Proceedings

of the Tomsk Technological Institute

of Emperor Nicholas II

Vol. 8. 1907. No. 5.

D. V. Frost.

Graphical Method for Solving Mine-Surveying Problems

1–34.

Graphical Method for Solving Mine-Surveying Problems

In mine-surveying practice it is often necessary to solve various problems in which one has to find the geometrical relationship between the elements of occurrence of sheet-like deposits and the workings driven in them.

We speak only of deposits of this kind, because only for them, in view of the known regularity of their occurrence, can any generalizations be made.

The determination of the above-mentioned relationship in each particular case may be made either from a geometrical consideration of the relation of the various elements of the deposit, or by a purely analytical method, using the formulas of analytic geometry.

As for the first method, it must be noted that deriving the geometrical relationship between the elements of a seam often requires complex explanatory drawings and therefore easily leads to various errors. The analytical method, although it does not have the latter shortcomings, is often rather laborious and not very visual.

Meanwhile, the solution of mine-surveying problems, at least here in Russia, is often carried out by persons who have not received sufficient mathematical training.

In view of the considerations stated, it is necessary to seek other methods which, while sufficiently accurate for mining practice, would be simple and visual.

Such a method is provided by descriptive geometry, which includes various methods for the graphical solution of problems.

One of these methods is the method of projection with numerical elevations (Cotierte Projectionsmethode), in which only one plane of projection is chosen, while the second is replaced by numerical elevations that give the distance of a point in space from the selected plane of projection.

The use of a single such plane of projection instead of two is especially valuable where a clear representation of the terrain and an exact designation of all objects located on this terrain are required.

For this reason, the application of the above-mentioned method of projection with numerical elevations may be of great importance, especially in mine surveying.

In this method, it is most convenient to take as the plane of projection the plane of some horizon.

The distance from this plane of points taken in space must be represented on some adopted scale. Since this distance may be laid off from the plane on either side of it, positive values should be assigned to distances on one side, and negative values on the other; points lying on the plane itself should be assigned the elevation 0. In view of the possibility of choosing any horizon as the plane of projection, we are able to choose it so that most of the points under consideration lie on one or the other side of this plane. In that case we may give a positive value to the elevations denoting the distance of the majority of the points from the adopted plane of the horizon and, for brevity, omit writing the plus sign; points lying on the other side of the chosen plane of projection are then given the minus sign.

As will be seen from what follows, the method of projection proposed here is very simple.

Unfortunately, it has been little touched upon both in foreign and in Russian literature.

In the former we may point to the following works: Peschka G.: “Darstellende und projective Geometrie,” 1889; “Cotierte Ebenen,” 1876; “Cotierte Projectionsmethode und deren Anwendung,” 1882; De la Gournerie: “Géométrie descriptive”; A. Leroy: “Plans cotés.” In Russian, the question touched upon is considered in Professor Kurnyumov’s course of descriptive geometry.

In view of the scantiness of the literature on the question of the method of projection under consideration, before proceeding to solve mine-surveying problems by this method, it seems to us not superfluous to give some theoretical concepts about the method itself. In doing so, we shall touch only upon those concepts that are necessary for the fulfillment of our main purpose.

I became acquainted with the application of the method of projection with numerical elevations for solving various mine-surveying problems at the Leoben Mining Academy (in Austria), where similar problems were proposed to students by Professor Lederer.

Unfortunately, these problems were not published there, and I was able to obtain only a few problems.

My efforts to find anything on this subject in the literature likewise were not crowned with success.* Considering, however, that such a method for solving mine-surveying problems deserves attention, I have decided to touch upon this question.

In offering this work, I by no means lay claim to a complete and perfected development of the given topic, but I shall be quite satisfied if the method I indicate for solving mine-surveying problems draws the attention of specialists in mine surveying and mining.

Brief concepts of the method of projection with numerical elevations.

A point in space will be fully determined if we give its projection on a definite horizontal plane and its numerical elevation (Cote), i.e., the length of the perpendicular from the point under consideration to the given plane of projection.

A straight line is determined by two points belonging to it, but it is more convenient to represent the latter by its horizontal projection and by the so-called scale of inclination (Gefälls—oder Böschungsmahsstab).

The latter consists in giving, on the straight line, the projections of all those points whose distances from the plane of projection are expressed in a given series of whole numbers.

In doing this, one has to find the distance between the projections of two points whose difference in heights is equal to the adopted unit of scale. This distance, which is a function of the inclination of the straight line to the plane of projection, will for every straight line be a constant quantity and is called its interval (Intervall).

Determination of the interval of a straight line, finding the projections of points on it from their given numerical elevations, and conversely.

Projecting the given straight line onto the horizontal plane passing through its lower point, and rotating this straight line about its projection until it coincides with the plane of projection, we obtain the right triangle \(abb_{0}\) (Fig. I). Taking on the given straight line a point whose distance from the plane of projection is equal to the adopted unit, we obtain a second right triangle \(acc_{0}\).

*) A graphical solution of mine-surveying problems is given in the recently published work of my colleague, mining engineer P. R. Leontovsky, Mine-Surveying Problems, which I partly used in choosing the conditions of the problems.

From the similarity of the triangles \(ab_0\) and \(ac_0\), we derive

\[ bb_0 : ab = cc_0 : ac, \]

or

\[ \Delta : m = 1 : J. \]

Hence the interval

\[ J=\frac{m}{\Delta}\ . . . . . . . . . . . . . . . . . . \tag{1} \]

and also

\[ m=J\cdot\Delta \ . . . . . . . . . . . . . . . . . . \tag{2} \]

\[ \Delta=\frac{m}{J}\ . . . . . . . . . . . . . . . . . . \tag{3} \]

where \(J\) is the interval of the straight line, \(m\) is the distance between the projections of two of its points, and \(\Delta\) is the difference in elevation of these points.

In our case

\[ J=\frac{m}{\Delta}=\frac{12.6}{10.5-4.9}=\frac{12.6}{5.6}=2.25. \]

By laying off the obtained value of the interval on the straight line in both directions from the given point, we obtain the projections of those points which lie one unit of height above or below the preceding ones.

It is much more convenient to determine those points of the straight line whose elevations would be expressed in whole numbers. It is clear that for this it is quite sufficient to find one such point, so that with the aid of the interval all the others may be determined. To determine the projection \(p\) of such a point, for example in our case with elevation 10, we find its distance from point \(b\), with elevation 10.5.

Applying formula (2), we obtain

\[ m=J\cdot\Delta=2.25\cdot(10.5-10.0)=1.125. \]

Laying off on the straight line from point \(b\), in the direction of decreasing elevations, the value \(m=1.125\), we find point \(p\) with elevation 10.

Then, laying off from it in both directions along the straight line the interval value \(J=2.25\), we obtain a series of points whose elevations will be expressed in whole numbers.

Instead of laying off, in order to avoid the accumulation of errors, it is more convenient and more accurate to find the two most distant points and divide the distance obtained between them into a number of parts equal to the difference of the elevations of the points found.

Let us be given a straight line by its horizontal projection and by the scale of its slope, and on it a point \(c\), whose elevation is to be determined. To do this, we measure by the scale the distances \(ac = 1.5\) (Fig. II) and \(ab = 5\), and determine by formula (3)

\[ \Delta = \frac{m}{J} = \frac{1.5}{5} = 0.3. \]

Consequently, the difference in height of points \(a\) and \(c\) will be equal to \(0.3\), and therefore the elevation of point \(c\) is

\[ h_c = h_a - 0.3 = 9 - 0.3 = 8.7. \]

If it is required to determine the projection of point \(d\), whose elevation is \(7.6\), then by formula (2) we compute the distance \(ed = m\)

\[ m = J \cdot \Delta = 5(7.6 - 7) = 3. \]

Thus, by setting off from point \(e\) a distance equal to 3, we obtain the required point \(d\). To find the true length of a straight line given by the projections of its end points (Fig. I), the elevations of these points are determined, if they are not given, and the length of the straight line is computed as the hypotenuse of the right triangle \(abb_0\), in which one leg is the horizontal projection of the straight line, i.e., the distance by scale of the straight line \(ab\), and the other leg is the difference in heights of points \(a\) and \(b\):

\[ L = \sqrt{\Delta^2 + m^2} = \sqrt{(12.6)^2 + (10.5 - 4.9)^2} = -13.8. \]

The same length of the straight line can be obtained by direct construction.

Slope of a Straight Line (Neigung, Böschung oder Fall).

By the slope of a straight line we shall agree to understand the tangent of the angle of inclination of the straight line to the horizon or, in other words, the ratio of the difference in height of two points to the horizontal projection of the distance between them.

Thus

\[ \frac{\Delta}{m} = tq\,\varphi = n. \]

Recalling equation (2),

\[ m = J\Delta, \]

we obtain

\[ n=\frac{\Delta}{m}=\frac{1}{J} \qquad \ldots \qquad (4) \]

or

\[ n \cdot J = 1. \qquad \ldots \qquad (5) \]

It follows from this that the slope of a straight line and its interval are mutually reciprocal quantities.

Mutual position of straight lines.

Parallel straight lines have parallel projections and one and the same slope to the horizon, and consequently equal intervals,

\[ n=n_1 \quad J=J_1 . \]

If straight lines intersect, then their projections also intersect, and the lines joining identical elevations of these straight lines, as the bisectors of the sides of the angle into equal parts, must be parallel to one another.

In the absence of the latter condition, the straight lines cannot lie in one and the same plane, and therefore will not intersect.

Representation and determination of planes.

