Associate Professor V. F. Turchinsky

Graphical Method for Determining the Elements of Attitude at a Given Point on a Stratum Surface

Tomsk
1929

Associate Professor V. F. Turchinskii.

Graphical Method for Determining the Elements of Attitude at a Given Point of a Stratum Surface

Introduction

The proposed graphical method for determining the directions of the lines of dip and strike at a given point of a stratum surface*) is based, first, on one of the branches of descriptive geometry—the branch of projections with numerical elevations—and, second, on the determination of the position of a point on a conical surface.

CHAPTER I.

Construction of the Conical Net.

§ 1.

In connection with what follows, it is necessary to recall the solution of the following problem:

A straight line is given by two of its points—\(A\) and \(B\), whose elevations are, respectively, \(z_A=-3,\ z_B=+7\); it is required to determine the point of intersection of the straight line \(AB\) with the horizontal plane of projections, whose elevation is to be taken as zero (drawing 1).

Solution: since the sought point must have elevation \(0\), we rotate the horizontally projecting plane of the straight line \(AB\) about its horizontal projection \(ab\) to the right or to the left until it coincides with the plane of projections (the plane of the drawing). We rotate to the left. Then the revolved positions \(A_0\) and \(B_0\) of the points \(A\) and \(B\) will be located on perpendiculars to the line \(ab\), respectively at distances \((-3)\) and \((+7)\). Now joining the points \(A_0\) and \(B_0\) by a straight line, we obtain the revolved position \(A_0B_0\) of the straight line \(AB\) (drawing 2). The point of intersection \(X\) of the straight lines \(A_0B_0\) and \(ab\) is the required point of meeting of the straight line \(AB\) with the plane of projections.

Remark 1. If the directions of the straight lines \(A_0B_0\) and \(ab\) are such that they intersect at an acute angle, then, for an exact determination of the position of the point \(X\) of their intersection, it is necessary, as is known, to increase the elevations of the points \(A\) and \(B\) by one and the same number of times.

Let us further note that, in determining the point \(X\), the straight lines \(bB_0\) and \(aA_0\) may be drawn, respectively, in any directions parallel to one another.

*) According to prospecting-shaft data: \((a_1, h_1)\), \((a_2, h_2)\) of two traces of the plane of the stratum on the vertical walls of the shaft and the elevation of the point of their convergence.

Remark 1. In \(a_1\) and \(a_2\)—azimuths.
\(h_1\) and \(h_2\)—angles of dip.

Remark 2. All other cases of exploration, when solving the question of determining the attitude elements of a stratum surface at a given point, are reduced to the case of a prospecting shaft.

Remark 2. A case is possible in which the indicated transformation of the elevations of the ends of the line is impossible (see § 11). The same problem, as is known, is easily solved by taking two mutually perpendicular planes H and V: it is sufficient only to draw in the plane of the drawing any axis of projections OX. For the solution of the problem, see Fig. 3.

Remark 3. In what follows we shall resort to one or another solution of the problem posed above.

§ 2.

Let us now dwell on determining the position of a point on a conical surface.

Take a conical surface (Fig. 4), the coordinates of whose center \(S\) \((S')\) are \((0,0,c)\) [where \(c \gtreqless 0\)] and whose directrix is the circle \(PQ\):

\[ \left. \begin{aligned} x^2+y^2-R^2&=0\\ z&=0 \end{aligned} \right\} \ldots\ldots\ldots (1) \]

Then the equation of the conical surface, as is known, will be:

\[ x^2+y^2-\operatorname{tg}^2\alpha\,(z-c)^2=0, \ldots\ldots\ldots (2) \]

where \(\alpha\) is the angle of conicity.

Express its equation in spherical coordinates, taking the origin of coordinates as the pole, the \(Z\)-axis as the polar axis, and the plane \(ZOX\) as the polar plane.

For an arbitrary point of it \((r,\theta,\varphi)\) we have:

\[ \begin{aligned} x&=r\sin\theta\cos\varphi\\ y&=r\sin\theta\sin\varphi\\ z&=r\cos\theta, \end{aligned} \]

where \(r\) is the radius vector; \(\theta\) and \(\varphi\) are, respectively, the zenith (polar) distance and the azimuth (polar angle) of the radius vector.

Then the equation of the conical surface will take the form:

\[ r^2\sin^2\theta-\operatorname{tg}^2\alpha\,(r\cos\theta-c)^2=0 \ldots\ldots\ldots (3) \]

Let us make the assumptions:

\[ \text{I. }\quad c>0 \]

and, moreover, let the coordinate \(z\) of the points of the conical surface satisfy the condition:

\[ c>z>0. \]

Then we obtain the part of the conical surface lying above the horizontal plane H and representing its lower half.

In this case each of the points, say \(A\), of the conical surface is determined by the intersection of three surfaces: the given conical surface (2), a vertical half-plane determined by the polar angle \(\varphi=a\), and the same conical surface whose center coincides with the pole \(O\) and whose equation is: \(\theta=\zeta\).

Taking, under our assumption, the points of the given conical surface as the ends of vectors issuing from the pole \(O\), we see directly from the drawing that the direction of any vector is determined identically by the same spherical elements as the points of the given conical surface. In the present case \(\varphi = a\) varies from \(0\) to \(2\pi\), while \(\theta = \zeta\) varies from \(0\) to \(+\dfrac{\pi}{2}\).

A given value of \(\theta\), with \(\varphi\) variable, determines on the conical surface a circle representing the locus of the ends of the vectors; the planes of the circles corresponding to constant \(\theta\) are perpendicular to the axis of the conical surface.

Taking \(\theta\) further as a parameter, we determine a family or system of circles on the selected part of the conical surface, corresponding to vectors whose zenith distances vary from \(0\) to \(+\dfrac{\pi}{2}\)

(or whose angles of dip vary from \(+\dfrac{\pi}{2}\) to \(0\)).

II. Let us now take another assumption: \(c < 0\), and let the coordinate \(z\) of the points of the conical surface satisfy the inequality:

\[ 0 > z > c. \]

We obtain a second cone, \(S'\), symmetric to the first \(S\) with respect to the horizontal plane \(H\), and intersecting it along the circle \(PQ\), corresponding to the points of the conical surfaces \(S\) and \(S'\) for which the zenith distances \(\theta = \dfrac{\pi}{2}\). [Both cones \(S\) and \(S'\) represent mirror images of each other].

By reasoning analogous to that relating to the first conical surface \(S\), we arrive at the conclusion: on the conical surface \(S'\) we also obtain a family or system of circles, symmetric to the circles on the conical surface \(S\), and corresponding to the locus of the ends of vectors whose zenith distances vary from \(+\dfrac{\pi}{2}\) to \(+\pi\) (and, correspondingly, whose angles of dip of the vectors are from \(0\) to \(-\dfrac{\pi}{2}\)).

Proceeding further, let us note that, with variable azimuth \(\varphi\) and zenith distance \(\theta = \mathrm{constans}\), the coordinate \(z\) of the corresponding points of the surfaces of the cones \(S\) and \(S'\) is a constant quantity.

Let us now project the two systems of circles corresponding to the first \(S\) and the second \(S'\) conical surfaces onto the horizontal plane \(H\), corresponding to the pole \(O\). They are projected as concentric circles in natural size. Since the two systems of circles on the surfaces \(S\) and \(S'\) are symmetric with respect to the plane \(H\), their projections will coincide, and each of the projections of the circles will have two marks corresponding to vectors whose zenith distances complement each other to \(\pi\).

It is clear that by the family of projections of concentric circles, expressed by marks of zenith distances, and by the family of traces of vertical planes (the so-called rose of azimuths), expressed by...

with marks of polar angles—azimuths, will be completely determined (by their mutual intersections), both as the position of the points of the ends of the vectors lying on the conical surfaces \(S\) and \(S'\), and as their coordinate—\(z\).

Indeed. Suppose we have constructed, for the given cones, the above-mentioned systems of projections of concentric circles and traces of vertical planes. Let there be given: the azimuth \(a\) and the zenith distance \(\zeta\) of a vector (which is what we have when sketching, for each of the traces of the bed plane on the vertical walls of the borehole).

We find the point of intersection \(b\) of the direction \(a\) with the circle bearing the mark \(\zeta\) (see Fig. 5).

We now rotate the horizontally projecting (vertical) plane of the given vector about its horizontal trace \(MM\) until it coincides with the plane of projections.

From the mark “\(c\)” of the vertices of the cones we find the coincident position of the generators of the cones. Having restored from the projection \(b\) of the end \(B\) of the vector \(OB\) a perpendicular to the axis of rotation \(OM\), and having found the point \(B_o\) of its intersection with the coincident position of the corresponding generator of the cone, we see that the required coordinate \(z\) of the end \(B\) of the vector \(OB\) is determined by the segment \(bB_o\).

Remark 1. It is easy to see that the coincident positions of the corresponding vectors can be obtained in another way, namely by first bringing the horizontally projecting (vertical) plane of the corresponding vector (and of the generators of the cones), by rotation about the axis \(OZ\), into a position coinciding with the principal vertical plane \(NS\)—the meridian plane (or with the plane \(OW\), perpendicular to \(NS\)).

Remark 2. It is not difficult to observe that the rays of the azimuth rose may at the same time be considered as coincident positions of vectors (or—as a rose of zenith distances, heights, angles of dip): one need only imagine the rotated horizontal plane \(H\) of the azimuth rose turned about the straight line \(NS\)—the trace of the meridian—(or the straight line \(OW\)) until it coincides with the vertical plane corresponding to that same straight line \(NS\) (or the straight line \(OW\)).

In what follows we shall make use precisely of the last two remarks.

§ 3.

Assuming the construction of the azimuth rose to be known, let us proceed to the construction of a family of concentric circles expressed by marks of zenith distances.

We take an arbitrary vertical axial section (corresponding, for example, to the straight line \(NS\)) of one of the cones, with given: the distance of its vertex—the center—from the plane \(H\) of projections and the angle of its conicity. Drawing through the pole \(O\) rays under various zenith distances, varying through a definite interval, and finding the points of their intersection with one of the generators, we project the latter orthogonally onto the horizontal trace of the vertical section we have taken. The distances of the obtained projections of the points from the pole \(O\) are the required radii of the projections of the circles, as the geometric loci of the projections of the ends of the vecto-

—vectors—that is, marks. It remains now to assign to each of the circles obtained the marks corresponding to the vectors whose zenith distances, as we know, complement one another to \(\pi\).

Let us dwell on this in somewhat more detail. Taking a vertical axial section of the cone and denoting by \(t\) the angle made by the generator with the base of the cone, draw from the pole \(O\) a system of straight lines—vectors, equally inclined to one another at an arbitrary angle \(\alpha\), and such that one of the straight lines \(OP\) is perpendicular to the generator \(KN\) (see Fig. 6).

Then the inclined straight lines drawn on each side of the perpendicular \(OP\) (\(OP'\)) to the generator \(KN(KS)\) will, as is known, cut off on the generator segments that increase successively in the direction from the base of the perpendicular \(P(P')\) toward the point \(N(S)\). Therefore:

\[ PP_1 < P_1P_2 < P_2P_3 < \ldots < P_nN \ldots . \tag{G} \]

Since the segments of the generator \(KN(KS)\) belong to one straight line, their orthogonal projections on the axis \(SN\) will likewise increase successively.

Let us find the projection \(l_x\) of the smallest segment \(l = PP_1\), corresponding to the angle \(\alpha\) chosen by us.

The projection \(l_x\) of the smallest segment \(l\) is equal to the difference of the radii of adjacent concentric circles with marks of zenith distances \(t\) and \(t \pm \alpha\), namely:

\[ l_x = \Delta r \begin{cases} \zeta = t \pm \alpha,\\ \zeta = t . \end{cases} \]

It remains natural now to pass to the determination of the radii \(r_n\) of the circles corresponding to the vectors whose zenith distances \(\zeta_n\) vary by \(\alpha\). Let us take the axis of the cone \(OK\) as the axis \(OZ\), and the horizontal projection of one of the two generators as the axis \(OX\). Draw the vectors (see Fig. 7). We note that the radii \(r_n\) of the successive circles are the coordinates \(x_n\) of the points \(P_n\), corresponding to the points of intersection of the successive radii-vectors with the generator.

The equation of the generator \(KN\) in Hesse normal form is:

\[ x\cos\beta + z\sin\beta - p = 0 \ldots . \tag{1} \]

From the triangle \(OPN\), right-angled at the point \(P\), we have \((\overline{ON}=R)\):

\[ p = R\sin t \]

\[ \beta = \frac{\pi}{2} - t . \]

Equation (1) will take the form:

\[ x\sin t+z\cos t-R\sin t=0 \ldots \ldots \ldots \ldots (2) \]

The equation of the vector \(P_n\):

\[ z=x\operatorname{tg}\delta \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

Equations (2) and (3) together determine the coordinates \(x_n\) and \(z_n\) of an arbitrary point \(P_n\) of the generator.

Substituting (3) into (2), we obtain:

\[ x_n(\sin t+\operatorname{tg}\delta\cos t)=R\sin t; \]

whence:

\[ x_n=\frac{R\sin t\cos\delta}{\sin(t+\delta)}=r_n, \ldots \ldots \ldots (I) \]

a formula determining the radii \(r_n\) of the successive concentric circles.

§ 4.

Determination of \(l_x=\Delta r\begin{matrix}\zeta=t\pm\alpha\\ \zeta=t.\end{matrix}\)

From drawings 6, 7 we find that

\[ l_x=\pm\left(\frac{x}{\delta=\beta\mp\alpha}-\frac{x}{\delta=\beta}\right)\ldots \ldots \ldots (1) \]

\[ \beta=\frac{\pi}{2}-t; \]

therefore:

\[ \beta\mp\alpha=\frac{\pi}{2}-(t\pm\alpha), \]

and hence, by virtue of formula (I), § 3, we have:

\[ \frac{x}{\delta=\beta\mp\alpha}= \frac{R\sin t\sin(t\pm\alpha)}{\cos\alpha}\ldots \ldots \ldots (2) \]

\[ \frac{x}{\delta=\beta}= \frac{R\sin t\sin t}{\sin\frac{\pi}{2}}=R\sin^2t \ldots \ldots (3) \]

Substituting (2) and (3) into (1), we obtain:

\[ l_x=\pm R\sin t\left\{\frac{\sin(t\pm\alpha)}{\cos\alpha}-\sin t\right\}= \]

\[ =\pm R\sin t\left\{\frac{\sin t\cos\alpha\pm\cos t\sin\alpha}{\cos\alpha}-\sin t\right\}= \]

\[ =\pm R\sin t(\sin t\pm\cos t\operatorname{tg}\alpha-\sin t)=R\sin t\cos t\operatorname{tg}\alpha, \]

and finally:

\[ l_x=\Delta r\begin{matrix}\zeta=t\pm\alpha\\ \zeta=t\end{matrix} =\frac{1}{2}R\sin 2t\,\operatorname{tg}\alpha \ldots \ldots \ldots (II) \]

For the given value of \(\alpha\), the greatest value of the smallest segment \(l_x\) corresponds to \(t=\dfrac{\pi}{4}\), whence:

\[ (\Delta r)_{\min}=(l_x)_{\max}=(\Delta r)^{\substack{\zeta=\frac{\pi}{4}\pm\alpha\\[2pt]\zeta=\frac{\pi}{4}}} =\frac{1}{2}R\operatorname{tg}\alpha,\ .\ .\ .\ (III) \]

and inequality (G) will give:

\[ (\Delta r)_{\min}=(\Delta r)^{\substack{\zeta=\frac{\pi}{4}\pm\alpha\\[2pt]\zeta=\frac{\pi}{4}}} < (\Delta r)^{\substack{\zeta=\frac{\pi}{4}\pm2\alpha\\[2pt]\zeta=\frac{\pi}{4}\pm\alpha}} < \]

\[ < (\Delta r)^{\substack{\zeta=\frac{\pi}{4}\pm3\alpha\\[2pt]\zeta=\frac{\pi}{4}\pm2\alpha}} < \ .\ .\ .\ < (\Delta r)^{\substack{\zeta=\frac{\pi}{4}\pm n\alpha\\[2pt]\zeta=\frac{\pi}{4}\pm(n-1)\alpha}} \ .\ .\ .\ (K) \]

With ordinary dimensions of Whatman paper one may freely take \(R=250\) mm. Then putting \(\alpha=15\) minutes [the accuracy of direct measurement of azimuths and zenith distances or dip angles], we find from formula (III):

\[ (\Delta r)_{\min}=\frac{1}{2}\cdot250\cdot\operatorname{tg}15'=0.54541\ .\ .\ .\ \text{mm}. \]

Consequently \((\Delta r)_{\min}\) satisfies the inequality:

\[ 0.6\ \text{mm}>(\Delta r)_{\min}>0.5\ \text{mm}. \]

By virtue of inequalities (K), the difference of the radii of all the remaining successive circles will be greater than \((\Delta r)_{\min}\). Since, with careful drawing, the width of the stroke corresponding to a line does not exceed \(0.1\) mm, it is clear that, within the limits of the accuracy of drawing lines, the proposed conical net can be freely constructed for all azimuths and zenith distances with an interval of 15 minutes.

§ 5.

Construction of roses of azimuths and zenith distances. The conical net.

According to Remark 2 to § 2, it is sufficient to construct only the rose of azimuths. In constructing the angles we shall apply the chord method; in doing so, as is known, it would be sufficient to construct the angles \(n\alpha\) within two arbitrary adjacent quadrants. The independent construction of the angles \(n\alpha\) within two adjacent quadrants will serve as a check on the correctness of the construction within the first.

In our case \(\alpha=15'\); we shall take the accuracy of laying off distances in the plane of the paper as \(\sigma=\pm0.1\) mm. The radius of the auxiliary cir-

circle, whose center coincides with point \(O\)—the point of intersection of the axis of the cone with the horizontal plane \(H\)—is determined, as is known, by the formula:

\[ r = 3438 \, \frac{\sigma \sqrt{1 + 4 \sin^2 \frac{A}{2}}}{\alpha \cos \frac{A}{2}} ; \]

\(r_{\max}\), as is seen from the formula \(\left[0 < A < \frac{\pi}{2}\right]\), will occur at \(A = \frac{\pi}{2}\).

Substituting into the formula the values:

\[ \alpha = 15'; \quad \sigma = 0.1 \text{ mm} \quad \text{and} \quad A = \frac{\pi}{2}, \]

we obtain:

\[ r_{\max} = 56 \text{ mm. (by defect).} \]

In our case \(r_{\max} = R = 250\) mm.

We conclude that the error in constructing the angles (azimuths and zenith distances) will not exceed a value equal to \(\pm 4'\) (by excess); the latter value will be determined from the formula:

\[ \alpha = \pm 3438 \, \frac{\sigma \sqrt{1 + 4 \sin^2 \frac{A}{2}}}{r \cos \frac{A}{2}}, \]

substituting into it the values:

\[ \sigma = 0.1 \text{ mm}; \quad A = \frac{\pi}{2} \quad \text{and} \quad r = R = 250 \text{ mm}. \]

We shall now proceed to the construction of the net of azimuths and zenith distances in connection with the construction of concentric circles as the loci of the projections of the ends of vectors (Fig. 8).

We draw the segment \(OS\), equal to 250 mm; from point \(O\), as center, we describe a circle of radius 250 mm. Further, taking \(OS\) as one of the sides of the angles sought, we construct, at the side \(OS\), successively angles varying through an interval of \(15'\).

Namely: from point \(S\), as center, we describe successive circles, with radii equal to the chords corresponding to \(OS = 250\) mm and to the angles \(n \alpha\);* (\(n\) varies by units from 1 to 360; and \(\alpha = 15'\)).

Drawing the straight line \(SK\) (the position of a generator of the cone, brought into coincidence with the vertical plane of the meridian \(SN\)) and finding the points of its intersection with the rays \(OZ_n\), we obtain the ordinates \((A_o a_o, B_o b_o, C_o c_o \ldots)\),

determining on the straight line \(OS\) the radii \(Oa_0, Ob_0, Oc_0 \ldots\) of the system of concentric circles, as the geometric loci of the projections of the ends of the vectors, and thus we arrive at the conical net.

Remark 1. The conical net is best drawn on millimeter graph paper, as will follow from the examples given below.

Remark 2. In constructing the net, it is precisely the radii \(r_n\) that are better determined by calculation.

We earlier adopted \(t = \dfrac{\pi}{4}\), therefore:

\[ r_n=\frac{\sqrt{2}}{2}\,R\, \frac{\cos n\alpha}{\sin\left(\frac{\pi}{4}+n\alpha\right)} \quad \ldots\ldots [\mathrm{cm}.*] \]

Fig. I.

The conical net is shown in Drawing I, in which the roses of azimuths and zenith distances (dip angles) are constructed at intervals of \(7^\circ .5\). The circle corresponding to \(h=+22^\circ .5\) is tangent, together with the horizontal plane of projections \(H\), to the generatrix of the cone at its middle (which is easy to prove).

Remark 3. In the drawing each circle has a mark corresponding to the angle of inclination of the corresponding vector to the horizontal plane \(H\).

CHAPTER II

The Most Advantageous Conditions for Constructing the Conical Net

§ 6.

We have seen that the determination of the radii of the circles corresponding to the given zenith distances \(\zeta_n\) is connected with the preliminary determination of the points of intersection of the directions \(\zeta_n\), issuing from the pole \(O\), with the corresponding generatrix of the cone.

Naturally, the question may be raised as to the accuracy of determining the position of the points \(P_n\) of intersection of the aforementioned directions issuing from the pole \(O\) with the corresponding generatrices of the cone.

The error in the position of an arbitrary point \(P\) of intersection will depend on the errors in the direction of the vector determined by the zenith distance \(\zeta\) and the azimuth \(a\), and in the direction \(t\) of the generatrix of the cone. In what follows we shall take these errors of direction—\(t\), \(\zeta\), and \(a\)—as equal:

\[ \delta_t=\delta_\zeta=\delta_a=\delta . \]

[\(\delta\) is taken in absolute value].

Taking into account for the time being only \(\delta_t\) and \(\delta_\zeta\), we easily see that any point \(P\) of intersection of the directions \(\zeta\) and \(t\) (Drawing 9) is determined by a “forward intersection” with respect to two fixed (given) points \(O\) and \(S\) (or, respectively, \(O\) and \(N\)).

Drawing 9

Denoting the variable angle \(SPO\) by \(\varphi\), the mean errors of the angles \(POS\) and \(PSO\) by \(\delta\), and the mean error of the position of the point \(P\) by \(M\), we shall have:

\[ M=\frac{\delta}{\sin \varphi}\sqrt{\overline{SP}^{2}+\overline{OP}^{2}} . \]

Remark. For the straight line determined by the points \(S\) and \(K\),

\[ \pi>\varphi>0 . \]

Further, the error \(\delta_a\) of the azimuth of the vector \((OP)\) will give a displacement \(PP_0\) of the point \(P\) along the arc of a circle in the plane passing through the point \(P\) and perpendicular to the axis \(OK\) of the cone.

In Drawing 10: \(OSK\) is the vertical axial section of the cone. \(OK\) is the axis of the cone. \(SK\) is a generatrix of the cone. \(O\) is the pole. \(OP\) is the radius vector \((R,\zeta)\).

Drawing 10

\[ \angle OSK=t;\qquad \angle SPO=\varphi;\qquad \angle PO_1P_0=\delta_a=\delta;\qquad \overset{\frown}{PP_0}=m . \]

As a result of the small displacement of the point \(P\), along the arc \(PP_o\) of the circle, to the point \(P_o\), we shall regard the error \(m\), caused by \(\delta_a\), as a parallel displacement of the plane \(SKO\) by the amount of the perpendicular \(P_oP_1\).

[For \(R = 250\) mm and the angle \(\delta_a = 15'\), the magnitudes of the arc, of the geometrical sine, and of the tangent are equal to \(1.0908\) mm, with an accuracy up to \(\frac{1}{2}\cdot\frac{1}{10^4}\).]

In this case it follows that the mean error (we take the absolute value) of the position of the point \(P\) is:

\[ M_P=\sqrt{M^2+m^2}\ldots\ldots\ldots\ldots\ldots\ldots\ldots (II) \]

Denoting the radius of the circle corresponding to the point \(P\) by \(r\), we shall have:

\[ m=r\delta; \]

and, consequently, (II), in connection with (I), will give:

\[ M_P^2=\frac{\delta^2}{\sin^2\varphi}\left(\overline{SP}^{\,2}+\overline{OP}^{\,2}\right)+r^2\delta^2 = \delta^2\left[\frac{\overline{SP}^{\,2}+\overline{OP}^{\,2}}{\sin^2\varphi}+r^2\right], \]

whence:

\[ \frac{M}{\delta} = \sqrt{\frac{\overline{SP}^{\,2}+\overline{OP}^{\,2}}{\sin^2\varphi}+r^2} \ldots\ldots\ldots\ldots (III) \]

Further, from the drawing we have:

\[ \zeta+\frac{\pi}{2}=t+\varphi, \]

whence:

\[ \zeta=-\left[\frac{\pi}{2}-(t+\varphi)\right]\ldots\ldots\ldots\ldots (1) \]

Denoting \(\overline{OP}=R\), we obtain:

\[ r=R\sin\zeta=-R\cos(t+\varphi)\ldots\ldots\ldots\ldots (2) \]

Putting \(\overline{SO}=b\), from triangle \(SPO\) we obtain:

\[ \overline{OP}=R=b\,\frac{\sin t}{\sin\varphi}\ldots\ldots\ldots\ldots (3) \]

(2), in connection with (3), gives:

\[ r=-b\,\frac{\sin t}{\sin\varphi}\cos(t+\varphi)\ldots\ldots\ldots\ldots (4) \]

From triangle \(SPO\) we have:

\[ \zeta'=\pi-(t+\varphi)\ldots\ldots\ldots\ldots (5) \]

\[ \overline{SP}=b\,\frac{\sin\zeta'}{\sin\varphi}\ldots\ldots\ldots\ldots (6) \]

(6), in connection with (5), gives:

\[ \overline{SP}=b\,\frac{\sin(t+\varphi)}{\sin\varphi}\ldots\ldots\ldots\ldots (7) \]

Substituting the values \(\overline{SP}\), \(\overline{OP}\), and \(r\) into formula (III), we obtain:

\[ M_p^2=\delta^2\left[ \frac{ \dfrac{b^2\sin^2(t+\varphi)}{\sin^2\varphi} + b^2\dfrac{\sin^2 t}{\sin^2\varphi} }{\sin^2\varphi} + \frac{b^2\sin^2 t\cos^2(t+\varphi)}{\sin^2\varphi} \right] = \]

\[ = \frac{b^2\delta^2}{\sin^2\varphi} \left\{ \frac{1}{\sin^2\varphi} \left[\sin^2(t+\varphi)+\sin^2 t\right] + \sin^2 t\cos^2(t+\varphi) \right\}, \]

whence:

\[ \left|\frac{M_p}{b\delta}\right| = \frac{1}{\sin\varphi} \sqrt{ \frac{1}{\sin^2\varphi} \left[\sin^2(t+\varphi)+\sin^2 t\right] + \sin^2 t\cos^2(t+\varphi) }. \]

Denoting

\[ \left|\frac{M_p}{b\delta}\right|=\mu \]

and calling the latter the measure of the accuracy of the determination of the position of the point \(P\),*) we obtain:

\[ \mu= \frac{1}{\sin\varphi} \sqrt{ \frac{1}{\sin^2\varphi} \left[\sin^2(t+\varphi)+\sin^2 t\right] + \sin^2 t\cos^2(t+\varphi) }. \tag{A} \]

—the desired equation of the curve of accuracy for determining the position of the point \(P\).

Taking in formula (A) the quantity \(t\) as a parameter, and \(\varphi\) as a variable changing within the limits \(\pi>\varphi>0\), we could find the extremum of \(\mu\) that interests us for each given \(t\), or the extremum of the same \(\mu\) by taking \(t\) and \(\varphi\) either as independent variables, or as satisfying the condition:

\[ \varphi+t=\frac{\pi}{2}+\zeta, \]

where \(\zeta\) is a parameter.

We omit the analytic solution of this problem.

Let us now turn to the graphical representation of the function \(\mu\), or, more simply, of the function \(b\mu\).

We have:

\[ \left|\frac{M_p}{\delta}\right|=b\mu\ldots\ldots\ldots\ldots (8) \]

On the other hand, by formula (III),

\[ \left|\frac{M_p}{\delta^2}\right| = \sqrt{ \frac{\overline{SP}^{\,2}+\overline{OP}^{\,2}}{\sin^2\varphi} +r^2 } \ldots\ldots\ldots (III_1) \]

Comparing (8) and \((III_1)\), we obtain:

\[ b\mu= \sqrt{ \frac{\overline{SP}^{\,2}+\overline{OP}^{\,2}}{\sin^2\varphi} +r^2 }, \ldots\ldots\ldots (IV) \]

) See Jordan, Handbook of Higher Geodesy*.

whence:

\[ \mu = \sqrt{ \frac{\overline{SP}^{\,2}+\overline{OP}^{\,2}}{\dfrac{b^2}{\sin^2\varphi}} +\left(\frac{r}{b}\right)^2 } \ .\ .\ .\ .\ .\ (V) \]

§ 7.

Graphical construction of the expression \(b\mu\).

Expression (IV) represents the hypotenuse of a right triangle whose legs are:

\[ \frac{\sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}}}{\sin\varphi} \quad \text{and } r. \]

Let us first construct the expression

\[ \frac{\sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}}}{\sin\varphi}, \]

and then (IV).

Construction of the expression
\[ \frac{\sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}}}{\sin\varphi}. \]

Fig. 11.

Suppose we have (Fig. 11) a vertical axial section of the cone. Take an arbitrary point \(P\) of the generator \(SK\) and connect it with the pole \(O\). Next, erect to the straight line \(SK\) at the point \(P\) the perpendicular \(PP_2\), and lay off on it from the point \(P\) the segment \(PO' = PO\). Then the segment

\[ \overline{SO'} = \sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}} \ .\ .\ .\ (1) \]

Then to the same straight line \(SK\) (the generator) we erect at the point \(S\) a second perpendicular \(SC\), and lay off on it from the point \(S\) the segment \(SC = SO'\). Through the point \(C\) we draw the straight line \(CD\), parallel to the straight line \(SK\). We find the point \(D\) of intersection of the straight line \(CD\) with the straight line \(PO\). The segment \(PD\) is the required one.

Indeed: drawing from the point \(D\) the straight line \(DE\), parallel to \(SC\), from the triangle \(PDE\) we find:

\[ \overline{ED}=\overline{PD}\sin\varphi, \]

whence:

\[ \overline{PD}= \frac{\sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}}}{\sin\varphi}. \]

It remains now to lay off the segment \(PD\) from the point \(P\) in the direction \(PO'\). Connecting the points \(P_1\) by a smooth curve, we obtain a graphical representation of the expression

\[ \frac{\sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}}}{\sin \varphi}. \]

Remark. As is easy to see, the curve represents precisely the function \((b\mu_0)\) of the accuracy of determining the position of the points of intersection of the vectors with the generator of the cone, when there exist only the errors \(\delta_\zeta\) and \(\delta_t\) in the directions of the vectors and of the generator of the cone.

