Or a sphere. Any segment connecting the center of the ball with a point on the spherical surface is called radius. A line segment connecting two points on a spherical surface and passing through the center of the sphere is called diameter. The ends of any diameter are called diametrically opposite points of the ball.Anything sphere section there is a plane a circle. The center of this circle is the base of the perpendicular dropped from the center to the cutting plane.The plane passing through the center of the sphere is called diametral plane. The cross section of the ball by the diametral plane is called big circle, and the section of the sphere - great circle. Any diametral plane of a ball is its plane of symmetry. The center of the ball is center of symmetry. The plane passing through a point on a spherical surface and perpendicular to the radius drawn to that point is called tangent plane. This point is called touch point. The tangent plane has only one common point with the ball - the point of contact.A straight line passing through a given point of a spherical surface perpendicular to the radius drawn to this point is called tangent. Through any point of the spherical surface there are infinitely many tangents, and all of them lie in the tangent plane of the ball.ball segment called the part of the ball cut off from it by a plane.ball layer called the part of the ball, located between two parallel planes intersecting the ball.Ball sector is obtained from a spherical segment and a cone.If the spherical segment is less than a hemisphere, then the spherical segment is complemented by a cone whose vertex is in the center of the ball and whose base is the base of the segment.If the segment is larger than a hemisphere, then the indicated cone is removed from it. Basic formulas Ball (R = OB - radius):S b \u003d 4πR 2; V = 4πR 3 / 3.Ball segment (R = OB - ball radius, h = SK - segment height, r = KV - segment base radius):V segm \u003d πh 2 (R - h / 3)or V segm \u003d πh (h 2 + 3r 2) / 6; S segment = 2πRh .Spherical sector (R = OB - ball radius, h = SK - segment height):V \u003d V segm ± V con, "+"- if the segment is less, "-" - if the segment is more than a hemisphere.or V \u003d V segm + V con \u003d πh 2 (R - h / 3) + πr 2 (R - h) / 3. Spherical layer (R 1 and R 2 - the radii of the bases of the spherical layer; h \u003d SC - the height of the spherical layer or the distance between the bases):V w/sl \u003d πh 3 / 6 + πh (R 1 2 + R 2 2 ) / 2; S w/sl = 2πRh.Example 1The volume of the ball is 288π cm 3. Find the diameter of the ball.SolutionV = πd 3 / 6288π = πd 3/6πd 3 = 1728πd3 = 1728d = 12 cm.Answer: 12.Example 2Three equal spheres of radius r touch each other and some plane. Determine the radius of the fourth sphere tangent to the three given data and the given plane.Solution Let O 1 , O 2 , O 3 be the centers of these spheres and O be the center of the fourth sphere touching the three data and the given plane. Let A, B, C, T be the points of contact of the spheres with the given plane. The points of contact of two spheres lie on the line of centers of these spheres, therefore O 1 O 2 \u003d O 2 O 3 \u003d O 3 O 1 \u003d 2r. The points are equidistant from the plane ABC, so AVO 2 O 1, AVO 2 O 3, AVO 3 O 1 are equal rectangles, therefore, ∆АВС is equilateral with side 2r . Let x is the desired radius of the fourth sphere. Then OT = x. Therefore, similar So T is the center of an equilateral triangle. Therefore From hereAnswer: r/3. Sphere inscribed in a pyramidA sphere can be inscribed in every regular pyramid. The center of the sphere lies at the height of the pyramid at the point of its intersection with the bisector of the linear angle at the edge of the base of the pyramid.Comment. If a sphere can be inscribed in a pyramid, which is not necessarily