When performing mechanical engineering drawings, the most common case is the intersection of two cylindrical surfaces, the axes of which are located at an angle of 90°. Let's look at an example of constructing a line of intersection of the surfaces of two straight circular cylinders, the axes of which are perpendicular to the projection planes (Figure 201). At the beginning of the construction, as is known, the projections of the obvious points /, 3 and 5 are found. The construction of the projection of intermediate points is shown in Figure 201. If in this example we apply the general method of constructing intersection lines using auxiliary mutually parallel planes intersecting both cylindrical surfaces along the generators, then at the intersection of these generators the required intermediate points of the intersection line will be found (for example, points 2, 4 in Figure 201). However, in this case there is no need to carry out such a construction for the following reasons. The horizontal projection of the desired line of intersection of the surfaces coincides with the circle - the horizontal projection of a large cylinder. The profile projection of the intersection line also coincides with the circle - the profile projection of a small cylinder. Thus, the frontal projection of the desired intersection line can be easily found by general rule constructing a curved line from points when two projections of points are known. For example, from the horizontal projection of point 2" we find the profile projection 2". Using two projections 2" and 2" we determine the frontal projection 2" of point 2, which belongs to the line of intersection of the cylinders. The construction of an isometric projection of intersecting cylinders (Figure 202) begins with the construction of an isometric projection of the vertical cylinder. Next, through point O, parallel to the axis l, the axis of the horizontal cylinder is drawn. The position of point 0) is determined by the value // taken from the complex drawing (Figure 201). A segment equal to L is laid off from point O up along the z axis. Laying off the segment / from point O along the axis of the horizontal cylinder, we obtain point 02 - the center of the base of the horizontal cylinder. The isometric projection of the line of intersection of the surfaces is constructed using points using three coordinates. However, in this example, the required points can be constructed somewhat differently. For example, a point. 2 is constructed as follows: from the center 02 up, parallel to the z> axis, a segment t taken from the complex drawing is laid out. Through the end of this segment, draw a straight line parallel to the >\ axis until it intersects with the base of the horizontal cylinder at point 2V. Then, from point 2, draw a straight line parallel to the x axis, and on it lay a segment equal to the distance from the base of the horizontal cylinder to the intersection line, taken from frontal or horizontal projection of a complex drawing. The end points of these segments will belong to the intersection line. Through the obtained points, a curve is drawn along the pattern, highlighting its visible and invisible parts. If the diameters of intersecting cylindrical surfaces are the same, then the frontal projection of the intersection line represents two intersecting straight lines. If intersecting cylindrical surfaces have axes located at an angle other than a right angle, then the line of their intersection is constructed using auxiliary cutting planes or other methods (for example, the method of spheres).

The method of auxiliary spherical surfaces is based on the following proposition: a sphere with any surface of revolution, the axis of which passes through the center of the sphere, intersects along a circle. If the axis of rotation is parallel to the projection plane, then onto this plane such circles are projected into straight lines perpendicular to the axis of rotation (Fig. 7.17).

Rice. 7.17. Intersection of bodies of rotation with a sphere

The method of auxiliary spherical surfaces is used to determine the line of intersection of bodies of rotation, the axes of which intersect and are parallel to the same projection plane.

The point of intersection of the rotation axes is taken as the center of the concentric spherical surfaces and a series of spheres are drawn intersecting both surfaces.

At the intersection of the contours of the resulting circles, points common to the two surfaces are found. The smallest auxiliary spherical surface will be inscribed in a larger body.

Task: Construct a line of intersection of two cylinders whose axes intersect and are parallel to the plane V(Fig. 7.18).

Solution:

We find the reference points - the intersection points of the outermost generatrices of the cylinder with an inclined axis with the rightmost generatrix of the vertical cylinder. These will be the highest and lowest points of the intersection line ( A v And IN v).

To construct intermediate points, a series of concentric spheres are drawn, the centers of which will lie at the intersection point of the axes of the given cylinders ( ABOUT v).

The smallest spherical surface here will be the surface inscribed in a vertical cylinder. This sphere touches a vertical cylinder along a circle, which is projected into a straight line 1v=2v, and an inclined cylinder intersects a circle projected into a straight line 3v=4v. The point of intersection of these lines (projections of circles) v WITH

and will be common to both cylinders.

Rice. 7.18. Line of intersection of two cylinders To construct random (intermediate) points, we draw a series of concentric spheres. Let's consider the construction of these points using the example of constructing a point v .