Let us now proceed to the graphical representation of the functions (IV). Dropping from the point \(P\) the perpendicular \(\overline{PQ}=r\) to the axis of the cone \(OK\), lay off the segment \(PQ' = PQ = r\) from the same point \(P\) in the direction \(SK\). Drawing through the points \(Q'\) and \(P_1\) straight lines parallel, respectively, to \(PP_1\) and \(SK\), we obtain the point \(P_0\). The segment \(PP_0\) is the required one.

Laying off on the perpendiculars to \(SK\) from the points \(P_n\) the segments \(\overline{PP}_2 = PP_0\), and connecting their ends by a smooth curve, we obtain a graphical representation of the variation of the function \(b\mu\), and consequently it is easy to obtain also \(\mu\) \((b=\text{constans})\).

A similar construction of the functions

\[ \frac{\sqrt{\overline{SP}^{\,2}+\overline{OP}^{\,2}}}{\sin \varphi} \quad \text{and} \quad b\mu \]

is indicated in Fig. II for the value \(t=\dfrac{\pi}{4}\). As we see, the course of the points of the curve is such that \(\mu_{\min}\) corresponds to the angles:

\[ \zeta_0 = 51^\circ 30' \quad \text{and} \quad \varphi_0 = 96^\circ 30'; \ [\varphi_0=\zeta_0+45^\circ]. \]

For the values: \(\zeta>\zeta_0,\ \varphi>\varphi_0\) and \(\zeta<\zeta_0,\ \varphi<\varphi_0\), the ordinates of the curve increase continuously.

§ 8.

From consideration of the graphical representation of the function \(b\mu\) (and consequently also \(\mu\)), corresponding to the angle \(t=\dfrac{\pi}{4}\), we see that the limiting values of this function, corresponding to the zenith distances \(\zeta=0\) and \(\zeta=\dfrac{\pi}{2}\), are not the same.

The same phenomenon will also be observed for other values of the angle \(t\). Naturally, therefore, one may pose the question of determining such a value of the angle \(t\) for which the measures of accuracy corresponding to the limiting values \(\zeta:\dfrac{\pi}{2}\) and \(0\), \(\mu_{\zeta=\frac{\pi}{2}}\) and \(\mu_{\zeta=0}\), would be equal or, at least, would differ little from one another.

Figure 11.

Let us turn to formula (V) of § 6 and consider the following assumptions.

\[ \text{I.}\quad \mu_{\zeta=\frac{\pi}{2}} \]

In the triangle \(SPO\) (Fig. 11) the point \(P\) coincides with \(S\); therefore:

\[ \overline{SP}^{\,2}+\overline{OP}^{\,2}=b^2;\quad r=b\quad\text{and}\quad \varphi=\pi-t. \]

Formula (V) takes the form:

\[ \mu_{\zeta=\frac{\pi}{2}} = \sqrt{\frac{1}{\sin^2 t}+1}, \ldots\ldots\ldots\ldots\ldots\ldots\ldots (I) \]

whence:

\[ \sin t = + \frac{1}{\sqrt{\mu_{\zeta=\frac{\pi}{2}}^{\,2}-1}} \]

\[ \left[ \text{in our case }\frac{\pi}{2}>t>0 \right]. \]

Since the inequality

\[ 0<\sin t<1, \]

must be satisfied, we have:

\[ \mu_{\zeta=\frac{\pi}{2}}^{\,2}-1>1, \]

\[ \zeta=\frac{\pi}{2} \]

and finally:

\[ \mu_{\zeta=\frac{\pi}{2}}>\sqrt{2}. \]

Thus, for any given \(t\) satisfying, in our case, the inequality

\[ \frac{\pi}{2}>t>0, \]

we always have:

\[ \mu_{\zeta=\frac{\pi}{2}}>1.4142\ldots\ldots \]

Let us now consider the assumption:

\[ \text{II.}\quad \mu_{\zeta}=0. \]

In triangle \(SKO\), the point \(P\) coincides with the point \(K\).

\[ \overline{OP}=b\tg t. \]

\[ \overline{SP}=b\sec t, \]

and

\[ OP^{2}+\overline{SP}^{2}=b^{2}(\tang^{2}t+\sec^{2}t)=b^{2}(1+2\tg^{2}t); \]

\[ r=0 \]

\[ \varphi=\frac{\pi}{2}-t; \]

\[ \xi=0 \]

therefore

\[ \sin\varphi=\cos t. \]

Formula (V) of § 6 is transformed into

\[ \mu_{\xi=0}=\frac{1}{\cos t}\sqrt{\,1+2\tg^{2}t\,}\ . . . . . . . . . . . . . . \text{(II)} \]

By the condition

\[ \frac{\pi}{2}>t>0, \]

and, consequently, from formula (II) it follows: the value \(\mu_{\xi=0}\) is such that we always have \(\mu_{\xi=0}>1\).

Thus, as a result:

\[ \mu_{\xi=0}>1 \quad \text{and} \quad \mu_{\xi=\frac{\pi}{2}}>1{,}4142 \ . . . . \]

In connection with what follows, let us transform formula (II). We have:

\[ \mu^{2}_{\xi=0} = \frac{1+2\dfrac{\sin^{2}t}{1-\sin^{2}t}}{\cos^{2}t} = \frac{1-\sin^{2}t+2\sin^{2}t}{(1-\sin^{2}t)^{2}} = \frac{1+\sin^{2}t}{(1-\sin^{2}t)^{2}} \ . . . . . . . . . . . . . . \text{(III)} \]

Let us now find that value of the angle \(t\) for which

\[ \mu_{\xi=\frac{\pi}{2}}=\mu_{\xi=0}. \]

Squaring expression (I) and equating it to expression (III), we obtain:

\[ \frac{1+\sin^{2}t}{\sin^{2}t} = \frac{1+\sin^{2}t}{(1-\sin^{2}t)^{2}} \ . . . . . . . . \text{(IV)} \]

Transforming (IV), we obtain:

\[ \sin^{4}t-3\sin^{2}t+1=0, \]

whence:

\[ \sin t=\pm \sqrt{\frac{3\pm\sqrt{5}}{2}}. \]

According to the condition

\[ \frac{\pi}{2}>t>0, \]

and the problem is satisfied by the \(t\) for which

\[ \sin t=+\sqrt{\frac{3-\sqrt{5}}{2}}. \]

Restricting the computation to seconds, we obtain

\[ t=38^\circ10'22''. \]

Remark. Dividing equation (IV) by

\[ 1+\sin^2 t \]

and extracting the square root from both sides, we obtain:

\[ \frac{1}{\sin t}=\frac{1}{1-\sin^2 t} \]

or

\[ \sin^2 t+\sin t-1=0, \]

whence:

\[ \sin t=\frac{-1\pm\sqrt{5}}{2}. \]

The problem is satisfied by the root

\[ \sin t=\frac{\sqrt{5}-1}{2} \]

and, as before,

\[ t=38^\circ10'22''. \]

Substituting the value of \(t\) thus obtained into each of formulas (I) and (II), we obtain:

\[ \mu_{\zeta=\frac{\pi}{2}}=\mu_{\zeta=0}=1.90\ \text{(with deficiency)}. \]

Thus, as a result of the investigation, we have found that for the angle \(t=38^\circ10'22''\) the measure of accuracy \(\mu\), corresponding to the extreme points of any generator of a right circular cone, is one and the same; namely:

\[ \mu_{\zeta=\frac{\pi}{2}}=\mu_{\zeta=0}=\mu_{s,k}=1.90\ \text{(with deficiency)}. \]

Remark. From formulas (I) and (II) it follows:

1) decreasing \(t\), the measures of accuracy are such that, for values of \(t<38^\circ 10'22''\),

\[ \mu_{\zeta=\frac{\pi}{2}}>\mu_{k,s}\quad \text{and}\quad \mu_{\zeta=0}<\mu_{k,s}. \]

2) increasing \(t\), the measures of accuracy are such that, for values of \(t>38^\circ 10'22''\),

\[ \mu_{\zeta=\frac{\pi}{2}}<\mu_{k,s}\quad \text{and}\quad \eta_{\zeta=0}>\mu_{k,s}. \]

With what has been said, our investigation is not yet finished, as will be confirmed by the following.

The smallest difference \(l_x\) of the radii of two adjacent concentric circles (for the chosen value of the angle \(\alpha\)) must be such that the latter do not merge when drawn. This, as is known, is possible only under the condition:

\[ l_x>0.2\ \mathrm{mm}. \]

Therefore [formula (II), § 4] the expression

\[ l_x=\frac{1}{2}R\operatorname{tg}\alpha\sin 2t \]

must satisfy the inequality:

\[ \frac{1}{2}R\operatorname{tg}\alpha\sin 2t>0.2, \]

whence:

\[ \sin 2t>0.4\,\frac{1}{R}\operatorname{ctg}\alpha. \]

Putting

\[ R=250\ \mathrm{mm}\quad \text{and}\quad \alpha=15', \]

we obtain:

\[ \sin 2t>0.4\,\frac{1}{250}\operatorname{ctg}15', \]

whence

\[ 2t>21^\circ 30'42''\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (1) \]

and

\[ 180^\circ-2t>21^\circ 30'42''\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (2) \]

Correspondingly to (1) and (2), we obtain:

\[ t_1>10^\circ 45'21'' \]

and

\[ t_2<79^\circ 14'39''. \]

Finally:

\[ 10^\circ 45'21''<t<79^\circ 14'39''\ .\ .\ .\ .\ .\ .\ .\ .\ (B) \]

—the condition under which the circles do not merge when drawn, for the given values: \(R=250\ \mathrm{mm}\) and \(\alpha=15'\).

We arrive at the conclusion that the value found above, \(t=38^\circ 10'22''\), satisfies inequality (B).

Let us also give the values:

1) of the least difference of radii \(l_x=(\Delta r)_{\min}\) [formula (II) § 4] and of the height \(h\) of the cone \((h=R\operatorname{tg}t)\), corresponding to the angle \(t=38^\circ10'22''\);

2) \(\mu_\zeta=\dfrac{\pi}{2}\) and \(\mu_\zeta=0\) [formulas (I) and (II)], corresponding to the angle \(t=\dfrac{\pi}{4}\).

We have:

\[ l_x=(\Delta r)_{\min}=0{,}5300037 \ldots \text{ mm.} \]

\[ h=196{,}538181 \ldots \text{ mm.} \]

\[ \mu_\zeta=\frac{\pi}{2}=1{,}73204 \ldots \]

\[ \mu_\zeta=0=2{,}4495 \ldots \]

Comparing the values of \(\mu_\zeta=\dfrac{\pi}{2}\) and \(\mu_\zeta=0\), corresponding, respectively, to the angles \(t=45^\circ\) and \(t=38^\circ10'22''\), we are immediately convinced that it is better to use the cone corresponding to the angle \(t=38^\circ10'22''\).

Remark. It is clear that the possibility is not excluded of using a net constructed for other values of the angle \(t\) [for example \(t=40^\circ, 35^\circ\ldots\)]; in these cases only the condition must be imposed that the values, corresponding to each case, \(\mu_\zeta=\dfrac{\pi}{2}\) and \(\mu_\zeta=0\), differ little from one another.

Drawing 12.

The derivation of formula (A) of § 6 can be made in another way.

Take (Drawing 12) the vertical axial section of a family of cones. Let \(ON\) be the radius of the base of the family of cones. \(OK\) is the direction of the common axis of the family of cones.

\(P\) is the point of intersection of one of the vectors with a generator of one of the cones. \(\angle OPN=\varphi\). Denoting the angles at the base \(ON\) by \(h\) and \(t\), further taking \(ON\) as the axis \(OX\), \(OK\) as the axis \(OY\), and calling the coordinates of the point \(P\) \(x\) and \(y\), from the corresponding triangles we obtain:

\[ \overline{OP}=\frac{b\sin t}{\sin\varphi}\ldots\ldots(3);\qquad \overline{NP}=\frac{b\sin h}{\sin\varphi}\ldots\ldots(4) \]

\[ y=x\operatorname{tg}h\ldots(5);\qquad y=(b-x)\operatorname{tg}t\ldots\ldots(6). \]

Comparing (5) and (6), we obtain:

\[ x\operatorname{tg}h=(b-x)\operatorname{tg}t, \]

whence:

\[ x=\frac{b\operatorname{tg}t}{\operatorname{tg}h+\operatorname{tg}t} =\frac{b\cos h\sin t}{\sin(h+t)}\ldots\ldots(7) \]

Further we have:

\[ h+t+\varphi=\pi\ldots\ldots(8). \]

Eliminating from (4) and (7), with the aid of (8), the angle \(h\), we obtain:

\[ \overline{NP}=b\,\frac{\sin(t+\varphi)}{\sin\varphi}\ldots\ldots\ldots\ldots(9) \]

\[ x=-b\,\frac{\sin t\cos(t+\varphi)}{\sin\varphi}\ldots\ldots\ldots(10) \]

The formulas obtained, (3), (9), and (10), are identical, respectively, to formulas (3), (7), and (4) of § 6.

Representing graphically or analytically the function \(\mu\) of the measure of accuracy [formulas: (IV) and (A) of § 6] for the values of \(t\) determined by inequality (B) of § 8, we find that to each value of \(t\) there will correspond a unique vector (and consequently also a unique value of \(\zeta\)), for which the value of \(\mu\) is a minimum, and hence the position of the point of its intersection with the corresponding conical surface is determined most accurately. Further possible investigation [which we have carried out, but do not present], from the standpoint specifically of its practical use in all cases of assignment, is superfluous (see the remark concerning the determination of the extremum of the measure of accuracy \(\mu\)).

CHAPTER III.

Solutions of auxiliary problems.

§ 9.

For the solution, on the proposed conical net, of the posed problem—determination of the directions of the dip and strike lines—it is necessary to premise the solutions of a number of auxiliary geometrical problems, on which the solution of our main problem rests:

I.

Let two vectors \(OI\) and \(OII\) be given in space. The first and second vectors are specified by the coordinates of the origin \(O\) of the vectors \((O,O,O)\), by azimuths \(a_{1,2}\), zenith distances \(\zeta_{1,2}\), and by the coordinates \(z_{1,2}\) of the ends of the vectors, where:

\[ 0<\zeta_1<\frac{\pi}{2}\quad \text{and}\quad z_1>0. \]

\[ \pi>\zeta_2>\frac{\pi}{2}\quad \text{and}\quad z_2<0. \]

We propose to find the intersection of the plane determined by the vectors with the horizontal plane \(H\) corresponding to the origin \(O\) of the vectors.

Solution. One of the points of the desired intersection, as is evident, is the point \(O\); the other, \(P\), we obtain by joining the ends of the vectors by a straight line and finding the intersection of the latter with the horizontal plane \(H\) (Fig. 13).

Fig. 13.

Remark 1. The reader will easily verify the following: if the \(\zeta_1\) and \(\zeta_2\) of the vectors are the same:

\[ 0<\zeta_{1,2}<\frac{\pi}{2} \]

or

\[ \pi>\zeta_{1,2}>\frac{\pi}{2}, \]

then it is first necessary to transform the azimuth and zenith distance of one of the vectors, without changing its magnitude (and, let us say, \(OI\)), into the directly opposite ones, with the same origin; namely, we take:

\[ a_1' = a_1 \pm \pi . \]

\[ \zeta_1' = \pi-\zeta_1 . \]

Remark 2. a) If the distance between the points \(O\) and \(P\) is small, then it is sufficient (assuming \(\zeta_1\) and \(\zeta_2\) to be unequal) to transform the vectors, without changing their magnitudes, with the same origin \(O\), into directly opposite ones. We shall find the third point \(p_1\) of the required intersection, symmetric to \(p\) with respect to the point \(O\);

b) if \(\zeta_1\) and \(\zeta_2\) are the same, then first we subject one of the vectors to the transformation indicated above (Remark 1).

II.

In the preceding problem it was assumed that the distance between the projections of the ends of the vectors is sufficient. Let this distance now be small. In order in the latter case to avoid a considerable error in drawing the straight line joining the projections of the ends of the vectors, we shall resort to the following construction.

Drawing 14.

Let us imagine two right circular cones with equal cone angles and having a common axis. Let them be specified by their directing circles lying in the horizontal plane \(H\), by the position and elevations of their centers: \(h_1\) and \(h_2\) \((h_1<h_2)\).

From the point \(O\)—the point of intersection of the common axis of the cones with the horizontal plane \(H\)—let us draw two directions \(OP\) and \(OQ\). From Drawing 14 (which represents the horizontal projections of the directions \(OP\), \(OQ\) and of the circles on the conical surfaces corresponding to the vectors of zenith distances \(\zeta_1\) and \(\zeta_2\)) we see directly that the straight line \(cd>ab\). Consequently, the problem is reduced, from the given vectors \(OA\), \(OB\), to determining the vectors \(OC\) and \(OD\). Let us now take the vertical axial section of the cones corresponding to one of the directions, for example \(OP\) (Drawing 15).

Denoting the projections of the points \(A\) and \(C\) of the direction \(OP\), respectively, by \(a\) and \(c\), and setting:

\[ \overline{Oa}=r_a;\quad \overline{Oc}=r_c;\quad \overline{aA}=h_a;\quad \overline{cC}=h_c, \]

from the similar triangles \(O_1AO\) and \(O_2CO\), \(OaA\) and \(OcC\) we obtain:

\[ \frac{h_1}{h_2}=\frac{\overline{OA}}{\overline{OC}}=\frac{\overline{Oa}}{\overline{Oc}}=\frac{\overline{aA}}{\overline{cC}}, \]

or:

\[ \frac{h_1}{h_2}=\frac{r_a}{r_c}=\frac{h_a}{h_c}=\frac{1}{n}, \]

whence:

\[ r_c=nr_a \quad \text{and} \quad h_c=nh_a . \]

We have taken an arbitrary vector; for another vector, consequently, we obtain:

\[ r_d=nr_b \quad \text{and} \quad h_d=nh_b . \]

Drawing 15.

We conclude that, using only one given cone, we can increase the distance between the ends of the vectors by increasing the radii of the corresponding circles (corresponding to the zenith distances \(\zeta_1\) and \(\zeta_2\)) by one and the same number of times.

In suitable cases the application of such a method is also possible.

Mark, on the extensions of the combined positions of the vectors \(OA\) and \(OB\) (for example, with the vertical plane \(SN\)), in a satisfactory manner, one point; the further solution presents no difficulty.

Remark 1. If the values \(r_a\) and \(r_b\) are very small, then it is better to determine them beforehand by formulas (1) of § 3.

Remark 2. In connection with Remark 1 of § 1 it is necessary to note that, for determining the point of intersection of the straight line joining the ends of the vectors with the horizontal plane \(H\), the values \(h_c\) and \(h_d\), when the need arises, are simultaneously increased or decreased an arbitrary number of times in a satisfactory manner.

III.

Determination of the projection of the line of dip.

The determination of the horizontal projection of the line of dip of the plane of a stratum is based on the following geometric image. Suppose we have the horizontal plane of projections \(H\) and some inclined plane \(P\). Take a point \(O\) outside the plane \(P\) (Drawing 16).

Through the point \(O\) draw two straight lines: the normal \(ON\) to the plane \(P\) and the vertical straight line \(OV\). The straight lines \(ON\) and \(OV\) determine the vertical plane \(S\); the traces of this plane on the planes \(P\) and \(H\) are, as is known, respectively the line of dip \(NV\) of the plane \(P\) and its horizontal projection \(MK\). It is easy to show,

Drawing 16.

that the projection \(MK\) of the dip line \(NV\) coincides with the projection of the normal passing through any point of the dip line \(NK\). Hence: the determination of the projection of the dip line reduces to the question of determining the projection of the normal to the plane \(P\).

Drawing 17.

Let us now turn to Drawing 17. Let \(V_1\) and \(V_2\) be the planes of the vertical walls of the shaft.

\(AI\) and \(AII\) are the traces of the bed plane \(P\) on the planes \(V_1\) and \(V_2\), whose azimuths and zenith distances are respectively:

\[ (a_1,\zeta_1)\ \text{and}\ (a_2,\zeta_2). \]

\(RS\) is a plumb line, as the intersection of the planes \(V_1\) and \(V_2\); the latter passes through the point \(A(x_A,y_A,z_A)\)—the point of convergence of the traces \(AI\) and \(AII\).

To determine the normal to the plane \(P\) of the bed, corresponding to the point \(A\),) we take on the vertical line \(RS\) an arbitrary point \(O\) (as in Drawing 16) and through it draw two planes \(Q_1\) and \(Q_2\), normal to the traces \(AI\) and \(AII\). The intersection of the planes \(Q_1\) and \(Q_2\) gives, as is easy to see, the normal \(ON\) to the plane \(P\),) passing through the point \(O\) and corresponding to the dip line \(AN\).**)

Further, the lines of intersection \(OB\) and \(OC\) of the auxiliary planes \(Q_1\) and \(Q_2\) with the vertical planes \(V_1\) and \(V_2\) represent the dip lines of the same auxiliary planes \(Q_1\) and \(Q_2\) [since the planes \(V_1\) and \(V_2\) are vertical and include the normals \(AI\) and \(AII\) to the planes \(Q_1\) and \(Q_2\), they must also include the dip lines of the planes \(Q_1\) and \(Q_2\), as the intersections of the latter with \(V_1\) and \(V_2\)].

From the drawing we immediately see, first, \((AI, OB\) and \(AII, OC\), respectively, lie in the planes \(V_1\) and \(V_2)\), that the dip lines of the auxiliary planes \(Q_1\) and \(Q_2\) are such that their horizontal projections coincide with the horizontal projections of the traces \(AI\) and \(AII\); and, second, \((OB \perp AI,\ OC \perp AII)\), for the plane \(Q_1\) we have \([a,h\ \text{and}\ \zeta\text{—respectively the azimuth, angle of dip, and zenith distance}]\):

\[ h_{BO}=-\left(\frac{\pi}{2}+h_{AI}\right),\quad \text{where } h_{AI}<0. \]

\[ a_{BO}=a_{AI}\pm\pi. \]

\[ \zeta_{BO}=\frac{3}{2}\pi-\zeta_{AI}, \]

*) and therefore the dip line passing through the point \(A\) (point \(V\) in Drawing 16).

**) the lines \(AI\) and \(AII\) are the lines of intersection of the plane \(P\) with the vertical planes \(V_1\) and \(V_2\).

***) compare with Drawing 16.

and, correspondingly, for the plane \(Q_2\):

\[ h_{\mathrm{co}}=-\left(\frac{\pi}{2}+h_{\mathrm{AII}}\right), \quad \text{where } h_{\mathrm{AII}}<0. \]

\[ a_{\mathrm{co}}=a_{\mathrm{AII}}\pm\pi \]

\[ \zeta_{\mathrm{co}}=\frac{3}{2}\pi-\zeta_{\mathrm{AII}}, \]

or, in words: for the planes \(Q_1\) and \(Q_2\), the corresponding lines of dip are such that their azimuths are opposite to the azimuths of the traces, and their angles of dip supplement the angles of dip of the traces to \(\frac{\pi}{2}\) (in absolute value).

Let us consider a special case: the point \(O\) coincides with the point \(A\). The auxiliary planes \(Q_1\) and \(Q_2\), consequently, will pass through the point \(A\). It is clear that we shall arrive at the same conclusions.

All that has been set out above leads us to the conclusion:

The normal to the plane of the bed \(P\) at its point \(A\,(x_A,y_A,z_A)\) is determined by the intersection of two planes specified by the same point \(A\,(x_A,y_A,z_A)\) and by their lines of dip (and, of course, of strike), perpendicular to the traces (in general, to any two arbitrary straight lines of the bed plane intersecting at the point \(A\)).

Remark 1. The solution of the latter problem*) is known. In our case it is sufficient to draw any horizontal plane and find the point of intersection of the obtained corresponding contour lines of the planes \(Q_1\) and \(Q_2\). Joining the point thus obtained by a straight line with the given point \(A\,(x_A,y_A,z_A)\), we obtain the required normal.

Remark 2. We have taken an example (in Fig. [1]) in which the traces AI and AII issuing from the point \(A\) are such that for them, respectively,

\[ h_{I,II}<0\left(\pi>\zeta_{I,II}>\frac{\pi}{2}\right). \]

For other combinations, possible and supplied by practice, namely (see Fig. 17):

\[ [2]\ —\ h_{I,II}>0. \]

\[ [3]\ —\ h_{I,II}\gtreqless 0. \]

\[ [4]\ —\ h_{I,II}\lesseqgtr 0. \]

we shall arrive, as the reader can easily verify, at the same conclusion and shall find, moreover, only one circumstance: the azimuths of the lines of dip of the planes \(Q_1\) and \(Q_2\) in some cases \((h_{I,II}>0)\) coincide with the azimuths of the traces; in others \((h_{I,II}<0)\) they differ by \((\pm\pi)\); and, finally, the third case \((h_I\gtreqless 0,\ h_{II}\lesseqgtr 0)\)—one of the azimuths coincides with the azimuth of the corresponding trace, while the other differs by \((\pm\pi)\) from the azimuth of that same corresponding trace.

Let us now consider solutions of the auxiliary geometrical problems for special cases of the assignment of the lines of dip of the auxiliary planes \(Q_1\) and \(Q_2\).

\[ \overline{\phantom{mmmmmmmmmmmm}} \]

*) problems of the intersection of planes.

IV.

The dip angles of the planes \(Q_1\) and \(Q_2\) are very steep. In this case, when solving by means of the proposed grid, the distances \(r_{s_1}\) and \(r'_{s_1}\) (Fig. 18) of the corresponding horizontals of the auxiliary planes \(Q_1\) and \(Q_2\) from the point \(O\) are small. For an exact determination of the projection of the line of dip of the bed plane \(P\), it is necessary to increase these distances—\(r_{s_1}\) and \(r'_{s_1}\)—by one and the same number of times.

Indeed.

Figure 18.

Take the vertical axial section of the cone corresponding to the line of dip of one of the auxiliary planes, say \(Q_1\). Let the distances of the cutting horizontal planes (whose traces on the vertical axial section are \(S_1\) and \(S_2\)) from the principal plane \(H\) be, respectively, \(h_1\) and \(h_2\). Denote by \(r_{s_1}\), \(r_{s_2}\) the radii of the circles lying, respectively, in the planes \(S_1\) and \(S_2\) and corresponding to the directions of the zenith distance \(\zeta\). From the corresponding triangles we obtain:

\[ \frac{r_{s_1}}{r_{s_2}}=\frac{h_1}{h_2}=\frac{1}{n}, \]

whence:

\[ r_{s_2}=n r_{s_1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

For the other directions of the zenith distance \(\zeta'\) we obtain analogously:

\[ r'_{s_2}=n r'_{s_1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

Remark. If the values \(r_{s_1}\) and \(r'_{s_1}\) are very small, it is better to determine them beforehand from the formulas:

\[ r_{s_1}=h_1 \operatorname{tg}\zeta_1 \]

\[ r'_{s_1}=h_1 \operatorname{tg}\zeta'. \]

Let us give still another proof of formulas (1) and (2). Let (Fig. 19) \(ON_1\) and \(ON_2\) be the projections of the normals to the traces (the projections of the lines of dip of the auxiliary planes \(Q_1\) and \(Q_2\)); \(11_1\), \(11_2\) and \(22_1\), \(22_2\) are the horizontals of the auxiliary planes \(Q_1\) and \(Q_2\), respectively, determining the projection of the normal (line of dip) \(01\). From the drawing it is clear that

\[ \frac{r_{s_1}}{r_{s_2}}=\frac{r'_{s_1}}{r'_{s_2}}=\frac{1}{n}, \]

whence formulas (1) and (2) also follow.

V.

The angles of dip \(h_{Q_1}\) and \(h_{Q_2}\) of the planes \(Q_1\) and \(Q_2\) are very gentle.

In this case the intersection of the traces \(S_n\) of the horizontal planes \(H_{\mathrm{ц}}\) with the coincident directions of the lines of dip is obtained at an acute angle. To avoid intersection at an acute angle, proceed as follows:

Fig. 19.

A). From Fig. 18*) we have, for an arbitrarily chosen \(h_n\),

\[ r_{s_n}=h_n \operatorname{tg}\zeta_{\mathrm{I}} \ . . . . . . . . . . (1) \]

\[ r'_{s_n}=h_n \operatorname{tg}\zeta_{\mathrm{II}} \ . . . . . . . . . (2) \]

\[ [h_1=h_n;\ \zeta=\zeta_{\mathrm{I}};\ \zeta'=\zeta_{\mathrm{II}}]. \]

Dividing (1) by (2), we obtain:

\[ \frac{r_{s_n}}{r'_{s_n}} = \frac{\operatorname{tg}\zeta_{\mathrm{I}}}{\operatorname{tg}\zeta_{\mathrm{II}}} = \frac{\operatorname{ctg}h_{Q_1}}{\operatorname{ctg}h_{Q_2}} = \frac{\operatorname{tg}h_{Q_2}}{\operatorname{tg}h_{Q_1}} \ . . . . . . . . . (3) \]

and finally:

\[ \frac{r_{s_n}}{r'_{s_n}} = \frac{\kappa \operatorname{tg}h_{Q_2}}{\kappa \operatorname{tg}h_{Q_1}} \ . . . . . . . . . (4) \]

\[ \left[\zeta_{\mathrm{I,II}}=\frac{\pi}{2}-h_{\mathrm{I,II}}\right]. \]

Having assigned arbitrarily, but in a reasonable way, a value to \(r'_{s_n}\) (or \(r_{s_n}\)), \(r_{s_n}\) (or \(r'_{s_n}\)) is determined as the fourth proportional, graphically or analytically. It goes without saying that in the graphical determination of \(r_{s_n}\) \((r'_{s_n})\) the numerator and denominator of the right-hand ratio of expression (3) are increased by one and the same number \(\kappa\) of times (see formula 4). The corresponding constructions are indicated in Fig. 20 \((OG=R)\).

Fig. 20.

*) In the drawing: the vertical axial sections of the cone corresponding to the lines of dip of the auxiliary planes \(Q_1\) and \(Q_2\), by rotation about the axis of the cone, are brought into one plane; the auxiliary plane \(S_{\mathrm{ц}}\) is arbitrary.

B). We bring both directions of the traces—by rotation about the axis of the cone—into coincidence with the vertical plane corresponding to the line \(SN\).

Let us take one of the traces. Let the elevation of the auxiliary horizontal plane be \(h\) (Fig. 21).

To the line \(OS(ON)\), at its point \(S(N)\) [the end of the radius \(R\)], we erect a perpendicular \(LS\), and on it, from the point \(S(N)\), lay off the segment \(SM=h\), taken an appropriate number of times, for example \(n\). We obtain the segment \(SS'\). Then, having found the point \(K\) of intersection of the given trace \(OK\) with the perpendicular \(LS\), we lay off the distance \(KS\) along the perpendicular \(LS\) from the point \(S\), taken the same number \(n\) of times. We obtain the segment \(SL\). We join the point \(L\) by a straight line with the point \(O\). Then through the point \(S'\) we draw the straight line \(S'X\), parallel to \(OS\). Having found the point \(X\) of intersection of the straight lines \(S'X\) and \(OL\), we project the latter onto the line \(OS\); the segment \(Oy\) is the required radius. The proof of the construction is left to the reader.