Since a plane is completely determined by three of its points, by two straight lines lying in it, and also by one straight line and a point outside the latter, then, on the basis of the preceding reasoning, the representation of planes by means of the above-mentioned elements presents no difficulty. However, in order to achieve the greatest simplicity, in the method of projection with numerical elevations it is better to introduce another method of representing planes. For this purpose, planes are represented by the lines of their intersection with horizontal planes, the distances of which from the horizon plane, taken as the plane of projection, are expressed in whole numbers. If a series of such planes is drawn, separated from one another by a distance equal to the adopted unit, then the lines of intersection of these planes with the given plane, as well as the projections of these lines—the horizontals of the plane—will be equidistant from one another.

By drawing a straight line that would intersect the obtained horizontals, and marking the points of intersection with the elevations corresponding to the horizontals, we shall represent a series of straight lines lying in one and the same plane (Fig. III). In this way we can find

an infinite set of straight lines lying in the plane under consideration and expressed by their slope scales.

The second elements for representing planes are taken to be those of these straight lines whose projections are perpendicular to the horizontals. Such straight lines will be the lines of greatest slope of the plane. In fact, lines drawn perpendicular to the horizontals will have the smallest intervals, and since, by formula (4), the slope

\[ n = \frac{1}{J}, \]

then, obviously, the greatest slope, or dip, will correspond to the smallest interval.

Thus the projection of the line of greatest dip of a plane, together with the scale of the slope of this line, is called the scale of dip of the plane.

Unlike other straight lines lying in the same plane and likewise given by their scales, the line of greatest dip—the scale of dip of the plane—will be represented by a double line, as is seen in Fig. III, where the line \(M\) represents the scale of dip of the plane (Böschungsmahstab).

It is clear that the tangent of the angle of inclination of the plane to the horizon (the tangent of the angle of dip) is equal to the reciprocal of the interval of the scale of dip of the plane.

These preliminary remarks, which explain the very method of projection with numerical marks, are sufficient for proceeding to the solution of various mine-surveying problems.

In mine-surveying practice, the plane of some bed or bedded deposit is characterized by the same elements as in the above-mentioned method, i.e., by the direction of the strike line of the bed, its horizontal, and by the line of dip and the magnitude of the latter.

In some problems, for comparison, both the graphical and the numerical solution will be given at the same time. It should also be noted that some problems, in numerical terms, do not fully correspond to mining practice; for example, the angles of inclination of adits and drifts have been taken larger than they are in reality.

The latter has been done only in order to reduce the size of the drawings, since the main purpose of this work is to indicate the method of solution itself, which will always remain unchanged.

Mine-surveying problems.

Problem No. 1. A seam is encountered by three boreholes \(A, B, C\), the horizontal positions of which and whose numerical elevations are given.

It is required to determine the strike and dip of the seam.

Numerical solution. Given: the numerical elevations of the boreholes, their position, and the direction of the meridian.

Passing through the point \(a\) (Fig. 1), which determines the position of the deepest borehole, a horizontal plane and projecting the remaining points onto the latter, we shall obtain:

\[ \omega = \omega_3 + z, \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

since

\[ \omega_1 + [(x+y)-\omega_3] = 180^\circ, \]

then

\[ x+y = 180^\circ - (\omega_1+\omega_3) \ldots \ldots \ldots (2) \]

From the right triangle \(A'B'E\) it follows:

\[ z = 90^\circ - x, \ldots \ldots \ldots \ldots \ldots (3) \]

and therefore

\[ \omega = \omega_3 + (90^\circ - x) \ldots \ldots \ldots \ldots 1. \]

Now determine the angles \(x\) and \(y\).

Since in the triangle \(A'B'C'\) (plan) all three sides \(a, b, c\) are known, it is easy to determine all the angles, for example the angle at \(B'\)

\[ tg \frac{x+y}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} , \ldots \ldots \ldots (4) \]

where

\[ s=\frac{a+b+c}{2}. \]

Thus

\[ x+y=m \]

is known.

To determine the difference of the same angles we obtain

\[ cs\,x=\frac{EB'}{A'B'}=\frac{EB'}{C} \]

\[ cs\,y=\frac{EB'}{B'D}=\frac{EB'}{b'D} = \frac{EB'}{a+(h_3-h_1)ctg\,\gamma} \]

\[ tg\,\gamma=\frac{h_2-h_3}{a} \]

\[ \frac{\operatorname{cs} x}{\operatorname{cs} y} = \frac{BD}{A'B'} = \frac{a+(h_3-h_1)\operatorname{ctg}\gamma}{C} = \operatorname{tg}\psi \ldots \tag{6} \]

\[ \frac{\operatorname{cs} x}{\operatorname{cs} y} = \frac{\operatorname{tg}\psi}{1}; \qquad \frac{\operatorname{cs} x+\operatorname{cs} y}{\operatorname{cs} x-\operatorname{cs} y} = \frac{\operatorname{tg}\psi+1}{\operatorname{tg}\psi-1} \]

\[ \frac{ 2\operatorname{cs}\frac{x+y}{2}\operatorname{cs}\frac{x-y}{2} }{ -2\operatorname{sn}\frac{x+y}{2}\operatorname{sn}\frac{x-y}{2} } = \frac{ \operatorname{sn}\psi\operatorname{sn}45^\circ+\operatorname{cs}\psi\operatorname{cs}45^\circ }{ \operatorname{sn}\psi\operatorname{sn}45^\circ-\operatorname{cs}\psi\operatorname{cs}45^\circ } \]

\[ -\operatorname{ctg}\frac{x+y}{2}\operatorname{ctg}\frac{x-y}{2} = \frac{\operatorname{cs}(\psi-45^\circ)}{\operatorname{sn}(\psi-45^\circ)} = \operatorname{ctg}(\psi-45^\circ) \]

\[ \operatorname{tg}\frac{x-y}{2} = -\operatorname{tg}(\psi-45^\circ)\operatorname{ctg}\frac{m}{2} \ldots \tag{7} \]

According to (5) and (7), having

\[ \frac{x+y}{2}=\frac{m}{2} \quad \text{and} \quad \frac{x-y}{2}=\frac{n}{2}, \]

one determines the angles \(x\) and \(y\), and, knowing the latter, it is easy to determine from equation I the angle \(\omega\), i.e., the strike of the deposit.

The dip angle is obtained from the right triangle \(Ebb'\)

\[ \operatorname{tg}\varphi=\frac{bb^1}{Eb^1} \]

where

\[ bb^1=h_2-h_1;\quad Eb^1=EB'=\operatorname{cs}nz=C\operatorname{cs}x \]

Thus,

\[ \operatorname{tg}\varphi=\frac{h_2-h_1}{C\operatorname{cs}x} \ldots \mathrm{II}. \]

Graphical solution (Fig. 2).

Dividing the line \(AC\) into four parts, we obtain the point \(b\) with elevation 50, equal to the elevation of point \(B\). The line \(Bb\) will give the direction of the strike of the bed. The scale of dip of the bed will be perpendicular to it. Drawing the horizontals of the bed, we obtain the interval of the dip scale, and from it, from the right triangle in which one leg is equal to the interval \(J\), and the other to the vertical distance between adjacent horizontals, determine the dip angle \(\varphi\).

Problem No. 2. A bedded deposit is given by two points \(A\) and \(B\), whose difference in heights is known, and, in addition, by the line \(Bx\),

the direction of which and inclination to the horizon have been measured. It is required to determine the strike and dip of the deposit.

Solution (Fig. 3). Given: A (30), B (60)

Direction of the line Bx: $\omega_1$, its inclination to the horizon: $\varphi_1$.

Required: strike of the bed: $\omega$, its dip: $\varphi$.

By means of the angle $\varphi_1$ we find the interval of the line Bx and, laying the latter off from point B, obtain on the line Bx the point $a$ with elevation 30. The line connecting points A and $a$ will be the strike line, and the line perpendicular to it will be the line of dip. If the direction of the meridian is given, then the angle of strike is determined directly from the drawing, and the angle of dip from the right triangle on the basis of the determined interval of the dip line.

Problem No. 3. The deposit has been encountered by a borehole C, whose position, numerical elevation at the surface, and depth are known, and by an adit $ab$, from which a crosscut has been driven with a given inclination to the horizon. It is required to determine the strike and dip of the deposit ($\omega$ and $\varphi$).

Solution (Fig. 4). Given: $b$ (80); $c$ (15); $h = 67$

$\varphi_1,\ \omega_1$.

Required: $\omega,\ \varphi$

Construct a right triangle for the given acute angle $\varphi_1$ and a leg equal to the chosen unit of scale; determine the interval $J_1$ of the line bd. Laying it off from point b, we obtain the scale of inclination of the straight line bd.

Since the elevation at the surface of point C = 15, and the depth of the borehole $h = 67$, the numerical elevation of point $c$ in the plane of the bed is $15 + 67 = 82$. Connecting point $c$ with the point of the same elevation on the straight line bd, we obtain the strike line. Drawing a straight line perpendicular to the latter, and finding on it the numerical elevations, we shall construct the scale of dip of the deposit and shall know its interval $J$. Having constructed a right triangle for the given interval $J$ and the unit of scale, we determine the angle $\varphi$.

Note. In the case of very acute angles, the interval is more conveniently determined by the formula

\[ J = \frac{1}{n} = \frac{1}{\operatorname{tg}\varphi} \]

Problem No. 4. A bed has been encountered by the crosscut AB and explored by the adits BC and BD. It is required to determine the strike and dip of the bed.

Solution (Fig. 5). Given: \(\omega_1, \zeta_1; \ \omega_2, \zeta_2\).

Required: \(\omega, \zeta\).

Determine the intervals of the lines \(BC\) and \(BD\) by the formula given in the preceding problem; construct the scales of the inclinations of the straight lines \(BC\) and \(BD\).

By connecting equal marks on these straight lines, obtain the line of strike. After this, construct in the known manner the scale of dip of the bed and determine its dip.

