THEORY OF THE METHOD OF MOST ADVANTAGEOUS PROJECTION ONTO A SINGLE PLANE AND ITS APPLICATION TO SOLVING PROBLEMS OF DESCRIPTIVE GEOMETRY

L. S. SKRIPOV

Introduction

The present article sets forth the method of most advantageous projection onto a single plane, proposed by the author for a simpler and faster solution of metric problems in the course of descriptive geometry. The method has been introduced into the lecture course on descriptive geometry at the Tomsk Polytechnic Institute and is not given in any of the existing textbooks.

The purpose of the present article is to eliminate this gap and to render assistance in studying the method.

The method of most advantageous projection is the basis of the Axonograph instrument, created by the author for the mechanical construction of axonometric images in 1940.

The theory of descriptive geometry offers three methods for solving metric problems, namely:

1. The method of rotation.
2. The method of superposition.
3. The method of changing the planes of projection.

With the aid of these methods, problems are solved for determining the true magnitudes of:

1) segments of straight lines,
2) areas of plane figures,
3) plane and dihedral angles,
4) distances between a point and a straight line, between straight lines, between a point and a plane, and between planes.

The author of the present article has proposed a fourth method, pursuing the same aims, but in many cases giving simpler and faster solutions of the enumerated problems than the three above-mentioned methods of descriptive geometry.

The method has been named by the author the method of most advantageous projection onto a single plane, because it is based on projecting the sought element onto one plane in such a way that the necessary answer is obtained on this plane.

The essence of the method is as follows.

1. Let us imagine the first octant (Fig. 1), with a plane \(Q\) of general position and a point \(A(a, a')\) located in it.

If the given point \(A(a, a')\) is projected orthogonally onto the plane \(Q\), i.e., if a perpendicular is dropped from point \(A\) to plane \(Q\), then the point \(a_1\), where it meets plane \(Q\), will be the required projection of point \(A\). If two planes \(R\) and \(T\) are passed through the perpendicular, perpendicular respectively to the planes of projection \(H\) and \(V\), then the planes thus drawn

intersect the planes \(H\) and \(V\), giving the traces \(R_h\) and \(T_v\) on the planes \(H\) and \(V\), and the traces \(R_q\) and \(T_q\) on the plane \(Q\).

The traces \(R_h\) and \(R_q\) will be perpendicular to the horizontal trace \(Q_h\) of the given plane \(Q\), as to the edge of the dihedral angle formed by the planes \(H\) and \(Q\). On the other hand, the trace \(R_h\) must coincide with the horizontal projection of the perpendicular \(Aa\), as the trace of the horizontally projecting plane \(R\) passed through the perpendicular \(Aa\).

Consequently, the lines \(aa_q\) and \(R_q\) are situated at angles of \(90^\circ\) to the trace \(Q_h\) (Fig. 1).

Carrying out analogous reasoning with respect to the vertically projecting plane \(T\), we conclude that the line \(a'a'_1\) (the direction of the vertical projection of the perpendicular \(Aa_1\)) and the trace \(T_q\) are situated at angles of \(90^\circ\) to the vertical trace \(Q_v\) of the plane \(Q\). Finally, we conclude that the traces \(R_q\) and \(T_q\) must intersect at the point \(a_1\)—the projection of the point \(A\) on the plane \(Q\).

From what has been said there follows a simple rule for constructing the projection \(a_1\) of the point \(A\) on the plane \(Q\):

In order to construct the orthogonal projection of a point \(A\) onto a plane \(Q\), arbitrarily situated in space, it is sufficient:

1. From the horizontal \(a\) and vertical \(a'\) projections of the point \(A\), to drop perpendiculars to the corresponding traces \(Q_h\) and \(Q_v\) of the plane \(Q\).

2. From the points of their intersection \(a_q\) and \(a'_1\) with the traces \(Q_h\) and \(Q_v\), to continue the perpendiculars, now on the plane \(Q\), until their mutual intersection.

3. The point \(a_1\) obtained in the intersection will be the required projection of the point \(A\) on the plane \(Q\).

Such a construction is carried out on an epure (Fig. 2), on which a plane in general position \(Q(Q_h, Q_v)\) and a point \((a, a')\) are given. Projecting the point \(A\) orthogonally onto the plane \(Q\), we draw from the projections of the point \(a\) and \(a'\) perpendiculars to the corresponding traces \(Q_h\) and \(Q_v\).