regular, then the radius r of this sphere can be calculated by the formula r \u003d 3V / S pp, where V is the volume of the pyramid, S pp is its total surface area.Example 3A conical funnel with base radius R and height H is filled with water. A heavy ball is dropped into the funnel. What should be the radius of the ball so that the volume of water displaced from the funnel by the immersed part of the ball is maximum?SolutionDraw a section through the center of the cone. This section forms an isosceles triangle. If there is a ball in the funnel, then the maximum size of its radius will be equal to the radius of the circle inscribed in the resulting isosceles triangle.The radius of a circle inscribed in a triangle is:r = S / p, where S is the area of ​​the triangle, p is its half-perimeter.The area of ​​an isosceles triangle is equal to half the height (H = SO) times the base. But since the base is twice the radius of the cone, then S = RH.The semi-perimeter is p = 1/2 (2R + 2m) = R + m.m is the length of each of the equal sides of an isosceles triangle;R is the radius of the circle constituting the base of the cone.Find m using the Pythagorean theorem: , whereBriefly it looks like this: Answer: Example 4In a regular triangular pyramid with a dihedral angle at the base equal to α, there are two balls. The first ball touches all the faces of the pyramid, and the second ball touches all the side faces of the pyramid and the first ball. Find the ratio of the radius of the first ball to the radius of the second ball if tgα = 24/7 .Solution
Let RABC is a regular pyramid and point H is the center of its base ABC. Let M be the midpoint of the edge BC. Then - the linear angle of the dihedral angle, which by condition is equal to α, and α< 90° . Центр первого шара, касающегося всех граней пирамиды, лежит на отрезке РН в точке его пересечения с биссектрисой . Let HH 1 is the diameter of the first ball and the plane passing through the point H 1 perpendicular to the straight line PH intersects the side edges RA, RV, PC, respectively, at points A 1 , B 1 , C 1 . Then H 1 will be the center of the correct ∆A 1 B 1 C 1, and the pyramid RA 1 B 1 C 1 will be similar to the pyramid RABC with the similarity coefficient k = PH 1 / PH. Note that the second ball, centered at the point O 1, is inscribed in the pyramid RA 1 B 1 C 1 and therefore the ratio of the radii of the inscribed balls is equal to the similarity coefficient: OH / OH 1 = PH / PH 1. From the equality tgα = 24/7 we find: Let AB = x. ThenHence the desired ratio OH / O 1 H 1 = 16/9.Answer: 16/9. Sphere inscribed in a prismDiameter D of a sphere inscribed in a prism is equal to the height H of the prism: D = 2R = H. Radius R of a sphere inscribed in a prism is equal to the radius of a circle inscribed in a perpendicular section of the prism.If a sphere is inscribed in a right prism, then a circle can be inscribed in the base of this prism. Radius R of a sphere inscribed in a straight prism is equal to the radius of a circle inscribed in the base of the prism.Theorem 1Let a circle be inscribed in the base of a straight prism, and the height H of the prism be equal to the diameter D of this circle. Then a sphere of diameter D can be inscribed in this prism. The center of this inscribed sphere coincides with the middle of the segment connecting the centers of the circles inscribed in the bases of the prism.Proof Let ABC ... A 1 B 1 C 1 ... - a direct prism and O - the center of a circle inscribed in its base ABC. Then point O is equidistant from all sides of the base ABC. Let O 1 be the orthogonal projection of the point O onto the base A 1 B 1 C 1 . Then O 1 is equidistant from all sides of the base A 1 B 1 C 1 , and OO 1 || AA 1 . It follows that the straight line OO 1 is parallel to each plane of the side face of the