I or II, respectively.

II.

Let us successively reduce to logarithmic form the numerator and denominator of expression (1).

A. Numerator:

\[ \operatorname{tg}\zeta_2 \cdot \cos a_2 - \operatorname{tg}\zeta_1 \cdot \cos a_1 = \]

\[ = \frac{1}{\cos\zeta_1 \cos\zeta_2} \left( \sin\zeta_2 \cdot \cos a_2 \cdot \cos\zeta_1 - \sin\zeta_1 \cos a_1 \cos\zeta_2 \right) = \]

\[ = \frac{1}{\cos\zeta_1 \cdot \cos\zeta_2}\cdot S, \ldots\ldots\ldots\ldots\ldots (2) \]

where:

\[ S = \sin\zeta_2 \cdot \cos a_2 \cos\zeta_1 - \sin\zeta_1 \cdot \cos a_1 \cos\zeta_2 . \]

Equating:

\[ \left. \begin{aligned} \cos a_2 \cdot \cos \zeta_1' &= M_2\\ -\sin \zeta_1' \cos a_1 &= N_2 \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

we find that the second factor \(S\) of expression (2) assumes the form:

\[ S = M_2 \sin \zeta_2' + N_2 \cos \zeta_2' \qquad \ldots \ldots \ldots \ldots \ldots (4) \]

As is known, the latter expression is easily brought to logarithmic form by introducing the auxiliary angle \(\varepsilon_2\), defined by the equality:

\[ \tg \varepsilon_2 = \frac{M_2}{N_2} \qquad \ldots \ldots \ldots \ldots \ldots (5) \]

Then formula (4) assumes the form:

\[ S = N_2 \frac{\cos(\zeta_2' - \varepsilon_2)}{\cos \varepsilon_2} \qquad \ldots \ldots \ldots \ldots \ldots (6) \]

Thus formula (4) has been replaced by the system of formulas (5) and (6). Formula (4) may also be replaced by another system of formulas:

\[ \left. \begin{aligned} \tg \varepsilon_2 &= \frac{N_2}{M_2}\\[6pt] S &= M_2 \frac{\sin(\zeta_2' + \varepsilon_2)}{\cos \varepsilon_2} \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots \ldots (7) \]

Remark. From formula (2) we also see that the factor \(S\) may be represented in the form:

\[ S = M_1 \sin \zeta_1' + N_1 \cos \zeta_1', \qquad \ldots \ldots \ldots \ldots (4_1) \]

where

\[ \left. \begin{aligned} M_1 &= -\cos a_1 \cos \zeta_2'\\ N_1 &= \sin \zeta_2' \cos a_2 \end{aligned} \right\}, \qquad \ldots \ldots \ldots \ldots (3_1) \]

and correspondingly in formulas (5), (6), and (7) the index 2 is replaced by the index 1.

B. Denominator:

\[ \tg \zeta_1' \cdot \sin a_1 - \tg \zeta_2' \sin a_2 = \]

\[ = \frac{1}{\cos \zeta_1' \cos \zeta_2'} \left( \sin \zeta_1' \sin a_1 \cos \zeta_2' - \sin \zeta_2' \sin a_2 \cos \zeta_1' \right) = \]

\[ = \frac{1}{\cos \zeta_1' \cdot \cos \zeta_2'} \cdot T, \qquad \ldots \ldots \ldots \ldots (8) \]

where

\[ T = \sin \zeta_1' \sin a_1 \cos \zeta_2' - \sin \zeta_2' \sin a_2 \cos \zeta_1'. \]

Similarly to the preceding, we find that

\[ T = K_1 \sin \zeta_1' + L_1 \cos \zeta_1', \qquad \ldots \ldots \ldots \ldots (9) \]

where

\[ \left. \begin{aligned} K_1 &= \sin a_1 \cos \zeta_2'\\ L_1 &= -\sin \zeta_2' \cdot \sin a_2 \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots (10) \]

Formula (9) is replaced by the systems of formulas:

\[ \left. \begin{aligned} \operatorname{tg}\varepsilon_1 &= \frac{K_1}{L_1} \\ T &= L_1 \frac{\cos(\zeta_1-\varepsilon_1)}{\cos\varepsilon_1} \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (11_1) \]

\[ \left. \begin{aligned} \operatorname{tg}\varepsilon_1 &= \frac{L_1}{K_1} \\ T &= K_1 \frac{\sin(\zeta_1+\varepsilon_1)}{\cos\varepsilon_1} \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (11_2) \]

Likewise, \(T\) may be represented in the form:

\[ T = K_2 \sin \zeta_2 + L_2 \cos \zeta_2, \]

where

\[ \begin{aligned} K_2 &= -\sin a_2 \cos \zeta_1 \\ L_2 &= \sin \zeta_1 \sin a_1, \end{aligned} \]

and, correspondingly, in formulas \((11_{1,2})\) the index 1 is replaced by the index (2).

Thus, as a result, \(\operatorname{tg}\varphi\) will be represented in the form:

\[ \operatorname{tg}\varphi = \frac{\dfrac{1}{\cos\zeta_1 \cdot \cos\zeta_2}\cdot S} {\dfrac{1}{\cos\zeta_1 \cdot \cos\zeta_2}\cdot T} \qquad \ldots \ldots \ldots \ldots \ldots \ldots (12) \]

Now, as we know, two cases must be considered:

\[ 1)\ \zeta_1 \text{ and } \zeta_2 \text{ are the same,} \]

that is:

\[ -\frac{\pi}{2} > \zeta_{1,2} > 0 \quad \bigg| \quad -\frac{\pi}{2} < \zeta_{1,2} < \pi \]

Then the product:

\[ \cos\zeta_1 \cdot \cos\zeta_2 > 0, \]

and formula (12) takes the form:

\[ \operatorname{tg}\varphi = \frac{S}{T} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots (13); \]

\[ 2)\ \zeta_1 \text{ and } \zeta_2 \text{ are not the same,} \]

that is:

\[ -\frac{\pi}{2} > \zeta_1 > 0 \quad \bigg| \quad -\frac{\pi}{2} < \zeta_1 < \pi \]

\[ -\frac{\pi}{2} < \zeta_2 < \pi \quad \bigg| \quad -\frac{\pi}{2} > \zeta_2 > 0. \]

Then the product:

\[ \cos\zeta_1 \cdot \cos\zeta_2 < 0, \]

and the product:

\[ -\cos\zeta_1 \cdot \cos\zeta_2 > 0 \qquad \ldots \ldots \ldots \ldots \ldots (14). \]

Let us put formula (12) in the form:

\[ \operatorname{tg}\varphi = \frac{ \dfrac{1}{-(-\cos \zeta_1 \cos \zeta_2)} \cdot S }{ \dfrac{1}{-(-\cos \zeta_1 \cdot \cos \zeta_2)} \cdot T } \]

and, finally, by virtue of (14), we obtain:

\[ \operatorname{tg}\varphi=\frac{-S}{-T}. \tag{15} \]

III.

The preceding method of reducing formula (I) to logarithmic form was carried out, as we have seen, by introducing two auxiliary angles: one respectively into the expression of the numerator, the other into the expression of the denominator.

The reduction of formula (I) to logarithmic form may also be carried out by introducing four auxiliary angles: respectively two into the expression of the numerator and the other two into the expression of the denominator.

IV.

Reduction to logarithmic form of the numerator of formula (I).
At first we proceed in a manner identical with the derivations of Section II, up to and including formula (4):

\[ S=M_{1,2}\sin \zeta_{1,2}+N_{1,2}\cos \zeta_{1,2}\ldots \tag{1} \]

Putting in the latter

\[ \left. \begin{aligned} M_{1,2}&=q_{1,2}\sin Q_{1,2}\\ N_{1,2}&=q_{1,2}\cos Q_{1,2} \end{aligned} \right\}, \tag{2} \]

we find:

\[ \begin{aligned} S&=q_{1,2}\left(\sin Q_{1,2}\sin \zeta_{1,2} +\cos Q_{1,2}\cos \zeta_{1,2}\right)=\\ &=q_{1,2}\cos\left(Q_{1,2}-\zeta_{1,2}\right)\ldots \end{aligned} \tag{3} \]

The auxiliary quantities \(q_{1,2}\) and \(Q_{1,2}\) are determined from formulas (2); namely, we have:

\[ \left. \begin{aligned} \operatorname{tg}Q_{1,2}&=\frac{M_{1,2}}{N_{1,2}}\\[6pt] q_{1,2}&=\frac{M_{1,2}}{\sin Q_{1,2}} =\frac{N_{1,2}}{\cos Q_{1,2}} \end{aligned} \right\} \tag{4} \]

If in formula (1) we put:

\[ \left. \begin{aligned} M_{1,2}&=p_{1,2}\cos P_{1,2}\\ N_{1,2}&=p_{1,2}\sin P_{1,2} \end{aligned} \right\}, \tag{5} \]

then we find that the same quantity \(S\) is determined by the formula:

\[ S=p_{1,2}\sin\left(\zeta_{1,2}+P_{1,2}\right)\ldots \tag{6} \]

The auxiliary quantities \(p_{1,2}\) and \(P_{1,2}\) are determined from formulas (5), giving:

\[ \left. \begin{aligned} \operatorname{tg} P_{1,2} &= \frac{N_{1,2}}{M_{1,2}},\\[4pt] p_{1,2} &= \frac{N_{1,2}}{\sin P_{1,2}}=\frac{M_{1,2}}{\cos P_{1,2}} \end{aligned} \right\}\ . . . . . . . . . . (7) \]

Reduction to logarithmic form of the denominator is analogous to the preceding one; namely, the corresponding formulas are obtained from (1), (2), (3), (4), (5), (6), and (7) by replacing in them \(M_{1,2}\), \(N_{1,2}\), and \(S\), respectively, by \(K_{1,2}\), \(L_{1,2}\), and \(T\).

§ 7.

I. Another derivation of the logarithmic formula for determining the azimuth of the line of true dip.

Let us pass from the system \(XYZ\) to the system \(X'Y'Z'\), rotating the former about the axis \(Z\) through the angle

\[ \alpha=\frac{1}{2}(a_1+a_2). \]

Then the azimuths of the directions—of the traces and of the line of true dip—relative to the axis \(X'\) will be, respectively:

\[ A_1=-\alpha+a_1;\quad A_2=-\alpha+a_2\quad \text{and}\quad \Phi=-\alpha+\varphi. \]

Substituting the value of the angle \(\alpha\) into the last formulas, we obtain:

\[ A_1=\frac{a_1-a_2}{2};\quad A_2=\frac{a_2-a_1}{2}\quad \text{and}\quad \Phi=\varphi-\frac{a_1+a_2}{2}. \]

Accordingly, with the new azimuths of the directions—of the traces—the azimuth of the line of true dip is determined by the formula known to us:

\[ \operatorname{tg}\Phi = \frac{\operatorname{tg}\zeta_2\cos A_2-\operatorname{tg}\zeta_1\cos A_1} {\operatorname{tg}\zeta_1\sin A_1-\operatorname{tg}\zeta_2\sin A_2} = \]

\[ = \frac{ \operatorname{tg}\zeta_2\cos\dfrac{a_2-a_1}{2} - \operatorname{tg}\zeta_1\cos\dfrac{a_1-a_2}{2} }{ \operatorname{tg}\zeta_1\sin\dfrac{a_1-a_2}{2} - \operatorname{tg}\zeta_2\sin\dfrac{a_2-a_1}{2} } = \frac{ \cos\dfrac{a_1-a_2}{2}\,(\operatorname{tg}\zeta_2-\operatorname{tg}\zeta_1) }{ \sin\dfrac{a_1-a_2}{2}\,(\operatorname{tg}\zeta_1+\operatorname{tg}\zeta_2) } \]

and, finally,

\[ \operatorname{tg}\left(\varphi-\frac{a_1+a_2}{2}\right) = \frac{ \cos\dfrac{a_1-a_2}{2}\cdot \dfrac{\sin(\zeta_2-\zeta_1)}{\cos\zeta_1\cdot\cos\zeta_2} }{ \sin\dfrac{a_1-a_2}{2}\cdot \dfrac{\sin(\zeta_2+\zeta_1)}{\cos\zeta_1\cdot\cos\zeta_2} }, \ . . . . . . . . . . (t) \]

for

\[ \operatorname{tg}\zeta_2+\operatorname{tg}\zeta_1 = \frac{\sin\zeta_2}{\cos\zeta_2} + \frac{\sin\zeta_1}{\cos\zeta_1} = \frac{\sin\zeta_2\cos\zeta_1+\cos\zeta_2\sin\zeta_1} {\cos\zeta_1\cdot\cos\zeta_2} = \frac{\sin(\zeta_2+\zeta_1)} {\cos\zeta_1\cdot\cos\zeta_2}. \]

Analysis of formula (t).

Consider the cases:

1) \[ 0<\zeta_{1,2}<\frac{\pi}{2}\quad \text{or}\quad \pi>\zeta_{1,2}>\frac{\pi}{2}. \]

Under these conditions the product \(\cos \zeta_1 \cos \zeta_2 > 0\), and formula (t) assumes, vectorially, the form:

\[ \tan\left(\varphi-\frac{a_1+a_2}{2}\right) = \frac{ \cos \dfrac{a_1-a_2}{2}\,\sin(\zeta_2-\zeta_1) }{ \sin \dfrac{a_1-a_2}{2}\,\sin(\zeta_2+\zeta_1) } \ldots\ldots\ldots\ldots\ldots\ldots (I) \]

\[ \begin{array}{c} \text{2) }\quad \dfrac{\pi}{2}>\zeta_1>0\\[4pt] \pi>\zeta_2>\dfrac{\pi}{2} \end{array} \quad \left\{ \begin{array}{c} \\[-16pt] \text{or}\\[12pt] \end{array} \right. \quad \begin{array}{c} \pi>\zeta_1>\dfrac{\pi}{2}\\[4pt] \dfrac{\pi}{2}>\zeta_2>0. \end{array} \]

Under these conditions the product \(\cos\zeta_1 \cdot \cos\zeta_2 < 0\). Multiplying, in formula (t), which determines \(\varphi\), the numerator and denominator of the expression

\[ \frac{\sin(\zeta_2\pm\zeta_1)}{\cos\zeta_1\cdot\cos\zeta_2} \]

by \((-1)\) and reducing by the product \((-\cos\zeta_1\cdot\cos\zeta_2>0)\), we obtain vectorially:

\[ \tan\left(\varphi-\frac{a_1+a_2}{2}\right) = \frac{ -\cos \dfrac{a_1-a_2}{2}\,\sin(\zeta_2-\zeta_1) }{ -\sin \dfrac{a_1-a_2}{2}\,\sin(\zeta_2+\zeta_1) } \ldots\ldots\ldots\ldots\ldots\ldots (II) \]

Since, when using logarithm tables, the angle taken is \(<\dfrac{\pi}{2}\), at the beginning—in all cases—the computations are carried out simply by the formula:

\[ \tan\left(\varphi-\frac{a_1+a_2}{2}\right) = \cot \frac{a_1-a_2}{2}\cdot \frac{\sin(\zeta_2-\zeta_1)}{\sin(\zeta_2+\zeta_1)} \ldots\ldots\ldots\ldots (III) \]

Having found the angle \(<\dfrac{\pi}{2}\), one determines the signs of the numerator and denominator of (I) [or, respectively, (II)] and computes vectorially the value of the angle

\[ \Phi=\varphi-\frac{a_1+a_2}{2}, \]

whence the sought angle is

\[ \varphi=\Phi+\frac{1}{2}(a_1+a_2). \]

II. Determination of the zenith distance of the line of dip.

Since the direction of the line of dip goes from the point with the higher elevation to the point with the lower elevation, its zenith distance \(\zeta\) must satisfy the inequality:

\[ \pi>\zeta>\frac{\pi}{2}, \]

which is immediately evident by taking the plane \(X'AZ'\), coinciding with the vertical plane of dip of the bed (see Fig. 12).

1. Let us take an arbitrary point \(N(x',z')\) of the line of dip, lying in the second quadrant; corresponding to it, we find:

\[ \tan \zeta=\frac{x'}{z'} \ldots\ldots\ldots\ldots\ldots\ldots (1) \]

Fig. 12.

The latter relation is easily found from the equations:

\[ z'=\pm \frac{x'}{\kappa}\sqrt{m^2+n^2}\quad \text{for } \kappa \gtrless 0 \text{ respectively.} \]

Therefore:

\[ \tg \zeta=\pm \frac{\kappa}{\sqrt{m^2+n^2}}\quad \text{respectively for } \kappa \gtrless 0. \]

By virtue of the condition: \(\pi>\zeta>\frac{\pi}{2}\), the quantity \(\tg \zeta<0\), and the zenith distance \(\zeta\) is determined uniquely. In what follows it is simpler to use the formula:

\[ \tg \zeta=-\frac{\kappa}{\sqrt{m^2+n^2}}, \]

therefore taking \(\kappa>0\).

1. Taking the point \(M(x',z')\) in the fourth quadrant, we find

\[ \tg \zeta=\frac{-x'}{-z'}. \]

By virtue of the condition that \(\pi>\zeta>\frac{\pi}{2}\), and consequently \(\tg \zeta<0\), the formula given above is simplified and becomes identical with (1).

1. Assuming \(\kappa>0\), take two points of the line of dip, \(N\) and \(M\), for which, respectively:

\[ x'_p>0;\ z'_n<0 \ . \ . \ . \ . \ . \ \text{(point N),} \]

and

\[ x'_n<0;\ z'_p>0 \ . \ . \ . \ . \ . \ \text{(point M).} \]

Then the zenith distance will be determined by the formula:

\[ \tg \zeta= \frac{x'_p-x'_n}{z'_n-z'_p} = \frac{x'_p-x'_n}{ -\dfrac{\sqrt{m^2+n^2}}{\kappa}(x'_p-x'_n)} = -\frac{\kappa}{\sqrt{m^2+n^2}}. \]

The formula obtained,

\[ \tg \zeta=-\frac{\kappa}{\sqrt{m^2+n^2}}, \]

is inconvenient for direct logarithming. Let us first transform the expression \(m^2+n^2\).

We have:

\[ m^2+n^2=(\tg \zeta_2\cos a_2-\tg \zeta_1\cos a_1)^2+ \]

\[ +(\tg \zeta_1\sin a_1-\tg \zeta_2\sin a_2)^2 = \tg^2\zeta_1+\tg^2\zeta_2-2\tg \zeta_1\,\tg \zeta_2\cos(a_1-a_2), \]

an expression analogous to the expression: \(b^2+c^2-2bc\cos A\).

It is known that if we have the expression:

\[ x=\sqrt{b^2+c^2-2bc\cos A}, \]

then it is easily reduced to the form:

\[ x=\left|\frac{b+c}{\cos \Delta}\sin\frac{A}{2}\right|, \]

where

\[ \Delta=\arc\tg\left[\frac{b-c}{b+c}\ctg\frac{A}{2}\right]. \]

In our case we may put:

\[ \left. \begin{array}{l} b=\tg \zeta_1\\ c=\tg \zeta_2 \end{array} \right\} \ . \ . \ . \ . \ . \ (2) \qquad \text{or} \qquad \left. \begin{array}{l} b=\tg \zeta_2\\ c=\tg \zeta_1 \end{array} \right\} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (3) \]

As in case (2), so also in case (3)

\[ A=a_1-a_2 \quad \text{or} \quad A=a_2-a_1. \]

Therefore:

\[ b\pm c=\operatorname{tg}\zeta_{1,2}\pm \operatorname{tg}\zeta_{2,1} =\frac{\sin(\zeta_{1,2}\pm \zeta_{2,1})}{\cos\zeta_1\cdot \cos\zeta_2}, \]

and

\[ \frac{b-c}{b+c}= \frac{\sin(\zeta_{1,2}-\zeta_{2,1})}{\sin(\zeta_{1,2}+\zeta_{2,1})}, \]

which we substitute into the formulas determining \(\Delta\) and \(x\).

As a result:

\[ x=\left| \frac{ \sin(\zeta_{1,2}+\zeta_{2,1})\sin\frac{\pm(a_1-a_2)}{2} }{ \cos\zeta_1\cdot \cos\zeta_2 \cos\left\{ \operatorname{arctg}\left[ \frac{\sin(\zeta_{1,2}-\zeta_{2,1})}{\sin(\zeta_{1,2}+\zeta_{2,1})} \operatorname{ctg}\frac{\pm(a_1-a_2)}{2} \right]\right\} } \right|. \]

\[ \operatorname{tg}z=-\frac{k}{x}. \]

The expression:

\[ x=\sqrt{b^2+c^2-2bc\cos A} \]

is also easily reduced to the form:

\[ x=(b+c)\sqrt{1-\frac{4bc\cos^2\frac{1}{2}A}{(b+c)^2}} \tag{1} \]

Since in our case \(x>0\), the latter, for identical \(b\) and \(c\), can easily be replaced by the system:

\[ \left\{ \begin{aligned} x&=\left|(b+c)\cos\varepsilon\right|,\\ \varepsilon&=\arcsin\left[\pm\frac{2\cos\frac{1}{2}A\sqrt{bc}}{b+c}\right] \end{aligned} \right\} \tag{2} \]

or—

\[ \left\{ \begin{aligned} x&=\left|(b+c)\sin\varepsilon\right|,\\ \varepsilon&=\arccos\left[\pm\frac{2\cos\frac{1}{2}A\sqrt{bc}}{b+c}\right] \end{aligned} \right\} \tag{3} \]

Further, putting in formula (1)

\[ \operatorname{tg}^2\varepsilon= \frac{4bc\cos^2\frac{1}{2}A}{(b+c)^2} \]

and taking into account that

\[ 1-\tg^{2}\varepsilon=\frac{2\tg\varepsilon}{\tg 2\varepsilon} =\frac{\cos 2\varepsilon}{\cos^{2}\varepsilon}, \]

formula \((1_1)\) is replaced by the systems:

\[ \left. \begin{aligned} x&=\left|(b+c)\sqrt{\frac{2\tg\varepsilon}{\tg 2\varepsilon}}\right|,\\ \varepsilon&=\arc\tg\left[\pm\frac{2\cos\frac{1}{2}A}{b+c}\sqrt{bc}\right] \end{aligned} \right\} \qquad \cdots\cdots\cdots\cdots (4) \]

\[ \left. \begin{aligned} x&=\left|\frac{b+c}{\cos\varepsilon}\sqrt{\cos 2\varepsilon}\right|,\\ \varepsilon&=\arc\tg\left(\pm\frac{2\cos\frac{1}{2}A}{b+c}\sqrt{bc}\right) \end{aligned} \right\} \qquad \cdots\cdots\cdots\cdots (5) \]

When \(b\) and \(c\) are unequal, formulas (2), (3), (4), (5), obviously, are not applicable, and therefore the radical expression in formula \((1_1)\) must be transformed differently.

Namely, we may write:

\[ 1-\frac{4bc\cos^{2}\frac{1}{2}A}{(b+c)^{2}} = 1+\left|\frac{4bc\cos^{2}\frac{1}{2}A}{(b+c)^{2}}\right| = 1+\tg^{2}\varepsilon, \]

where

\[ \tg^{2}\varepsilon= \left|\frac{4bc\cos^{2}\frac{1}{2}A}{(b+c)^{2}}\right|. \]

Taking into account that

\[ 1+\tg^{2}\varepsilon=\frac{1}{\cos^{2}\varepsilon} =\frac{2\tg\varepsilon}{\sin 2\varepsilon}, \]

formula \((1_1)\), when \(b\) and \(c\) are unequal, is replaced by the system of formulas:

\[ \left. \begin{aligned} x&=\left|\frac{b+c}{\cos\varepsilon}\right| = \left|(b+c)\sqrt{\frac{2\tg\varepsilon}{\sin 2\varepsilon}}\right|,\\ \tg^{2}\varepsilon&= \left|\frac{4bc\cos^{2}\frac{1}{2}A}{(b+c)^{2}}\right| \end{aligned} \right\} \qquad \cdots\cdots (6),\ (7) \]

Putting further:

\[ \tg\varepsilon_{1}= \left|\frac{4bc\cos^{2}\frac{1}{2}A}{(b+c)^{2}}\right| \]

or

\[ \cos \varepsilon_2= \left| \frac{4bc\cos^2 \frac{1}{2}A}{(b+c)^2} \right| \]

and, taking into account that, respectively,

\[ 1+\operatorname{tg}\varepsilon_1 = \frac{\sin(45^\circ+\varepsilon_1)} {\cos45^\circ\cos\varepsilon_1} \]

\[ 1+\cos\varepsilon_2 = 2\cos^2\frac{1}{2}\varepsilon_2 = \frac{\sin\varepsilon_2} {\operatorname{tg}\frac{1}{2}\varepsilon_2}, \]

formula \((1_1)\) is replaced, respectively, by the systems:

\[ \left\{ \begin{aligned} x&= \left| (b+c)\sqrt{ \frac{\sin(45^\circ+\varepsilon_1)} {\cos45^\circ\cos\varepsilon_1} } \right|,\\[6pt] \varepsilon_1&= \operatorname{arctg} \left| \frac{4bc\cos^2\frac{1}{2}A}{(b+c)^2} \right| \end{aligned} \right. \qquad . . . . . . . .\ (8) \]

\[ \left\{ \begin{aligned} x&= \left| (b+c)\sqrt{2\cos\frac{1}{2}\varepsilon_2} \right| =\\[6pt] &= \left| (b+c)\sqrt{ \frac{\sin\varepsilon_2} {\operatorname{tg}\frac{1}{2}\varepsilon_2} } \right|,\\[6pt] \varepsilon_2&= \arccos \left| \frac{4bc\cos^2\frac{1}{2}A}{(b+c)^2} \right| \end{aligned} \right. \qquad . . . .\ (9),\ (10) \]

Finally, assuming, for any \(b\) and \(c\),

\[ \cos\varepsilon_1= \frac{4bc\cos^2\frac{1}{2}A}{(b+c)^2} = \operatorname{tg}\varepsilon_2 \]

and taking into account that

\[ 1-\cos\varepsilon_1 = 2\sin^2\frac{1}{2}\varepsilon_1 = \sin\varepsilon_1\operatorname{tg}\frac{1}{2}\varepsilon_1 \]

\[ 1-\operatorname{tg}\varepsilon_2 = \frac{\sin(45^\circ-\varepsilon_2)} {\cos45^\circ\cos\varepsilon_2}, \]

formula (1) is replaced respectively by the systems:

\[ \left. \begin{aligned} x &= \left|(b+c)\sqrt{\,2\sin \frac{1}{2}\zeta_1\,}\right| \\ &= \left|(b+c)\sqrt{\,\sin \zeta_1 \operatorname{tg}\frac{1}{2}\zeta_1\,}\right| \\[6pt] \zeta_1 &= \arccos\left[ \frac{4bc\cos^2 \frac{1}{2}A}{(b+c)^2} \right] \end{aligned} \right\} \qquad \ldots\ (11),\ (12) \]

\[ \left. \begin{aligned} x &= \left|(b+c)\sqrt{ \frac{\sin(45^\circ-\zeta_2)} {\cos45^\circ\cos\zeta_2} }\right| \\[6pt] \zeta_2 &= \operatorname{arctg}\left[ \frac{4bc\cos^2 \frac{1}{2}A}{(b+c)^2} \right] \end{aligned} \right\} \qquad \ldots\ (13) \]

It remains now to insert the values of \(b\), \(c\), and \(A\) into the formulas \([(2)\ldots(13)]\).

We have:

\[ b=\operatorname{tg}\zeta_{1,2},\quad c=\operatorname{tg}\zeta_{2,1},\quad A=\pm(a_1-a_2), \]

and consequently:

\[ b+c=\frac{\sin(\zeta_1+\zeta_2)} {\cos\zeta_1\cdot\cos\zeta_2} \]

\[ \frac{4bc\cos^2 \frac{1}{2}A}{(b+c)^2} = \frac{ 4\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2 \cos^2 \frac{1}{2}(a_1-a_2) }{ \left[ \frac{\sin^2(\zeta_1+\zeta_2)} {\cos^2\zeta_1\cdot\cos^2\zeta_2} \right] } = \]

\[ = \frac{ 4\sin\zeta_1\cos\zeta_1\sin\zeta_2\cos\zeta_2 \cos^2 \frac{1}{2}(a_1-a_2) }{ \sin^2(\zeta_1+\zeta_2) } = \]

\[ = \frac{ \sin 2\zeta_1\sin 2\zeta_2 \cos^2 \frac{1}{2}(a_1-a_2) }{ \sin^2(\zeta_1+\zeta_2) }. \]

Let us also note that the radicand in formula \((1_1)\) may be represented in the form: \(A - B\),

where

\[ A = 1; \qquad B = \frac{4bc\cos^2 \frac{1}{2}A}{(b+c)^2} = \]

\[ = \frac{\sin 2\zeta_1 \sin 2\zeta_2 \cos^2 \frac{1}{2}(a_1-a_2)} {\sin^2(\zeta_1+\zeta_2)} . \]

For equal \(b\) and \(c\) we have:

\[ A > 0; \qquad B > 0; \qquad A > B. \]

For unequal \(b\) and \(c\) we have:

\[ A > 0; \qquad B < 0; \qquad A > |B|; \]

and therefore the further transformations of the first and second cases are, respectively, identical with the transformations of § 9 \([I, a_1); II, a)]\).