We mark the points of their meeting, \(a_q\) and \(a'_1\), with the traces. We then bring the plane \(Q\) into coincidence with the horizontal plane \(H\) by rotation about the horizontal trace \(Q_h\), by the known method of descriptive geometry.

In such a superposition the vertical trace will occupy the position \(Q_{v1}\). Together with the plane \(Q\), the point \(a'_1\) is also brought into coincidence with \(H\), occupying the position \(a'_2\), obtained at the intersection of \(Q_{v1}\) with the arc of a circle drawn from the center \(Q_x\) (the point of convergence of the traces), with radius equal to \(Q_x a'_1\). We then continue the perpendiculars on the coincident plane from the points \(a_q\) and \(a'_2\) to their mutual intersection at the point \(a_1\), which will be the required projection of the point \(A\) on the epure of the planes \(Q\) and \(H\).

The simple rule set out for projecting a point should be made the basis of the analogous projection of a line, a figure, and a solid.

Let us consider the application of the method to a number of examples usually solved by one of the three known methods of descriptive geometry, and let us be convinced of the simplicity and rapidity of the results obtained, which prove the advantage of the proposed method.

Example 1. Determine the true size of the triangle \(ABC(abc, a'b'c')\) (Fig. 3).

The figure of the triangle can be projected in true size onto a plane \(Q\) parallel to the plane of the triangle \(ABC\). To construct such a plane, in the plane of the triangle \(ABC\) we draw a horizontal \((a1, a'1')\) and a frontal \((c2, c'2')\).

From any point \(Q_x\) taken on the axis \(OX\), we draw the traces \(Q_h\) and \(Q_v\) parallel to the corresponding projections of the horizontal and frontal of the triangle.

The plane \(Q\), as is known, will be parallel to the plane of the triangle \(ABC\).

We project orthogonally onto the plane \(Q\) the vertices of the triangle \(A, B, C\), according to the rule of the proposed method set forth above.

Having brought the plane \(Q\) into coincidence with the plane \(H\), we construct the projections of the vertices of the triangle on the coincident plane \(Q\), as was done above (Fig. 2).

Joining the obtained projections \(a_1, b_1, c_1\) of the vertices of the triangle on the plane \(Q\) (or, what is the same, on the plane \(H\)) by straight lines, we obtain the desired answer—the true size \(a_1b_1c_1\) of the triangle \(ABC\).

Example 2. Determine the true size of the dihedral angle of the pyramid \(SABC(sabc, s'a'b'c')\) at the edge \(SC(sc, s'c')\), (Fig. 4).

As is known, a dihedral angle is projected in its true size onto a plane \(Q\) if the plane \(Q\) is positioned perpendicular to the edge of the given dihedral angle.

Taking an arbitrary point \(Q_x\) on the axis \(OX\), we draw through it the horizontal \(Q_h\) and vertical \(Q_v\) traces of the plane \(Q\), perpendicular to the corresponding projections \(sc\) and \(s'c'\) of the edge \(SC\) of the dihedral angle \(ASCB\) of the given pyramid.

We drop perpendiculars to the constructed traces from the corresponding projections of the vertices of the dihedral angle \(a, s, c, b\) and \(a', s', c', b'\).

On the trace \(Q_v\) we mark the points of intersection of the perpendiculars drawn, namely \(a'_1, b'_1, c'_1\).

We bring the plane \(Q\) into coincidence with the plane \(H\) and find the coincident position \(Q_{v1}\) of the vertical trace \(Q_v\).

The points \(a'_1, b'_1, c'_1\) belonging to the trace \(Q_v\) are transferred onto \(Q_{v1}\) as the points \(a'_2, b'_2, c'_2\).

We extend the perpendiculars in the plane \(Q\) from the obtained points until they meet the perpendiculars to the trace \(Q_h\) drawn from the horizontal projections of the vertices of the pyramid \(s, a, b, c\). We mark the points of intersection of like-named perpendiculars \(a_1, s_1, c_1, b_1\) and join them by straight lines. We note that the edge \(SC\) has been projected into one point \(s_1(c_1)\), the faces \(ASC\) and \(BSC\) into straight lines, and the required dihedral angle \(ASCB\) into the angle

\[ a_1s_1c_1b_1=\alpha, \]

which expresses the true size of the required angle.