prism, and the length of the segment OO 1 is equal to the height of the prism and, by condition, the diameter of the circle inscribed in the base of the prism. This means that the points of the segment OO 1 are equidistant from the side faces of the prism, and the middle F of the segment OO 1, equidistant from the planes of the bases of the prism, will be equidistant from all the faces of the prism. That is, F is the center of a sphere inscribed in a prism, and the diameter of this sphere is equal to the diameter of a circle inscribed in the base of the prism. The theorem has been proven.Theorem 2Let a circle be inscribed in a perpendicular section of an inclined prism, and the height of the prism be equal to the diameter of this circle. Then a sphere can be inscribed in this inclined prism. The center of this sphere bisects the height passing through the center of a circle inscribed in a perpendicular section.Proof
Let АВС…А 1 В 1 С 1 … be an inclined prism and F be the center of a circle with radius FK inscribed in its perpendicular section. Since the perpendicular section of the prism is perpendicular to each plane of its side face, the radii of a circle inscribed in the perpendicular section, drawn to the sides of this section, are perpendicular to the side faces of the prism. Therefore, point F is equidistant from all side faces.Let us draw a straight line OO 1 through point F, perpendicular to the plane of the bases of the prism, intersecting these bases at points O and O 1. Then OO 1 is the height of the prism. Since according to the condition OO 1 = 2FK, then F is the midpoint of the segment OO 1:FK \u003d OO 1 / 2 \u003d F0 \u003d F0 1, i.e. point F is equidistant from the planes of all the faces of the prism without exception. This means that a sphere can be inscribed in a given prism, the center of which coincides with the point F - the center of the circle inscribed in that perpendicular section of the prism that divides the height of the prism passing through the point F in half. The theorem has been proven.Example 5A ball of radius 1 is inscribed in a rectangular parallelepiped. Find the volume of the parallelepiped.Solution Draw a top view. Or on the side. Or in front. You will see the same thing - a circle inscribed in a rectangle. Obviously, this rectangle will be a square, and the box will be a cube. The length, width and height of this cube is twice the radius of the sphere.AB \u003d 2, and therefore, the volume of the cube is 8.Answer: 8.Example 6In a regular triangular prism with base side equal to , there are two balls. The first ball is inscribed in the prism, and the second ball touches one base of the prism, two of its side faces and the first ball. Find the radius of the second ball.Solution
Let ABCA 1 B 1 C 1 be a regular prism and the points P and P 1 be the centers of its bases. Then the center of the ball O inscribed in this prism is the midpoint of the segment PP 1 . Consider the plane РВВ 1 . Since the prism is correct, then РВ lies on the segment BN, which is the bisector and height ΔАВС. Therefore, the plane and is the bisector plane of the dihedral angle at the side edge BB 1 . Therefore, any point of this plane is equidistant from the side faces AA 1 BB 1 and SS 1 B 1 B . In particular, the perpendicular OK , dropped from the point O to the face ACC 1 A 1 , lies in the plane RVV 1 and is equal to the segment OR .Note that KNPO is a square whose side is equal to the radius of the sphere inscribed in the given prism. Let About 1 - the center of the ball touching the inscribed ball with the center O and the side faces AA 1 BB 1 and CC 1 B 1 B of the prism. Then the point O 1 lies on the plane RVV 1, and its projection P 2 onto the plane ABC lies on the segment RV.According to the condition, the side of the base is equal to