Fig. 21.

CHAPTER IV.

Solution of the fundamental problem by means of a conical net.

§ 10.

In what follows, the solution of the problem is given for the various, four possible, cases of the position of the traces on the vertical walls of the shaft; moreover, independent determinations are presented—first, of the strike line, in the order: the strike line and, respectively, the projection of the dip line of the bed plane (A), then—the projections of the dip line, in the order: the projection of the dip line and, respectively, the strike line, and the dip angle of the bed plane (B).

Remark 1. It is assumed that the conical net is constructed on squared millimeter paper for directions, azimuths \(a\), and zenith distances \(\zeta\) (or dip angles \(h\)), which vary at intervals of 15 minutes.

Remark 2. The azimuths of the strike and dip lines shall be denoted, respectively, by \(a_p\) and \(a_f\).

Drawing III.

The elements determining the positions of the traces shall be, respectively, as follows:

1. Point \(O(x_0,y_0,z_u)\)

\[ a_1 = 225^\circ \]

\[ h_1 = +15^\circ. \]

1. Point \(O(x_0,y_0,z_c)\)

\[ a_2 = 315^\circ \]

\[ h_2 = +30^\circ. \]

A.

Determination of the Strike Line.

1. We determine, brought into coincidence with the horizontal plane \(H\), corresponding to the point \(O\), the positions \(04\) and \(06\) of directions 1 and 2, rotating their horizontal-projecting planes first about the vertical axis \(Z\) until they coincide with the vertical plane \(SN\), and then rotating this latter plane—\(SN\)—with the directions 1 and 2 contained in it, to the right about the trace \(SN\).

1. We determine the points 4 and 6 of intersection of the brought-into-coincidence directions with the brought-into-coincidence position—\(l_{10}, l_{12}\)—of the generator of the cone, and project them orthogonally onto the line \(SN\) (points 3 and 5).

2. From the point \(O\), as center, we describe circles with radii equal to \(03\) and \(05\), respectively.

1. Having changed the direction of the vector \((a_2, h_2)\) to the diametrically opposite one \((a'_2, h'_2)\), we find the points of intersection \(a\) and \(b_2\) of the directions \((a_1, h_1)\) and \((a'_2, h'_2)\) with the corresponding circles.

2. We find the point of meeting \(P(p, p_{01})\) of the line \(AB_2\) of the stratum plane with the horizontal plane \(H\) corresponding to the point \(O\). The horizontal projection of the line is \(ab_2\), the vertical one is \(a''b''_2\).

3. The line \(pO\) is the required strike line (and its projection). For checking and for an exact drawing of the line \(pO\), we also find a second point \(P_1(p_1, p'_{01})\) of the strike line, symmetric to \(P\) (as indicated in the drawing, or independently); \(a_p = \angle NOp_1\).

Determination of the projection of the dip line.

The line \(f_1 f\), perpendicular to \(pp_1\), is the required projection of the dip line; \(a_f = \angle NOf\).

Determination of the angle of dip of the stratum plane.

We project one of the vectors of the stratum plane, for example \(OB_2\) (whose horizontal projection is \(Ob_2\)), orthogonally onto the dip plane of the stratum (whose trace on \(H\) is \(f_1 f\)). The angle formed by the latter projection \(OB_{20}\) with the horizontal plane \(H\), as is known, is the required angle of dip of the stratum plane. Therefore we project the point \(b_2\) onto the projection of the dip line \(f_1 f\) and obtain the point \(b_{20}\). Next—the first method—rotating to the left the vertical dip plane of the stratum \(V_F\) about the vertical axis \(Z\) until it coincides with the vertical plane \(OstW\), and then turning the latter to the left (with the dip line \(OF\) contained in it) about its trace \(OstW\) until it coincides with the horizontal plane \(H\), corresponding to the point \(O\), we find the superposed position \(B_{02}\) of the point \(B_2\) of the dip line (the elevation of the point \(B_{20}\) is equal to the elevation of the point \(B_2\)—the end of the vector \(OB_2\): \(h_{\text{obs.}} = 56\)). The angle \(b_{02}OB_{02}\) is the required angle of dip of the dip line,

or—the second method—we rotate the dip plane \(V_F\) about the vertical axis \(Z\) to the left until it coincides with the vertical plane \(SN\), and this latter (with the line \(OF\) contained in it) we rotate to the left about its trace \(SN\) until it coincides with the horizontal plane \(H\), corresponding to the point \(O\). The point \(B'_{02}\) is the superposed position of the point \(B_{20}\) of the dip line \(OF\) (the elevation of the point \(B_{20}\) is equal to the elevation of the point \(B_2\)). The angle \(NOB'_{02}\) is the required angle of dip of the stratum plane.

Remark. It is clear that, for determining the angle of dip of the stratum plane, one may use any point (chosen in a suitable manner) of either of the two directions. Finally—the third method—by virtue of the remark just made, the inclination angle \(\varphi\) of the dip line may be determined by taking the point \(L_1\) of the same vector \(OB_2\).

Namely, from the point \(l_1\) we drop the perpendicular \(l_1l\) to the projection of the dip line \(f_1 f\), then rotate to the right about the vertical axis \(Z\) the horizontal-projecting plane of the vector \(OL\) until it coincides with the vertical plane \(SN\), and this latter (with the line \(OL\) contained in it) we rotate to the right about the trace \(SN\) until it coincides with the horizontal plane \(H\). The point \(L_0\) is the superposed position of the point \(L\) of the dip line (the elevation of the point \(L\) is equal to the elevation of the point \(L_1\)). The angle \(SOL_0\) is the required angle of dip of the stratum plane.

Let us indicate one more modification of the first method. Namely: we rotate the vertical plane \(V_F\) of the line of dip of the bed plane to the right about the vertical axis \(Z\) until it coincides with the vertical plane \(SN\), and then rotate this latter plane, together with the line of dip contained in it, about the trace \(SN\) until it coincides with the horizontal plane \(H\). In the construction we use the point \(B_2\). The angle \(SOB_{30}\) is the required angle of dip of the line of dip of the bed plane.

B.

Determination of the projection of the line of dip.

Above we showed (see III § 9) that the orthogonal projections onto the horizontal plane \(H\) of the normals to the traces and of the normal to the bed plane at the point \(O\) coincide—each, respectively—with the projections of the traces and of the line of dip.

Therefore, by determining the projection of the normal to the bed plane, we thereby also determine the projection of the line of dip.

We find the positions of the normals to the traces at the point \(O\), brought into coincidence with the vertical plane \(SN\), by rotating each of the horizontal-projecting planes of the traces (normals) first about the vertical axis \(Z\), and then, having turned \(SN\), together with the normals contained in it, about the trace \(SN\) to the right until it coincides with the horizontal plane \(H\). The coincident positions of the normals are the lines \(OU_0\) and \(OT_0\), respectively perpendicular to the coincident positions \(04\) and \(06\) of the trace-vectors (the zenith distances of the normals to the traces are equal to the angles of dip of the traces).

On the basis of what was proved above (see III § 9), to determine the horizontals of the planes \(Q_1\) and \(Q_2\) (for which the normals to the traces are their lines of dip), we draw—chosen in an appropriate manner—an arbitrary horizontal plane \(S\), whose trace on the position of the vertical plane \(SN\) coincident (with the horizontal plane \(H\)) is \(S_v S_v\).

The points \(C_0\) and \(D_0\) are the coincident positions of the points of intersection of the plane \(S\) with the normals to the traces. We find their horizontal projections \(c\) and \(d\).*)

Drawing through the points \(c\) and \(d\) straight lines perpendicular to the projections of the normals \(Oa_1\) and \(Ob_1\), we obtain the required horizontals \(df\) and \(cf\) of the planes \(Q_2\) and \(Q_1\).

The mutual intersection \(f\) of the horizontals \(df\) and \(cf\) is the projection of the point \(F\) of the normal, passing through \(O\), to the bed plane.

In an analogous manner, the point \(f_1\) has also been determined, corresponding to the auxiliary plane \(S_1\), symmetric to the plane \(S\) with respect to the horizontal plane \(H\).**) The points \(f_1\), \(O\), and \(f\) are the points of projection of the corresponding points \(F_1\), \(O\), and \(F\) of the normal to the bed plane. Joining the latter by a straight line, we obtain the required, mutually coincident, orthogonal projections of the normal to the bed plane at the point \(O\) and of the line of dip.

The line \(pp_1\), perpendicular to the line \(f_1 f\), is the line of strike.

*) The points \(C\) and \(D\) are located above the horizontal plane \(H\).

**) The points \(F_1\) and \(F\) are located, respectively, below and above the horizontal plane \(H\).

Determination of the dip angle of the stratum plane.

Let us first determine the dip angle of the normal to the stratum plane at the point \(O\). For this it is sufficient to find the superposed position \(F_0\) of the point \(F\) of the normal. Note that the elevation of the point \(F\)—the elevation of the auxiliary horizontal plane \(S\)—is equal to \(\overline{OK_0}\). Therefore:

1) we rotate the horizontal-projecting plane of the normal to the stratum plane at the point \(O\) about the vertical axis \(Z\) to the left until it coincides with the vertical plane \(OstW\), and then,

2) we rotate this latter plane, with the straight line \(OF\) contained in it, about its trace—\(OstW\)—until it coincides with the horizontal plane \(H\).

The point \(F_0\) is the superposed position of the point \(F\) of the normal to the stratum plane at the point \(O\).

The angle \(NOF_0\) is the required dip angle of the stratum plane [the angle \(F_0, O, Ost\) is the angle of inclination of the normal \(OF\) to the horizontal plane \(H\)].

Remark. The superposed position of the normals to the traces at the point \(O\) may also be obtained in another way.

Indeed: by rotating the horizontal-projecting planes of the normals to the traces about the vertical axis \(Z\) until they coincide with the vertical plane \(OstO\), and then rotating this latter plane, with the normals contained in it, about its trace \(OstO\) to the left until it coincides with the horizontal plane \(H\), we obtain the superposed positions \(04\) and \(06\) of the normals to the traces \(S_v'\), \(S_v'\)—the traces of the auxiliary plane \(S'\) (the elevation of the plane \(S'\) is equal to the elevation of the plane \(S\)). The construction just indicated is obviously preferable to the one given above.

Drawing IV.

A case in which the dip angles of the traces are very gentle. Let the elements determining the positions of the traces be, respectively, as follows:

\[ \begin{aligned} &1.\ \text{Point } O\ (x_0, y_0, z_0)\\ &a_1 = 137^\circ\\ &h_1 = +3^\circ. \end{aligned} \]

\[ \begin{aligned} &2.\ \text{Point } O\ (x_0, y_0, z_0)\\ &a_2 = 44^\circ 30'.\\ &h_2 = -1^\circ. \end{aligned} \]

A.

Determination of the line of strike.

Construction stages 1, 2, 3 are entirely analogous to construction stages 1, 2, 3 of the preceding case.

At stage 4 we immediately see that the directions of none of the vectors need be changed.

At stage 5, in order to determine as accurately as possible the position of the point \(P\), where the straight line \(AB\) meets the horizontal plane \(H\) corresponding to the point \(O\), we increase the elevations of the points \(A\) and \(B\) (see § 1) ten (10) times. Clearly, it is better to carry out the increase of elevations analytically, using the formula:

\[ r = \frac{R \sin t \cos \delta}{\sin(t+\delta)}, \]

given in § 3.

Otherwise the solution is analogous to the preceding case. The determination of the projection of the line of dip is also analogous to the preceding case.

Determination of the angle of dip of the plane of the stratum.

For determining the angle of dip, the best method in the present case is the third method, applied in solving the first case of the problem. The angle \(E_0O l_{01}\) is the required angle of dip of the plane of the stratum.

B.

Determination of the projection of the line of dip.

In the present case, the horizontally projecting planes of the normals to the traces at the point \(O\) are first rotated about the vertical axis \(Z\) until they coincide with the vertical plane \(OOst\), and then this latter plane, together with the normals contained in it, is rotated about

of the trace \(OOst\), to the left, until coincidence with the horizontal plane \(H\), corresponding to point \(O\). It is clear that the superposed positions of the normals will coincide with the found, already superposed, positions of the traces. As in the preceding case, here too, in order to determine the horizontals of the auxiliary planes \(Q_1\) and \(Q_2\), we draw an auxiliary horizontal plane \(S\) (in the drawing, \(S_v S_v\)).

Since in this process circles are obtained (the locus of points corresponding to the ends \(E\) of the normals, respectively having the same elevations and zenith distances) of small radii, then—according to what was proved above (see II § 9)—we increase the latter (radii) by the same number of times (10); namely, we have:

\[ r_1 = (250 \cdot \operatorname{tg} 1^\circ)\,10\ \mathrm{mm}. \]

\[ r_2 = (250 \cdot \operatorname{tg} 3^\circ)\,10\ \mathrm{mm}. \]

Having found the points of intersection \(c, d\) and \(c_1, d_1\)*) of the circles of radii \(r_1\) and \(r_2\) with the corresponding projections of the normals, we draw the required horizontals: \(df\) and \(cf\) determine the horizontal projection \(f\) of one point \(F\) of the normal to the plane of the stratum at the point \(O\), while \(d_1 f_1\) and \(c_1 f_1\) determine the horizontal projection \(f_1\) of another point \(F_1\) of the same normal at the same point \(O\).

Let us note that, in order to determine the dip angle of the plane of the stratum, one cannot use the points \(F\) and \(F_1\), since their elevations, respectively, are equal to \(\pm 2.5\) meters.

The straight line \(f_1 Of\) is the required projection of the dip line. The points \(F\) and \(F_1\) are situated, respectively, above and below the horizontal plane \(H\).

Drawing V.

A case in which one of the dip angles of the traces is steep, the other—gentle.

Let the elements determining the positions of the traces be, respectively, as follows:

\[ \begin{aligned} &1.\ \text{Point } O\,(x_0, y_0, z_0)\\ &a_1 = 328^\circ.\\ &h_1 = -75^\circ. \end{aligned} \]

\[ \begin{aligned} &2.\ \text{Point } O\,(x_0, y_0, z_0)\\ &a_2 = 55^\circ.\\ &h_2 = -3^\circ. \end{aligned} \]

A.

The determination of the projections of the strike line and of the dip line is the same as in the first case.

B.

Let us make several remarks on Drawing V.

1. The horizontally projecting planes of the normals to the traces at the point \(O\) are first rotated about the vertical axis \(Z\) until they coincide with the vertical plane \(OW\), and this latter—\(OW\)—together with the normals contained in it, is rotated about the trace \(WOst\) to the left until it coincides with the horizontal plane \(H\), corresponding to point \(O\). The superposed positions of the normals are the straight lines \(OD_0\) and \(OC_0\).

*) The points \(C\) and \(D\) are situated above the horizontal plane \(H\); the points \(C_1, D_1\), respectively, below the horizontal plane \(H\).

1. The auxiliary plane \(S\) (in the drawing its trace \(S_v S_v\)) is drawn in such a way that, to the superposed position \(D_0\) of the point \(D\)—the intersection of \(S\) with the normal to the trace \((a_1, h_1)\)—there corresponds a circle with an elevation equal to zero. In general, when drawing—in the present

![Drawing I]

case of specifying—the trace \(S_v S_v\) of the auxiliary plane \(S\), one first specifies it by an arbitrary, but suitably chosen, point on the superposed position of the gently inclined normal.

1. The radius of the small circle is best determined by the formula:

\[ r = K_0 C_0 = \overline{OK_0}\,\operatorname{tg}\zeta, \]

where \(\zeta\) is the zenith distance of the normal corresponding to the gentle trace; \(\zeta = |h|\) of the gentle trace; in the present case \(\zeta = 3^\circ\).

1. To determine the angle of dip of the stratum plane, it is better to use either the point \(F(F_1)\) of the normal to the stratum plane (the elevation of point \(F\) is equal to \(d_0D_0\)), or the first method (indicated in the solution of the first case), taking vector 1.

The angle \(F_0OOst\) is the required angle of dip of the stratum plane.

Note. The elevation of point \(F\)—when determining its projection \(f\) and the angle of dip \(\varphi\) of the stratum plane—is best determined from the formula:

\[ H_F = R \operatorname{tg}\zeta', \]

where

\[ R \geq 250\ \text{mm.} \quad \text{and} \quad \zeta' = 90^\circ - |h|; \]

\([h\) is the angle of dip of the steep trace].

Drawing VI.

A case in which the angles of dip of the traces are close to \(\dfrac{\pi}{2}\).

Let the elements determining the positions of the traces be, respectively, as follows:

1. Point \(O(x_0, y_0, z_0)\)

\[ a_1 = 145^\circ \]

\[ h_1 = -84^\circ . \]

1. Point \(O(x_0, y_0, z_0)\)

\[ a_2 = 236^\circ 30' . \]

\[ h_2 = -87^\circ . \]

A.

In the present case:

1. The horizontally projecting planes of the trace-vectors are first rotated about the vertical axis \(Z\) until they coincide with the vertical plane \(OW\), and then this latter plane, together with the traces contained in it, is rotated about the trace \(OW\) to the right until it coincides with the horizontal plane \(H\) corresponding to the point \(O\).

\(OT_0\) and \(OU_0\) are the required positions of the superposed traces, and the angles of their inclination to the horizon are, respectively: \(\hat{h}_1 = WOT_0\) and \(\hat{h}_2 = WOU_0\).

1. Since the radii of the circles corresponding to the trace-vectors are small, then—for a sufficiently accurate determination of the direction of the strike line—they are enlarged eight (8) times (see II § 9); thereafter it is sufficient (see § 1) to reduce the marks of the points \(U\) and \(T\), the ends of the vectors \(OU\) and \(OT\), by a factor of two (2) when determining the strike line (however, the more precise position of the point \(p_{01}\) is determined by taking the marks of the points \(U\) and \(T\) without alteration). The line \(p_1p\) is the required strike line, and the line \(f_1f\) is the required projection of the line of dip of the stratum plane.

B.

The determination of the projections \(f\) and \(f_1\) of the points \(F\) and \(F_1\) (situated respectively above and below the horizontal plane, \(H\)) of the normal to the stratum plane at the point \(O\) is preceded by the constructions set forth above.

The horizontally projecting planes of the normals to the trace-vectors at the point \(O\) are first rotated about the vertical axis \(Z\) until they coincide with the vertical plane \(SN\), and this latter plane, together with the normals contained in it, is rotated about the trace \(SN\) from the observer forward until it coincides with the horizontal plane \(H\).

The angles \(SOU_0\) and \(SOT_0\) are the angles of inclination of the corresponding normals to the horizon.

The constructions corresponding to method (A) are denoted by the same letters as in drawing 20, V, § 9; \(k = 8R\); \(R = 250\) mm.

The trace \(C_0D_0\) \((S_v', S_v')\) corresponds to the line \(S'X\) of drawing 21 [method (B), V, § 9]; \(n = 8\).

Having determined the points \(c_0\) and \(d_0\), we then—for determining the points \(f\) and \(f_1\)—proceed in the same way as in the preceding cases of setting the traces.

The line \(f_1Of\) is the required projection of the dip line; the line \(p_1Op\), perpendicular to the line \(f_1Of\) at the point \(O\), is the strike line. The determination of the dip angle of the bed plane has been carried out using the point \(f_1\). The angle \(F_{01}OOst\) is the required dip angle of the bed plane.

Remark. In all cases of assigning the dip angle of the bed plane, it may be obtained by determining, in consecutive order, the strike line and the projection of the dip line; then, having chosen on the latter, in a suitable manner, an arbitrary point \(f\) or \(f_1\) lying on one of the already drawn circles of equal marks (heights, zenith distances) (the projections of the points \(F\) and \(F_1\)), determine the mark \(F\) (\(F_1\)), knowing the superposed position of the normals.

Indeed, for the latter purpose it is sufficient to drop the perpendiculars \(fc\) and \(fd\) (\(f_1c_1\) and \(f_1d_1\)), draw circles with radii \(Od\) and \(Oc\) (\(Od_1\), \(Oc_1\)), and find the points of their intersection \(c_0\) and \(d_0\) with the lines \(SN\) or \(WO\) (depending on which plane the normals have been superposed with).

Then the segments of the perpendiculars erected at the points \(c_0\) and \(d_0\) to \(SN\) (\(WO\)) and enclosed within the dip angles of the normals will be equal to the mark of the point \(F\).

In conclusion, let us note that with the conical net proposed here, problems relating to displacements*) are readily solved; the solutions of the latter will form the subject of a separate article.

§ 11.

In Remark 2 to § 1 it was indicated that a case is possible in which the transformation of the marks of the ends of a line is impossible; this arises because the mark of one end of the line is large, while that of the other is small, and the dimensions of the plane of the paper are limited.

Then—in order to determine the point of intersection of the line with the horizontal plane \(H\)—one must resort to a grapho-analytic method of solution.

Namely: we compose the equation of the bed plane determined by the traces and by their point of convergence; then we pass to the equation of the strike line passing through the point of convergence of the traces and, finally, having arbitrarily, but suitably, assigned the coordinate \(x_f\) of some point \(F\) of the strike line, we determine from the equation of the latter the other coordinate \(y_f\). The point \(F\) and the point of convergence of the traces graphically determine the strike line.

Composition of the equation of the bed plane.

Let us take the point \(O\) of intersection of the traces (determined as directions) 1 and 2 of the bed plane \(P\) on the vertical walls of the shaft \(V_1\) and \(V_2\) as the origin of coordinates, defined by the intersection of three mutually perpendicular planes: the horizontal \(H\) and two vertical ones, of which one—\(ZOX\)—coincides with the meridian plane. The position of each of the directions, in the coordinate system adopted in observation—the spherical one—is determined, as is known, by the azimuth \(a\) and the zenith distance \(\zeta\). Let us denote the azimuths and zenith distances of the first and second directions, respectively, by \(a_1, \zeta_1\) and \(a_2, \zeta_2\) (Plate 22).

*) on questions of crystallography, and in general on questions connected with directions.

Let us dwell on one of the methods for composing the equation of the plane of a stratum. From the origin \(O\), 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 \(O(0,0,0)\), and the other two are the points where the directions meet the surface of the sphere; the coordinates of the latter two points, as is known, are determined as the projections of the corresponding vectors \((R,a_1,\zeta_1)\) and \((R,a_2,\zeta_2)\) on the coordinate axes.

Fig. 22.

We have:

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

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:

\[ \begin{vmatrix} x & y & z & 1\\ x_1 & y_1 & z_1 & 1\\ x_2 & y_2 & z_2 & 1\\ 0 & 0 & 0 & 1 \end{vmatrix} =0, \]

or:

\[ \begin{vmatrix} y_1 & z_1\\ y_2 & z_2 \end{vmatrix}x - \begin{vmatrix} x_1 & z_1\\ x_2 & z_2 \end{vmatrix}y + \begin{vmatrix} x_1 & y_1\\ x_2 & y_2 \end{vmatrix}z =0 \quad . . . . (2) \]

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:

\[ \begin{vmatrix} \sin\zeta_1\sin a_1,\cos\zeta_1\\ \sin\zeta_2\sin a_2,\cos\zeta_2 \end{vmatrix}x - \begin{vmatrix} \sin\zeta_1\cos a_1,\cos\zeta_1\\ \sin\zeta_2\cos a_2,\cos\zeta_2 \end{vmatrix}y + \]

\[ + \begin{vmatrix} \sin\zeta_1\cos a_1,\sin\zeta_1\sin a_1\\ \sin\zeta_2\cos a_2,\sin\zeta_2\sin a_2 \end{vmatrix}z =0. \]

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

We obtain:

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

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

\[ (\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2 \sin a_2)x - (\operatorname{tg}\zeta_1 \cos a_1-\operatorname{tg}\zeta_2 \cos a_2)y + \operatorname{tg}\zeta_1 \operatorname{tg}\zeta_2 \sin(a_2-a_1)z =0 \qquad \ldots \ldots \ldots (3) \]

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

Equation of the strike line.

We obtain the equation of the strike line of the plane of the stratum by putting \(z=0\) in equation (3). We have:

\[ (\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2 \sin a_2)x - (\operatorname{tg}\zeta_1 \cos a_1-\operatorname{tg}\zeta_2 \cos a_2)y =0 \qquad \ldots \ldots (4) \]

—the required equation of the strike line.

From (4) we determine \(y\):

\[ y= \frac{\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2 \sin a_2} {\operatorname{tg}\zeta_1 \cos a_1-\operatorname{tg}\zeta_2 \cos a_2} \,x, \]

and subsequently we must proceed as indicated at the beginning.

However, the last formula is inconvenient for computation. We shall now reduce it to a form convenient for computation.

Derivation of the logarithmic formula.

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

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

Then the azimuths \(A_1\) and \(A_2\) of the traces, with respect to the new axis \(X'\), will be, respectively:

\[ A_1=-\alpha+a_1;\quad A_2=-\alpha+a_2. \]

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

\[ A_1=\frac{a_1-a_2}{2};\quad A_2=\frac{a_2-a_1}{2}. \]

The equation of the strike line will take the form:

\[ y'= \frac{\operatorname{tg}\zeta_1 \sin A_1-\operatorname{tg}\zeta_2 \sin A_2} {\operatorname{tg}\zeta_1 \cos A_1-\operatorname{tg}\zeta_2 \cos A_2} \,x'. \]

Substituting the values of \(A_1\) and \(A_2\) into the last expression, we obtain:

\[ y'= \frac{ \tg\zeta_1 \sin \frac{1}{2}(a_1-a_2) - \tg\zeta_2 \sin \frac{1}{2}(a_2-a_1) }{ \tg\zeta_1 \cos \frac{1}{2}(a_1-a_2) - \tg\zeta_2 \cos \frac{1}{2}(a_2-a_1) }\,x', \]

or:

\[ y'=Kx'. \]

Let us transform the coefficient \(K\). We have:

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

\[ \left[ \begin{aligned} \text{for: }\quad \tg\zeta_1+\tg\zeta_2 &= \frac{\sin\zeta_1}{\cos\zeta_1} + \frac{\sin\zeta_2}{\cos\zeta_2} = \frac{\sin\zeta_1\cos\zeta_2+\cos\zeta_1\sin\zeta_2}{\cos\zeta_1\cos\zeta_2} \\ &= \frac{\sin(\zeta_1+\zeta_2)}{\cos\zeta_1\cos\zeta_2} \end{aligned} \right]. \]

Thus, finally, the equation of the strike line takes the form:

\[ y'=\tg \frac{1}{2}(a_1-a_2)\, \frac{\sin(\zeta_1+\zeta_2)}{\sin(\zeta_1-\zeta_2)}\,x' \qquad\qquad (I) \]

Directly, graphically, the construction of the \(X'\) axis presents no difficulty, if we recall that, relative to the \(X\) axis, it is determined by the angle

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

Having assigned \(x'\), from equation (I) we determine \(y'\) and construct the point \(F\). The straight line determined by the points \(O\) and \(F\) is the required strike line.

Remark. Introducing the notation:

\[ x''= \frac{\sin(\zeta_1+\zeta_2)}{\sin(\zeta_1-\zeta_2)}\,x' = \tg\varphi\,x', \qquad\qquad (II) \]

we obtain:

\[ y'=\tg \frac{1}{2}(a_1-a_2)\,x'' \qquad\qquad (III) \]

and, consequently, the determination of \(y'\) graphically presents no difficulty. It is easy to see, however, that it is simpler to compute \(y'\) from equation (I) than to construct successively expressions (II) and (III) [the equations of straight lines].

The determination of the attitude elements at a given point of the surface of a stratum can be carried out by an analytical method, which is set forth separately in my article[^1].

Chapter V

Appendix. Analytical Methods for Determining the Attitude Elements at a Given Point of the Stratum Surface

In the present work we shall allow ourselves to present those analytical methods for determining the attitude elements at a given point of the stratum surface which were not included in my article.¹

§ 12.

Determination of the azimuth \(\varphi\) of the line of greatest dip of the stratum plane.

We rotate the rectangular coordinate system \(XYZ\) about the axis \(Z\) through the angle \(\varphi\), corresponding to the azimuth of the line of greatest dip of the stratum plane (Fig. 23). Denoting, respectively, by \((x, y, z)\) and \((x', y', z')\) the coordinates of any point of the line of greatest dip in the systems \(XYZ\) and \(X'Y'Z'\), we have:

\[ \left. \begin{aligned} x &= x'\cos\varphi - y'\sin\varphi,\\ y &= x'\sin\varphi + y'\cos\varphi,\\ z &= z'. \end{aligned} \right\} \qquad \ldots\ldots (1_{1,2}) \]

In the system \(X'Y'Z'\), the coordinate \(y'\) of any point of the line of greatest dip is equal to zero: \(y'=0\) (the axis \(X'\) coincides with the direction of the projection of the line of greatest dip).

Therefore the system \((1_{1,2})\) passes into the system (2):

\[ \left. \begin{aligned} x &= x'\cos\varphi,\\ y &= x'\sin\varphi,\\ z &= z'. \end{aligned} \right\}; \qquad \ldots\ldots (2) \]

whence

\[ \sin\varphi = \frac{y}{x'} \]

\[ \cos\varphi = \frac{x}{x'} \]

and, consequently, the signs of \(\sin\varphi\) and \(\cos\varphi\), taken simultaneously, determine the quadrant corresponding to the sought azimuth \(\varphi\).

Therefore, in vectorial form, the sought azimuth \(\varphi\) of the line of greatest dip is determined by the formula:

\[ \tg\varphi = \frac{\left(\dfrac{y}{x'}\right)} {\left(\dfrac{x}{x'}\right)} \qquad \ldots\ldots (3) \]

Formula (3), as is easy to see, may be simplified; namely, for points of the line of greatest dip corresponding to

\[ z' \geqq 0 \;(\text{or } z \geqq 0), \]

formula (3) passes, respectively, into formulas (4) and (5):

\[ \operatorname{tg}\varphi=\frac{y}{x},\ldots\ldots\ldots\ldots\ldots (4) \]

or \(x' > 0\);

\[ \operatorname{tg}\varphi=\frac{-y}{-x},\ldots\ldots\ldots\ldots\ldots (5) \]

or \(x' < 0\).

Remark. Since we determine the angle \(\varphi\) vectorially, in formulas (4) and (5) the numerator and denominator may be reduced only by a positive quantity.

It remains now, in formulas (4) and (5), to express the coordinates \((x,y)\) of an arbitrary point of the line of dip as functions of \(z\), \((a_1,\zeta_1)\), \((a_2,\zeta_2)\), and, in some cases, also as functions of parameters introduced in advance in the solution.

§ 13.

Determination of the coordinates \((x,y,z)\) of an arbitrary point of the line of dip of the bed plane.

1.

Let us describe, from the origin of coordinates as center, a sphere of arbitrary radius \(R\). The equation of its surface is:

\[ x^2+y^2+z^2-R^2=0\ldots\ldots\ldots\ldots\ldots (1) \]

Next let us imagine a cylindrical surface, mutually tangent with the sphere (1), whose generators are parallel to the line of strike of the bed plane.

The required point of the line of dip will evidently be determined by the intersection of three surfaces: the bed plane, the sphere, and the cylindrical surface.

The derivation of the equation of the cylindrical surface may be carried out in various ways. We shall dwell on two.