Problem No. 5. In sinking a shaft, a bed has been encountered, and the directions and inclinations to the horizon of the lines of intersection of the hanging wall of this bed with the shaft walls have been determined. It is required to determine the strike and dip of this bed.

Solution (Fig. 6). Given: \(\omega_1, \zeta_1; \ \omega_2, \zeta_2\).

Required: \(\omega, \zeta\).

Determine the azimuths of the lines of intersection of the hanging wall of the bed with the shaft walls and the angles of inclination of these straight lines to the horizon; find, from the given angles \(\zeta_1\) and \(\zeta_2\), the intervals \(J_1\) and \(J_2\). Having connected equal marks of the straight lines \(AB\) and \(AC\), and thus obtained the line of strike, and, perpendicular to it, the line of dip, we find the interval of the latter; after this we determine the angle of dip of the bed.

Note. In this, it is wholly immaterial in which direction the lines of outcrop of the bed are directed; it is only necessary to place increasing numerical marks in the direction of rise of the lines of outcrop.

Problem No. 6. With respect to a bedded deposit, its strike \(\omega\) and the angle of inclination \(\zeta_1\) of a diagonal crosscut having direction \(\omega_1\) are known. It is required to find the angle of dip of this deposit.

Solution (Fig. 7). Given: \(\omega, \omega_1\) and \(\zeta_1\).

Required: \(\zeta\).

Drawing through point \(A\) the meridian direction, lay off, according to the given angles from it, from point \(A\) the line of strike of the deposit \(AB\), the line of dip perpendicular to it, and also the horizontal projection of the inclined crosscut \(AC\). Then, having determined from the right triangle, by the given angle \(\zeta_1\) of inclination of the line \(AC\), its interval \(= J_1\), lay the latter off on the line \(AC\), and, drawing horizontals through its ends, we obtain the interval of the line of dip \(= J\), and from the corresponding right triangle the angle of dip of the deposit \(\zeta\).

Problem No. 7. In driving a horizontal working, a bed has been encountered, and the directions and inclinations to the horizon of the lines of intersection of the hanging wall of this bed have been determined. It is required to determine the strike and dip of the bed.

Solution (Fig. 8). Given: \(\omega_1, \varphi_1, \omega\)

Required: \(\varphi\).

In the present case the line of intersection of the hanging wall of the seam with the floor of the horizontal working will be a horizontal line, and consequently the strike line of the given seam. Determine the interval of the line \(AC\), its numerical elevations, and, drawing the adjacent horizontals, obtain the interval of the dip line; and, knowing it, it is easy to determine also the angle of dip of the seam.

Note. In the present case the direction of the line of outcrop is entirely immaterial. This problem is a particular case of Problem No. 5. If the seam is encountered by an inclined working and the lines of outcrop of the hanging wall of the seam in the floor and wall of the working are measured, then the determination of the strike and dip of the seam is carried out in the same way as in Problem No. 5.

Problem No. 8. To connect two horizons of a seam, the strike and dip of which are known, it is necessary to drive, from a given point \(A\), a diagonal drift of a given inclination to the horizon \(\varphi_1\). It is required to determine its direction.

Solution (Fig. 9). Given: \(\omega, \varphi, \varphi_1\).

Required: \(\omega_1\).

Having drawn, at the given angle to the meridian direction, the strike line \(AD\) and, perpendicular to it, the dip line, we lay off on the latter from point \(A\) the interval \(J\), obtained from the given angle of dip \(\varphi\). Through point \(C\) we draw the horizontal of the seam and then cut the latter with an arc of radius equal to the interval \(J_1\), corresponding to the angle \(\varphi_1\) of inclination of the projected drift. By joining the obtained points \(B\) and \(B'\) with point \(A\), we find two directions of the drift satisfying the conditions of the problem.

Problem No. 9. It is required to determine the inclination to the horizon of the diagonal drift \(ab\), if its direction is given and the dip and strike of the seam itself are known.

Solution (Fig. 10). Given: \(\omega, \varphi, \omega_1\).

Required: \(\varphi_1\).

Constructing the scale of dip of the seam and drawing the direction of the diagonal drift and the horizontals, we obtain the interval of the straight line \(ab\), and from the right triangle the angle of inclination to the horizon.

Problem No. 10. A seam is encountered by a crosscut \(a\), driven from shaft \(s\). On the basis of exploratory work, the strike \(I\) on the horizon of crosscut \(a\) and the dip of this seam were determined. It is required to determine the line of emergence of the deposit on the surface and the strike on the horizon of crosscut \(b\), if the vertical ...

the distance of horizon I from the surface and from the required lower horizon.

Solution (Fig. 11). Given: $\omega$, $\varphi$, $h_1 = 40$, $h_2 = 20$.

Required: the line of outcrop, the line of strike of the lower horizon.

Having plotted on the map the strike I and the scale of dip of the seam, we measure off along the angle of dip the corresponding interval $u$, and, laying it off the proper number of times on both sides of strike I along the scale of dip, we obtain the required elevations; the horizontals drawn through them will give the line of outcrop and the strike on the horizon of the adit $b$.

Problem No. 11. On horizon I the points $a$, $b$, $c$ of a seam are known. On horizon II, whose vertical height above the first is known, it is required to set out the initial point of a crosscut, which is to serve as a counter-heading when driving from point $a$ along the seam of an inclined shaft.

Solution (Fig. 12). Given: $a$ (20), $b$ (50), $c$ (70), $h = 40$.

Constructing, from the three points of the seam, its strike and, through point $a$, the scale of dip, we find on the latter the numerical elevation corresponding to the second horizon. This will be the initial point of the crosscut. Having determined, from the interval of the scale of dip, the angle of dip and constructed the right triangle $aa_0x$, in which the acute angle $= \varphi$, and one leg equals the vertical height of horizon I above the second, we shall determine the true length of this crosscut as the hypotenuse of the triangle constructed.

Problem No. 12. It is required to determine the depth of a shaft sunk to the meeting with a seam whose strike and dip are known.

Solution (Fig. 13). Given: $\omega$, $\varphi$, $a$ (22.5).

Required: $h$.

Since the strike and dip of the seam are given, we can construct the scale of dip of the given seam. Having determined the numerical elevation of the horizontal corresponding to point $a$, and subtracting from it the altitude of point $a$, we obtain the depth of the shaft

\[ H = 142.5 - 22.5 = 120. \]

Problem No. 13. A seam is encountered at point $a$ on the surface and by two boreholes $b$ and $c$. What is the depth of the third borehole, sunk at point $d$?

Solution (Fig. 14). Given: altitudes of points $a$ (105), $b$ (130), $c$ (110), $d$ (125); depth of the borehole at $b$, $h_1 = 100$, at $c$, $h_2 = 90$.

Required: depth of the borehole at $d$, $h_3 = ?$

Determine the numerical elevations of the points at which the boreholes meet the seam, construct the scale of dip of the latter, and find the numerical elevation of the point at which the seam meets the third borehole. Subtracting from it the altitude of point \(d\), obtain the depth of the third borehole.

Problem No. 14. A bedded deposit is given by the points \(a, b, c\); \(c\) lies in the first level, and \(a\) in the second horizontal drift. These two main drifts must be connected by a working, the horizontal projection of which forms an angle \(\alpha\) with the horizontal drift. It is required to determine the direction and length of the working. What would be the length of a perpendicular working?

Solution (Fig. 15). Given: \(a\ (120)\), \(b\ (100)\), \(c\ (50)\), \(\alpha = 30\).

Required: \(\omega_1, l\).

Having plotted all the points on the map and constructed the scale of dip, draw from \(a\) a straight line at the angle \(\alpha = 30^\circ\). At its intersection with the upper drift we obtain the terminal point of the working. The true magnitude \(ea\) is determined as the hypotenuse of the right triangle \(aef\). The true magnitude of the perpendicular working is determined as the hypotenuse of the right triangle \(adk\).

Problem No. 15. Between two horizontal drifts, the vertical distance between which is known, a diagonal drift is designed from a given point \(A\) in a known direction. It is required to determine the angle formed by the diagonal drift with the lower horizontal one.

Numerical solution (Fig. 16). With steep angles of dip of coal seams, both for haulage purposes and for other reasons, it is sometimes necessary to drive drifts that would have the most convenient angle of inclination to the horizon. If such drifts are driven along the lines of dip, they will be too steep for haulage, and therefore these drifts have to be driven diagonally. Let \(E\) be the plane of the seam, \(H\) the plane of the horizon, \(FS\) the strike of the seam and at the same time a horizontal drift; \(fs\) the same horizontal drift in the upper level.

From point \(A\) one must drive a diagonal drift \(x = AB\) to the upper level. Let us further denote: \(BC\) the line of dip of the seam, \(\angle BAD = \varphi_1\) the angle formed by the diagonal drift with the horizon, \(\angle BAC = \gamma\) the angle formed by the diagonal with the line of strike, \(\angle BCD = \varphi\) the angle of dip of the seam. The angles \(\varphi, \varphi_1\), and \(\gamma\) are in a known relation to one another.

The angle \(\gamma\) cannot be very acute, since in the latter case the pressure of the hanging side of the seam crushes the coal at this place.

Consequently, in selecting a certain angle $\varphi_1$, it is necessary to make sure that the angle $\gamma$ does not become very acute.

From the corresponding right triangles we have:

\[ \begin{array}{ll} \triangle BAD & H = x \sin \varphi_1 \\ \triangle BCD & H = a \sin \varphi \end{array} \left\} \quad x \sin \varphi_1 = a \sin \varphi \right. \]

\[ x = \frac{a \sin \varphi}{\sin \varphi_1}. \]

Thus, the length of the diagonal is directly proportional to the sine of the angle of dip of the seam and inversely proportional to the sine of the angle formed by the diagonal with the horizon.