1. The image of the ball. Let F 0 is a ball. We choose the direction of projection and consider the tangents to the ball that belong to the chosen direction. These tangents form a cylindrical surface and pass through points of the great circle of the ball, the plane of which is perpendicular to the design direction.

Let's choose the image plane. In general, a cylindrical surface will intersect this plane in an ellipse, and the projection F 1 ball F 0 will be part of the plane bounded by this ellipse. Such an image of the ball is not visual (Fig. 59). If the image plane is chosen perpendicular to the design direction, then the image of the ball will be a circle F. The circle, of course, gives a more visual representation of the ball, but both a circle equal to it and a cylinder can be projected into a circle (if the projection is carried out parallel to its generators).

Before continuing the conversation on how to make the image of the ball visual, let's recall the concepts associated with the ball known from school. The section of a sphere by a plane passing through the center of the sphere is called big circle, and its circumference is equator. The points of intersection of a straight line perpendicular to the plane of the equator with the surface of the ball are called poles, corresponding to this equator, and the diameter connecting them is polar axis.

If any equator and the corresponding poles are depicted on the projection drawing of the ball, then the image will have three-dimensionality. It will become visible.

Which equator to represent? Firstly, it is desirable that the segment connecting the images of the poles be vertical in the drawing. This wish will be fulfilled if the image plane p will be vertical and the plane a passing through the poles N 0 ,S 0 ball, - perpendicular to it and also vertical. (Recall that we agreed to use orthogonal projection.) Moreover, we can assume that the image plane p passes through the center of the ball, and, therefore, crosses it along the circumference of the great circle. This circle is usually called essay the circumference of the ball.

Let us denote the points of intersection of the straight line with the surface of the ball by the letters P 0 and Q 0 . If the equatorial plane is also chosen perpendicular to the plane p, then the equator and the diameter connecting the poles will be depicted as perpendicular diameters of the circle (Fig. 60) and the image of the ball will not become clearer. Therefore, the equatorial plane should not be perpendicular to the image plane. On fig. 61 given a section of a ball by a plane a. In this figure P 0 Q 0 - line of intersection of planes a and p; C 0 D 0 - intersection a and the equatorial circle N 0 S 0 is the diameter connecting the poles. When designing on a plane p poles N 0 and S 0 is projected into points N and S respectively, the diameter C 0 D 0 equator - in the minor axis of the ellipse depicting this equator.

The major axis of the ellipse (Fig. 62) will be the projection of the diameter of the equator, perpendicular to the diameter and, therefore, parallel to the plane.

To indicate the position of the poles, let us return to Fig. 61. Right-angled triangles and in this figure are equal in hypotenuse and acute angle (angles with respectively perpendicular sides). That's why . But in turn, where is a segment of the tangent to the ellipse representing the equator (Fig. 62).

So, a visual image of the ball can be constructed as follows:

1) We build an ellipse, which we take as an image of the equator, and its axes.

2) We draw a circle centered in the center of the ellipse, the radius of which is equal to the major semiaxis of the ellipse.

3) We build a segment of the tangent to the ellipse, parallel to its major axis, and then the images of the poles.

On fig. 63 shows a fairly typical error when the poles are depicted on a sketch circle, while the equator is depicted as an ellipse.

2. Image of parallels and meridians. Consider the image of the poles and meridians of the sphere, which is the surface of the ball. Recall that the parallels of a sphere are its sections by planes parallel to the plane of the equator. Sections of the sphere by planes passing through the polar axis are called meridians.

Through each point of the sphere, other than the pole, there passes exactly one meridian and one parallel. Each meridian passes through both poles.

Parallels and meridians are circles, so they are also depicted as ellipses.

Let's start with drawing parallels. A parallel will be defined by specifying the point at which its plane intersects the polar axis. Since the plane of the parallel is parallel to the plane of the equator, the image of the parallel will be an ellipse, similar to the ellipse representing the equator.

To construct this ellipse, consider a section of a sphere (ball) by a plane passing through the polar axis perpendicular to the image plane (right side of Fig. 64). The constructed auxiliary section makes it easy to find the minor axis of the ellipse representing the equator and the images of the corresponding poles.

Let the parallel be given by a point, then the plane of the parallel intersects the ball along a segment perpendicular to the axis. This segment is equal to the major axis of the ellipse, which is the image of the parallel. The minor axis is found by projecting points onto a straight line. Finally, with the help of a straight line, points are found that touch the image of the parallel with the outline circle. Points separate the visible and invisible parts of the parallel image.

When constructing an ellipse, which is an image of a parallel, it is not at all necessary to build an ellipse, which is an image of the equator, to which it is similar. Moreover, it is possible not to perform separately the construction of an auxiliary section (Fig. 65).

As can be seen from fig. 66, in each of the hemispheres, it is possible to construct an ellipse-parallel that touches the outline circle at only one point. In the upper hemisphere, images of parallels lying to the north of such a parallel will be completely visible, and in the lower hemisphere, images of parallels lying to the south of such a parallel will be completely invisible.

