Application of the Most Advantageous Design Method to the Construction of Axonometric Projections

L. S. Skripov

(Presented by the scientific seminar of the Department of Descriptive Geometry and Graphics)

As is known from the theory of descriptive geometry, axonometric projections are obtained by projecting onto a single plane by central, orthogonal, and oblique methods.

In descriptive geometry the laws of such projections are analyzed in detail and are, in essence, rather laborious. The method of most advantageous design makes it possible to simplify the construction of axonometric images, reducing them to simple geometric constructions while using only two orthogonal projections (horizontal and frontal) of the given object.

The present article considers techniques for such constructions by means of the method of most advantageous design; moreover, the axonometric projections obtained theoretically correspond to the requirements with allowance for the necessary distortion coefficients, that is, for orthogonal isometry \(p=q=r=0.82\), for orthogonal dimetry \(p=r=0.94,\ q=0.47\), for frontal oblique dimetry \(p=r=1,\ q=\frac{1}{2}\), and for military perspective \(p=q=r=1\).

I. Orthogonal Isometric Projection

The isometric projection of any object is obtained by parallel-orthogonal projection of it onto a plane \(P\), equally inclined to the projection planes \(H, V, W\) and intercepting equal segments on the coordinate axes, i.e. \(OX_1=OY_1=OZ_1\) (Fig. 1).

The axonometric triangle formed by the traces of the axonometric plane \(P_H, P_V, P_W\) is, in this case, equilateral, with angles of \(60^\circ\) at the vertices, i.e. \(Z_1X_1Y_1=X_1Y_1Z_1=Y_1Z_1X_1=60^\circ\), and the traces of the plane \(P\) form angles of \(45^\circ\) with the corresponding coordinate axes, i.e.

\[ \angle OX_1Y_1=\angle OY_1X_1=\angle OY_1Z_1=\angle OZ_1X_1=\angle OX_1Z_1=45^\circ . \]

The application of the method of most advantageous design to the case under consideration is easier to analyze first for the projection of a single point onto the plane \(P\), and then to extend the regularities already found to all points and lines of the body being projected.

Let there be some point \(A\) in space which is to be projected orthogonally onto the plane \(P\), and let its projection \(a_1\) be found on this plane.

The ray projecting the point \(A\) will be perpendicular to the plane \(P\), and consequently its projections are perpendicular to the corresponding traces of the plane \(P\), i.e. \(\angle a'2'Z_1=\angle a1Y_1=90^\circ\). For convenience, the point \(A\) has been chosen so that the ray projecting it is directed toward the center of the coordinate axes \(O\), although, of course, it may occupy any position in space.

Having imagined a projecting plane passing through the ray projecting the point \(A\) and perpendicular to the coordinate plane \(H\) (the hatched area), we conclude that this plane will be perpendicular to the planes \(H\) and \(P\), which form a dihedral angle with edge \(P_H\). By the laws of geometry, the line of intersection of this plane with \(P\) will be perpendicular to the edge \(P_H\) (the trace), that is, the angle \(Y_1 1 Z_1\) will be equal to \(90^\circ\). The point \(a_1\) of intersection of the line \(1Z_1\) with the ray will be the required projection of the given point \(A\) on the axonometric plane \(P\).

Fig. 1.

The same reasoning could have been carried out by imagining a projecting plane passing through the ray, perpendicular to the frontal plane of projections \(V\); one would likewise arrive at the conclusion that the line of intersection of the new projecting plane with \(P\) will be perpendicular to the edge (trace) \(P_V\) of the second dihedral angle, i.e. \(\angle Y_1 2' Z_1=90^\circ\).

It follows from Fig. 1 that the projection \(a_1\) of the point \(A\) can be found by the following simple construction. From point \(1\), the point where the horizontal projection of the ray meets the trace \(P_H\), a perpendicular to \(P_H\) should be erected in the plane \(P\); and from point \(2'\), the point where the frontal projection of the ray meets the trace \(P_V\), a perpendicular to \(P_V\) should be erected in the plane \(P\). The point \(a_1\) of intersection of the two perpendiculars will be the required axonometric projection of the given point \(A\).

