Principles of the Method of Numerical Marks in the Simplest Form of Parallel Projections

P. A. Maslennikov

Purpose and essence of the method

The principal purpose of the method of projections with numerical marks is to construct images of spatial forms on a single projection plane and, from the images obtained, to solve various geometrical problems. Among the problems solved by this method in its practical application, one must first of all include problems of constructing images of the earth’s surface, of surface earth structures and workings, of underground workings, and the solution of problems connected with determining the true magnitude of various geometrical forms from their distorted images.

Fig. 1

In order to clarify the essence of the method with numerical marks, let us imagine a projection plane coinciding with some horizontal plane \(H\). Let us take in space some point \(T\), belonging to the form being represented, and, parallel to a previously specified direction \(S\), project the spatial point \(T\) onto the projection plane \(H\). The point of intersection \(t\) of the straight line passing through the spatial point \(T\) parallel to the specified direction \(S\) with the projection plane \(H\) will be the image (projection) of the spatial point \(T\). Next to the point \(t\) is written a number \(n\), expressing the elevation or depression of the represented point \(T\) relative to the projection plane \(H\). It is not difficult to see (Fig. 1) that, from the given projection \(t\), the known direction of projection, and the given number \(n\) (the mark of the point \(T\)), it will be possible to reconstruct the spatial position of the point \(T\).

Projections of a system of rectangular axes

Let us imagine a system of spatial rectangular axes \(OX\), \(OY\), and \(OZ\). Let the coordinate axes \(OX\) and \(OY\) lie in the horizontal projection plane \(H\), and let the axis \(OZ\) be perpendicular to it.

If one projects, parallel to a given direction \(S\), a system of rectangular axes \(OX\), \(OY\), and \(OZ\) onto the horizontal plane of projections \(H\) (Fig. 2), then, since the coordinate axes \(OX\) and \(OY\) lie, by assumption, in the plane of projections \(H\), their images in the plane of projections \(H\) will obviously be the coordinate axes themselves. Consequently, segments taken on the coordinate axes \(OX\) and \(OY\) will be represented in their true magnitude; that is, the images of these segments will undergo no distortion, and likewise all segments of lines and the angles between them lying in the plane \(XOY\), or in a plane parallel to it, will undergo no distortion. The projection \(OZ_1\) of the coordinate axis \(OZ\), as is not difficult to see from Fig. 2, will in general make with the image of the coordinate axis \(OX\) not a right angle, but a certain angle \(\alpha_z\), depending on the angle that the given direction of projection \(S\) makes with the coordinate axis \(OX\). From the right triangle \(ZOZ_1\) it is seen that \(OZ_1 = OZ \cotg \varphi\); that is, the linear magnitude of the image \(OZ_1\) of a segment of a line on the axis \(OZ\) is equal to the true magnitude of the segment on the axis \(OZ\), multiplied by the cotangent of the angle which the direction of projection makes with the horizontal plane of projections. Thus, the cotangent of the angle \(\varphi\) serves as the scale of distortion of segments of lines situated on the axis \(OZ\), or parallel to it.

Fig. 2

Denoting the scale of distortion along the axis \(OZ\) by \(m_z\), that is, putting

\[ m_z = \cotg \varphi \]

and assigning various values to the angle \(\varphi\), we shall obtain various values of the scale of distortion \(m_z\).

In the general case, for \(90^\circ \ne \varphi \ne 45^\circ\), there will be obtained the so-called oblique projections of the simplest kind of the dimetric class, for which the scales along the coordinate axes \(OX\) and \(OY\) are equal to unity, while along the axis \(OZ\), \(m_z \ne 1\). For the special values \(\varphi = 90^\circ\) and \(\varphi = 45^\circ\), the scale of distortion \(m_z\) along the axis \(OZ\) will have the values \(m_z = 0\) and \(m_z = 1\); in the first case an orthogonal projection will be obtained, and in the second—an oblique projection of the simplest kind of the isometric class.

Projections of a Point

If, for some spatial point \(T\), its rectangular coordinates \(x\), \(y\), and \(z\) are given, then the construction of its image \(t\) in projections with numerical marks is, in the general case, carried out in the following manner. From the image \(O_1\) of the origin of coordinates \(O\) we lay off along the axes \(O_1X_1\) and \(O_1Y_1\) the segments \(O_1a = x\) and \(O_1b = y\); from the obtained points \(a\) and \(b\) we draw straight lines parallel to the axes \(O_1Y_1\) and \(O_1X_1\), respectively, until they intersect at the point \(t_0\); from the point \(t_0\) we draw a straight line parallel to the axis \(O_1Z_1\), of length \(t_0t = z_1 = m_z \cdot z\). The point \(t\) obtained by this construction will be the dimetric projection of the point \(T\).