D 5 v -6 v And 7 v -8 v We draw a sphere whose radius is greater than the radius of the circle of the base of the vertical cylinder. This sphere intersects the cylinders along circles that project into straight lines To construct random (intermediate) points, we draw a series of concentric spheres. Let's consider the construction of these points using the example of constructing a point v.

The point of intersection of these lines (

) and will be a point belonging to the line of intersection of two cylinders.

The remaining points are constructed similarly.

7.6. Development of a surface of revolution Industry uses a large number of different structures made from sheet material by bending, for example, various tanks, the outer skin of an airplane wing, a bus body, etc. Therefore, the construction of surface developments is of great practical importance.

Developable surfaces include cylindrical and conical surfaces of revolution, and non-developable surfaces include surfaces of a sphere, torus, ellipsoid of revolution, paraboloid of revolution and other surfaces of revolution, both regular and general.

In practice, conditional (approximate) developments of non-developable surfaces are very often constructed, approximating them with developable surfaces (faceted, cylindrical, conical).

The development of a cylinder of rotation is a rectangle, one side of it is equal to  d (d – cylinder diameter), and the other - h(cylinder height).

Task: Construct a development of a horizontally projecting cylinder cut by a frontally projecting plane R.

Solution:

To construct a development of a cylindrical surface, the rolling method was used (Fig. 7.19).

The circumference of the base is divided into 12 equal parts. The generators of the cylinder are drawn through the points.

When constructing a development, the cylindrical surface is “cut” along the generatrix 1-1  and aligned with the plane V. Moreover, the length of the line 1 ABOUT ,2 ABOUT …12 ABOUT ,1 ABOUT should theoretically be equal to the circumference of the base, but practically 12 segments equal to  12  . The cylindrical surface is approximated by a prismatic (faceted) surface inscribed into it.

Finding points 1  ABOUT ,2  ABOUT …12  ABOUT ,1   ABOUT on the scan can be seen from the constructions in Fig. 7.19.

Rice. 7.19. Development of a truncated cylinder

The development of a cylinder consists of a development of a cylindrical surface and two figures: a circle (base) and an ellipse (section lying in the plane R). An ellipse can be constructed as shown in Fig. 7.19 or along two axes (the major axis of the ellipse is equal to the segment 1  ’ V 7’ V, and the minor axis is 4 N 10 N).

Task: Construct a development of a cone of rotation cut off by a frontally projecting plane R(Fig. 7.20).

Solution:

The development of its lateral surface represents a circular sector, the radius of which is equal to the length of the generatrix of the conical surface S V 1 V, and the central angle  =360 o r/(S V 1 V ) , Where r– radius of the circle of the base of the cone.

To construct a development, the circle of the base of the cone is divided into 12 equal parts and the generators of the cone are drawn through the points: S V 2 V , S V 3 V , S V 5 V etc., which intersect with the plane R at points 2’ V , 3’ V , 5’ V etc. When constructing a development, it is necessary to determine the actual size of those cut off by the plane R segments forming S V 2’ V , S V 3’ V , S V 5’ V etc., which are determined by rotating the generatrix to the position of the frontal straight line, i.e. until aligned with the straight line S V 1 V .

Rice. 7.20. Development of a truncated cone

From an arbitrary point S ABOUT draw a circle with radius S V 1 V, on which 12 equal segments are laid out: 1 0 2 0 =1 H 2 H ; 2 0 3 0 =2 H 3 H etc. Points 1 0 , 2 0 , 3 0 etc. connect to a point S ABOUT and points are marked on these straight lines 1’ 0 , 2’ 0 , 3’ 0 etc. over distances S ABOUT 1’ 0 = S V 1’ V ; S ABOUT 2’ 0 = S V 2’’ V ; S ABOUT 3’ 0 = S V 3’’ V etc. The resulting points are smoothly connected by a line, which represents the line of section of the cone by the plane R.

The development of a cone also includes a circle at the base of the cone and an ellipse (section of the cone by a plane R), which can be constructed as shown in Fig. 6.6 or along two axes (the major axis of the ellipse is equal to the segment 1’ V 7’ V, and the small one is located in the middle between the points 1 V And 7 V).

The construction of the development of the lateral surface of a truncated cone, the surface of a sphere and a torus can be found in the literature /1/ and /2/.

Algorithm for solving the problem The method of auxiliary concentric spheres is used if:

Both surfaces are surfaces of revolution;

The axes of the surfaces intersect;

The general plane of symmetry of bodies is parallel to any projection plane.