III.

We shall give another derivation of the logarithmic formula for determining the zenith distance of the line of dip.

We have:

\[ \tg \zeta = -\frac{k}{\sqrt{m^2+n^2}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

Put in formula \((1)\):

\[ -m = n\tg\varphi, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

whence—

\[ \tg\varphi = \frac{-m}{n}. \]

The latter expression, as is known, determines the azimuth \(\varphi\) of the line of dip.

Substituting the value of \(m\) from \((2)\) into the expression \(m^2+n^2\), we obtain:

\[ m^2+n^2 = n^2(1+\tg^2\varphi) = \frac{n^2}{\cos^2\varphi} \]

and, consequently:

\[ \sqrt{m^2+n^2} = \pm \frac{n}{\cos\varphi} > 0, \ldots \ldots \ldots \ldots (2_1) \]

in accordance with formula \((1)\); therefore the signs must be taken according to \(k>0\).

Further, determining \(n\) from \((2)\) and substituting the latter in the expression \(m^2+n^2\), we obtain:

\[ m^2+n^2 = m^2\left(1+\frac{1}{\tg^2\varphi}\right) = \frac{m^2}{\sin^2\varphi} \]

and, consequently:

\[ \sqrt{m^2+n^2}=\pm \frac{m}{\sin\varphi}>0,\ldots\ldots\ldots (2_2) \]

in accordance with formula (1); therefore the signs \((k>0)\) must be taken accordingly.

Thus we have:

\[ \tg \zeta = -\frac{k}{\sqrt{m^2+n^2}} = -\frac{k}{\pm\left(\frac{n}{\cos\varphi}\right)} = -\frac{k}{\pm\left(\frac{m}{\sin\varphi}\right)} \]

and, finally:

\[ \tg \zeta = -\frac{\tg\zeta_1\,\tg\zeta_2\,\sin(a_2-a_1)} {\pm\left(\frac{n}{\cos\varphi}\right)} = \]

\[ = -\frac{\tg\zeta_1\,\tg\zeta_2\,\sin(a_2-a_1)} {\pm\left(\frac{m}{\sin\varphi}\right)} \ldots\ldots (3_1,\,3_2) \]

Let us now transform the expressions \(n\) and \(m\).

We have (see § 5, I, 1):

\[ 1)\quad n=\tg\zeta_1\sin a_1-\tg\zeta_2\sin a_2 = \]

\[ = \tg\zeta_2\cdot \cos a_2 \left( \frac{\tg\zeta_1\sin a_1}{\tg\zeta_2\cos a_2} -\tg a_2 \right). \]

Denoting:

\[ \frac{\tg\zeta_1\cdot \sin a_1}{\tg\zeta_2\cos a_2} = \tg u \ldots\ldots\ldots\ldots\ldots (4) \]

[see formula \((1_1)\), § 6, II], we obtain—

\[ n=\tg\zeta_2\cos a_2(\tg u-\tg a_2) \]

and, finally, in accordance with formula \((1_4)\), § 6, we have:

\[ n=\tg\zeta_2\frac{\sin(u-a_2)}{\cos u} \ldots\ldots\ldots\ldots (5) \]

\[ 2)\quad m=\tg\zeta_1\cos a_1-\tg\zeta_2\cos a_2 = \]

\[ = \tg\zeta_2\cos a_2 \left( \frac{\tg\zeta_1\cos a_1}{\tg\zeta_2\cdot \cos a_2} -1 \right); \]

denoting:

\[ \frac{\tg\zeta_1\cdot \cos a_1}{\tg\zeta_2\cdot \cos a_2} = \tg u\,\ctg a_1 \ldots\ldots\ldots\ldots (6) \]

[see formulas \((1_2)\), \((1_3)\), § 6, II], we obtain—

\[ m=\tg\zeta_2\cos a_2(\tg u\,\ctg a_1-1) \]

and, finally,

\[ m=\tg\zeta_2\cos a_2 \frac{\sin(u-a_1)}{\sin a_1\cos u} \ldots\ldots\ldots (7) \]

A.

In the computations: a) determine successively \(u, m, n\); then b) by means of \((2_1)\) and \((2_2)\), respectively, test the signs of \(n, \cos\varphi, m, \sin\varphi\); and, finally, c) determine \(\zeta\) with checking; or:

B.

Discarding in formulas \((3_1)\) and \((3_2)\) the signs \((\pm)\) and substituting successively the values of \(n\) and \(m\), respectively, from (5) and (7), we obtain:

\[ \tg(180^\circ-\zeta)= \frac{\tg\zeta_1 \sin(a_2-a_1)\cos\varphi\cos u}{\sin(u-a_2)} \ldots\ldots\ldots (I) \]

or

\[ \tg(180^\circ-\zeta)= \frac{\tg\zeta_1 \sin(a_2-a_1)\sin\varphi\sin a_1\cos u}{\cos a_2 \sin(u-a_1)} \ldots\ldots (II) \]

Noting that

\[ 90^\circ < \zeta < 180^\circ \]

and, consequently,

\[ \tg(180^\circ-\zeta)>0, \]

one must, as a result of taking logarithms of the right-hand side of formulas (I) and (II), discard the sign \(n^*)\) (negatif).

Remark. The quantities \(n\) and \(m\) [in formulas \((3_1, 3_2)\)], as is known to us, are represented in the form of differences:

\[ \begin{aligned} n &= A_1-B_1,\\ m &= A_2-B_2, \end{aligned} \]

and these latter can easily be brought to logarithmic form, analogously to what is indicated below (see § 9).

§ 8.

1. Second method for determining the zenith distance of the line of dip.

Let us pass from the system \(XYZ\) to the system \(X'Y'Z'\), rotating the former about the \(Z\)-axis through the angle \(\varphi\), corresponding to the azimuth of the line of dip. Then, as we found earlier, for points of the line of dip we shall have [(see § 5, 2, formula (4)]:

\[ \begin{aligned} x &= x'\cos\varphi\\ y &= x'\sin\varphi . \end{aligned} \]

The equation of the plane of the bed in the new coordinate system is represented in the form:

\[ (n\cos\varphi-m\sin\varphi)x' + \kappa z' = 0, \ldots\ldots (1) \]

whence—

\[ \frac{x'}{z'}=-\frac{\kappa}{n\cos\varphi-m\sin\varphi}; \]

but, by what was proved earlier,

\[ \frac{x'}{z'}=\tg\zeta, \]

and therefore:

\[ \tg\zeta=-\frac{\kappa}{n\cos\varphi-m\sin\varphi} \ldots\ldots (2) \]

\[ \text{*) in the event of its occurrence.} \]

We had (see § 5, 2):

\[ \left. \begin{aligned} \sin \varphi &= \pm \frac{m}{\sqrt{m^{2}+n^{2}}},\\[4pt] \cos \varphi &= \pm \frac{n}{\sqrt{m^{2}+n^{2}}} \end{aligned} \right\} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \tag{3} \]

Substituting (3) into (2), we obtain the final expression for determining the zenith distance:

\[ \tan \zeta=\pm \frac{\kappa}{\sqrt{m^{2}+n^{2}}}. \]

The sign \((-)\) corresponds to the assumption \(\kappa>0\), the sign \((+)\) to the assumption \(\kappa<0\); this follows directly from the analysis of the determination of the azimuth of the line of dip.

2. Third method for determining the zenith distance of the line of dip.

Into equation (1) of the preceding method we substitute the values of \(m\) and \(n\). We obtain:

\[ \bigl[(\tan \zeta_{1}\sin a_{1}-\tan \zeta_{2}\sin a_{2})\cos \varphi - (\tan \zeta_{1}\cos a_{1}-\tan \zeta_{2}\cos a_{2})\sin \varphi\bigr]x' +\kappa z'=0. \]

Or, transforming,

\[ \bigl[\tan \zeta_{1}(\sin a_{1}\cos \varphi-\cos a_{1}\sin \varphi) - \tan \zeta_{2}(\sin a_{2}\cos \varphi-\cos a_{2}\sin \varphi)\bigr]x' +\kappa z'=0, \]

or, finally,

\[ \bigl[\tan \zeta_{1}\sin (a_{1}-\varphi)-\tan \zeta_{2}\sin (a_{2}-\varphi)\bigr]x' +\kappa z'=0. \]

Hence:

\[ \frac{x'}{z'}= -\frac{\kappa}{\tan \zeta_{1}\sin (a_{1}-\varphi)-\tan \zeta_{2}\sin (a_{2}-\varphi)} \]

or, noting that \(\dfrac{x'}{z'}=\tan \zeta\), we obtain the final expression, in another form, determining the zenith distance \(\zeta\):

\[ \tan \zeta = -\frac{\tan \zeta_{1}\tan \zeta_{2}\sin (a_{2}-a_{1})} {\tan \zeta_{1}\sin (a_{1}-\varphi)-\tan \zeta_{2}\sin (a_{2}-\varphi)} = -\frac{\sin (a_{2}-a_{1})} {\operatorname{ctg}\zeta_{2}\sin (a_{1}-\varphi)-\operatorname{ctg}\zeta_{1}\sin (a_{2}-\varphi)}. \]

As we see, it is first necessary to determine \(\varphi\)—the azimuth of the line of dip.

Remarks: 1. Since \(\tan \zeta<0\), the numerator and denominator of the preceding expression for \(\tan \zeta\) are of the same sign.

1. The last formula can easily be put into a form convenient for logarithming. However, it is simpler to find such a form directly.

3. Fourth method for determining the zenith distance of the dip line.

We found earlier that each of the two directions—traces—is determined by the points: \((0,0,0)\) and, respectively,

\[ \begin{aligned} x_{1,2} &= R \sin \zeta_{1,2}\cos a_{1,2}\\ y_{1,2} &= R \sin \zeta_{1,2}\sin a_{1,2}\\ z_{1,2} &= R \cos \zeta_{1,2} \end{aligned} \tag{1} \]

Therefore the equations of the directions—traces—will be, respectively:

\[ \frac{x}{R\sin \zeta_{1,2}\cos a_{1,2}} = \frac{y}{R\sin \zeta_{1,2}\sin a_{1,2}} = \frac{z}{R\cos \zeta_{1,2}} . \tag{2} \]

or, dividing the denominators of the ratios by \(R\cos \zeta_{1,2}\), we obtain:

\[ \frac{x}{\operatorname{tg}\zeta_{1,2}\cos a_{1,2}} = \frac{y}{\operatorname{tg}\zeta_{1,2}\sin a_{1,2}} = \frac{z}{1}. \tag{3} \]

Combining the first and second ratios with the third of expression (3), we obtain:

\[ x-\operatorname{tg}\zeta_{1,2}\cos a_{1,2}\,z=0 \tag{4} \]

\[ y-\operatorname{tg}\zeta_{1,2}\sin a_{1,2}\,z=0, \tag{5} \]

i.e., each of the two directions—traces—is determined as the intersection of two planes, (4) and (5), respectively.

Let us now pass from the system \(XYZ\) to the system \(X'Y'Z'\), rotating the former about the axis \(Z\) through the angle \(\varphi\), corresponding to the azimuth of the dip line. Substituting the transformation formulas into (4) and (5), we obtain:

\[ x'\cos\varphi+y'\sin\varphi-\operatorname{tg}\zeta_{1,2}\cos a_{1,2}\,z'=0 \tag{6} \]

\[ x'\sin\varphi+y'\cos\varphi-\operatorname{tg}\zeta_{1,2}\sin a_{1,2}\,z'=0 . \tag{7} \]

Multiplying (6) by \(\cos\varphi\), (7) by \(\sin\varphi\), and adding, we obtain:

\[ x'-\operatorname{tg}\zeta_{1,2}\left(\cos a_{1,2}\cos\varphi+\sin a_{1,2}\sin\varphi\right)z'=0, \]

or—

\[ x'-\operatorname{tg}\zeta_{1,2}\cos(\varphi-a_{1,2})\,z'=0; \]

whence:

\[ \frac{x'}{z'}=\operatorname{tg}\zeta_{1,2}\cos(\varphi-a_{1,2}), \]

and finally we have:

\[ \operatorname{tg}\zeta=\operatorname{tg}\zeta_{1,2}\cos(\varphi-a_{1,2}), \tag{8} \]

two formulas mutually checking the determination of \(\zeta\).

Below we shall give a further series of methods for determining the zenith distance \(\zeta\).

§ 9.

Proceeding to reduce to logarithmic form the formula (see 2 § 8)

\[ \operatorname{tg}(180^\circ-\zeta)= - \frac{\operatorname{tg}\zeta_1\,\operatorname{tg}\zeta_2\sin(a_2-a_1)} {\operatorname{tg}\zeta_1\sin(a_1-\varphi)-\operatorname{tg}\zeta_2\sin(a_2-\varphi)} \tag{1} \]

we note that:

\[ 1)\quad 90^\circ<\zeta<180^\circ \]

and \(2)\) by condition

\[ \operatorname{tg}\zeta_1\cdot \operatorname{tg}\zeta_2\sin(a_2-a_1)>0. \]

Therefore the denominator \(d\) of expression (1):

\[ d=\tg z_1 \cdot \sin(a_1-\varphi)-\tg z_2 \sin(a_2-\varphi)>0. \ldots \tag{2} \]

Denoting the minuend and subtrahend of the denominator (2) of expression (1) respectively by \(A\) and \(B\), we see that the question reduces to putting the expression \(d\) into logarithmic form:

\[ d=A-B. \ldots \ldots \ldots \ldots \ldots \tag{2_1} \]

under condition (2)—

\[ A>B. \ldots \ldots \ldots \ldots \ldots \tag{3} \]

Let us consider the possible assumptions concerning \(A\) and \(B\) in expression \((2_1)\) under condition (3).

I. \(B\) and \(A\) have the same signs.

In this case:

\[ \text{a}_1)\quad B>0 \ \text{and}\ A>0, \]

or—

\[ \text{b}_1)\quad B<0 \ \text{and}\ A<0. \]

Under condition (3), assumptions \(\text{a}_1)\) and \(\text{b}_1)\) entail, respectively, the consequences:

\[ A>B;\qquad |A|<|B|. \]

II. \(B\) and \(A\) have different signs.

In this case:

\[ \text{a}_2)\quad B>0 \ \text{and}\ A<0, \]

or—

\[ \text{b}_2)\quad B<0 \ \text{and}\ A>0. \]

Under condition (3), assumption \(\text{a}_2)\) is eliminated; assumption \(\text{b}_2)\) gives:

\[ A>|B| \quad \text{and} \quad A<|B|. \]

Let us now proceed successively to putting the expression \(A-B\) into logarithmic form.

I. \(A\) and \(B\) have the same sign.

\(\text{a}_1)\) We have:

\[ A>0;\quad B>0,\quad \text{with } A>B. \]

1. Transforming \(A-B\), we obtain:

\[ d=A-B=A\left(1-\frac{B}{A}\right). \ldots \ldots \ldots \tag{3_1} \]

Putting the ratio \(\dfrac{B}{A}\) equal to:

\[ \frac{B}{A}=\sin^2 z_1=\cos^2 z_2=\cos z_3=\cos 2z_4=\tg^2 z_5, \]

where \(z_{1\ldots 5}\) are the corresponding auxiliary angles, we obtain, respectively:

\[ d=A-B=A(1-\sin^2 z_1)=A\cos^2 z_1. \]

\[ d=A-B=A(1-\cos^2 z_2)=A\sin^2 z_2. \]

\[ d=A-B=A(1-\cos z_3)=2A\sin^2 \frac{1}{2}z_3 =A\sin z_3 \tg \frac{1}{2}z_3. \]

\[ d=A-B=A(1-\cos 2z_4)=2A\sin^2 z_4 =A\sin 2z_4 \tg z_4. \]

\[ d=A-B=A(1-\tg^2 z_5) =A\frac{2\tg z_5}{\tg 2z_5} =A\frac{\cos 2z_5}{\cos^2 z_5}. \]

Thus, formula \((3_1)\) is replaced by the system of formulas (4), (5), (6), (7), (8):

\[ \begin{gathered} A-B=A\cos^2 \varepsilon_1,\\ \varepsilon_1=\arcsin\left(\pm\sqrt{\frac{B}{A}}\right) \end{gathered} \qquad\qquad\qquad\qquad (4) \]

\[ \begin{gathered} A-B=A\sin^2 \varepsilon_2,\\ \varepsilon_2=\arccos\left(\pm\sqrt{\frac{B}{A}}\right) \end{gathered} \qquad\qquad\qquad\qquad (5) \]

\[ \begin{gathered} A-B=2A\sin^2\frac{1}{2}\varepsilon_3 =A\sin\varepsilon_3\,\tg\frac{1}{2}\varepsilon_3,\\ \varepsilon_3=\arccos\frac{B}{A} \end{gathered} \qquad\qquad\qquad\qquad (6) \]

\[ \begin{gathered} A-B=2A\sin^2\varepsilon_4=A\sin2\varepsilon_4\,\tg\varepsilon_4,\\ \varepsilon_4=\frac{1}{2}\arccos\frac{B}{A} \end{gathered} \qquad\qquad\qquad\qquad (7) \]

\[ \begin{gathered} A-B=A\frac{2\tg\varepsilon_5}{\tg2\varepsilon_5} =A\frac{\cos2\varepsilon_5}{\cos^2\varepsilon_5},\\ \varepsilon_5=\arctg\left(\pm\sqrt{\frac{B}{A}}\right) \end{gathered} \qquad\qquad\qquad\qquad (8) \]

1. We shall now transform the difference \(A-B\) in another way.
   We have:

\[ A-B=-A\left(\frac{B}{A}-1\right). \]

Putting:

\[ \frac{B}{2A}=\cos^2\varepsilon_1=\sin^2\varepsilon_2, \]

we obtain—

\[ A-B=-A(2\cos^2\varepsilon_1-1)=-A\cos2\varepsilon_1, \]

\[ A-B=-A(2\sin^2\varepsilon_2-1)=A\cos2\varepsilon_2. \]

As a result we have two transformations:

\[ \begin{gathered} A-B=-A\cos2\varepsilon_1,\\ \varepsilon_1=\arccos\left(\pm\sqrt{\frac{B}{2A}}\right) \end{gathered} \qquad\qquad\qquad\qquad (9) \]

\[ \begin{gathered} A-B=A\cos2\varepsilon_2,\\ \varepsilon_2=\arcsin\left(\pm\sqrt{\frac{B}{2A}}\right) \end{gathered} \qquad\qquad\qquad\qquad (10) \]

1. Finally, transform the difference \(A-B\) as follows:

\[ A-B=A\left(1-\frac{B}{A}\right)=A\left[1+\left(-\frac{B}{A}\right)\right]. \]

Taking the ratio \(\left(-\frac{B}{A}\right)\) to be equal to:

\[ -\frac{B}{A}=\cos \varepsilon_1=\cos 2\varepsilon_2=\operatorname{tg}\varepsilon_3, \]

we obtain—

\[ A-B=A(1+\cos\varepsilon_1)=2A\cos^2\frac{1}{2}\varepsilon_1 = A\,\frac{\sin\varepsilon_1}{\operatorname{tg}\frac{1}{2}\varepsilon_1}, \]

\[ A-B=A(1+\cos 2\varepsilon_2)=2A\cos^2\varepsilon_2 = A\,\frac{\sin 2\varepsilon_2}{\operatorname{tg}\varepsilon_2}, \]

\[ A-B=A(1+\operatorname{tg}\varepsilon_3) = A\,\frac{\sin(45^\circ+\varepsilon_3)}{\cos45^\circ\cos\varepsilon_3}. \]

Thus, the formula \(d=A-B\) is replaced by the system of formulas (11), (12), (13):

\[ \left. \begin{aligned} A-B&=2A\cos^2\frac{1}{2}\varepsilon_1 = A\,\frac{\sin\varepsilon_1}{\operatorname{tg}\frac{1}{2}\varepsilon_1},\\ \varepsilon_1&=\arc\cos\left(-\frac{B}{A}\right) \end{aligned} \right\} \qquad \ldots\ldots\ldots (11) \]

\[ \left. \begin{aligned} A-B&=2A\cos^2\varepsilon_2 = A\,\frac{\sin 2\varepsilon_2}{\operatorname{tg}\varepsilon_2},\\ \varepsilon_2&=\frac{1}{2}\arc\cos\left(-\frac{B}{A}\right) \end{aligned} \right\} \qquad \ldots\ldots\ldots (12) \]

\[ \left. \begin{aligned} A-B&=A\,\frac{\sin(45^\circ+\varepsilon_3)}{\cos45^\circ\cos\varepsilon_3},\\ \varepsilon_3&=\arc\operatorname{tg}\left(-\frac{B}{A}\right) \end{aligned} \right\} \qquad \ldots\ldots\ldots (13) \]

Let us now take the second assumption \(b_1\).

\(b_1)\) We have:

\[ A<0;\quad B<0;\quad \text{moreover}\quad |A|<|B|. \]

1. Transforming the difference \(A-B\), we obtain:

\[ A-B=-B\left(1-\frac{A}{B}\right) =-B\left[1-\frac{|A|}{|B|}\right], \]

and the further reduction of the difference to logarithmic form is analogous to the reduction corresponding to case \(1_{a_1}\); as a result we obtain:

\[ \left. \begin{aligned} A-B&=-B\cos^{2}\varepsilon_{1},\\[4pt] \varepsilon_{1}&=\arcsin\left(\pm\sqrt{\frac{A}{B}}\right) \end{aligned} \right\} \tag{14} \]

\[ \left. \begin{aligned} A-B&=-B\sin^{2}\varepsilon_{2},\\[4pt] \varepsilon_{2}&=\arccos\left(\pm\sqrt{\frac{A}{B}}\right) \end{aligned} \right\} \tag{15} \]

\[ \left. \begin{aligned} A-B&=-2B\sin^{2}\frac{1}{2}\varepsilon_{3} =-B\sin\varepsilon_{3}\operatorname{tg}\frac{1}{2}\varepsilon_{3},\\[4pt] \varepsilon_{3}&=\arccos\frac{A}{B} \end{aligned} \right\} \tag{16} \]

\[ \left. \begin{aligned} A-B&=-2B\sin^{2}\varepsilon_{3} =-B\sin 2\varepsilon_{3}\operatorname{tg}\varepsilon_{3},\\[4pt] \varepsilon_{3}&=\frac{1}{2}\arccos\frac{A}{B} \end{aligned} \right\} \tag{17} \]

\[ \left. \begin{aligned} A-B&=-B\frac{2\operatorname{tg}\varepsilon_{4}}{\operatorname{tg}2\varepsilon_{4}} =-B\frac{\cos 2\varepsilon_{4}}{\cos^{2}\varepsilon_{4}},\\[4pt] \varepsilon_{4}&=\operatorname{arctg}\left(\pm\sqrt{\frac{A}{B}}\right) \end{aligned} \right\} \tag{18} \]

1. Transform the difference \(A-B\) in the following way:

\[ A-B=B\left(\frac{A}{B}-1\right) =B\left[\frac{|A|}{|B|}-1\right]. \]

Analogously to the second transformation of case \(a_{1}\), we obtain:

\[ \left. \begin{aligned} A-B&=B\cos 2\varepsilon_{1},\\[4pt] \varepsilon_{1}&=\arccos\left(\pm\sqrt{\frac{|A|}{2|B|}}\right) \end{aligned} \right\} \tag{19} \]

\[ \left. \begin{aligned} A - B &= -B\cos 2\varepsilon_2,\\ \varepsilon_2 &= \arcsin\left(\pm \sqrt{\frac{|A|}{2|B|}}\right) \end{aligned} \right\} \qquad \cdots\cdots\cdots (20) \]

1. Finally, representing the difference \(A - B\) in the form:

\[ A - B = -B\left(1 - \frac{A}{B}\right) = -B\left[1 + \left(-\frac{|A|}{|B|}\right)\right] \]

we see that the result of the transformation is analogous to the third transformation of case \(I_{3a_1}\).

We obtain:

\[ \left. \begin{aligned} A - B &= -2B\cos^2 \frac{1}{2}\varepsilon_1 = -B\,\frac{\sin \varepsilon_1}{\operatorname{tg}\frac{1}{2}\varepsilon_1},\\ \varepsilon_1 &= \arccos\left(-\frac{|A|}{|B|}\right) \end{aligned} \right\} \qquad \cdots\cdots\cdots (21) \]

\[ \left. \begin{aligned} A - B &= -2B\cos^2 \varepsilon_2 = -B\,\frac{\sin 2\varepsilon_2}{\operatorname{tg}\varepsilon_2},\\ \varepsilon_2 &= \frac{1}{2}\arccos\left(-\frac{|A|}{|B|}\right) \end{aligned} \right\} \qquad \cdots\cdots\cdots (22) \]

\[ \left. \begin{aligned} A - B &= -B\,\frac{\sin(45^\circ + \varepsilon_3)}{\cos 45^\circ \cos \varepsilon_3},\\ \varepsilon_3 &= \operatorname{arctg}\left(-\frac{|A|}{|B|}\right) \end{aligned} \right\} \qquad \cdots\cdots\cdots (23) \]

Thus the difference \(d = A - B\) is replaced by the system of formulas (21), (22), and (23).