Let us now repeat the constructions just carried out on a plane drawing, when the axonometric plane \(P\) is brought into coincidence with the plane \(H\) by rotation about its horizontal trace \(P_H\), and let us find the superposed position of the projection \(a_1\) of the given point \(A\).

Let two projections of some point \(A\) be given—the horizontal projection \(a\) and the frontal projection \(a'\)—and let the axonometric plane \(P\) be given by the traces \(P_H\) and \(P_V\), which meet at the point \(P\) on the axis of projections \(OX\) (Fig. 2). Since the projection is orthogonal, we construct the projections of the ray passing through \(a\) and \(a'\), perpendicular to the corresponding traces of the planes \(P_H\) and \(P_V\). From the drawing we conclude that the projections of the ray will be inclined to the axis \(OX\) at the angles \(\varphi_0\) and \(\psi_0\) of \(45^\circ\) (or to lines parallel to the axis \(OX\)), since the traces \(P_H\) and \(P_V\) are also inclined to it at an angle of \(45^\circ\).

We mark the points \(1\) and \(2'\) where the projections of the ray intersect the traces \(P_H\) and \(P_V\). We bring the plane \(P\) into coincidence with the plane \(H\) and construct the superposed position \(P_{V_1}\) of the frontal trace \(P_V\). We determine the new position \(2_1\) of the point \(2'\) on the superposed trace \(P_{V_1}\) by an arc drawn from \(P_X\) as center, with radius equal to \(P_X2'\). From the points \(1\) and \(2_1\), on the superposed plane \(P\), we draw perpendiculars to the traces \(P_H\) and \(P_{V_1}\) until they mutually

intersection at point \(a_1\), which will be the desired axonometric projection of point \(A\) on the superposed plane \(P\).

In this way one can find on \(P\) the projection of any points of space, as well as of the lines connecting them. However, with such projection the image of the object on plane \(P\) is obtained rotated clockwise relative to the observer, and for a clear understanding of it the drawing has to be rotated in the opposite direction so that the vertical lines occupy a vertical position before the observer.

For eliminating this drawback it is sufficient to rotate in the drawing the superposed plane counterclockwise, and together with it the projections of all points.

About point \(P_x\) the traces \(P_H\) and \(P_V\) are rotated through an angle \(\alpha\) so that the trace \(P_H\) coincides with the projection axis \(OX\) and takes the position \(P_{H_1}\). Through the same angle \(\alpha\) the trace \(P_{V_1}\) is rotated to the position \(P_{V_2}\). Under such a rotation, points 1 and \(2_1\) on the traces \(P_H\) and \(P_{V_1}\) pass into the positions \(1_1\) and \(2_2\). From them perpendiculars are drawn (as before) to the traces \(P_{H_1}\) and \(P_{V_2}\) until they intersect at point \(a_2\), which will be the new position of the axonometric projection of the given point \(A\). The new position of the projection could also have been found by rotating the projection \(a_1\) through the angle \(\alpha\).

Fig. 2.

The construction carried out can be simplified, avoiding the rotation of the superposed plane, on the basis of the following considerations.

Let us extend the horizontal projection of the ray to its mutual intersection with the extension of the perpendicular \(1_1a_2\) at point 3. Through point 3 and the convergence point of the traces \(P_x\) we draw the line \(B_1\) and obtain two right triangles \(3—1_1—P_x\) and \(3—1—P_x\), equal to one another, since they have the common hypotenuse \(3P_x\) and equal legs \(1—P_x\) and \(1_1—P_x\) by construction. Consequently, the corresponding acute angles of the triangles are equal, i.e. \(\angle 1_1P_x3=\angle 1P_x3=\varphi\) and \(\angle 13P_x=\angle 1_13P_x\); hence the line \(B_1\) is the bisector of the angle \(\alpha\) through which the trace \(P_H\) was rotated into the new position \(P_{H_1}\).