The point of intersection of the rotation axes is taken as the center of the concentric spherical surfaces and a series of spheres are drawn intersecting both surfaces.

At the intersection of the contours of the resulting circles, points common to the two surfaces are found. The smallest auxiliary spherical surface will be inscribed in a larger body.

The sphere of the largest radius should not extend beyond the most distant point of intersection of the bodies.

Intermediate spheres are constructed with arbitrary radii and must be located between the smallest and largest auxiliary spheres.

When solving this problem:

1 Find the reference points - the intersection points of the outermost generatrices of the cylinder with an inclined axis with the rightmost generatrix of the vertical cylinder. These will be the highest and lowest points of the intersection line ( A v And In v).

2 To construct intermediate points, a series of concentric spheres are drawn, the centers of which will lie at the point of intersection of the axes of the given cylinders ( About v).

3 The smallest spherical surface here will be the surface inscribed in a vertical cylinder. This sphere touches a vertical cylinder along a circle, which is projected into a straight line 1v=2v, and an inclined cylinder intersects a circle projected into a straight line 3v=4v. The point of intersection of these lines (projections of circles) C v and will be common to both cylinders.

4 To construct random (intermediate) points, we will draw a series of concentric spheres. Let's consider the construction of these points using the example of constructing a point Dv.

5 Draw a sphere whose radius is greater than the radius of the circle of the base of the vertical cylinder. This sphere intersects the cylinders along circles that project into straight lines 5v-6v And 7v -8v. The point of intersection of these lines ( Dv) and will be a point belonging to the line of intersection of two cylinders.

6 The remaining points are constructed similarly.

Figure 14 – Intersection of two cylinders

Karpova Irina Evgenievna

Karpov Egor Konstantinovich

DESCRIPTIVE GEOMETRY

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Related information:

1. II. Dynamics of rotational motion of a material point (rigid body) (problem 2)
2. III. Indicate the numbers of sentences in which the predicate verb is in the group of completed tenses
3. IX. Indicate the numbers of sentences in which the ing form is translated into Russian by a participle ending in –schey, -schaya, shchey

To construct a curved line obtained when a cylindrical surface intersects a plane, one should generally find the points of intersection of the generatrices with the cutting plane, as was said on p. 170 regarding ruled surfaces in general. But this does not exclude the possibility of using auxiliary planes that intersect each time the surface and the plane.

First of all, we note that any cylindrical surface is intersected by a plane located parallel to the generatrix of this surface along straight lines (generators). In Fig. 360 shows the intersection of a cylindrical surface with a plane. In this case, this surface is an auxiliary element in constructing the point of intersection of a curved line with a plane: a cylindrical surface is drawn through the given curve (see Fig. 360, left) DMNE, projecting the curve onto the square. π 1. Further, the plane (in Fig. 360 - a triangle) intersects the cylindrical surface along a flat curve M 1 ... N 1. The desired point of intersection of the curve with the plane - point K - is obtained at the intersection of the curves - given and constructed.

This scheme for solving the problem of the intersection of a curved line with a plane coincides with the scheme for solving problems of the intersection of a straight line with a plane(see §§ 23

and 25); in both cases, an auxiliary surface is drawn through the line, which for a straight line is a plane.

The horizontal projection of the curve M 1 ... N 1 along which the cylindrical surface intersects with the plane coincides with the horizontal projection of the curve D ... E, since this curve is a guide for the cylindrical surface when perpendicular to the square. π 1 that form it. Therefore, from the point M" 1 on the projection A"C" we can find the projection of M" 1 on A"C" and from the point N" 1 - the projection N" 1. Next, in Fig. 360 on the right shows the auxiliary square. α, intersecting ABC along the straight line CF, and the cylindrical surface along its generatrix with a horizontal projection at point 1". At the intersection of this generatrix with the straight line CF, a point with projections 1" and 1" is obtained, belonging to the curve M 1 ... N 1 Obviously , you can not indicate a trace of the plane, but simply draw a line in the triangle, as shown in relation to the CG line, on which a point with projections 2" and 2" is obtained.

The examples below will show sweeps. In the general case, the unfolding of a cylindrical surface can be carried out according to the scheme of unfolding the surface of a prism. The cylindrical surface is, as it were, replaced by an inscribed or described prismatic surface, the edges of which correspond to the generatrices of the cylindrical surface. The deployment itself is similar to that shown in Fig. 283, is produced using a normal section. But instead of a broken line, a smooth curve is drawn.