Further, from the right triangle $\triangle BCA$ we have

\[ a = x \sin \gamma \]

Substituting this value into the preceding equation, we obtain

\[ x = \frac{x \sin \gamma \cdot \sin \varphi}{\sin \varphi_1} \]

or

\[ \sin \gamma = \frac{\sin \varphi_1}{\sin \varphi}. \]

Thus, the sine of the angle formed by the diagonal with the line of strike of the seam is inversely proportional to the sine of the angle of dip.

If the angle of inclination of the diagonal drift to the horizon $\varphi_1$ is not given, and only the direction of the diagonal drift is known, then we determine it in the following way.

From the right triangles we have:

\[ \triangle ABD:\qquad \tg \varphi_1 = \frac{BD}{AD}. \]

\[ \triangle ADC:\qquad AD = \frac{CD}{\sin \alpha}. \]

\[ \triangle CBD:\qquad CD = H \ctg \varphi. \]

Consequently

\[ AD = \frac{CD}{\sin \alpha} = \frac{H \ctg \varphi}{\sin \alpha} \]

and

\[ \tg \varphi_1 = \frac{H \sin \alpha}{H \ctg \varphi} = \tg \varphi \cdot \sin \alpha, \]

but

\[ \alpha = \omega - \omega_1. \]

Thus, we shall finally obtain

\[ tg\,\varphi_1 = tg\,\varphi\,\sin(\omega-\omega_1). \]

Graphical solution (Fig. 17).

Given: \(\omega, \varphi, \omega_1, H\).

Required: \(\varphi_1, \gamma\).

Having drawn the projections of the upper and lower horizontal drifts, and also, from point \(A\), the projection of the diagonal drift in the given direction, at the intersection of the latter with the projection of the upper horizontal drift we obtain point \(B\). The angle \(\gamma_1\) will be the angle between the projections of the diagonal and horizontal drifts. In order to find the angle \(\gamma\) between the drifts themselves in space, rotate the plane of the seam about the horizontal axis \(AD\) until it coincides with the plane of the horizon of the lower drift. Points \(A\) and \(D\), as lying on the axis of rotation, will remain fixed, while point \(B\), remaining on the perpendicular to the axis of rotation, will describe an arc whose radius is equal to the hypotenuse of a right triangle in which one leg is the horizontal distance of point \(B\) from the axis of rotation, i.e., \(BD\), and the other leg is the difference in elevations of points \(B\) and \(D = 3h_0\).

Having constructed this right triangle \(CBD\), we mark off the line \(BD\) with an arc of radius \(DC\).

By joining the obtained point \(B_0\) with \(A\), we find the required magnitude of the angle \(\gamma\) between the diagonal and the lower horizontal drift.

Problem No. 16. From a given point \(P\) it is proposed to drive, along a seam whose strike and dip are known, a diagonal drift to another given point \(Q\). It is required to determine the length, direction, and inclination to the horizon of this diagonal drift.

Solution (Fig. 18). Given: \(\omega, \varphi, P, Q\).

Required: \(l, \omega_1, \varphi_1\).

Since the strike and dip of the seam are given, construct in the known manner the scale of dip of the seam. Then, having found the elevations of points \(P\) and \(Q\)—for which horizontals are drawn through these points—and also having determined the interval of the line \(PQ\), from the corresponding right triangle obtain the angle of its inclination to the horizon. The true length of this diagonal will be determined as the hypotenuse of a right triangle in which one leg is the horizontal projection of the required drift, and the other leg is the difference in elevations of points \(P\) and \(Q\).

Problem No. 17. A seam is encountered by three boreholes, and its strike and dip have been determined. For preparing the deposit on the hanging-wall side of the seam, a shaft has been sunk to a depth of 50 m, and from it

across the strike a crosscut is to be driven until it meets the seam. It is required to determine the point at which the crosscut will meet the seam and the length of the latter.

Solution (Fig. 19). Given: \(a\) (60), \(b\) (90), \(c\) (70), D (30), \(h = 50\).

Required: \(l\); E.

Having determined, by the known method, the strike line and constructed the scale of dip, draw, parallel to the latter, from point D a line until it meets the horizontal of the seam corresponding to the elevation of the point at which the seam is met by the shaft. We shall determine this elevation if we add together the numbers expressing the elevation of point D on the surface and the depth of the shaft. The length of the crosscut is determined to scale directly from the drawing.

Problem No. 18. For the exploration of a bedded deposit encountered by three boreholes, it is required to drive an adit running across the strike of the seam and at a known inclination to the horizon.

Solution (Fig. 20). Given: \(a\) (150), \(b\) (200), \(c\) (170), \(\mathfrak{c}_1\).

Required: the point at which the adit meets the seam.

In order to determine the point of meeting of the line \(df\) (the adit) with the plane M (the seam), through the given straight line draw any plane whatever and find the line of its intersection with the given plane. The intersection of the resulting straight line with the given \(df\) will give the required point at which the plane meets the line.

To find the line of intersection of the planes, from any two elevations, for example (150) and (170), draw on the plane \(df\) horizontals parallel to any direction, \(fk\) and \(dl\), and from the corresponding elevations of the plane M draw horizontals \(pk\) and \(ql\).

The intersection of the line joining the points \(k\) and \(l\) with the given straight line will give us the required point \(m\).

Problem No. 19. A seam whose strike and dip are known is intersected by some working; moreover, the points of intersection of the hanging wall and the footwall of the seam and their numerical elevations have been determined.

It is required to determine the thickness of the seam.

Solution (Fig. 21). Given: \(\omega\), \(\mathfrak{c}\), \(a\) (20), \(b\) (21).

Required: thickness \(m\).

If from point \(b\) of the hanging wall of the seam we drop a perpendicular onto the plane of the other wall, then the length of this perpendicular will give us the required thickness. The direction of the latter will coincide with the direction of the line of dip of the plane, and its interval, in relation to the interval of the scale of dip of this plane, will be a reciprocal quantity and with the opposite sign. To find the length

of this perpendicular it is necessary to determine the point of its intersection with the plane of the footwall of the seam. Having found this point \(c\), as the intersection of a straight line with a plane, as was discussed in the preceding problem, we find the true magnitude of this perpendicular.

For this purpose construct the right triangle \(bdd_0\), in which one leg \(bd\) is equal to the distance between the points \(b\) and \(d\), and the other \(bb_0\) is equal to the difference in elevations of these points.

Now, projecting the segment \(bc\) onto the hypotenuse of the right triangle obtained, determine the thickness of the seam \(c_0b_0 = m\), which remains only to be measured to scale.

Problem No. 20. A seam is penetrated by a horizontal working, the length \(l\) and direction of which are known.

In addition, the numerical elevation of the intersection of the working with the hanging wall of the seam has been determined. What is the thickness of the seam, if its strike and dip are known?

Note. By the intersection of a seam by a working is always meant the intersection of it by a line whose direction coincides with the direction of the working.

Solution (Fig. 22). Given: \(\omega\), \(\varphi\), \(l = ab\), \(a\) (20).

Required: the thickness \(m\).

Since the working is horizontal, the elevations \(a\) and \(b\) of the hanging and footwall are the same. In all other respects the solution is the same as in the preceding problem. The true magnitude of the seam thickness \(m = b_0c_0\) will be obtained if we project the segment \(bc\) onto the hypotenuse of the right triangle \(def\), in which the leg \(de\) is equal to the horizontal distance between the points \(d\) and \(e\), and the leg \(ef\) represents the difference in elevations of those same points.

Problem No. 21. In a known direction along the seam a diagonal drift has been driven. It is required to determine the point where it emerges at the ground surface, which represents a horizontal plane at height \(h\) above the zero horizon.

The dip and strike of the seam are known.

Solution (Fig. 23). Given: \(\omega\), \(\varphi\), \(\omega_1\), \(h = 50\).

Required: the point of emergence \(a\) (50).

Since the direction of the working, and the strike and dip of the seam, are known, we plot on the map the direction of this line and the scale of dip of the seam.

Having found on the scale of dip the elevation corresponding to the ground surface, and drawing through it the horizontal of the seam, we obtain, at the intersection of the latter with the direction of the working, the required point of emergence.

Problem No. 22. It is required to determine the point at which a working reaches the ground surface, represented by a horizontal plane at a known height above the zero horizon, if the direction of this working and its inclination to the horizon are known.

Solution (Fig. 24). Given: $\omega$, $\zeta_1$, $h = 220$.

Required: the point of emergence on the surface $a(220)$.

Since the direction and inclination of the working to the horizon are given, we can construct the scale of inclination of the working. The mark of the latter corresponding to the surface horizon will give the required point of emergence.

Problem No. 23. It is required to determine the point of emergence of a working, the direction and inclination of which are given, onto a ground surface of uniform slope, on which the position of three points is known.

Solution (Fig. 25). Given: $\omega_1$, $\zeta_1$, A (5), B (11), C (8).

Required: the point of emergence P.

Having constructed, in the known manner, the scale of the dip of the surface and the scale of the inclination of the working, we shall find the point of intersection of the working with the ground surface. For this we must determine the intersection of the straight line $ab$ with the plane M, as was done in Problem No. 17. Having found the point P, we determine its numerical mark. Construct the triangle $abc$, in which the leg $ab$ is the horizontal distance between the points $a$ and $b$, while the leg $bc$ is the difference in height of the latter; the difference in heights of the points $a$ and $p$ is determined as the segment $pd$, drawn parallel to the line $bc$. Adding its value to the mark of point $a$, we obtain the mark of point $p$.

Problem No. 24. It is required to determine the line of emergence of a seam, the strike, dip, and one point of which are known, onto the ground surface, represented by a horizontal plane at a known distance from the zero horizon.