Putting next to the point \(t\) the number \(n=z\), expressing the magnitude of the elevation or lowering of the spatial point \(T\) relative to the plane \(XOY\), which is still taken to lie in the horizontal plane, we obtain a projection with the numerical mark \(t_{(n)}\) of the spatial point \(T\) (Fig. 3).

Fig. 3

It is not difficult to see that, in order to determine the spatial position of the point \(T\) from its projection with the numerical mark \(t_{(n)}\), precisely the same constructions must be carried out, but in the reverse order.

Fig. 4

For constructing the image of a point in the simplest form of projection of the isometric class (Fig. 4), instead of \(z_1=m_z\cdot z\), we lay off from the point \(t_0\) on the straight line \(t_0t\) the undistorted magnitude of the mark \(z=n\).

Projections of a straight line

The projection of a straight line in the general case is a straight line, but, unlike orthogonal projections with numerical marks, in which the length of a straight-line segment on the projection is less than or equal to the length of the straight line being represented, in the present form of projection the image of a segment of a straight line may be less than, equal to, or greater than the represented segment of the straight line. Thus:

Since oblique and orthogonal projection belong to the same kind of projection (parallel), it follows that the rules for graduating a straight line and for determining the relative position of a point with respect to a straight line and of two straight lines in oblique parallel projections with numerical marks will be exactly the same as in projection in orthogonal projections with numerical marks. The difference will lie only in the method of determining the elements of the position of a straight line (the spread \(\alpha\) and the fall \(\delta\)) and in the method of determining the true length of a segment of a straight line from its oblique-angle image with numerical marks.

Thus, if \(t_n u_m\) (Fig. 5) is the oblique-angle image with numerical marks of the segment \(TU\) of some straight line, in which the point \(T\) is at a distance of \(n\) units from the plane \(XOY\) (horizontal), and the point \(U\) is at a distance of \(m\) units, and if, moreover, \(m>n\), then, in order to determine the elements of the position and the true length of this segment, one must:

1) from the point \(u_m\) draw the straight line \(u_m u_n\), equal in length to the difference \((m-n)\) of the marks of the end points of the segment of the straight line and parallel to the axis \(Z_1O_1\);

2) join by the straight line \(t_n u_n\) the points \(t_n\) and \(u_n\);

3) from the point \(u_n\) to the straight line \(t_n u_n\) erect the perpendicular \(u_n u_n'\), of length equal to the difference \((m-n)\) of the marks of the end points of the segment;

4) join by the straight line \(t_n u_n'\) the points \(t_n\) and \(u_n'\).

The length of the straight line \(t_n u_n'\) will express the true length of the segment \(TU\); the angle at the point \(t_n\) between the straight lines \(t_n u_n'\) and \(t_n u_n\) will be the angle of fall \(\delta\) of the straight line \(TU\), while the angle measured clockwise from the positive direction of the axis \(O_1X_1\) to the straight line \(u_n t_n\) will be the spread \(\alpha\) of the straight line (as the direction of the straight line, the direction toward its fall is taken).

Fig. 5

To construct a straight line (Fig. 5) passing through a given point \(T\), at a distance of \(n\) units from the plane \(XOY\), from its elements of position (the spread \(\alpha\) and the fall \(\delta\)), one must:

1) from the point \(t_n\) draw the straight line \(t_n u_n\) according to the given spread \(\alpha\) and the straight line \(t_n u_n'\), making the angle \(\delta\) with the former;

2) on the straight line \(t_n u_n'\), from the point \(t_n\), set off a segment equal to the true length \(L\) of the segment of the straight line \(TU\);

3) from the point \(u_n'\), drop the perpendicular \(u_n'u_n\) to the line \(t_nu_n\);

4) from the foot of the perpendicular \(u_n\), draw the line \(u_nu_m\), parallel to the axis \(O_1Z_1\), and lay off on it the segment \((m-n)=u_nu_m\);

5) connect the point \(u_m\), obtained by this construction, by a straight line with the point \(t_n\); the segment \(t_nu_m\) will be the required representation.

In the case when the length of a segment of a line is unknown, the construction of the oblique projection with numerical marks for the line \(TU\), passing through the given point \(T\), is carried out from the elements of inclination (\(\alpha\) and \(\delta\)) of this line in the manner just described, but with the difference that on the line \(t_nu_n\) there is laid off from the point \(t_n\) not the given length of the segment, but an arbitrary length \(L\), which makes it possible to find a second arbitrary point of the line \(TU\) and its mark.