1. We have:

\[ nx-my+kz=0\ldots\ldots\ldots\ldots\ldots (2) \]

—the equation of the bed plane.

Putting \(z=0\) in (2), we obtain the equation of the line of strike of the bed plane:

\[ \frac{x}{m}=-\frac{y}{n}=\frac{z}{0}\ldots\ldots\ldots (3) \]

Now take on the surface of the sphere an arbitrary point \(P\) \((\alpha,\beta,\gamma)\).

Through the point \(P\) draw a straight line parallel to (3). Its equation is:

\[ \frac{x-\alpha}{m}=-\frac{y-\beta}{n}=\frac{z-\gamma}{0}\ldots\ldots (4) \]

Since the point \(P\) lies on the surface of the sphere, then

\[ \alpha^2+\beta^2+\gamma^2-R^2=0\ldots\ldots\ldots (5) \]

Let us find the points of intersection of the straight line (4) with the sphere (1).
From (4) we have:

\[ y=\frac{n}{m}(x-\alpha)+\beta \]

\[ z=\gamma . \]

Substituting the obtained values of \(y\) and \(z\) into (1), we obtain:

\[ x^2+\frac{n^2}{m^2}(x-\alpha)^2+2\frac{n}{m}\beta(x-\alpha)+\beta^2+\gamma^2-R^2=0 \]

or

\[ \left(1+\frac{n^2}{m^2}\right)x^2+2n\left(\frac{\beta}{m}-\frac{n}{m^2}\alpha\right)x+\frac{n^2}{m^2}\alpha^2- \]

\[ -2\frac{n}{m}\alpha\beta+\beta^2+\gamma^2-R^2=0 \]

and, finally,

\[ (m^2+n^2)x^2+2n(m\beta-n\alpha)x+\left[n^2\alpha^2-2mn\alpha\beta+ m^2(\beta^2+\gamma^2-R^2)\right]=0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (6) \]

Let us impose the condition that the straight line (4) be tangent to the surface of the sphere (1). Then in equation (6) we must set \(x_1=x_2\); therefore we obtain:

\[ n^2(m\beta-n\alpha)^2-(m^2+n^2)\left[n^2\alpha^2-2mn\alpha\beta+m^2(\beta^2+\gamma^2-R^2)\right]=0=A. \]

We transform the last expression.

\[ A=n^2(n\alpha-m\beta)^2-(m^2+n^2)\left[(n\alpha-m\beta)^2+m^2(\gamma^2-\right. \]

\[ \left.-R^2)\right]=-m^2(n\alpha-m\beta)^2-m^2(m^2+n^2)(\gamma^2-R^2)=0 \]

and, finally,

\[ (n\alpha-m\beta)^2+(m^2+n^2)(\gamma^2-R^2)=0 \ . \ . \ . \ . \ . \ . \ . \ . (7) \]

—the first condition that the parameters \(\alpha,\beta\), and \(\gamma\) must satisfy in order that the straight line (4) be tangent to the sphere.

The second condition, obviously, is expression (5).
The system of equations:

\[ \left. \begin{aligned} &(n\alpha-m\beta)^2+(m^2+n^2)(\gamma^2-R^2)=0\\ &\alpha^2+\beta^2+\gamma^2-R^2=0\\ &\frac{x-\alpha}{m}=\frac{y-\beta}{n}=\frac{z-\gamma}{0} \end{aligned} \right\} \ . \ . \ . \ . \ . \ . \ . \ . (8) \]

determines any generator parallel to a straight line in the plane of the stratum, of the desired cylindrical surface mutually tangent to the sphere (1).

Eliminating the parameters \(\alpha,\beta\), and \(\gamma\) in the system of equations \((8_{1,3})\), we obtain the desired equation of the cylindrical surface.

From \((8_3)\) we have:

\[ (x-\alpha)n=(y-\beta)m \]

\[ z=\gamma \]

or

\[ n\alpha-m\beta=nx-my \ldots\ldots\ldots (9) \]

\[ \gamma=z \ldots\ldots\ldots (10) \]

Substituting (9) and (10) into \((8_1)\), we obtain:

\[ (nx-my)^2+(m^2+n^2)(z^2-R^2)=0 \ldots\ldots (11) \]

—the required equation of the cylindrical surface.

1. Let us take in the plane \(YOZ\) an arbitrary point \(Q(0,\beta,\gamma)\) and draw through it a straight line parallel to (3).

The equation of the straight line passing through the point \(Q\) and parallel to (3) will be:

\[ \frac{x}{m}=\frac{y-\beta}{n}=\frac{z-\gamma}{0}, \ldots\ldots\ldots (12) \]

where the parameters \(\beta,\gamma\) must be chosen so that (12), while remaining parallel to (3), is tangent to the surface of the sphere (1).

Let us find the points of intersection of the straight line (12) with the sphere (1). From (12) we have:

\[ z=\gamma \]

\[ y=\beta+\frac{n}{m}x . \]

Substituting the latter into (1), we obtain:

\[ x^2+\left(\beta+\frac{n}{m}x\right)^2+(\gamma^2-R^2)=0 \]

or

\[ \left(1+\frac{n^2}{m^2}\right)x^2+2\frac{n}{m}\beta x+(\beta^2+\gamma^2-R^2)=0 \]

and, finally,

\[ (m^2+n^2)x^2+2mn\beta x+m^2(\beta^2+\gamma^2-R^2)=0. \]

From the last equation we find two values of \(x\), corresponding to the required points of intersection. Since, by hypothesis, (12) must be tangent to the surface of the sphere (1), then, for the two points of intersection to coincide in one point—the point of tangency—it is necessary to satisfy:

\[ m^2n^2\beta^2-m^2(m^2+n^2)(\beta^2+\gamma^2-R^2)=0 \]

or

\[ n^2\beta^2-(m^2+n^2)(\beta^2+\gamma^2-R^2)=0 \]

and, finally,

\[ m^2\beta^2+(m^2+n^2)(\gamma^2-R^2)=0 \]

—the condition relating the parameters \(\beta\) and \(\gamma\), under which (12) is tangent to the surface of the sphere (1).

Thus the equations:

\[ \left. \begin{aligned} \frac{x}{m}&=\frac{y-\beta}{n}=\frac{z-\gamma}{0}\\ m^2\beta^2+(m^2+n^2)(\gamma^2-R^2)&=0 \end{aligned} \right\} \ldots\ldots (13_{1,2}) \]

taken together, determine any generator of the required cylindrical surface.

Eliminating from them the parameters \(\beta\) and \(\gamma\), we obtain the equation of the required cylindrical surface.

From \((13_1)\) we have:

\[ \beta = y - \frac{n}{m}x \]

\[ \gamma = z . \]

Substituting the latter into \((13_2)\), we obtain:

\[ m^2 \left( y - \frac{n}{m}x \right)^2 + (m^2+n^2)(z^2-R^2)=0 \]

and, finally,

\[ (nx-my)^2 + (m^2+n^2)(z^2-R^2)=0 \]

—the equation of the required cylindrical surface tangent to the surface of the sphere \((1)\).

The coordinates of the required point of the line of dip of the bed plane are the solution of the system of equations:

\[ \left. \begin{aligned} nx-my+kz&=0\\ x^2+y^2+z^2-R^2&=0\\ (nx-my)^2+(m^2+n^2)(z^2-R^2)&=0 \end{aligned} \right\} \qquad \ldots\ldots (14) \]

From \((14_1)\) we have:

\[ nx-my=-kz; \]

substituting the latter into \((14_3)\), we obtain:

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

whence

\[ z_{1,2}=\pm R\sqrt{\frac{m^2+n^2}{m^2+n^2+k^2}} \qquad \ldots\ldots (15) \]

We now determine the values of the coordinates \(x\) and \(y\), which alone we need; for it is geometrically clear that we shall obtain two values \(z_{1,2}\) equal in absolute magnitude.

From \((14_2)\) we have:

\[ z^2-R^2=-(x^2+y^2). \]

Substituting the latter into \((14_3)\), we obtain:

\[ (nx-my)^2-(m^2+n^2)(x^2+y^2)=0 \qquad \ldots\ldots (16) \]

From \((14_1)\) we have:

\[ y=\frac{1}{m}(nx+kz) \qquad \ldots\ldots (17) \]

Substituting the latter into \((16)\), we obtain:

\[ m^2k^2z^2-m^2(m^2+n^2)x^2-(m^2+n^2)(n^2x^2+2nkzx+k^2z^2)=0 \]

or

\[ \bigl[m^2(m^2+n^2)+(m^2+n^2)n^2\bigr]x^2 +2n\kappa z(m^2+n^2)x -\bigl[m^2\kappa^2 z^2-(m^2+n^2)\kappa^2 z^2\bigr]=0 \]

and, finally,

\[ (m^2+n^2)^2x^2+2n\kappa z(m^2+n^2)x+n^2\kappa^2z^2=0 \]

or

\[ \bigl[(m^2+n^2)x+n\kappa z\bigr]^2=0, \]

whence

\[ x_{1,2}=-\frac{n\kappa}{m^2+n^2}\,z_{1,2}; \qquad . . . . . . . . . (18) \]

substituting further the value found for \(x_{1,2}\) into (17), we obtain:

\[ y_{1,2}=-\frac{m\kappa}{m^2+n^2}\,z_{1,2} \qquad . . . . . . . . . (19) \]

It remains now to apply formulas (4) and (5) of § 12.

Remark. In formulas (18) and (19) there enters the quantity

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

which we may take to be either positive or negative.

In my article¹ (§ 5) it is indicated that the sign of the quantity “\(\kappa\)” determines the order of designation of the traces \([(a_1,\zeta_1); (a_2,\zeta_2)]\).

Let us suppose that \(\kappa>0\).

Then, for \(z_{1,2}\lessgtr 0\), we shall have, respectively, according to formulas (4) and (5) of § 12:

\[ \operatorname{tg}\varphi = \frac{-y_1}{-x_1} = \frac{ -\dfrac{m\kappa}{m^2+n^2}z_1 }{ -\dfrac{n\kappa}{m^2+n^2}z_1 } = -\frac{m}{n}; \]

\[ \operatorname{tg}\varphi = \frac{y_2}{x_2} = \frac{ -\dfrac{m\kappa}{m^2+n^2}(-z_2) }{ -\dfrac{n\kappa}{m^2+n^2}(-z_2) } = -\frac{m}{n}, \]

for \((-z_2)>0\).

It is easy to verify that, taking \(\kappa<0\), we obtain:

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

II.

Determination of an arbitrary point of the line of dip of the plane of the stratum by the intersection of three planes.

The equation of the plane of the stratum:

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

The equation of an arbitrary horizontal plane

\[ z=H . . . . . . . . . . (2) \]

The equation of the plane of dip of the stratum (§ 5¹):

\[ \left. \begin{array}{r} mx+ny=0\\ z=z \end{array} \right\} \ldots\ldots\ldots\ldots\ldots (3_{1,2}) \]

The coordinates \((x_0,y_0)\) of the point of the line of dip are obtained by substituting (2) in (1), and then solving the system of equations:

\[ \begin{aligned} nx-my+\kappa H&=0\\ mx+ny&=0 \end{aligned} \]

We have:

\[ x_0=-\,\frac{ \begin{vmatrix} \kappa H,&-m\\ 0,&n \end{vmatrix} }{ \begin{vmatrix} n,&-m\\ m,&n \end{vmatrix} } =-\frac{\kappa n}{m^2+n^2}\,H \]

\[ y_0=-\,\frac{ \begin{vmatrix} n,&\kappa H\\ m,&0 \end{vmatrix} }{ \begin{vmatrix} n,&-m\\ m,&n \end{vmatrix} } =+\frac{\kappa m}{m^2+n^2}\,H . \]

\[ z_0=H. \]

It remains now to apply formulas (4) and (5) of § 12.

III.

Let us describe, from the origin of coordinates as center, a sphere of arbitrary radius \(R\).

The equation of the surface of the sphere:

\[ x^2+y^2+z^2-R^2=0\ldots\ldots\ldots\ldots (1) \]

Take the equation of the normal plane (see § 10 of article ¹):

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

The equation of the pencil of planes parallel to (2) will be:

\[ nx-my-\frac{1}{\kappa}\left(m^2+n^2\right)z+c=0\ldots\ldots (3) \]

From the pencil of planes parallel to (2), choose that plane which is at a distance equal to \(R\) from (2).

This latter plane will be tangent to the surface of the sphere at the sought point—the point of the line of dip.

By the formula determining the distance between parallel planes, we have:

\[ R=\left|\frac{c}{\sqrt{m^{2}+n^{2}+\frac{1}{\kappa^{2}}\left(m^{2}+n^{2}\right)^{2}}}\right|, \]

whence

\[ c=\pm \frac{R}{\kappa}\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}. \]

Thus, the equation of the plane tangent to the surface of the sphere at the desired point will be:

\[ nx-my-\frac{1}{\kappa}\left(m^{2}+n^{2}\right)z \pm \frac{R}{\kappa}\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}=0. \tag{4} \]

We can determine the desired point in two ways:
1, by solving the system of equations—

\[ \left. \begin{aligned} &nx-my+\kappa z=0\\ &nx-my-\frac{1}{\kappa}\left(m^{2}+n^{2}\right)z \pm \frac{R}{\kappa}\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}=0\\ &x^{2}+y^{2}+z^{2}-R^{2}=0 \end{aligned} \right\}\ . . (5), \]

or 2, proceeding as follows:

The equation of the plane tangent to the surface of the sphere at an arbitrary point of it \((x_{0}, y_{0}, z_{0})\), as is known, is

\[ x_{0}x+y_{0}y+z_{0}z-R^{2}=0 . . . . . . . . (6) \]

Next, multiplying equation (4) by

\[ \mp \frac{R\kappa}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}}, \]

we bring it to the form:

\[ \pm \frac{Rn\kappa}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}}\,x \mp \frac{Rm\kappa}{\sqrt{(m^{2}+n^{2})(m^{2}+n^{2}+\kappa^{2})}}\,y + \]

\[ \pm R\sqrt{\frac{m^{2}+n^{2}}{m^{2}+n^{2}+\kappa^{2}}}\,z -R^{2}=0 . . . . . . . . (7) \]

Comparing now (6) and (7), we obtain the coordinates \((x_0, y_0, z_0)\) of the sought point:

\[ x_0=\pm \frac{R n \kappa}{\sqrt{(m^2+n^2)(m^2+n^2+\kappa^2)}} \]

\[ y_0=\pm \frac{R m \kappa}{\sqrt{(m^2+n^2)(m^2+n^2+\kappa^2)}} \]

\[ z_0=\pm R \sqrt{\frac{m^2+n^2}{m^2+n^2+\kappa^2}}. \]

IV.

Determination of the coordinates of an arbitrary point of the line of dip of the stratum plane by the intersection of the circle

\[ x^2+y^2-R^2=0 \ldots \ldots \ldots \ldots \ldots (1) \]

and the projection of the line of dip [see II, \((3_1)\)]

\[ \frac{x}{-n}=\frac{y}{m} \ldots \ldots \ldots \ldots \ldots (2) \]

From (2) we have:

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

Substituting the latter into (1), we obtain:

\[ \left(1+\frac{m^2}{n^2}\right)x^2-R^2=0, \]

whence

\[ x_{1,2}=\pm \frac{Rn}{\sqrt{m^2+n^2}} \]

and, correspondingly,

\[ y_{1,2}=\mp \frac{Rm}{\sqrt{m^2+n^2}}. \]

We determine the corresponding value \(z_{1,2}\) from the equation of the stratum plane:

\[ nx-my+\kappa z=0 \ldots \ldots \ldots \ldots \ldots (3) \]

Substituting in (3) the values \(x_{1,2}\) and \(y_{1,2}\), we obtain:

\[ z_{1,2}=\frac{-nx+my}{\kappa} = \frac{1}{\kappa} \left\{ \mp \frac{Rn^2}{\sqrt{m^2+n^2}} \mp \frac{Rm^2}{\sqrt{m^2+n^2}} \right\} = \]

\[ =\mp \frac{R}{\kappa}\sqrt{m^2+n^2}. \]

V.

Determination of the horizontal projection of the point of the line of dip of the bed plane, as the point of tangency with the circle

\[ x^2 + y^2 - R^2 = 0 \]

of a line parallel to the line of strike, lying in the plane \(XOY\).

Let us describe, from the origin of coordinates as center, in the plane \(XOY\), the circle:

\[ x^2 + y^2 - R^2 = 0 \ldots \ldots \ldots \ldots \ldots (1) \]

The equation of the line of strike is

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

The equation of the pencil of lines parallel to (2) will be:

\[ nx - my + c = 0 \ldots \ldots \ldots \ldots \ldots (3) \]

From the pencil of parallel lines (3) we choose that line which is at a distance from (2) equal to \(R\).

The latter is tangent to the circle at the required point—the horizontal projection of the point of the line of dip of the bed plane.

By the formula for determining the distance between parallel straight lines, we have:

\[ R = \left| \frac{c}{\sqrt{m^2+n^2}} \right|, \]

whence

\[ c = \pm R\sqrt{m^2+n^2}. \]

Thus, we have two equations of one and the same tangent to the circle (1) at the required point:

\[ nx - my \pm R\sqrt{m^2+n^2} = 0 \ldots \ldots \ldots \ldots (4) \]

\[ x_0x + y_0y - R^2 = 0 \ldots \ldots \ldots \ldots (5) \]

Let us present equation (4) in another form, multiplying (4) by

\[ \mp \frac{R}{\sqrt{m^2+n^2}}. \]

We obtain:

\[ \mp \frac{Rn}{\sqrt{m^2+n^2}}\,x \pm \frac{Rm}{\sqrt{m^2+n^2}}\,y - R^2 = 0 \ldots \ldots (6) \]

Comparing (6) and (5), we have:

\[ x_0 = \mp \frac{Rn}{\sqrt{m^2+n^2}} \ldots \ldots \ldots \ldots (7) \]

\[ y_0 = \pm \frac{Rm}{\sqrt{m^2+n^2}} \ldots \ldots \ldots \ldots (8) \]

The obtained formulas (7) and (8), as was to be expected, are identical with the formulas of the preceding method.

VI.

Determination of the horizontal projection of a point of the line of dip, as the pole of an arbitrary straight line parallel to the strike line of the bed plane, lying in the \(XOY\) plane, with respect to the circle

\[ x^2 + y^2 - R^2 = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

The equation of the strike line is

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

The equation of an arbitrary straight line parallel to (2):

\[ nx - my + c = 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

The required point—the pole—as is known, is determined by the coordinates:

\[ x_0 = - \frac{nR^2}{c} \ldots \ldots \ldots \ldots \ldots \ldots \ldots (4) \]

\[ y_0 = \frac{mR^2}{c} \ldots \ldots \ldots \ldots \ldots \ldots \ldots (5) \]

The coordinate \(z_0\) of the bed plane

\[ nx - my + \kappa z = 0, \ldots \ldots \ldots \ldots \ldots \ldots (6) \]

corresponding to \(x_0\) and \(y_0\), is determined by substituting the values (4) and (5) into (6). We obtain:

\[ z_0 = \frac{R^2}{\kappa c}\left(m^2+n^2\right). \]

VII.

Determination of a point of the line of dip as the point of tangency, to the given bed plane, of a sphere whose center lies on the \(Z\)-axis.

Take on the \(Z\)-axis an arbitrary point \((0,0,z_c)\) and, from it as center, describe a sphere of radius \(R\) tangent to the bed plane.

The equation of the surface of the sphere is

\[ x^2 + y^2 + (z-z_c)^2 - R^2 = 0 \ldots \ldots \ldots \ldots \ldots (1) \]

The equation of the bed plane is

\[ nx - my + \kappa z = 0 \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

Let us write the equation of the plane tangent to (1) at an arbitrary point of it \((x_0,y_0,z_0)\).

Its equation is:

\[ x_0x + y_0y + (z_0-z_c)(z-z_c) - R^2 = 0 \ldots \ldots (3) \]

Now let plane (3) coincide with plane (2). Let us write the condition under which plane (3) coincides with (2). From this condition, obviously, we determine the required point \((x_0,y_0,z_0)\).

We have:

\[ x_0x + y_0y + (z_0-z_c)(z-z_c) - R^2 = \lambda(nx - my + \kappa z). \]

Whence

\[ x_0 = \lambda n \ldots \ldots \ldots \ldots \ldots \ldots \ldots (4) \]

\[ y_0 = -\lambda m \ldots \ldots \ldots \ldots \ldots \ldots \ldots (5) \]

\[ z_0 - z_c = \lambda \kappa \ldots \ldots \ldots \ldots \ldots \ldots (6) \]

\[ z_c(z_0-z_c) + R^2 = 0 \ldots \ldots \ldots \ldots \ldots (7) \]

Equation (7) serves, obviously, only to determine the radius \(R\) of the spherical surface.

The required point is determined by solving the system: (2), (4), (5), (6). Substituting into (2) the values \(x_0, y_0, z_0\) from (4), (5), (6), we obtain:

\[ \lambda(m^2+n^2+\kappa^2)=-\kappa z_c, \]

whence

\[ \lambda=-\frac{\kappa}{m^2+n^2+\kappa^2}z_c \]

and, consequently,

\[ x_0=-\frac{\kappa n}{m^2+n^2+\kappa^2}z_c \]

\[ y_0=\frac{\kappa m}{m^2+n^2+\kappa^2}z_c. \]

Without determining \(z_c\) as a function of \(z_0\), it is easy to see that the signs of \(z_0\) and \(z_c\) are the same.

Therefore, putting \(\kappa>0\), for \(z_{c(o)}<0\), according to (4) § 12, we obtain:

\[ \operatorname{tg}\varphi=\frac{y_0}{x_0} =\frac{m\kappa z_c}{-n\kappa z_c} =\frac{-m\kappa(-z_c)}{n\kappa(-z_c)} =\frac{-m}{n}, \]

or \((-z_c)>0\);

for \(z_{c(o)}>0\), according to (5) § 12, we have:

\[ \operatorname{tg}\varphi=\frac{-y_0}{-x_0} =\frac{-m\kappa z_c}{n\kappa z_c} =\frac{-m}{n}, \]

or \(z_c>0\).

VIII.

Determination of the point of the line of dip of the plane of the stratum, as the pole of a polar plane parallel to the plane of the stratum, with respect to a sphere whose center lies on the \(Z\)-axis \((0,0,z_c)\).

The equation of the surface of the sphere

\[ x^2+y^2+(z-z_c)^2-R^2=0 \ldots \ldots \ldots \ldots (1) \]

The equation of the plane of the stratum

\[ nx-my+\kappa z=0 \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

The equation of a plane parallel to the plane of the stratum:

\[ nx-my+\kappa z+c=0 \ldots \ldots \ldots \ldots \ldots (3) \]

We take (3) as the polar plane.

Denoting by \((x_0,y_0,z_0)\) the coordinates of the pole, we shall have:

\[ x_0x+y_0y+(z_0-z_c)(z-z_c)-R^2 =\lambda(nx-my+\kappa z+c)\ldots \ldots (4) \]

Whence

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

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

\[ z_0-z_c=\lambda\kappa \ldots \ldots \ldots \ldots \ldots \ldots (7) \]

\[ z_c(z_0-z_c)+R^2=-\lambda c \ldots \ldots \ldots \ldots \ldots (8) \]

Since, by hypothesis, the pole \(P_0(x_0,y_0,z_0)\) lies on the given plane (2), we have:

\[ nx_0-my_0+\kappa z_0=0. \]

Substituting here the values \((x_0, y_0, z_0)\) from (5), (6), (7), we determine the parameter

\[ \lambda = - \frac{\kappa}{m^2+n^2+\kappa^2}\, z_c , \]

and, respectively, (5), (6),

\[ x_0 = - \frac{\kappa n}{m^2+n^2+\kappa^2}\, z_c \]

\[ y_0 = \frac{\kappa m}{m^2+n^2+\kappa^2}\, z_c . \]

IX.

Determination of an arbitrary point of the line of dip as the center of the circle of intersection, by the given plane of the stratum, of the surface of an auxiliary sphere whose center is located on the \(Z\)-axis.

Take an arbitrary point on the \(Z\)-axis \((0,0,z_c)\). Let us take this point as the center of a sphere of radius \(R\).

The equation of the surface of the sphere is

\[ x^2 + y^2 + (z-z_c)^2 - R^2 = 0 . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (1) \]

The equation of the given plane is

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

Together, (1) and (2) determine a circle whose center is a point of the line of dip.

Eliminating the variable \(z\) from (1) and (2), we find the projection—an ellipse—onto the plane \(z=0\) \((XOY)\).

From (2) we have:

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

Substituting \(z\) into (1), we obtain:

\[ x^2 + y^2 + \frac{(nx-my)^2}{\kappa^2} - 2\,\frac{-nx+my}{\kappa}\,z_c + z_c^2 - R^2 = 0, \]

or

\[ (x^2+y^2)\kappa^2 + n^2x^2 - 2mnxy + m^2y^2 + 2n\kappa z_c x - 2m\kappa z_c y + \kappa^2(z_c^2-R^2)=0; \]

and, finally,

\[ (n^2+\kappa^2)x^2 - 2mnxy + (m^2+\kappa^2)y^2 + 2n\kappa z_c x - 2m\kappa z_c y + \kappa^2(z_c^2-R^2)=0 \]

—the desired equation of the ellipse. As is known, for a curve of the second order

\[ a_{11}x^2 + 2a_{12}xy + a_{22}y^2 + 2a_{13}x + 2a_{23}y + a_{33}=0, \]

the coordinates \((x_0, y_0)\) of the center of the curve are determined by the formulas:

\[ x_0 = -\,\frac{ \begin{vmatrix} a_{13} & a_{12}\\ a_{23} & a_{22} \end{vmatrix} }{ \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} }; \qquad y_0 = -\,\frac{ \begin{vmatrix} a_{11} & a_{13}\\ a_{21} & a_{23} \end{vmatrix} }{ \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} }. \]

where

\[ a_{21}=a_{12}. \]

In the present case

\[ x_0=-\frac{ \begin{vmatrix} n\kappa z_c,\ -mn\\ -m\kappa z_c,\ m^2+\kappa^2 \end{vmatrix} }{ \begin{vmatrix} n^2+\kappa^2,\ -mn\\ -mn,\ m^2+\kappa^2 \end{vmatrix} } = \]

\[ =-\frac{ n\kappa z_c \begin{vmatrix} 1,\ -m\\ -m,\ m^2+\kappa^2 \end{vmatrix} }{ (m^2+\kappa^2)(n^2+\kappa^2)-m^2n^2 } = -n\kappa z_c\frac{\kappa^2}{n^2\kappa^2+\kappa^2(m^2+\kappa^2)} = \]

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

\[ y_0=-\frac{ \begin{vmatrix} n^2+\kappa^2,\ n\kappa z_c\\ -mn,\ -m\kappa z_c \end{vmatrix} }{ \kappa^2(m^2+n^2+\kappa^2) } = -\frac{ -m\kappa z_c \begin{vmatrix} n^2+\kappa^2,\ n\\ n,\ 1 \end{vmatrix} }{ \kappa^2(m^2+n^2+\kappa^2) } = \]

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

Remark. The reader may convince himself a priori that the last three solutions must give precisely identical results.

X.

Determination of the point of the line of dip of the plane of the stratum, as the center of the circle of section of the surface of the auxiliary sphere

\[ x^2+y^2+z^2-R^2=0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (1) \]

by an arbitrary plane parallel to the normal plane

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

The equation of an arbitrary plane parallel to (2):

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

Eliminating the variable \(z\) from (1) and (3), we obtain the projection—an ellipse—on the plane \(XOY\).

From (3) we have:

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

Substituting (4) into (1), we obtain:

\[ x^2+y^2+\frac{\kappa^2(nx-my+c)^2}{(m^2+n^2)^2}-R^2=0 \]

or

\[ \bigl[(m^2+n^2)^2+\kappa^2n^2\bigr]x^2+ \bigl[(m^2+n^2)^2+\kappa^2m^2\bigr]y^2- \]

\[ -2mn\kappa^2xy+2n\kappa^2cx-2m\kappa^2cy+ \bigl[\kappa^2c^2-R^2(m^2+n^2)^2\bigr]=0 \qquad \ldots\ldots (5) \]

—the equation of an ellipse.

From (5) we determine the coordinates \((x_0,y_0)\) of the center of the ellipse:

\[ x_0=-\frac{n\kappa^2c}{(m^2+n^2)(m^2+n^2+\kappa^2)} \qquad \ldots\ldots (6) \]

\[ y_0=\frac{m\kappa^2c}{(m^2+n^2)(m^2+n^2+\kappa^2)} \qquad \ldots\ldots (7) \]

Finally, the coordinate \(z_0\) is more simply determined from the equation of the plane of the stratum:

\[ nx-my+\kappa z=0 \qquad \ldots\ldots (8) \]

Substituting (6) and (7) into (8), we obtain:

\[ z_0=\frac{\kappa c}{m^2+n^2+\kappa^2}. \]

XI.

Let us imagine a right conical surface whose center and axis coincide, respectively, with the origin of coordinates and the \(Z\)-axis; the directrix of the conical surface is the circle:

\[ x^2+y^2-R^2=0 \]

\[ z=c. \]

The given plane,

\[ nx-my+\kappa z=0, \]

of the stratum intersects the conical surface along two generators, symmetric with respect to the line of dip of the plane of the stratum.

An arbitrary point \(P\) of the line of dip of the plane of the stratum is determined, for a given value \(Z=H\), evidently by the coordinates:

\[ x_p=\frac{1}{2}(x_1+x_2) \]

\[ y_p=\frac{1}{2}(y_1+y_2) \]

\[ z_p=z_{1,2}=H, \]

where \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\), corresponding to \(z_{1,2}\), are the coordinates of the points of the two generators, respectively.

We have:

\[ \frac{x^2+y^2}{R^2}-\frac{z^2}{c^2}=0 \ldots \ldots \ldots \ldots \ldots (1) \]

—the equation of a conical surface.

\[ nx-my+\kappa z=0 \ldots \ldots \ldots \ldots \ldots (2) \]

—the equation of the plane of the stratum.

Eliminating the variable \(z\) from equations (1) and (2), we obtain the projection of the curve of intersection onto the plane \(XOY\).

From (2) we have:

\[ z=\frac{-nx+my}{\kappa} \]

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

\[ c^2(x^2+y^2)-R^2\frac{(nx-my)^2}{\kappa^2}=0 \]

or

\[ (\kappa^2c^2-R^2n^2)x^2+2mnR^2xy+(\kappa^2c^2-R^2m^2)y^2=0 \ldots \ldots (3) \]

—the equation of a curve of intersection of the second order

\[ a_{11}x^2+2a_{12}xy+a_{22}y^2=0 \ldots \ldots \ldots \ldots (4_1) \]

We see directly [the discriminant of the left-hand side of (3) is equal to zero] that the curve of the second order, as should be expected, decomposes into two straight lines:

\[ x+\frac{a_{12}\pm\sqrt{-\Delta}}{a_{11}}\,y=0, \ldots \ldots \ldots (4_2) \]

which are the projections of the generators of the conical surface and are symmetric with respect to the projection of the line of dip of the plane of the stratum.