Let us likewise extend the frontal projection of the ray to its meeting at point 4 with the extension of the perpendicular to the trace \(P_{V_2}\) from point \(2_2\), to which point 2 moved after the rotation through the angle \(\alpha\) of the trace \(P_{V_1}\). Through points 4 and \(P_x\) we draw the straight line \(B_2\) and obtain two equal right triangles \(4—2'—P_x\) and \(4—2_2—P_x\), since they have the common hypotenuse \(4P_x\) and one equal leg \(2'P_x=2_2P_x\) by construction. Consequently, their corresponding acute angles are equal; for example, at point 4, \(\angle 2_24P_x=\angle 2'4P_x=\psi_1\). Hence the line \(B_2\) will be the bisector of the angle between the traces \(P_V\) and \(P_{V_2}\) and is inclined to the axis \(OX\) at the angle \(\psi\).

Thus, finding the isometric projection of the given point \(A\) reduces to the following constructions:

1. On the axis \(OX\), the point of convergence of the traces \(P_x\) is chosen at any place, but in such a way that the construction does not go beyond the limits of the first octant.

2. From this point the lines \(B_1\) and \(B_2\) are drawn at the angles \(\varphi\) and \(\psi\) to the projection axis \(OX\).

3. From the projections of the given point \(a\) and \(a'\), the projections of the projecting ray are drawn at the angles \(\varphi_0=\psi_0=45^\circ\) to the axis \(OX\), up to their intersection with the bisectors \(B_1\) and \(B_2\) at points 3 and 4.

4. From point 3 a vertical line is drawn, perpendicular to \(P_H\) (or, otherwise, to the axis \(OX\)); and from point 4—a line at the same angle \(\psi_1\) to the bisector \(B_2\) at which the frontal projection of the ray was drawn.

5. At the intersection of the constructed lines, the required isometric projection of the given point \(A\)—\(a_2\)—will be found.

It is therefore necessary to know the magnitudes of the angles \(\varphi\) and \(\psi\). The angle

\[ \varphi=\frac{45^\circ}{2}=22^\circ 30'. \]

The angle \(\psi\) is determined from the equality

\[ \psi=\frac{60^\circ-45^\circ}{2}+45^\circ=52^\circ 30'. \]

Example 1. Construct the rectangular isometric projection of the prism 1, 2, 3, 4, 5, 6, 7, 8, given by its two projections (horizontal and frontal), Fig. 3.

The problem is solved in the following order:

1. The projections \(s\), \(s'\) of the projecting ray \(S\) are constructed at angles of \(45^\circ\) to the axis \(OX\) (or to lines parallel to it).

2. An arbitrary point \(P_x\) is chosen on the axis \(OX\) so that the isometry is obtained in the first octant.

3. From the point \(P_x\), the bisectors \(B_1\) and \(B_2\) are drawn at the angles \(\varphi=22^\circ 30'\) and \(\psi=52^\circ 30'\) to the axis \(OX\).

4. From the frontal projections of the vertices of the prism \(1'\), \(2'\), \(3'\), \(4'\), \(5'\), \(6'\), \(7'\), \(8'\), projecting rays are drawn parallel to \(s'\) until they intersect the bisector \(B_2\) at the points \(1_0'\), \(2_0'\), \(3_0'\), \(4_0'\), \(5_0'\), \(6_0'\), \(7_0'\), \(8_0'\).

Fig. 3.

1. From the points thus obtained on the bisector \(B_2\), projecting lines are drawn at the same angle \(\psi\) to the bisector as the projecting rays from the points \(1'\), \(2'\), \(3'\), \(4'\), \(5'\), \(6'\), \(7'\), \(8'\).

2. From the horizontal projections of the vertices of the prism 1, 2, 3, 4, 5, 6, 7, 8, projections of rays are drawn parallel to \(s\), until they meet the bisector \(B_1\) at the points \(1_0\), \(2_0\), \(3_0\), \(4_0\), \(5_0\), \(6_0\), \(7_0\), \(8_0\).

3. From these points one should draw vertical lines until they meet the projecting lines drawn from the corresponding projections of the points on the bisector \(B_2\), but a convenient placement of the axonometric image is not always obtained, since sometimes it is superposed on the given projections or is located very close to them; therefore it is desirable to move it to the right, onto the free field of the drawing.

For this purpose, the lower bisector is moved parallel to itself upward in the drawing to the required position \(B_3\), and the horizontal projections of the rays from the vertices of the prism are brought to intersection with it at the points \(1_0, 2_0, 3_0, 4_0, 5_0, 6_0, 7_0, 8_0\).