A task. Construct an image of a cylinder inscribed in a sphere if the height of the cylinder is equal to the radius of the sphere.

Solution. Let's build an image of the outline circle of the ball and mark the images of the poles on its vertical diameter (Fig. 67).

On the same diameter, we build images of the centers and bases of the cylinder. From the condition of the problem , where is the radius of the ball, equal to the radius of the outline circle. That's why . This sets the position of the parallels. In accordance with the considered rules, we build an ellipse-image of the upper base. An ellipse depicting the lower base can be obtained using parallel translation by a vector.

In conclusion, let's consider how the image of meridians is constructed if the image of a sphere, its equator and its corresponding poles are given.

Let the image of the point be given through which the depicted equator passes (Fig. 68). In the original, the diameter is perpendicular to the polar axis , so the segments , are the conjugate diameters of the ellipse representing the meridian in question. This means that an ellipse - an image of a meridian - can be constructed from these conjugate diameters.

When constructing a meridian "by hand", they usually additionally look for points, touching the ellipse with the outline circle (Fig. 68). The diameter of the outline circle for the ellipse will be the major axis, and , which means that the diameter of the sphere is parallel to the projection plane.

The points and can be found from the following considerations. Let us construct the diameter of the ellipse-equator conjugate to the diameter . In the original, , , so the diameter is perpendicular to the plane of the considered meridian. This implies that , but then and (the projection is orthogonal). Points and separate the visible and invisible parts of the meridian image.

Image of shadows

Sometimes shadows are used to make the drawing more visible. In addition, the construction of shadows is an interesting geometric problem that contributes to the development of spatial thinking, the essence of which is as follows.

Let rays of light propagate from a luminous point in all directions in a straight line. If a ray meets an opaque body on its way, then it lingers on it and does not reach a certain screen. At the same time, a dark region is formed on the latter, which is called falling shadow from the body (Fig. 69).

The body itself is also divided into two parts: illuminated and dark (unlit). The dark part of the body is called it own shadow.

The boundary of the falling shadow is formed by the points of intersection with the screen of rays touching the surface of the body and forming light cone with top point . The line along which these rays touch the surface of the body is called dividing line light and shadow.

In the case shown in Fig. 69, the lighting is called torch, the corresponding shadow has the same name. This kind of lighting occurs when artificial lighting sources are used: an electric light bulb in a room, a street lamp, a candle flame, etc.

We can assume that natural sources (sun, moon) are at infinity and the rays from them are parallel. Therefore, the illumination produced by a beam of parallel rays is called solar. Solar illumination is shown in fig. 70.

In order to proceed to the tasks of building shadows, we will agree on how we will set the rays of light in the projection drawing. Under sunlight, such a light beam can be set by a straight line and its projection onto the main plane (Fig. 71). Let it be required to construct a falling shadow from a point on the main plane (screen). In order for the point itself to be defined, it is necessary to specify its projection onto the main plane. The construction of the shadow is reduced to finding the point of intersection of the line passing through the point parallel to , and the line passing through the point parallel to . Note that in this case the segment is a falling shadow of the segment .

With torch lighting on the projection drawing, you must specify a point that is a light source. It is determined by a point and its projection onto the main plane (Fig. 72). Here the falling shadow of the point is the point of intersection of the lines and .
It is clear that not only the main plane can be chosen as a screen. The most interesting cases of building shadows take place exactly when you have to build falling shadows on other planes. (For example, the falling shadow of one polyhedron on the surface of another.)

Task 1. In fig. 73 shows a triangular pyramid, its height and a parallelepiped. Construct own and drop shadows of these opaque shapes under the given lighting.