Example 3. Determine the size of the dihedral angle \(R_xMNS_x\), formed by two intersecting planes \(R\) and \(S\) (Fig. 5).

The planes \(R\) and \(S\), intersecting along the line \(MN\), form the dihedral angle \(R_xMNS_x\) with edge along the line \(MN(mn, m'n')\) and faces \(R_xMN\) and \(S_xMN\).

The dihedral angle \(R_xMNS_x\) is projected in its true size onto a plane \(Q\), if the plane \(Q\) is perpendicular to the edge \(MN\) of the given angle.

Consequently, from an arbitrary point on the axis \(OX\) we draw the traces \(Q_h\) and \(Q_v\) of the plane \(Q\), perpendicular to the corresponding projections of the edge \(MN\), namely

\[ Q_h \perp mn,\qquad Q_v \perp m'n'. \]

We project the points \(R_x(m, m')\), \((n, n')\), \(S_x\) onto the constructed plane \(Q\), dropping perpendiculars to the corresponding traces, for example, from \(R_x\) to \(Q_h\) and \(Q_v\) (point \(R_{x1}\)).

We bring the plane \(Q\) into coincidence with the plane \(H\) and construct the coincident trace \(Q_{v1}\), onto which we transfer all the endpoints of the perpendiculars, obtaining the points \(R_{x2}, m'_2, n'_2, S_{x2}\).

From the obtained points we extend the perpendiculars in the plane \(Q\) (or, what is the same, in the plane \(H\)) until they meet the corresponding perpendiculars drawn from the horizontal projections of the same points to the horizontal trace \(Q_h\).

We mark the points \(R_{x3}, m_1, n_1, S_{x3}\) of their intersection, which we join by straight lines, noting that the edge of the dihedral angle \((mn, m'n')\) is projected...

was projected into the point \((m_1, n_1)\), while the faces \(R_xMN\) and \(S_xMN\)—into the straight lines \(R_{x3}m_1(n_1)\) and \(S_{x3}m_1(n_1)\).

Thus, the dihedral angle was projected onto the plane \(Q\) in its true magnitude, in the form of the plane angle
\(R_{x3}m_1(n_1)S_{x3}=\alpha^\circ\).

Example 4. Determine the distance from the point \(D(d,d')\) to the straight line \(AB(ab,a'b')\) (Drawing 6).

The distance from a point to a straight line is determined by the length of the perpendicular dropped from the point to the line. To solve the problem, we project the straight line \((ab,a'b')\) and the point \((d,d')\) onto a plane \(Q(Q_h,Q_v)\) perpendicular to the given straight line \((ab,a'b')\), as a result of which the point \(D\) and the straight line \(AB\) are projected into the points \(d_1\) and \(a_1\).

The distance \(d_1a_1\) between the projections thus found will be the distance between the point and the straight line in true magnitude. All constructions are clear from Drawing 6.

Example 5. Determine the distance from the point \(D(d,d')\) to the plane triangle \(ABC(abc,a'b'c')\), (Drawing 7).

To determine the required distance, it is sufficient to drop from the point \(D\) a perpendicular to the plane of the triangle \(ABC\) and to find the point of its intersection with the plane of the triangle.

To solve the problem posed, we construct a plane \(Q(Q_h,Q_v)\) perpendicular to the plane of the triangle \((abc,a'b'c')\).

We choose on the axis \(OX\) an arbitrary point \(Q_x\), through which we draw the two traces of the plane \(Q_h\) and \(Q_v\) perpendicular to one (any) of the sides of the triangle, for example \((bc,b'c')\).

This position of the plane \(Q\) must be chosen so that the plane triangle is projected onto it as a single straight line, as onto a plane perpendicular to it.

We project the point \(D\) and the vertices of the triangle \(ABC\) onto the constructed plane \(Q\), dropping perpendiculars to the traces \(Q_h\) and \(Q_v\) from the corresponding projections of the points \(d,a,b,c\) and \(d',a',b',c'\).