1. From the points obtained, vertical lines are drawn as far as their intersection with the corresponding lines of the vertical projections, and the points of their intersection \(1_1, 2_1, 3_1, 4_1, 5_1, 6_1, 7_1, 8_1\) are marked.

2. The points obtained are connected by straight lines: continuous lines for visible edges, and dashed lines for invisible ones (in axonometry, invisible lines are usually not shown).

2. Rectangular dimetric projection

A rectangular dimetric projection of an object is obtained by parallel projection of it onto a plane \(P\), inclined at equal angles to the horizontal and profile planes of projection and at a different angle to the frontal plane of projection. As is known from the theory of descriptive geometry, in this case the axonometric plane cuts off equal segments \(OX_1\) and \(OZ_1\) on the coordinate axes \(OX\) and \(OZ\) (let us assume them equal to 1), and on the axis \(OY\) a segment \(OY_1\) equal to \(0.377\,OX_1\) (or \(OZ_1\)) (Fig. 4). The traces of the plane \(P_H, P_V, P_W\) in this case form an isosceles triangle \(X_1Y_1Z_1\) with vertex \(Y_1\) and with angles at the vertices equal to
\(\angle Z_1X_1Y_1=48^\circ35'\),
\(\angle X_1Z_1Y_1=48^\circ35'\),
\(\angle X_1Y_1Z_1=82^\circ50'\).
The traces of the plane \(P_H, P_V, P_W\) are inclined to the coordinate axes at angles equal to
\(\angle OX_1Y_1=20^\circ42'\),
\(\angle OY_1X_1=69^\circ18'\),
\(\angle OZ_1X_1=45^\circ\),
\(\angle OX_1Z_1=45^\circ\),
\(\angle OZ_1Y_1=20^\circ42'\),
\(\angle OY_1Z_1=69^\circ18'\).

Fig. 4.

The projection of some point \(A\) onto the dimetric plane is carried out in exactly the same way as onto the isometric plane and requires no repetition.

Let us consider finding the dimetric projection of some point \(A\) on the plane \(P\), brought into coincidence with \(H\), given by the traces \(P_H\) and \(P_V\), inclined to the axis \(OX\), respectively, at angles \(20^\circ42'\) and \(45^\circ\) (Fig. 5).

Since the projection is performed in parallel, the projections of the ray will be respectively perpendicular to the traces \(P_H\) and \(P_V\), and, consequently, the horizontal projection of the ray will be inclined to the axis \(OX\) at an angle of \(69^\circ18'\) (the angle complementary to \(20^\circ42'\) to \(90^\circ\)), while the frontal projection of the ray will be inclined to the axis \(OX\) at an angle of \(45^\circ\). Next, just as in isometry, the bisector \(B_1\) of the angle \(P_H P_x P_{H_1}\), inclined to the axis \(OX\) at an angle \(\varphi\), and the bisector \(B_2\) of the angle \(P_V P_x P_{V_2}\), inclined to the axis \(OX\) at an angle \(\psi\), are constructed.

The projections of the ray are extended to intersection with these bisectors: the horizontal one—to point 3, and the frontal one—to point 4, from which the construction of the projecting lines is continued; from point 3—a verti-

vertically (at an angle \(\varphi_1\) to \(B_1\)), and from point 4—at an angle \(\psi_1\) to the bisector \(B_2\), at which the frontal projection of the ray meets it. The point \(a_2\) of intersection of the constructed lines is marked; this point will be the required dimetric projection of the given point after the axonometric plane has been rotated through an angle \(\alpha\) about the point \(P_x\). It remains to determine the magnitudes of the angles \(\varphi\) and \(\psi\) of inclination of the bisectors \(B_1\) and \(B_2\) to the axis \(OX\).

The angle \(\varphi\), as is to be understood from the drawing, will be equal to:

\[ \varphi=\frac{20^\circ42'}{2}=10^\circ21', \quad \text{and the angle } \psi= \]

\[ =48^\circ35'-\frac{48^\circ35'-45^\circ}{2} =48^\circ35'- \]

\[ -\frac{3^\circ35'}{2} =48^\circ35'-1^\circ47.5'= \]

\[ =46^\circ47.5'. \]

Knowing the angles \(\varphi\) and \(\psi\), one can construct the rectangular dimetry of some object, which is done in the following example.