Solution. We are dealing with solar lighting. First of all, let's find the falling shadow of the parallelepiped on the main plane. The drop shadow of the edge is the segment , where , . Similarly, there are falling shadows , edges , respectively. It follows that is the drop shadow of the face , and is the drop shadow of the face (partially covered by the image of the parallelepiped). In passing, we note that is the own shadow of the parallelepiped.

To find the falling shadows of the pyramid on the faces of the parallelepiped, we first find its falling shadow on the main plane. This is a triangle ( , ), the triangle will be the pyramid's own shadow. The projecting plane of the straight line intersects the face of the parallelepiped along the segment . Drawing a straight line parallel to , we find the falling shadow of the vertex on the upper base of the parallelepiped. The lines , , passing through the point parallel to the lines , respectively, determine the falling shadow of the pyramid on the upper base of the parallelepiped.

It remains to find the drop shadow on the side face of the parallelepiped. To do this, note that is the trace of the plane on the main plane. The face intersects the trace at the point , and the point belongs to the planes and . Hence we conclude that the plane intersects the lateral edge of the parallelepiped at the point , and we build the falling shadow of the pyramid on the face .

Represents a plane curve - a circle that belongs to a cutting plane.
Build section of a sphere by a plane general position β

Since the cutting plane is in general position, this circle is projected onto the projection planes in the form of ellipses. To construct an ellipse, you need to know the dimensions of the ellipse along its major and minor axes.
For bodies of revolution, which include a cylinder, a cone and a sphere, a section line can be constructed with characteristic points of the curve, which include:
- points at which the sign of visibility changes;
- points at which its coordinates take the maximum and minimum values:
- x max ; x min ;
- y max ; y min ;
- z max ; z min ;
The use of characteristic points allows you to perform a more accurate construction of the line of intersection of the surface of revolution and the plane.

Solving the problem on section of a sphere by a plane is greatly simplified if the cutting plane occupies the projecting position.

Using the method of changing projection planes, we translate the plane β from a general position to a particular one - frontally projecting. On the frontal projection plane V 1 construct a trace of the plane β and projection of the ball. On the trail of the plane β V take an arbitrary point 3" measure its distance from the projection plane H and postpone it along the communication line already on the plane V 1, getting a point 3" 1 . A trace will pass through it. Ball section line - points A" 1, B" 1 coincides here with the trace of the plane. Further on the frontal projection plane V 1 construct the center of the section circle - a point C" 1 which we obtain by restoring the perpendicular from the center of the ball (point 0" 1 ) to [ A" 1 B" 1] at their intersection. Next, turn on the back projection: through points A" 1, B" 1 and C" 1 draw horizontal lines h belonging to the plane β , and on the projection plane H through the center of the ball we draw an auxiliary horizontally projecting plane γ 1. Horizontal plane trace γ 1 will stop the projection of the horizontal h and define a point at this point C`- the center of the circumference of the section. Horizontal h` intersects the ball projection at points D' and E`, thereby determining the real value of the segment [ DE] - major axis of the ellipse. Points are constructed in the same way. A` and B`, defining the value of the segment [ A`B`] - minor axis of the ellipse.

Projections of the major and minor axes of the ellipse onto the horizontal projection plane H found, which means that the ellipse is the projection of the section circle onto H can be built, see article: Circle

Repeat the same steps for the frontal projection plane V and construct another ellipse - the projection of the section circle onto V.

To find points indicating the boundaries of visibility of the horizontal projection of the section circle

we draw a front-projecting plane through the center of the ball γ 2 ⊥ V β horizontally h(h`, h"). Line h` intersects with the horizontal projection of the section circle by points 5,6 indicating the limit of visibility. Points of the section circle located on the frontal projection below the trace of the plane γ 2, on the horizontal projection plane H 5`, 6` ] - and will be invisible on it.

For finding points indicating the boundaries of visibility of the frontal projection of the section circle. We draw a horizontally projecting plane through the center of the ball γ 1 ⊥ H, which intersects the plane β frontal f(f`, f"). Line f" intersects with the frontal projection of the section circle by points 7", 8" indicating the limit of visibility. Points of the section circle located on the horizontal projection above the trace of the plane γ 1, on the frontal projection plane V will be located to the left of the segment [ 7", 8" ] - and will be invisible on it.