Solution (Fig. 26). Given: $\omega$, $\zeta$, $h = 15$.

Required: the line of emergence.

Since the intersection of any plane with a horizontal plane is a horizontal line, to solve this problem it is necessary to construct the scale of the dip of the seam, find its horizontal corresponding to the horizon of the ground surface. This straight line will be the line of emergence of the seam on the ground surface.

Problem No. 25. It is required to determine the line of emergence of a seam, the strike, dip, and one point of which are known, onto a ground surface of uniform slope, on which the position of three points is known.

Solution (Fig. 27). Given: \(\omega\), \(\varphi\), \(A\) (160), \(B\) (220), \(C\) (190). Required: the line of outcrop.

Having constructed the scales of dip of the plane of the seam and of the ground surface, we find the intersection of these planes, for which it is necessary to connect the points of intersection of the corresponding contours of these planes. The obtained line \(\alpha \beta\) will be the line of outcrop of the seam on the surface. In order to find the inclination of this line to the horizon, we determine its interval, for which we extend the contours of one or the other plane until they intersect it. Knowing the interval of this straight line, we easily find the angle of its inclination to the horizon.

Remark. In solving this problem it is entirely immaterial what the directions of dip of the seam and of the surface are. The course of the solution changes somewhat in the case where the directions of strike are parallel.

Problem No. 26. It is required to determine the line of outcrop of a seam on the ground surface, if its strike is perpendicular to the strike of the ground surface.

Solution (Fig. 28). Given: \(\omega\), \(\varphi\), \(\omega_1 = 90 - \omega\), \(\varphi_1\).
Required: the line of outcrop.

The given problem is a special case of the preceding one, and its solution does not differ in any way from the solution of that problem.

Problem No. 27. It is required to determine the line of outcrop of a seam on the ground surface, if their strikes are parallel, while the angles of dip are arbitrary.

Solution (Fig. 29). Given: \(\omega\), \(\varphi\), \(\omega_1 = \omega\), \(\varphi_1\).
Required: the line of outcrop.

Since the lines of strike of both planes are parallel, the line of intersection of these planes is also parallel to their strike. In the present case it is not possible to find directly the intersection of the planes; therefore we take, in the plane \(M_1\), an arbitrary straight line \(ab\) and find the point of its intersection with the plane of the seam \(M\). Since the line of intersection must be parallel to the line of strike, it remains only to draw through the found point \(P\) a line perpendicular to the scales of dip of the planes.

Remark. In this problem the planes have the same direction of dip.

The solution will be the same if the planes are inclined in opposite directions.

Problem No. 28. It is required to determine the line of outcrop of a vertical seam on the surface, if its strike and one point are known, as well as the strike and dip of the ground surface.

Solution (Fig. 30). Given: \(\omega, P, \omega_1, \varphi_1, \varphi = 90^\circ\).

Required: the outcrop line.

Since the seam is vertical, its dip line will be represented by a single point \(P\). The strike line will have the given direction. Obviously, the seam outcrop line will have the same direction as the strike line of this seam; and since the outcrop line will lie on the surface represented at scale \(M\), by drawing the contours of the latter we obtain the interval of the outcrop line, and consequently determine also its inclination to the horizon.

Problem No. 29. A seam has been intersected by three boreholes. It is required to determine its outcrop line on the surface represented by contours.

Solution (Fig. 31). Given: \(A\) (400), \(B\) (325), \(C\) (300).

Required: the outcrop line.

Having constructed, from the three given points of the seam, the scale of its dip, we draw the contours of this seam until they intersect the like-named contours of the surface. By joining the points of intersection, we obtain the outcrop line of the seam on the surface.

Crossing of Stratiform Deposits

If two stratified deposits intersect, this is called crossing. Since stratiform deposits may be represented by bounding planes, then, instead of the intersection of the deposits themselves, we may consider the intersection of the planes that bound them.

Thus, by the line of crossing is meant the line of intersection of two planes belonging to different deposits. Depending on the position of the planes bounding the deposit relative to the horizon, the line of intersection may be either vertical, or horizontal, or, finally, arbitrarily inclined to the horizon.

1. A vertical line of crossing occurs as a result of the intersection of two deposits that are arranged vertically but have different strikes.

2. A horizontal line of crossing is obtained from the intersection of two deposits whose strikes are parallel but whose dip angles are different.

3. An inclined line of crossing is obtained in all other cases.

The line of intersection will be completely determined in space if its direction, one of its points, and its inclination to the horizon are known.

Determining the line of intersection of seams, in essence, does not differ from determining the outcrop of a bed at the surface.

Problem No. 30. Given the strike and dip of two intersecting beds or deposits. It is required to determine the strike and dip of their line of intersection.

Numerical solution.

From Fig. 32 we have

\[ \omega=\omega_1-(P+180^\circ)\ldots\ldots\ldots (1) \]

\[ \omega=\omega_1-(90^\circ-x+180^\circ)=\omega_1-(270^\circ-x) \]

\[ \tg \zeta=\frac{h}{a}\ldots\ldots\ldots (2) \]

where \(\omega\) is the direction of the line of intersection, \(\zeta\) is the angle of its inclination to the horizon.

Further,

\[ \left. \begin{aligned} m&=180^\circ-(x+y)\\ m&=\omega_1-\omega_2 \end{aligned} \right\}\ldots\ldots\ldots (3) \]

Consequently,

\[ x+y=180^\circ-m=180^\circ-(\omega_1-\omega_2)\ldots\ldots\ldots (4) \]

To find \(x\) and \(y\), we determine their difference. From the right triangles:

\[ \begin{array}{ll} \triangle\ EE'G & a_1=h\,ctg\,\zeta_1\\[2mm] \triangle\ FF'G & a_2=h\,ctg\,\zeta_2\\[2mm] \triangle\ E'H'G & \cos x=\dfrac{a_1}{a}\\[3mm] \triangle\ F'H'G & \cos y=\dfrac{a_2}{a} \end{array} \left\}\ldots\ldots\ldots (4') \right. \]

Hence

\[ \frac{\cos x}{\cos y}=\frac{a_1}{a_2}=\frac{ctg\,\zeta_1}{ctg\,\zeta_2}=\tg \psi\ldots\ldots\ldots (5) \]

Transforming this expression, as we did in Problem No. 1, we obtain

\[ \tg \frac{x-y}{2}=-\tg(\psi-45^\circ)\tg\frac{x+y}{2}\ldots\ldots\ldots (6) \]

Knowing now the sum and difference of \(x\) and \(y\), we determine each of them separately, and consequently:

\[ P = 90^\circ - x. \]

Thus the direction of the line of intersection, by equation (1), is

\[ \omega = \omega_1 - (P - 180^\circ) \ldots\ldots\ldots\ldots\ldots I. \]

The inclination to the horizon of the line of intersection, by equation (2), is

\[ tg\,\varphi = \frac{h}{a}, \]

or, taking into account expressions \((4')\), we obtain

\[ \left. \begin{aligned} h &= a_1\, tg\,\varphi_1 = a \cos x\, tg\,\varphi_1 \\ h &= a_2\, tg\,\varphi_2 = a \cos y\, tg\,\varphi_2 \end{aligned} \right\} \]

Consequently, the inclination of the line of intersection to the horizon is

\[ tg\,\varphi = \cos x\, tg\,\varphi_1 = \cos y\, tg\,\varphi_2 \ldots\ldots\ldots II. \]

Graphical solution (Fig. 33). Given: \(\omega_1, \varphi_1, \omega_2, \varphi_2\).
Required: \(\omega, \varphi\).

Having laid off the strike lines of the deposits and constructed the scales of their dip at the intersections of like-named horizontals, we find the second point of intersection, and, joining it with the first common point \(C\) of the deposits, we obtain the line of intersection, whose direction is determined directly from the drawing, while the angle of inclination is found by means of its interval \(J\) from the corresponding right triangle.

Problem No. 31. It is required to determine the dip angles of two intersecting seams on the basis of their known strike, and also of the direction and angle of inclination of their line of intersection.

Solution (Fig. 34). Given: \(\omega_1, \omega_2, \omega, \varphi\).
Required: \(\varphi_1, \varphi_2\).

Drawing through the common point \(C\) the strike lines of the seams and the line of their intersection, and also, perpendicular to the strike lines, the scales of dip, we find the interval of the line of intersection.

Then drawing, through the marks of the line of intersection, the horizontals of the one and the other seam, we obtain the intervals of their dip scales, and consequently determine also the dip angles.

Problem No. 32. Given are the dip angles of the seams, as well as the direction and angle of inclination of the line of their intersection. It is required to determine the strike of the seams.

Graphical solution (Fig. 35). Given: \(\varphi_1,\ \varphi_2,\ \omega,\ \varphi\).
Required: \(\omega_1,\ \omega_2\).

We first draw, through some point, the line of intersection and, finding along the angle \(\varphi\) its interval, lay off the latter from the initial point \(C\) along the line of intersection. Having then determined, from the angles \(\varphi_1,\ \varphi_2\), the intervals \(J_1\) and \(J_2\), we describe on the interval \(J\), taken on the line of intersection, a circle as on a diameter, and intersect the latter from the end of the diameter \(K\) by radii equal to the intervals \(J_1\) and \(J_2\). Now drawing through the point \(C\) lines parallel to the straight lines \(PL\) and \(PN\), we obtain the strike lines of the seams, and perpendicular to them we construct the scales of dip. Drawing from the marks of the line of intersection the horizontals of the one and the other seam, we obtain on the dip scales all their numerical marks.

Problem No. 33. It is required to determine the angle which the line of intersection forms in space with the strike line of one of the intersecting seams, if the strikes and dips of the latter are known.

Numerical solution.