Fig. 5.

Example 2. It is required to construct the rectangular dimetry of the prism \(1, 2, 3, 4, 5, 6, 7, 8\), given by two projections: the frontal and the horizontal (Fig. 6).

The order of solving the problem is as follows:

1. An arbitrary point \(P_x\) is chosen on the axis \(OX\) so that the drawing is executed in the first octant.

2. From the point \(P_x\), the lines of the bisectors \(B_1\), at an angle \(\varphi=10^\circ21'\) to the axis \(OX\), and \(B_2\), at an angle \(\psi=46^\circ47.5'\) to the axis \(OX\), are drawn.

3. From the frontal projection of the vertices of the prism \(1', 2', 3', 4', 5', 6', 7', 8'\), projecting lines are drawn at an angle \(\psi_0=45^\circ\) to the axis \(OX\), as far as their intersection with the bisector \(B_2\) at the points \(1_0', 2_0', 3_0', 4_0', 5_0', 6_0', 7_0', 8_0'\), at an angle \(\psi_1\) to it.

4. From these points, projecting lines are drawn at the same angle \(\psi_1\) to the bisector.

5. From the horizontal projections of the prism (its vertices), horizontal projections are drawn

Fig. 6.

rays at an angle \(\varphi_0=69^\circ 18'\) to the axis \(OX\) until they meet the bisector \(B_1\), or the line \(B_3\), parallel to \(B_1\), chosen in the drawing so that the axonometric image is located in the free field of the drawing.

1. The points of intersection of these rays with \(B_3\) are marked—\(1_{01}, 2_{01}, 3_{01}, 4_{01}, 5_{01}, 6_{01}, 7_{01}, 8_{01}\); from these points vertical lines are then drawn until they meet the lines of the same names drawn from the points \(1'_0, 2'_0, 3'_0, 4'_0, 5'_0, 6'_0, 7'_0, 8'_0\).

2. The points of intersection of the indicated lines are marked \(1_1, 2_1, 3_1, 4_1, 5_1, 6_1, 7_1, 8_1\); these will be the dimetric projections of the vertices of the given prism.

3. The points obtained are joined to one another by straight lines, taking into account their visibility on the projections, and thus a dimetric image of the given body is obtained in a position convenient for viewing it.

3. Oblique frontal dimetry

The construction of parallel frontal oblique dimetry, strictly speaking, is not based on the method of most advantageous projection, but in essence it is one of the types of projection onto a single plane and has become widely used in constructing pictorial spatial images. Such images can be constructed fairly simply from two orthogonal projections of the given object and on the basis of the following considerations.

Let some point \(S\) \((s, s')\) be projected obliquely onto the axonometric plane \(P\), located parallel to the frontal plane of projections \(V\). For simplicity of reasoning, the direction of projection is chosen so that the projecting ray is directed toward the center of the coordinate axes \(O\) (Fig. 7), which is projected onto the plane \(P\)

Fig. 7.

at the point \(O_1\). As is known from descriptive geometry, under such projection a right angle formed by the coordinate axes \(XOZ=90^\circ\) is projected onto the plane \(P\) in its true size, i.e. the angle \(X_1O_1Z_1\) will be equal to \(90^\circ\). Segments of lines on the coordinate axes \(OX\) and \(OZ\) will also be projected onto \(P\) in their true size, i.e.

the distortion coefficients along the axonometric axes \(O_1X_1\) and \(O_1Z_1\) will be equal to \(p=1\) and \(r=1\). Segments along the \(OY\) axis may be projected in different magnitudes depending on the direction of projection, and the ratio of the segments \(O_1Y_1 : OY_1\) may be taken arbitrarily. In drafting practice, however, it is customary to take it equal to \(1:2\), which, in addition to simplifying the constructions, gives an image of the object that is close to the image perceived by the human eye.

Let us further assume that the chosen direction of projection gives this value of the distortion coefficient along the axonometric axis \(O_1Y_1\), i.e., \(q=0.5\).