In Fig. 361 shows the intersection of a right circular cylinder with a frontally projecting plane. The cross-sectional figure is an ellipse, the minor axis of which is equal to the diameter of the base of the cylinder; the magnitude of the major axis depends on the angle between the cutting plane and the axis of the cylinder.

Since the axis of the cylinder is perpendicular to the square. π 1 then the horizontal projection of the cross-sectional figure coincides with the horizontal projection of the cylinder.

Usually, to construct the points of the section contour, uniformly spaced generatrices are drawn, i.e., those whose projections onto the square. π 1 are points equidistant from each other. This “marking” is convenient to use not only for constructing section projections, but also for developing the side surface of the cylinder, as will be shown below.

Projection of the section figure onto the square. π 3 is an ellipse, the major axis of which in this case is equal to the diameter of the cylinder, and the minor axis is the projection of the segment 1"7". In Fig. 361 on pl. π 3 image is constructed as if top part the cylinder is removed after the plane intersects it.

If in Fig. 361 plane α made an angle of 45° with the cylinder axis, then the projection of the ellipse onto π 3 would be a circle. In this case, the segments 1""7"" and 4""10"" would be equal.

If the same cylinder is intersected by a plane in general position, which also makes an angle of 45° with the axis of the cylinder, then the projection of a sectional figure (ellipse) in the form of a circle can be obtained on an additional projection plane parallel to the axis of the cylinder and the horizontals of the cutting plane.

Obviously, with an increase in the angle of inclination of the cutting plane to the axis, the segment 1""7"" decreases; if this angle is less than 45°, the segment 1""7"" increases and becomes the major axis of the ellipse on the square. π 3, the minor axis of this ellipse becomes the segment 4""10".

The natural type of section is, as mentioned above, an ellipse. Its axes are obtained in the drawing: the major one is the segment 1 0 7 0 = 1"7", the minor one is the segment 4 0 10 0, equal to the diameter of the cylinder. An ellipse can be constructed along these axes.

In Fig. 362 shows a full development of the lower part of the cylinder.

The unfolded circle of the base of the cylinder is divided into equal parts according to the divisions in Fig. 361; the segments of the generatrices are laid out on perpendiculars drawn at the dividing points of the unfolded circle of the base of the cylinder. The ends of these segments correspond to the points of the ellipse. Therefore, by drawing a curved line through them, we obtain a developed ellipse (this line is a sinusoid) - the upper edge of the development of the side surface of the cylinder.

To the development of the side surface in Fig. 362 a circle of base and an ellipse are attached - a natural type of section, which makes it possible to make a model of a truncated cylinder.

In Fig. 363 shows an elliptical cylinder with a circular base; its axis is parallel to square. π 2. To determine the normal section of this cylinder, it must be dissected by a plane perpendicular to the generatrices, in this case by a frontally projecting plane. The figure of a normal section is an ellipse with a major axis equal to the segment 3 0 7 0 and a minor axis equal to 1 0 5 0 = 1 "5".

If it is necessary to expand the lateral surface of a given cylinder, then, having a normal section, unfold the curve limiting it into a straight line and at the corresponding points of this straight line, perpendicular to it, lay down the generatrix segments, taking them from the frontal projection. To mark the generatrices, divide the circle of the base into equal parts. In this case, the ellipse (normal section) will be divided into the same number of parts, but not all of these parts turn out to be equal

length. The expansion of an ellipse into a straight line can be done by sequentially laying down sufficiently small parts of the ellipse on a straight line.

In Fig. 364 shows a right circular cylinder intersected by a general plane. The cross-section results in an ellipse: the cutting plane makes a certain acute angle with the axis of the cone.

Just like it was in Fig. 361, the horizontal projection of the section coincides with the horizontal projection of the cylinder. Therefore, the position of the horizontal projection of the intersection point of any of the generatrices of the cylinder with the square. α is known (for example, point A" in Fig. 365). To find the corresponding frontal projection, you can draw a horizontal line or frontal in the area α, on which the desired point should be located. In Fig. 365, a frontal is drawn; in the place where the frontal projection frontal intersects the frontal projection of the corresponding generatrix, the projection A lies. The same frontal defines two points of the curve, A and B (Fig. 365). If we construct a frontal corresponding to point C, then

this line will define only one point of the intersection curve. The frontal, constructed from points D and E, determines the extreme points D" and E".

Continuing similar constructions, one can find enough points to draw the frontal projection of the intersection line.