The angle \(\delta\), which is to be determined, is obtained from the right triangle \(EGH\) (Fig. 36).

\[ \sin \delta = \frac{EG}{GH}\ .\ .\ .\ .\ .\ .\ .\ .\ (1) \]

But from the right triangles \(E'H'G\), \(EE'G\), and \(HH'G\) we have:

\[ EG=\frac{h}{\sin \varphi_1}=\frac{a_1}{\cos \varphi_1} \]

\[ HG=\frac{h}{\sin \varphi}=\frac{a}{\cos \varphi} \]

\[ a_1=a\cos x \]

Consequently:

\[ \sin \delta=\frac{a_1\cos \varphi}{a\cos \varphi_1} =\frac{a\cos x\cos \varphi}{a\cos \varphi_1} =\frac{\cos x\cdot \cos \varphi}{\cos \varphi_1}\ .\ .\ .\ I. \]

The unknown quantities in the last formula, \(\cos x\) and \(\cos \varphi\), are determined from equations (4), (6), and (II), derived in the solution of Problem No. 30.

Graphical solution (Fig. 37). Given: \(\omega_1,\ \varphi_1,\ \omega_2,\ \varphi_2\).
Required: \(\omega,\ \varphi\).

Since we know the angle between the projection of the line of intersection and the line of strike, in order to find the angle formed by these lines in space, we must bring the plane passing through the line of strike and the line of intersection into a horizontal position, rotating it about the horizontal \(AC\) as about an axis.

In doing so, as we have already seen in Problem No. 15, the points \(C\) and \(F\), as lying on the axis, do not change their position; while point \(D\), remaining during the rotation in a plane perpendicular to the axis, describes an arc whose radius will be equal to the hypotenuse of the right triangle \(DFH\), in which one leg is the length of the perpendicular \(DF\) to the axis \(AC\), and the other leg is the difference in heights of the points \(C\) and \(D\). Having constructed the said triangle and marking off, with an arc of radius \(FH\), equal to the hypotenuse of this triangle, the extension of this perpendicular \(DF\), we obtain point \(G\), i.e., the new position of point \(D\), when the plane \(ACD\) has been brought into a horizontal position.

Joining the obtained point \(G\) with point \(C\), we obtain the magnitude of the angle \(ACG\) in space between the line of strike and the line of intersection.

Displacements of bedded deposits.

In mining practice one often encounters various displacements of bedded deposits.

The task of the mine surveyor in this case is the numerical or graphical solution of the question of finding the displaced parts of the deposit and determining the character of movement of the separated parts.

If, in the displacements, there is no mutual penetration of the displaced part into the fault, then we may consider the following kinds of motion:

1. Rectilinear motion in any direction in the plane of the fault (Fig. 38):

a) a downthrow, if the motion occurs along the line of dip downward,

b) an upthrow, if the motion is upward,

c) a diagonal downthrow or upthrow, if the displacement occurs not along the line of dip;

1. Rotational motion, the axis of rotation being perpendicular to the plane of the fault;

2. A combination of several kinds of motion.

The difference between these three kinds of motion consists in the fact that in the first case the displacement of the bed or vein occurs parallel to itself, and consequently the angle formed

of the part of the deposit remaining at rest and of the part displaced together with the fault remains constant. The latter condition is very important both for the numerical determination of displaced parts and, especially, for the graphical solution of this question.

On the basis of this condition, the kind of movement is easily recognized from the relative position of the lines of intersection of the part remaining at rest and of the displaced part of the deposit with the fault.

1. If the lines of intersection of one and the other part of the deposit with the fault are parallel to each other, then the movement will be rectilinear;

2. If these lines are not parallel, then the movement is rotational or a combination of two kinds of movement.

In mine-surveying practice, the problem of finding displaced parts of a deposit plays a very large role, since, in the event that such displacements are encountered, it becomes necessary to determine the direction in which work should be continued in order to meet the displaced part.

It is impossible to derive any exact rules for finding displaced parts of deposits; however, certain guiding principles can greatly assist in solving this difficult problem.

1. In sedimentary formations, whose hanging wall and footwall are different, the sequence of stratification of the various rocks must be preserved during the displacements that occur.

Bearing this in mind, we can decide whether we are in the displaced part of the deposit or in the part that has remained undisturbed, understanding by the former the lower-lying part of the deposit.

Let us suppose that the roof of the bed consists of shaly rocks, and the floor of sandstones. In such a case, if at point \(A_0\) (Fig. 39) a displacement of the bed is encountered and beyond the fault, after point \(A_0\), shale is found, this will be proof that we are in the undisplaced part of the deposit. Conversely, if during workings beyond point \(B_0\) sandstone is found, this will indicate that the given part of the bed is displaced.

Among other signs indicating the direction of displacement are:

1. mechanically admixed particles of the deposit occurring in the fissures of the fault,

2. the so-called slickensides and mirror-like surface in the rock itself,

4. Bends of a deposit at the point of displacement.

Without dwelling further on an examination of the enumerated features, we assume that, on the basis of certain data, it is possible to say precisely whether, upon displacement, a known part of a deposit is higher or lower in relation to another, sought part.

Despite the large number of rules given for finding unknown parts of a bed or vein, most of them are either extremely unclear, or else are not general, but require certain limitations.

Moreover, all the authors of these rules, such as Schmidt, Carnall, Zimmermann, and others, in their derivations proceed from the assumption that, upon displacement, the deposit slides along the line of dip of the fault.

Reality, however, does not justify the latter, since movement during dislocation may occur at various angles to the line of dip of the fault.

Before proceeding to the solution of problems on finding unknown parts in the displacement of parts of deposits, let us agree on certain designations.

Suppose that a part of the deposit \(AB\) (Fig. 39), together with all the overlying rocks, has been displaced along the line of dip of the fault into the position \(A_1B_1\), while the part \(CD\) has remained in its original position.

Such a phenomenon is called a normal fault.

\(B'b = h\) vertical throw,

\(B'b = s\) floor of the fault,

\(BB' = f\) inclined height of the fault.

The part of the deposit adjoining the hanging side of the fault, in some cases during displacements, may be situated higher than that adjoining the footwall side.

In this case there results the so-called double position of the deposit, and such displacements are called upthrows or overthrusts.

Double positions of this kind are the cause of many uncertainties.

For example, by means of a borehole one may establish the presence of two beds or veins, whereas in reality there will be only two parts of one and the same deposit (Fig. 40).

Let us take another case. Let there be (Fig. 41)

\(L_1\) the part of the deposit that has remained in its former position,

\(L_2\) the displaced part of it,

$VV$ — fault,

$BA$ — the line of intersection of the remaining, undisplaced part of the deposit with the fault,

$BB'$ — the line of sliding along the dip line of the fault,

$B'C$ — the line of intersection of the displaced part of the deposit with the fault,

$B$ and $B'$ — points of both parts of the deposit that formerly corresponded to one point,

$BB' = f$ — the inclined height of the displacement,

$BD = h$ — the vertical height of the latter,

$B'D = s$ — the footwall, and

$AB' = g$ — the magnitude of the displacement.

If the strike and dip of the deposit and of the fault, as well as one of the quantities $f$, $h$, $s$, $g$, are known, then all the others can be determined from the right triangles $BB'D$ and $ABB'$.

Such a displacement can be constructed; moreover, a definite point is obtained, in the direction toward which, from the initial point, the mine surveyor should conduct exploration in order to reach the other, sought part of the deposit.

The construction is best carried out by the method of projection with numerical elevations, since the latter, as in solving all the other problems, gives the mine surveyor a very simple and general rule for finding the unknown parts.

In doing this, it is only necessary always to keep in mind that if rocks of the footwall are encountered by a working beyond the fault, then the displaced part of the deposit is before us.

If, beyond the fault, rocks of the hanging wall are encountered, it is necessary to conclude that the workings are being driven in the undisplaced part.

The rule derived below remains valid also in the case when the displacement of the deposit occurred not along the dip line of the fault, but in some other direction.

We shall regard the dip of the seam and of the fault as concordant if the arrows indicating the direction of dip in the acute angle between the strike lines of the intersecting planes are directed to one side, and as discordant when the arrows are directed to different sides, as may be seen in Figs. 41-a and 41-b.

In the following drawings, the direction of exploration workings for finding the unknown parts when parts of the deposit have been displaced will likewise be indicated by arrows.

Rectilinear Displacement

Problem No. 34. In a bedded deposit, whose strike and dip are known, a displacement was encountered at point A, and, in addition, the strike and dip of the fault were determined. It is required to determine in what direction the work should be continued in order to reach the unknown part of the deposit, if its dip and that of the fault are concordant and if, after driving through the latter, rocks of the hanging wall were encountered.

Solution (Fig. 42). Given: $\omega_1, \varphi_1, \omega, \varphi$.

Since, beyond the fault, rocks of the hanging wall were encountered, we are in the undisplaced part of the lode. Consequently, the sought displaced part will be located lower; therefore the construction should be made on the footwall side of the deposit and of the fault.

If the height of the displacement is unknown to us, then we can determine only one direction in which the unknown part of the deposit should be sought.

Having plotted on the plan the strikes of the lode and of the fault, on the basis of their dip angles construct their dip scales and, by the known rule, the line of intersection $S_1$ of the undisplaced part of the lode with the fault.

If the slipping of the deposit in the given case occurred along the direction of the line of dip of the fault, then point B will correspond to point A. Through point B, parallel to the line $S_1$, there must pass the line of intersection $S_2$ of the displaced part with the fault.

At the intersection of this line with the latter, point D is obtained, which lies at the same elevation as point A. The line AD gives the direction of exploration, with point D corresponding to the displaced part of the deposit.

In practice, the angles at A and D are usually rounded.