1. Let us give still further transformations, respectively for cases a) and b):

\[ \begin{aligned} \text{a)}\quad A - B &= A\left(1 - \frac{B}{A}\right) = A\left(1 - \operatorname{tg}^2\varepsilon\right) = \\ &= A\,\frac{\cos 2\varepsilon}{\cos^2\varepsilon} = A\,\frac{2\operatorname{tg}\varepsilon}{\operatorname{tg}2\varepsilon}. \end{aligned} \]

Thus we have:

\[ \left. \begin{aligned} A-B &= A\,\frac{\cos 2\varepsilon}{\cos^2\varepsilon} = A\,\frac{2\operatorname{tg}\varepsilon}{\operatorname{tg}2\varepsilon},\\[6pt] \varepsilon &= \operatorname{arctg}\left(\pm\sqrt{\frac{B}{A}}\right) \end{aligned} \right\} \qquad \ldots\ldots\ (24),\ (25) \]

b)
\[ A-B=-B\left(1-\frac{A}{B}\right)= \]

\[ =-B\left(1-\left|\frac{A}{B}\right|\right) =-B(1-\operatorname{tg}^2\varepsilon)= \]

\[ =-B\,\frac{\cos 2\varepsilon}{\cos^2\varepsilon} =-B\,\frac{2\operatorname{tg}\varepsilon}{\operatorname{tg}2\varepsilon}. \]

Thus:

\[ \left. \begin{aligned} A-B &=-B\,\frac{\cos 2\varepsilon}{\cos^2\varepsilon} =-B\,\frac{2\operatorname{tg}\varepsilon}{\operatorname{tg}2\varepsilon},\\[6pt] \varepsilon &= \operatorname{arctg}\left(\pm\sqrt{\left|\frac{A}{B}\right|}\right) \end{aligned} \right\} \qquad \ldots\ldots\ (26),\ (27) \]

II. \(A\) and \(B\) are of different signs.

a) We have:

\[ A>0;\quad B<0,\quad \text{with } A>|B|. \]

1. Transforming, we obtain:

\[ A-B=A\left(1-\frac{B}{A}\right). \]

Taking the ratio \(\dfrac{B}{A}\) equal to:

\[ \frac{B}{A}=\cos\varepsilon_1=\cos 2\varepsilon_2, \]

where \(\varepsilon_{1,2}\) are auxiliary angles, we obtain respectively—

\[ A-B=A(1-\cos\varepsilon_1) =2A\sin^2\frac{1}{2}\varepsilon_1 =A\sin\varepsilon_1\operatorname{tg}\frac{1}{2}\varepsilon_1 \]

\[ A-B=A(1-\cos 2\varepsilon_2) =2A\sin^2\varepsilon_2 =A\sin 2\varepsilon_2\operatorname{tg}\varepsilon_2. \]

Thus the difference \(d=A-B\) is replaced by the system of formulas (28), (29)

\[ \left. \begin{aligned} A-B &= 2A\sin^2\frac{1}{2}\varepsilon_1 = A\sin\varepsilon_1\operatorname{tg}\frac{1}{2}\varepsilon_1,\\[6pt] \varepsilon_1 &= \arccos\frac{B}{A} \end{aligned} \right\} \qquad \ldots\ldots\ (28) \]

\[ \begin{gathered} A-B=2A\sin^2\varepsilon_2=A\sin 2\varepsilon_2\,\tg \varepsilon_2,\\ \varepsilon_2=\frac{1}{2}\arccos\frac{B}{A} \end{gathered} \tag{29} \]

1. Representing the difference \(A-B\) otherwise, we obtain:

\[ A-B=A\left(1-\frac{B}{A}\right)=A\left(1+\frac{|B|}{A}\right). \]

Taking the ratio \(\dfrac{B}{A}\) to be equal to:

\[ \frac{|B|}{A}=\cos\varepsilon_1=\cos 2\varepsilon_2=\tg^2\varepsilon_3=\tg\varepsilon_4, \]

we obtain, respectively,—

\[ A-B=A(1+\cos\varepsilon_1)=2A\cos^2\frac{1}{2}\varepsilon_1 =A\frac{\sin\varepsilon_1}{\tg\frac{1}{2}\varepsilon_1} \]

\[ A-B=A(1+\cos 2\varepsilon_2)=2A\cos^2\varepsilon_2 =A\frac{\sin 2\varepsilon_2}{\tg\varepsilon_2}. \]

\[ A-B=A(1+\tg^2\varepsilon_3)=\frac{A}{\cos^2\varepsilon_3} =A\frac{2\tg\varepsilon_3}{\sin 2\varepsilon_3}. \]

\[ A-B=A(1+\tg\varepsilon_4) =A\frac{\sin(45^\circ+\varepsilon_4)}{\cos45^\circ\cos\varepsilon_4}. \]

Thus the difference \(d=A-B\) is replaced by the system of formulas (30), (31), (32), (33).

\[ \left. \begin{gathered} A-B=2A\cos^2\frac{1}{2}\varepsilon_1 =A\frac{\sin\varepsilon_1}{\tg\frac{1}{2}\varepsilon_1}\\[6pt] \varepsilon_1=\arccos\frac{|B|}{A} \end{gathered} \right\} \tag{30} \]

\[ \left. \begin{gathered} A-B=2A\cos^2\varepsilon_2 =A\frac{\sin 2\varepsilon_2}{\tg\varepsilon_2}\\[6pt] \varepsilon_2=\frac{1}{2}\arccos\frac{|B|}{A} \end{gathered} \right\} \tag{31} \]

\[ \left. \begin{gathered} A-B=\frac{A}{\cos^2\varepsilon_3} =A\frac{2\tg\varepsilon_3}{\sin 2\varepsilon_3}\\[6pt] \varepsilon_3=\arctg\left(\pm\sqrt{\frac{|B|}{A}}\right) \end{gathered} \right\} \tag{32} \]

\[ \left. \begin{aligned} A-B &= A\,\frac{\sin(45^\circ+\varepsilon_4)}{\cos45^\circ\cos\varepsilon_4},\\ \varepsilon_4 &= \operatorname{arctg}\frac{|B|}{A} \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots (33) \]

b) We have:

\[ A>0;\quad B<0,\quad \text{with } A<|B|. \]

1. Let us represent the difference \(A-B\) in the form:

\[ A-B=-B\left(1-\frac{A}{B}\right) \]

Taking the ratio \(\dfrac{A}{B}\) equal to:

\[ \frac{A}{B}=\cos\varepsilon_1=\cos2\varepsilon_2, \]

we obtain, respectively,—

\[ A-B=-B(1-\cos\varepsilon_1)=-2B\sin^2\frac{1}{2}\varepsilon_1= \]

\[ =-B\sin\varepsilon_1\,\operatorname{tg}\frac{1}{2}\varepsilon_1. \]

\[ A-B=-B(1-\cos2\varepsilon_2)=-2B\sin^2\varepsilon_2= \]

\[ =-B\sin2\varepsilon_2\,\operatorname{tg}\varepsilon_2. \]

Thus the difference \(d=A-B\) is replaced by the systems of formulas (34), (35):

\[ \left. \begin{aligned} A-B&=-2B\sin^2\frac{1}{2}\varepsilon_1\\ &=-B\sin\varepsilon_1\,\operatorname{tg}\frac{1}{2}\varepsilon_1\\ \varepsilon_1&=\arccos\frac{A}{B} \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots (34) \]

\[ \left. \begin{aligned} A-B&=-2B\sin^2\varepsilon_2\\ &=-B\sin2\varepsilon_2\,\operatorname{tg}\varepsilon_2\\ \varepsilon_2&=\frac{1}{2}\arccos\frac{A}{B}. \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots (35) \]

1. Let us now represent the difference \(A-B\) in another form:

\[ A-B=-B\left(1-\frac{A}{B}\right)=-B\left(1+\frac{A}{|B|}\right). \]

Taking the ratio \(\dfrac{A}{|B|}\) equal to:

\[ \frac{A}{|B|}=\cos\varepsilon_1=\cos2\varepsilon_2=\operatorname{tg}^2\varepsilon_3=\operatorname{tg}\varepsilon_4, \]

we obtain, respectively—

\[ A-B=-B(1+\cos\varepsilon_1)=-2B\cos^2\frac{1}{2}\varepsilon_1 =-B\,\frac{\sin 2\varepsilon_1}{\tg \frac{1}{2}\varepsilon_1} \]

\[ A-B=-B(1+\cos 2\varepsilon_2)=-2B\cos^2\varepsilon_2 =-B\,\frac{\sin 2\varepsilon_2}{\tg \varepsilon_2} \]

\[ A-B=-B(1+\tg^2\varepsilon_3) =-B\,\frac{2\tg \varepsilon_3}{\sin 2\varepsilon_3} =-\frac{B}{\cos^2\varepsilon_3} \]

\[ A-B=-B(1+\tg\varepsilon_4) =-B\,\frac{\sin(45^\circ+\varepsilon_4)}{\cos45^\circ\cos\varepsilon_4}. \]

Thus the difference \(d=A-B\) is replaced by the systems of formulas (36), (37), (38), (39):

\[ \left. \begin{aligned} A-B&=-2B\cos^2\frac{1}{2}\varepsilon_1 =-B\,\frac{\sin\varepsilon_1}{\tg\frac{1}{2}\varepsilon_1},\\[4pt] \varepsilon_1&=\arccos\frac{A}{|B|} \end{aligned} \right\} \ldots\ldots\ldots (36) \]

\[ \left. \begin{aligned} A-B&=-2B\cos^2\varepsilon_2 =-B\,\frac{\sin 2\varepsilon_2}{\tg\varepsilon_2},\\[4pt] \varepsilon_2&=\frac{1}{2}\arccos\frac{A}{|B|} \end{aligned} \right\} \ldots\ldots\ldots (37) \]

\[ \left. \begin{aligned} A-B&=-B\,\frac{1}{\cos^2\varepsilon_3} =-B\,\frac{2\tg\varepsilon_3}{\sin 2\varepsilon_3},\\[4pt] \varepsilon_3&=\arctg\left(\pm\sqrt{-\frac{A}{|B|}}\right) \end{aligned} \right\} \ldots\ldots\ldots (38) \]

\[ \left. \begin{aligned} A-B&=-B\,\frac{\sin(45^\circ+\varepsilon_4)}{\cos45^\circ\cos\varepsilon_4},\\[4pt] \varepsilon_4&=\arctg\frac{A}{|B|} \end{aligned} \right\} \ldots\ldots\ldots (39) \]

1. We shall now give transformations that are equally applicable to the inequalities:

\[ A>|B|\quad \text{and}\quad A<|B|. \]

1) We have:

\[ A-B=A\left(1-\frac{B}{A}\right) =A\left(1+\frac{|B|}{A}\right) =A(1+\tg^2\varepsilon_1) =\frac{A}{\cos^2\varepsilon_1} =A\,\frac{2\tg\varepsilon_1}{\sin 2\varepsilon_1}. \]

\[ A-B=A(1+\tg\varepsilon_2) =A\,\frac{\sin(45^\circ+\varepsilon_2)}{\cos45^\circ\cos\varepsilon_2}. \]

Thus the difference \(d=A-B\) is replaced by the systems of formulas (40), (41):

\[ \left. \begin{aligned} A-B&=-\frac{A}{\cos^2 \varepsilon_1} =A\,\frac{2\,\operatorname{tg}\varepsilon_1}{\sin 2\varepsilon_1},\\[4pt] \varepsilon_1&=\operatorname{arc\,tg} \left(\pm\sqrt{-\frac{|B|}{A}}\right) \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (40) \]

\[ \left. \begin{aligned} A-B&=A\,\frac{\sin(45^\circ+\varepsilon_2)} {\cos45^\circ\cos\varepsilon_2},\\[4pt] \varepsilon_2&=\operatorname{arc\,tg}\frac{|B|}{A} \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (41) \]

2) We have:

\[ A-B=-B\left(1-\frac{A}{B}\right) =-B\left(1+\frac{A}{|B|}\right) =-B(1+\operatorname{tg}^2\varepsilon_1)= \]

\[ =-\frac{B}{\cos^2\varepsilon_1} =-B\,\frac{2\,\operatorname{tg}\varepsilon_1}{\sin2\varepsilon_1} \]

\[ A-B=-B(1+\operatorname{tg}\varepsilon_2) =-B\,\frac{\sin(45^\circ+\varepsilon_2)} {\cos45^\circ\cos\varepsilon_2}. \]

Thus the difference \(d=A-B\) is replaced by the systems of formulas (42), (43):

\[ \left. \begin{aligned} A-B&=-B\,\frac{\sin(45^\circ+\varepsilon_2)} {\cos45^\circ\cos\varepsilon_2},\\[4pt] \varepsilon_2&=\operatorname{arc\,tg}\frac{A}{|B|} \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (42) \]

\[ \left. \begin{aligned} A-B&=-\frac{B}{\cos^2\varepsilon_1} =-B\,\frac{2\,\operatorname{tg}\varepsilon_1}{\sin2\varepsilon_1},\\[4pt] \varepsilon_1&=\operatorname{arctg} \left(\pm\sqrt{\frac{A}{|B|}}\right) \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (43) \]

III. The reduction to logarithmic form of the difference \(A-B\) for arbitrary \(A\) and \(B\) is known, and comes down to replacing the difference by the system of formulas (44):

\[ \left. \begin{aligned} A-B&=A\,\frac{\sin(45^\circ-\varepsilon)} {\cos45^\circ\cos\varepsilon},\\[4pt] \varepsilon&=\operatorname{arctg}\frac{B}{A} \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (44) \]

We now pass to other methods for determining the direction of the line of true dip—the azimuth \(\varphi\).

§ 10.

Second method for determining the direction of the line of dip (azimuth φ).

The second method is based on determining the coordinates of an arbitrary point of the line of dip, starting from the equations of the plane of the bed and the vertical plane of the line of dip.

The equation of the plane of the bed:

\[ nx - my + kz = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (1) \]

Putting \(z = 0\), we find the equation of the strike line:

\[ nx - my = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (2) \]

The equation of the horizontal projection of the line of dip, perpendicular to (2), will be:

\[ mx + ny = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (3) \]

The equations:

\[ \left. \begin{aligned} mx + ny &= 0 \\ z &= z \end{aligned} \right\} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (4) \]

determine the vertical plane of the line of dip.

The system of equations:

\[ nx - my + kz = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (5) \]

\[ \left. \begin{aligned} mx + ny &= 0 \\ z &= z \end{aligned} \right\} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (6) \]

determines the line of dip.

We shall find the coordinates of an arbitrary point of the line of dip by solving the system of two homogeneous equations with three variables, namely:

\[ \left. \begin{aligned} nx - my + kz &= 0 \\ mx + ny + 0 \cdot z &= 0 \end{aligned} \right\} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (7) \]

The solution of system (7) will be:

\[ x = \begin{vmatrix} -m & k \\ n & 0 \end{vmatrix} t;\quad y = \begin{vmatrix} k & n \\ 0 & m \end{vmatrix} t;\quad z = \begin{vmatrix} n & -m \\ m & n \end{vmatrix} t, \]

or—

\[ x = -nkt;\quad y = mkt;\quad z = (m^2 + n^2)t, \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (8) \]

where \(t\) is an arbitrary parameter varying from \(-\infty\) to \(+\infty\).

We now proceed to determine the azimuth of the line of dip.

1. Take the points of the line of dip for which \(z_n < 0\); then, by virtue of (8), \(t < 0\).

The azimuth of the line of dip, determined by the points \((0, 0, 0)\) and \((x_{z_n}, y_{z_n}, z_n)\), is found vectorially by the formula:

\[ \tg \varphi = \frac{y}{x} = \frac{mkt}{-nkt} = \frac{mk(-t)}{nk(-t)} = -\frac{mk}{nk}, \]

for \((-t) > 0\).

For the quantity “\(k\)” two assumptions are equally possible: \(k > 0\) and \(k < 0\) (for the proof see 2, § 5).

I. Let \(k > 0\).

Then:

\[ \tg \varphi = -\frac{m}{n}. \]

2. Let \(\kappa < 0\).

Then:

\[ \tg \varphi = \frac{-mk}{nk} = \frac{m(-k)}{-n(-k)} = \frac{m}{-n}, \]

or \((-k)>0\).

II. Let us take points of the line of dip for which \(z_p>0\); then, by virtue of (8), \(t>0\).

The azimuth of the line of dip determined by the points \((0,0,0)\) and \((x_{z_p},y_{z_p},z_p>0)\) is found analogously from the formula:

\[ \tg \varphi=\frac{-m}{n}, \quad \text{under the condition } \kappa>0. \]

\[ \tg \varphi=\frac{m}{-n}, \quad \text{under the condition } \kappa<0. \]

III. Let us take two points of the line of dip for which, respectively, \(z_p>0\) and \(z_n<0\). We find that the azimuth of the line of dip is determined from the formula:

\[ \tg \varphi = \frac{y_{z_n}-y_{z_p}}{x_{z_n}-x_{z_p}} = \frac{mkt_n-mkt_p}{-nkt_n+nkt_p} = \frac{-mk(t_p-t_n)}{nk(t_p-t_n)} = \frac{-mk}{nk}, \]

or \((t_p-t_n)>0\), where the indices \(p\) and \(n\) correspond to positive and negative values of \(t\), respectively.

Further, the solution is analogous to assumptions I and II.

In all cases, respectively, one obtains the same form of the formula determining \(\varphi\).

For an analysis of the formulae determining \(\varphi\), see the rotation method.

Remark. Equations (7) are easily brought to the form:

\[ \frac{x}{-nk} = \frac{y}{mk} = \frac{z}{m^2+n^2}, \tag{9} \]

which represents the equation of the line of dip.

From the latter, the coordinates of an arbitrary point of the line of dip are easily determined.

Let us use the last equation to derive the equation of the plane perpendicular to the bed plane and passing through its line of strike, corresponding to the plane \(z=0\).

The latter (which we shall call the normal plane) is the plane passing through the point \((0,0,0)\) and perpendicular to the line (9).

We have:

\[ -nkx+mky+(m^2+n^2)z=0, \]

or:

\[ nx-my-\frac{1}{k}(m^2+n^2)z=0 \]

—the equation of the normal plane.

Third method.

We found that the equation of the plane normal to the given one,

\[ nx-my+kz=0, \]

and corresponding to the line of strike of the latter in the plane \(z=0\), is:

\[ nx-my-\frac{1}{k}(m^2+n^2)z=0. \tag{1} \]

Describe, with the origin of coordinates as center, a sphere of arbitrary radius \(R\).

Its equation is:

\[ x^{2}+y^{2}+z^{2}-R^{2}=0. \]

The equation of the plane tangent to the sphere at an arbitrary point \((x_{0}, y_{0}, z_{0})\) of it will be:

\[ x_{0}x+y_{0}y+z_{0}z-R^{2}=0 . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (2) \]

Let us impose the condition that plane (2) be parallel to plane (1). From this condition, as is easy to see, we determine the coordinates \((x_{0}, y_{0}, z_{0})\) of the required point of the line of dip.

The condition for parallelism of planes (1) and (2) is:

\[ \frac{n}{x_{0}}=\frac{-m}{y_{0}}= \frac{-\frac{1}{k}(m^{2}+n^{2})}{z_{0}} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (3) \]

On the other hand [since the point \((x_{0}, y_{0}, z_{0})\) lies on the sphere] we have:

\[ x_{0}^{2}+y_{0}^{2}+z_{0}^{2}-R^{2}=0 . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (4) \]

Solving the system of equations (3) and (4) gives the required point. From (3) we have:

\[ y_{0}=-\frac{m}{n}x_{0}, \]

\[ z_{0}=-\frac{m^{2}+n^{2}}{kn}x_{0}. \]

Substituting the obtained values \(y_{0}\) and \(z_{0}\) into equation (4), we obtain:

\[ x_{0}^{2}+\frac{m^{2}}{n^{2}}x_{0}^{2}+ \frac{(m^{2}+n^{2})^{2}}{k^{2}n^{2}}x_{0}^{2} -R^{2}=0, \]

or:

\[ (m^{2}+n^{2})(m^{2}+n^{2}+k^{2})x_{0}^{2}-R^{2}k^{2}n^{2}=0, \]

whence:

\[ \left. \begin{aligned} x_{0}&=\pm \frac{Rkn}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+k^{2})}},\\[6pt] y_{0}&=\mp \frac{Rkm}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+k^{2})}},\\[6pt] z_{0}&=\mp R\sqrt{\frac{m^{2}+n^{2}}{m^{2}+n^{2}+k^{2}}} \end{aligned} \right\} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (5) \]

As indicated above, for the quantity “\(k\)” two assumptions are equally possible: \(k \gtrless 0\).

Putting \(k>0\) and denoting the common factor of the expressions \((5_{1,2})\) by \(c\), we obtain:

\[ c=\frac{Rk}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+k^{2})}}>0 \]

\[ x_{0}=\pm cn. \]

\[ y_{0}=\mp cm. \]

Remark. The coordinates \(y_0\) and \(z_0\) are readily determined as functions of \(x_0\), and as the point of intersection of the line of dip (9) [see III, § 10] with the sphere.

We proceed to the determination of the azimuth of the line of dip.

1. Let \(z_0 < 0\); accordingly it is necessary to take \((x_0)_p\) and \((y_0)_n\), according to \((5_3)\) [where \(p\) and \(n\) indicate, respectively, the signs plus \((+)\) and minus \((-)\)].

The azimuth of the line of dip \(\varphi\), determined by the points \((0,0,0)\) and

\[ (x_0, y_0, z_0 < 0), \]

we find vectorially by the formula:

\[ \tg \varphi = \frac{(y_0)_n}{(x_0)_p} = \frac{-cm}{cn} = -\frac{m}{n}, \]

for \(c > 0\).

1. Let \(z_0 > 0\); then we take \((x_0)_n\) and \((y_0)_p\), according to \((5_3)\). For the points of the line of dip corresponding to \((0,0,0)\) and \((x_0,y_0,z_0>0)\), we obtain:

\[ \tg \varphi = \frac{-(y_0)_p}{-(x_0)_n} = \frac{-cm}{cn} = -\frac{m}{n}, \]

for \(c > 0\).

Fourth method.

Let us take a conical surface whose center is the origin of coordinates; let the directrix be the circle defined by the equations:

\[ \left. \begin{aligned} x^2 + y^2 - R^2 &= 0,\\ z &= c. \end{aligned} \right\} \tag{1} \]

Then the equation of the conical surface will be:

\[ \frac{x^2+y^2}{R^2} - \frac{z^2}{c^2} = 0. \ldots \ldots \ldots \ldots \ldots \tag{2} \]

The equation of the plane tangent to the conical surface at an arbitrary point of it \((x_0,y_0,z_0)\) is:

\[ \frac{x_0x+y_0y}{R^2} - \frac{z_0z}{c^2} = 0. \ldots \ldots \ldots \ldots \tag{3} \]

Let the given plane of the bed,

\[ nx - my + \kappa z = 0, \ldots \ldots \ldots \ldots \ldots \tag{4} \]

be tangent to the conical surface (2); it is easy to see that the generator of tangency is the line of dip of the given plane (4).

Further, let the generator of tangency of plane (3) coincide with the generator of tangency of plane (4).

Then plane (3) merges with plane (4), and therefore:

\[ \frac{x_0}{R^2}x + \frac{y_0}{R^2}y - \frac{z_0}{c^2}z = \lambda (nx - my + \kappa z). \ldots \ldots \ldots \tag{5} \]

From the last identity (5) we shall determine the coordinates of an arbitrary point of the line of dip.

We have:

\[ \left. \begin{aligned} x_0&=\lambda R^2 n,\\ y_0&=-\lambda R^2 m,\\ z_0&=-\lambda c^2 k. \end{aligned} \right\} \tag{6} \]

Let us note, in connection with what follows, that for the quantity “\(k\)” two assumptions, \(k \gtrless 0\), are equally possible (see 2, § 5).

We proceed to determine the azimuth of the line of dip, putting \(k>0\).

1. Take points of the line of dip for which \(z_{0_n}<0\).

Then, by virtue of \((6_3)\), \(\lambda>0\).

The azimuth of the line of dip, determined by the points \((0,0,0)\) and \((x_0,y_0,z_0<0)\), is found vectorially from the formula:

\[ \tg\varphi=\frac{y_0}{x_0} =\frac{-\lambda R^2 m}{\lambda R^2 n} =\frac{-m}{n}, \]

for \(\lambda>0\).

1. Take points of the line of dip for which \(z_{0_p}>0\).

Then, by virtue of \((6_3)\), \(\lambda<0\).

The azimuth of the line of dip, determined by the points \((0,0,0)\) and \(x_0,y_0,z_0>0\), is found vectorially from the formula:

\[ \tg\varphi=\frac{-y_0}{-x_0} =\frac{\lambda R^2 m}{-\lambda R^2 n} =\frac{-(-\lambda)R^2m}{(-\lambda)R^2 n} =\frac{-m}{n}, \]

for \((-\lambda)>0\).

1. Take points of the line of dip for which, respectively,

\[ z_{0_p}>0 \quad \text{and} \quad z_{0_n}<0 \]

Then:

\[ \tg\varphi= \frac{(y_0)_{z_n}-(y_0)_{z_p}} {(x_0)_{z_n}-(x_0)_{z_p}} = \frac{-\lambda_p R^2 m+\lambda_n R^2 m} {\lambda_p R^2 n-\lambda_n R^2 n} = \]

\[ =\frac{-m(\lambda_p-\lambda_n)R^2} {n(\lambda_p-\lambda_n)R^2} = \frac{-m}{n}, \]

for \((\lambda_p-\lambda_n)>0\).

In all cases one and the same form of the formula is obtained, vectorially determining \(\varphi\).

Fifth method.

On the \(Z\)-axis take an arbitrary point \(P(0,0,z_c)\). From point \(P\) drop a perpendicular to the given plane:

\[ nx-my+kz=0 \ldots \ldots \ldots \ldots \ldots \tag{1} \]

The point of intersection of the latter with (1) will obviously determine the required point of the line of dip.

The equation of the perpendicular passing through the point \(P\) will be:

\[ \frac{x}{n}=-\frac{y}{m}=\frac{z-z_c}{\kappa}\, . \tag{2} \]

We determine the point of intersection of (2) with (1).
From (2) we have:

\[ y=-\frac{m}{n}x \, . \tag{3} \]

\[ z=\frac{\kappa}{n}x+z_c \, . \tag{4} \]

and, substituting the latter into (1), we obtain:

\[ x=-\frac{n\kappa z_c}{m^2+n^2+\kappa^2}\, . \tag{5} \]

Substituting (5) into (3) and (4), we obtain:

\[ y=\frac{m\kappa z_c}{m^2+n^2+\kappa^2}\, . \]

\[ z=\frac{m^2+n^2}{m^2+n^2+\kappa^2}\,z_c . \]

Putting \(\kappa>0\), for \(z_c\ge 0\) the required azimuth of the line of dip is determined vectorially by the formula:

\[ \tan \varphi=\frac{-m}{n}. \]

Putting instead \(\kappa<0\), for \(z_c\ge 0\) we find:

\[ \tan \varphi=\frac{m}{-n}. \]

The further analysis is identical with the analysis in the method of rotation set forth above.

Sixth method.

Determination of an arbitrary point of the line of dip as the pole \(P\) of a plane parallel to the normal plane (corresponding to the line of strike of the given plane and lying in the plane \(z=0\)) with respect to a sphere whose center is at the origin.

Let the point \(P(x_0,y_0,z_0)\) be an arbitrary point of the line of dip. We shall take it as the pole of a certain plane (the polar plane) with respect to the sphere. Then it is easy to see that the polar plane will be parallel to the normal plane.

The equation of the sphere:

\[ x^2+y^2+z^2-R^2=0 \, . \tag{1} \]

The equation of the polar plane of the point \(P(x_0,y_0,z_0)\):

\[ x_0x+y_0y+z_0z-R^2=0 . \tag{2} \]

The equation of the normal plane:

\[ nx-my-\frac{1}{\kappa}(m^2+n^2)z=0 . \tag{3} \]

The equation of the plane parallel to the normal (3):

\[ nx - my - \frac{1}{\kappa}(m^2+n^2)z + c = 0 . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (4) \]

Since (4) is the plane polar to the point \(P\) with respect to the sphere, we have

\[ x_0x + y_0y + z_0z - R^2 \equiv \lambda \left| nx - my \right. \]

\[ \left. -\frac{1}{\kappa}(m^2+n^2)z + c \right|; \]

whence:

\[ x_0 = n\lambda . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (5) \]

\[ y_0 = -m\lambda . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (6) \]

\[ z_0 = -\frac{1}{\kappa}(m^2+n^2)\lambda . \ . \ . \ . \ . \ . \ . \ . (7) \]

\[ R^2 = -c\lambda . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (8) \]

Let us consider the quantities \(\kappa\) and \(c\) entering into expressions (7) and (8). For the quantity “\(\kappa\)” two assumptions are equally possible: \(\kappa \gtrless 0\) (see the analysis of “\(\kappa\)”).

The quantity \(c\) is arbitrary.

In what follows we shall take \(\kappa > 0\).

I. Let \(c > 0\).

Then from (8) it follows that \(\lambda < 0\); (7) gives \(z_0 > 0\).

The sought azimuth of the line of dip, determined by the points \((0,0,0)\) and

\[ (x_0,\ y_0,\ z_0 > 0), \]

is found vectorially from the formula:

\[ \tan\varphi = \frac{-y_0}{-x_0} = \frac{m\lambda}{-n\lambda} = \frac{-m(-\lambda)}{n(-\lambda)} = \frac{-m}{n}, \]

for \((-\lambda)>0\).

II. Let \(c < 0\).

We find that \(\lambda > 0\) and, correspondingly, \(z_0 < 0\).

The required azimuth is determined by the formula:

\[ \tan\varphi = \frac{-m}{n}. \]

Subsequently, a number of other assumptions and the analysis will lead to the results indicated for the method of rotation.

Seventh method.

The equation of the bed plane:

\[ nx - my + \kappa z = 0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (1) \]

The equation of the vertical plane of dip:

\[ \begin{aligned} mx + ny &= 0 . \\ z &= z . \end{aligned} \Bigg\} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (2) \]

Let us express the coordinates \((x,y)\) of the points of intersection of (1) and (2) as functions of the coordinate \(z\) of the same points.

Determining \(y\) from (2) and substituting its value into (1), we find:

\[ (m^2+n^2)x=-nkz, \]

whence:

\[ x=\frac{-nkz}{m^2+n^2}, \]

and, accordingly, from (2),

\[ y=\frac{mkz}{m^2+n^2}. \]

Assuming \(k>0\), let \(z<0\). Then:

\[ \tg \varepsilon=\frac{y}{x} = \frac{\dfrac{mkz}{m^2+n^2}}{\dfrac{-nkz}{m^2+n^2}} = \frac{-m(-z)}{n(-z)} = \frac{-m}{n}, \]

for \((-z)>0\).

A number of other possible assumptions will lead to the results obtained by us above.

Eighth method.

The equation of the plane of the bed is:

\[ nx-my+kz=0 \ldots\ldots\ldots\ldots\ldots (1) \]

Let us reduce equation (1) to the normal Hesse form:

\[ \frac{n}{\pm\sqrt{m^2+n^2+k^2}}\,x + \frac{-m}{\pm\sqrt{m^2+n^2+k^2}}\,y + \frac{k}{\pm\sqrt{m^2+n^2+k^2}}\,z = 0 \ldots\ldots (2) \]

Proceeding further, let us note that in equation (2), as is well known, the coefficients of the variables \(x\), \(y\), and \(z\) express respectively the cosines of the angles formed by the normal \(AP\) to plane (1) with the coordinate axes. Therefore:

\[ \left. \begin{aligned} \cos(AP,AX)&=\frac{n}{\pm\sqrt{m^2+n^2+k^2}},\\[4pt] \cos(AP,AY)&=\frac{-m}{\pm\sqrt{m^2+n^2+k^2}},\\[4pt] \cos(AP,AZ)&=\frac{k}{\pm\sqrt{m^2+n^2+k^2}}. \end{aligned} \right\} \ldots\ldots (3) \]

The projections onto the plane \(z=0\) of the line of dip and of the normal \(AP\), issuing from the origin \(A\), coincide; consequently, the azimuth

of the line of dip, as we know, is determined by also finding the coordinates \((x,y)\) of some point of the normal.