Next suppose that the chosen direction of projection gives an angle of inclination of the axis \(O_1Y_1\) to the extension of the axis \(O_1X_1\) equal to \(45^\circ\) (again, this is recommended by drafting practice).

The axonometric center \(O_1\) is then projected onto the plane \(H\) at the point \(O_2\) on the horizontal projection of the ray \(S\).

From the right triangle \(O_1Y_1O_2\) it follows that \(O_2Y_1 = O_1Y_1 \cdot \cos 45^\circ\), and from the right triangle \(OY_1O_2\), that \(OY_1 = O_2Y_1 \cdot \tg \varphi_0\). Substituting into this last equation the expression for \(O_2Y_1\) from the first equality, we have \(OY_1 = O_1Y_1 \cdot \cos 45^\circ \cdot \tg \varphi_0\). Hence we obtain

\[ \tg \varphi_0 = \frac{OY_1}{O_1Y_1 \cdot \cos 45^\circ}. \]

Since it was assumed that

\[ \frac{O_1Y_1}{OY_1}=\frac{1}{2}, \]

we have

\[ \tg \varphi_0 = \frac{2}{\cos 45^\circ} = \frac{2}{\frac{1}{2}\sqrt{2}} = \frac{4}{\sqrt{2}} \simeq 2.825, \]

whence we obtain for the angle

\[ \varphi_0 = 70^\circ 35'. \]

Thus, in order to construct the oblique dimetric projection of some point \(S\) on the plane \(P\), it is sufficient:

1. From the horizontal projection of the point \(s\), draw the horizontal projection of the ray at an angle \(\varphi_0 = 70^\circ 35'\) to the line of trace of the plane \(P_H\) as far as the point \(O_2\).
2. From the point \(O_2\), erect a perpendicular to the line \(P^H\).
3. From the frontal projection of the point \(s'\), draw the line of the frontal projection of the ray at an angle \(\psi_0 = 45^\circ\) to the axis \(O_1X_1\) until it meets the perpendicular from \(O_2\) at the point \(S_1\), which will be the required frontal dimetric projection of the point \(S\).

On the basis of the foregoing, let us show in Example 3 the construction of the frontal oblique dimetric projection of a right prism.

Example 3. Two rectangular projections of a right prism \(1, 2, 3, 4, 5, 6, 7, 8\) are given. It is required to construct its frontal oblique dimetric projection (Fig. 8).

The construction is carried out according to the following plan:

1. The trace \(P_H\) is taken as the coordinate axis \(OX\).
2. From the frontal projections of the vertices of the prism \(1', 2', 3', 4', 5', 6', 7', 8'\), the frontal projections of the projecting rays are drawn at an angle \(\psi_0 = 45^\circ\) to the axis \(OX\).
3. From the horizontal projections of the same vertices, the horizontal projections of the projecting rays are drawn at an angle \(\varphi_0 = 70^\circ 35'\) to the axis \(OX\) until they meet the latter. From the points of intersection, vertical lines should be drawn until they intersect the corresponding frontal projections of the rays, thereby obtaining the required dimetric image of the vertices of the prism.

However, the image constructed in this way may be superimposed on the given projections. To avoid this and to obtain the image to the right ...

Fig. 8.

in the free field of the drawing, the axis \(OX\) should be moved parallel to itself into a position, for example \(O_2X_2\), and the horizontal projections of the rays should be extended to their intersections with it at the points \(1_0, 2_0, 3_0, 4_0, 5_0, 6_0, 7_0, 8_0\).

1. From the points obtained, drop perpendiculars downward to the axis \(O_2X_2\) until they intersect the corresponding frontal projections of the rays at the points \(1_1, 2_1, 3_1, 4_1, 5_1, 6_1, 7_1, 8_1\).

2. Connect the points obtained by lines, taking into account their visibility in the axonometric image; this will give the required frontal dimetry of the prism.

4. Military Perspective

Military perspective is one of the varieties of projection onto a single plane, and in this case the axonometric plane \(P\) is arranged parallel to the plane \(H\).

The image on such a plane can also be constructed quite simply on the basis of the following considerations.