We mark the points of intersection of the perpendiculars with the trace \(Q_v\): \(d_1^{\,1}, a_1^{\,1}, b_1^{\,1}, c_1^{\,1}\).

We then superpose the plane \(Q\) with the plane \(H\), find the superposed position of the vertical trace \(Q_{v1}\), onto which we transfer the found endpoint points of the perpendiculars, obtaining the points \(d_2', a_2', b_2', c_2'\).

From the points thus found we continue the perpendiculars in the plane \(Q\) at an angle of \(90^\circ\) to \(Q_{v1}\), until they meet the like-named perpendiculars drawn from the horizontal projections of the same points to the trace \(Q_h\).

We mark the intersection points \(d_1, a_1, b_1, c_1\) and note that the points \(b_1\) and \(c_1\) merge into one \((b_1,c_1)\).

We connect the obtained projections of the vertices of the triangle \(a_1,b_1,c_1\) by straight lines and obtain the projection of the given triangle in the form of the straight line \(a_1b_1(c_1)\), as was to be expected under the conditions of the choice of the projection plane \(Q\).

Next we drop a perpendicular from the point \(d_1\) to the projection of the triangle \(a_1b_1c_1\) and mark the point \(k\) at which it meets the plane of the triangle. The segment of the perpendicular \(d_1k\) will be the required distance from the point \(D\) to the plane of the triangle \(ABC\).

Example 6. Determine the distance from the point \(A(a,a')\) to the plane \(Q\), given by its traces \(Q_h, Q_v\), (Drawing 8).

The distance from a point to a plane is determined by the length of the perpendicular dropped from the point \(A\) to the plane \(Q\).

Through the given point \((a,a')\) we pass a horizontally projecting plane \((R_h,R_v)\), at the same time perpendicular also to the given plane \((Q_h,Q_v)\).

We project onto the selected plane \(R\) the given point \(A\) and the plane \(Q\).

With such a choice of the plane $R$, the plane $Q$ is projected onto $R$ in the form of a straight line, by virtue of their mutual perpendicularity.

We drop a perpendicular from the point $a'$ to the trace $R_h$ and mark the point of its intersection $a_1'$ with the trace. We also mark the points of intersection of the like-named traces: $m$—the intersection of the traces $R_h$ and $Q_h$, and $n'$—the intersection of the traces $R_v$ and $Q_v$.

We then bring the plane $R$ into coincidence with the plane $H$ by rotation about the trace $R_h$, determine the superposed position of the trace $R_{v1}$ together with the points on it $a_2'$ and $n_1'$. From the point $a_2'$ we extend a perpendicular to $R_{v1}$ until it meets the like-named perpendicular drawn from the point $a$ to the trace $R_h$, and mark their point of intersection $a_1$ as the projection of the point $A$ on the chosen plane $R$.

We join the point $n_1'$ with the point $m$, which has remained fixed, and obtain the projection (trace) of the plane $Q$ on the plane $R$.

From the point $a_1$ we drop a perpendicular to the line (trace) $mn_1'$ and mark the point $k$ where it meets the line $mn'$ (with the plane $Q$).

The segment of the perpendicular $a_1k$ will be the required magnitude of the distance from the point $A$ to the plane $Q$.

Example 7. Determine the distance between two parallel straight lines $AB$ and $CD$ (Fig. 9).

To determine the distance between the given straight lines, we project them onto a plane $Q$ perpendicular to them, as a result of which the straight lines are projected into points, and the distance between them—in its true magnitude.

The order of solving the problem and the course of the construction are clear from Fig. 9 and do not require special explanation.

Example 8. Determine the shortest distance between two skew straight lines $AB$ and $CD$ (Fig. 10).

The shortest distance between two skew straight lines is measured by the length of the perpendicular to both straight lines.

To solve the problem it is sufficient to project both straight lines $AB$ and $CD$ onto a plane $Q$ perpendicular to one of the straight lines, for example, $AB$.

With such a position of the plane $Q$, the straight line $AB$ is projected onto it as the point $(a_1—b_1)$, and the other straight line $CD$—as some straight line $c_1d_1$. Now it is enough to drop from the point $(a_1—b_1)$ a perpendicular to the projection of the line $c_1d_1$, mark the point of intersection $k$ with the line $c_1d_1$, and the segment $a_1(b_1)k$ will determine, in true magnitude, the shortest distance between the straight lines. The straight line $a_1(b_1)k$ will indeed be the required perpendicular, because it proves to be parallel to the plane $Q$, as the side of a right angle formed with the straight line $a_1b_1$, and consequently will give a right angle at the point $k$ with the other straight line $c_1d_1$ as well.