Of the two directions of the normal, let us first take that for the points of which \(z>0\) (the normal above the plane \(z=0\)). In this case \(\cos(AP, AZ)>0\) and, consequently, in equation (2) the coefficient of \(z\) must be positive.

Under this condition it is necessary to put, in the latter: either 1) the sign before the radical plus \((+)\) and, correspondingly, \(k>0\), or 2) the sign before the radical minus \((-)\) and, correspondingly, \(k<0\).

Having made the first assumption, let us take an arbitrary point \(P\) on the chosen direction of the normal and find its coordinates \((x_p, y_p)\)—the end of the normal-vector.

We obtain:

\[ x_p=\overline{AP}\cos(AP, AX). \]

\[ y_p=\overline{AP}\cos(AP, AY). \]

Whence, taking the points \(A\) and \(P\), we find:

\[ \operatorname{tg}\zeta=\frac{y_p}{x_p} = \frac{\cos(AP, AY)}{\cos(AP, AX)}. \ldots\ldots\ldots\ldots \tag{4} \]

(where \(\zeta\) is the azimuth of the line of dip).

Taking into consideration that, by the condition, the sign before the radical is plus \((+)\) and, correspondingly, \(k>0\), we substitute in expression (4) the corresponding values (3).

We obtain:

\[ \operatorname{tg}\zeta = \frac{\dfrac{-m}{+\sqrt{m^{2}+n^{2}+k^{2}}}} {\dfrac{n}{+\sqrt{m^{2}+n^{2}+k^{2}}}} = \frac{-m}{n}. \]

Similarly, taking the second assumption, we find:

\[ \operatorname{tg}\zeta=\frac{m}{-n}. \]

Now let us take the direction of the normal corresponding to

\[ \cos(AP', AZ)<0. \]

Then, as is easy to see, either 1) the sign before the radical is minus \((-)\) and, correspondingly, \(k>0\), or 2) the sign before the radical is plus \((+)\) and, correspondingly, \(k<0\).

Carrying out the corresponding calculations, we shall arrive at the conclusions indicated by us when considering the method of rotation.

Ninth method.

Since the projections onto the plane \(z=0\) of the normal to the given plane and of the line of dip of the latter coincide, it is easy to see that the azimuth of the line of dip coincides with the azimuth of the normal for the points for which \(z>0\), and is opposite to the azimuth of the normal for the points for which \(z<0\).

Let us first find the expression for the azimuth of an arbitrary vector, knowing the angles which it makes with the coordinate axes (Fig. 13).

Let the length of the vector be \(l\), and let the angles which it makes with the coordinate axes \(OX\), \(OY\), and \(OZ\) be, respectively, \(\alpha, \beta, \gamma\); denote the azimuth of the vector with respect to the \(Z\) axis by \(a_z\).

Find the projections of the vector \(l\) on the coordinate axes. Let their magnitudes be, respectively, \(l_x, l_y\), and \(l_z\).

We have:

\[ l_x = l\sin\gamma \cos a_z = l\cos\alpha. \]

\[ l_y = l\sin\gamma \sin a_z = l\cos\beta. \]

\[ l_z = l\cos\gamma. \]

Whence:

\[ \begin{aligned} \cos\alpha &= \sin\gamma \cos a_z,\\ \cos\beta &= \sin\gamma \sin a_z, \end{aligned} \qquad . . . . . . . . . . . . (1) \]

and, consequently, vectorially, for any position of the vector:

\[ \operatorname{tg} a_z = \frac{\sin a_z}{\cos a_z} = \frac{\dfrac{\cos\beta}{\sin\gamma}}{\dfrac{\cos\alpha}{\sin\gamma}} \qquad . . . . . . . . . . . . (2) \]

The angle \(\gamma\) lies within the limits:

\[ \pi > \gamma > 0, \]

and therefore

\[ \sin\gamma > 0 \qquad . . . . . . . . . . . . . . . . (3) \]

In view of (3), expression (2), which determines the azimuth \(a_z\) of the vector \(l\), takes the form:

\[ \operatorname{tg} a_z = \frac{\cos\beta}{\cos\alpha} \qquad . . . . . . . . . . . . (4) \]

We found earlier that the angles made by the normal to the given plane of the bed with the coordinate axes \(AX\), \(AY\), and \(AZ\) are determined respectively by the formulas:

\[ \left. \begin{aligned} \cos\alpha_{1,2} &= \frac{n}{\pm\sqrt{m^2+n^2+\kappa^2}}, \\[6pt] \cos\beta_{1,2} &= \frac{-m}{\pm\sqrt{m^2+n^2+\kappa^2}}, \\[6pt] \cos\gamma_{1,2} &= \frac{\kappa}{\pm\sqrt{m^2+n^2+\kappa^2}}. \end{aligned} \right\} \qquad . . . . . . . (5) \]

Let us take the direction of the normal for points for which \(z>0\). We have seen that in this case its azimuth is identical with the azimuth of the line of dip. Assuming \(\kappa>0\), by virtue of the condition \(z>0\), we must take \(\alpha_1, \beta_1\) (i.e., by condition, \(\cos\gamma_1>0\)).

The desired azimuth of the line of dip, by virtue of (4), is determined by the formula:

\[ \operatorname{tg}\varphi = \frac{\cos \beta_1}{\cos \alpha_1} = \frac{ \frac{-m}{\sqrt{m^2+n^2+k^2}} }{ \frac{n}{\sqrt{m^2+n^2+k^2}} } = \frac{-m}{n}. \]

Let us now take the direction of the normal for points for which \(z<0\). It is easy to see that in this case (for \(k>0\))

\[ \operatorname{tg}\varphi = \frac{-\cos \beta_2}{-\cos \alpha_2} = \frac{ -\frac{-m}{-\sqrt{m^2+n^2+k^2}} }{ -\frac{n}{-\sqrt{m^2+n^2+k^2}} } = \frac{-m}{n}. \]

Accordingly, assuming \(k<0\), we find:

\[ \operatorname{tg}\varphi=\frac{m}{-n}. \]

For an analysis of the quantity “\(k\),” see the rotation method.

Tenth method.

Let us intersect the given plane of the bed,

\[ nx-my+kz=0, \ldots\ldots\ldots\ldots\ldots (1) \]

by the plane

\[ z=H, \ldots\ldots\ldots\ldots\ldots (2) \]

where \(H<0\).

Next, from the origin of coordinates let us drop a perpendicular to the straight line determined by equations (1) and (2). It is easy to see that this perpendicular—vector—is the line of dip. Denote the magnitude of the vector by \(P\). The azimuth of the line of dip is identical to the azimuth of its horizontal projection—the vector \(p\).

Eliminate the variable \(z\) from equations (1) and (2). We obtain:

\[ nx-my+kH=0. \ldots\ldots\ldots\ldots\ldots (3) \]

Reducing equation (3) to Hesse’s normal form, we directly find the desired azimuth.

We have:

\[ \frac{n}{\pm\sqrt{m^2+n^2}}\,x + \frac{-m}{\pm\sqrt{m^2+n^2}}\,y + \frac{kH}{\pm\sqrt{m^2+n^2}} =0. \ldots (4) \]

As is known, the signs before the root must be chosen so that the condition be satisfied:

\[ -p=\frac{kH}{\pm\sqrt{m^2+n^2}}<0 \ldots\ldots\ldots (5) \]

By the condition \(H<0\); consequently, as is evident, two assumptions are possible further: 1) \(\kappa>0\), in which case the sign before the radical must be taken as plus \((+)\), and 2) \(\kappa<0\), in which case the sign before the radical must be taken as minus \((-)\).

Both assumptions, as we have seen, are equally possible. In accordance with the choice in which \(\kappa>0\), we obtain the value of the sought azimuth \(\varphi\), as the azimuth of the vector \(p\)—the projection of the vector \(P\). Namely:

\[ \sin \varphi=\frac{-m}{\sqrt{m^{2}+n^{2}}}, \]

\[ \cos \varphi=\frac{n}{\sqrt{m^{2}+n^{2}}}, \]

and, consequently, vectorially:

\[ \tg \varphi=\frac{-m}{n}. \]

Similarly, for \(\kappa<0\), we obtain:

\[ \sin \varphi=\frac{-m}{-\sqrt{m^{2}+n^{2}}}, \]

\[ \cos \varphi=\frac{n}{-\sqrt{m^{2}+n^{2}}}, \]

and, consequently, vectorially:

\[ \tg \varphi=\frac{m}{-n}. \]

Remark. For an analysis of the quantity “\(\kappa\),” see the rotation method.

§ 11.

Determination of the zenith distance \(\zeta\).

The determination of the zenith distance \(\zeta\) of the line of fall, proceeding from the absolute value of the free term (5) of equation (4), representing the horizontal projection \(p\) of the vector \(P\).

Let us rotate the coordinate system \(XYZ\) about the axis \(Z\) through the angle \(\varphi\), consequently until the new axis \(X'\) coincides with the direction of the projection of the vector \(P\) (the projection of the line of fall).

Denote the coordinates of the end of the vector \(P\) by \(x'\) and \(z'\) \((x'>0,\ z'<0)\). Then, according to (5), for \(\kappa>0\), we have:

\[ -x'=\frac{\kappa z'}{+\sqrt{m^{2}+n^{2}}}, \]

or, since the ratio

\[ \frac{x'}{z'}=\tg \zeta, \]

then:

\[ \tg \zeta=-\frac{\kappa}{\sqrt{m^{2}+n^{2}}}. \]

For \(\kappa<0\) we obtain:

\[ \tg \zeta=-\frac{\kappa}{\sqrt{m^{2}+n^{2}}}. \]

Remark. It is easy to see that the same formulas for \(\zeta\) and \(\zeta'\) are obtained by setting \(H>0\).

Another method for determining the zenith distance \(\zeta\) of the line of dip (Fig. 14).

1. The case in which the direction of the normal satisfies the inequality:
   \[ \cos(AP,AZ)>0. \]

Denoting \((AP,AZ)=\theta\), directly from the drawing we find:

\[ \zeta=\frac{\pi}{2}+\theta \quad\text{and, consequently,}\quad \tg\zeta=-\ctg\theta=-\frac{\cos\theta}{\sin\theta}= \]

\[ =-\frac{\cos\theta}{\sqrt{1-\cos^{2}\theta}} =-\frac{\dfrac{\kappa}{\sqrt{m^{2}+n^{2}+\kappa^{2}}}} {\sqrt{\dfrac{m^{2}+n^{2}}{m^{2}+n^{2}+\kappa^{2}}}} =-\frac{\kappa}{\sqrt{m^{2}+n^{2}}} \]

[taking \(\kappa>0\) and the sign before the root as plus \((+)\)].

Accordingly, for \(\kappa<0\) and the sign before the root \((-)\), we find:

\[ \tg\zeta=-\frac{\kappa}{\sqrt{m^{2}+n^{2}}}. \]

1. The case in which the direction of the normal satisfies the inequality:
   \[ \cos(AP',AZ)<0. \]

Denoting \((AP',AZ)=\theta'\), directly from the drawing we find:

\[ \zeta=\frac{3}{2}\pi-\theta'; \]

and, consequently,

\[ \tg\zeta=\ctg\theta'=-\frac{\cos\theta'}{\sqrt{1-\cos^{2}\theta'}} =-\frac{\kappa}{\sqrt{m^{2}+n^{2}}} \]

(for \(\kappa>0\)) and, for \(\kappa<0\),

\[ \tg\zeta=\frac{\kappa}{\sqrt{m^{2}+n^{2}}}. \]

For the analysis of the quantity “\(\kappa\),” see the method of rotation.

Remark. The angle of dip \(h\) of the bed plane is determined by the formula:

\[ h=-\frac{\pi}{2}-\zeta, \]

and, by its sign, will be negative, as it should be.

§ 12.

Solutions by the method of differential calculus.

The method of relative maxima and minima.

First method.

Take a sphere of arbitrary radius \(R\), whose center coincides with the origin of coordinates.

Its equation is:

\[ x^2 + y^2 + z^2 - R^2 = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

The equation of the plane of the bed:

\[ nx - my + kz = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

Together, (1) and (2) give a curve—a circle. As is known, the points of the circle \([(1), (2)]\) that are at the greatest and least distance from the horizontal plane \(z = 0\) belong to the line of dip. The question is therefore reduced to determining the maxima and minima of the coordinate \(z\) of the circle, defined from (2) by the expression:

\[ z = \frac{1}{k}(-nx + my). \ldots \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

In the latter, by virtue of the compatibility of (1) and (2), the coordinates \(x\) and \(y\) are the coordinates of the points of the circle.

Let us find the conditions jointly satisfied by the coordinates \(x\) and \(y\) of the circle \([(1), (2)]\).

Eliminating the variable \(z\) from equations (1) and (2), we find the required condition:

\[ x^2 + y^2 + \frac{1}{k^2}(nx - my)^2 - R^2 = 0. \ldots \ldots (4) \]

We find the extremum of \(z\) under condition (4).

We form the function:

\[ W = \frac{1}{k}(-nx + my) + \lambda \left[ x^2 + y^2 + \frac{1}{k^2}(nx - my)^2 \right], \]

where \(\lambda\) is an undetermined multiplier or coefficient.

We find \(\dfrac{\partial W}{\partial x}\), \(\dfrac{\partial W}{\partial y}\) and set them equal to zero.

\[ \frac{\partial W}{\partial x} = -\frac{n}{k} + 2\lambda \left[ x + \frac{n}{k^2}(nx - my) \right] = 0, \]

or

\[ -nk + 2\lambda \left[k^2x + n(nx - my)\right] = 0 \ldots \ldots \ldots (5) \]

\[ \frac{\partial W}{\partial y} = \frac{m}{k} + 2\lambda \left[ y - \frac{m}{k^2}(nx - my) \right] = 0, \]

or

\[ mk + 2\lambda \left[k^2y - m(nx - my)\right] = 0 \ldots \ldots \ldots (6) \]

The system of equations (4), (5), and (6) will determine those values of \(x\) and \(y\) for which the function \(z\) is an extremum.

Transposing in equations (5) and (6) the terms not containing \(\lambda\) to the right-hand sides and dividing the resulting equations by one another, we find:

\[ -\frac{n}{m}=\frac{\kappa^{2}x+n(nx-my)}{\kappa^{2}y-m(nx-my)}; \]

or, transforming, we finally obtain:

\[ y=-\frac{m}{n}x. \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (7) \]

Substituting now (7) into (4):

\[ x^{2}+\frac{m^{2}}{n^{2}}x^{2}+\frac{1}{\kappa^{2}n^{2}}(m^{2}+n^{2})^{2}x^{2}-R^{2}=0; \]

or

\[ \left[\kappa^{2}(m^{2}+n^{2})+(m^{2}+n^{2})^{2}\right]x^{2}-\kappa^{2}n^{2}R^{2}=0; \]

whence:

\[ x_{1,2}=\pm\frac{\kappa nR}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}}; \]

and, by virtue of (7),

\[ y_{1,2}=\mp\frac{\kappa mR}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}}. \]

By virtue of (3):

\[ z_{1,2}=\frac{1}{\kappa\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}}(\mp\kappa n^{2}R\mp\kappa m^{2}R)= \]

\[ =\mp R\sqrt{\frac{m^{2}+n^{2}}{m^{2}+n^{2}+\kappa^{2}}}. \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (10) \]

It is easy to see directly from (10) that the upper sign—minus—corresponds to the minimum, and the lower—plus—to the maximum of the function defining \(z\).

The formulas obtained, (8), (9), and (10), are identical with the formulas obtained by us in § 10 [third method, formulas (5)], as was to be expected.

Second method.

On the \(Z\)-axis take an arbitrary point \((0,0,z_c)\).

We shall now find in the given plane of the bed

\[ nx-my+\kappa z=0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (1) \]

the point at the smallest, in absolute value, distance from the point \((0,0,z_c)\). The sought point is, obviously, the point of the line of dip.

The problem reduces to determining the values \(x\), \(y\), and \(z\) of the point of the plane for which the function

\[ F(x,y,z)=x^{2}+y^{2}+(z-z_c)^{2} \]

is a minimum and equation (1) is satisfied.

We form the function:

\[ W=x^{2}+y^{2}+(z-z_c)^{2}+\lambda(nx-my+\kappa z), \]

where \(\lambda\) is an undetermined multiplier.

We find $\dfrac{\partial W}{\partial x}$, $\dfrac{\partial W}{\partial y}$, $\dfrac{\partial W}{\partial z}$ and set them equal to zero.

We have:

\[ \frac{\partial W}{\partial x}=2x+\lambda n=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

\[ \frac{\partial W}{\partial y}=2y-\lambda m=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

\[ \frac{\partial W}{\partial z}=2(z-z_c)+\lambda k=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots (4) \]

We determine the required coordinates $(x,y,z)$.

Transferring in (3) and (2) the terms containing $\lambda$ to the right-hand side of the equations and dividing them, we obtain:

\[ y=-\frac{m}{n}x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (5) \]

Correspondingly, from (4) and (2), we obtain:

\[ z=\frac{k}{n}x+z_c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (6) \]

Substituting (5) and (6) into (1), we determine $x$:

\[ x=-\frac{nk}{m^2+n^2+k^2}z_c \ldots \ldots \ldots \ldots \ldots \ldots (7) \]

Substituting (7) into (5) and (6), respectively, we obtain:

\[ y=\frac{mk}{m^2+n^2+k^2}z_c. \]

\[ z=\frac{m^2+n^2}{m^2+n^2+k^2}z_c. \]

Having determined the coordinates $(x,y,z)$ of the point of the line of dip, it is easy to find also the azimuth of the latter.

Assuming $k>0$, with $z_c \gtreqless 0$ we find, vectorially,

\[ \operatorname{tg}\varphi=\frac{-m}{n}. \]

Assuming instead $k<0$, with $z_c \gtreqless 0$ we obtain, vectorially,

\[ \operatorname{tg}\varphi=\frac{m}{-n}. \]

The further analysis is identical with the analysis in the rotation method set forth above.

Method of channel-like surfaces.

From the origin of coordinates, as center, describe a sphere of arbitrary radius $R$.

Its equation is:

\[ x^2+y^2+z^2-R^2=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

The equation of the plane of the bed is:

\[ nx - my + kz = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

The system of equations:

\[ \left. \begin{array}{r} nx - my + kz = 0 \\ z = 0 \end{array} \right\} \]

defines the strike line of the plane (2), lying in the plane \(z = 0\).

Its equation is:

\[ nx - my = 0, \]

or

\[ y = \frac{n}{m}x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

Let us denote by \((\alpha, \beta, 0)\) the coordinates of an arbitrary point of the strike line (3).

Then the equation of the family of spheres of one and the same radius \(R\), having their centers on the line (3), will be expressed in the form:

\[ (x - \alpha)^2 + (y - \beta)^2 + z^2 - R^2 = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots (4) \]

Since \(\alpha\) and \(\beta\) satisfy equation (3), we have

\[ \beta = \frac{n}{m}\alpha \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (5) \]

By virtue of (5), the equation of the family of spheres takes the form:

\[ (x - \alpha)^2 + \left(y - \frac{n}{m}\alpha\right)^2 + \]

\[ + z^2 - R^2 = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (6) \]

Differentiating equation (6) with respect to \(\alpha\), we obtain:

\[ (x - \alpha) + \frac{n}{m}\left(y - \frac{n}{m}\alpha\right) = 0 \ldots \ldots \ldots \ldots \ldots (7) \]

Equations (6) and (7) jointly determine the so-called characteristic of the family of spheres (4). Eliminating further from equations (6) and (7) the parameter \(\alpha\), we obtain the equation of the geometric locus of the characteristics—the equation of the envelope of the family of spheres of one and the same radius \(R\) with centers on the line (3).

It is easy to see (the axis of the envelope is a straight line, and the radii of the family of spheres are equal) that the envelope is a cylindrical surface whose generatrices are parallel to the line (3).

Equation (7) gives:

\[ \alpha \frac{m^2 + n^2}{m^2} = \frac{mx + ny}{m}, \]

whence:

\[ \alpha = \frac{m}{m^2 + n^2}(mx + ny) \ldots \ldots \ldots \ldots \ldots \ldots \ldots (8) \]

Substituting (8) into (6), we obtain:

\[ \left[x-\frac{m}{m^2+n^2}(mx+ny)\right]^2+ \left[y-\frac{n}{m^2+n^2}(mx+ny)\right]^2+ z^2-R^2=0 \tag{9} \]

—the equation of the envelope.

We transform (9).
Let

\[ \begin{aligned} mx+ny&=P,\\ m^2+n^2&=Q \end{aligned} \tag{10}, (11) \]

Then (9) becomes the formula:

\[ \left(x-m\frac{P}{Q}\right)^2+ \left(y-n\frac{P}{Q}\right)^2+ z^2-R^2=0, \tag{\(\Delta\)} \]

or:

\[ (Qx-mP)^2+(Qy-nP)^2+Q^2(z^2-R^2)=0; \]

or, after transformation,

\[ Q^2(x^2+y^2+z^2-R^2)-2PQ(mx+ny)= -(m^2+n^2)P^2. \]

In view of (10) and (11), we obtain:

\[ Q(x^2+y^2+z^2-R^2)-P^2=0 \]

and, finally:

\[ (m^2+n^2)(x^2+y^2+z^2-R^2)- (mx+ny)^2=0 \tag{12} \]

—the equation of the desired cylindrical surface—the envelope of the family of spheres of radius \(R\).

The desired point of the line of dip is determined as the intersection of three surfaces: the sphere, the cylindrical surface, and the plane of the bed, whose equations are respectively:

\[ \left. \begin{aligned} x^2+y^2+z^2-R^2&=0\\ (m^2+n^2)(x^2+y^2+z^2-R^2)&\\ \quad -(mx+ny)^2=0\\ nx-my+\kappa z&=0 \end{aligned} \right\} \begin{array}{l} \ldots\ldots\ldots (13)\\ \ldots\ldots\ldots (14)\\ \ldots\ldots\ldots (15) \end{array} \]

Substituting (13) into (14), we obtain:

\[ mx+ny=0, \]

whence

\[ y=-\frac{m}{n}x \tag{16} \]

Substituting (16) into (15), we obtain:

\[ nx+\frac{m^2}{n}x+\kappa z=0, \]

whence

\[ z=-\frac{m^2+n^2}{\kappa n}x. \tag{17} \]

Substituting further (16) and (17) into (13), we obtain:

\[ x^2+\frac{m^2}{n^2}x^2+\frac{(m^2+n^2)^2}{\kappa^2 n^2}x^2-R^2=0; \]

or

\[ x^2(m^2+n^2)(m^2+n^2+\kappa^2)=R^2n^2\kappa^2 \]

and, finally,

\[ x=\pm \frac{Rn\kappa}{\sqrt{(m^2+n^2)(m^2+n^2+\kappa^2)}} \ldots \ldots \ldots \ldots \ldots \ldots (18) \]

Substituting (18) into (16) and (17), we obtain:

\[ y=\mp \frac{Rm\kappa}{\sqrt{(m^2+n^2)(m^2+n^2+\kappa^2)}} \ldots \ldots \ldots \ldots \ldots (19) \]

\[ z=\mp R\sqrt{\frac{m^2+n^2}{m^2+n^2+\kappa^2}} \ldots \ldots \ldots (20) \]

The formulas obtained are identical with the formulas of the first method, as was to be expected.

Remark 1. Equation (16), as is easy to see, is the equation of the plane of dip, and therefore the sought point of the line of dip is determined by the solution of the system of equations:

\[ \left. \begin{array}{r} x^2+y^2+z^2-R^2=0\\ mx+ny=0\\ nx-my+\kappa z=0 \end{array} \right\} \]

Remark 2. Equation (12) is easily obtained by describing about the sphere

\[ x^2+y^2+z^2-R^2=0 \]

a cylindrical surface whose generators are parallel to the straight line of strike

\[ y=\frac{n}{m}x, \]

or

\[ \frac{x}{m}=\frac{y}{n}=\frac{z}{0}\ldots \ldots \ldots \ldots \ldots \ldots (1) \]

The line of contact of the cylindrical surface with the surface in general, or the directrix, as is known, is determined by the system of equations:

\[ F(x,y,z)=0 \ldots \ldots \ldots (\text{surface}) \]

and

\[ m\frac{\partial F}{\partial x}+n\frac{\partial F}{\partial y}+p\frac{\partial F}{\partial z}=0, \]

where \(m, n, p\) are the directions of the generators.

In our case, this is the system:

\[ \left. \begin{array}{r} x^2+y^2+z^2-R^2=0\\ mx+ny=0 \end{array} \right\} \begin{array}{l} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (2)\\ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (3) \end{array} \]

The equation of the generators of the cylindrical surface parallel to (1) will be:

\[ \frac{X-x}{m}=\frac{Y-y}{n}=\frac{Z-z}{0}\ldots\ldots\ldots\ldots\ldots (4) \]

where \(X, Y, Z\) are the current coordinates of the system of generators, and \(x, y, z\) are the current coordinates of the line of contact.

Eliminating the parameters \(x, y, z\) from (2), (3), and (4), we obtain the required equation.

Namely, denoting

\[ \frac{X-x}{m}=\frac{Y-y}{n}=\frac{Z-z}{0}=\lambda,\ldots\ldots (5) \]

we obtain:

\[ \frac{m(X-x)}{m^2}=\frac{n(Y-y)}{n^2}=\frac{Z-z}{0}=\lambda; \]

or

\[ \frac{mX+nY-(mx+ny)}{m^2+n^2}=\lambda; \]

and, finally, according to (3),

\[ \frac{mX+nY}{m^2+n^2}=\lambda. \]

Denoting

\[ mX+nY=P;\quad m^2+n^2=Q, \]

we obtain:

\[ \frac{P}{Q}=\lambda. \]

or, by virtue of (5),

\[ \frac{X-x}{m}=\frac{Y-y}{n}=\frac{Z-z}{0}=\frac{P}{Q}. \]

Whence:

\[ \left. \begin{array}{r} x=X-m\,\dfrac{P}{Q}\\[6pt] y=Y-n\,\dfrac{P}{Q}\\[6pt] z=Z \end{array} \right\} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (6) \]

Substituting (6) into (2), we obtain:

\[ \left(X-m\frac{P}{Q}\right)^2+\left(Y-n\frac{P}{Q}\right)^2+Z^2-R^2=0. \]

—an equation identical to equation \((\Delta)\).

The further transformations are as indicated above.

§ 13.

Solutions by the method of integral calculus.

I. Determination of the coordinates \((x_c, y_c)\) of a point of the dip line, as the center of inertia of plane curves.

In the plane \(z = 0\), from the origin as center, describe a circle of arbitrary radius \(r\). Its equation in polar coordinates is:

\[ x = r\cos\varphi. \]

\[ y = r\sin\varphi. \]

where \(r\) is the radius vector, and \(\varphi\) is the polar angle.

The system of equations:

\[ \left. \begin{aligned} nx - my + kz &= 0\\ z &= 0 \end{aligned} \right\} \]

determines the strike line of the plane of the bed

\[ nx - my + kz = 0, \]

lying in the plane \(z = 0\).

Its equation is:

\[ nx - my = 0. \]

Let us now take successively each semicircle as the result of dividing the circle by the strike line. Then, as is easy to see, the center of inertia of each of the semicircles is the horizontal projection of one of the points of the dip line. Having found the horizontal projection of the desired point, it is then easy to find the desired point of the dip line itself.

The coordinates of the center of inertia of an arc of a plane curve*), as is known, are determined by the formulas:

\[ x_c = \int \frac{x\,ds}{ds}, \quad y_c = \int \frac{y\,ds}{ds}; \]

or, expressing \(x\), \(y\), and \(ds\) in polar coordinates of the given curve—the circle—we obtain:

\[ \left. \begin{aligned} ds &= r\,d\varphi;\quad x_c = r \int \frac{\cos\varphi\,d\varphi}{d\varphi};\\[4pt] y_c &= r \int \frac{\sin\varphi\,d\varphi}{d\varphi} \end{aligned} \right\} \qquad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \tag{1} \]

We found earlier (the method of rotation of analytic geometry) that the azimuth \(A\) of the dip line is determined by the formula:

\[ \tg A = -\frac{m}{n}. \]

*) provided it is homogeneous.

Let us take the same assumptions as in the solution by the method of rotation of analytic geometry.

Namely:

\[ \text{I. }\tg A=\frac{-m}{n}, \]

and assuming that the azimuth of the line of dip belongs to the third quadrant; consequently,

\[ \sin A<0 \quad \text{and} \quad \cos A<0. \]

The values of the latter will be:

\[ \left. \begin{aligned} \sin A&=-\frac{m}{\sqrt{m^2+n^2}}\\[6pt] \cos A&=+\frac{n}{\sqrt{m^2+n^2}} \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots (2) \]

1. Let us take the semicircle to which the variations of \(\varphi\) correspond:

lower limit

\[ \varphi_{\alpha}=\arctg\left(\frac{-m}{n}\right)-\frac{\pi}{2}, \]

upper limit

\[ \varphi_{\omega}=\arctg\left(\frac{-m}{n}\right)+\frac{\pi}{2}. \]

Then it is easy to see (see Fig. 7) that the coordinates \(z\) of the given plane of the bed must satisfy the inequality: \(z<0\).