Problem No. 35. Under the conditions of the preceding problem, it is required to find the displaced part of the deposit if the slipping of the latter occurred in any direction.

Solution (Fig. 43). Given: $\omega_1, \varphi_1, \omega, \varphi$.

The construction is to be carried out analogously to that in the preceding problem, except that the line of intersection $S_2$ of the displaced part of the lode with the fault will pass through point B, lying on the line of slipping, which in the present case does not coincide with the line of dip of the fault.

Problem No. 36. At point A of the deposit a displacement was encountered. The strike and dip both of the fault and of the deposit are known. It is required to indicate the direction in which the unknown part of the deposit should be sought, if it has been determined that it is the displaced part, and if the dip of the deposit and of the fault are discordant.

Solution (Fig. 44). Given: $\omega_1$, $\varphi_1$, $\omega$, $\varphi$.

Since the displaced part of the deposit is being sought, the construction should be carried out on the footwall side of the deposit and of the fault. In all other respects the course of the solution is the same as before.

Problem No. 37. In deposit $L_2$, whose strike and dip are known, a displacement has been encountered. The strike and dip of the fault have been determined. It is required to find the second unknown part of the deposit, if the dip of the latter and of the fault is concordant and the rocks of the footwall occur beyond the latter.

Solution (Fig. 45). Given: $\omega_2$, $\varphi_2$, $\omega$, $\varphi$.

Since in the given case the rocks of the footwall have been encountered beyond the fault, the work is therefore being carried out in the displaced part of the deposit.

The construction should therefore be made on the hanging-wall side of the deposit and of the fault.

Having constructed the line of intersection $S_2$ of the displaced part of the deposit with the fault, draw through point B a line of intersection of the undisplaced part parallel to it. Point D will correspond to the latter.

Problem No. 38. At point A of a deposit, whose strike and dip are known, a displacement has been encountered.

It is required to find the unknown part of the deposit, if it has been determined that the work is being carried out in the displaced part of the deposit, and if the dip of the latter and of the fault is discordant.

Solution (Fig. 46). Given: $\omega_2$, $\varphi_2$, $\omega$, $\varphi$.

Since here the undisplaced part of the deposit is being sought, then, as in the preceding problem, the construction should be carried out on the hanging-wall side of the deposit and of the fault.

Faults and Upthrows.

Having constructed, in each of the preceding cases of displacement, its profile, we shall see whether in the given case there is a fault or an upthrow. However, this can easily be ascertained directly as well, without constructing profiles, if one is guided by the following rule.

Having obtained, by construction, the strike lines of both parts of the deposit and of the fault, from the point of intersection of one or the other part of the deposit with the fault we drop a perpendicular to the strike line of the other part of the deposit.

1. If this perpendicular does not intersect the latter, then, when the dips of the bed and the fault are concordant, there will be a downthrow;

2. If it does intersect, then an upthrow;

3. When the dips of the bed and the fault are discordant, the rule is the reverse.

Rule of searching in the case of displacement of unknown parts of a deposit.

In order to derive a general rule for finding, in the case of displacement, the unknown parts of beds and veins, let us agree upon certain notations.

We shall call the point of meeting of the fault through \(A\), its dip line through \(AB\), the line of intersection of the corresponding part of the deposit with the fault through \(AC\), and the line of search for the unknown part through \(AD\).

The line of search and the line of intersection of the known part of the deposit with the fault will always be situated on different sides of the line of slip (the dip line of the fault).

In practice, when encountering displacement of a bed or vein, one should proceed as follows:

1. determine the strike and dip of the fault;

2. intersect the latter;

3. construct the line of intersection between the known part of the deposit and the fault; moreover, if the work is being carried out in the undisplaced part, then the construction is made on the footwall side of the deposit; if, however, in the displaced part, then, conversely, on the hanging-wall side;

4. determine the direction of the line of slip (the dip line of the fault).

Having done all this, the search should be conducted along the strike of the fault so that the preceding rule is observed, i.e., so that the line of search and the line of intersection are situated on different sides of the line of slip.

In the case where the deposit and the fault have parallel strikes, the search should be conducted along the dip of the fault if we are in the undisplaced part of the deposit, and conversely, along the rise of the fault if the work is being carried out in the displaced part.

Rotational Motion

If, during the displacement of a bedded deposit, a rotational motion of it takes place and no penetration of the displaced part into the fault is observed, then the axis of rotation must be normal to the plane of the fault.

Knowing the strike and dip of both parts of the deposit, as well as of the fault, we can easily determine the lines of intersection \(S_1\) and \(S_2\) of one and the other part of the deposit with the fault. These lines will not be parallel to each other, but will intersect.

Theoretically, it is not difficult to determine the position of the axis of rotation. First, it must lie in the bisector plane between the two parts of the deposit.

In addition, it must be perpendicular to the plane of the fault.

If we know the position of two corresponding points \(A\) and \(B\) of both parts of the deposit, then the axis of rotation can be determined as the line of intersection of the aforementioned bisector plane and a plane perpendicular to the plane of the fault and, moreover, passing through the midpoint of the straight line \(AB\).

Problem No. 39. By means of a drift driven along the strike of bed \(L_1\), a displacement was encountered at point \(A\), and at point \(B\) the displaced part was encountered.

The strike and dip of one and the other part of the deposit, as well as of the fault, are given. It is required to drive a horizontal drift to a horizon lying 50 m below the upper one.

Solution (Fig. 47). Given: \(\omega_1, \varphi_1, \omega_2, \varphi_2\).

\[ \omega, \varphi. \]

Required: the length of the drift.

To solve the proposed problem it is necessary to determine:

1. the horizontal length of the inclined working \(CD\),
2. the length of the drift \(DE\) up to the meeting with the displacement of the bed,
3. the length and direction of the part of the drift driven in barren rock until meeting the displaced part of bed \(L_2\).

Having constructed the lines of intersection of one and the other part of the bed with the fault, and drawing the corresponding horizontals of one and the other part of the bed and of the fault, we obtain the direction of the projected drift. Taking the lengths \(DE\) and \(EF\) to scale, we shall also know the length of our drift.

Problem No. 40. Given the strike and dip of deposit \(L_1\). At point \(A\) a displacement was encountered, and in doing so the strike was determined

and dip of the fault. In the lower horizon the deposit \(L_2\) was encountered, whose strike and dip were likewise determined. At point B of the latter deposit a new displacement was encountered, and since in this case the strike and dip of the fault were the same as in the displacement at point A of the first deposit, the question arises whether both deposits do not belong to one and the same one.

Solution (Fig. 48). Given: \(\omega_1, \varphi_1, \omega_2, \varphi_2\).

\[ \omega,\ \varphi. \]

To verify our supposition, construct, by the method indicated earlier, the lines of intersection \(S_1\) and \(S_2\) of both deposits with the fault. By extending both lines of intersection to their mutual intersection at point \(O\), we shall obtain at the latter the intersection of the corresponding horizontals of deposits \(L_1\) and \(L_2\).

It is obvious that, in the space between the lines of intersection on the drawing upward from point \(O\), we shall have a throw, while downward from point \(O\), between the same lines of intersection, there will be an upthrow. A mine shaft or borehole sunk at point \(C\) will encounter neither the one nor the other deposit.

If, however, it is sunk at point \(D\), we shall first encounter deposit \(L_1\), then the fault, and finally deposit \(L_2\). If above point \(O\) and below it we draw a series of horizontals of one and the other part of the deposit, with the horizontals obviously extending only as far as the corresponding line of intersection, then we shall see that in the upper part the horizontals, not coinciding with one another, will be separated by a certain space.

Within the limits of the latter, when sinking, therefore, neither one nor the other part of the deposit can be encountered.

In the lower part of the drawing the picture is entirely different. We see that the horizontals of one part of the deposit are projected onto the horizontals of the other part, i.e., one part of the deposit lies above the other, and therefore, when sinking in this place, we shall encounter both parts of the deposit.

The picture becomes still clearer if we draw profiles along known directions, i.e., mentally cut the ground along these directions by vertical planes or by planes inclined in some way to the horizon.

In Fig. 48 we have given vertical profiles along the directions \(MN\) and \(PQ\).

The construction of profiles by means of the projection method with numerical elevations is carried out with remarkable simplicity, especially for vertical profiles.

In the latter case it is necessary only to determine the numerical elevations of the intersections of the vertical plane with all the principal lines.

Profiles acquire very great importance for mines with strongly branched workings.

By giving on the mine-surveying plan the contours of the terrain and expressing seams or veins by the scales of their dip, we shall possess everything necessary for solving various mine-surveying problems.

As an example of the application of the projection method with numerical elevations to the solution of mine-surveying problems, one may point to the article by mining engineer V. Graunig, recently published in No. 10 of the Oesterreichische Zeitschrift für Berg- und Hüttenwesen, “Über die Ausbisse der ‘Hangendlagerstätte’ am Schneeberg bei Sterzing in Tirol.”

In this article the author theoretically determines the outcrop line of the seam at the surface, the lines of faults, and gives a profile of the deposit. The correctness of the theoretical conclusions is confirmed by the author by actually carried out exploration.

The accuracy of the agreement of the results of the exploration with the conclusions based on the application of the projection method with numerical elevations fully confirms the suitability of this method for mine-surveying purposes.

Persons who have read this article will readily be convinced of the simplicity of solution that is achieved by this method.

Continuation.

Fig. 1.

Diagram with the following visible annotations: \(b(100)\), \(p(100)\), \(K\), \(y=55\), \(c\), \(m=72{,}6\), \(c_0\), \(c(8{,}5)\), \(a=49{,}5\), \(b_0\),

\[ \Delta=-10{,}5-49{,}5=-31{,}6 . \]

Fig. II.