In Fig. 366 the upper part of the cylinder seems to have been cut off. If the frontal projection is shown in full, then the intersection line is drawn as shown in Fig. 364.

In Fig. 365 shows auxiliary frontal planes β, γ, δ intersecting the cylinder along the generators, and pl. α on the fronts. This corresponds to what was said at the beginning of the paragraph. Auxiliary square δ only touches the cylinder, which makes it possible to define only one point for the curve.

When constructing a frontal projection of the intersection line, in addition to points D" and E" (Fig. 365), two more extreme points should be found, namely M" and N" - the highest and lowest points of the section projection on the square. π 2. To construct them, you need to select an auxiliary plane perpendicular to the trace h" 0α and passing through the axis of the cylinder (Fig. 366). This plane is the common plane of symmetry of the data of the cylinder and the secant plane a. Having found the intersection line of the planes α and β, we mark the points M " and N", constructing them on the frontal projection along the points M" and N".

Another way to find the points M" and N" is to draw two planes tangent to the cylinder, the horizontal traces of which are parallel to the trace h" 0α. These planes will intersect with the plane α along the horizontals of the latter (Fig. 364, auxiliary planes β and γ); Having noted the points M" and N", we will construct the points M" and N" on the frontal projections of the horizontal lines.

The segment MN represents the major axis of the ellipse - the cross-sectional figure of a given cylinder square. α. This can also be seen in Fig. 366, where it was built in combination with the square. π 1 ellipse - natural type of section. But the segment M"N" in the same figure is by no means the major axis of the ellipse - the frontal projection of the cross-sectional figure. This major axis can be found from the conjugate diameters M"N" and F"G" (Fig. 364) using the construction indicated in § 21, or a special construction given in § 76.

The natural view of the section can be found by combining the cutting plane with one of the projection planes, π 1 or π 2.

In Fig. 366 ellipse in a combined position is constructed along the major and minor axes (point D" is also obtained by combining the frontal).

The development of the side surface is shown in Fig. 364. Please note that the marking of points - horizontal projections of the generators - on the base circle was made from point N". This simplified the construction, since using the same horizontal line two points are obtained on the frontal projection

tions of the ellipse. In addition, the sweep figure has an axis of symmetry. But at the same time, points D" and E" were not included in the number of points marked on the circle.

Another example of constructing a cross-section of a cylinder of rotation by a plane is given in Fig. 367. This construction was made using the method of changing projection planes. The cutting plane is defined by intersecting straight lines - the frontal line (AF) and the profile straight line (AP). Since the profile projection of the frontal and the frontal projection of the profile straight line lie on the same straight line A"≡A"", A""F"" = А"Р", then these straight lines lie respectively in the planes π 2 and π 3 (see Fig. . 367, top left). The π 2 /π 3 axis passes through A""F""(A"P").

We introduce a new square. π 4 so that π 4 ⊥π 3, and π 4 ⊥AP. The cutting plane turns out to be perpendicular to π 4, and the projection onto π 4 of the section figure is obtained in the form of a straight line segment 2 IV 6 IV, equal to the major axis of the ellipse - the section figure. The position of the straight line A IV 6 IV is determined by constructing the projections of points A and 1 onto the square. π 4.

Let's trace the construction of some points. To avoid unnecessary constructions, the projection 1"" was taken on the continuation of the perpendicular drawn from O"" to π 3 / π 4. At point 1"" projection 1" was obtained; the segment 1"1"", laid off from the π 3 /π 4 axis, determined point IV and the coinciding point O 1 - the projection of the center of the ellipse. Knowing the projections 0 IV and O"", one can obtain O" - the center of the ellipse - the desired frontal projection of the cross-sectional figure.

Using points 2 IV and 2"" we found point 2", the least distant from π 3, and using points 6 IV and 6"" - point 6", the furthest from π 3.

Using point 5"" we took point 5 IV, and now using points 5 IV and 5"" we found point 5" - one of the points that determines the division of the ellipse on the frontal projection of the cylinder into “visible” and “invisible” parts. The second point is located symmetrically point 5" in relation to O".

The rest is clear from the drawing. The natural view of the cross-sectional figure (ellipse in Fig. 367, on the right) is built along axes - large, equal to 2 IV 6 IV, and small, equal to the diameter of the cylinder.