The rules given for constructing and determining the elements of inclination of a line apply to projections of the isometric class; if, however, the projection is dimetric, then the distance between the points \(u_m\) and \(u_n\) should be taken equal to \(m_z(m-n)\).

Projection of a Plane

If one adheres to the accepted method of representing a plane in the form of horizontals, then, in order to construct their oblique parallel representation with numerical marks from the given elements of inclination of the plane—fall \(\delta_p\) and strike \(\alpha_p\)—under the condition that the plane passes through some point of space \(T\), standing away from the plane \(XOY\) at a distance of \(n\) units, one proceeds as follows. Through the projection \(t_n\) of the point \(T\) (Fig. 6)

Fig. 6

draw the line \(t_nv_n\), making the given angle \(\alpha_p\) with the positive direction of the axis \(O_1X_1\), and the line \(t_nu\) (the line of fall), making in space with the plane \(XOY\) the angle \(\delta_p\) and having a strike azimuth equal to \(\alpha_p+90^\circ\). Then, after graduating the line \(t_nu\), through the obtained division points \(n-1\), \(n-2\), \(n-3\), etc., draw straight lines which are oblique-angle representations of the horizontals of the plane, parallel to the line \(t_nv_n\) drawn.

The horizontals of the plane constructed in oblique projection completely determine its spatial position. It is also true that if the lines \(t_nv_n\), \(t_{n+1}v_{n+1}\), \(t_{n+2}v_{n+2}\ldots\) are the representation in oblique projection

horizontals of the plane (Fig. 7), then, in order to determine the strike, it will suffice to measure the angle formed by these horizontals with the positive direction of the axis \(O_1X_1\). To determine the angle of dip \(\delta_p\) from some point \(k_{n+2}\) of a certain horizontal \(t_{n+2}v_{n+2}\), parallel to the axis \(O_1Z_1\), we draw the straight line \(k_{n+2}k_{n+1}\), whose length \(s\) is equal to the difference of the elevations of two adjacent horizontals (in the present case \(s=1\)). From the obtained point \(k_{n+1}\), having an elevation one unit less than the elevation of the chosen point \(k_{n+2}\), we drop to the horizontal \(t_{n+1}v_{n+1}\) the perpendicular \(k_{n+1}p_{n+1}\). Since the segment \(k_{n+1}p_{n+1}\) lies in a horizontal plane, it therefore expresses, in projection, the true magnitude of the setting-out of a segment of length \(l\), whose difference of elevations is equal to unity. Finally, from the point \(k_{n+1}\) to the straight line \(k_{n+1}p_{n+1}\), we erect the perpendicular \(k_{n+1}q_{n+1}\), whose length is equal to unity, and join the obtained point \(q_{n+1}\) with the point \(p_{n+1}\) by the straight line \(q_{n+1}p_{n+1}\). The angle at the point \(p_{n+1}\), formed by the straight lines \(p_{n+1}q_{n+1}\) and \(p_{n+1}k_{n+1}\), will be the required angle of dip of the plane \(\delta_p\).

Fig. 7

All the remaining problems, such as: the construction of a plane 1) from three given points not lying on one straight line; 2) from a straight line and a point not lying on this straight line; 3) from two parallel straight lines; 4) from two intersecting straight lines; the determination of the mutual position of planes; the construction of the line of intersection of planes; the determination of the point of meeting of a straight line with a plane, etc., except for perpendicular planes and lines, are solved in exactly the same way as in orthogonal projections with numerical marks.

The construction of the horizontals of a plane and the determination of its setting-out elements in the present case were carried out in isometric projections; in the case where these problems must be solved in dimetric projections, the distortion along the axis \(OZ\) must be taken into account in the same way as when solving problems involving a straight line.

Construction of a Perpendicular to a Plane

Let us imagine, in an oblique-angled projection, a plane given by the horizontals \(u_nv_n\), \(u_{n+1}v_{n+1}\), \(u_{n+2}v_{n+2}\), etc., and a point \(t_n\) lying outside this plane (Fig. 8).

If from the point \(t_n\) a perpendicular \(t_nk_n\) is dropped to the horizontal \(u_nv_n\), then the corresponding line in space \(TK\) will also be perpendicular to the corresponding horizontal of the plane in space, since

horizontal straight lines \(t_n k_n\) and \(u_n v_n\) are projected in this case without distortion. The angle of incidence of the perpendicular to the plane will be equal to the complement to \(90^\circ\) of the angle of incidence of the plane; that is, if the angle of incidence of the plane is \(\delta_p\), then the angle of incidence of the perpendicular to the plane must be equal to \(90^\circ-\delta_p\) (Fig. 9).