In Fig. 9 such a plane \(P\) is shown, and onto it are projected the point \(S\), as well as the coordinate axes \(OX, OY, OZ\). For simplicity of reasoning, the direction of the projecting ray is chosen as going to the center of the coordinate axes \(O\), which is also projected onto the plane \(P\) at the point \(O_1\). The right angle between the coordinate axes \(XOY\) is projected in its true size, i.e., the angle \(X_1O_1Y_1 = 90^\circ\), and segments on these same axes are likewise projected in true size; consequently, the distortion coefficients along the axes \(O_1X_1\) and \(O_1Y_1\) are equal to \(p = 1\) and \(q = 1\).

Fig. 9.

The direction of projection is usually chosen so that segments on the axis \(OZ\) will also be projected in true size, i.e., the third distortion coefficient will be \(r = 1\).

Let us suppose that the chosen direction of projection in Fig. 9 also satisfies this last condition, and that the segment \(OZ_1\) is projected into an equal segment \(O_1Z_1\) on the plane \(P\).

Finally, the axonometric axis \(O_1Z_1\) is usually chosen in military perspective so that its extension bisects the right angle between the axonometric axes \(O_1X_1\) and \(O_1Y_1\), i.e., the angle \(X_1O_1Y_1\).

Let us assume that the chosen direction of projection for the point \(S\) also satisfies this condition. Thus, the line of the axonometric axis \(O_1Z_1\) bisects the angle between the axonometric traces \(P_V\) and \(P_W\), i.e., the angles \(O_1Z_1P_V\) and \(O_1Z_1P_W\) are equal to \(45^\circ\).

If the plane \(P\) is taken as a new horizontal plane of projection \(H_1\), parallel to \(H\), and the point \(s_1\) as the new horizontal projection of the given point \(S\), then the military perspective \(s_2\) of this point can be found in the following way, given the frontal \((s')\) and horizontal \((s)\) projections of the point.

From the frontal projection of the point \(s'\), the frontal projection of the ray is drawn at a certain angle \(\psi_0\) to the trace \(P_V\) until it intersects it at the point \(O_1'\), from which a perpendicular to \(P_V\) is erected. From the horizontal projection of the point \(S\), the horizontal projection of the ray is drawn at the angle \(\varphi_0 = 45^\circ\) to the line of the trace \(P_V\) (or to the axonometric axis \(O_1X_1\)) until it meets the preceding perpendicular at the required point \(s_2\).

It remains to determine the magnitude of the angle \(\psi_0\). From the triangle \(O_1Z_1O_1'\) we have \(O_1'Z_1 = O_1Z_1 \cdot \sin 45^\circ\). From the triangle \(O_1'Z_1O\) it follows that

\[ O_1'Z_1 = OZ_1 \cdot \operatorname{ctg}\psi_0,\quad \text{whence}\quad \operatorname{ctg}\psi_0=\frac{O_1'Z_1}{OZ_1}. \]

Substituting into this equality the value for \(O_1'Z_1\) from the preceding equality, we obtain

\[ \operatorname{ctg}\psi_0=\frac{O_1Z_1\cdot \sin45^\circ}{OZ_1}. \]

Since by construction \(O_1Z_1 = OZ_1\), it follows that \(\operatorname{ctg}\psi_0=\sin 45^\circ=0.70711\), and the angle \(\psi_0=54^\circ44'\). We shall henceforth take the line of the trace \(P_V\) as the projection axis \(O_2X_2\), and we shall show, by an example, the application of the reasoning set forth above.

Example 4. Construct the military perspective of the right prism \(1, 2, 3, 4, 5, 6, 7, 8\), given by its two projections—horizontal and frontal (Fig. 10).

The order of solution of the problem is as follows:

1. The line of the trace \(P_V\) is taken as the new projection axis \(O_2X_2\), and the plane \(P\)—as the new horizontal plane of projection \(H_1 \equiv P\).

2. The given solid is projected orthogonally onto this plane \(H_1\), and a new horizontal projection of the object is obtained, completely identical with the same projection of the solid on the plane \(H\).

3. The plane \(H_1\) is rotated about the axis \(O_2X_2\) until it coincides with the plane \(V\), and the given projections of the prism take positions corresponding to the new combination of the projections on the planes \(V\) and \(H_1\) obtained by this rotation.

Fig. 10.