In expression \((4_{1,2})\):

\[ \begin{aligned} a_{11}&=\kappa^2c^2-R^2n^2\\ a_{12}&=mnR^2\\ a_{22}&=\kappa^2c^2-R^2m^2; \end{aligned} \]

and

\[ \Delta= \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} \]

—the discriminant of the terms of the second degree.

We determine the value of \(\Delta\), assuming that the inequality is satisfied:

\[ \Delta<0. \]

We have:

\[ \begin{aligned} \Delta &=(\kappa^2c^2-R^2n^2)(\kappa^2c^2-R^2m^2)-m^2n^2R^4=\\ &=\kappa^4c^4-\kappa^2c^2R^2n^2-\kappa^2c^2R^2m^2=\\ &=\kappa^2c^2(\kappa^2c^2-R^2n^2-R^2m^2)=\kappa^2c^2\,[\kappa^2c^2-R^2(m^2+n^2)]. \end{aligned} \]

The equations \((4_2)\) of the projections of the generators take the form:

\[ x+\frac{mnR^2 \pm \kappa c\sqrt{R^2(m^2+n^2)-\kappa^2c^2}}{\kappa^2c^2-R^2n^2}\,y=0 \quad \ldots (5), (6) \]

Introducing the notation:

\[ t_{1,2}= \frac{mnR^2 \pm \kappa c\sqrt{R^2(m^2+n^2)-\kappa^2c^2}} {\kappa^2c^2-R^2n^2}, \]

we determine the required generators by the system of equations—

\[ \left. \begin{aligned} nx-my+\kappa z&=0\\ x+t_{1,2}y&=0 \end{aligned} \right\} \quad \ldots\ldots\ldots (7) \]

We replace system (7) by the system:

\[ (nt_{1,2}+m)y-\kappa z=0 \]

\[ x+t_{1,2}y=0, \]

whence

\[ \frac{x}{-t_{1,2}}=\frac{y}{1}= \frac{z}{\left(\dfrac{m+nt_{1,2}}{\kappa}\right)}. \]

Finding

\[ m+nt_{1,2} = m+ \frac{n\left(mnR^2 \pm \kappa c\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right)} {\kappa^2c^2-R^2n^2} = \]

\[ = \frac{m\kappa^2c^2 \pm \kappa cn\sqrt{R^2(m^2+n^2)-\kappa^2c^2}} {\kappa^2c^2-R^2n^2} = \]

\[ = \frac{\kappa c\left(m\kappa c \pm n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right)} {\kappa^2c^2-R^2n^2}, \]

the required equations of the generators are represented in the form:

\[ \frac{x}{ -\left( \dfrac{mnR^2 \pm \kappa c\sqrt{R^2(m^2+n^2)-\kappa^2c^2}} {\kappa^2c^2-R^2n^2} \right)} = \frac{y}{1} = \]

\[ = \frac{z}{ \left[ \dfrac{ c\left(m\kappa c \pm n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right)} {\kappa^2c^2-R^2n^2} \right]} \quad \ldots\ldots (8) \]

Transform the numerators and denominators of the ratios, multiplying each of them, respectively, by

\[ m\kappa c \mp n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}. \]

We have:

\[ \begin{aligned} 1.\;&\left(mnR^2 \pm \kappa c\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right) \left(m\kappa c \mp n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right) \\ &=m^2nR^2\kappa c \pm m\kappa^2c^2\sqrt{R^2(m^2+n^2)-\kappa^2c^2} \mp mn^2R^2\sqrt{R^2(m^2+n^2)-\kappa^2c^2} \\ &\quad -n\kappa c\left[R^2(m^2+n^2)-\kappa^2c^2\right] \\ &=n\kappa c(-n^2R^2+\kappa^2c^2) \pm m\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\,(\kappa^2c^2-n^2R^2) \\ &=\kappa cn(\kappa^2c^2-n^2R^2) \pm m\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\,(\kappa^2c^2-n^2R^2) \\ &=(\kappa^2c^2-n^2R^2) \left(\kappa cn\pm m\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right). \end{aligned} \]

\[ 2.\quad m\kappa c\mp n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}. \]

\[ \begin{aligned} 3.\;&c\left\{m^2\kappa^2c^2-n^2\left[R^2(m^2+n^2)-\kappa^2c^2\right]\right\} \\ &=c\left[\kappa^2c^2(m^2+n^2)-n^2R^2(m^2+n^2)\right] \\ &=c(m^2+n^2)(\kappa^2c^2-n^2R^2). \end{aligned} \]

Finally,

\[ \frac{x}{-\left(\kappa cn\pm m\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right)} = \frac{y}{m\kappa c\mp n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}} = \frac{z}{c(m^2+n^2)}\ldots\ldots (9) \]

—the required equations of the generators of the conical surface; whence, putting \(z_{1,2}=H\), we obtain:

\[ x_{1,2} = -\left(\kappa cn\pm m\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right) \frac{H}{c(m^2+n^2)}. \]

\[ y_{1,2} = \left(m\kappa c\mp n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}\right) \frac{H}{c(m^2+n^2)}. \]

The coordinates \((x_p,y_p)\) of the required point of the line of dip are determined by the formulas:

\[ x_p=\frac{1}{2}(x_1+x_2)=-\frac{\kappa n}{m^2+n^2}\,H \]

\[ y_p=\frac{1}{2}(y_1+y_2)=\frac{\kappa m}{m^2+n^2}\,H. \]

\[ z_p=H. \]

XII.

Let us give a somewhat different method of solution.

We had:

The equation of the conical surface:

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

or

\[ c^2(x^2+y^2)-R^2z^2=0 \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . (1) \]

The equation of the plane of the stratum

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

The required generators are determined by the intersection of (1) and (2). Dividing (1) by \(z^2\) and (2) by \(z\), we obtain:

\[ c^2\left[\left(\frac{x}{z}\right)^2+\left(\frac{y}{z}\right)^2\right]-R^2=0 \ . \ . \ . \ . \ . \ . \ . \ . \ . (3) \]

\[ n\left(\frac{x}{z}\right)-m\left(\frac{y}{z}\right)+\kappa=0 \ . \ . \ . \ . \ . \ . \ . \ . \ . (4) \]

From (4) we have:

\[ \frac{x}{z}=\frac{1}{n}\left[m\left(\frac{y}{z}\right)-\kappa\right] \ . \ . \ . \ . \ . \ . \ . \ . \ . (5) \]

Substituting the latter into (3), we obtain:

\[ \frac{1}{n^2}\left[m\left(\frac{y}{z}\right)-\kappa\right]^2+\left(\frac{y}{z}\right)^2-\left(\frac{R}{c}\right)^2=0 \]

and, finally,

\[ (m^2+n^2)\left(\frac{y}{z}\right)^2-2m\kappa\left(\frac{y}{z}\right)+\left[\kappa^2-\left(\frac{Rn}{c}\right)^2\right]=0, \]

whence

\[ \frac{y}{z}= \frac{ m\kappa\pm\sqrt{m^2\kappa^2-(m^2+n^2)\left[\kappa^2-\left(\frac{Rn}{c}\right)^2\right]} }{ m^2+n^2 }. \]

Finally:

\[ \frac{y}{z}= \frac{m\kappa c\pm n\sqrt{R^2(m^2+n^2)-\kappa^2c^2}} {c(m^2+n^2)} \ . \ . \ . \ . \ . \ . \ . \ . \ . (6) \]

Substituting \(\frac{y}{z}\) into (5), we obtain:

\[ \frac{x}{z}= -\frac{\kappa cn\mp m\sqrt{R^2(m^2+n^2)-\kappa^2c^2}} {c(m^2+n^2)} \ . \ . \ . \ . \ . \ . \ . \ . \ . (7) \]

Now (6) and (7) together also determine the required equations of the generators, namely:

\[ \frac{x}{-\left(\kappa c n \mp m\sqrt{R^2(m^2+n^2)-\kappa^2 c^2}\right)} = \frac{y}{m\kappa c \pm n\sqrt{R^2(m^2+n^2)-\kappa^2 c^2}} = \frac{z}{c(m^2+n^2)} \ldots (8) \]

Remark 1. If we impose the condition that the plane of the stratum be tangent to the conical surface, then, obviously, the radical quantity in the expressions for \(\dfrac{x}{z}\) and \(\dfrac{y}{z}\) must be equal to zero.

Then equations (8) pass directly into the equation of the line of dip of the plane of the stratum.

Indeed. We have:

\[ \frac{x}{-\kappa c n} = \frac{y}{m\kappa c} = \frac{z}{c(m^2+n^2)} \]

or

\[ \frac{x}{-n} = \frac{y}{m} = \frac{z}{\left(\dfrac{m^2+n^2}{\kappa}\right)}; \]

whence the required coordinates are:

\[ x=-\frac{\kappa n}{m^2+n^2}z \]

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

Remark 2. Passing to other methods, let us note that the cosines of the angles formed by the given traces \([(a_1,\zeta_1);(a_2,\zeta_2)]\) with the coordinate axes \(X,Y,Z\), as is known, are determined, respectively, by the formulas:

\[ \begin{aligned} \cos \alpha_{1,2} &= \sin \zeta_{1,2}\cos a_{1,2}\\ \cos \beta_{1,2} &= \sin \zeta_{1,2}\sin a_{1,2}\\ \cos \gamma_{1,2} &= \cos \zeta_{1,2} \end{aligned} \qquad \ldots \ldots \ldots (1) \]

The equations of the traces will be:

\[ \frac{x}{\cos \alpha_{1,2}} = \frac{y}{\cos \beta_{1,2}} = \frac{z}{\cos \gamma_{1,2}} \ldots \ldots \ldots (2) \]

Taking formulas (1) into account, we obtain:

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

or

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

—the equations of the traces.

XIII.

The system of equations—of the traces and of the horizontal plane—

\[ \left. \begin{gathered} \frac{x}{\operatorname{tg}\zeta_{1,2}\cos a_{1,2}} = \frac{y}{\operatorname{tg}\zeta_{1,2}\sin a_{1,2}} = \frac{z}{1} \\ z=H \end{gathered} \right\} \qquad \cdots \cdots \cdots (4) \]

determines two points \((1,2)\) of a straight line in space:

\[ \left. \begin{aligned} x_{1,2}&=H\operatorname{tg}\zeta_{1,2}\cos a_{1,2}\\ y_{1,2}&=H\operatorname{tg}\zeta_{1,2}\sin a_{1,2}\\ z_{1,2}&=H \end{aligned} \right\} \qquad \cdots \cdots \cdots (4_1) \]

The equation of the straight line in space, consequently, will be:

\[ \frac{x-x_{1,2}}{x_{1,2}-x_{2,1}} = \frac{y-y_{1,2}}{y_{1,2}-y_{2,1}} = \frac{z-H}{0} \]

or

\[ \frac{x-H\operatorname{tg}\zeta_{1,2}\cos a_{1,2}} {\operatorname{tg}\zeta_{1,2}\cos a_{1,2}-\operatorname{tg}\zeta_{2,1}\cos a_{2,1}} = \frac{y-H\operatorname{tg}\zeta_{1,2}\sin a_{1,2}} {\operatorname{tg}\zeta_{1,2}\sin a_{1,2}-\operatorname{tg}\zeta_{2,1}\sin a_{2,1}} = \frac{z-H}{0} \qquad \cdots \cdots \cdots (5) \]

The line of dip corresponding to the point \((0,0,0)\), as is known, is determined by the equations \((6_{1,2})\):

\[ (\operatorname{tg}\zeta_{1,2}\cos a_{1,2}-\operatorname{tg}\zeta_{2,1}\cos a_{2,1})x + (\operatorname{tg}\zeta_{1,2}\sin a_{1,2}-\operatorname{tg}\zeta_{2,1}\sin a_{2,1})y =0. \]

\[ \left| \begin{array}{ccc} x-H\operatorname{tg}\zeta_{1,2}\cos a_{1,2},& y-H\operatorname{tg}\zeta_{1,2}\sin a_{1,2},& z-H\\ \operatorname{tg}\zeta_{1,2}\cos a_{1,2},& \operatorname{tg}\zeta_{1,2}\sin a_{1,2},& 1\\ \operatorname{tg}\zeta_{1,2}\cos a_{1,2}-\operatorname{tg}\zeta_{2,1}\cos a_{2,1},& \operatorname{tg}\zeta_{1,2}\sin a_{1,2}-\operatorname{tg}\zeta_{2,1}\sin a_{2,1},& 0 \end{array} \right|=0. \]

Transforming \((6_2)\), we obtain:

\[ \left| \begin{array}{ccc} x&y&z\\ \operatorname{tg}\zeta_{1,2}\cos a_{1,2},& \operatorname{tg}\zeta_{1,2}\sin a_{1,2},& 1\\ \operatorname{tg}\zeta_{2,1}\cos a_{2,1},& \operatorname{tg}\zeta_{2,1}\sin a_{2,1},& 1 \end{array} \right|=0 \]

or, expanding the determinant by the elements of the first row,

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

and, finally,

\[ (\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)x - (\operatorname{tg}\zeta_1\cos a_1-\operatorname{tg}\zeta_2\cos a_2)y + \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)z =0. \]

Introducing the notations:

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

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

\[ \operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)=K, \]

the equations \((6_{1,2})\) take the form:

\[ \begin{gathered} mx+ny=0\\ nx-my+kz=0; \end{gathered} \]

whence

\[ \begin{gathered} x=-nkt\\ y=mkt\\ z=(m^2+n^2)t. \end{gathered} \]

XIV.

Let us take, on the given traces \((a_{1,2}, \zeta_{1,2})\), respectively, two points \((1,2)\) corresponding to one and the same value of the coordinate \(z=z_1=z_2=H\), and consequently determining the strike line. We shall take the strike line as the line of centers of a pencil of spherical surfaces.

From points 1 and 2, as centers, we shall describe two spherical surfaces of arbitrary radii, respectively, \(R_1\) and \(R_2\).

The equations of the spherical surfaces will be:

\[ (x-x_{1,2})^2+(y-y_{1,2})^2+(z-z_{1,2})^2-R^2_{1,2}=0 \ . . . . (1) \]

Equation:

\[ (1+\lambda)(x^2+y^2+z^2)-2(x_1+\lambda x_2)x-2(y_1+\lambda y_2)y- \]
\[ -2(z_1+\lambda z_2)z+(M_1+\lambda M_2)=0, . . . . . . (2) \]

where

\[ M_{1,2}=x^2_{1,2}+y^2_{1,2}+z^2_{1,2}-R^2_{1,2}, \]

and \(\lambda\)—an arbitrary parameter, represents, as is known, the general equation of a pencil of spherical surfaces whose centers lie, in our case, on the strike line 12.

\[ \frac{x-x_{1,2}}{x_{1,2}-x_{2,1}} = \frac{y-y_{1,2}}{y_{1,2}-y_{2,1}} = \frac{z-H}{0} \ . . . . . (3) \]

The equation of the radical plane [from (2), for \(\lambda=-1\)], coinciding with one of the planes of dip of the plane of the stratum, will be

\[ \left. \begin{gathered} (x_1-x_2)x+(y_1-y_2)y-\frac{1}{2}(M_1-M_2)=0\\ z=z \end{gathered} \right\} \ . . . . . . (4) \]

Now impose the condition that the radical plane (4) pass through the point \((0,0,0)\).

Then (4) gives:

\[ M_1=M_2. \]

The equation of the radical plane takes the form:

\[ \left. \begin{gathered} (x_{1,2}-x_{2,1})x+(y_{1,2}-y_{2,1})y=0\\ z=z \end{gathered} \right\} \ . . . . . . (5) \]

The sought point of the dip line, as is easily seen, is determined by the system of equations (5) and (3).

We have:

\[ \left. \begin{aligned} z&=H,\\ (x_{1,2}-x_{2,1})x+(y_{1,2}-y_{2,1})y&=0,\\ (y_{1,2}-y_{2,1})x-(x_{1,2}-x_{2,1})y+x_{1,2}y_{2,1}-x_{2,1}y_{1,2}&=0 \end{aligned} \right\}\, \ldots (6) \]

whence

\[ z=H \]

\[ x=-\frac{ \left| \begin{array}{cc} 0 & y_{1,2}-y_{2,1}\\ x_{1,2}y_{2,1}-x_{2,1}y_{1,2}, & -(x_{1,2}-x_{2,1}) \end{array} \right| }{ \left| \begin{array}{cc} x_{1,2}-x_{2,1}, & y_{1,2}-y_{2,1}\\ y_{1,2}-y_{2,1}, & -(x_{1,2}-x_{2,1}) \end{array} \right| } = \]

\[ =-\frac{(y_{1,2}-y_{2,1})(x_{1,2}y_{2,1}-x_{2,1}y_{1,2})} {(x_{1,2}-x_{2,1})^{2}+(y_{1,2}-y_{2,1})^{2}} \ldots (7) \]

\[ y=-\frac{ \left| \begin{array}{cc} x_{1,2}-x_{2,1}, & 0\\ y_{1,2}-y_{2,1}, & x_{1,2}y_{2,1}-x_{2,1}y_{1,2} \end{array} \right| }{ \left| \begin{array}{cc} x_{1,2}-x_{2,1}, & y_{1,2}-y_{2,1}\\ y_{1,2}-y_{2,1}, & -(x_{1,2}-x_{2,1}) \end{array} \right| } = \]

\[ =\frac{(x_{1,2}-x_{2,1})(x_{1,2}y_{2,1}-x_{2,1}y_{1,2})} {(x_{1,2}-x_{2,1})^{2}+(y_{1,2}-y_{2,1})^{2}} \ldots (8) \]

Substituting in (7) and (8) the values:

\[ \begin{aligned} x_{1,2}-x_{2,1}&=H(\operatorname{tg}\zeta_{1,2}\cos a_{1,2}-\operatorname{tg}\zeta_{2,1}\cos a_{2,1}),\\ y_{1,2}-y_{2,1}&=H(\operatorname{tg}\zeta_{1,2}\sin a_{1,2}-\operatorname{tg}\zeta_{2,1}\sin a_{2,1}),\\ x_{1,2}y_{2,1}-x_{2,1}y_{1,2}&=H^{2}\operatorname{tg}\zeta_{1}\operatorname{tg}\zeta_{2}\sin(a_{2,1}-a_{1,2}), \end{aligned} \]

we obtain:

\[ x=-\frac{ \operatorname{tg}\zeta_{1}\operatorname{tg}\zeta_{2}\sin(a_{2}-a_{1}) (\operatorname{tg}\zeta_{1}\sin a_{1}-\operatorname{tg}\zeta_{2}\sin a_{2}) }{ (\operatorname{tg}\zeta_{1}\cos a_{1}-\operatorname{tg}\zeta_{2}\cos a_{2})^{2} + (\operatorname{tg}\zeta_{1}\sin a_{1}-\operatorname{tg}\zeta_{2}\sin a_{2})^{2} }H= \]

\[ =-\frac{nk}{m^{2}+n^{2}}H \]

\[ y=\frac{ \operatorname{tg}\zeta_{1}\operatorname{tg}\zeta_{2}\sin(a_{2}-a_{1}) (\operatorname{tg}\zeta_{1}\cos a_{1}-\operatorname{tg}\zeta_{2}\cos a_{2}) }{ (\operatorname{tg}\zeta_{1}\cos a_{1}-\operatorname{tg}\zeta_{2}\cos a_{2})^{2} + (\operatorname{tg}\zeta_{1}\sin a_{1}-\operatorname{tg}\zeta_{2}\sin a_{2})^{2} }H= \]

\[ =\frac{mk}{m^{2}+n^{2}}H. \]

XV.

The preceding solution may be presented in a somewhat different form. Let us impose the condition that the radical plane (5) be tangent to both spherical surfaces.

As is known, we shall obtain the point of tangency common to the radical plane and to both spherical surfaces; this point will be the required point of the line of greatest dip of the bed plane.

The equation of the tangent plane to the spherical surfaces at the required point \((x_0, y_0, z_0)\), as is known, will be:

\[ (x_0 - x_{1,2})(x - x_{1,2}) + (y_0 - y_{1,2})(y - y_{1,2}) + (z_0 - H)(z - H) - R_{1,2}^{2} = 0 \]

or

\[ (x_0 - x_{1,2})x + (y_0 - y_{1,2})y + (z_0 - H)z \]

\[ - \bigl[x_{1,2}(x_0 - x_{1,2}) + y_{1,2}(y_0 - y_{1,2}) + H(z_0 - H) + R_{1,2}^{2}\bigr] = 0 . \ . \ . \ (9) \]

By assumption, (5) coincides with (9).

Therefore:

\[ (x_0 - x_{1,2})x + (y_0 - y_{1,2})y + (z_0 - H)z - \bigl[x_{1,2}(x_0 - x_{1,2}) + y_{1,2}(y_0 - y_{1,2})+ \]

\[ + H(z_0 - H) + R_{1,2}^{2}\bigr] = \lambda \bigl[(x_{1,2} - x_{2,1})x + (y_{1,2} - y_{2,1})y\bigr], \]

whence

\[ x_0 - x_{1,2} = \lambda (x_{1,2} - x_{2,1}) . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (10) \]

\[ y_0 - y_{1,2} = \lambda (y_{1,2} - y_{2,1}) . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (11) \]

\[ z_0 - H = 0 . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (12) \]

and, in connection with (12),

\[ x_{1,2}(x_0 - x_{1,2}) + y_{1,2}(y_0 - y_{1,2}) + R_{1,2}^{2} = 0 . \ . \ . \ . \ . \ . \ (13) \]

Expression (13), evidently, serves only for determining those values of the radii for which the spherical surfaces are tangent.

The required point is determined from relations (10), (11) and the equation (5) of the radical plane.

Dividing (10) by (11), we obtain:

\[ \frac{x_0 - x_{1,2}}{y_0 - y_{1,2}} = \frac{x_{1,2} - x_{2,1}}{y_{1,2} - y_{2,1}} . \]

Transforming the latter, we find:

\[ (y_{1,2} - y_{2,1})x_0 - (x_{1,2} - x_{2,1})y_0 + x_{1,2}y_{2,1} - x_{2,1}y_{1,2} = 0. \]

Thus we have arrived at the previously given system of equations (6).

Remark. For the solution one may use the beginning of the simple ratio of three points: the given points 1, 2 and the required point.

As is known, we may write:

\[ x = x_1 + \mu (x_2 - x_1) = \frac{x_1 - \lambda x_2}{1 - \lambda} \]

\[ y = y_1 + \mu (y_2 - y_1) = \frac{y_1 - \lambda y_2}{1 - \lambda}, \]

where \(\mu, \lambda\) are parameters corresponding to \(x\) and \(y\). To determine the parameters \(\mu, \lambda\), as is easily seen, the values of \(x\) and \(y\) must be substituted into the equation of the radical plane.

XVI.

Let us draw through the origin two planes perpendicular to the traces.

The equation of a plane passing through the point \((x_0, y_0, z_0)\) and perpendicular to the straight line

\[ \frac{x-a}{m}=\frac{y-b}{n}=\frac{z-c}{p}, \ldots \ldots \ldots (1) \]

as is known, is

\[ m(x-x_0)+n(y-y_0)+p(z-z_0)=0. \ldots \ldots \ldots (2) \]

In our case: \(x_0=0;\ y_0=0;\ z_0=0\).
Therefore, (2) assumes the form:

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

Now, in accordance with formula (3), Remark 2 of Section XII, we obtain:

\[ \tg \zeta'_{1,2}\cos a_{1,2}\,x+\tg \zeta'_{1,2}\sin a_{1,2}\,y+z=0. \ldots \ldots \ldots (4) \]

—the equations of the planes determining the normal at the point \((0,0,0)\) of the plane of the stratum.

The equation of the normal, expressed as a function of the parameter \(t\), will be:

\[ x= \begin{vmatrix} \tg \zeta_1 \sin a_1,\ 1\\ \tg \zeta_2 \sin a_2,\ 1 \end{vmatrix}t =(\tg \zeta_1\sin a_1-\tg \zeta_2\sin a_2)t=nt. \]

\[ y= \begin{vmatrix} 1,\ \tg \zeta_1 \cos a_1\\ 1,\ \tg \zeta_2 \cos a_2 \end{vmatrix}t =(\tg \zeta_2\cos a_2-\tg \zeta_1\cos a_1)t=-mt. \]

\[ z= \begin{vmatrix} \tg \zeta_1\cos a_1,\ \tg \zeta_1\sin a_1\\ \tg \zeta_2\cos a_2,\ \tg \zeta_2\sin a_2 \end{vmatrix}t =\tg \zeta_1\tg \zeta_2 \begin{vmatrix} \cos a_1,\ \sin a_1\\ \cos a_2,\ \sin a_2 \end{vmatrix}t= \]

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

Eliminating the parameter \(t\) from the obtained values \(x,y,z\), we obtain the required equation of the normal:

\[ \frac{x}{n}=\frac{y}{-m}=\frac{z}{k}. \ldots \ldots \ldots (5) \]

Remark. In passing to what follows, let us note that the coordinates \((x_n,y_n)\) of points of the normal are identical to the coordinates \((x_p,y_p)\) of the corresponding points of the plane of the stratum; however, the sign of the coordinate \(z_n\) of points of the normal is opposite to the sign of the coordinate \(z_p\) of the corresponding points of the plane of the stratum.

From (5) we have:

\[ x=\frac{n}{k}z_n \]

\[ y=-\frac{m}{k}z_n, \]

whence, according to the remark, we obtain:

\[ \text{1. }\operatorname{tg}\varphi=\frac{-m}{n},\quad \text{taking } k>0 \]

\[ \text{2. }\operatorname{tg}\varphi=\frac{m}{-n},\quad \text{taking } k<0. \]

XVII.

Let us take the strike and dip lines of the stratum plane corresponding to the point \((0,0,0)\) as bisectors; then the bisectors will correspond to an infinite set of pairs of straight lines of the stratum plane passing through the point \((0,0,0)\) and, accordingly, equally inclined both to the former and to the horizontal plane \(XOY\).

Let us replace the two given straight lines \([(0,0,0); (a_1,\zeta_1)\) and \((0,0,0); (a_2,\zeta_2)]\), which determine the stratum plane, by two arbitrary straight lines 1 and 2 of the same stratum plane, but equally inclined to the horizontal plane \(XOY\) (or to the \(Z\)-axis).

Let their equations, respectively, be:

\[ \frac{x}{m_{1,2}}=\frac{y}{n_{1,2}}=\frac{z}{p}\ . . . . . . . . . . (1_{1,2}) \]

Since the straight lines 1 and 2 are equally inclined to \(XOY\), then

\[ \pm\frac{p}{\sqrt{m_1^2+n_1^2+p^2}} = \pm\frac{p}{\sqrt{m_2^2+n_2^2+p^2}}; \]

whence

\[ m_1^2+n_1^2=m_2^2+n_2^2. . . . . . . . . . . (2) \]

On the other hand, the straight lines 1 and 2 lie in the given plane

\[ nx-my+kz=0 . . . . . . . . . . (3) \]

Taking into account that the straight lines 1 and 2 pass through the point \((0,0,0)\), the condition for their coincidence with plane (3) is expressed, respectively, by the formulas:

\[ nm_{1,2}-mn_{1,2}+kp=0 . . . . . . . . . . (4_{1,2}) \]

Equations \((4_{1,2})\), upon eliminating \(p\), give

\[ mn_1-nm_1=mn_2-nm_2; \]

whence

\[ (n_1-n_2)m=-(m_2-m_1)n \]

or

\[ \frac{n_1-n_2}{m_2-m_1}=-\frac{n}{m}. . . . . . . . . . . (5) \]

The equations of the horizontally projecting planes of the straight lines 1 and 2, respectively, will be:

\[ \left. \begin{array}{c} n_{1,2}x-m_{1,2}y=0\\ z=z \end{array} \right\}. . . . . . . . . . . (6_{1,2}) \]

or, in normal form,

\[ \left. \frac{n_{1,2}}{\pm\sqrt{m_{1,2}^{2}+n_{1,2}^{2}}}\,x - \frac{m_{1,2}}{\pm\sqrt{m_{1,2}^{2}+n_{1,2}^{2}}}\,y=0 \quad \right\} \ldots\ldots (7_{1,2}) \]

\[ z=z \]

The equations of the bisecting planes corresponding to the planes \((7_{1,2})\) will be:

\[ \left. \begin{aligned} &\left( \frac{n_1}{\pm\sqrt{m_1^2+n_1^2}} + \frac{n_2}{\pm\sqrt{m_2^2+n_2^2}} \right)x - \left( \frac{m_1}{\pm\sqrt{m_1^2+n_1^2}} + \frac{m_2}{\pm\sqrt{m_2^2+n_2^2}} \right)y=0,\\ &z=z \end{aligned} \right\} \ldots\ldots (8) \]

\[ \left. \begin{aligned} &\left( \frac{n_1}{\pm\sqrt{m_1^2+n_1^2}} - \frac{n_2}{\pm\sqrt{m_2^2+n_2^2}} \right)x + \left( \frac{m_2}{\pm\sqrt{m_2^2+n_2^2}} - \frac{m_1}{\pm\sqrt{m_1^2+n_1^2}} \right)y=0,\\ &z=z \end{aligned} \right\} \ldots\ldots (9) \]

By virtue of (2), the equations of the bisecting planes take the form

\[ \left. \begin{aligned} &(n_1+n_2)x-(m_1+m_2)y=0,\\ &z=z \end{aligned} \right\} \ldots\ldots (10) \]

\[ \left. \begin{aligned} &(n_1-n_2)x+(m_2-m_1)y=0,\\ &z=z \end{aligned} \right\} \ldots\ldots (11) \]

or

\[ \left. \begin{aligned} &x-\frac{m_1+m_2}{n_1+n_2}\,y=0,\\ &z=z \end{aligned} \right\} \ldots\ldots (12) \]

\[ \left. \begin{aligned} &\frac{n_1-n_2}{m_2-m_1}\,x+y=0,\\ &z=z \end{aligned} \right\} \ldots\ldots (13) \]

Further, from (2) we have:

\[ (n_1-n_2)(n_1+n_2)=(m_2-m_1)(m_2+m_1), \]

whence, in connection with (6),

\[ \frac{m_1+m_2}{n_1+n_2} = \frac{n_1-n_2}{m_2-m_1} = -\frac{\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2} {\operatorname{tg}\zeta_1\cos a_1-\operatorname{tg}\zeta_2\cos a_2} = -\frac{n}{m} \ldots (14) \]

Substituting into (12) and (13) the corresponding expressions (14), we obtain:

\[ \left. \begin{aligned} x+\frac{n}{m}y&=0\\ z&=z \end{aligned} \right\} \ . . . . . . . . . . . . . (15) \]

\[ \left. \begin{aligned} -\frac{n}{m}x+y&=0\\ z&=z \end{aligned} \right\} \ . . . . . . . . . . . . . (16) \]

and, finally,

\[ \left. \begin{aligned} mx+ny&=0\\ z&=z \end{aligned} \right\} \ . . . . . . . . . . . . . (17) \]

—the equation of the first bisecting plane;

\[ \left. \begin{aligned} nx-my&=0\\ z&=z \end{aligned} \right\} \ . . . . . . . . . . . . . (18) \]

—the equation of the second bisecting plane.

Now finding, successively, the intersections of the planes (17) and (18) with the given plane (3), we determine the corresponding lines of dip and strike.