Fig. 8

In order to draw a perpendicular to the plane, meeting the plane at the given point \(k_n\), as is not difficult to see from Fig. 9, it is necessary to raise the point \(t_n\) along the \(OZ\) axis by the height \(\Delta z\), determined by the formula

\[ \Delta z = l_0 \operatorname{tg}(90^\circ-\delta_p), \]

where \(l_0\) is the horizontal projection of the perpendicular, equal to the length of the straight line \(t_n k_n\).

Fig. 9

In the projection this raising will be expressed by displacement of the point \(t_n\) (Fig. 8) by the amount \(m_z \Delta z\), if the projection is dimetric, and by \(\Delta z\), if the projection is isometric, along the straight line passing through this point parallel to the axis \(Z_1O_1\).

After the preceding considerations it will not be difficult to derive also the corresponding rule for constructing a perpendicular to a plane in the simplest form of parallel projections of the dimetric and isometric classes with numerical marks.

Thus, if it is required to draw through the point \(r_m\) a perpendicular to a plane given by the horizontals \(u_n v_n,\ u_{n-1}v_{n-1},\ u_{n-2}v_{n-2}\), etc., it is necessary (Fig. 10):

1) through the point \(r_m\), parallel to the axis \(O_1Z_1\), to draw the straight line \(r_m t_n\), whose length is equal to the product of the distortion scale \(m_z\) by the difference of the marks \((n-m)\) of the given point and of any one of the horizontals of the plane;

2) from the obtained point \(t_n\), drop to the horizontal \(u_n v_n\) the perpendicular \(t_n k_n\) and measure its magnitude \(l_0\);

3) determine the angle of fall of the plane \(\delta_p\);

4) lay off from the point \(t_n\), in the direction from \(t_n\) toward \(r_m\), the segment \(t_n r_{n-\Delta z} = m_z \Delta z\), where \(\Delta z = l_0 \operatorname{tg}(90^\circ-\delta_p)\);

5) join by a straight line the points \(r_{n-\Delta z}\) and \(k_n\);

Fig. 10

6) parallel to the straight line \(r_{n-\Delta z} k_n\), through the point \(r_m\), draw the straight line \(r_m q_p\), which will express the perpendicular to the plane drawn through the point \(r_m\).

Instead of computing the magnitude \(\Delta z\), in the case of an isometric projection one may use the following graphical determination of it. At the point \(k_n\) of the straight line \(t_n k_n\) construct the straight line \(k_n r'_{n-\Delta z}\), forming with the straight line \(t_n k_n\) the angle \(90^\circ-\delta_p\), and from the point \(t_n\) erect a perpendicular to the straight line \(t_n k_n\); the right triangle \(t_n k_n r'_{n-\Delta z}\) thus constructed will have one leg equal to \(l_0\), and the other, \(t_n r'_{n-\Delta z}\), equal to the sought magnitude \(\Delta z\).

Superposition

To determine the true size of plane figures, as is known, the most suitable method is that of superposing the plane containing the given figure with the horizontal plane or with some other plane of projection.

In the present case, in order to obtain the true size of the figure, the superposition will have to be carried out with the horizontal plane, since the plane of projection of parallel projecting has been chosen by us parallel to the horizontal plane.

By superposition of a plane is usually meant its rotation about one of the horizontals until it coincides with the horizontal plane passing through the horizontal that serves as the axis of rotation. Since the horizontals of a plane are horizontal straight lines and at the same time are parallel to one another, it follows that, under parallel projecting, they are represented likewise by parallel straight lines. When the plane is rotated about one of the horizontals, the latter will remain in the same place throughout, while all the other horizontals of the plane will move parallel to themselves.

Thus, in the superposed position the horizontals of the plane remain parallel to their original position, but the distances between them will, obviously, be different; thus, if \(l_1\) is the distance between two adjacent horizontals on the projection, then in the superposed position these horizontals must be separated from one another by a quantity \(l\), equal to

Fig. 11

the distance between two adjacent horizontals in space, and consequently, when superposing the plane, one must first of all determine the quantity \(l\).

Let the plane (Fig. 11) be given by its horizontals \(u_k v_k\), \(u_{k+n} v_{k+n}\), \(u_{k+2n} v_{k+2n}\), etc., and let it be required to superpose it with the horizontal plane passing through the horizontal \(u_k v_k\).