We find the values of the integrals entering into formulas (1), for the given limits of variation of \(\varphi\):

\[ \int_{\varphi_{\alpha}}^{\varphi_{\omega}}\cos\varphi\,d\varphi = \left[\sin\varphi\right]_{\varphi_{\alpha}}^{\varphi_{\omega}} = \sin\left(\arctg\frac{-m}{n}+\frac{\pi}{2}\right) - \]

\[ -\sin\left(\arctg\frac{-m}{n}-\frac{\pi}{2}\right) = \cos\left(\arctg\frac{-m}{n}\right) + \]

\[ +\cos\left(\arctg\frac{-m}{n}\right) = 2\cos\left(\arctg\frac{-m}{n}\right); \]

or, by virtue of (2), finally:

\[ \int_{\varphi_{\alpha}}^{\varphi_{\omega}}\cos\varphi\,d\varphi = 2\frac{n}{\sqrt{m^2+n^2}}; \]

\[ \int_{\varphi_\alpha}^{\varphi_\omega} \sin \varphi\, d\varphi = [-\cos \varphi]_{\varphi_\alpha}^{\varphi_\omega} = \]

\[ = -\cos\left(\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}\right) + \cos\left(\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}\right) = \]

\[ = 2\sin\left(\operatorname{arctg}\frac{-m}{n}\right). \]

Finally:

\[ \int_{\varphi_\alpha}^{\varphi_\omega} \sin \varphi\, d\varphi = -2\,\frac{m}{\sqrt{m^2+n^2}} \]

\[ \int_{\varphi_\alpha}^{\varphi_\omega} d\varphi = \pi. \]

Substituting the obtained values of the integrals into formulas (1), we obtain:

\[ x_c = 2\,\frac{r}{\pi}\, \frac{n}{\sqrt{m^2+n^2}} \]

\[ y_c = -2\,\frac{r}{\pi}\, \frac{m}{\sqrt{m^2+n^2}}. \]

Let us now find the expression for the coordinate \(z\) of the bed plane corresponding to the point \((x_c, y_c)\).

From the equation of the plane we have:

\[ z = \frac{1}{\kappa}(-n x_c + m y_c) = \]

\[ = \frac{1}{\kappa} \left\{ -n\,2\,\frac{r}{\pi}\,\frac{n}{\sqrt{m^2+n^2}} - m\,2\,\frac{r}{\pi}\,\frac{m}{\sqrt{m^2+n^2}} \right\} = \]

\[ = -2\,\frac{r}{\kappa\pi}\sqrt{m^2+n^2} \ldots \ldots \ldots \ldots \ldots \ldots \tag{3} \]

Since, by assumption, \(z < 0\), it follows from (3) that \(\kappa > 0\).

The azimuth of the line of dip, determined by the points \((0,0,0)\) and \((x_c,y_c,z<0)\), will be found vectorially by the formula:

\[ \tg A=\frac{y_c}{x_c} = \frac{ -2\frac{r}{\pi}\frac{m}{\sqrt{m^2+n^2}} }{ 2\frac{r}{\pi}\frac{n}{\sqrt{m^2+n^2}} } = \frac{-m}{n}. \]

1. Let us take the semicircle to which the changes of \(\varphi\) correspond:

lower limit

\[ \varphi_{\alpha}=\arctg\frac{-m}{n}+\frac{\pi}{2}, \]

upper limit

\[ \varphi_{\omega}=\arctg\frac{-m}{n}+\frac{3}{2}\pi . \]

Then it is easy to see (see Fig. 7) that the coordinates \(z\) of the given plane of the bed satisfy the inequality: \(z>0\).

Carrying out the calculations analogously to 1, we find that:

\[ \left. \begin{aligned} x_c&=-2\frac{r}{\pi}\frac{n}{\sqrt{m^2+n^2}}\\[6pt] y_c&=+2\frac{r}{\pi}\frac{m}{\sqrt{m^2+n^2}}\\[6pt] z&=2\frac{r}{k\pi}\sqrt{m^2+n^2} \end{aligned} \right\} \qquad\qquad\ldots\ldots\ldots\ldots (4) \]

Since, by the condition, \(z>0\), it follows from \((4_3)\) that \(k>0\), and

\[ \tg A=\frac{-y_c}{-x_c}=\frac{-m}{n}. \]

II. Let us take the assumption:

\[ \tg A=\frac{m}{-n}>0, \]

under the condition that \(\sin A<0\) and \(\cos A<0\).

Then (see the method of rotation in analytical geometry):

\[ \sin A=+\frac{m}{\sqrt{m^2+n^2}} \quad \text{and} \quad \cos A=-\frac{n}{\sqrt{m^2+n^2}}. \]

In what follows the solution proceeds analogously to case I. As a result we obtain: the azimuth of the line of dip is determined vectorially by the formula \(\tg A=\frac{m}{-n}\), under the condition \(k<0\).

Thus we shall arrive at the conclusion that, taking \(k>0\), one must set

\[ \tg A=\frac{-m}{n}, \]

whereas taking \(k<0\), one must set

\[ \tg A=\frac{m}{-n}. \]

In what follows, the results of the analysis will be the same as in the method of rotation of analytic geometry.

The consideration of other possible cases is left to the reader.

2. Determination of the coordinates \((x_c, y_c)\) of the point of the line of fall as the center of inertia of surfaces

Let us imagine a right circular conical surface whose center is the origin of coordinates, whose axis is the \(Z\)-axis, and which is cut by the planes

\[ z=\pm H. \]

The equation of the conical surface, as is known, will be:

\[ x^2+y^2-z^2\tg^2\beta=0,\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (1) \]

where \(\beta\) is the angle of conicity.

Let us take successively each quarter of the conical surface bounded by the planes:

\[ \begin{aligned} z&=0,\\ z&=l,\\ nx-my&=0 \end{aligned} \]

and

\[ z=\pm H. \]

Then, as is easy to see, the horizontal projection of the center of inertia of each of the quarters of the conical surface is at the same time the horizontal projection of the point of the line of fall. Having found the horizontal projection of the point—its coordinates \(x_c, y_c\)—it is then easy to find the required point of the line of fall as well—its coordinates \(x_c, y_c, z\).

The coordinates \((x_c, y_c)\) of the center of inertia of curved surfaces*), as is known, are determined by the formulas:

\[ x_c= \frac{ \iint x\sqrt{1+p^2+q^2}\,dx\,dy }{ \iint \sqrt{1+p^2+q^2}\,dx\,dy }, \]

\[ y_c= \frac{ \iint y\sqrt{1+p^2+q^2}\,dx\,dy }{ \iint \sqrt{1+p^2+q^2}\,dx\,dy }, \]

where \(p=\dfrac{\partial z}{\partial x}\) and \(q=\dfrac{\partial z}{\partial y}\), and they are determined from the equation of the surface.

*) For homogeneous surfaces.

Differentiating equation (1) with respect to \(x\), we obtain:

\[ x - z \operatorname{tg}^{2}\beta \frac{\partial z}{\partial x} = 0, \]

whence

\[ \frac{\partial z}{\partial x} = \frac{x}{z}\operatorname{ctg}^{2}\beta. \]

Differentiating equation (1) with respect to \(y\), we obtain:

\[ \frac{\partial z}{\partial y} = \frac{y}{z}\operatorname{ctg}^{2}\beta. \]

Therefore:

\[ \sqrt{1+p^{2}+q^{2}} = \sqrt{\frac{z^{2}+(x^{2}+y^{2})\operatorname{ctg}^{4}\beta}{z^{2}}} = \sqrt{1+\operatorname{ctg}^{2}\beta} = \frac{1}{\sin\beta}. \tag{2} \]

Expressing \(x\) and \(y\) in polar coordinates:

\[ x = r\cos\varphi, \]

\[ y = r\sin\varphi, \]

we obtain, taking (2) into account,

\[ \left. \begin{aligned} x_c &= \frac{\iint r^{2}\cos\varphi\,dr\,d\varphi} {\iint r\,dr\,d\varphi} \\[6pt] y_c &= \frac{\iint r^{2}\sin\varphi\,dr\,d\varphi} {\iint r\,dr\,d\varphi} \end{aligned} \right\} \tag{3} \]

We shall take the same assumptions as in the preceding solution of the problem.

Namely:

\[ 1.\quad \operatorname{tg} A = \frac{-m}{n} > 0, \]

where

\[ \sin A < 0 \quad \text{and} \quad \cos A < 0. \]

The values of the latter are:

\[ \left. \begin{aligned} \sin A &= -\frac{m}{\sqrt{m^{2}+n^{2}}} \\[6pt] \cos A &= +\frac{n}{\sqrt{m^{2}+n^{2}}} \end{aligned} \right\} \tag{4} \]

1. Let us consider that quadrant of the conical surface for whose points \(z > 0\), and the limits of variation of \(\varphi\) are:

lower limit

\[ \varphi_{\alpha}=\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}, \]

upper limit

\[ \varphi_{\omega}=\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}. \]

The limits of integration for \(r\) are:

\[ r_{\alpha}=0;\qquad r_{\omega}=H\operatorname{tg}\beta \quad (\text{where } H>0). \]

It is easy to see that, in this case, the coordinate \(z\) of the bed plane satisfies the condition: \(z<0\).

We find the values of the integrals entering into formulas (3), for the given limits of variation of the variables.

\[ \int_{\varphi_{\alpha}}^{\varphi_{\omega}} \int_{r_{\alpha}}^{r_{\omega}} r\,dr\,d\varphi = \int_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} d\varphi \int_{0}^{H\operatorname{tg}\beta} r\,dr = \frac{1}{2}\pi H^{2}\operatorname{tg}^{2}\beta . \]

\[ \begin{aligned} \int_{\varphi_{\alpha}}^{\varphi_{\omega}} \int_{r_{\alpha}}^{r_{\omega}} r^{2}\cos\varphi\,dr\,d\varphi &= \int_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} \cos\varphi\,d\varphi \int_{0}^{H\operatorname{tg}\beta} r^{2}\,dr \\ &= \left[\sin\varphi\right]_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} \frac{1}{3}H^{3}\operatorname{tg}^{3}\beta \\ &= 2\cos\left(\operatorname{arctg}\frac{-m}{n}\right)\frac{1}{3}H^{3}\operatorname{tg}^{3}\beta \\ &= \frac{2}{3}H^{3}\operatorname{tg}^{3}\beta\, \frac{n}{\sqrt{m^{2}+n^{2}}}. \end{aligned} \]

\[ \begin{aligned} \int_{\varphi_{\alpha}}^{\varphi_{\omega}} \int_{r_{\alpha}}^{r_{\omega}} r^{2}\sin\varphi\,dr\,d\varphi &= \int_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} \sin\varphi\,d\varphi \int_{0}^{H\operatorname{tg}\beta} r^{2}\,dr \\ &= -2\,\frac{m}{\sqrt{m^{2}+n^{2}}}\,\frac{1}{3}H^{3}\operatorname{tg}^{3}\beta \\ &= -\frac{2}{3}H^{3}\operatorname{tg}^{3}\beta\, \frac{m}{\sqrt{m^{2}+n^{2}}}. \end{aligned} \]

Therefore, finally:

\[ x_c = \frac{ \frac{2}{3} H^3 \tg^3 \beta \, \frac{n}{\sqrt{m^2+n^2}} }{ \frac{1}{2}\pi H^2 \tg^2 \beta } = \frac{4}{3\pi} H \tg \beta \, \frac{n}{\sqrt{m^2+n^2}} . \]

\[ y_c = \frac{ -\frac{2}{3} H^3 \tg^3 \beta \, \frac{m}{\sqrt{m^2+n^2}} }{ \frac{1}{2}\pi H^2 \tg^2 \beta } = -\frac{4}{3\pi} H \tg \beta \, \frac{m}{\sqrt{m^2+n^2}} . \]

From the equation of the plane we obtain:

\[ z=\frac{1}{\kappa}(-nx_c+my_c) = -\frac{4}{3\pi \kappa} H \tg \beta \sqrt{m^2+n^2}. \]

By condition: \(z<0\), corresponding to changes of \(\varphi\) and \(H>0\); consequently, \(\kappa>0\).

The azimuth of the line of dip, determined by the points \((0,0,0)\) and \((x_c,y_c,z<0)\), is found vectorially by the formula:

\[ \tg A=\frac{y_c}{x_c}=\frac{-m}{n}. \]

In what follows—when considering other possible assumptions—we shall obtain formulas determining the azimuth of the line of dip, and the results of their analysis, identical to those considered above under other methods of solving the problem. The latter is left to the reader to derive.

3. Determination of the coordinates \((x_c,y_c)\) of a point on the line of dip as the projection of the center of inertia of the volume of bodies.

Let us describe, from the origin of coordinates as center, a sphere of arbitrary radius \(R\).

The equation of the sphere:

\[ x^2+y^2+z^2-R^2=0 \qquad \ldots\ldots\ldots\ldots\ldots (1) \]

The equation of the vertical plane corresponding to the strike line is—

\[ \left. \begin{aligned} nx-my&=0\\ z&=z \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots\ldots (2) \]

Let us take successively each quarter of the volume of the sphere, bounded by the planes \(z=0\) and (2).

Then, as is easy to see, the horizontal projection of the center of inertia of the volume of each quarter of the sphere is the horizontal projection of a point of the line of greatest dip. Having found the horizontal projection of the desired point, it is not difficult to find the required point of the line of greatest dip as well.

The coordinates of the center of inertia \((x_c, y_c)\) of the volume of a homogeneous body, as is known, are determined by the formulas:

\[ x_c=\frac{\iiint x\,dx\,dy\,dz}{\iiint dx\,dy\,dz}, \]

\[ y_c=\frac{\iiint y\,dx\,dy\,dz}{\iiint dx\,dy\,dz}. \]

Or, expressing \(x\) and \(y\) in spherical coordinates,

\[ x=\rho\sin\theta\cos\varphi \]

\[ y=\rho\sin\theta\sin\varphi, \]

we obtain:

\[ \left. \begin{aligned} x_c&=\frac{\iiint \rho^3\sin^2\theta\cos\varphi\,d\rho\,d\theta\,d\varphi} {\iiint \rho^2\sin\theta\,d\rho\,d\theta\,d\varphi},\\[6pt] y_c&=\frac{\iiint \rho^3\sin^2\theta\sin\varphi\,d\rho\,d\theta\,d\varphi} {\iiint \rho^2\sin\theta\,d\rho\,d\theta\,d\varphi}. \end{aligned} \right\} \qquad\cdots\cdots\cdots (3) \]

Let us adopt assumptions corresponding to the assumptions of the preceding solution.

\[ \text{I. }\operatorname{tg}A=\frac{-m}{n}>0, \]

where

\[ \sin A<0 \quad \text{and} \quad \cos A<0. \]

The values of the latter are:

\[ \sin A=-\frac{m}{\sqrt{m^2+n^2}} \]

\[ \cos A=+\frac{n}{\sqrt{m^2+n^2}}. \]

1. Let us consider a quarter of the volume of a sphere, for the points of which \(z > 0\), and the limits of variation of \(\varphi\) are:

lower limit

\[ \varphi_\alpha = \operatorname{arctg}\frac{-m}{n} - \frac{\pi}{2} \]

upper limit

\[ \varphi_\omega = \operatorname{arctg}\frac{-m}{n} + \frac{\pi}{2} \tag{4} \]

It is easy to see that, in this case, the coordinate \(z\) of the plane of the bed satisfies the condition: \(z < 0\).

The limits of integration will be—

for \(\varphi\): expressions (4),

for \(\rho\): \(\rho_\alpha = 0,\ \rho_\omega = R\),

for \(\theta\): \(\theta_\alpha = 0,\ \theta_\omega = \dfrac{\pi}{2}\).

We find the values of the integrals:

\[ \int_{\rho_\alpha}^{\rho_\omega} \int_{\theta_\alpha}^{\theta_\omega} \int_{\varphi_\alpha}^{\varphi_\omega} \rho^2 \sin\theta\, d\rho\, d\theta\, d\varphi = \]

\[ = \int_{0}^{R} \rho^2 d\rho \int_{0}^{\frac{\pi}{2}} \sin\theta\, d\theta \int_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} d\varphi = \]

\[ = \frac{1}{3} R^3 \left[-\cos\theta\right]_{0}^{\frac{\pi}{2}} \pi = \frac{1}{3}\pi R^3, \]

as was to be expected.

\[ \int_{\rho_\alpha}^{\rho_\omega} \int_{\theta_\alpha}^{\theta_\omega} \int_{\varphi_\alpha}^{\varphi_\omega} \rho^3 \sin^2\theta \cos\varphi\, d\rho\, d\theta\, d\varphi = \]

\[ = \int_{0}^{R} \rho^3 d\rho \int_{0}^{\frac{\pi}{2}} \sin^2\theta\, d\theta \int_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} \cos\varphi\, d\varphi = \]

\[ = \frac{1}{4} R^4 \left[-\frac{1}{4}\sin 2\theta + \frac{\theta}{2}\right]_{0}^{\frac{\pi}{2}} \left[\sin \varphi\right]_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} \]
\[ = \frac{1}{4}R^4 \frac{\pi}{4}\, 2\cos\left(\operatorname{arctg}\frac{-m}{n}\right) = \frac{\pi R^4}{8}\,\frac{n}{\sqrt{m^2+n^2}}. \]

\[ \int_{\rho_\alpha}^{\rho_\omega} \int_{\theta_\alpha}^{\theta_\omega} \int_{\varphi_\alpha}^{\varphi_\omega} \rho^3 \sin^2\theta \sin\varphi\, d\rho\, d\theta\, d\varphi = \]
\[ = \int_{0}^{R} \rho^3\, d\rho \int_{0}^{\frac{\pi}{2}} \sin^2\theta\, d\theta \int_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} \sin\varphi\, d\varphi = \]
\[ = \frac{1}{4}R^4 \frac{\pi}{4} \left[-\cos\varphi\right]_{\operatorname{arctg}\frac{-m}{n}-\frac{\pi}{2}}^{\operatorname{arctg}\frac{-m}{n}+\frac{\pi}{2}} = \frac{\pi R^4}{16}\,2\sin\left[\operatorname{arctg}\frac{-m}{n}\right] = \]
\[ = -\frac{\pi R^4}{8}\,\frac{m}{\sqrt{m^2+n^2}}. \]

Substituting the obtained values of the integrals into the expressions for \(x_c\) and \(y_c\), we finally obtain:

\[ x_c = \frac{3}{8}R\,\frac{n}{\sqrt{m^2+n^2}}, \]

\[ y_c = -\frac{3}{8}R\,\frac{m}{\sqrt{m^2+n^2}}. \]

We determine the coordinate \(z\) of the point of the line of dip from the equation of the plane:

\[ z = \frac{1}{\kappa}(-nx_c + my_c) = -\frac{3R}{8\kappa}\sqrt{m^2+n^2}. \ldots\ldots (5) \]

Since, according to the condition, \(z<0\), it follows from (5) that \(\kappa>0\).

The azimuth of the line of dip, determined by the points \((0,0,0)\) and \((x_c, y_c, z<0)\), is found vectorially from the formula:

\[ \tg A=\frac{y_c}{x_c} = \frac{-\dfrac{3}{8}R\,\dfrac{m}{\sqrt{m^2+n^2}}} {\dfrac{3}{8}R\,\dfrac{n}{\sqrt{m^2+n^2}}} = -\frac{m}{n}, \]

under the condition \(\kappa>0\).

Let us take assumption II, under which

\[ \tg A=\frac{m}{-n}>0,\quad \text{whereas:} \]

\[ \sin A<0 \quad \text{and} \quad \cos A<0. \]

The values of the latter are:

\[ \sin A=+\frac{m}{\sqrt{m^2+n^2}};\qquad \cos A=-\frac{n}{\sqrt{m^2+n^2}}. \]

Let us consider the quadrant of the volume of the sphere corresponding to the preceding assumption 1.

The limits of integration will remain the same; similarly we shall find that:

\[ \left. \begin{aligned} x_c&=-\frac{3}{8}R\,\frac{n}{\sqrt{m^2+n^2}}\\[6pt] y_c&=+\frac{3}{8}R\,\frac{m}{\sqrt{m^2+n^2}}\\[6pt] z&=+\frac{3}{8\kappa}R\sqrt{m^2+n^2} \end{aligned} \right\} \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (6) \]

According to the condition, \(z<0\), and from \((6_3)\) we find that \(\kappa<0\).

The required azimuth of the line of dip is determined vectorially from the formula:

\[ \tg A=\frac{y_c}{x_c}=\frac{m}{-n}, \quad \text{under the condition } \kappa<0. \]

The solution of the problem under the other possible assumptions is analogous to those considered, and we arrive at the conclusions indicated in the method of rotation of analytic geometry.

§ 14.

We shall present another method for determining the zenith distance \(\zeta\) of the line of dip.

We pass from the system \(XYZ\) to the system \(X'Y'Z'\), rotating the former about the axis \(Z\) through the angle \(\varphi\), corresponding to the azimuth of the line of dip. Then, as we found earlier, for points of the line of dip:

\[ x = x'\cos\varphi \]

\[ y = x'\sin\varphi . \]

Squaring the last two equalities and adding them, we obtain, determining \(x'\),

\[ x'=\pm\sqrt{x^2+y^2}. \]

We substitute the latter value of \(x'\) into the formula determining

\[ \operatorname{tg}\zeta=\pm\frac{x'}{z}; \]

we obtain:

\[ \operatorname{tg}\zeta=\pm\frac{\sqrt{x^2+y^2}}{z}, \ldots\ldots\ldots\ldots (1) \]

where the plus sign \((+)\) is taken for values \(z<0\), and the minus sign \((-)\)—for values \(z>0\) (as follows from Fig. 12).

It remains now to substitute into formula (1) the coordinates of an arbitrary point of the line of dip; these coordinates were determined by us when considering the question of determining the azimuth of the line of dip.

After minor transformations we obtain:

\[ \operatorname{tg}\zeta=\mp\frac{k}{\sqrt{m^2+n^2}}, \]

respectively for \(k\gtreqless 0\).

§ 15.

At the beginning we pointed out that all cases of exploration (pitting, drilling, etc.), when solving the question of determining the attitude elements of a bed at a given point—the lines of dip and strike—reduce to a single case: the case of pitting.

Without going into details (which would constitute a separate work), we shall give only one example in connection with the proposition mentioned at the beginning concerning the necessity—in certain cases of practice—of giving the problem of determining the attitude elements of a bed specifically an analytical solution.

Let three boreholes be given for determining the position of a bed (see Fig. 15); consequently, in the final result, the rectangular coordinates of the points \(A, B, C\) of the bed will be known, corresponding to the points of intersection of the axes of the boreholes with the latter.

Let us denote the coordinates of the points \(A, B, C\) of the bed respectively:

\[ (x_a,\ y_a,\ z_a);\ (x_b,\ y_b,\ z_b);\ (x_c,\ y_c,\ z_c). \]

Let us take any one of the points, for example \(A\), as the origin of two vectors, whose endpoints are the points \(B\) and \(C\). Then the azimuths of the vectors will be found from the formulas:

\[ \left. \begin{aligned} (AB)&=\operatorname{arctg}\frac{y_b-y_a}{x_b-x_a}\\[4pt] (AC)&=\operatorname{arctg}\frac{y_c-y_a}{x_c-x_a} \end{aligned} \right\} \tag{1} \]

The zenith distances of the vectors, as is easy to see, are determined by the formulas:

\[ \operatorname{tg}\zeta_{(AB),(AC)} = \frac{\sqrt{(x_{b,c}-x_a)^2+(y_{b,c}-y_a)^2}}{z_{b,c}-z_a}, \tag{2} \]

and the problem is reduced to that considered above.

Remark. The obtained formula determining \(\zeta\) is inconvenient for logarithming. Let us bring it to a form convenient for logarithming.

The expression:

\[ \sqrt{(x_{b,c}-x_a)^2+(y_{b,c}-y_a)^2} \]

represents, as is easy to see, the projection of the vector \((AB)\) [or \((AC)\)] onto the plane \(z=0\).

Therefore, as is known, we may write:

\[ \sqrt{(x_b-x_a)^2+(y_b-y_a)^2} = \frac{\Delta x_a^b}{\cos(AB)} = \frac{\Delta y_a^b}{\sin(AB)}. \]

\[ \sqrt{(x_c-x_a)^2+(y_c-y_a)^2} = \frac{\Delta x_a^c}{\cos(AC)} = \frac{\Delta y_a^c}{\sin(AC)} \]

and, consequently,

\[ \operatorname{tg}\zeta_{(AB)} = \frac{\Delta x_a^b}{\Delta z_a^b}\cdot\frac{1}{\cos(AB)} = \frac{\Delta x_a^b}{\Delta z_a^b}\sec(AB), \tag{3} \]

or

\[ \operatorname{tg}\zeta_{(AB)} = \frac{\Delta y_a^b}{\Delta z_a^b}\cdot\frac{1}{\sin(AB)} = \frac{\Delta y_a^b}{\Delta z_a^b}\operatorname{cosec}(AB). \tag{4} \]

Similarly:

\[ \operatorname{tg}\zeta_{(AC)} = \frac{\Delta x_a^c}{\Delta z_a^c}\sec(AC), \tag{5} \]

or

\[ \operatorname{tg}\zeta_{(AC)} = \frac{\Delta y_a^c}{\Delta z_a^c}\operatorname{cosec}(AC). \tag{6} \]

Comparing pairwise (3) and (4), (5) and (6), we see that for each of the zenith distances \(\zeta_{(AB)}\) and \(\zeta_{(AC)}\) we have obtained two formulas that mutually check the correctness of the determination of the zenith distances.

The problem posed at the beginning could have been solved in another way as well, starting from the strike line of the bed plane. The solution in that case turns out to be somewhat complicated; therefore we do not give it; this is one reason. On the other hand—as indicated—it is natural to start precisely from the dip line.

§ 16.

Appendix.

Numerical examples.

A. Examples of determining the azimuth \(\varphi\) of the dip line.

I.

Formulas (P) and (Q) § 6:

\[ \operatorname{tg}\varphi = \frac{\displaystyle \pm \frac{\sin(a_1-u)}{\sin a_1}} {\displaystyle \pm \frac{\sin(u-a_2)}{\cos a_2}} \ldots\ldots\ldots (P) \]

\[ \operatorname{tg}\varphi = \frac{\displaystyle \pm \frac{\sin(u-a_1)}{\sin a_1}} {\displaystyle \pm \frac{\sin(a_2-u)}{\cos a_2}} \ldots\ldots\ldots (Q) \]

We have:

\[ \begin{array}{rcl|rcl} a_1 &=& 13^\circ 30', & a_2 &=& 281^\circ 15',\\ \zeta_1 &=& 65^\circ 45', & \zeta_2 &=& 105^\circ 30'. \end{array} \]

1. Determination of the auxiliary angle \(u\).

\[ \operatorname{tg}u = \frac{\operatorname{tg}\zeta_1 \sin a_1} {\operatorname{tg}\zeta_2 \cos a_2} \ldots\ldots\ldots (1.) \]

\[ \begin{aligned} \lg \operatorname{tg}\zeta_1 &= 0\cdot 34634\\ \lg \sin a_1 &= 9\cdot 36819\\ c'\lg \operatorname{tg}\zeta_2 &= 9\cdot 44299\\ c'\lg \cos a_2 &= 0\cdot 70976^{n}\\ \hline \lg \operatorname{tg}u &= 9\cdot 86728_{n} \end{aligned} \]

\[ \begin{aligned} u_0 &= 36^\circ 22' 42''\\ u_1 &= -36^\circ 22' 42''\\ u_2 &= 143^\circ 37' 18''. \end{aligned} \]

1. Determination of the angle \(\varphi_0\).

In the subsequent computations we shall take formula (B); the angle

\[ u=u_1=-36^\circ 22' 42''. \]

\[ \operatorname{tg}\varphi = \frac{\cos a_2 \sin(u-a_1)} {\sin a_1 \sin(a_2-u)} \ldots\ldots\ldots (B) \]

\[ u-a_1=-49^\circ 52' 42'', \]

\[ a_2-u=317^\circ 37' 42''. \]

\[ \begin{aligned} \lg \cos a_2 &= 9\cdot 29024,\\ \lg \sin (u-a_1) &= 9\cdot 88348_{\text{n}},\\ \operatorname{c}\lg \sin a_1 &= 0\cdot 63181,\\ \operatorname{c}\lg \sin (a_2-u) &= 0\cdot 17138_{\text{n}}, \end{aligned} \]

\[ \lg \operatorname{tg}\varphi=9\cdot 97691. \]

\[ \varphi_0=43^\circ 28' 39''. \]

3. Determination of signs.

The sign of \((\operatorname{tg}\zeta_2 \cos a_2)\) is the sign of \((\operatorname{tg}105^\circ 30' \cos 281^\circ 15')<0\).