Questions for §§ 55 -56

1. How is a curved line constructed when a curved surface intersects a plane?
2. Along what lines does a cylindrical surface intersect with a plane drawn parallel to the generatrix of this surface?
3. What technique is generally used to find the point of intersection of a curved line with a plane?
4. What lines are obtained when planes intersect a cylinder of rotation?
5. In what case is the ellipse obtained by intersecting a cylinder of revolution whose axis is perpendicular to the square? π 1, frontally projecting plane, is projected onto the square. π 3 in the form of a circle?
6. How should the additional projection plane be positioned so that the ellipse obtained by intersecting the cylinder of rotation, the axis of which is perpendicular to the square? π 1, a plane of general position making an angle of 45° with the axis of the cylinder, was projected onto this plane of projections in the form of a circle?

Mutual intersection of bodies of revolution

In Fig. Figure 4.21 shows the construction of the line of intersection of two cylinders of different diameters. The axes of the cylinders are mutually perpendicular and intersect.

In Fig. 4.21, and a part is shown (a tee used to connect pipes and its model), which represents two intersecting cylinders. Intersecting, the cylindrical surfaces form a spatial curved line. The horizontal projection of the intersection line coincides with the horizontal projection of a vertically located cylinder, i.e. with a circle (Fig. 4.21, b). The profile projection of the intersection line coincides with the circle, which is the profile projection of a horizontally located cylinder. Mark characteristic points on horizontal and profile projections 1, 2, 3. By horizontal and profile projections of points 1 , 2, 3 find their frontal projections 1", 2", 3". In this way, the projections of the points defining the transition line are found.

Rice. 4.21.

b– intersection line: b", b, b"– projections of the intersection line

In some cases, this number of points is not enough. To obtain additional points, you can use the method of auxiliary cutting planes.

Method of auxiliary cutting planes

This method consists in intersecting the surfaces of bodies with an auxiliary plane that forms sectional figures whose contours intersect. The points obtained as a result of the intersection of section contours are located on the intersection line.

In this case, both cylinders are intersected by an auxiliary cutting plane R(Fig. 4.21, a, c). When a vertical cylinder intersects, a circle is formed, and a horizontal cylinder intersects a rectangle.

Intersection points 4 and 5 circles and rectangles belong to both cylinders and, therefore, are located on the line of intersection of both bodies (Fig. 4.21, A).

Marking the profile and then horizontal projections of the points 4 and 5, which lie on the circles, find their frontal projections using connection lines, as shown by arrows in Fig. 4.21, V.

The resulting five points are connected by a smooth curve.

If it is necessary to increase the number of points defining the intersection line, several more parallel cutting planes are drawn.

If both cylinders have the same diameters, then one of the projections of their intersection lines represents intersecting straight lines (Fig. 4.21, G, d), and in space the intersection lines are ellipses.

The intersection line of a ball and a right circular cylinder, the axis of which passes through the center of the ball, is shown in Fig. 4.22. As can be seen from the drawing, on one projection the intersection line is depicted as a circle 1, and on the other it is projected into a straight line 1".

Rice. 4.22.

1 – intersection line; 1" And 1 – projections of the intersection line

Projecting solids with holes

In technology there are many parts that have holes of cylindrical, rectangular, triangular or mixed shapes (Fig. 4.23). When holes intersect with the surfaces of parts, intersection lines are formed that must be constructed in the drawing. This problem is solved in general by the same methods as constructing intersection lines of geometric bodies. In each case, the hole can be considered as a body passing through the part.

Rice. 4.23.

In Fig. 4.24, A shows a cylinder having a cylindrical shaped hole. The axes of the cylinder and the hole intersect at right angles. The intersection line is depicted as a curve. The construction of such a line was shown in Fig. 4.21. In Fig. 4.24, A shows how to obtain characteristic points of a given curve.

Rice. 4.24.

The line of intersection of a cylinder with a rectangular hole in the case of the intersection of their axes at right angles is shown in Fig. 4.24, b. To construct it on the horizontal projection, characteristic points were selected 1, 2, 3, 4, 5, 6. Their profile projections 1", 2", 3", 4", 5" , 6" lie on a circle that is a projection of the cylinder. Frontal projections 1", 2", 3", 4", 5" , 6" found from the obtained horizontal and profile ones. Connecting the dots 1", 2", 3", 4", 5", 6" straight lines, we obtain a projection of the intersection line in the form of a rectangular depression. The projection of the intersection line on the other side of the hole has the same shape.

In Fig. 4.24, V shows the line of intersection of the cylinder with the hole, which is a combination of the first two. The hole is formed by a quadrangular prism and two half-cylinders. The keyway has this shape.