The details of the construction are clear from Fig. 10.

Example 9. Determine the distance between two parallel planes $Q$ and $R$, given by the traces $(Q_h, Q_v)$ and $(R_h, R_v)$ (Fig. 11).

The distance between parallel planes is determined by the length of a perpendicular dropped from a point taken on one plane to the point of intersection of this perpendicular with the other plane.

To solve the problem, we project the given planes $(Q_h, Q_v)$ and $(R_h, R_v)$ onto a plane $T(T_h, T_v)$ perpendicular to the given planes and to the plane of projections $H$.

We mark the points of intersection of the like-named traces: $m$ (of the traces $Q_h$ and $T_h$), $m_1$ (of the traces $R_h$ and $T_h$), $n_1$ (of the traces $Q_v$ and $T_v$), $n_1'$ (of the traces $R_v$ and $T_v$).

We then bring the plane $T$ into coincidence with the plane $H$, construct the superposed trace $T_v$, together with the points on it $n_2'$ and $m_3'$, which we join, respectively, with the fixed points $m$ and $m_1$ by straight lines.

The resulting straight lines $mn_2$ and $m_1n_3'$ are the projections of the plane $Q$ (traces) on the chosen plane $T$.

Taking an arbitrary point \(k\) on the line \(mn_2'\) and dropping from it a perpendicular to the line \(mn_3'\) at the point \(p\), we find the required distance between the planes \(Q\) and \(R\) in true size, equal to \(kp\).

Example 10. Rotate the given point \(k\) about the axis \(AB\), arbitrarily situated in space, to the position \(k_3\), located at the given distance \(E\) from the vertical plane of projection \(V\) (Fig. 12).

To solve the problem we construct an auxiliary plane \(Q(Q_h, Q_v)\), passing through the given point \(K(k,k')\) and at the same time perpendicular to the given axis of rotation \(AB(ab,a'b')\).

We project the given point \((k,k')\) and the axis of rotation \((ab,a'b')\) onto the chosen plane \(Q\), as a result of which we have two points \((a_1—b_1)\) and \(k_1\).

We rotate the point \(k_1\) about the point \((a_1—b_1)\), as about a center, in the direction of the arrow (Fig. 12), to the position \(k_2\) on the frontal of the plane \(Q\), drawn at a distance \(E(e)\) from the plane \(V\). Then we return the plane \(Q\) to its former position \((Q_h,Q_v)\); the point \(k_2\) will move in a direction perpendicular to the trace \(Q_h\) to the point \(k_3\), located on the horizontal projection of the given frontal.

The vertical projection of the point \(k_3'\) is found on the vertical projection of the frontal.

The solution of a similar problem by other methods of descriptive geometry is more complicated.

Example 11. Rotate the given point \(K\) about the axis \(AB\), arbitrarily situated in space, to the position \(K_3\), located at the given distance \(D\) from the horizontal plane of projection \(H\) (Fig. 13).

The procedure for solving the problem is analogous to the solution of the preceding example, with the sole difference that the rotation of the point is carried out until it coincides with the horizontal of the plane \(Q\), drawn at the given distance \(D(d')\) from the plane of projection \(H\). The details of the construction performed are clear from Fig. 13.

Example 12. Rotate the segment of the straight line \(AB\) about the given axis \(JJ\), arbitrarily situated in space, through the given angle \(\alpha_0^\circ\) (Fig. 14).

Let us suppose that the axis of rotation \(JJ\) passes through the point \(A\) of the straight line \(AB\).

We draw the plane \(Q(Q_h,Q_v)\), perpendicular to the direction of the axis of rotation \(JJ\) and passing through the point \(B\).

We project onto the chosen plane \(Q_h\) the axis \(JJ\) and the straight line \(AB\).

On the plane \(Q\) we obtain, as projected, the axis \(JJ\) and the point \(A(a,a')\), coinciding in one point \(a_1, i_1, i_1\), and the point \(B(b,b')\) at the point \(b_1\).