Linear diagram with intervals and labels: \(a(0)\), \(c(0)\), \(d(10{,}8)\), \(e(0)\); \(15\), \(5\), \(20\).

Fig. 3.

Diagram with point labels and annotations including: \(B(60)\), \(40\), \(50\), \(30\), \(20\), \(10\), \(S\), \(\omega\), \(\gamma\), \(3\), \(3{,}1\).

Auxiliary triangle: \(x\), \(y\), \(5\), \(5_1\).

Fig. 2.

Diagram with point labels and annotations including: \(A(0)\), \(B(10)\), \(C(40)\), \(S\), \(N\), \(a\), \(b\), \(10\), \(5\), \(\omega\), \(\delta\).

Auxiliary triangle: \(x\), \(y\), \(h=10\).

Fig. III.

Longitudinal diagram with center line \(M\) and points/sections labeled \(l_0\), \(l_1\), \(l_2\), \(l_3\), \(l_4\), \(l_5\); corresponding verticals and paired labels on the upper and lower boundary lines.

Fig. 1.

Diagram with points and annotations including: \(a\), \(b\), \(c\), \(d\), \(e\), \(d'\), \(e'\), \(A\), \(B\), \(C\), \(D\), \(E\), \(S\), \(\omega\), \(N\), \(M\), \(\alpha\), \(\beta\), \(30^\circ\), \(90^\circ\), and other construction lines.

Bed 222.2 g [[unclear]]

Fig. 6.

Annotations visible in the drawing:

• Scale
• Strike
• Dip
• \(A\)
• \(B\)
• \(C\)
• \(y\)
• \(y_1\)
• \(y_2\)
• \(h=1\)
• \(\beta\)
• \(\gamma\)
• \(\omega\)
• \(s\)

Fig. 8.

Annotations visible in the drawing:

• Dip
• Strike
• \(A\)
• \(C\)
• \(y\)
• \(y_1\)
• \(y_2\)
• \(h=1\)
• \(\beta\)
• \(\omega\)
• \(s\)

Fig. 5.

Annotations visible in the drawing:

• \(A\)
• \(C\)
• \(y\)
• \(y_1\)
• \(y_2\)
• \(h=1\)
• \(\beta\)
• \(\gamma\)
• \(\omega\)
• \(s\)

Visible numerical notes:

\[ y_1 = 18.1 \]

\[ y_2 = 37.4 \]

\[ y_0 = 6.05 \]

\[ y_2 = 0.105 \]

\[ \frac{y_2}{y_1} = 3.40 \]

Additional note:

\[ y = 9.2 \]

Fig. 7.

Annotations visible in the drawing:

• Dip
• Strike
• \(A\)
• \(C\)
• \(y\)
• \(y_1\)
• \(y_2\)
• \(h=1\)
• \(\beta\)
• \(\gamma\)
• \(\omega\)
• \(s\)

Fig. 4.

Annotations visible in the drawing:

• Scale of sections
• Strike
• \(A\)
• \(C\)
• \(y\)
• \(y_1\)
• \(y_2\)
• \(h=1\)
• \(\beta\)
• \(\omega\)
• \(s\)

Appendix 5

Fig. 9

Diagram with labels: \(B\), \(N\), \(O\), \(O'\), \(A\), \(S\), \(13\), \(a\), \(b\), \(\omega\), \(\omega'\), \(\gamma\), \(\gamma_1\).

Auxiliary inset: \(h_0 = 1\), \(\gamma\), \(\gamma_1\).

Fig. 10

Diagram with labels: \(a\), \(b\), \(\omega\), \(\gamma\), \(O\), \(S\).

Auxiliary inset: \(\gamma\), \(\gamma_1\).

Fig. 11

Diagram with labels: \(S\), \(s\), \(a\), \(b_a\), \(a_1\), \(b_1\), \(N^y\), \(I\), \(\overline{I}\), \(A\), \(\overline{I}\).

Scale marks visible: \(10\), \(20\), \(30\), \(40\), \(50\), \(60\), \(70\).

Auxiliary inset: \(h_0 = 10\), \(\gamma\), \(\gamma_1\).

Fig. 12

Diagram with labels: \(a(\omega)\), \(b(\omega)\), \(a_\omega\), \(S\), \(S_2\), \(a\), \(b\), \(\gamma\), \(30\), \(40\), \(50\), \(60\), \(H = 40\), \(50\), \(40\), \(30\), \(20\), \(10\), \(10\), \(20\), \(50\).

Auxiliary inset: \(h_0 = 10\), \(\gamma\), \(\gamma_1\).

Handwritten note: [[unclear: handwritten Russian note near the lower left of Fig. 12]].

Fig. 13

Diagram with labels: \(M\), \(a(\omega)\), \(b(\omega)\), \(h = 10\), \(\beta\), \(\gamma\), \(M'\).

Scale marks visible: \(120\), \(130\), \(140\), \(142.5\), \(150\), \(170\).

Fig. 14

Diagram with labels: \(a(105)\), \(b(105)\), \(a_{110}\), \(b_{110}\), \(b(13)\), \(\gamma 105 = 30^\circ\), \(\gamma 110 = 29^\circ 30'\), \(\alpha = 120^\circ\), \(\beta = 120^\circ\), \(n\), \(m\).

Scale marks visible: \(120\), \(130\), \(140\), \(150\), \(160\), \(170\), \(180\), \(190\), \(200\), \(210\), \(220\), \(240\), \(260\), \(280\), \(300\).

[[unclear: handwritten heading, possibly “Scale …”]]

Figure 15.

Figure 16.

Figure 17.

Figure 18.

Figure 19.

Figure 20.

Plane $\alpha$ — 5

Figure 22.2.

Figure 22.2a.

Figure 22.2b.

Figure 22.2c.

Figure 22.2d.

Figure 22.2e.

Figure 22.2f.

Figure 22.2g.

Figure 22.2h.

Figure 22.25.

Construction 6.

Fig. 28.

Fig. 29.

Fig. 30.

Fig. 31.

Fig. 32.

Fig. 33.

Fig. 34.

Fig. 35.

Fig. 36.

Visible labels in the diagram: \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), \(G\), \(h\), \(y\), \(\delta\), \(\delta_1\), \(\delta_2\), \(\alpha_1\), \(\alpha_2\).

Fig. 37.

Visible labels in the diagram: \(A\), \(C\), \(D\), \(F\), \(H\), \(S\), \(M_1\), \(M_2\), \(h_0=10\), \(y_1\), \(y_2\), \(y_3\), \(y\), \(z\), \(\gamma\), \(\gamma_1\), \(\gamma_2\), \(\omega_1\), \(\omega_2\), \(\omega_3\), \(\omega_4\), \(\omega_5\), \(24\,\mathrm{m}\).

Fig. 38.

Visible labels in the diagram: “[[unclear: plane/working label]]”, “[[unclear: plane/working label]]”, \(A\), \(S\), \(S_1\), \(\alpha\), \(\delta\), \(\gamma\).

Fig. 39.

Visible labels in the diagram:

• “crosscut”
• “synclines” [[unclear]]
• “clayey shale”
• “limestone”
• “sandstone”
• “[[unclear: rock label]]”
• “[[unclear: rock label]]”
• “[[unclear: rock label]]”
• \(A\), \(A'\), \(B\), \(B'\), \(C\), \(h\), \(g\), \(\delta\)

Fig. 40.

Visible labels in the diagram: \(A\), \(B\), \(B_1\), \(C\), \(D\), \(\delta\), \(\varepsilon\).

Fig. 41.

Fig. 41a. Skew.

Fig. 41b. Non-skew.

Fig. 42.

Visible labels: \(M_1\), \(M_2\), \(Vg_1\), \(Vg_2\), \(S_1\), \(S_2\), \(a\), \(b\), \(c\), \(20\), \(30\), \(40\), \(50\), \(60\), \(70\); “Profile by marks.”

Fig. 43.

Visible labels: \(A\), \(B\), \(C\), \(a\), \(b\), \(c\), \(d\), \(e\), \(f\), \(g\), \(h\), \(M\), \(N\), \(V\).

Fig. 44. Construction of \(g\).

Visible labels: \(B\), \(M_1\), \(M_2\), \(S_1\), \(S_2\), \(Vg_1\), \(Vg_2\), \(a\), \(b\), \(c\), \(d\), \(e\), \(20\), \(30\), \(40\), \(50\), \(60\); “Profile by marks.”

Fig. 45. Determination of [[unclear: handwritten word]].

Visible labels: \(A\), \(B\), \(M_1\), \(M_2\), \(S_1\), \(S_2\), \(Vg_1\), \(Vg_2\), \(a\), \(b\), \(c\), \(d\), \(e\), \(20\), \(30\), \(40\), \(50\), \(60\); “Profile by marks.”

Table 10.

Fig. 47

(diagram)

Visible inscription: Table 10.

Fig. 45

(diagram)

Visible inscription: Projection according to scale

Fig. 46

(diagram)

Visible inscription: Projection according to scale

Table II

Fig. 48.

Visible handwritten annotations in the drawing include:

• \(M_2\)
• \(M_1\)
• \(M_3\)
• \(S_1\)
• \(S_2\)
• longitudinal [[unclear]]
• transverse [[unclear]]
• \(a(0)\), \(a(1)\), \(a(2)\), \(a(n)\)
• \(m(0)\), \(m(1)\), \(m(2)\), \(m(n)\)
• \(n(0)\), \(n(1)\), \(n(2)\), \(n(n)\)
• \(l(0)\), \(l(n)\)
• \(b\)
• \(c\)
• \(d\)
• \(e\)
• \(f\)
• \(g\)
• \(h\)
• \(i\)
• \(k\)
• \(u\)
• \(v\)
• \(x\)
• \(y\)
• \(z\)