The sign of \(\cos u\) is the sign of \([\cos(-36^\circ 22' 42'')]>0\).

In this particular case we apply the formula:

\[ \operatorname{tg}\varphi= \frac{\dfrac{\sin(u-a_1)}{\sin a_1}} {\dfrac{\sin(a_2-u)}{\cos a_2}}. \]

Therefore:

\[ \text{sign }\operatorname{tg}\varphi= \frac{ \dfrac{(-)}{(+)} }{ \dfrac{(-)}{(+)} } = \frac{(-)}{(-)}. \]

Consequently, the required azimuth is

\[ \varphi=180^\circ+43^\circ 28' 39''=223^\circ 28' 39''. \]

We take \(\varphi=223^\circ 30'\).

II.

We have:

\[ \begin{array}{c|c} a_1=193^\circ 30'. & a_2=101^\circ 15'.\\ \zeta_1=114^\circ 15'. & \zeta_2=74^\circ 30'. \end{array} \]

a. Formulas II, § 6.

In the present case \(\zeta_1\) and \(\zeta_2\) are not equal; therefore:

\[ \operatorname{tg}\varphi=\frac{-S}{-T}. \]

(1). Determination of \(S\).

\[ S=M_1\frac{\sin(\zeta_1+\varepsilon_1)}{\cos\varepsilon_1} \]

\[ \operatorname{tg}\varepsilon_1=\frac{N_1}{M_1} \]

\[ M_1=-\cos a_1\cos\zeta_2 \]

\[ N_1=\sin\zeta_2\cos a_2. \]

1. Determination of \(\lg M_1\) and \(\lg N_1\).

\[ \begin{aligned} \lg(-\cos a_1) &= 9 \cdot 98783\\ \lg\cos \zeta_2' &= 9 \cdot 42690\\ \hline \lg M_1 &= 9 \cdot 41473\\ c'\lg M_1 &= 0 \cdot 58527. \end{aligned} \]

\[ \begin{aligned} \lg\sin \zeta_2 &= 9 \cdot 98391\\ \lg\cos a_2 &= 9 \cdot 29024_{n}\\ \hline \lg N_1 &= 9 \cdot 27415_{n}. \end{aligned} \]

1. Determination of \(\varepsilon_1\) and \((\zeta_1+\varepsilon_1)\).

\[ \begin{aligned} \lg N_1 &= 9 \cdot 27415_{n}\\ c'\lg M_1 &= 0 \cdot 58527\\ \hline \lg\tg \varepsilon_1 &= 9 \cdot 85942_{n} \end{aligned} \]

\[ \varepsilon_1 = -35^\circ 53' 04'' \]

\[ \zeta_1+\varepsilon_1 = 78^\circ 21' 56''. \]

1. Determination of \(\lg S\).

\[ \begin{aligned} \lg M_1 &= 9 \cdot 41473\\ \lg\sin(\zeta_1+\varepsilon_1) &= 9 \cdot 99099\\ c'\lg\cos\varepsilon_1 &= 0 \cdot 09141\\ \hline \lg S &= 9 \cdot 49713. \end{aligned} \]

(II). Determination of \(T\).

\[ T=L_2\frac{\cos(\zeta_2-\varepsilon_2)}{\cos\varepsilon_2} \]

\[ \tg\varepsilon_2=\frac{K_2}{L_2} \]

\[ K_2=-\sin a_2\cos\zeta_1' \]

\[ L_2=\sin\zeta_1\sin a_1 \]

1. Determination of \(\lg K_2\) and \(\lg L_2\).

\[ \begin{aligned} \lg(-\sin a_2) &= 9 \cdot 99157_{n}\\ \lg\cos \zeta_1' &= 9 \cdot 61354_{n}\\ \hline \lg K_2 &= 9 \cdot 60511. \end{aligned} \]

\[ \begin{aligned} \lg\sin\zeta_1 &= 9 \cdot 95988\\ \lg\sin a_1 &= 9 \cdot 36819_{n}\\ \hline \lg L_2 &= 9 \cdot 32807_{n}\\ c'\lg L_2 &= 0 \cdot 67193_{n}. \end{aligned} \]

1. Determination of \(\varepsilon_2\) and \((\zeta_2-\varepsilon_2)\).

\[ \begin{aligned} \lg K_2 &= 9\cdot 60511,\\ c'\lg L_2 &= 0\cdot 67193_{n},\\ \hline \lg \tg \varepsilon_2 &= 0\cdot 27704_{n} \end{aligned} \]

\[ \varepsilon_2=-62^\circ 08'54''. \]

\[ \zeta_2-\varepsilon_2=136^\circ 38'54''. \]

1. Determination of \(\lg T\).

\[ \begin{aligned} \lg L_2 &= 9\cdot 32807_{n},\\ \lg \cos(\zeta_2-\varepsilon_2) &= 9\cdot 86163_{n},\\ c'\lg \cos \varepsilon_2 &= 0\cdot 33052,\\ \hline \lg T &= 9\cdot 52022. \end{aligned} \]

\[ c'\lg T=0\cdot 47978. \]

(III). Determination of \(\zeta_0\).

\[ \begin{aligned} \lg S &= 9\cdot 49713,\\ c'\lg T &= 0\cdot 47978,\\ \hline \lg \tg \zeta_0 &= 9\cdot 97691. \end{aligned} \]

\[ \varphi_0=43^\circ 28'39''. \]

\[ \varphi=180^\circ+\zeta_0=223^\circ 28'39''. \]

We take \(\varphi=223^\circ 30'\).

b. Formulas IV § 6.

\[ \tg \zeta=\frac{-S}{-T}. \]

(I). Determination of \(S\).

\[ S=q_1\cos(Q_1-\zeta_1) \]

\[ q_1=\frac{M_1}{\sin Q_1}=\frac{N_1}{\cos Q_1} \]

\[ \tg Q_1=\frac{M_1}{N_1}. \]

1. On the basis of the preceding solution we have:

\[ \begin{aligned} \lg M_1 &= 9\cdot 41473,\\ \lg N_1 &= 9\cdot 27415_{n},\\ c'\lg N_1 &= 0\cdot 72585_{n}. \end{aligned} \]

1. Determination of \(Q_1\).

\[ \begin{aligned} \lg M_1 &= 9\cdot 41473,\\ c'\lg N_1 &= 0\cdot 72585_{n},\\ \hline \lg \tg Q_1 &= 0\cdot 14058_{n}. \end{aligned} \]

\[ Q_1=-54^\circ 06'56''. \]

1. Determination of \(\lg q_1\).

\[ \begin{array}{c|c} \lg M_1 = 9\cdot 41473 & \lg N_1 = 9\cdot 27415_n \\ c'\lg\sin Q_1 = 0\cdot 09141_n & c'\lg\cos Q_1 = 0\cdot 23199 \\ \hline \lg q_1 = 9\cdot 50614_n. & \lg q_1 = 0\cdot 50614_n. \end{array} \]

1. Determination of \((Q_1-\zeta_1)\) and \(\lg S\).

\[ Q_1-\zeta_1 = 168^\circ 21'56''. \]

\[ \begin{aligned} \lg q_1 &= 9\cdot 50614_n \\ \lg\cos(Q_1-\zeta) &= 9\cdot 99099_n \\ \hline \lg S &= 9\cdot 49713. \end{aligned} \]

(II). Determination of \(\lg T\).

\[ T=q_2\cos(Q_2-\zeta_2) \]

\[ q_2=\frac{K_2}{\sin Q_2}=\frac{L_2}{\cos Q_2} \]

\[ \tg Q_2=\frac{K_2}{L_2}. \]

1. On the basis of solution “a” we have:

\[ Q_2=-62^\circ 08'54''. \]

1. Determination of \(\lg q_2\).

\[ \begin{array}{c|c} \lg K_2 = 9\cdot 60511 & \lg L_2 = 9\cdot 32807_n \\ c'\lg\sin Q_2 = 0\cdot 05347_n & c'\lg\cos Q_2 = 0\cdot 33052 \\ \hline \lg q_2 = 9\cdot 65858_n. & \lg q_2 = 9\cdot 65859_n. \end{array} \]

1. Determination of \((Q_2-\zeta_2)\) and \(\lg T\).

\[ Q_2-\zeta_2=-136^\circ 38'54''. \]

\[ \begin{aligned} \lg q_2 &= 9\cdot 65859_n \\ \lg\cos(Q_2-\zeta_2) &= 9\cdot 86163_n \\ \hline \lg T &= 9\cdot 52022 \\ c'\lg T &= 0\cdot 47978. \end{aligned} \]

(III). Determination of \(\varphi_0\).

\[ \begin{aligned} \lg S &= 9\cdot 49713 \\ c'\lg T &= 0\cdot 47978 \\ \hline \lg\tg\varphi_0 &= 9\cdot 97691. \end{aligned} \]

\[ \varphi_0=43^\circ 28'39''. \]

\[ \varphi=223^\circ 28'39''. \]

We take

\[ \varphi=223^\circ 30'. \]

III.

Formula (III) § 7.

\[ \tg \Phi = \tg\left(\varphi-\frac{a_1+a_2}{2}\right) = \ctg \frac{a_1-a_2}{2}\cdot \frac{\sin(\zeta_2-\zeta_1)}{\sin(\zeta_2+\zeta_1)}. \]

Let us denote one of the traces (it makes no difference which) by the symbol \(P\), the other by \(Q\).

a) We have:

\[ \begin{array}{c|c} a_P=13^\circ 30' & a_Q=281^\circ 15'\\ h_P=+24^\circ 15' & h_Q=-15^\circ 30' \end{array} \]

Having constructed the horizontal projections of the traces and, noting that the two traces are unequal, we draw (see § 5, 4) an arcuate arrow to the right (clockwise) outside the angle corresponding to the adit; we conclude that the direction \((OP)\) must be denoted by the number 1, and the direction \((OQ)\) by the number 2.

Therefore:

\[ \begin{array}{c|c} a_1=13^\circ 30' & a_2=281^\circ 15'\\ \zeta_1=65^\circ 45' & \zeta_2=105^\circ 30' \end{array} \]

Using formula (III), we have:

\[ \frac{1}{2}(a_1+a_2)=147^\circ 22'30'' \]

\[ \frac{1}{2}(a_1-a_2)=-133^\circ 52'30'' \]

\[ \zeta_2-\zeta_1=39^\circ 45'00'' \]

\[ \zeta_2+\zeta_1=171^\circ 15'00'' \]

\[ \lg \ctg \frac{1}{2}(a_1-a_2)=9\cdot98294 \]

\[ \lg \sin(\zeta_2-\zeta_1)=9\cdot80580 \]

\[ c'\lg \sin(\zeta_2+\zeta_1)=0\cdot81780 \]

\[ \lg \tg \Phi=0\cdot60654. \]

\[ \Phi_0=76^\circ 06'08'' \]

Since the zenith distances \(\zeta_1\) and \(\zeta_2\) of the traces are unequal, in order to determine the signs of the numerator and denominator of the expression

\[ \tg\left(\varphi-\frac{a_1+a_2}{2}\right) \]

we use formula (II), § 7.

We have:

\[ \operatorname{sign}\tg\left(\varphi-\frac{a_1+a_2}{2}\right) = \frac{(-)(-)(+)}{(-)(-)(+)} = \frac{+}{+} \]

and, consequently,

\[ \Phi=\Phi_0=76^\circ 06'08''. \]

According to the formula

\[ \varphi=\Phi+\frac{1}{2}(a_1+a_2) \]

the required azimuth of the line of dip is

\[ \varphi=223^\circ 28' 38''. \]

We take

\[ \varphi=223^\circ 30'. \]

b) We have:

\[ \begin{array}{c|c} a_P=193^\circ 30' & a_Q=281^\circ 15'\\ h_P=-24^\circ 15' & h_Q=-15^\circ 30' \end{array} \]

In the present case both traces are the same \((h_{P,Q}<0)\), and, in accordance with § 5, the arc arrow is inside the angle corresponding to the borehole; the directions \((OP)\) and \((OQ)\) must be designated respectively by the numerals 1 and 2.

Therefore:

\[ \begin{array}{c|c} a_1=193^\circ 30' & a_2=281^\circ 15'\\ \zeta_1=114^\circ 15' & \zeta_2=105^\circ 30' \end{array} \]

We have:

\[ \frac{1}{2}(a_1+a_2)=237^\circ 22' 30'' \]

\[ \frac{1}{2}(a_1-a_2)=-43^\circ 52' 30'' \]

\[ \zeta_2-\zeta_1=-8^\circ 45' 00'' \]

\[ \zeta_2+\zeta_1=219^\circ 45' 00''. \]

\[ \lg \operatorname{ctg}\frac{1}{2}(a_1-a_2)=0\cdot01706_n \]

\[ \lg \sin(\zeta_2-\zeta_1)=9\cdot18220_n \]

\[ c'\lg \sin(\zeta_2+\zeta_1)=0\cdot19420_n \]

\[ \lg \tg \Phi=9\cdot39346_n. \]

\[ \Phi_0=13^\circ 53' 52''. \]

The zenith distances \(\zeta_1\) and \(\zeta_2\) of the traces are the same; therefore, to determine the signs of the numerator and denominator of the expression

\[ \tg\left(\varphi-\frac{a_1+a_2}{2}\right) \]

we use formula (I) of § 7.

We have:

\[ \text{sign }\tg\Phi=\frac{(+)(-)}{(-)(-)}=\frac{-}{+}; \]

and, consequently,

\[ \Phi=360^\circ-13^\circ 53' 52''=346^\circ 06' 08''. \]

According to the formula,

\[ \varphi=\Phi+\frac{1}{2}(a_1+a_2), \]

the required azimuth of the line of dip is

\[ \varphi=223^\circ 28' 38''. \]

We take

\[ \varphi=223^\circ 30'. \]

Remark. Geometrically, an azimuth equal to \(360^\circ n+\varphi\) is identical to an azimuth equal to \(\varphi\) (\(n\) is an integer).

c) We have:

\[ \begin{array}{c|c} a_P=13^\circ 30' & a_Q=101^\circ 15'\\ h_P=+24^\circ 15' & h_Q=+15^\circ 30' \end{array} \]

In the present case both traces are of the same kind \((h_{P,Q}>0)\), and, in accordance with § 5, the arc arrow—inside the angle corresponding to the shaft—the directions \((OP)\) and \((OQ)\) must be denoted respectively by the numbers 1 and 2.

Therefore:

\[ \begin{array}{c|c} a_1=13^\circ 30' & a_2=101^\circ 15'\\ \zeta_1=65^\circ 45' & \zeta_2=74^\circ 30' \end{array} \]

We have:

\[ \frac{1}{2}(a_1+a_2)=57^\circ 22' 30'' \]

\[ \frac{1}{2}(a_1-a_2)=-43^\circ 52' 30'' \]

\[ \zeta_2-\zeta_1=8^\circ 45' 00'' \]

\[ \zeta_2+\zeta_1=140^\circ 15' 00''. \]

\[ \lg \operatorname{ctg}\frac{1}{2}(a_1-a_2)=0\cdot01706_n \]

\[ \lg \sin(\zeta_2-\zeta_1)=9\cdot18220 \]

\[ c'\lg \sin(\zeta_2+\zeta_1)=0\cdot19420 \]

\[ \lg \operatorname{tg}\Phi=9\cdot39346_n \]

\[ \Phi_0=13^\circ 53' 52''. \]

The zenith distances \(\zeta_1\) and \(\zeta_2\) of the traces are the same; therefore, to determine the signs of the numerator and denominator of the expression \(\operatorname{tg}\Phi\), we use formula (I), § 7.

We have:

\[ \text{sign }\operatorname{tg}\Phi= \frac{(+)(+)}{(-)(+)} =\frac{+}{-} \]

and, consequently,

\[ \Phi=180^\circ-13^\circ 53' 52''=166^\circ 06' 08''. \]

According to the formula

\[ \varphi = \Phi + \frac{1}{2}(a_1 + a_2), \]

the required azimuth of the dip line is

\[ \varphi = 223^\circ 28' 38''. \]

We take

\[ \varphi = 223^\circ 30'. \]

d) We have:

\[ \begin{array}{c|c} a_p = 101^\circ 15' & a_Q = 193^\circ 30' \\ h_p = +15^\circ 30' & h_Q = -24^\circ 15' \end{array} \]

In the present case the two traces are unlike \((h_{P,Q} \gtrless 0)\), and, in accordance with § 5, the arc arrow—outside the angle corresponding to the pit; the directions \((OP)\) and \((OQ)\) must be denoted respectively by the numbers 2 and 1.

Therefore:

\[ \begin{array}{c|c} a_1 = 193^\circ 30' & a_2 = 101^\circ 15' \\ \zeta_1 = 114^\circ 15' & \zeta_2 = 74^\circ 30' \end{array} \]

We have:

\[ \frac{1}{2}(a_1 + a_2) = 147^\circ 22' 30'' \]

\[ \frac{1}{2}(a_1 - a_2) = 46^\circ 07' 30'' \]

\[ \zeta_2 - \zeta_1 = -39^\circ 45' 00'' \]

\[ \zeta_2 + \zeta_1 = 188^\circ 45' 00''. \]

\[ \lg \operatorname{ctg} \frac{1}{2}(a_1 - a_2) = 9 \cdot 98294 \]

\[ \lg \sin(\zeta_2 - \zeta_1) = 9 \cdot 80580_n \]

\[ c'\lg \sin(\zeta_2 + \zeta_1) = 0 \cdot 81780_n \]

\[ \lg \operatorname{tg}\Phi = 0 \cdot 60654. \]

\[ \Phi_0 = 76^\circ 06' 08''. \]

The zenith distances of the traces are unlike; therefore, to determine the signs of the numerator and denominator of the expression \(\operatorname{tg}\Phi\), we use formula (II) of § 7.

We have:

\[ \text{sign } \operatorname{tg}\Phi = \frac{(-)(+)(-)}{(-)(+)(-)} = \frac{+}{+} \]

and, consequently,

\[ \Phi = \Phi_0 = 76^\circ 06' 08''. \]

According to the formula

\[ \varphi = \Phi + \frac{1}{2}(a_1 + a_2) \]

the sought azimuth of the dip line

\[ \varphi = 223^\circ 28' 38''. \]

We take

\[ \varphi = 223^\circ 30'. \]

The above four examples (III: a, b, c, d), as is easy to see, refer to one and the same plane of the bed and have been given by us only for mastering the order of designation of traces in connection with the use of the formulae: (III), (II), (I) § 7.

B. Examples of determining the zenith distance \(\zeta\) of the dip line.

I.

We have:

\[ \begin{array}{rl|rl} a_1 &= 13^\circ 30' & a_2 &= 281^\circ 15' \\ \zeta_1 &= 65^\circ 45' & \zeta_2 &= 105^\circ 30', \end{array} \]

\[ \zeta_1 - \zeta_2 = -39^\circ 45' \]

\[ \zeta_1 + \zeta_2 = 171^\circ 15' \]

\[ \frac{1}{2}(a_1-a_2)=-133^\circ 52' 30''. \]

1. Determination of \(\Delta\) (see II § 7).

\[ \tg \Delta = \frac{\sin(\zeta_1-\zeta_2)}{\sin(\zeta_1+\zeta_2)} \ctg \frac{1}{2}(a_1-a_2) \]

\[ \lg \sin(\zeta_1-\zeta_2)=9 \cdot 80580_{n} \]

\[ c'\lg \sin(\zeta_1+\zeta_2)=0 \cdot 81780. \]

\[ \lg \ctg \frac{1}{2}(a_1-a_2)=9 \cdot 98294. \]

\[ \lg \tg \Delta=0 \cdot 60654_{n} \]

\[ \Delta_1=-76^\circ 06' 08'' \]

\[ \Delta_2=103^\circ 53' 52'' \]

In the subsequent calculations we shall take

\[ \Delta_1=-76^\circ 06' 08''. \]

1. Determination of \(x\).

\[ x= \frac{\sin(\zeta_1+\zeta_2)} {\cos\zeta_1\cos\zeta_2\cos\Delta_1} \sin \frac{1}{2}(a_1-a_2). \]

\[ \lg \sin(\zeta_1+\zeta_2)=9 \cdot 18220 \]

\[ c'\lg \cos\zeta_1=0 \cdot 38646 \]

\[ c'\lg \cos\zeta_2=0 \cdot 57310_{n} \]

\[ c'\lg \cos\Delta_1=0 \cdot 61945 \]

\[ \lg \sin \frac{1}{2}(a_1-a_2)=9 \cdot 85785_{n} \]

\[ \lg x=0 \cdot 61906. \]

3. Determination of \(\zeta\).

\[ \tg(180^\circ-\zeta)= \frac{\tg\zeta_1\,\tg\zeta_2\,\sin(a_2-a_1)}{x}. \]

\[ \begin{aligned} \lg\tg\zeta_1&=0\cdot34634\\ \lg\tg\zeta_2&=0\cdot55701_{n}\\ \lg\sin(a_2-a_1)&=9\cdot99967_{n}\\ c'\lg x&=9\cdot38094\\ \hline \lg\tg(180^\circ-\zeta)&=0\cdot28396 \end{aligned} \]

\[ 180^\circ-\zeta=62^\circ31'25'' \]

\[ \zeta=117^\circ28'35''. \]

We take

\[ \zeta=117^\circ30'. \]

II.

We have:

\[ \begin{array}{r@{\quad}c@{\quad}l} a_1=193^\circ30' & \big| & a_2=101^\circ15'\\ \zeta_1=114^\circ15' & \big| & \zeta_2=74^\circ30'. \end{array} \]

In the present case \(\zeta_1\) and \(\zeta_2\) are unequal; we shall apply formulas (6), (7) of § 7.

\[ x=\left|\frac{\sin(\zeta_1+\zeta_2)}{\cos\zeta_1\cos\zeta_2}\cdot\frac{1}{\cos\varepsilon}\right| \]

\[ \tg^2\varepsilon= \left| \frac{\sin2\zeta_1\,\sin2\zeta_2\,\cos^2\frac{1}{2}(a_1-a_2)} {\sin^2(\zeta_1+\zeta_2)} \right| \]

\[ 2\zeta_1=228^\circ30';\quad 2\zeta_2=149^\circ;\quad \zeta_1+\zeta_2=188^\circ45';\quad \frac{1}{2}(a_1-a_2)=46^\circ07'30''. \]

1. Determination of \(\varepsilon\).

\[ \begin{aligned} \lg|\sin2\zeta_1|&=9\cdot87446\\ \lg\sin2\zeta_2&=9\cdot71184\\ 2\lg\cos\frac{1}{2}(a_1-a_2)&=9\cdot68156\\ c'\lg\sin^2(\zeta_1+\zeta_2)&=1\cdot63560\\ \hline \lg\tg^2\varepsilon&=0\cdot90346\\ \lg\tg\varepsilon&=0\cdot45173 \end{aligned} \]

\[ \varepsilon=70^\circ32'12''. \]

2. Determination of \(x\).

\[ \begin{aligned} \lg|\sin(\zeta_1+\zeta_2)|&=9\cdot18220\\ c'\lg|\cos\zeta_1|&=0\cdot38646\\ c'\lg\cos\zeta_2&=0\cdot57310\\ c'\lg\cos\varepsilon&=0\cdot47730\\ \hline \lg x&=0\cdot61906. \end{aligned} \]

Subsequently the solution is identical to the preceding case.

III.

We have:

\[ \begin{array}{ll} a_1=13^\circ 30' & a_2=281^\circ 15'\\ \zeta_1=65^\circ 45' & \zeta_2=105^\circ 30'. \end{array} \]

Above we found that

\[ u=143^\circ 37'18'' \]

\[ \varphi=223^\circ 28'38''. \]

Further, we have:

\[ \begin{aligned} a_2-a_1&=267^\circ 45'00''\\ u-a_1&=130^\circ 07'18''\\ u-a_2&=-137^\circ 37'42''. \end{aligned} \]

Solution A (III, § 7).

1) We have:

\[ \lg \sqrt{m^2+n^2}=\lg n+c'\lg \cos\varphi. \]

a) Determination of \(\lg n\) and \(c'\lg \cos\varphi\).

\[ \begin{aligned} \lg \tg \zeta_2'&=0\cdot55701_n\\ \lg \sin(u-a_2)&=9\cdot82362_n\\ c'\lg \cos u&=0\cdot09414_n\\ \hline \lg n&=0\cdot47977_n \end{aligned} \]

\[ c'\lg \cos\varphi=0\cdot13928_n. \]

Therefore:

\[ \begin{aligned} \lg n&=0\cdot47977_n\\ c'\lg \cos\varphi&=0\cdot13928_n\\ \hline \lg \sqrt{m^2+n^2}&=0\cdot61905. \end{aligned} \]

\[ c'\lg \sqrt{m^2+n^2}=9\cdot38095. \]

b) Determination of \(\zeta\).

\[ \begin{aligned} \lg \tg \zeta_1&=0\cdot34634\\ \lg \tg \zeta_2&=0\cdot55701_n\\ \lg \sin(a_2-a_1)&=9\cdot99967_n\\ c'\lg \sqrt{m^2+n^2}&=9\cdot38095\\ \hline \lg \tg(180^\circ-\zeta)&=0\cdot28397. \end{aligned} \]

\[ 180^\circ-\zeta=62^\circ31'27'' \]

\[ \zeta=117^\circ28'33'' \]

We take

\[ \zeta=117^\circ30'. \]

2) We have:

\[ \operatorname{lg}\sqrt{m^{2}+n^{2}}=\operatorname{lg}m+c'\operatorname{lg}\sin\varphi . \]

a) Determination of \(\operatorname{lg}m\) and \(c'\operatorname{lg}\sin\varphi\).

\[ \begin{aligned} \operatorname{lg}\operatorname{tg}\zeta_{2}&=0\cdot55701_{n}\\ \operatorname{lg}\cos a_{2}&=9\cdot29024\\ \operatorname{lg}\sin(u-a_{1})&=9\cdot88348\\ c'\operatorname{lg}\sin a_{1}&=0\cdot63181\\ c'\operatorname{lg}\cos u&=0\cdot09414_{n}\\ \hline \operatorname{lg}m&=0\cdot45668. \end{aligned} \]

\[ c'\operatorname{lg}\sin\varphi=0\cdot16237_{n}. \]

We take (see § 7, III, A)

\[ c'\operatorname{lg}\sin\varphi_{1}=0\cdot16237. \]

Therefore:

\[ \begin{aligned} \operatorname{lg}m&=0\cdot45668\\ c'\operatorname{lg}\sin\varphi_{1}&=0\cdot16237\\ \hline \operatorname{lg}\sqrt{m^{2}+n^{2}}&=0\cdot61905. \end{aligned} \]

The further calculations are identical to case 1).

Solution B (III, § 7).

Computations by formulas (I), (II).

Formula (I). We have:

\[ \begin{aligned} \operatorname{lg}\operatorname{tg}\zeta_{1}&=0\cdot34634\\ \operatorname{lg}\sin(a_{2}-a_{1})&=9\cdot99967_{n}\\ \operatorname{lg}\cos\varphi&=9\cdot86072_{n}\\ \operatorname{lg}\cos u&=9\cdot90586_{n}\\ c'\operatorname{lg}\sin(u-a_{2})&=0\cdot17138_{n}\\ \hline \operatorname{lg}\operatorname{tg}(180^\circ-\zeta)&=0\cdot28397 \end{aligned} \]

Further, see b) of solution A.

Formula (II). We have:

\[ \begin{aligned} \operatorname{lg}\operatorname{tg}\zeta_{1}&=0\cdot34634\\ \operatorname{lg}\sin(a_{2}-a_{1})&=9\cdot99967_{n}\\ \operatorname{lg}\sin\varphi&=9\cdot83763_{n}\\ \operatorname{lg}\sin a_{1}&=9\cdot36819\\ \operatorname{lg}\cos u&=9\cdot90586_{n}\\ c'\operatorname{lg}\cos a_{2}&=0\cdot70976\\ c'\operatorname{lg}\sin(u-a_{1})&=0\cdot11652\\ \hline \operatorname{lg}\operatorname{tg}(180^\circ-\zeta)&=0\cdot28397_{n} \end{aligned} \]

Dropping the symbol \(n\), we obtain:

\[ \operatorname{lg}\operatorname{tg}(180^\circ-\zeta)=0\cdot28397. \]

Further, see b) of solution A.

IV.