The straight line \(AB\) has been projected onto the plane \(Q\) in the form of the segment \(a_1b_1\).

We rotate the segment \(a_1b_1\) through the angle \(\alpha_0^\circ\), equal to the given angle of rotation, in the direction of the arrow, about the point \(a_1, i_1, i_1\), as the center of rotation, whereby the point \(b_1\) moves to the position \(b_2\). To find the new position of the point \(B\) after rotation through the angle \(\alpha_0^\circ\) on the original epure, we return the plane \(Q\) to its initial position. The point \(b_2\) will then also move to the new position \(b_3\), lying on the corresponding horizontal \(b_2n_1'\), drawn through the point \(b_2\), and occupying the position \(b_3, n, b_3'n'\) in the original position of the plane \(Q\).

Under such a rotation the point \(b_2\) moves along the line \(b_2b_3\), perpendicular to the horizontal trace \(Q_h\).

The projections of the angle of rotation will be \(\alpha\) and \(\alpha'(<bo_1b_3,\ b'o_1'b_3')\), where \((O_1,O_1')\) is the point of intersection of the axis of rotation \(JJ(ii,i'i')\) with the plane \(Q\).

Example 13. Construct the base of a right circular cone for its given height \(SO(so,s'o')\) (Fig. 15).

We construct the plane \(Q(Q_h,Q_v)\), passing through the base of the height of the cone \(O(o,o')\) and perpendicular to the height of the cone \((so,s'o')\).

We bring the plane \(Q\) into coincidence with the plane of projection \(H\) and find the projections

of the superposed center \(O_1\), about which we describe a circle of the given radius, as the true size of the base of the right circular cone.

We divide the circle into equal parts, for example 12; through the points of division we draw horizontals and perpendiculars to the superposed position of the vertical trace \(Q_v\) of the plane \(Q\).

We return the plane \(Q\) to its original position, and in it draw the constructed horizontals and perpendiculars to the original position of the trace \(Q_v\), through the same points.

We mark the points of intersection of the vertical projections of the horizontals with the perpendiculars to \(Q_v\)—\(1', 2', 3', 4', 5', 6', 7', 8', 9', 10', 11', 12'\), which we then join by a smooth curve, obtaining an ellipse as the vertical projection of the base of the cone.

Next we mark the points of intersection of the horizontal projections of the horizontals with the perpendiculars to the trace \(Q_h\), drawn from the same points of division of the circle \(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\).

We join the points found by a smooth curve, obtaining the horizontal projection (ellipse) of the base of the cone.

Example 14. Construct the axonometric representation of a prismatic body given in orthogonal projections in Fig. 16.

The construction of axonometric projections with arbitrary, preassigned distortion coefficients, the choice of the direction of projection, and the arrangement of the axonometric axes for given coefficients are discussed in detail in the author’s work Axonograph (an instrument for the mechanical construction of axonometric representations mechanically from two orthogonal projections of a body).

In the work indicated, the theory is given of an instrument by means of which the construction of axonometric projections is accelerated up to 5 times in comparison with ordinary construction.

The description and use of the instrument are set forth in the TESKhO library under No. 939/21—1942, series 26. In the present article only an example is given of the construction of axonometric representations on an arbitrarily chosen plane \(Q\).

The construction of such a representation is clear from Fig. 16, which presents two projections of an angle section. From all points of the given body, projecting rays have been drawn orthogonally to the plane \(Q\), as in the preceding examples.

The projections of the points on the plane \(Q\) are obtained as the points of intersection of the perpendiculars to the corresponding traces, dropped from the points of the given projections of the body.

The construction of the axonometric representation of the body is easily clarified by considering Fig. 16, using as an example the construction of the projections of the points \(1, 2, 3, 4, 5, 6, 7, 8\). By joining the obtained projections of the points with straight lines, we obtain the required axonometric projection of the body on the plane \(Q\).

Example 15. Construct the axonometric projection of a body of arbitrary outline (Fig. 17).

In the case of a body of arbitrary outline, for example of cylindrical form with edges, the axonometric representation is constructed by points taken in sufficient number on the orthogonal projections of the body. In Fig. 17 the construction of the axonometry of the circle of the base of a cylinder is shown in detail.