For this purpose we choose on the horizontal \(u_{k+n} v_{k+n}\) some point \(t_{k+n}\), and through it, parallel to the axis \(O_1 Z_1\), draw the straight line \(t_{k+n} t_k\), whose length is equal to the difference of the marks of two adjacent horizontals (in the given case this length

Fig. 12

in space is equal to \(n\), and on the projection—\(m_z n\), if it is dimetric, and \(n\), if the projection is isometric). From the obtained point \(t_k\) on the horizontal \(u_k v_k\) we drop the perpendicular \(t_k r_k\), from the foot of which \(r_k\) we lay off on the horizontal \(u_k v_k\) a segment of length \(m_z n = r_k q_k\). The distance between the points-

marks \(t_k\) and \(q_k\) will express the true value \(l\) of the distance between two adjacent horizontals. Having erected from the point \(q_k\) to the horizontal \(u_kv_k\) a perpendicular, we lay off on the latter equal segments \(l\), and through the obtained division points \(q_{k+n}\), \(q_{k+2n}\), etc., we draw straight lines parallel to the axis of rotation; these will express the superposed position of the horizontals of the plane.

Example. It is required to determine the true value of the angle between two straight lines \(ac\) and \(bc\) (Fig. 12).

Solution. Having graduated the given straight lines, we draw the horizontals of the plane. We superpose the plane of the angle by rotation about the horizontal passing through the point \(c\). From the points \(a\) and \(b\) we draw perpendicular lines to the horizontals, until they meet at the points \(a_0\) and \(b_0\) the superposed position of the horizontals having marks equal to the marks of the points \(a\) and \(b\). We connect the points \(a_0\) and \(b_0\) with the point \(c\) by straight lines. We measure the angle \(w\) at the point \(c\) between the straight lines \(ca_0\) and \(cb_0\). The angle \(w\) will be the required one.

Conclusion

As can be seen from the basic examples given for the solution of various geometric problems and for the construction of an image with numerical marks under parallel projection in the simplest form of the dimetric and isometric classes, the majority of problems are solved in exactly the same way as in the case of applying the method with numerical marks in orthogonal projections. The difference between the method presented here with numerical marks and the method of numerical marks in orthogonal projections arises only in determining the true value of distorted forms; moreover, in this case the final operations for solving the given problems are identical with the method of numerical marks in orthogonal projections.

Comparing the method presented here with the method of numerical marks in orthogonal projections, it is not difficult to see that, when constructing an image in the simplest form of projections of the dimetric and isometric classes, first of all the orthogonal projection of the spatial form being represented is constructed, and then, from the obtained orthogonal projection, the parallel dimetric or isometric projection is constructed. Conversely, when determining the true value of a form from its parallel projection, first of all a transition is made to the orthogonal representation of the form, and then, from the obtained orthogonal projection, the sought true value is determined.

It is not difficult to see that projections of the simplest form and of the dimetric and isometric classes, with a distortion scale along the coordinate axis \(OZ\) equal to zero, that is, with \(m_z = 0\), turn into orthogonal projections, and, consequently, the method of numerical marks in orthogonal projections is a special case of the method presented in the present work.

The existing view of axonometric images of spatial forms as expensive ones, not permitting the solution of metric problems, is explained first of all by the absence of a method convenient for them that would make it possible, from an axonometric image, quickly to determine the dimensions and mutual arrangement of the projected forms.

The method of numerical marks in axonometric images proposed in the present work solves all the basic metric problems and, consequently, it can be applied in constructing pictorial axonometric images.

The principal field in which this method may be used is the mapping of mineral workings, especially of thick and steeply dipping deposits. The plan of mineral workings

of mineral deposits, constructed in axonometric representation on a horizontal plane with numerical marks, will be easy to measure and sufficiently clear. In this connection it should be taken into account that the method proposed by the author presupposes oblique parallel projection onto the horizontal plane of projections, that is, a projection which will not distort the horizontal extents of the lengths of workings, the horizontal extents of the lengths of sides, and the horizontal angles of the polygons being laid out. Therefore such axonometric representations of mineral-deposit workings will be not an expensive supplement to the existing orthogonal representation (plan), but the principal plan, since no special calculations are required for its construction and supplementation. The horizontal extents of lines and measured horizontal angles are constructed on such a plan by the usual methods, with the aid of a scale ruler and a protractor, while direction angles retain their magnitude. In constructing an axonometric plan of workings from a finished orthogonal one, a pantograph may be used.