Formula (see 2 § 8):

\[ \operatorname{tg}(180^\circ-\zeta)= \frac{\operatorname{tg}\zeta_1\,\operatorname{tg}\zeta_2\,\sin(a_2-a_1)} {\operatorname{tg}\zeta_1\sin(a_1-\varphi)-\operatorname{tg}\zeta_2\sin(a_2-\varphi)} \qquad . . . . . . (1) \]

We have:

\[ \begin{array}{rlrl} a_1&=193^\circ 30' \quad & a_2&=101^\circ 15'\\ \zeta_1&=114^\circ 15' \quad & \zeta_2&=74^\circ 30'. \end{array} \]

\[ \begin{aligned} \varphi&=223^\circ 28'38'',\\ a_1-\varphi&=-29^\circ 58'38'',\\ a_2-\varphi&=-122^\circ 13'38''. \end{aligned} \]

In the present case:

\[ A=\operatorname{tg}\zeta_1\sin(a_1-\varphi)>0. \]

\[ B=\operatorname{tg}\zeta_2\sin(a_2-\varphi)<0, \]

therefore, they are of different signs.

We shall first determine the logarithms: \(\log A\) and \(\log B\).

We have:

\[ \begin{array}{rl|rl} \log\operatorname{tg}\zeta_1&=0\cdot34634_n & \lg\operatorname{tg}\zeta_2&=0\cdot55701 \\ \lg\sin(a_1-\varphi)&=9\cdot69867_n & \lg\sin(a_2-\varphi)&=9\cdot92734_n \\ \hline \log A&=0\cdot04501 & \log B&=0\cdot48435_n. \end{array} \]

We see directly that

\[ A<|B|. \]

Therefore, to determine the logarithm of the denominator of the right-hand side of formula (1), we apply the formulas [II, b), § 9].

Let us take, for example, the formulas:

1. Formula (34) § 9.

\[ A-B=-2B\sin^2\frac{1}{2}\varepsilon \]

\[ \varepsilon=\arccos\frac{A}{B}. \]

Determination of \(\varepsilon\).

\[ \begin{array}{r} \lg A=0\cdot04501\\ -\ \lg B=0\cdot48435_n\\ \hline \lg\cos\varepsilon=9\cdot56066_n. \end{array} \]

\[ \varepsilon_1=111^\circ19'24'';\quad \varepsilon_2=248^\circ40'36''. \]

\[ \frac{1}{2}\varepsilon_1=55^\circ39'42''. \]

Determination of $\log(A-B)$.

\[ \begin{aligned} \log(-2)&=0\cdot 30103_{n}\\ \lg B&=0\cdot 48435_{n}\\ 2\lg\sin \frac{1}{2}\varepsilon&=9\cdot 83366\\ \hline \lg(A-B)&=0\cdot 61904. \end{aligned} \]

In what follows, the solution is identical to case 1) of Example III.

2. Formula (36), § 9:

\[ A-B=-2B\cos^{2}\frac{1}{2}\varepsilon \]

\[ \varepsilon=\arccos\frac{A}{|B|}. \]

Determination of $\varepsilon$.

\[ \begin{aligned} \log A&=0\cdot 04501\\ -\lg|B|&=0\cdot 48435\\ \hline \lg\cos\varepsilon&=9\cdot 56066. \end{aligned} \]

\[ \varepsilon=68^\circ40'36''. \]

\[ \frac{1}{2}\varepsilon=34^\circ20'18''. \]

Determination of $\log(A-B)$.

\[ \begin{aligned} \lg(-2)&=0\cdot 30103_{n}\\ \lg B&=0\cdot 48435_{n}\\ 2\lg\cos\frac{1}{2}\varepsilon&=9\cdot 83366\\ \hline \lg(A-B)&=0\cdot 61904. \end{aligned} \]

3. Formula (40), § 9.

\[ A-B=\frac{A}{\cos^{2}\varepsilon} \]

\[ \varepsilon=\operatorname{arctg}\left(\pm\sqrt{\frac{|B|}{A}}\right). \]

Determination of $\varepsilon$.

\[ \begin{aligned} \lg|B|&=0\cdot 48435\\ \lg A&=0\cdot 04501\\ \hline 2\lg\operatorname{tg}\varepsilon&=0\cdot 43934. \end{aligned} \]

\[ \lg\operatorname{tg}\varepsilon=0\cdot 21967 \]

\[ \varepsilon_{1}=58^\circ54'33'';\qquad \varepsilon_{2}=238^\circ54'33''. \]

Determination of \(\log(A-B)\).

\[ \begin{aligned} \lg A &= 0\cdot 04501\\ -\,2\lg\cos\varepsilon &= 9\cdot 42596\\ \hline \lg(A-B) &= 0\cdot 61905. \end{aligned} \]

V.

Formula \([(8),\,3,\,\S 8]\).

\[ \tg \zeta'=\tg \zeta_{1,2}\cos(\varphi-a_{1,2}). \]

We have:

\[ \begin{array}{c|c} a_1=13^\circ 30' & a_2=281^\circ 15'\\ \zeta_1=65^\circ 45' & \zeta_2=105^\circ 30' \end{array} \]

\[ \begin{aligned} \varphi &= 223^\circ 28'38'',\\ \varphi-a_1 &= 209^\circ 58'38'',\\ \varphi-a_2 &= -57^\circ 46'22''. \end{aligned} \]

\[ \begin{array}{c|c} \lg\tg\zeta_1=0\cdot 34634 & \lg\tg\zeta_2=0\cdot 55701_n\\ \lg\cos(\varphi-a_1)=9\cdot 93763_n & \lg\cos(\varphi-a_2)=9\cdot 72696\\ \hline \lg\tg\zeta'=0\cdot 28397_n & \lg\tg\zeta'=0\cdot 28397_n \end{array} \]

and the value satisfying the problem is

\[ \zeta'=117^\circ 28'33''. \]

We take:

\[ \zeta'=117^\circ 30'. \]

§ 17. Solutions by the method of spherical trigonometry.

Describe, from the point \(O\) (the point of intersection of the traces \(OI\) and \(OII\)), as center, a sphere of arbitrary radius \(R\). Let us represent this sphere on the horizontal plane \(H\), projecting it orthogonally onto the latter. The sphere will be represented by the circle \(ABCDA\) of radius \(R\) (Fig. 16).

Next, the plane of the bed, in its intersection with the surface of the sphere, will give a great circle, which in projection onto the plane \(H\) will be represented by the ellipse \(FBF_1DF\).

The points of intersection of the traces—the vectors \(OI\), \(OII\), and the line of dip \(F_1OF\)—with the surface of the sphere we shall denote, respectively, by the symbols \(I\), \(II\), \(F_1\), \(F\).

The three vertical planes corresponding to the traces—the vectors and the line of dip—and the plane of the bed, in their intersection with the surface of the sphere, form spherical triangles; their sides, in projection onto the plane \(H\), will give, respectively, three straight lines and the arcs of an ellipse.

Fig. 16.

Let us denote the azimuths and zenith distances of the traces—vectors and of the line of dip, respectively, by

\[ (a_1,\zeta'_1);\ (a_2,\zeta'_2);\ (\varphi,\zeta). \]

From the spherical triangles OAFBIO and OAFDИO, right-angled at the point F, respectively, we obtain:

\[ \operatorname{tg}\zeta=\operatorname{tg}\zeta'_1\cos(a_1-\varphi) =\operatorname{tg}\zeta'_1\cos(\varphi-a_1) \tag{1} \]

\[ \operatorname{tg}\zeta=\operatorname{tg}\zeta'_2\cos(\varphi-a_2). \tag{2} \]

From formulas (1) and (2) it is easy to determine the required \(\varphi\) and \(\zeta\).

Determination of the azimuth \(\varphi\).

First represent (1) and (2) in the form:

\[ \operatorname{tg}\zeta\,\operatorname{ctg}\zeta'_1 =\cos\varphi\cos a_1+\sin\varphi\sin a_1, \tag{3} \]

\[ \operatorname{tg}\zeta\,\operatorname{ctg}\zeta'_2 =\cos\varphi\cos a_2+\sin\varphi\sin a_2. \tag{4} \]

Multiplying equations (3) and (4), respectively, by \(\cos a_2\), \(\cos a_1\), and subtracting the first from the second, we obtain:

\[ \operatorname{tg}\zeta\left(\operatorname{ctg}\zeta'_2\cos a_1 -\operatorname{ctg}\zeta'_1\cos a_2\right) =\sin\varphi(\cos a_1\sin a_2-\sin a_1\cos a_2), \]

whence

\[ \sin\varphi =\operatorname{tg}\zeta\, \frac{\operatorname{tg}\zeta'_2\cos a_2-\operatorname{tg}\zeta'_1\cos a_1} {\operatorname{tg}\zeta'_1\operatorname{tg}\zeta'_2\sin(a_1-a_2)}. \tag{5} \]

Multiplying further equations (3) and (4), respectively, by \(\sin a_2\), \(\sin a_1\), and subtracting the first from the second, we obtain:

\[ \operatorname{tg}\zeta\left(\operatorname{ctg}\zeta'_2\sin a_1 -\operatorname{ctg}\zeta'_1\sin a_2\right) =\cos\varphi(\sin a_1\cos a_2-\cos a_1\sin a_2), \]

whence

\[ \cos\varphi =\operatorname{tg}\zeta\, \frac{\operatorname{tg}\zeta'_1\sin a_1-\operatorname{tg}\zeta'_2\sin a_2} {\operatorname{tg}\zeta'_1\operatorname{tg}\zeta'_2\sin(a_1-a_2)}. \tag{6} \]

Let us assume for the moment that in formulas (5) and (6) the zenith distance \(\zeta\) is known; we see directly that formulas (5) and (6), together, namely vectorially, determine the required azimuth \(\varphi\).

But \(\zeta\) is unknown; however, it is easy to eliminate it by determining \(\varphi\) as a function of tangents of \(a\).

Namely:

\[ \operatorname{tg}\varphi =\frac{\sin\varphi}{\cos\varphi} = \frac{ \operatorname{tg}\zeta\, \dfrac{\operatorname{tg}\zeta'_2\cos a_2-\operatorname{tg}\zeta'_1\cos a_1} {\operatorname{tg}\zeta'_1\operatorname{tg}\zeta'_2\sin(a_1-a_2)} }{ \operatorname{tg}\zeta\, \dfrac{\operatorname{tg}\zeta'_1\sin a_1-\operatorname{tg}\zeta'_2\sin a_2} {\operatorname{tg}\zeta'_1\operatorname{tg}\zeta'_2\sin(a_1-a_2)} } \tag{7} \]

— a formula that determines \(\varphi\) vectorially.

It represents the ratio of the trigonometric sine to the trigonometric cosine; consequently, together the signs of the numerator and denominator determine precisely the quadrant corresponding to the required azimuth \(\varphi\).

Therefore the further transformations of formula (7) must be such as not to change the signs of the numerator and denominator.

Proceeding further, we note that the angle \(\zeta\) satisfies the condition:

\[ \frac{\pi}{2}<\zeta<\pi \]

and therefore \(\operatorname{tg}\zeta<0\).

We shall now write formula (7) as follows:

\[ \operatorname{tg}\varphi= \frac{ (-\operatorname{tg}\zeta)\, \dfrac{\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1} {\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)} }{ (-\operatorname{tg}\zeta)\, \dfrac{\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2} {\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)} }. \]

Denoting

\[ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)=\kappa, \]

and observing that, by condition, \((-\operatorname{tg}\zeta)>0\), formula (7) may be represented in two ways:

\[ \operatorname{tg}\varphi= \frac{\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1} {\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2} \ldots\ldots\ldots\ldots (A) \]

\[ \operatorname{tg}\varphi= \frac{-(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)} {-(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)} \ldots\ldots\ldots\ldots (B) \]

Formula (A) corresponds to the assumption \(\kappa>0\), formula (B) to the assumption \(\kappa<0\).

Obviously, it is simpler to adopt formula (A), assuming, consequently, \(\kappa>0\). The choice of the sign of “\(\kappa\),” as we showed earlier, determines the order of designation of the trace-vectors.

Remark. The derivation of the logarithmic formulas for \(\operatorname{tg}\varphi\) is identical to that indicated earlier.

Determination of the zenith distance \(\zeta\).

1. Having found \(\varphi\), the zenith distance \(\zeta\) is determined directly by formulas (1) and (2) [mutual check in computations].

2. The zenith distance \(\zeta\) is readily determined as a function depending only on the given arguments.

Namely, taking the relation

\[ \sin^2\varphi+\cos^2\varphi=1 \]

and substituting in the latter from formulas (5) and (6) the value of \(\varphi\), we obtain:

\[ \frac{\operatorname{tg}^2\zeta} {\operatorname{tg}^2\zeta_1\operatorname{tg}^2\zeta_2\sin^2(a_1-a_2)} \left[ (\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)^2+ \right. \]

\[ \left. +(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)^2 \right]=1, \]

whence, putting

\[ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)=\kappa>0 \]

and, by virtue of the condition, \(\operatorname{tg}\zeta<0\), we obtain:

\[ \operatorname{tg}\zeta= -\frac{ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1) }{ \sqrt{(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)^2+ (\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)^2} } \ldots (C) \]

—the formula which we previously obtained by other methods.

Remark. With another arrangement of the principal axis \(NS\), the formulas (A), (B), and (C), as the reader can readily verify, retain their form.

In deriving the formulas we assumed that \(\zeta_1\) and \(\zeta_2\) satisfy the condition:

\[ 0<\zeta_{1,2}<\frac{\pi}{2}. \]

It is easy, however, to show that, for any arrangement of the vectors—traces with respect to the plane \(H\), the form of the formulas does not change.

We shall confine ourselves to one example.

Indeed. Take one of the vectors, say \(OI(a_1,\zeta_1)\); change its direction to the directly opposite one. Then the new and the old azimuth and zenith distance are determined from the relations:

\[ \zeta_1=\pi-\zeta'_1;\quad a_1=a'_1\pm\pi. \]

Substituting the latter into formulas (A), (B), and (C), we see that the form of the formulas is invariant.

The formula for determining the zenith distance \(\zeta\) may be given in another form.

Take the spherical triangle \(OII O\) and write, on the one hand, the relation between the spherical angle at the point \(O\) and the amplitude of the triangle, and, on the other hand, the relation between the height \(OF_1\) and the amplitude of the same triangle.

Denoting the amplitude of the triangle by \(S\) and observing that the arc \(OF_1=\pi-\zeta\), we obtain:

\[ \sin(a_2-a_1)=\frac{2S}{\sin\zeta_1\sin\zeta_2}\ldots\ldots\ldots\ldots\ldots (8) \]

\[ \sin(\pi-\zeta)=\frac{2S}{\sin c},\ldots\ldots\ldots\ldots\ldots (9) \]

where

\[ c=\overset{\frown}{III}=\overset{\frown}{IF_1}+\overset{\frown}{F_1II}=c_1+c_2. \]

Determining \(S\) from (8) and (9) and comparing, we obtain:

\[ \sin\zeta_1\sin\zeta_2\sin(a_2-a_1)=\sin c\sin\zeta\ldots\ldots\ldots (10) \]

Let us now determine \(\sin c\).

We have:

\[ c=c_1+c_2 \]

\[ \sin c=\sin c_1\cos c_2+\cos c_1\sin c_2\ldots\ldots\ldots (11) \]

From the right spherical triangles at the point \(F_1\), \(OIF_1O\) and \(OIIF_1O\), we have:

\[ \cos\zeta_{1,2}=\cos(\pi-\zeta)\cos c_{1,2}, \]

whence

\[ \cos c_{1,2}=-\frac{\cos\zeta_{1,2}}{\cos\zeta}\ldots\ldots\ldots (12_{1,2}) \]

and, further,

\[ \sin c_1=\sin\zeta_1\sin(a_1-\varphi)\ldots\ldots\ldots (13) \]

\[ \sin c_2=-\sin\zeta_2\sin(a_2-\varphi)\ldots\ldots\ldots (14) \]

Substituting first the obtained values of \(\sin c_{1,2}\), \(\cos c_{1,2}\) from formulas (13), (14) and \((12_{1,2})\) into formula (11), and this latter into formula (10), we obtain:

\[ \sin\zeta_1 \sin\zeta_2 \sin(a_2-a_1)=\tg\zeta\left[-\sin\zeta_1\cos\zeta_2\sin(a_1-\varphi)+ \right. \]

\[ \left. +\sin\zeta_2\cos\zeta_1\sin(a_2-\varphi)\right], \]

whence

\[ \tg\zeta=-\frac{\tg\zeta_1\,\tg\zeta_2\,\sin(a_2-a_1)} {\tg\zeta_1\sin(a_1-\varphi)-\tg\zeta_2\sin(a_2-\varphi)} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (15) \]

—the formula that we had earlier.

It would have been possible to give other derivations of the formulas indicated above, but lack of time and space restricts us.

Remark. It is easy to verify that the numerator and denominator of expression (A) [or (B)] are quantities corresponding, respectively, to the geometric sine and cosine of the direction \(\varphi\).

Indeed. Let us take the center of the sphere as the origin of a rectangular coordinate system in which the \(X\)-axis coincides with the trace NS of the meridian, and the \(Z\)-axis is vertical.

Let us denote the coordinates of the points of intersection I, II, F of the directions—vectors

\[ (a_1,\zeta_1);\ (a_2,\zeta_2);\ (\varphi,\zeta) \]

with the sphere, respectively, by

\[ (x_1,y_1,z_1);\ (x_2,y_2,z_2);\ (x,y,z). \]

By hypothesis, the points I, II, F and O lie in the plane of the bed. The condition that four points lie in one plane is expressed by the determinant:

\[ \left| \begin{array}{cccc} 0&0&0&1\\ x&y&z&1\\ x_1&y_1&z_1&1\\ x_2&y_2&z_2&1 \end{array} \right|=0,\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (16) \]

or

\[ \left| \begin{array}{ccc} x&y&z\\ x_1&y_1&z_1\\ x_2&y_2&z_2 \end{array} \right|=0 \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (17) \]

Expanding the determinant by the elements of the third column, we obtain:

\[ z\left| \begin{array}{cc} x_1&y_1\\ x_2&y_2 \end{array} \right| -z_1\left| \begin{array}{cc} x&y\\ x_2&y_2 \end{array} \right| +z_2\left| \begin{array}{cc} x&y\\ x_1&y_1 \end{array} \right|=0 \ .\ .\ .\ .\ .\ .\ .\ .\ (18) \]

Earlier we found that the coordinates of the point of intersection of any vector \(J\) with the sphere are determined by the formulas:

\[ \begin{aligned} x_i&=R\sin\zeta_i\cos a_i,\\ y_i&=R\sin\zeta_i\sin a_i,\\ z_i&=R\cos\zeta_i. \end{aligned} \left\} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (19) \right. \]

Therefore the preceding condition is rewritten as follows:

\[ R^3 \cos \zeta \begin{vmatrix} \sin \zeta_1 \cos a_1, & \sin \zeta_1 \sin a_1\\ \sin \zeta_2' \cos a_2, & \sin \zeta_2' \sin a_2 \end{vmatrix} - R^2 \cos \zeta_1 \begin{vmatrix} x & y\\ \sin \zeta_2 \cos a_2, & \sin \zeta_2 \sin a_2 \end{vmatrix} + R^2 \cos \zeta_2' \begin{vmatrix} x & y\\ \sin \zeta_1 \cos a_1, & \sin \zeta_1 \sin a_1 \end{vmatrix} =0. \]

Taking into account that

\[ \cos \zeta=-\frac{1}{\sqrt{1+\operatorname{tg}^{2}\zeta}}, \]

we obtain:

\[ -R\frac{\sin \zeta_1 \sin \zeta_2'}{\sqrt{1+\operatorname{tg}^{2}\zeta}} \begin{vmatrix} \cos a_1 & \sin a_1\\ \cos a_2 & \sin a_2 \end{vmatrix} - x\operatorname{tg}\zeta_2' \cos \zeta_1 \cos \zeta_2' \begin{vmatrix} 1 & \dfrac{y}{x}\\ \cos a_2 & \sin a_2 \end{vmatrix} + x\operatorname{tg}\zeta_1 \cos \zeta_1 \cos \zeta_2' \begin{vmatrix} 1 & \dfrac{y}{x}\\ \cos a_1 & \sin a_1 \end{vmatrix} =0 \]

or, finally,

\[ x\left\{(\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2) + (\operatorname{tg}\zeta_2' \cos a_2-\operatorname{tg}\zeta_1 \cos a_1)\frac{y}{x}\right\} = R\frac{\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2'\sin(a_2-a_1)} {\sqrt{1+\operatorname{tg}^{2}\zeta}} \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \tag{20} \]

But

\[ \frac{y}{x}=\operatorname{tg}\varphi = \frac{\operatorname{tg}\zeta_2' \cos a_2-\operatorname{tg}\zeta_1 \cos a_1} {\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2}, \]

therefore equation (20) takes the form:

\[ x\frac{ (\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2)^2 + (\operatorname{tg}\zeta_2' \cos a_2-\operatorname{tg}\zeta_1 \cos a_1)^2 }{ \operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2 } = R\frac{\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2'\sin(a_2-a_1)} {\sqrt{1+\operatorname{tg}^{2}\zeta}}; \]

whence

\[ x= R\frac{ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2'\sin(a_2-a_1) (\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2) }{ \sqrt{ 1+\operatorname{tg}^{2}\zeta \left[ (\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2)^2 + (\operatorname{tg}\zeta_2' \cos a_2-\operatorname{tg}\zeta_1 \cos a_1)^2 \right] } } \ldots \tag{21} \]

and, correspondingly,

\[ y= R\frac{ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2'\sin(a_2-a_1) (\operatorname{tg}\zeta_2' \cos a_2-\operatorname{tg}\zeta_1 \cos a_1) }{ \sqrt{ 1+\operatorname{tg}^{2}\zeta \left[ (\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2' \sin a_2)^2 + (\operatorname{tg}\zeta_2' \cos a_2-\operatorname{tg}\zeta_1 \cos a_1)^2 \right] } } \ldots \tag{22} \]

Taking now the formula \(\operatorname{tg}\varphi=\dfrac{y}{x}\), we may assert the proposition of the remark.

As a consequence of the results obtained, there also follows a new method of vectorial determination of the azimuth \(\varphi\); it is necessary only to have an expression for \(\operatorname{tg}\varphi\).

Namely, comparing formulas (1) and (2), we obtain:

\[ \operatorname{tg}\zeta_1(\cos\varphi\cos a_1+\sin\varphi\sin a_1) = \operatorname{tg}\zeta_2(\cos\varphi\cos a_2+\sin\varphi\sin a_2), \]

or

\[ \sin\varphi(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2) = \cos\varphi(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1) \]

and, finally,

\[ \operatorname{tg}\varphi= \frac{\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1} {\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2}. \]

It is then easy to show that \(x\) and \(y\) represent the projections of the vector \(OF\) (the line of dip) respectively on the \(X\) and \(Y\) axes.

Indeed: it is enough to indicate the identity of the formulas

\[ x=R\sin\zeta\cos\varphi \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (23) \]

\[ y=R\sin\zeta\sin\varphi \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (24) \]

with expressions (21) and (22), after substituting into formulas (23) and (24) the values of \(\zeta\) and \(\varphi\).

Note. The idea of solving the proposed problem by the method of spherical trigonometry was suggested by the model—a stereogram for determining the elements of attitude of a bed—constructed by Prof. P. K. Sobolevsky and kept in the mine-surveying laboratory of the S.T.I.

I have not, however, encountered the actual solution by the indicated method in the literature, nor have I received it from anyone either in writing or orally.

Remark 1. In the preceding paragraph we saw that the azimuth \(\varphi\) is determined through the tangent, as the ratio of the trigonometric sine to the trigonometric cosine.

By this last proposition we introduce a new—the second—principle for determining the azimuth \(\varphi\).

The formulas derived earlier lead us directly to the second principle.

Indeed. For determining the azimuth \(\varphi\) we have

\[ \operatorname{tg}\varphi=-\frac{m}{n}. \]

We represented the last formula (see 2 § 5) in two ways:

\[ \operatorname{tg}\varphi=\frac{-m}{n} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (1) \]

\[ \operatorname{tg}\varphi=\frac{m}{-n} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (2) \]

Accordingly, under the assumptions that \(\varphi\) belongs to the third quadrant and by applying first formula (1), and then formula (2), we have obtained:

\[ \begin{gathered} m>0;\ n<0\\[4pt] \sin\varphi=-\frac{m}{\sqrt{m^{2}+n^{2}}}\\[6pt] \cos\varphi=+\frac{n}{\sqrt{m^{2}+n^{2}}}\\[6pt] z=-\frac{x'}{\kappa}\sqrt{m^{2}+n^{2}} \end{gathered} \tag{3} \]

\[ \begin{gathered} m<0;\ n>0\\[4pt] \sin\varphi=+\frac{m}{\sqrt{m^{2}+n^{2}}}\\[6pt] \cos\varphi=-\frac{n}{\sqrt{m^{2}+n^{2}}}\\[6pt] z=\frac{x'}{\kappa}\sqrt{m^{2}+n^{2}} \end{gathered} \tag{4} \]

Whence, according to \((3_{1,2,3})\) and \((4_{1,2,3})\), we conclude that formulas (1) and (2), namely vectorially, determine the required azimuth \(\varphi\): formula (1)—under assumption \((3_1)\); formula (2)—under assumption \((4_1)\); the signs of the numerator and denominator of formulas (1) and (2) correspond, respectively, to the signs of the sine and cosine.

Replacing now \(m\) and \(n\) by the corresponding values, we obtain:

\[ \begin{array}{c|c} \operatorname{tg}\zeta_1\cos a_1-\operatorname{tg}\zeta_2\cos a_2>0 & \operatorname{tg}\zeta_1\cos a_1-\operatorname{tg}\zeta_2\cos a_2<0 \\ \operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2<0 & \operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2>0, \end{array} \]

whence

\[ \begin{array}{c|c} \operatorname{tg}\zeta_1\cos a_1>\operatorname{tg}\zeta_2\cos a_2 & \operatorname{tg}\zeta_1\cos a_1<\operatorname{tg}\zeta_2\cos a_2 \\ \operatorname{tg}\zeta_2\sin a_2>\operatorname{tg}\zeta_1\sin a_1 & \operatorname{tg}\zeta_2\sin a_2<\operatorname{tg}\zeta_1\sin a_1 \end{array} \tag{5} \]

Taking the product of the left-hand and right-hand sides for each of the columns of (5), we obtain, for each of the multiplication results, one expression:

\[ \operatorname{tg}\zeta_1\cos a_1\,\operatorname{tg}\zeta_2\sin a_2 \gtreqless \operatorname{tg}\zeta_1\sin a_1\,\operatorname{tg}\zeta_2\cos a_2 \]

or, transforming, we obtain

\[ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)\gtreqless 0. \]

Finally, since

\[ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)=\kappa, \]

then

\[ \kappa\gtreqless 0. \]

Now, in order finally to determine the sign of “\(\kappa\)” for formulas (1) and (2), it is necessary to refer, respectively, to expressions \((3_4)\) and \((4_4)\). In the latter, as we know, respectively,

\[ x' \gtreqless 0 \quad \text{and} \quad z \gtreqless 0. \]

Therefore, for formula (1) we obtain \(\kappa > 0\), and for formula (2), \(\kappa < 0\). The sign of “\(\kappa\),” as we have seen, determines the order of designation of the vectors—traces.

Assigning \(\varphi\) to the other quadrants, we arrive at identical results of the analysis.

Remark 2. The tenth method (see p. 52) gives a direct solution of the question according to the second principle.

§ 18. Graphical-Analytical Method.

In conclusion, we shall present a graphical-analytical method for determining the azimuth \(\varphi\) and the zenith distance \(\zeta\) of the line of dip.

From consideration of the corresponding formulas it follows that, for graphical construction, it is simplest to use the formulas:

\[ \text{I.} \]

\[ \left. \begin{aligned} \tg \Phi &= \frac{ \cos \frac{1}{2}(a_1-a_2)\bigl(\tg \zeta_2-\tg \zeta_1\bigr) }{ \sin \frac{1}{2}(a_1-a_2)\bigl(\tg \zeta_2+\tg \zeta_1\bigr) },\\[6pt] \varphi &= \Phi+\frac{1}{2}(a_1+a_2), \end{aligned} \right\} \ldots\ldots (I_1) \]

or

\[ \left. \begin{aligned} \tg \Phi &= \frac{ \pm \cos \frac{1}{2}(a_1-a_2)\sin(\zeta_2-\zeta_1) }{ \pm \sin \frac{1}{2}(a_1-a_2)\sin(\zeta_2+\zeta_1) },\\[6pt] \varphi &= \Phi+\frac{1}{2}(a_1+a_2), \end{aligned} \right\} \ldots\ldots (I_2) \]

and

\[ \text{II.} \]

\[ \tg \zeta = \tg \zeta_{1,2}\cos(\varphi-a_{1,2}) \ldots\ldots (II_1) \]

The system of formulas I, among others, is convenient in that in it the condition is always approximately fulfilled:

\[ \left|\cos \frac{1}{2}(a_1-a_2)\right| \cong \left|\sin \frac{1}{2}(a_1-a_2)\right| \cong \sin 45^\circ\;(\cong \cos 45^\circ). \]

[for the angle between adjacent vertical walls of the shaft is approximately equal to \(90^\circ\)].