For this purpose a series of points is taken on the circle and, according to the rules already known to us, is projected onto an arbitrarily chosen plane \(Q\).

The projections of the points on the plane \(Q\) are then joined by a smooth curve giving an ellipse as the axonometric projection of the circle.

Those interested in the theory and details of the construction of such representations are referred to the above-mentioned work of the author—Axonograph.

Example 16. Construct the self-shadow on a sphere for a given direction of the light rays \(S(s,s')\) (Fig. 18).

To construct the self-shadow on a sphere for the given direction of the light rays \((s,s')\), construct a plane \(Q(Q_h,Q_v)\), perpendicular to the direction of the rays, onto which the center of the given sphere \((o_1,o_1')\) is projected at the point \(o_2\).

About the point \(o_2\) describe a circle with the radius of the given sphere; this circle will be the projection of the line on the epure that separates the illuminated half of the sphere from the unilluminated half, in other words, the line of tangency of the projecting cylinder with the surface of the sphere.

Divide the circle into several equal parts, for example 12, and transfer the points obtained onto the projections of the sphere.

Among the points taken it is necessary to have points 1 and 7, at the ends of a diameter of the circle parallel to the vertical trace \(Q_v\), since they give the extreme upper and lower transition points of the shadow of the vertical projection of the sphere. Likewise, points 3 and 9 on the diameter of the circle parallel to the horizontal trace of the plane \(Q_h\) give the extreme left and right transition points on the horizontal projection of the contour of the sphere.

Bearing in mind that the contour of the shadow on the sphere will be represented on the projections in the form of ellipses, one may, in addition to the four characteristic points already found \(1,1'\), \(3,3'\), \(7,7'\), \(9,9'\), readily mark on the projections of the sphere the symmetric points, namely \(11,11'\), \(5,5'\).

The points found are already sufficient for constructing the contours of the ellipses of the shadow line.

If it is desirable also to determine the minor diameters of the shadow ellipses, then one may resort to constructing an additional plane \(R\), which we recommend taking as a horizontally projecting plane; onto it project the sphere in the form of a circle of diameter \(6',12'\), equal to the diameter of the given sphere, with center at the point \(o_2\).

On the plane \(R\) one should also project the direction of the light ray at \(s_1'\) and, parallel to it, draw tangent lines to the circles \(6'\) and \(12_1'\) at the points \(6_1'\) and \(12_1'\), which are then transferred to the horizontal projection of the sphere as the points 6 and 12 of intersection with the continuation of the line \(6_1\) and \(12_1\), parallel to the direction of the horizontal projection of the ray \(s\). In a similar way we find the points \(4'\) and \(10'\) on the vertical projection of the shadow ellipse.

If it is necessary to obtain a further series of points of the outline of the shadow ellipses, the sphere should be cut by a series of horizontally projecting planes passing through the points previously marked on the circle of the plane \(Q\). Project the circles obtained from such sections of the sphere onto the plane \(R\), draw tangent lines to them parallel to the direction \(s_1'\), and transfer the points of tangency onto the horizontal and vertical projections of the sphere, as shown in Fig. 18.

Example 17. Construct the line of intersection of the prism \(DEFD_1E_1F_1\) and the pyramid \(SABC\) (Fig. 19).

To solve the problem posed, construct a plane \(Q(Q_h,Q_v)\) perpendicular to the edges of the prism \(DEFD_1E_1F_1\).

The point of convergence \(Q_x\) of the traces may be chosen arbitrarily on the axis \(OX\).

Project the given prism and pyramid onto the selected plane \(Q\), and obtain the projection of the prism in the form of the triangle \(d_2e_2f_2\), and the pyramid in the form of the figure \(s_1a_1b_1c_1\).

On the obtained figures mark the points of entry and exit of the edges of the pyramid in the body of the prism:

at the edge \(s_1a_1\) — points \(1_1\) and \(2_1\),

at the edge \(s_1b_1\) — points \(3_1\) and \(4_1\),

at the edge \(s_1c_1\) — points \(5_1\) and \(6_1\).

We transfer the points found onto the horizontal projection of the pyramid at points 1, 2, 3, 4, 5, 6.