Namely:

1. The system:

\[ \left. \begin{aligned} nx-my&=0\\ z&=z \end{aligned} \right\} \ . . . . . . . . . . . . . (19) \]

\[ nx-my+kz=0 . . . . . . . . . . . . . (20) \]

determines the strike line, as is immediately evident.

1. The system:

\[ \left. \begin{aligned} mx+ny&=0\\ z&=z \end{aligned} \right\} \ . . . . . . . . . . . . . (21) \]

\[ nx-my+kz=0 . . . . . . . . . . . . . (22) \]

determines the dip line.

Let us reduce the equation of the dip line to normal form. Representing (21) in the form:

\[ \frac{x}{-n}=\frac{y}{m}; . . . . . . . . . . . . . (23) \]

then determining from the latter one of the variables, say \(x\), and substituting it into (22), we obtain:

\[ -\frac{n^{2}}{m}y-my+kz=0, \]

whence

\[ \frac{y}{m}=\frac{z}{\left(\frac{m^{2}+n^{2}}{\kappa}\right)} \ldots \ldots \ldots \ldots (24) \]

The last equation (24), in connection with (23), gives:

\[ \frac{x}{-n}=\frac{y}{m}=\frac{z}{\left(\frac{m^{2}+n^{2}}{\kappa}\right)} \ldots \ldots \ldots (25) \]

—the required equation of the line of dip in normal form.

Finally, from (25) we determine \(x\) and \(y\) as functions of \(z\). We have:

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

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

XVIII.

The equation of the plane of the stratum

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

The equation of the pencil of straight lines passing through the point \((0,0,0)\):

\[ \frac{x}{m_i}=\frac{y}{n_i}=\frac{z}{p_i} \ldots \ldots \ldots (2) \]

The condition for coincidence of the pencil of straight lines (2) with the plane (1):

\[ n\cdot 0+m\cdot 0+\kappa\cdot 0+0=0, \ldots \ldots \ldots (3) \]

as it should be;

\[ nm_i-mn_i+\kappa p_i=0 \ldots \ldots \ldots (4) \]

For the line of strike

\[ m_i=m_p;\quad n_i=n_p;\quad p_i=0. \]

Therefore, (2) and (4) take, respectively, the form:

\[ \frac{x}{m_p}=\frac{y}{n_p}=\frac{z}{0} \ldots \ldots \ldots (5) \]

\[ nm_p-mn_p=0 \ldots \ldots \ldots (6) \]

Let us represent (5) in the form:

\[ \frac{x}{\left(\frac{m_p}{n_p}\right)}=\frac{y}{1}=\frac{z}{0} \ldots \ldots \ldots (7) \]

From (6) we have:

\[ \frac{m_p}{n_p}=\frac{m}{n} \ldots \ldots \ldots (8) \]

Substituting (8) into (7), we obtain:

\[ \frac{x}{m}=\frac{y}{n}=\frac{z}{0}\ldots\ldots\ldots\ldots (9) \]

—the equation of the strike line, which is also easily obtained from (1) by putting \(z=0\).

Further, the dip line is perpendicular to the strike line. Therefore the condition for coincidence of the line

\[ \frac{x}{m_i}=\frac{y}{n_i}=\frac{z}{p_i}\ldots\ldots\ldots (10) \]

with the dip line is given by the relations:

\[ nm_i-mn_i+\kappa p_i=0 \]

\[ mm_i+nn_i=0, \]

whence

\[ m_i= \begin{vmatrix} -m,\ \kappa\\ n,\ 0 \end{vmatrix} t=-n\kappa t. \]

\[ n_i= \begin{vmatrix} \kappa & n\\ 0 & m \end{vmatrix} t=m\kappa t. \]

\[ p_i= \begin{vmatrix} n,\ -m\\ m,\ n \end{vmatrix} t=(m^2+n^2)t. \]

Substituting the values \(m_i\), \(n_i\), and \(p_i\) into (10), we obtain the equation of the dip line of the stratum plane:

\[ \frac{x}{-n}=\frac{y}{m}=\frac{z}{\left[\dfrac{m^2+n^2}{\kappa}\right]}, \]

whence

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

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

Remark. Other methods for determining the coordinates \((x,y,z)\) of an arbitrary point of the dip line of the stratum plane are set forth in my article[^1].

Finally, let us give one more method, deserving attention, for determining the azimuth \(\varphi\) of the dip line of the stratum plane.

§ 14.

A special method for determining the azimuth \(\varphi\) of the line of dip of the stratum plane

We have:

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

—the equation of the stratum plane.

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

—the equation of the surface of a sphere.

Eliminating the variable \(z\) from (1) and (2), we obtain, in projection onto the plane \(XOY\), an ellipse, whose short axis precisely coincides with the horizontal projection of the line of dip of the stratum plane.

Namely, from (1) we have:

\[ z=\frac{my-nx}{\kappa}. \]

Substituting \(z\) into equation (2), we obtain:

\[ \kappa^{2}x^{2}+\kappa^{2}y^{2}+m^{2}y^{2}-2mnxy+n^{2}x^{2}-R^{2}\kappa^{2}=0 \]

or, finally,

\[ (n^{2}+\kappa^{2})x^{2}-2mnxy+(m^{2}+\kappa^{2})y^{2}-R^{2}\kappa^{2}=0 \ . \ . \ . \ . \ (3) \]

—the equation of the desired ellipse, where, in accordance with the general equation of curves of the second order (after transformation to the center),

\[ a_{11}=n^{2}+\kappa^{2} \]

\[ a_{12}=-mn \]

\[ a_{22}=m^{2}+\kappa^{2} \]

\[ a_{33}=-R^{2}\kappa^{2}. \]

It remains to transform equation (3) to the principal axes of the ellipse, rotating the system \(XYZ\) about the origin \(0\) through the angle \(\varphi\), corresponding to the azimuth of the line of dip of the stratum plane.

The angle \(\varphi\), as is known, is determined by the relation:

\[ \tan 2\varphi=\frac{2a_{12}}{a_{11}-a_{22}}; \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (4) \]

this last relation leaves wholly indeterminate the question of in which of the four directions of the principal axes of the ellipse the new (initial, principal) direction—the axis \(X'\)—lies.

Let us agree, provisionally, to regard the angle of rotation \(\varphi\) as acute: \(\varphi=\varphi_{0}\).

In connection with what follows, let us pass to \(\sin 2\varphi_{0}\), and, as is known,

\[ \sin 2\varphi_{0}=\frac{2a_{12}}{\rho}, \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (5) \]

where

\[ \rho=f(a_{11},a_{12},a_{22}). \]

By virtue of the condition

\[ 0<\varphi=\varphi_{0}<\frac{1}{2}\pi, \]

in formula (5) we must take the signs of \(a_{12}\) and \(\varphi\) to be the same; therefore two assumptions follow:

\[ a_{12}>0 \quad \text{and} \quad a_{12}<0 . \]

We now proceed to the determination of the invariants \(s,\rho\) and of the coefficients \(a'_{11}, a'_{22}\) and \(a'_{33}\) of the equation transformed to the principal axes of the ellipse:

\[ a'_{11}x'^2+a'_{22}y'^2+a'_{33}=0, \]

where \(x'\) and \(y'\) are the coordinates of the points of the ellipse in the new coordinate system \(X'Y'Z'\).

A. Determination of the invariants \(s\) and \(\rho\).

\[ \begin{aligned} 1.\quad &s=a_{11}+a_{22}=n^2+2\kappa^2+m^2.\\ 2.\quad &\rho^2=(a_{11}-a_{22})^2+4a_{12}^2=(n^2-m^2)^2+4m^2n^2=\\ &=n^4+m^4+2m^2n^2=(n^2+m^2)^2, \end{aligned} \]

whence

\[ \rho_{p,n}=\pm(m^2+n^2). \]

B. Determination of the coefficients \(a'_{11}, a'_{22}, a'_{33}\).

I. Let \(a_{12}>0\),

therefore, \(\rho>0\).

\[ 1.\quad a'_{11}=\frac{1}{2}(s+\rho_p)=\frac{1}{2}(n^2+2\kappa^2+m^2+n^2+m^2)=m^2+n^2+\kappa^2. \]

\[ a'_{22}=\frac{1}{2}(s-\rho_p)=\kappa^2. \]

\[ a'_{33}=a_{33}=-R^2\kappa^2. \]

Finally, we reduce the equation of the ellipse, in the system \(X'Y'Z'\), to normal form:

\[ \frac{x'^2}{p_1}+\frac{y'^2}{p_2}-1=0, \]

where

\[ p_1=-\frac{a'_{33}}{a'_{11}};\qquad p_2=-\frac{a'_{33}}{a'_{22}}. \]

In the present case:

\[ p_1=\frac{\kappa^2}{m^2+n^2+\kappa^2}\,R^2. \]

\[ p_2=R^2. \]

We see directly that:

\[ p_1>0;\quad p_2>0 \quad \text{and} \quad p_1<p_2. \]

We conclude that, after a rotation to the right (corresponding to the mutual disposition of the axes \(OX\) and \(OY\)) through the angle

\[ 0<\varphi_0<\frac{1}{2}\pi, \]

the new axis \(X'\) coincides with the minor axis of the ellipse (the projection of the line of dip of the bed plane).

Since the new axis \(X'\) must coincide precisely with the direction of the projection of the line of dip, it is not difficult to see that, vectorially, the axis \(X\) must be rotated to the right through an angle \(\varphi\) satisfying either the condition (Figs. 24, 25):

\[ 0<\varphi<\frac{1}{2}\pi,\ldots\ldots\ldots\ldots\ldots\ldots (P) \]

or the condition (Figs. 26, 27):

\[ \pi<\varphi<\frac{3}{2}\pi.\ldots\ldots\ldots\ldots\ldots\ldots (Q) \]

Under these conditions, \((P)\) and \((Q)\),

\[ \operatorname{tg}\varphi>0. \]

Corresponding to condition \((P)\)—

\[ \varphi=\varphi_0. \]

and, as is known, the values

\[ \left. \begin{aligned} \sin\varphi &= + \sqrt{\frac{1}{2}\left(1-\frac{a_{11}-a_{22}}{\rho_p}\right)} \\[6pt] \cos\varphi &= + \sqrt{\frac{1}{2}\left(1+\frac{a_{11}-a_{22}}{\rho_p}\right)} \end{aligned} \right\} \qquad \ldots\ldots (P_1) \]

Corresponding to condition \((Q)\)—

\[ \varphi=\varphi_0+\pi, \]

and the values

\[ \left. \begin{aligned} \sin\varphi &= -\sin\varphi_0 = -\sqrt{\frac{1}{2}\left(1-\frac{a_{11}-a_{22}}{\rho_p}\right)} \\[6pt] \cos\varphi &= -\cos\varphi_0 = -\sqrt{\frac{1}{2}\left(1+\frac{a_{11}-a_{22}}{\rho_p}\right)} \end{aligned} \right\} \qquad \ldots\ldots (Q_1) \]

Let us proceed to the direct determination of \(\sin\varphi\) and \(\cos\varphi\) as functions of the given quantities.

\[ 1.\quad a_{11}-a_{22}=n^2-m^2. \]

\[ 2.\quad \frac{1}{2}\,\frac{\rho_p-(a_{11}-a_{22})}{\rho_p} = \frac{m^2}{m^2+n^2}. \]

\[ 3.\quad \frac{1}{2}\,\frac{\rho_p+(a_{11}-a_{22})}{\rho_p} = \frac{n^2}{m^2+n^2}. \]

Substituting 1, 2, 3 into \((P_1)\) and \((Q_1)\), respectively, we obtain:

\[ \left. \begin{aligned} \sin\varphi &= \pm \frac{m}{\sqrt{m^2+n^2}} \\[6pt] \cos\varphi &= \pm \frac{n}{\sqrt{m^2+n^2}} \end{aligned} \right\} \qquad \ldots\ldots (6), (7) \]

We now proceed to the determination of the signs of expressions (6), (7), or, what is the same thing, to the vector determination of the angle \(\varphi\), satisfying the condition:

\[ 0<\varphi<\frac{1}{2}\pi, \]

or the condition:

\[ \pi<\varphi<\frac{3}{2}\pi, \]

in connection with the determination of the signs of the quantities \(m\) and \(n\).

We have:

\[ a_{12}>0 \quad \text{or} \quad (-mn)>0, \]

whence four assumptions are equally possible.

1.

\[ -m>0 \]
\[ n>0 \]

and, in accordance with the condition

\[ 0<\varphi<\frac{1}{2}\pi, \]

we have:

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

\[ \cos\varphi=+\frac{n}{\sqrt{m^{2}+n^{2}}}. \]

2.

\[ -m<0 \]
\[ n<0 \]

and, in accordance with the condition

\[ \pi<\varphi<\frac{3}{2}\pi, \]

we have:

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

\[ \cos\varphi=+\frac{n}{\sqrt{m^{2}+n^{2}}}. \]

3.

\[ m>0 \]
\[ -n>0 \]

and, in accordance with the condition

\[ 0<\varphi<\frac{1}{2}\pi, \]

we obtain:

\[ \sin\varphi=+\frac{m}{\sqrt{m^{2}+n^{2}}} \]

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

4.

\[ m<0 \]
\[ -n<0 \]

and, in accordance with the condition

\[ \pi<\varphi<\frac{3}{2}\pi, \]

we obtain:

\[ \sin\varphi=+\frac{m}{\sqrt{m^2+n^2}} \]

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

From considering assumptions 1, 2, 3, and 4 it follows that there are only two independent assumptions, and these latter are the ones we must take: either 1 and 2, or 3 and 4.

Further, we establish that, for each of the inequalities:

\[ 0<\varphi<\frac{1}{2}\pi \ . . . . . . . . . . . . . (P) \]

\[ \pi<\varphi<\frac{3}{2}\pi, . . . . . . . . . . . . . (Q) \]

assumptions 1 and 2 determine, vectorially, the sought azimuth \(\varphi\) of the line of dip of the bed plane, by the formula:

\[ \tg\varphi=\frac{\sin\varphi}{\cos\varphi}=\frac{-m}{n}; . . . . . . . . . . . . . (8) \]

assumptions 3 and 4 determine, vectorially, the very same sought azimuth \(\varphi\) of the line of dip, but by another formula:

\[ \tg\varphi=\frac{\sin\varphi}{\cos\varphi}=\frac{m}{-n} . . . . . . . . . . . . . (9) \]

We arrive at the conclusion that it is necessary and natural to pose the question: in precisely which cases does one or the other of formulas (8), (9) hold?

The answer to the question posed must follow from the consideration that we have rotated the \(X\)-axis through the angle \(\varphi\), corresponding to the azimuth precisely of the line of dip.

Indeed. After rotating the system \(XYZ\) about the \(Z\)-axis through the angle \(\varphi\), we have (for points of the line of dip):

\[ \left. \begin{array}{l} x=x'\cos\varphi\\ y=x'\sin\varphi\\ z=z' \end{array} \right\} \ . . . . . . . . . . . . . (10) \]

Let us now express \(z\) (or \(z'\)) as a function of \(x'\) and \(\varphi\).

From equation (1) of the bed plane we have:

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

Substituting here relations (10), we obtain:

\[ z=\frac{x'}{\kappa}(-n\cos\varphi+m\sin\varphi)\ldots\ldots\ldots\ldots (11) \]

According to assumptions 1 and 2, formula (11) gives:

\[ z=\frac{x'}{\kappa}\left(-\frac{n^2}{\sqrt{m^2+n^2}}-\frac{m^2}{\sqrt{m^2+n^2}}\right) =-\frac{x'}{\kappa}\sqrt{m^2+n^2}\ldots (12) \]

According to assumptions 3 and 4, formula (11) gives:

\[ z=\frac{x'}{\kappa}\sqrt{m^2+n^2}\ldots\ldots\ldots\ldots (13) \]

Let us now take the points of the dip line for which

\[ z\geq 0; \]

then, as we know,

\[ x'\geq 0, \]

and formulas [(12), (13)] give, respectively, \(\kappa\geq 0\).

Thus we obtain the final answer to the question posed above.

Namely:

The formula

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

determines vectorially the sought azimuth \(\varphi\), taking \(\kappa>0\); the formula

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

determines, vectorially, precisely the same sought azimuth \(\varphi\) of the dip line, taking \(\kappa<0\).

The sign of the quantity

\[ \kappa=\tg\zeta_1\,\tg\zeta_2\,\sin(a_2-a_1) \]

determines, as indicated in my article,^1 the order of designating the directions—traces.

II. Let \(a_{12}<0\),

therefore \(\rho<0\).

\[ 1.\quad a'_{11}=\frac{1}{2}(S+\rho_n)=\frac{1}{2}(S-\rho_p) =\tg^2\zeta_1\,\tg^2\zeta_2\,\sin^2(a_2-a_1)=\kappa^2. \]

\[ 2.\quad a'_{22}=\frac{1}{2}(S-\rho_n)=\frac{1}{2}(S+\rho_p) =m^2+n^2+\kappa^2. \]

\[ 3.\quad a'_{33}=a_{33}=-R^2\kappa^2. \]

We reduce the equation of the ellipse (3), in the system \(X'Y'Z'\), to normal form:

\[ \frac{x'^2}{p'_1}+\frac{y'^2}{p'_2}-1=0, \]

where

\[ p'_1=-\frac{a'_{33}}{a'_{11}};\qquad p'_2=-\frac{a'_{33}}{a'_{22}}. \]

In the present case:

\[ p'_1=p_2=R^2. \]

\[ p'_2=p_1=R^2\frac{\kappa^2}{m^2+n^2+\kappa^2}. \]

We see directly that

\[ p'_1>0;\qquad p'_2>0\quad \text{and}\quad p'_1>p'_2. \]

We conclude that, after rotation to the right through the angle

\[ 0<\varphi_0<\frac{1}{2}\pi, \]

the new axis \(X'\) coincides with the major axis of the ellipse (the strike line of the plane of the stratum).

Since the new axis \(X'\) must coincide with the projection of the direction of the dip line, it is not difficult to see that, vectorially, the axis \(X\) must be turned to the right through an angle \(\varphi\) satisfying either the condition (Figs. 28, 29)

\[ \frac{1}{2}\pi<\varphi<\pi \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (R) \]

or the condition (Figs. 30, 31)

\[ 2\pi>\varphi>\frac{3}{2}\pi \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (S) \]

Under these conditions, \((R)\) and \((S)\),

\[ \tg \varphi<0. \]

Corresponding to condition \((R)\),

\[ \varphi=\varphi_0+\frac{1}{2}\pi, \]

and the values

\[ \left. \begin{aligned} \sin\varphi &= \cos\varphi_0 = +\sqrt{\frac{1}{2}\left(1+\frac{a_{11}-a_{22}}{\rho_n}\right)},\\[6pt] \cos\varphi &= -\sin\varphi_0 = -\sqrt{\frac{1}{2}\left(1-\frac{a_{11}-a_{22}}{\rho_n}\right)} \end{aligned} \right\} \tag{14} \]

According to condition \((S)\),

\[ \varphi=\frac{3}{2}\pi+\varphi_0, \]

and the values

\[ \left. \begin{aligned} \sin\varphi &= -\cos\varphi_0 = -\sqrt{\frac{1}{2}\left(1+\frac{a_{11}-a_{22}}{\rho_n}\right)},\\[6pt] \cos\varphi &= \sin\varphi_0 = +\sqrt{\frac{1}{2}\left(1-\frac{a_{11}-a_{22}}{\rho_n}\right)} \end{aligned} \right\} \tag{15} \]

We proceed to the direct determination of \(\sin\varphi\) and \(\cos\varphi\) as functions of the given quantities.

\[ \begin{aligned} 1.\quad & a_{11}-a_{22}=n^2-m^2,\\[6pt] 2.\quad & \frac{1}{2}\,\frac{\rho_n-(a_{11}-a_{22})}{\rho_n} = \frac{n^2}{m^2+n^2},\\[6pt] 3.\quad & \frac{1}{2}\,\frac{\rho_n+(a_{11}-a_{22})}{\rho_n} = \frac{m^2}{m^2+n^2}. \end{aligned} \]

Therefore, (14) and 15) pass into

\[ \left. \begin{aligned} \sin \varphi &= \pm \frac{m}{\sqrt{m^{2}+n^{2}}} \\[4pt] \cos \varphi &= \pm \frac{n}{\sqrt{m^{2}+n^{2}}} \end{aligned} \right\} \ . . . . . . . . . (16), (17) \]

We now proceed to the determination of the signs of the expressions (16), (17), or, what is the same thing, to the vectorial determination of the angle \(\varphi\), satisfying the condition

\[ \frac{1}{2}\pi < \varphi < \pi, \]

or the condition

\[ 2\pi > \varphi > \frac{3}{2}\pi, \]

in connection with the determination of the signs of the quantities \(m\) and \(n\).

We have:

\[ a_{12}<0 \quad \text{or} \quad (-mn)<0, \]

whence four cases are equally possible,

\[ 1. \]

\[ -m>0 \]
\[ n<0 \]

and, correspondingly to the condition

\[ \frac{1}{2}\pi < \varphi < \pi, \]

we obtain:

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

\[ \cos \varphi = + \frac{n}{\sqrt{m^{2}+n^{2}}}. \]

\[ 2. \]

\[ -m<0 \]
\[ n>0 \]

and, correspondingly to the condition

\[ 2\pi > \varphi > \frac{3}{2}\pi, \]

we obtain:

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

\[ \cos\varphi=+\frac{n}{\sqrt{m^{2}+n^{2}}}. \]

3.

\[ m>0 \]

\[ -n<0 \]

and, in accordance with the condition

\[ \frac{1}{2}\pi<\varphi<\pi, \]

we obtain:

\[ \sin\varphi=+\frac{m}{\sqrt{m^{2}+n^{2}}} \]

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

4.

\[ m<0 \]

\[ -n>0 \]

and, in accordance with the condition

\[ 2\pi>\varphi>\frac{3}{2}\pi, \]

we obtain:

\[ \sin\varphi=+\frac{m}{\sqrt{m^{2}+n^{2}}} \]

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

In what follows the analysis is completely identical to the analysis when the inequality \(a_{12}>0\) holds; it is only necessary to replace conditions (P) and (Q) by conditions (R) and (S).

Thus, for any

\[ a_{12}\lesseqgtr 0, \]

we have:

1. Having put \(k>0\), the azimuth of the line of dip of the stratum plane is determined vectorially by the formula

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

1. Taking \(k < 0\), the azimuth of the line of dip of the bed plane is determined vectorially by the formula

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

§ 15.

Method for the simultaneous determination of the azimuth \(\varphi\) and the zenith distance \(\zeta\) of the line of dip of the bed plane.

Let us take, on the given traces \((a_{1,2}, \zeta_{1,2})\), respectively, two points 1 and 2 corresponding to one and the same value of the coordinate \(z\)—\(z_1 = z_2 = H\)—and therefore defining the strike line.

We now rotate the coordinate system \(XYZ\) about the \(Z\)-axis through an angle \(\varphi\), corresponding to the azimuth of the line of dip of the bed plane.

Then, in the new coordinate system \(X'Y'Z'\), the coordinates \(x'\) of points 1 and 2 will be equal:

\[ x'_1 = x'_2, \]

for the plane of dip will be perpendicular to the straight line determined by points 1 and 2 (the strike line).

We take the coordinate-transformation formula for \(x'_{1,2}\):

\[ x'_{1,2} = x_{1,2}\cos\varphi + y_{1,2}\sin\varphi . \ . \ . \ . \ . \ . \ . \ . \ (1) \]

Substituting into (1) the values derived earlier:

\[ x_{1,2} = H\tg \zeta_{1,2}\cos a_{1,2} \]

\[ y_{1,2} = H\tg \zeta_{1,2}\sin a_{1,2} \]

we obtain:

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

whence, putting

\[ x'_1 = x'_2 = x', \]

we find:

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

But

\[ \frac{x'}{H} = \tg \zeta; \]

therefore, finally,

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

Further, the determination of \(\varphi\) and \(\zeta\) is set forth in my article¹ (§ 17).

§ 16.

Methods for determining the zenith distance \(\zeta\) of the line of dip of the plane of the stratum.

I.

Rotating the system \(XYZ\) about the axis \(Z\) through an angle \(\varphi\), corresponding to the azimuth of the line of dip, we see that, for the point \((x' > 0, z < 0)\), respectively,

\[ \operatorname{tg}\zeta=\frac{x'}{z}, \qquad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

where

\[ \frac{1}{2}\pi<\zeta<\pi. \]

Express \(x'\) as a function of \(z\).

Take the invariant of expressions \((1_{1,2})\) of § 12:

\[ x^2+y^2=x'^2+y'^2. \]

In our case

\[ x'>0;\ y'=0; \]

therefore

\[ x'=+\sqrt{x^2+y^2}. \]

Substituting here the values of \(x\) and \(y\), expressed as functions of \(z\), and taking \(k>0\), we obtain:

\[ x'=-\frac{k}{\sqrt{m^2+n^2}}\,z; \]

and, finally, formula (1) takes the form:

\[ \operatorname{tg}\zeta=-\frac{k}{\sqrt{m^2+n^2}}. \]

II.

Drawing through the origin the line of dip of the plane of the stratum \(P\) and the normal to the latter, let us denote the zenith distances of the line of dip, of the normal above and below the plane \(P\), respectively, by \(\zeta, h_1, h_2\).

It is easy to see that:

1. \(\zeta=\dfrac{1}{2}\pi+h_1\), taking \(h_1\) satisfying the condition

\[ 0<h_1<\frac{1}{2}\pi. \]

1. \(\zeta=\dfrac{3}{2}\pi-h_2\), taking \(h_2\) satisfying the condition

\[ \frac{1}{2}\pi<h_2<\pi. \]

In the first case

\[ \operatorname{tg}\zeta=-\operatorname{ctg}h_1=-\frac{\cos h_1}{\sqrt{1-\cos^2 h_1}}\ldots\ldots\ldots (1_1) \]

In the second case

\[ \operatorname{tg}\zeta=\operatorname{ctg}h_2=+\frac{\cos h_2}{\sqrt{1-\cos^2 h_2}}\ldots\ldots\ldots (1_2) \]

We note that in both cases the form of the formula is one and the same and, therefore, dropping the subscripts 1 and 2, we obtain:

\[ \operatorname{tg}\zeta=\pm\frac{\cos h}{\sqrt{1-\cos^2 h}}\ldots\ldots\ldots (2) \]

Since the normal is perpendicular to the traces, then, as is known,

\[ \cos h=\pm\frac{\Delta_\gamma}{\sin\varphi},\ldots\ldots\ldots (3) \]

where \(\varphi\) is the angle between the traces, and is determined by the formula:

\[ \sin^2\varphi=\Delta_\alpha^2+\Delta_\beta^2+\Delta_\gamma^2 \ldots\ldots\ldots (4) \]

Substituting (3) and (4) into (2), we obtain:

\[ \operatorname{tg}\zeta=\pm\frac{\Delta_\gamma}{\sqrt{\Delta_\alpha^2+\Delta_\beta^2}}\ldots\ldots\ldots (5) \]

Taking into account formula (1) of the remark to Section XII, we obtain:

\[ \Delta_\alpha = \left| \begin{array}{cc} \cos\beta_1, & \cos\gamma_1\\ \cos\beta_2, & \cos\gamma_2 \end{array} \right| = \left| \begin{array}{cc} \sin\zeta_1\sin a_1, & \cos\zeta_1\\ \sin\zeta_2\sin a_2, & \cos\zeta_2 \end{array} \right| = \]

\[ =\cos\zeta_1\cos\zeta_2 \left| \begin{array}{cc} \operatorname{tg}\zeta_1\sin a_1, & 1\\ \operatorname{tg}\zeta_2\sin a_2, & 1 \end{array} \right| = \cos\zeta_1\cos\zeta_2(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2), \]

\[ \Delta_\beta = \left| \begin{array}{cc} \cos\gamma_1, & \cos\alpha_1\\ \cos\gamma_2, & \cos\alpha_2 \end{array} \right| = \left| \begin{array}{cc} \cos\zeta_1, & \sin\zeta_1\cos a_1\\ \cos\zeta_2, & \sin\zeta_2\cos a_2 \end{array} \right| = \]

\[ =\cos\zeta_1\cos\zeta_2 \left| \begin{array}{cc} 1, & \operatorname{tg}\zeta_1\cos a_1\\ 1, & \operatorname{tg}\zeta_2\cos a_2 \end{array} \right| = \cos\zeta_1\cos\zeta_2(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1). \]

\[ \Delta_\gamma = \begin{vmatrix} \cos \alpha_1, & \cos \beta_1\\ \cos \alpha_2, & \cos \beta_2 \end{vmatrix} = \begin{vmatrix} \sin \zeta_1 \cos a_1, & \sin \zeta_1 \sin a_1\\ \sin \zeta_2 \cos a_2, & \sin \zeta_2 \sin a_2 \end{vmatrix} = \]

\[ = \sin \zeta_1 \sin \zeta_2 \begin{vmatrix} \cos a_1, & \sin a_1\\ \cos a_2, & \sin a_2 \end{vmatrix} = \sin \zeta_1 \sin \zeta_2 \sin (a_2 - a_1). \]

Substituting the obtained values \(\Delta_\alpha, \Delta_\beta, \Delta_\gamma\) into (5), we obtain:

\[ \operatorname{tg}\zeta = \pm \frac{\sin \zeta_1 \sin \zeta_2 \sin (a_2-a_1)} {\cos \zeta_1 \cos \zeta_2 \sqrt{(\operatorname{tg}\zeta_1 \sin a_1-\operatorname{tg}\zeta_2 \sin a_2)^2+ (\operatorname{tg}\zeta_2 \cos a_2-\operatorname{tg}\zeta_1 \cos a_1)^2}} = \]

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

Taking \(\operatorname{tg}\zeta_1 \operatorname{tg}\zeta_2 \sin (a_2-a_1)=\kappa>0\) and the condition

\[ \frac{1}{2}\pi<\zeta<\pi, \]

we finally obtain:

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

III.

We had:

\[ \operatorname{tg}\zeta = \pm \frac{\cos h}{\sqrt{1-\cos^2 h}}, \]

where \(h\) is one of the angles formed by the direction of the \(Z\)-axis and one of the directions of the normal.

But it is known that

\[ \cos h = \pm \frac{m_1 m_2+n_1 n_2+p_1 p_2} {\sqrt{m_1^2+n_1^2+p_1^2}\cdot \sqrt{m_2^2+n_2^2+p_2^2}}, \qquad \ldots \qquad (1) \]

where

\[ m_1=0;\quad n_1=0;\quad p_1=1 \]

\[ m_2=n;\quad n_2=-m;\quad p_2=\kappa \]

are the direction coefficients, respectively, of the \(Z\)-axis and of the normal to the plane of the stratum:

\[ \frac{x}{0}=\frac{y}{0}=\frac{z}{1}. \qquad \ldots \qquad (2) \]

\[ \frac{x}{n}=\frac{y}{-m}=\frac{z}{\kappa}. \qquad \ldots \qquad (3) \]

Therefore, (1) takes the form:

\[ \cos h=\pm \frac{\kappa}{\sqrt{m^2+n^2+\kappa^2}}, \]

\[ 1-\cos^2 h=\frac{m^2+n^2}{m^2+n^2+\kappa^2}, \]

and

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

Putting \(\kappa>0\), by virtue of the condition

\[ \frac{1}{2}\pi<\zeta<\pi, \]

we finally obtain:

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

IV.