Therefore the absolute values of the geometric sine and cosine,

\[ R\left|\cos \frac{1}{2}(a_1-a_2)\right| \quad \text{and} \quad R\left|\sin \frac{1}{2}(a_1-a_2)\right|, \]

are approximately equal to \(0.7R\).

I₁.

Construction of the angle \(\Phi\), determined by the expression:

\[ \operatorname{tg}\Phi = \frac{ \cos \frac{1}{2}\left(a_1-a_2\right)\left(\operatorname{tg}\zeta_2-\operatorname{tg}\zeta_1\right) }{ \sin \frac{1}{2}\left(a_1-a_2\right)\left(\operatorname{tg}\zeta_2+\operatorname{tg}\zeta_1\right) } \ldots\ldots (1) \]

We know that formula (1) expresses the ratio of the geometric sine to the geometric cosine.

Thus, multiplying the numerator and denominator of the right-hand side of (1) by an arbitrary positive number \(R\), we obtain the required geometric sine \(S\) and cosine \(C\), corresponding to the angle \(\Phi\) and determining it vectorially.

Thus:

\[ S = R\cos \frac{1}{2}\left(a_1-a_2\right) \left(\operatorname{tg}\zeta_2-\operatorname{tg}\zeta_1\right) \ldots\ldots (2) \]

\[ C = R\sin \frac{1}{2}\left(a_1-a_2\right) \left(\operatorname{tg}\zeta_2+\operatorname{tg}\zeta_1\right) \ldots\ldots (3) \]

Now it is evident that the construction of \(S\) and \(C\) (and consequently also of the angle \(\Phi\)) must be preceded by transformations of (2) and (3).

Namely, we may write:

\[ S = R\left|\cos \frac{1}{2}\left(a_1-a_2\right)\right| \left|\pm\left(\operatorname{tg}\zeta_2-\operatorname{tg}\zeta_1\right)\right| \ldots\ldots (4) \]

\[ C = R\left|\sin \frac{1}{2}\left(a_1-a_2\right)\right| \left|\pm\left(\operatorname{tg}\zeta_2+\operatorname{tg}\zeta_1\right)\right| \ldots\ldots (5) \]

Further, denoting

\[ R_s = R\left|\cos \frac{1}{2}\left(a_1-a_2\right)\right| \]

\[ R_c = R\left|\sin \frac{1}{2}\left(a_1-a_2\right)\right|, \]

we may write:

\[ S=\pm\left|R_s\operatorname{tg}\zeta_2-R_s\operatorname{tg}\zeta_1\right| \]

\[ C=\pm\left|R_c\operatorname{tg}\zeta_2+R_c\operatorname{tg}\zeta_1\right| \]

or

\[ S=R_s\operatorname{tg}(\pm\zeta_2)-R_s\operatorname{tg}(\pm\zeta_1) \]

\[ C=R_c\operatorname{tg}(\pm\zeta_2)+R_c\operatorname{tg}(\pm\zeta_1), \]

where the signs \((\pm)\) coincide with the signs of the expressions

\[ \cos \frac{1}{2}\left(a_1-a_2\right) \quad \text{and} \quad \sin \frac{1}{2}\left(a_1-a_2\right). \]

We conclude: the geometric sine \(S\) (cosine \(C\)) corresponding to the angle \(\Phi\) is the algebraic sum, corresponding to the radius \(R_s\) \((R_c)\), of the geometric tangents of angles that are functions of the given angles \(\zeta_1\) and \(\zeta_2\).

I\(_2\).

Construction of the angle \(\Phi\), determined by the expression:

\[ \operatorname{tg}\Phi = \frac{ \pm \cos \frac{1}{2}(a_1-a_2)\sin(\zeta_2-\zeta_1) }{ \pm \sin \frac{1}{2}(a_1-a_2)\sin(\zeta_2+\zeta_1) } \ldots \tag{6} \]

As we know, the sign \((+)\) in the numerator and denominator is taken in the case when \(\zeta_1\) and \(\zeta_2\) are alike:

\[ \cos\zeta_1\cos\zeta_2>0; \]

the sign \((-)\)—when \(\zeta_1\) and \(\zeta_2\) are unlike:

\[ \cos\zeta_1\cos\zeta_2<0. \]

Transforming expression (6) for the construction of the angle \(\Phi\), we proceed analogously to the preceding case.

Namely, we have:

\[ \operatorname{tg}\Phi = \frac{ \pm R\cos \frac{1}{2}(a_1-a_2)\sin(\zeta_2-\zeta_1) }{ \pm R\sin \frac{1}{2}(a_1-a_2)\sin(\zeta_2+\zeta_1) } = \]

\[ = \frac{ R\left|\cos\frac{1}{2}(a_1-a_2)\right| \sin\left[\pm(\zeta_2-\zeta_1)\right] }{ R\left|\sin\frac{1}{2}(a_1-a_2)\right| \sin\left[\pm(\zeta_2+\zeta_1)\right] } \]

and, finally,

\[ \operatorname{tg}\Phi = \frac{ R_s\sin\left[\pm(\zeta_2-\zeta_1)\right] }{ R_c\sin\left[\pm(\zeta_2+\zeta_1)\right] }, \]

where

\[ R_s=R\left|\cos\frac{1}{2}(a_1-a_2)\right| \]

\[ R_c=R\left|\sin\frac{1}{2}(a_1-a_2)\right|. \]

III₁.

Construction of the zenith distance \(\zeta\), determined correspondingly by the expressions:

\[ \operatorname{tg}\zeta=\operatorname{tg}\zeta_{1,2}\cos(\varphi-a_{1,2}), \ldots\ldots\ldots (7) \]

where

\[ \frac{\pi}{2}<\zeta<\pi. \]

1. Multiplying both sides of equation (7) by an arbitrary positive number \(R\), we obtain:

\[ R\operatorname{tg}\zeta = R\left|\operatorname{tg}\zeta_{1,2}\right| \left[\pm\cos(\varphi-a_{1,2})\right], \ldots\ldots\ldots (8) \]

Introducing the notation

\[ R_{t_{1,2}}=R\left|\operatorname{tg}\zeta_{1,2}\right|, \ldots\ldots\ldots (9) \]

we obtain:

\[ R\operatorname{tg}\zeta = R_{t_{1,2}}\cos(\varphi-a_{1,2}), \ldots\ldots\ldots (10) \]

corresponding to the sign \((+)\) before the cosine in expression (8); or

\[ R\operatorname{tg}\zeta = R_{t_{1,2}}\cos\left[180^\circ\pm(\varphi-a_{1,2})\right], \ldots\ldots\ldots (11) \]

corresponding to the sign \((-)\) before the cosine in expression (8).

Remark: In formula (11), obviously, either sign may be taken before the binomial \((\varphi-a_{1,2})\): \((+)\) or \((-)\).

From consideration of formula (10) we conclude: the geometric tangent \(R\operatorname{tg}\zeta\) of the angle \(\zeta\) corresponding to the radius \(R\) is equal to the geometric cosine \(R_{t_{1,2}}\cos(\varphi-a_{1,2})\) of the angle \((\varphi-a_{1,2})\), corresponding to the radius \(R_{t_{1,2}}=R|\operatorname{tg}\zeta_{1,2}|\).

Consideration of formula (11) leads to an analogous conclusion.

Thus, the construction of the required angle \(\zeta\) includes three steps:
1) construction of the radius \(R_{t_{1,2}}\); 2) construction of the geometric cosine of the right-hand side of expression (10) or (11), respectively; and, finally, 3) construction of the angle \(\zeta\).

For construction it is simpler and more natural to represent formula (7) in the form:

\[ R\operatorname{tg}\zeta = R\left|\cos(\varphi-a_{1,2})\right| \operatorname{tg}(\pm\zeta_{1,2}) = R_c\operatorname{tg}(\pm\zeta_{1,2}). \]

II₂.

The formula

\[ \operatorname{tg}\zeta=\operatorname{tg}\zeta_{1,2}\cos(\varphi-a_{1,2}) \]

can be transformed differently and—in accordance with the transformation—another method of constructing the angle \(\zeta\) can be indicated.

We have:

\[ \operatorname{tg}\zeta = \frac{\sin\zeta_{1,2}\cos(\varphi-a_{1,2})}{\cos\zeta_{1,2}} \ldots\ldots\ldots (12) \]

Multiplying the numerator and denominator of the right-hand side of (12) by an arbitrary positive number \(R\), we obtain:

\[ \operatorname{tg}\zeta = \frac{R\sin\zeta_{1,2}\cos(\varphi-a_{1,2})}{R\cos\zeta_{1,2}}. \]

It now remains to apply, in successive order, the constructions:

\[ R_{1,2}=R\sin \zeta_{1,2}; \qquad R_{\mathrm{I,II}}=R_{1,2}\cos(\varphi-a_{1,2}); \qquad R_{\mathrm{III,IV}}=R\cos \zeta_{1,2} \]

and, finally, knowing that the ratio

\[ \frac{R_{\mathrm{I,II}}}{R_{\mathrm{III,IV}}} \]

is precisely the ratio of the geometric sine and cosine (second quadrant), we determine the angle \(\zeta\) by construction.

Remark. It is clear that it is superfluous to remind the reader that graphical constructions are practically limited by the limit as a function of the given numerical values of the quantities.

§ 19. Examples of the construction of the angles \(\Phi\) and \(\zeta\).

I. Construction of the angle \(\Phi\).

Formula \((I_1)\).

First example of the transformation of formula \((I_1)\).

We have:

\[ \begin{array}{c|c} a_1=13^\circ 30' & a_2=281^\circ 15' \\ \zeta_1=65^\circ 45' & \zeta_2=105^\circ 30' \end{array} \]

\[ \frac{1}{2}(a_1-a_2)=-133^\circ 52'30''. \]

and

\[ \operatorname{tg}\Phi = \frac{R(-\cos 46^\circ 07'30'')(-\operatorname{tg}74^\circ 30'-\operatorname{tg}65^\circ 45')} {R\sin(-46^\circ 07'30'')(-\operatorname{tg}74^\circ 30'+\operatorname{tg}65^\circ 45')} = \]

\[ = \frac{R\cos 46^\circ 07'30''(\operatorname{tg}74^\circ 30'+\operatorname{tg}65^\circ 45')} {R\sin 46^\circ 07'30''(\operatorname{tg}74^\circ 30'-\operatorname{tg}65^\circ 45')} = \]

\[ = \frac{R_s\operatorname{tg}74^\circ 30' + R_s\operatorname{tg}65^\circ 45'} {R_c\operatorname{tg}74^\circ 30' - R_c\operatorname{tg}65^\circ 45'}, \]

where

\[ R_s=R\cos 46^\circ 07'30'' \]

\[ R_c=R\sin 46^\circ 07'30'', \]

and \(R\) is an arbitrary positive number.

The transformation of the formula may be represented in an elegant and natural form.

Namely:

\[ \operatorname{tg}\Phi = \frac{R\cos(-133^\circ 52'30'')(\operatorname{tg}105^\circ 30'-\operatorname{tg}65^\circ 45')} {R\sin(-133^\circ 52'30'')(\operatorname{tg}105^\circ 30'+\operatorname{tg}65^\circ 45')} = \]

\[ = \frac{ R\left|\cos(-133^\circ 52'30'')\right|(-\operatorname{tg}105^\circ 30' + \operatorname{tg}65^\circ 45') }{ R\left|\sin(-133^\circ 52'30'')\right|(-\operatorname{tg}105^\circ 30' - \operatorname{tg}65^\circ 45') }. \]

Second example of the transformation and construction of formula \((I_1)\).

We have:

\[ \begin{array}{c|c} a_1=13^\circ 30' & a_2=101^\circ 30' \\ \zeta_1=50^\circ 30' & \zeta_2=23^\circ 15' \end{array} \]

\[ \frac{1}{2}(a_1-a_2)=-44^\circ; \qquad \frac{1}{2}(a_1+a_2)=57^\circ 30'. \]

\[ \operatorname{tg}\Phi = \frac{R\cos(-44^\circ)(\operatorname{tg}23^\circ 15' - \operatorname{tg}50^\circ 30')} {R\sin(-44^\circ)(\operatorname{tg}23^\circ 15' + \operatorname{tg}50^\circ 30')} = \]

\[ = \frac{-\left[R_s\operatorname{tg}50^\circ 30' - R_s\operatorname{tg}23^\circ 15'\right]} {-\left[R_c\operatorname{tg}50^\circ 30' + R_c\operatorname{tg}23^\circ 15'\right]} \ldots\ldots\ldots\ldots (13) \]

where

\[ \begin{aligned} R_s &= R\cos 44^\circ \\ R_c &= R\sin 44^\circ \end{aligned} \Bigg\} \ldots\ldots\ldots\ldots (14) \]

and \(R\) is an arbitrary positive number.

Fig. 17.

Figure 17 shows the construction of the required angle \(\Phi\), determined by expressions (13) and (14).

In the plane of the paper we take an arbitrary, suitably chosen point \(O\), and adopt it as the origin of a plane rectangular coordinate system, whose axes are: the principal direction \(SN\) (the meridian direction) and the direction \(OO_{st}\); consequently, the vectorial reckoning of angles is positive (clockwise).

From point \(O\), as center, we describe a circle of arbitrary radius \(R = OI\). We construct the angle \(NOA = 44^\circ\). The segment \(OB = R_s = R\cos 44^\circ\);

segment \(\overline{BA}=R_c=R\sin44^\circ\). Next we construct the angles: \(NOT=\zeta_2=23^\circ15'\) and \(NOR=\zeta_1=50^\circ30'\).

1. Now, from point \(O\), as center, we describe a circle of radius \(\overline{OB}=R_s\); then: \(\overline{BQ}=R_s\operatorname{tg}23^\circ15'\); \(\overline{BK}=R_s\operatorname{tg}50^\circ30'\). The segment

\[ \overline{KQ}=-(\overline{BK}-\overline{BQ})=-\left[R_s\operatorname{tg}50^\circ30'-R_s\operatorname{tg}23^\circ15'\right] \]

is the required geometric sine of the angle \(\Phi\).

1. Next—from point \(O\), as center, we describe a circle of radius:

\[ \overline{Oa}=\overline{BA}=R\sin44^\circ=R_c. \]

Then: the segment \(\overline{LP_2}=R_c\operatorname{tg}23^\circ15'\); the segment \(\overline{LM}=R_c\operatorname{tg}50^\circ30'\), and, finally, the segment

\[ \overline{MP_1}=\overline{p_3L}=-\left[R_c\operatorname{tg}50^\circ30'+R_c\operatorname{tg}23^\circ15'\right] \]

is the required geometric cosine of the angle \(\Phi\).

Remark. From the drawing it is evident that, in order to determine the absolute magnitude of the last sum, we may proceed in two ways:

1) on the continuation of the straight line \(ML\) from point \(L\) to the left, set off \(\overline{EP_1}=\overline{LP_2}\). Then the segment \(\overline{MP_1}\) is the required one; or

2) from point \(E\) along the axis \(SN\) we set off downward the segment \(\overline{Ep_1}=\overline{LP_2}\), then through the end \(p_1\) of the segment \(\overline{Lp_1}\) we draw a straight line parallel to \(OOst\); next, from point \(M\) we drop a perpendicular to \(\overline{p_1p_2}\); the segment \(\overline{Mp_2}=\overline{Ep_1}=\overline{EP_2}=R_c\operatorname{tg}23^\circ15'\); this latter we set off from point \(M\) along the straight line \(LM\) to the right \((\overline{Mp_3}=\overline{Mp_2})\).

It is now clear that

\[ \overline{p_3L}=-(\overline{ML}+\overline{P_2L}). \]

1. Finally, it remains to construct the required angle \(\Phi\) from its geometric cosine and sine as found.

a) In accordance with the sign of the denominator [see formula (13)], from point \(O\) we set off, in the negative direction of the axis \(SN\), the segment

\[ \overline{OS}=\overline{MP_1}=\overline{pO}=\overline{p_3L}. \]

b) In accordance with the sign of the numerator [see formula (13)], at point \(S\) we erect the perpendicular \(SJ\), and to the left of point \(S\) on this latter we set off the segment \(\overline{SJ}=\overline{KQ}\) directly, or by a continuous construction:

\[ \overline{SJ}=\overline{Ok_3}=\overline{Ok_2}=\overline{kk_1}=\overline{kq}=\overline{KQ}. \]

c) Joining points \(O\) and \(J\) by a straight line, we obtain vectorially the required angle \(\Phi=206^\circ30'\) (indicated in the drawing by an arc arrow). The azimuth \(\varphi\) of the dip line of the plane of the bed is

\[ \varphi=\Phi+\frac{1}{2}(a_1+a_2)=264^\circ. \]

Remark. The transformation of the formula may be represented in another, elegant and natural, form.

Namely:

\[ \operatorname{tg}\Phi= \frac{R|\cos(-44^\circ)|\left(\operatorname{tg}23^\circ15'-\operatorname{tg}50^\circ30'\right)} {R|\sin(-44^\circ)|\left(-\operatorname{tg}23^\circ15'-\operatorname{tg}50^\circ30'\right)}. \]

The construction of the angle \(\Phi\) is carried out analogously to the preceding one.

Formula \((I_2)\).

First example of the transformation of formula \((I_2)\).
We have:

\[ \begin{array}{c|c} a_1 = 13^\circ 30' & a_2 = 281^\circ 15' .\\ \zeta_1 = 65^\circ 45' & \zeta_2 = 105^\circ 30' . \end{array} \]

\[ \frac{1}{2}(a_1 - a_2) = -133^\circ 52' 30'';\qquad \frac{1}{2}(a_1 + a_2) = 147^\circ 22' 30'' . \]

\[ \zeta_2 - \zeta_1 = 39^\circ 45';\qquad \zeta_2 + \zeta_1 = 171^\circ 15' . \]

\[ \sin \frac{1}{2}(a_1 - a_2) = -\sin 46^\circ 07' 30'' . \]

\[ \sin(\zeta_2 + \zeta_1) = \sin 8^\circ 45' . \]

\[ \cos \frac{1}{2}(a_1 - a_2) = -\cos 46^\circ 07' 30'' . \]

In the present case \(\zeta_1\) and \(\zeta_2\) are not the same.
Thus:

\[ \operatorname{tg}\Phi = \frac{-[-\cos 46^\circ 07' 30'']\sin 39^\circ 45'} {-[-\sin 46^\circ 07' 30'']\sin 8^\circ 45'} = \]

\[ = \frac{\cos 46^\circ 07' 30''\sin 39^\circ 45'} {\sin 46^\circ 07' 30''\sin 8^\circ 45'} = \frac{R_s\sin 39^\circ 45'}{R_c\sin 8^\circ 45'}, \]

where

\[ R_s = R\cos 46^\circ 07' 30'' \]

\[ R_c = R\sin 46^\circ 07' 30'', \]

and \(R\) is an arbitrary positive number.

Transformation of formula \((I_2)\) into another form:

\[ \operatorname{tg}\Phi = \frac{-\cos(-133^\circ 52' 30'')\sin 39^\circ 45'} {-\sin(-133^\circ 52' 30'')\sin 171^\circ 15'} = \]

\[ = \frac{R\left|\cos(-133^\circ 52' 30'')\right|\sin 39^\circ 45'} {R\left|\sin(-133^\circ 52' 30'')\right|\sin 171^\circ 15'} . \]

Second example of the transformation and construction of the angle \(\Phi\), determined by formula \((I_2)\).

Let us take the data of the second example of the transformation and construction of the angle \(\Phi\), determined by formula \((I_1)\).
We have:

\[ \frac{1}{2}(a_1 - a_2) = -44^\circ . \]

\[ \zeta_2 - \zeta_1 = -27^\circ 15' . \]

\[ \zeta_2 + \zeta_1 = 73^\circ 45' . \]

In the present case \(r_1\) and \(r_2\) are identical.

Thus:

\[ \operatorname{tg}\Phi = \frac{\cos(-44^\circ)\sin(-27^\circ 15')}{\sin(-44^\circ)\sin 73^\circ 45'} = \]

\[ = \frac{\cos 44^\circ\sin(-27^\circ 15')}{\sin 44^\circ\sin(-73^\circ 45')} = \frac{R_s\sin(-27^\circ 15')}{R_c\sin(-73^\circ 45')}, \ldots \tag{15} \]

where

\[ \begin{aligned} R_s &= R\cos 44^\circ \\ R_c &= R\sin 44^\circ \end{aligned} \Bigg\}, \ldots \tag{16} \]

and \(R\) is an arbitrary positive number.

In drawing I the construction of the angle \(\Phi\) is shown. As in the preceding example of the construction of formula \((I_1)\), from point \(O\), as center, we describe a circle of radius \(R\).

We construct the angle \(NOA = 44^\circ\). The segment

\[ OB = R\cos 44^\circ = R_s, \]

the segment

\[ BA = R\sin 44^\circ = R_c. \]

Next we construct the angles:

\[ NOC_1 = -27^\circ 15' \quad \text{and} \quad NOC_2 = -73^\circ 45'. \]

Then:

1. From point \(O\), as center, we describe a circle of radius

\[ \overline{OB} = R\cos 44^\circ = R_s; \]

then the segment

\[ ED = R_s\sin(-27^\circ 15') \]

is the geometric sine of the sought angle \(\Phi\).

1. From point \(O\), as center, we describe a circle of radius

\[ \overline{OL} = \overline{Oa} = BA = R_c; \]

then the segment

\[ GF = R_c\sin(-73^\circ 45') \]

is the geometric cosine of the sought angle \(\Phi\).

1. Finally, it remains to construct the sought angle \(\Phi\) from its found geometric cosine \(GF\) and sine \(ED\).

a. In accordance with the sign of the denominator [see formulas (15) and (16)], from point \(O\) we lay off, in the negative direction \(SN\), the segment

\[ Of_1 = Of = GF. \]

b. Next—in accordance with the sign of the numerator [see formulas (15) and (16)]—at point \(f_1\) we erect to the axis \(SN\) the perpendicular \(f_1 f_2\), and to the left of point \(f_1\) on the latter we lay off the segment \(f_1 f_2 = ED\).

c. Joining the points \(O\) and \(f_2\) by a straight line, we obtain, vectorially, the sought angle \(\Phi = 206^\circ 30'\).

Remark. The transformation of the formula for construction may be represented in the form:

\[ \operatorname{tg}\Phi = \frac{R\lvert \cos(-44^\circ)\rvert \sin(-27^\circ 15')} {R\lvert \sin(-44^\circ)\rvert \sin(-73^\circ 45')} . \]

II. Construction of the angle \(\zeta\).

Formula

\[ \operatorname{tg}\zeta = \operatorname{tg}\zeta_{1,2}\cos(\varphi-a_{1,2}). \]

First example of transformation of the formula.

Given:

\[ a_2=28^\circ 15', \]

\[ \zeta_2=105^\circ 30', \]

\[ \varphi=223^\circ 30'. \]

Therefore,

\[ \varphi-a_2=-57^\circ 45', \]

and

\[ R\operatorname{tg}\zeta = R\operatorname{tg}105^\circ 30'\cos(-57^\circ 45') = \]

\[ = R\lvert \operatorname{tg}105^\circ 30'\rvert \cos(180^\circ \pm 57^\circ 45'); \]

or

\[ R\operatorname{tg}\zeta = R(-\operatorname{tg}74^\circ 30')\cos(-57^\circ 45') = \]

\[ = R\operatorname{tg}74^\circ 30'\cos(180^\circ \pm 57^\circ 45'), \]

or (which is better):

\[ R\operatorname{tg}\zeta = R\lvert \cos(-57^\circ 45')\rvert \operatorname{tg}105^\circ 30'. \]

Second example of transformation and construction of the angle \(\zeta\).

We have:

\[ \begin{array}{c|c} a_1=13^\circ 30' & a_2=101^\circ 30'\\ \zeta_1=50^\circ 30' & \zeta_2=23^\circ 15' \end{array} \]

In accordance with the latter data we found (see above) that

\[ \varphi=264^\circ. \]

Next we have:

\[ \varphi-a_1=250^\circ 30' \]

\[ \varphi-a_2=162^\circ 30'. \]

Therefore,

\[ 1.\quad R\operatorname{tg}\zeta = R\operatorname{tg}50^\circ 30'\cos250^\circ 30' = R_{t_1}\cos250^\circ 30', \]

where

\[ R_{t_1}=R\operatorname{tg}50^\circ 30', \]

and \(R\) is an arbitrary positive number.

\[ 2.\quad R\operatorname{tg}\zeta = R\operatorname{tg}23^\circ 15'\cos162^\circ 30' = R_{t_2}\cos162^\circ 30', \]

where

\[ R_{t_2}=R\operatorname{tg}23^\circ 15'. \]

We take (see Fig. 17) the principal direction \(OZ\), from which the angles are counted vectorially, to coincide with the axis \(SN\).

A. Construction according to the data:

\[ (a_1,\ \zeta_1,\ \varphi-a_1). \]

1. Construct the angles:

\[ ZOR=\zeta_1=50^\circ 30' \]

and

\[ NOA_1=\varphi-a_1=250^\circ 30'. \]

Take \(R=\overline{OL}\); then

\[ \overline{LM}=R_{t_1}=\overline{OL}\tg 50^\circ 30'. \]

1. Next, from point \(O\), as center, describe a circle of radius

\[ R_{t_1}=\overline{Om}=\overline{LM}, \]

find the point \(M_1\) of intersection of the circle drawn with the direction

\[ (OA_1)=\varphi-a_1=250^\circ 30'; \]

drop from point \(M_1\) the perpendicular \(M_1M_2\) to the principal direction \(SN\) \((SZ)\), and then the segment

\[ OM_2=R_{t_1}\cos 250^\circ 30'=R\tg \zeta. \]

1. On the continuation of the straight line \(ML\), from point \(L\) to the left, lay off the segment

\[ Lm_1=\overline{Om_2}=\overline{OM_2}=R\tg \zeta \]

and, finally, join the points \(m_1\) and \(O\) by a straight line; the direction \((m_1O)\) is the one sought.

\[ (m_1O)=(OV)=\zeta=157^\circ 30'. \]

B. Construction according to the data:

\[ (a_2,\ \zeta_2,\ \varphi-a_2). \]

Construct the angles \(ZOT=23^\circ 15'\) and \(NOA_2=162^\circ 30'\).
Take \(R=\overline{Ol}\), and in the subsequent construction proceed analogously to the first case. The direction \((t_1O)\) is the one sought.

\[ (t_1O)=(m_1O)=(OV)=\zeta=157^\circ 30'. \]

Remark 1. The formula for the construction is better represented in the forms:

\[ R\tg\zeta = R\left|\cos 250^\circ 30'\right|\tg(-50^\circ 30') \]

\[ R\tg\zeta = R\left|\cos 162^\circ 30'\right|\tg(-23^\circ 15'). \]

Remark 2. The formulas determining the azimuth \(\varphi\) and the zenith distance \(\zeta\) of the line of dip may be represented in the form of nomograms, which would constitute a separate work.

CONTENTS

	Page
§ 1. Statement of the question in connection with the representation of the plane of the bed.	1
§ 2. Principles of the proposed analytical methods.	2
§ 3. Literature on the question; absence of a solution of the question in general form. Necessity of an analytical solution in special cases of exploration practice; connection of the analytical solution with the question of “determining the general strike and dip of a bed.”	3
§ 4. Geometric elements of prospecting pits in connection with determining the line of dip.	4
§ 5. Solutions by the method of analytic geometry.	4
Compilation of the equation of the plane of the bed.
Determination of the azimuth $\varphi$ of the line of dip by the method of rotation.
Determination of the order for designating the directions of traces. Geometric interpretation of the conditions.
§ 6. Derivation of logarithmic formulas for determining the azimuth $\varphi$ of the line of dip.	14
§ 7. The same—the method of rotation. Determination of the zenith distance $\zeta$ of the line of dip. Transformation formulas.	20
§ 8. Other methods (3) for determining the zenith distance $\zeta$ of the line of dip.	29
§ 9. Reduction to logarithmic form of the formula $\tg^{2}\zeta$.	31
§ 10. Other methods (9) for determining the azimuth $\varphi$ of the line of dip.	42
§ 11. Fifth and sixth methods for determining the zenith distance.	53
§ 12. Solutions by the method of differential calculus (3 solutions).	55
§ 13. Solutions by the method of integral calculus (3 solutions).	62
§ 14. One of the methods for determining the zenith distance $\zeta$ of the line of dip.	73
§ 15. One of the examples relating to the note to § 4: the case of specifying three boreholes.	74
§ 16. Appendix. Numerical examples.	76
§ 17. Solutions by the method of spherical trigonometry.	91
§ 18. Graphical-analytical method for determining the elements of attitude at a given point of the surface of a bed.	99
§ 19. Examples of constructing the angles $\Phi$ and $\zeta$.	103