1. From the horizontal projections of the vertices of the prism 1, 2, 3, 4, 5, 6, 7, 8, the horizontal projections of the projecting rays are drawn at an angle \(\varphi_0 = 45^\circ\) to the axis \(O_2X_2\).

2. From the frontal projections of the same vertices \(1', 2', 3', 4', 5', 6', 7', 8'\), the frontal projections of the projecting rays are drawn to the axis \(O_2X_2\), and from their points of intersection with it perpendiculars are erected to the axis \(O_2X_2\) until they meet the rays from the horizontal projections of the vertices. However, with such a construction the axonometric image may be superposed on the given projections, and, to avoid this, it is recommended to move the axis \(O_2X_2\) parallel to itself downward into such a position that the required projection is located at the right on a free area of the drawing, for example, in the position \(O_3X_3\).

3. The projecting lines are extended to this axis and the points of their intersection with \(O_3X_3\) are marked—\(1'_0, 2'_0, 3'_0, 4'_0, 5'_0, 6'_0, 7'_0, 8'_0\); from these points perpendiculars are then erected to \(O_3X_3\) until they meet the horizontal projecting lines at the points \(1_1, 2_1, 3_1, 4_1, 5_1, 6_1, 7_1, 8_1\).

4. The points obtained are joined by straight lines, taking into account their visibility in the drawing, and the required military perspective of the prism is obtained: \(1_1, 2_1, 3_1, 4_1, 5_1, 6_1, 7_1, 8_1\). But the image has turned out to be rotated by \(45^\circ\) clockwise and is inconvenient for viewing; therefore the image should be rotated in the opposite direction by \(45^\circ\). Such a rotation need not be made, and the construction just described may be performed somewhat differently (Fig. 11):

Fig. 11.

1. A point is chosen at an arbitrary place on the axis \(O_2X_2\) so that the construction of the image is obtained on a free area of the drawing.

2. The frontal projections of the rays are brought to the line \(O_2X_2\), and the points of their intersection with \(O_2X_2\) are marked—\(1'_0, 2'_0, 3'_0, 4'_0, 5'_0, 6'_0, 7'_0, 8'_0\).

3. The line \(O_2X_2\) is rotated about the point \(Z_1\), counterclockwise, through an angle of \(45^\circ\) into the position \(O_3X_3\), together with the points lying on it, which, after the rotation, will occupy the positions \(1'_0, 2'_0, 3'_0, 4'_0, 5'_0, 6'_0, 7'_0, 8'_0\).

4. From the points obtained, perpendiculars are erected to \(O_3X_3\).

5. The horizontal projections of the rays are intersected by a line perpendicular to them at any place, for example I—I, which is then likewise rotated by \(45^\circ\) counterclockwise and may occupy any position on a free area of the drawing, for example II—II; then perpendicular lines are drawn to it at the same distances from one another as they were located before the rotation, i.e., \(a, b, c\).

6. The points of intersection of such perpendiculars with the corresponding perpendiculars to \(O_3X_3\) are marked—\(1_1, 2_1, 3_1, 4_1, 5_1, 6_1, 7_1, 8_1\), and then

these points are joined by straight lines, giving the required military perspective of the proposed prism.

1. The resulting image is already located in a free area of the drawing and is convenient for viewing by the observer.

In addition, a table is given indicating the magnitudes of the angles \(\varphi_0\) and \(\psi_0\), at which the projections of the projecting rays should be directed when constructing rectangular isometric, dimetric, oblique frontal axonometric images and military perspective, and the angles of inclination of the bisectors \(B_1\) and \(B_2\) to the axis \(OX\).

Table

REFERENCES

1. L. S. Skripov. Axonograph—an instrument for the mechanical drawing of axonometric images. TEKhSSO card file, series 26, No. 282/2, Moscow, 1942.

2. L. S. Skripov. The method of the most advantageous design—the projective correspondence of two superposed planes. Tomsk. Publishing House of TSU, Proceedings of TPI, vol. 83, 1956.

3. L. S. Skripov. Theory of the method of the most advantageous design onto one plane and its application to the solution of problems of descriptive geometry. Tomsk, Publishing House of TSU, Proceedings of TPI, vol. 78, 1955.