Let us denote by \(h\) the dip angle of the plane of the stratum. The angle \(h\) is the angle between the planes: the stratum

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

and an arbitrary horizontal plane

\[ z-H=0 \ldots\ldots\ldots\ldots\ldots\ldots (2) \]

It is easy to see that the zenith distance \(\zeta\) of the line of dip satisfies the condition:

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

and, consequently,

\[ \tg \zeta=\ctg h=\frac{\sqrt{1-\sin^2 h}}{\sin h}=-\frac{\cos h}{\sqrt{1-\cos^2 h}}, \]

for

\[ -\frac{1}{2}\pi<h<0. \]

1. In general we have:

\[ \sin^2 h= \frac{ \left|\begin{matrix} A_1 & B_1\\ A_2 & B_2 \end{matrix}\right|^2 + \left|\begin{matrix} B_1 & C_1\\ B_2 & C_2 \end{matrix}\right|^2 + \left|\begin{matrix} C_1 & A_1\\ C_2 & A_2 \end{matrix}\right|^2 } {(A_1^2+B_1^2+C_1^2)(A_2^2+B_2^2+C_2^2)}. \]

In our case, respectively for (1) and (2),

\[ A_1=n;\quad B_1=-m;\quad C_1=\kappa \]

\[ A_2=0;\quad B_2=0;\quad C_2=1. \]

\[ \left| \begin{matrix} A_1 & B_1\\ A_2 & B_2 \end{matrix} \right|^2 = \left| \begin{matrix} n, & -m\\ 0 & 0 \end{matrix} \right|^2 =0. \]

\[ \left| \begin{matrix} B_1 & C_1\\ B_2 & C_2 \end{matrix} \right|^2 = \left| \begin{matrix} -m, & \kappa\\ 0 & 1 \end{matrix} \right|^2 =m^2. \]

\[ \left| \begin{matrix} C_1 & A_1\\ C_2 & A_2 \end{matrix} \right|^2 = \left| \begin{matrix} \kappa, & n\\ 1 & 0 \end{matrix} \right|^2 =n^2. \]

\[ A_1^2+B_1^2+C_1^2=m^2+n^2+\kappa^2 \]

\[ A_2^2+B_2^2+C_2^2=1. \]

\[ \sin^2 h=\frac{m^2+n^2}{m^2+n^2+\kappa^2} \]

and

\[ \sqrt{1-\sin^2 h}=\frac{\kappa}{\sqrt{m^2+n^2+\kappa^2}} \]

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

1. We have, in general:

\[ \cos h=\pm \frac{A_1A_2+B_1B_2+C_1C_2} {\sqrt{A_1^2+B_1^2+C_1^2}\cdot\sqrt{A_2^2+B_2^2+C_2^2}}. \]

In our case the sign \((-)\) drops out, and

\[ A_1A_2=0;\quad B_1B_2=0;\quad C_1C_2=\kappa. \]

\[ \sqrt{A_1^2+B_1^2+C_1^2}=\sqrt{m^2+n^2+\kappa^2} \]

\[ \cdot\sqrt{A_2^2+B_2^2+C_2^2}=1. \]

Therefore

\[ \cos h=\frac{\kappa}{\sqrt{m^2+n^2+\kappa^2}} \]

\[ \sqrt{1-\cos^2 h}= \sqrt{\frac{m^2+n^2}{m^2+n^2+\kappa^2}}, \]

and

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

§ 17.

Solution by the method of spherical trigonometry.

From the point (Fig. 32) \(O\), the point of convergence of the traces, as center, describe a sphere of arbitrary radius. Next mark the points \(I\), \(II\), \(Z\) of intersection of the traces \(OI\), \(OII\) and of the vertical straight line \(OZ\) with the surface of the sphere. Join the points obtained by arcs of great circles, as indicated in the drawing (the drawing represents an orthogonal projection of the sphere onto the horizontal plane).

Next draw from the point \(O\) the arc \(OF_1 = \pi - \zeta\) (the arc \(OAF = \zeta\)), perpendicular to the arc \(IF_1II = c\). Let \(\zeta_{1,2}\) satisfy the condition:

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

1. From the spherical triangle \(OI\,II\,O\) we have:

\[ \sin(a_2-a_1)=\frac{2S}{\sin\zeta_1\sin\zeta_2} \tag{1} \]

\[ \sin(\pi-\zeta)=\frac{2S}{\sin c}, \tag{2} \]

where \(S\) is the amplitude of the spherical triangle.

Determining \(2S\) from (1) and (2) and comparing, we obtain:

\[ \sin\zeta= \frac{\sin(a_2-a_1)\sin\zeta_1\sin\zeta_2}{\sin c}. \tag{3} \]

Further we have:

\[ \cos c=\cos\zeta_1\cos\zeta_2+\sin\zeta_1\sin\zeta_2\cos(a_2-a_1). \tag{4} \]

and, correspondingly,

\[ \sin c=+\sqrt{1-\cos^2 c}, \tag{5} \]

or

\[ \pi > c > 0. \]

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

\[ \sin\zeta= \frac{\sin(a_2-a_1)\sin\zeta_1\sin\zeta_2} {\sqrt{1-\cos^2 c}}. \tag{6} \]

\[ \cos\zeta= -\sqrt{ 1- \frac{\sin^2(a_2-a_1)\sin^2\zeta_1\sin^2\zeta_2} {1-\cos^2 c} }; \tag{7} \]

or

\[ \pi>\zeta>\frac{1}{2}\pi. \]

Dividing (6) by (7), we obtain:

\[ \operatorname{tg}\zeta = -\frac{\sin(a_2-a_1)\sin\zeta_1\sin\zeta_2} {\sqrt{\,1-\cos^2 c-\sin^2(a_2-a_1)\sin^2\zeta_1\sin^2\zeta_2\,}} = \]

\[ = -\frac{\sin(a_2-a_1)\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2} {\sqrt{\dfrac{1}{\cos^2\zeta_1\cos^2\zeta_2} \left[1-\cos^2 c-\sin^2(a_2-a_1)\sin^2\zeta_1\sin^2\zeta_2\right]}} . \]

But, in view of (4),

\[ \frac{1}{\cos^2\zeta_1\cos^2\zeta_2} \left[1-\cos^2 c-\sin^2(a_2-a_1)\sin^2\zeta_1\sin^2\zeta_2\right] = \]

\[ =(1+\operatorname{tg}^2\zeta_1)(1+\operatorname{tg}^2\zeta_2) -1-2\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\cos(a_2-a_1) - \]

\[ -\operatorname{tg}^2\zeta_1\operatorname{tg}^2\zeta_2\cos^2(a_2-a_1) -\sin^2(a_2-a_1)\operatorname{tg}^2\zeta_1\operatorname{tg}^2\zeta_2 = \]

\[ =\operatorname{tg}^2\zeta_1+\operatorname{tg}^2\zeta_2 -2\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\cos(a_2-a_1) = \]

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

Finally we have:

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

—the formula determining the zenith distance \(\zeta\) of the line of dip.

Remark. For an analysis of the formula see article 1.

1. Continue the arcs \(0(Z)I\) and \(0(Z)II\) until they intersect the circle \(ABCDA\), respectively at the points \(K_1\) and \(K_2\).

From the spherical triangles \(BIK_1B\) and \(DIIK_2D\), right-angled respectively at the points \(K_1\) and \(K_2\), we obtain:

\[ \operatorname{tg}\left(\frac{1}{2}\pi-\zeta_1\right) = \sin\left[a_1-\left(\frac{1}{2}\pi+\varphi\right)\right] \operatorname{tg}\left(\zeta-\frac{1}{2}\pi\right) \]

\[ \operatorname{tg}\left(\frac{1}{2}\pi-\zeta_2\right) = \sin\left[\varphi-\left(a_2+\frac{1}{2}\pi\right)\right] \operatorname{tg}\left(\zeta-\frac{1}{2}\pi\right), \]

whence

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

1. The formulas \((1_{1,2})\) can be obtained in another way.

Through the point \(O\) draw the normal to the plane of the stratum, determined by the traces \(OI\) and \(OII\), and mark the point \(N_1\) where it meets the surface of the sphere. Then connect the point \(N_1\) with the points \(O, I\), and \(II\) by arcs of great circles. It is easy to see that the spherical triangles \(ON_1IO\) and \(ON_1IIO\) are right-sided:

\[ \smile N_1I=\smile N_1II=\frac{1}{2}\pi . \]

We have:

\[ \cos(a_1-\varphi) = -\operatorname{ctg}\left(\zeta-\frac{1}{2}\pi\right)\operatorname{ctg}\zeta_1 \]

\[ \cos(\varphi-a_2) = -\operatorname{ctg}\left(\zeta-\frac{1}{2}\pi\right)\operatorname{ctg}\zeta_2, \]

whence

\[ \tg \zeta=\tg \zeta_{1,2}\cos(\varphi-a_{1,2}) \ldots\ldots\ldots (1_{1,2}) \]

Further transformations of formulas \((1_{1,2})\)—the determination of \(\varphi\) and \(\zeta\)—are indicated in § 17 of article \(^{1}\).

1. From the spherical triangle \(OIIO\) we have:

\[ \cos c=\cos\zeta_1\cos\zeta_2+\sin\zeta_1\sin\zeta_2\cos(a_2-a_1) \ldots\ldots (1) \]

But

\[ c=c_1+c_2, \]

therefore

\[ \cos c=\cos c_1\cos c_2-\sin c_1\sin c_2 \ldots\ldots (2) \]

Comparing (1) and (2), we obtain:

\[ \cos c_1\cos c_2-\sin c_1\sin c_1=\cos\zeta_1\cos\zeta_2+\sin\zeta_1\sin\zeta_2\cos(a_2-a_1) \ldots (3) \]

We shall now determine the values \(c_1\) and \(c_2\) entering into formula (3).
From the spherical triangles \(OIF_1O\) and \(OIIF_1O\), right-angled at the point \(F_1\), respectively, we have:

\[ \left. \begin{aligned} \cos c_1&=-\frac{\cos\zeta_1}{\cos\zeta}\\ \sin c_1&=-\sin\zeta_1\sin(\varphi-a_1) \end{aligned} \right\} \ldots\ldots (4) \]

\[ \left. \begin{aligned} \cos c_2&=-\frac{\cos\zeta_2}{\cos\zeta}\\ \sin c_2&=\sin\zeta_2\sin(\varphi-a_2) \end{aligned} \right\} \ldots\ldots (5) \]

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

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

\[ \left. -\sin(\varphi-a_1)\sin(\varphi-a_2) \right] \ldots\ldots (6) \]

But

\[ \frac{1}{\cos^2\zeta}-1=\tg^2\zeta \ldots\ldots (7) \]

Substituting (7) into (6) and dividing the latter by \((\cos\zeta_1\cos\zeta_2)\), we obtain:

\[ \tg^2\zeta=\tg\zeta_1\tg\zeta_2 \left[\cos(a_2-a_1)-\sin(\varphi-a_1)\sin(\varphi-a_2)\right] \ldots (8) \]

But

\[ \cos(a_2-a_1)=\cos\left[(a_2-\varphi)+(\varphi-a_1)\right]= \]

\[ =\cos(\varphi-a_2)\cos(\varphi-a_1)+\sin(\varphi-a_2)\sin(\varphi-a_1) \ldots (9) \]

Substituting (9) into (8), we have:

\[ \tg^2\zeta=\tg\zeta_1\tg\zeta_2\cos(\varphi-a_1)\cos(\varphi-a_2); \ldots\ldots (10) \]

whence, taking the condition

\[ \frac{1}{2}\pi < \zeta < \pi, \]

we obtain:

\[ \operatorname{tg}\zeta = -\sqrt{\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2 \cos(\varphi-a_1)\cos(\varphi-a_2)}. \]

Formula (10), obviously, is a consequence of the formulas derived earlier:

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

1. Let us denote the spherical excesses of the triangles \(OIIO\), \(OIF_1O\), \(OIIF_1O\), respectively, by \(\varepsilon\), \(\varepsilon_1\), and \(\varepsilon_2\).

We have:

\[ \frac{1}{2}\varepsilon = \frac{1}{2}\varepsilon_1 + \frac{1}{2}\varepsilon_2 \tag{1} \]

\[ \operatorname{tg}\frac{1}{2}\varepsilon = \frac{ \operatorname{tg}\frac{1}{2}\varepsilon_1 + \operatorname{tg}\frac{1}{2}\varepsilon_2 }{ 1-\operatorname{tg}\frac{1}{2}\varepsilon_1 \operatorname{tg}\frac{1}{2}\varepsilon_2 } \tag{2} \]

\[ \operatorname{tg}\frac{1}{2}\varepsilon = \frac{ \sin \widehat{IOII} }{ \operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{ctg}\frac{1}{2}\zeta_2 + \cos \widehat{IOII} } \tag{3} \]

\[ \operatorname{tg}\frac{1}{2}\varepsilon_1 = \frac{ \sin \widehat{IOF_1} }{ \operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta + \cos \widehat{IOF_1} } \tag{4} \]

\[ \operatorname{tg}\frac{1}{2}\varepsilon_2 = \frac{ \sin \widehat{IIOF_1} }{ \operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta + \cos \widehat{IIOF_1} } \tag{5} \]

But

\[ \widehat{IOII}=2\pi-(a_1-a_2). \]

\[ \sin \widehat{IOII}=\sin(a_2-a_1);\qquad \cos \widehat{IOII}=\cos(a_2-a_1). \]

\[ \widehat{IOF_1}=-a_1+\varphi+\pi. \]

\[ \sin \widehat{IOF_1}=-\sin(\varphi-a_1);\qquad \cos \widehat{IOF_1}=-\cos(\varphi-a_1). \]

\[ \widehat{IIOF_1}=-\varphi+a_2+\pi. \]

\[ \sin \widehat{IIOF_1}=\sin(\varphi-a_2);\qquad \cos \widehat{IIOF_1}=-\cos(\varphi-a_2). \]

Substituting now (3), (4), and (5) into (2) and transforming the right-hand side of the latter, we obtain:

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

\[ = \frac{ \frac{-\sin(\varphi-a_1)} {\operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_1)} + \frac{\sin(\varphi-a_2)} {\operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_2)} } { 1- \frac{-\sin(\varphi-a_1)} {\operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_1)} \cdot \frac{\sin(\varphi-a_2)} {\operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_2)} } = \]

\[ = \frac{ -\sin(\varphi-a_1) \left[ \operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_2) \right] +\sin(\varphi-a_2) } { \left[ \operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_1) \right] \left[ \operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_2) \right] + } \]

\[ \frac{ \times \left[ \operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta-\cos(\varphi-a_1) \right] } { +\sin(\varphi-a_1)\sin(\varphi-a_2) } = \frac{A}{B}, \ldots \ldots (6) \]

denoting by \(A\) the numerator, and by \(B\) the denominator.
We rewrite expression (6) as follows:

\[ \frac{\sin(a_2-a_1)} {\operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{ctg}\frac{1}{2}\zeta_2+\cos(a_2-a_1)} = \frac{A}{B}, \]

or

\[ \sin(a_2-a_1)B = \left[ \operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{ctg}\frac{1}{2}\zeta_2+\cos(a_2-a_1) \right]A . \ldots (7) \]

Let us now transform \(A\) and \(B\).

\[ A = -\sin(\varphi-a_1)\operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta +\sin(\varphi-a_1)\cos(\varphi-a_2) + \]

\[ +\sin(\varphi-a_2)\operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta -\sin(\varphi-a_2)\cos(\varphi-a_1) = \]

\[ = -\sin(\varphi-a_1)\operatorname{ctg}\frac{1}{2}\zeta_2 \operatorname{tg}\frac{1}{2}\zeta +\sin(\varphi-a_2)\operatorname{ctg}\frac{1}{2}\zeta_1 \operatorname{tg}\frac{1}{2}\zeta + \]

\[ +\sin(a_2-a_1) = -\operatorname{tg}\frac{1}{2}\zeta \left[ \operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_1) - \operatorname{ctg}\frac{1}{2}\zeta_1\sin(\varphi-a_2) \right] + \]

\[ +\sin(a_2-a_1), \ldots \ldots \ldots (8) \]

or

\[ \sin(\varphi-a_1)\cos(\varphi-a_2)-\cos(\varphi-a_1)\sin(\varphi-a_2)= \]

\[ =\sin\bigl[(\varphi-a_1)-(\varphi-a_2)\bigr]=\sin(a_2-a_1). \]

\[ \begin{aligned} B={}&\operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2\,\operatorname{tg}^2\frac12\zeta -\cos(\varphi-a_1)\operatorname{ctg}\frac12\zeta_2\,\operatorname{tg}\frac12\zeta\\ &-\operatorname{ctg}\frac12\zeta_1\,\operatorname{tg}\frac12\zeta\,\cos(\varphi-a_2) +\cos(\varphi-a_1)\cos(\varphi-a_2)\\ &+\sin(\varphi-a_1)\sin(\varphi-a_2) =\operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2\,\operatorname{tg}^2\frac12\zeta\\ &-\cos(\varphi-a_1)\operatorname{ctg}\frac12\zeta_2\,\operatorname{tg}\frac12\zeta -\operatorname{ctg}\frac12\zeta_1\,\operatorname{tg}\frac12\zeta\,\cos(\varphi-a_2)\\ &+\cos(a_2-a_1) =\operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2\,\operatorname{tg}^2\frac12\zeta\\ &-\operatorname{tg}\frac12\zeta\left[ \operatorname{ctg}\frac12\zeta_2\cos(\varphi-a_1) +\operatorname{ctg}\frac12\zeta_1\cos(\varphi-a_2) \right]\\ &+\cos(a_2-a_1). \qquad\qquad\qquad (9) \end{aligned} \]

or

\[ \cos(\varphi-a_1)\cos(\varphi-a_2)+\sin(\varphi-a_1)\sin(\varphi-a_2)= \]

\[ =\cos\bigl[(\varphi-a_1)-(\varphi-a_2)\bigr]=\cos(a_2-a_1). \]

Substituting (8) and (9) into (7), we obtain:

\[ \sin(a_2-a_1)\operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2\,\operatorname{tg}^2\frac12\zeta- \]

\[ -\sin(a_2-a_1)\operatorname{tg}\frac12\zeta \left[ \operatorname{ctg}\frac12\zeta_2\cos(\varphi-a_1) +\operatorname{ctg}\frac12\zeta_1\cos(\varphi-a_2) \right]= \]

\[ =-\left[ \operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2+\cos(a_2-a_1) \right]\operatorname{tg}\frac12\zeta \left[ \operatorname{ctg}\frac12\zeta_2\sin(\varphi-a_1)- \right. \]

\[ \left. -\operatorname{ctg}\frac12\zeta_1\sin(\varphi-a_2) \right] +\operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2\sin(a_2-a_1), \qquad (10) \]

after reducing both sides of the equation by

\[ \sin(a_2-a_1)\cos(a_2-a_1). \]

Transforming (10), we obtain:

\[ \sin(a_2-a_1)\operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2 \left(\operatorname{tg}^2\frac12\zeta-1\right)= \]

\[ =\operatorname{tg}\frac12\zeta\left\{ \sin(a_2-a_1)\left[ \operatorname{ctg}\frac12\zeta_2\cos(\varphi-a_1)+ \right.\right. \]

\[ \left. +\operatorname{ctg}\frac12\zeta_1\cos(\varphi-a_2) \right] -\left[ \operatorname{ctg}\frac12\zeta_1\,\operatorname{ctg}\frac12\zeta_2+ \right. \]

\[ \left.\left. +\cos(a_2-a_1) \right]\left[ \operatorname{ctg}\frac12\zeta_2\sin(\varphi-a_1) -\operatorname{ctg}\frac12\zeta_1\sin(\varphi-a_2) \right] \right\} \]

or, after division by \(\operatorname{tg}\dfrac{1}{2}\zeta\),

\[ \begin{aligned} 2\operatorname{ctg}\zeta \sin(a_2-a_1)\operatorname{ctg}\frac{1}{2}\zeta_1\operatorname{ctg}\frac{1}{2}\zeta_2 ={}&-\sin(a_2-a_1)\left[\operatorname{ctg}\frac{1}{2}\zeta_2\cos(\varphi-a_1)+\operatorname{ctg}\frac{1}{2}\zeta_1\cos(\varphi-a_2)\right]+\\ &+\operatorname{ctg}\frac{1}{2}\zeta_1\operatorname{ctg}^{2}\frac{1}{2}\zeta_2\sin(\varphi-a_1) +\operatorname{ctg}\frac{1}{2}\zeta_2\cos(a_2-a_1)\sin(\varphi-a_1)-\\ &-\operatorname{ctg}^{2}\frac{1}{2}\zeta_1\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_2) -\operatorname{ctg}\frac{1}{2}\zeta_1\cos(a_2-a_1)\sin(\varphi-a_2). \end{aligned} \tag{11} \]

Transform the right-hand side of equation (11), taking into account that

\[ \sin(a_2-a_1)=\sin(\varphi-a_1)\cos(\varphi-a_2)-\cos(\varphi-a_1)\sin(\varphi-a_2). \]

\[ \cos(a_2-a_1)=\cos(\varphi-a_1)\cos(\varphi-a_2)+\sin(\varphi-a_1)\sin(\varphi-a_2). \]

We have:

\[ \begin{aligned} &-\left[\sin(\varphi-a_1)\cos(\varphi-a_2)-\cos(\varphi-a_1)\sin(\varphi-a_2)\right] \left[\operatorname{ctg}\frac{1}{2}\zeta_2\cos(\varphi-a_1)+\right.\\ &\left.\qquad+\operatorname{ctg}\frac{1}{2}\zeta_1\cos(\varphi-a_2)\right] +\left[\cos(\varphi-a_1)\cos(\varphi-a_2)+\sin(\varphi-a_1)\sin(\varphi-a_2)\right]\cdot\\ &\qquad\cdot\left[\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_1)-\operatorname{ctg}\frac{1}{2}\zeta_1\sin(\varphi-a_2)\right] +\operatorname{ctg}\frac{1}{2}\zeta_1\operatorname{ctg}^{2}\frac{1}{2}\zeta_2\sin(\varphi-a_1)-\\ &-\operatorname{ctg}^{2}\frac{1}{2}\zeta_1\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_2) =\cos^{2}(\varphi-a_1)\sin(\varphi-a_2)\operatorname{ctg}\frac{1}{2}\zeta_2-\\ &-\operatorname{ctg}\frac{1}{2}\zeta_1\cos^{2}(\varphi-a_2)\sin(\varphi-a_1) +\operatorname{ctg}\frac{1}{2}\zeta_2\sin^{2}(\varphi-a_1)\sin(\varphi-a_2)-\\ &-\operatorname{ctg}\frac{1}{2}\zeta_1\sin(\varphi-a_1)\sin^{2}(\varphi-a_2) +\operatorname{ctg}\frac{1}{2}\zeta_1\operatorname{ctg}^{2}\frac{1}{2}\zeta_2\sin(\varphi-a_1)-\\ &-\operatorname{ctg}^{2}\frac{1}{2}\zeta_1\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_2) =\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_2)-\\ &-\operatorname{ctg}\frac{1}{2}\zeta_1\sin(\varphi-a_1) +\operatorname{ctg}\frac{1}{2}\zeta_1\operatorname{ctg}^{2}\frac{1}{2}\zeta_2\sin(\varphi-a_1)-\\ &-\operatorname{ctg}^{2}\frac{1}{2}\zeta_1\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_2)=\\ &=\operatorname{ctg}\frac{1}{2}\zeta_2\sin(\varphi-a_2)\left(1-\operatorname{ctg}^{2}\frac{1}{2}\zeta_1\right) -\operatorname{ctg}\frac{1}{2}\zeta_1\sin(\varphi-a_1)\left(1-\operatorname{ctg}^{2}\frac{1}{2}\zeta_2\right). \end{aligned} \]

Thus, finally, equation (11) takes the form:

\[ 2\operatorname{ctg}\zeta \sin(a_2-a_1)\operatorname{ctg}\frac{1}{2}\zeta_1\operatorname{ctg}\frac{1}{2}\zeta_2 = \]

\[ = -\operatorname{ctg}\frac{1}{2}\zeta_2 \sin(\varphi-a_2) \left(1-\operatorname{ctg}^2\frac{1}{2}\zeta_1\right) -\operatorname{ctg}\frac{1}{2}\zeta_1 \sin(\varphi-a_1) \left(1-\operatorname{ctg}^2\frac{1}{2}\zeta_2\right); \]

whence

\[ \operatorname{ctg}\zeta = \frac{ -\sin(\varphi-a_2) \dfrac{\operatorname{ctg}^2\frac{1}{2}\zeta_1-1} {2\operatorname{ctg}\frac{1}{2}\zeta_1} +\sin(\varphi-a_1) \dfrac{\operatorname{ctg}^2\frac{1}{2}\zeta_2-1} {2\operatorname{ctg}\frac{1}{2}\zeta_2} } {\sin(a_2-a_1)} = \]

\[ = -\frac{\sin(\varphi-a_2)\operatorname{ctg}\zeta_1 -\sin(\varphi-a_1)\operatorname{ctg}\zeta_2} {\sin(a_2-a_1)} \]

and, finally,

\[ \operatorname{tg}\zeta = -\frac{\operatorname{tg}\zeta_1 \operatorname{tg}\zeta_2 \sin(a_2-a_1)} {\operatorname{tg}\zeta_2 \sin(\varphi-a_2)-\operatorname{tg}\zeta_1 \sin(\varphi-a_1)} . \]

1. Let us take on the arc \(IF_1II\) an arbitrary point \(P\), which we join with the point \(O\) by the arc of a great circle \(OP=p\).

Introduce the notations:

\[ \angle OIII=\alpha;\quad \angle OII I=\beta;\quad \angle OPI=x_1;\quad \angle OPII=x_2; \]

\[ \angle IOP=\gamma_1;\quad \angle IIOP=\gamma_2;\quad \angle IOII=\gamma;\quad \smile IF_1II=c_1+c_2=c. \]

Denoting by \(S\) the amplitude of the spherical triangle \(OIIIO\), we have:

\[ 2S=\sin\gamma\sin\zeta_1\sin\zeta_2 \ldots \ldots \ldots \ldots \ldots (1) \]

\[ 2S=\sin\alpha\sin c\sin\zeta_1 \ldots \ldots \ldots \ldots \ldots (2) \]

\[ 2S=\sin\beta\sin c\sin\zeta_2 \ldots \ldots \ldots \ldots \ldots (3) \]

We now find the values of \(\sin\alpha\) and \(\sin\beta\).

\[ \sin\alpha=\sin x_1 \frac{\sin p}{\sin\zeta_1}; \ldots \ldots \ldots \ldots (4) \]

\[ \sin\beta=\sin x_1 \frac{\sin p}{\sin\zeta_2}; \ldots \ldots \ldots \ldots (5) \]

for

\[ x_1+x_2=\pi. \]

Substituting (4) and (5), respectively, into (2) and (3), we obtain:

\[ 2S=\sin x_1\sin p\sin c \ldots \ldots \ldots \ldots \ldots (6) \]

Comparing (1) and (6), we obtain:

\[ \sin\gamma\sin\zeta_1\sin\zeta_2=\sin x_1\sin p\sin c = \]

\[ =\sin x_1\sin p(\sin c_1\cos c_2+\cos c_1\sin c_2)\ldots \ldots \ldots (7) \]

We determine the values of \(\sin c_1, \sin c_2, \cos c_1, \cos c_2\).

From the spherical triangle \(OIPO\), respectively, we have:

\[ \sin c_{1,2}=\sin \zeta_{1,2}\frac{\sin \gamma_{1,2}}{\sin x_1}\ . \ . \ . \ . \ . \ . \ . \ . (8),\ (9) \]

\[ \cos c_{1,2}=\cos p\cos \zeta_{1,2}+\sin p\sin \zeta_{1,2}\cos \gamma_{1,2}\ . \ . \ . (10),\ (11) \]

We now substitute (8), (9), (10), (11) into (7). We have:

\[ \sin \gamma \sin \zeta_1 \sin \zeta_2 = \sin p\left[\sin \zeta_1 \sin \gamma_1 \sin \zeta_2 \left(\cos p\,\operatorname{ctg}\zeta_2+\sin p\cos \gamma_2\right) +\sin \zeta_2 \sin \gamma_2 \sin \zeta_1 \left(\cos p\,\operatorname{ctg}\zeta_1+\sin p\cos \gamma_1\right)\right], \]

whence

\[ \frac{\sin \gamma}{\sin p} = \sin \gamma_1\cos p\,\operatorname{ctg}\zeta_2 +\sin \gamma_1\sin p\cos \gamma_2 +\sin \gamma_2\cos p\,\operatorname{ctg}\zeta_1 +\sin \gamma_2\sin p\cos \gamma_1 = \]

\[ = \cos p\left(\sin \gamma_1\operatorname{ctg}\zeta_2 +\sin \gamma_2\operatorname{ctg}\zeta_1\right) +\sin p\left(\sin \gamma_1\cos \gamma_2 +\sin \gamma_2\cos \gamma_1\right) = \cos p\left(\sin \gamma_1\operatorname{ctg}\zeta_2 +\sin \gamma_2\operatorname{ctg}\zeta_1\right) \]

\[ +\sin p\sin(\gamma_1+\gamma_2) = \cos p\left(\sin \gamma_1\operatorname{ctg}\zeta_2 +\sin \gamma_2\operatorname{ctg}\zeta_1\right) +\sin p\sin \gamma \]

or

\[ \frac{\sin \gamma}{\sin p}-\sin p\sin \gamma = \cos p\left(\sin \gamma_2\operatorname{ctg}\zeta_2 +\sin \gamma_2\operatorname{ctg}\zeta_1\right)\ . \ . \ . \ . (12) \]

Transform the left-hand side of equation (12). We have:

\[ \frac{\sin \gamma}{\sin p}-\sin p\sin \gamma = \sin \gamma\left(\frac{1}{\sin p}-\sin p\right) = \]

\[ = \sin \gamma\frac{\cos^2 p}{\sin p} = \sin \gamma\,\operatorname{ctg}p\cos p. \]

Finally, equation (12) takes the form:

\[ \sin \gamma\,\operatorname{ctg}p = \sin \gamma_1\operatorname{ctg}\zeta_2 +\sin \gamma_2\operatorname{ctg}\zeta_1, \]

whence

\[ \operatorname{tg}p = \frac{\sin \gamma} {\sin \gamma_1\operatorname{ctg}\zeta_2+\sin \gamma_2\operatorname{ctg}\zeta_1} \ . \ . \ . \ . \ . (13) \]

In our case the point \(P\) is such that it must coincide with the point \(F_1\). Therefore

\[ \left. \begin{aligned} p&=\pi-\zeta,\\ \gamma&=\angle IOII=2\pi-(a_1-a_2),\\ \gamma_1&=\angle IOF_1=-a_1+\varphi+\pi,\\ \gamma_2&=\angle IIOF_1=-\varphi+a_2+\pi \end{aligned} \right\} \ . \ . \ . \ . \ . (14) \]

Substituting expressions (14) into (13), we obtain:

\[ \operatorname{tg}\zeta = -\frac{\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)} {\operatorname{tg}\zeta_1\sin(a_1-\varphi)-\operatorname{tg}\zeta_2\sin(a_2-\varphi)} \]

—a formula determining the zenith distance \(\zeta\) of the line of dip of the bed plane.