Next, from the projections of the bodies on the plane \(Q\), we note that the edge of the prism \(DD\) (point \(d_2\)) intersects the faces of the pyramid \(SAB\) \((s_1a_1b_1)\) and \(SAC\) \((s_1a_1c_1)\). To find the points at which the edge \(DD_1\) meets the indicated faces of the pyramid, we extend the plane of the prism face \(DD_1E_1E\) to its intersection with the pyramid edge \(s_1a_1\) at point 7, which is done on the plane \(Q\).

We transfer the obtained point onto the horizontal projection of the edge \(sa\) and join point 7 with points 3 and 5 of the edges \(sa\), \(sc\), thereby obtaining the lines of intersection of the prism face \(DD_1E_1E\) with the pyramid faces \(SAB\) and \(SAC\)—37 and 57.

On these lines we mark points 8 and 9, where the edge meets the sides 37 and 57 of the section figure; these will be the desired points of entry and exit of the prism edge \(DD_1\) in the faces of the pyramid.

Now, according to the generally known rules, we join the obtained entry and exit points of the edges of each body on the faces of the other, and obtain two closed figures of the lines of intersection of the bodies: \(2—4—6\) and \(1—8—3—5—9\).

It remains to transfer the points found onto the vertical projections of the figures and to obtain the outlines \(2'—4'—6'\) and \(1'—8'—3'—5'—9'\) of the line of intersection on the plane \(V\).

Example 18. Construct the line of intersection of a cone and a cylinder, given by their projections on the planes \(H\) and \(V\) (Fig. 20).

As in the preceding example, we construct an auxiliary plane \(Q(Q_h,Q_v)\), placing it perpendicular to the axis of the cylinder \(O_1O_2\). Onto the plane \(Q\) we project the cylinder and the cone, obtaining the projection of the cylinder in the form of the ellipse \(r_1 m_1 u_1 q_1 t_1 g_1 n_1\), the projection of the circle of the base of the cone in the form of the ellipse \(a_1 k_1 d_1 f_1 c_1 l_1 b_1 e_1\), and the projection of the vertex of the cone in the form of the point \(s_1\).

To construct the line of intersection of the surfaces of the cylinder and cone, on the new projection of the cone we draw a number of generators intersecting the figure of the ellipse \(r_1 m_1 u_1 q_1 t_1 g_1 n_1\).

We then transfer these generators onto the horizontal projection of the cone, for example the generators \(s_1 1_0\), \(s_1 b_1\), \(s_1 l_1\), \(s_1 c_1\), \(s_1 f_1\), \(s_1 d_1\), \(s_1 k_1\); moreover, it is essential to take the generators on the cone that are tangent to the contour of the projection of the cylinder on the plane \(Q\), such as, for example, the generator \(s_1 b_0\), tangent to the ellipse \(r_1 m_1 u_1 q_1 t_1 g_1 n_1\) at point \(1_1\).

Point 1 is important because at it the curve of the line of intersection must turn (point 1—on the horizontal projection of the cone).

It is not necessary to draw any further points of the generators on the cone, since they no longer intersect the body of the cylinder.

Thus the entry and exit points of a number of generators of the cone on the surface of the cylinder have been found, for example, \(1_1—2_1—3_1\), etc.; these are then transferred onto the corresponding generators of the cone on the horizontal projection, and afterwards also onto the vertical projection.

The obtained points are joined by a smooth curve, taking into account its visibility and on the assumption that the cylinder remains uncut.

The application of the method of most advantageous projection onto a single plane considerably simplifies the construction of lines of intersection of bodies in many practical cases.

The examples presented are quite sufficient to make it possible to judge the positive aspects of the proposed method, which has advantages over those previously existing in the theory of descriptive geometry.

It remains only to apply the proposed method to the solution of other problems not yet examined in the present article and to investigate its usefulness more broadly.

Table I

Drawing 1.

Drawing 2.

Drawing 3.

Drawing 4.

Drawing 5.

Table 11

Drawing 6.

Drawing 7.

Drawing 8.

Drawing 9.

Drawing 10.

Drawing 11.

Table III

Drawing 12.

Drawing 13.

Drawing 14.

Drawing 15.

Drawing 16.

Drawing 17.

Table IV

Table V

![Geometric construction diagram labeled “5”]
Scale 4:20.