7. A). Let us apply the cotangent formula to each of the spherical triangles \(OIPO\) and \(OIIPO\).
We have:

\[ \operatorname{ctg}\zeta_{1,2}\sin p = \operatorname{ctg}x_{1,2}\sin\gamma_{1,2} + \cos p\cos\gamma_{1,2} \ldots\ldots (1_{1,2}) \]

In view of the condition

\[ x_1+x_2=\pi, \]

formulas \((1_{1,2})\) take the form:

\[ \cos p\bigl(\operatorname{ctg}\zeta_{1,2}\operatorname{tg}p-\cos\gamma_{1,2}\bigr) = \pm \operatorname{ctg}x_1\sin\gamma_{1,2}, \ldots\ldots (2_{1,2}) \]

where the plus and minus signs correspond, respectively, to the indices 1 and 2.
Dividing \((2_2)\) by \((2_1)\), we obtain:

\[ \frac{\operatorname{ctg}\zeta_2\operatorname{tg}p-\cos\gamma_2} {\operatorname{ctg}\zeta_1\operatorname{tg}p-\cos\gamma_1} = -\frac{\sin\gamma_2}{\sin\gamma_1}, \]

whence

\[ \operatorname{ctg}\zeta_2\operatorname{tg}p\sin\gamma_1-\cos\gamma_2\sin\gamma_1 = -\operatorname{ctg}\zeta_1\operatorname{tg}p\sin\gamma_2+\cos\gamma_1\sin\gamma_2 \]

or

\[ \operatorname{tg}p\bigl(\operatorname{ctg}\zeta_2\sin\gamma_1+\operatorname{ctg}\zeta_1\sin\gamma_2\bigr) = \cos\gamma_1\sin\gamma_2+\cos\gamma_2\sin\gamma_1, \]

whence

\[ \operatorname{tg}p = \frac{\cos\gamma_1\sin\gamma_2+\cos\gamma_2\sin\gamma_1} {\operatorname{ctg}\zeta_2\sin\gamma_1+\operatorname{ctg}\zeta_1\sin\gamma_2} \ldots\ldots (3). \]

Substituting in \((3)\) the values (see 6)

\[ \operatorname{tg}p=-\operatorname{tg}\zeta, \]

\[ \cos\gamma_1=-\cos(\varphi-a_1), \]

\[ \sin\gamma_2=\sin(\varphi-a_2), \]

\[ \cos\gamma_2=-\cos(\varphi-a_2), \]

\[ \sin\gamma_1=-\sin(\varphi-a_1), \]

we obtain:

\[ \cos\gamma_1\sin\gamma_2+\cos\gamma_2\sin\gamma_1 = -\cos(\varphi-a_1)\sin(\varphi-a_2) + \]

\[ +\cos(\varphi-a_2)\sin(\varphi-a_1) = \sin\bigl[(\varphi-a_1)-(\varphi-a_2)\bigr] = \sin(a_2-a_1) \]

and

\[ \operatorname{tg}\zeta = - \frac{\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)} {\operatorname{tg}\zeta_2\sin(\varphi-a_2)-\operatorname{tg}\zeta_1\sin(\varphi-a_1)} = \]

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

B). Putting (as indeed must be the case for our problem)

\[ x_1=x_2=\frac{1}{2}\pi, \]

formulas \((1_{1,2})\) take the form:

\[ \operatorname{ctg}\zeta_{1,2}\sin p=\cos p\cos\gamma_{1,2}, \]

whence

\[ \operatorname{tg} p=\frac{\cos \gamma_{1,2}}{\operatorname{ctg}\zeta_{1,2}}; \]

taking into account the values of \(p\) and \(\gamma_{1,2}\), we finally obtain:

\[ \operatorname{tg}\zeta_1=\operatorname{tg}\zeta_2\cos(\varphi-a_{1,2}). \]

From the last two formulas we shall determine, as functions of the given quantities \((a_{1,2},\zeta_{1,2})\), the required \(\zeta\) and \(\varphi\) (see article 1, § 17).

1. A). Let us take the formula

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

which determines the zenith distance \(\zeta\) of the line of dip of the plane of the stratum.

Transforming (1), we obtain:

\[ \operatorname{tg}\zeta\,[\sin\varphi(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)+ \cos\varphi(\operatorname{tg}\zeta_1\sin a_1- \operatorname{tg}\zeta_2\sin a_2)] +\operatorname{tg}\zeta_1\operatorname{tg}\zeta_2\sin(a_2-a_1)=0. \tag{2} \]

Let us solve the last equation with respect to \(\varphi\). Denoting

\[ \operatorname{tg}\zeta(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)=a \]

\[ \operatorname{tg}\zeta(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)=b \]

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

equation (2) takes the form:

\[ a\sin\varphi+b\cos\varphi+\kappa=0. \tag{3} \]

Equation (3), as is known, can be transformed into the equation:

\[ (a^2-\kappa^2)\operatorname{tg}^2\varphi+2ab\operatorname{tg}\varphi+(b^2-\kappa^2)=0, \tag{4} \]

whence

\[ \operatorname{tg}\varphi= \frac{-ab\pm\sqrt{a^2b^2-(b^2-\kappa^2)(a^2-\kappa^2)}}{a^2-\kappa^2}. \tag{5} \]

Let us transform the quantity under the radical in expression (5). We have:

\[ a^2b^2-(b^2-\kappa^2)(a^2-\kappa^2)=\kappa^2(a^2+b^2-\kappa^2). \]

Thus

\[ \operatorname{tg}\varphi= \frac{-ab\pm\kappa\sqrt{a^2+b^2-\kappa^2}}{a^2-\kappa^2}. \tag{6} \]

Now put into expression (6)

\[ a^2+b^2-\kappa^2= \operatorname{tg}^2\zeta\left[(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)^2+ (\operatorname{tg}\zeta_1\sin a_1- \operatorname{tg}\zeta_2\sin a_2)^2\right] -\operatorname{tg}^2\zeta_1\operatorname{tg}^2\zeta_2\sin^2(a_2-a_1)=0. \tag{7} \]

Then

\[ \operatorname{tg}\varphi_0=-\frac{ab}{a^2-\kappa^2}= \]

\[ = -\frac{\operatorname{tg}^2\zeta(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)} {\operatorname{tg}^2\zeta(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)^2-\operatorname{tg}^2\zeta_1\operatorname{tg}^2\zeta_2\sin^2(a_2-a_1)}. \tag{8} \]

From (7) we have:

\[ \tg^2 \zeta\,(\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1)^2 -\tg^2 \zeta_1 \tg^2 \zeta_2 \sin^2(a_2-a_1) = -\tg^2 \zeta\,(\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2)^2 . \tag{9} \]

Substituting (9) into (8), we obtain:

\[ \tg \varphi_0= \frac{\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1} {\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2}. \tag{10} \]

But from comparing formulas \([3;\ (1_{1,2})]\),

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

we have (see § 17 of the article\(^1\)):

\[ \tg \varphi= \frac{\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1} {\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2}. \tag{11} \]

Comparing (10) and (11), we conclude that

\[ \tg \varphi_0 \equiv \tg \varphi . \]

This result, of course, was to be expected.

Indeed: geometrically it is clear that only a single value of \(\tg \varphi\) corresponds to our problem, corresponding to the two directions of the line of dip of the plane of the stratum.

As a consequence, from (7) it follows that

\[ \tg \zeta = \pm \frac{\tg \zeta_1 \tg \zeta_2 \sin(a_2-a_1)} {\sqrt{(\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1)^2+ (\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2)^2}}, \]

whence, putting

\[ \tg \zeta_1 \tg \zeta_2 \sin a_2=\kappa>0 \]

and taking into account that, by hypothesis,

\[ \frac{1}{2}\pi<\zeta<\pi, \]

we finally obtain:

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

We have thus obtained a new derivation of the formula determining the zenith distance \(\zeta\) of the line of dip of the plane of the stratum.

B). Let us now pass to the vectorial determination of the azimuth \(\varphi\).

Transform equation (2) in two ways:

\[ 1.\quad \tg \zeta \sin \varphi \left[ (\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1) +\ctg \varphi(\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2) \right] +\tg \zeta_1 \tg \zeta_2 \sin(a_2-a_1)=0, \tag{12} \]

\[ 2.\quad \tg \zeta \cos \varphi \left[ \tg \varphi(\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1) +(\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2) \right] +\tg \zeta_1 \tg \zeta_2 \sin(a_2-a_1)=0 . \tag{13} \]

From formulas (12) and (13), respectively, we obtain:

\[ \sin \varphi= -\frac{\tg \zeta_1 \tg \zeta_2 \sin(a_2-a_1)} {\tg \zeta\left[(\tg \zeta_2 \cos a_2-\tg \zeta_1 \cos a_1) +\ctg \varphi(\tg \zeta_1 \sin a_1-\tg \zeta_2 \sin a_2)\right]} . \tag{14} \]

\[ \cos \varphi = \]

\[ = \frac{\operatorname{tg}\zeta_1\,\operatorname{tg}\zeta_2 \sin(a_2-a_1)} {\operatorname{tg}\zeta\,[\operatorname{tg}\varphi(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)+(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)]}. \tag{15} \]

Eliminating \(\operatorname{tg}\varphi\) from (14) and (15) by means of formula (11), we obtain:

\[ \sin \varphi = \]

\[ = \frac{\operatorname{tg}\zeta_1\,\operatorname{tg}\zeta_2 \sin(a_2-a_1)(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)} {(-\operatorname{tg}\zeta)\,[(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)^2+(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)^2]}. \tag{16} \]

\[ \cos \varphi = \]

\[ = \frac{\operatorname{tg}\zeta_1\,\operatorname{tg}\zeta_2 \sin(a_2-a_1)(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)} {(-\operatorname{tg}\zeta)\,[(\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1)^2+(\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2)^2]}. \tag{17} \]

Putting

\[ \operatorname{tg}\zeta_1\,\operatorname{tg}\zeta_2 \sin(a_2-a_1)=k>0 \]

and noting that

\[ \frac{1}{2}\pi<\zeta<\pi, \]

we obtain vectorially:

\[ \operatorname{tg}\varphi= \frac{\operatorname{tg}\zeta_2\cos a_2-\operatorname{tg}\zeta_1\cos a_1} {\operatorname{tg}\zeta_1\sin a_1-\operatorname{tg}\zeta_2\sin a_2}. \]

CHAPTER VI.

Plotting on the plan the line of dip and the isogyre of the stratum plane.

In my article\(^1\) (§ 1) the usual method was indicated for drawing on the plan the line of dip from the data: \(x\), \(y\), azimuth, interval, and elevation of one of its points.

Let us indicate another method, in which we make use of the coordinate grid of the plan.

§ 18.

A). The case of shafting: a point \(A\) is given and two directions issuing from it—the traces I and II (of the stratum plane on the vertical walls of the shaft; see drawing 33).

We mark on the plan the point \(A(x_a, y_a, z_a)\) where the traces meet, through which the line of dip of the stratum plane passes.

Another point (B or C) of the line of dip is determined as the intersection of the latter with some suitably chosen plane parallel to one of the coordinate planes: \(XOZ\) (the meridian plane) or \(YOZ\).

The points A and B (or C) determine the required line of dip.

Taking into account the coordinate system of the plan, in accordance with equation [(25), XVII], we have:

\[ \left. \begin{aligned} \frac{x-x_a}{-n} &= \frac{y-y_a}{m} = \frac{z-z_a}{\left(\dfrac{m^2+n^2}{\kappa}\right)},\\[4pt] y &= y_{t_1},\\ z &= z, \end{aligned} \right\} \ldots\ldots (1_{1,2}) \]

or

\[ \left. \begin{aligned} \frac{x-x_a}{-n} &= \frac{y-y_a}{m} = \frac{z-z_a}{\left(\dfrac{m^2+n^2}{\kappa}\right)},\\[4pt] x &= X_{t_2},\\ z &= z, \end{aligned} \right\} \ldots\ldots (2_{1,2}) \]

where \([(1_1),(2_1)]\) are the equations of the line of dip of the plane of the bed, and \([(1_2),(2_2)]\) are the equations of planes parallel, respectively, to the meridian planes and to the plane perpendicular to it \((t=1,2,3,4\ldots,\ \text{respectively})\).

Fig. 33.

From equations \((1_{1,2})\) and \((2_{1,2})\) we determine the required points \(B\) and \(C\), namely:

\[ \text{1.}\quad x_b=x_a-\frac{n}{m}(y_{t_1}-y_a);\quad y_b=y_{t_1};\quad z_b=z_a+\frac{m^2+n^2}{m\kappa}(y_{t_1}-y_a)\ldots(3_{1,2,3}) \]

\[ \text{2.}\quad x_c=X_{t_2};\quad y_c=y_a-\frac{m}{n}(X_{t_2}-x_a);\quad z_c=z_a-\frac{m^2+n^2}{n\kappa}(X_{t_2}-x_a)\ldots(4_{1,2,3}) \]

Subsequently, in drawing the strike lines of the plane of the stratum, one may proceed in two ways: either as set forth in the article² [a) p. 247 and b) p. 180] by Prof. P. K. Sobolevsky, or as indicated below.

Namely:

In the equation of the line of dip of the plane of the stratum

\[ \frac{x-x_a}{-n}=\frac{y-y_a}{m}=\frac{z-z_a}{\left(\dfrac{m^2+n^2}{k}\right)} \qquad\ldots\ldots\ldots (5) \]

we change the value of \(z\) by a prescribed interval (for example, by \(0.5\ \mathrm{m}\), \(1\ \mathrm{m}\), \(2\ \mathrm{m}\ldots\)) and determine from (5), for the corresponding \(z_i\), the values of the coordinates \((i=1,2,3\ldots)\),

\[ \left. \begin{aligned} x_i&=x_a-\frac{nk}{m^2+n^2}\,(z_i-z_a)\\[4pt] y_i&=y_a+\frac{mk}{m^2+n^2}\,(z_i-z_a) \end{aligned} \right\} \qquad\ldots\ldots\ldots (6_{1,2}) \]

It remains now to plot the system of points \((x_i,y_i,z_i)\) of the line of dip on the plan, and through them to draw the required strike lines of the plane of the stratum, perpendicular to the line of dip.

As a result we see that each of the strike lines is determined by a point and a direction.

In suitable cases, when the strike lines may turn out to be of great length, each of the latter should be determined by two points, one of which—\(J\) \((x_i,y_i,z_i)\)—is known.

The other (\(K\) or \(L\)) we determine as the intersection of the strike line of the point \(J\) with some suitably chosen plane parallel to the plane \(XOZ\) or \(YOZ\).

We have:

\[ z_{k,l}=z_i \]

\[ \frac{x-x_a}{-n}=\frac{y-y_a}{m} \qquad\ldots\ldots\ldots (7) \]

—the equation of the projection of the line of dip,

\[ x=X_{t_1};\quad y=Y_{t_2} \qquad\ldots\ldots\ldots (8_{1,2}) \]

—the equations of straight lines parallel, respectively, to the axes \(OY\) and \(OX\) \((t_{1,2}=1,2,3,4,\ldots\ \text{respectively})\).

The equation of the projection of the strike line, as perpendicular to (7) and passing through the point \((x_i,y_i)\), will be:

\[ y=y_i+\frac{n}{m}(x-x_i)\qquad\ldots\ldots\ldots (9) \]

Equations \((8_{1,2})\) and (9) jointly determine, respectively, the coordinates \([(x_k,y_k),(x_l,y_l)]\) of the points \(K\) and \(L\):

\[ \left. \begin{aligned} x_k&=X_{t_1}\\ y_k&=y_i+\frac{n}{m}(X_{t_1}-x_i) \end{aligned} \right\} \qquad\ldots\ldots\ldots (10_{1,2}) \]

The geometrical meaning of the subtraction is clear: the coordinate plane \(XOY\) has been shifted to the new position \(X'O'y'\) in the direction of the plumb line by the amount \(z_x = z_m\).

It is now clear that the point of intersection \(X\) of the line \(PN\) with the plane \(X'O'y'\) is the desired one, and the problem has been reduced to the one considered earlier (Chapter I, § 1).

The solution of the problem corresponding to inequalities (4), (5), (6) is shown, respectively, in drawings 34, 35, 36 (rotation of the horizontal-projecting plane of the line \(PN\) to the left, if one looks in the direction \(pn\)).

II. Let \(z_p > z_n\).

Analogously to case I, we obtain:

\[ 0 > z_p - z_x > z_n - z_x . \tag{7} \]

\[ z_p - z_x > 0 > z_n - z_x . \tag{8} \]

\[ z_p - z_x > z_n - z_x > 0 . \tag{9} \]

The solution of the problem corresponding to inequalities (7), (8), (9) is shown, respectively, in drawings 37, 38, 39 (rotation,

as in case I).

III. Let \(z_p = z_n\).

Then the line \(PN\) is the desired strike line.

We note in passing that the perpendicular dropped from either of the other given points (\(N\) or \(P^*\)), as is known, is the desired dip line, determined by two points whose elevations are \(z_n\) (or \(z_p\)) and \(z_m\).

Remark 1. In determining the desired point \(X\), one chooses such one of the given points that \(X\) lies within the limits of the drawing.

Remark 2. Both in case I and in case II there remain, evidently, by virtue of remarks 1 and 2 of § 1 of Chapter I.

In the case where the transformation of the elevations of the ends of the line \(PN\) does not lead to the goal (see § 11, beginning), one may then resort to the graphico-analytic method set forth below.

* To the line \(MX\).

Let us take the vertical plane corresponding to the straight line PN. Let us take one of the horizontal projections of the points P and N, for example \(p\), as the origin of a rectangular coordinate system; let the axis \(OX'\) be placed horizontally in the direction from point \(p\) to point \(n\), and the axis \(Z\) vertically.

Since we have three points—two given points P and N and the sought point X—it is natural to apply the principle of the simple ratio of three points.

As is known, we may write:

\[ x'_x = x'_m = x'_p + \mu (x'_n - x'_p), \ldots \ldots \ldots (10) \]

\[ z_x = z_m = z_p + \mu (z_n - z_p), \ldots \ldots \ldots (11) \]

where, in the coordinate system \(X'O'Z\) adopted for the solution, \((x'_m, z_m)\), \((x'_p, z_p)\), \((x'_n, z_n)\) are, respectively, the coordinates of the points X, P, and N.

Drawing 38.                Drawing 39.

Eliminating the parameter \(\mu\) from (10) by means of (11), we obtain:

\[ x'_x = x'_m = x'_p + \frac{z_m - z_p}{z_n - z_p}\,(x'_n - x'_p) \ldots \ldots \ldots (12) \]

In our case \(x'_p = 0\); therefore (12) takes the form

\[ x'_x = x'_m = \frac{z_m - z_p}{z_n - z_p}\,x'_n, \ldots \ldots \ldots (13) \]

where \(x'_n\)—the horizontal distance between the points P and N—may be determined in two ways: directly from the plan, or analytically by the formula

\[ x'_n = \overline{pn} = \frac{x_n - x_p}{\cos(PN)} = \frac{y_n - y_p}{\sin(PN)};\ldots \ldots \ldots (14) \]

in the last expression \(\overline{pn}\) and \((PN)\) are, respectively, the horizontal projection and the azimuth of the vector PN.

It remains now, on the plan, from the point \(p\)—the projection of P—to lay off the segment \(x'_x = x'_m\); the end of this segment will be the sought point X.

The graphical determination, by formula (13), of the sought \(x'_x = x'_m\), as is easy to see, likewise presents no difficulty.

Determination of contour lines from the sides of the exploration triangle.

1. Applying formulas (13) and (14), respectively, to each of the sides of the exploration triangle, we may assert:

By varying the variable \(z_m\) through a prescribed interval, it is easy to determine, on each of the straight lines PN, NM, MP (and their projections \(pn\), \(nm\), \(mp\)), a system

of points equally spaced from one another, corresponding to the contour lines (and their projections) of the stratum plane (the contour lines are straight lines determined by points \(z\)).

Let us now give another method for determining, on the straight lines \(PN\), \(NM\), \(MP\), a system of points equally spaced from one another.

1. Take one of the sides, for example \(PN\), determined by the points
   \(P(x_p, y_p, z_p)\) and \(N(x_n, y_n, z_n)\). . . . .

The equation of the straight line \(PN\):

\[ \frac{x-x_p}{x_p-x_n} = \frac{y-y_p}{y_p-y_n} = \frac{z-z_p}{z_p-z_n}, \qquad \ldots \ldots (15) \]

whence

\[ x=x_p+\frac{x_p-x_n}{z_p-z_n}(z-z_p)\, . \ldots \ldots \ldots (16) \]

\[ y=y_p+\frac{y_p-y_n}{z_p-z_n}(z-z_p)\, . \ldots \ldots \ldots (17) \]

Varying in (16) and (17) the variable \(z\) by the prescribed interval, we obtain a system of points equally spaced from one another.

Remark 1. We know that, in order to draw the contour lines of the stratum plane, it is sufficient, as a first step, to mark on the dip line a system of points equally spaced from one another, corresponding to a change of the coordinate \(z\) by a prescribed interval.

In accordance with what has been set forth, we find the equation of the strike line of the stratum plane corresponding to some point, for example \(M(x_m, y_m, z_m)\). Then we form the equation of the dip line corresponding to the point \(P\) or \(N\).

We have:

\[ z=z_m\, . \ldots \ldots \ldots \ldots \ldots (18) \]

—the equation of the horizontal plane corresponding to the point \(M\).

We now find the point \(K\) of intersection of the straight line \(PN\) (15) with (18), and obtain

\[ x_k=x_p+\frac{x_p-x_n}{z_p-z_n}(z_m-z_p) \]

\[ y_k=y_p+\frac{y_p-y_n}{z_p-z_n}(z_m-z_p) \]

\[ z_k=z_m . \]

The equation of the strike line determined by the points \(M\) and \(K\) will be:

\[ \frac{x-x_k}{x_k-x_m} = \frac{y-y_k}{y_k-y_m} = \frac{z-z_k}{z_k-z_m} \qquad \ldots \ldots (19) \]

The dip line corresponding to the point \(N\) \((P)\) and perpendicular to (19) is determined by the system of equations:

\[ \left. \begin{aligned} &(x_k-x_m)(x-x_{n(p)})+(y_k-y_m)(y-y_{n(p)}) \\ &\qquad +(z_k-z_m)(z-z_{n(p)})=0, \\[4pt] & \begin{vmatrix} x-x_k,\, y-y_k,\, z-z_k \\ x_{n(p)}-x_k,\, y_{n(p)}-y_k,\, z_{n(p)}-z_k \\ x_k-x_m,\, y_k-y_m,\, z_k-z_m \end{vmatrix} =0 \end{aligned} \right\} \qquad \ldots \ldots (20), \]

Introducing the notation:

\[ x_k-x_m=A_1;\quad y_k-y_m=B_1;\quad z_k-z_m=C_1 \]

\[ -\bigl[x_{n(p)}(x_k-x_m)+y_{n(p)}(y_k-y_m)+z_{n(p)}(z_k-z_m)\bigr]=D_1 \]

\[ \left| \begin{array}{cc} y_{n(p)}-y_k,\ z_{n(p)}-z_k\\ y_k-y_m,\ z_k-z_m \end{array} \right|=A_2;\quad -\left| \begin{array}{cc} x_{n(p)}-x_k,\ z_{n(p)}-z_k\\ x_k-x_m,\ z_k-z_m \end{array} \right|=B_2 \]

\[ \left| \begin{array}{cc} x_{n(p)}-x_k,\ y_{n(p)}-y_k\\ x_k-x_m,\ y_k-y_m \end{array} \right|=C_2;\quad -(x_kA_2+y_kB_2+z_kC_2)=D_2, \]

we obtain the equations of the line of dip in general form:

\[ \left. \begin{array}{l} A_1x+B_1y+C_1z+D_1=0\\ A_2x+B_2y+C_2z+D_2=0 \end{array} \right\}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (21) \]

Finally, it remains to reduce equation of the line (21) to normal form and then proceed further in accordance with A).

Remark 2. The method set forth for determining isotachs from the sides of a reconnaissance triangle is also applicable to the traces of a reconnaissance shaft.

In fact: it is sufficient to mark an arbitrary point on each of the traces and determine its coordinates. Then the point \(A(x_a,y_a,z_a)\) of convergence of the traces and the points \(I, II\) marked on the traces determine the reconnaissance triangle \(IIIAI\).

The coordinates of the sought points are determined, respectively, by the formulas [see formula (1), item 1, § 5 of article¹]:

\[ x_{I,II}=x_a+x_{1,2}=x_a+R\sin\zeta_{1,2}\cos a_{1,2} \]

\[ y_{I,II}=y_a+y_{1,2}=y_a+R\sin\zeta_{1,2}\sin a_{1,2} \]

\[ z_{I,II}=z_a+z_{1,2}=z_a+R\cos\zeta_{1,2}, \]

where \(R\) is an arbitrary, suitably chosen positive number, and \(a_{1,2}\) and \(\zeta_{1,2}\) are respectively the azimuths and zenith distances of the directions of the traces.

Remark. The essential points in the geometry of surveying the surfaces of bodies of occurrence are set forth in the article² by Prof. P. K. Sobolevsky and in my article³.

List of literature cited in the article

1. V. F. Turchinsky. Analytical and grapho-analytical methods for determining the elements of occurrence at a given point on the surface of a stratum.—Proceedings of the Siberian Technological Institute, vol. 50, issue 4. Tomsk, 1929.

2. P. K. Sobolevsky. Modern mine-surveying art as a methodology for solving the fundamental problems of mining art. a) Proceedings of the Ural Polytechnic Institute, vol. V. Sverdlovsk, 1926.
   b) Proceedings of the First All-Union Mining Scientific and Technical Congress, April 14–27, 1926, vol. VII. Publication of the Scientific and Technical Administration of the Supreme Council of the National Economy of the USSR. Moscow, 1928.

3. V. F. Turchinsky. The geometry of surveying the surfaces of bodies of occurrence in connection with the principle of least work.—Proceedings of the First All-Union Mining Scientific and Technical Congress, April 14–27, 1926, vol. VII.—Publication of the Scientific and Technical Administration of the Supreme Council of the National Economy of the USSR. Moscow, 1928.

CONTENTS

	Page
Introduction. Statement of the question	1

CHAPTER I

Construction of the conical net.

	Page
§ 1. Solution of the problem: determining the point of intersection of a straight line with a plane	1
§ 2. Determining the position of a point on a conical surface	2
§ 3. Construction of a family of concentric circles expressed by marks of zenith distances	4
§ 4. Determination of \(l_x\) and \((\Delta r)_{\min}\)	6
§ 5. Construction of roses of azimuths and zenith distances. Conical net	7

CHAPTER II.

The most advantageous conditions for constructing the conical net.

	Page
§ 6. Accuracy of determining the position of the points of intersection of directions with the generators of the conical surface	10
§ 7. Graphical construction of the expression \(b\mu\)	13
§ 8. Determination of the angle \(t\), for which \(\mu_\zeta = \dfrac{\pi}{2} = \mu'_\zeta = 0\) and \(l_x > 0.2\ \mathrm{mm}\).	14

CHAPTER III.

Solutions of auxiliary problems.

	Page
§ 9. Solutions of 5 auxiliary problems when using the conical net	21

CHAPTER IV.

Solution of the main problem by means of the conical net.

	Page
§ 10. Graphical solution, by means of the conical net, of the problem corresponding successively to four cases of the position of the traces of the bed plane on the vertical walls of a prospecting shaft	28
§ 11. Consideration of the special case indicated in note 2 to § 1	38

CHAPTER V.

Appendix. Analytical methods for determining the attitude elements at a given point of a stratum surface.

	Page
§ 12. Determination of the azimuth \(\varphi\) of the dip line of the stratum plane—the method of analytic geometry	42
§ 13. Determination of the coordinates of an arbitrary point of the dip line of the stratum plane (18 solutions)	43
§ 14. A special method for determining the azimuth \(\varphi\) of the dip line of the stratum plane	72
§ 15. Method of joint determination of the azimuth \(\varphi\) and the zenith distance \(\zeta\) of the dip line of the stratum plane	83
§ 16. Methods for determining the zenith distance \(\zeta\) of the dip line of the stratum plane (4 solutions)—the method of analytic geometry	84
§ 17. Solutions by the method of spherical trigonometry	89

CHAPTER VI.

Plotting on the plan the dip line and the isohypses of the stratum plane.

	Page
§ 18. Case of shaft sinking; case of drilling	101
List of references	108

ERRATA NOTED

in the article by Assoc. Prof. V. F. Turchinskii, “Graphical Method for Determining Attitude Elements at a Given Point on a Stratum Surface.”

Vol. 51, Issue III, “Proceedings of the Siberian Technological Institute.”

(lines marked with an asterisk are counted from the bottom)

Page	Line	Printed	Should read
1	4*	$\mathbf{b}_1$	$\mathbf{a}_1$
4	18, 19	from the projection $b$, of the end $B$	at the point $b$, the projection of the end $B$
7	4*	adjacent quadrants	adjacent vertical quadrants
10	21	azimuths	azimuth
11	15	$\dfrac{M}{\delta}$	$\dfrac{M_p}{\delta}$
12	3*	$\left\lvert\dfrac{M_p}{\delta^2}\right\rvert$	$\left\lvert\dfrac{M_p}{\delta}\right\rvert$
14	11	$PQ = PQ = r$	$PQ' = PQ = \xi$
19	4	$\eta$	$\mu$
24		in drawing 17 connect points $A$ and $N$ by a straight line	in drawing 17 connect points $A$ and $N$ by a straight line
25	6*	$(h_{1,\mathrm{II}}\geqq 0, h_{1,\mathrm{II}}\geqq 0)$	$(h_{1,\mathrm{II}}\geqq 0, h_{1,\mathrm{II}}\geqq 0)$
26	2*	$r_{s_2}$	$r'_{s_2}$
27	3	$H_u$	$H_n$
27	1*	$S_u$	$S_n$
30	8*	Namely from the point	Namely: from the point
32	22	normals to the traces $S'_vS'_v$—traces	normals to the traces. $S'_vS'_v$—traces
33	7	Angle $E_0O1_{0.1}$	$E_0Oe_{01}$
47	16	determines	determines
50	1	$(x_0, y_0, z)$	$(x_0, y_0, z_0)$
63	9	$(a_{1,2}, \zeta_{1,2})$	$[(0,0,0);(a_1,\zeta_1)\ \text{and}\ (0,0,0);(a_2,\zeta_2)]$
78	8	$z \geqq 0$	$z \leqq 0$
83	7	$(a_{1,2}, \zeta_{1,2})$	$[(0,0,0);(a_1,\zeta_1)\ \text{and}\ (0,0,0);(a_2,\zeta_2)]$
90	5	the bracket $]$ is missing at the end of the expression	the bracket $]$ is missing at the end of the expression
91	12	$\sin c_1 \sin c_1$	$\sin c_1 \sin c_2$