Introduction

A ball is a body that consists of all points in space that are at a distance not greater than a given distance from a given point. This point is called the center of the ball, and this distance is called the radius of the ball.

The boundary of a sphere is called a spherical surface, or sphere. Thus, the points of the sphere are all points of the ball that are at a distance from the center equal to the radius. Any line segment that connects the center of a ball to a point on the ball surface, also called a radius.

The segment connecting two points of the spherical surface passing through the center of the ball is called the diameter. The ends of any diameter are called diametrically opposite points of the ball.

A ball, like a cylinder and a cone, is a body of revolution. It is obtained by rotating a semicircle around its diameter as an axis.

Section of a sphere by a plane

Any section of a sphere by a plane is a circle. The center of this circle is the base of the perpendicular dropped from the center of the ball to the cutting plane.

Proof: Let be the cutting plane and O - the center of the ball (Fig. 1) Let us drop the perpendicular from the center of the ball to the plane and denote the base of this perpendicular by O ".

Let X be an arbitrary point of the ball belonging to the plane. According to the Pythagorean theorem, OX2 \u003d OO "2 + O" X2. Since OX is not greater than the radius R of the ball, then O "X?, i.e. any point of the section of the ball by a plane is from the point O" at a distance not greater, therefore, it belongs to a circle with center O "and radius. Conversely: any the point X of this circle belongs to the ball, which means that the section of the ball by the plane is a circle centered at the point O". The theorem has been proven.

The area passing through the center of the sphere is called the diametrical plane. The cross section of a ball with a diametral plane is called a great circle, and the cross section of a sphere is called a great circle.

On fig. 11 shows the construction of projections of some points.

Projections C" and D" built on a horizontal projection of the parallel of radius 0"1", built with

projection 1 ".Projection C"" and D"" built on a profile projection of a circle drawn on a sphere through the projections C"(D") so that the plane of the circle is parallel to the plane of projections.

Projection E" is the tangent point of the ellipse (horizontal projection of the slice circle) and the equator of the sphere. It is built in projection connection on the horizontal projection of the equator on the frontal projection E".

Horizontal projection M" an arbitrary point on the cut line is constructed using a radius parallel About "2", the frontal projection of which passes through the projections M"and 2 " . Projection F "is the point of contact of the ellipse (profile projection of the cut circle) and the profile projection of the outline of the sphere.

If the plane intersecting the sphere is a plane in general position, then the problem is solved by changing the projection planes. The additional projection plane is chosen so as to ensure that it is perpendicular to the cutting plane. it allows you to simplify the construction of the intersection line.

12. Construction of torus sections

In the example in fig. 12 shows the use of auxiliary planes γ 1 (γ 1 ") and γ 2 (γ 2 "), perpendicular to the axis of the torus, to construct the line of intersection and the natural form of the figure of the section of the surface of the torus by the plane α (α ""). The torus in Fig. 12 has two images - a frontal projection and a half profile projection.

Semicircle radius R 2 (profile projection of the line of intersection of the torus of the auxiliary

plane γ 2 ) touches the projection of the plane α (trace α ""). This defines the profile projection 3"" and along it a frontal projection 3"" one of the projection points of the desired intersection line. Semicircle radius R 1 - profile projection of the line of intersection of the torus by the auxiliary plane γ 1 . It intersects the profile projection of the plane α (the trace α "") at two points 5"" and 7"" - profile projections of points of the intersection line. Carrying out similar constructions, you can get the required number of point projections for the desired intersection line. We use the found points to build a natural view of the sectional figure. The figure of a section of a torus by a plane parallel to its axis has axes and a center of symmetry. When constructing it, the distances l 1 and l 2 on the frontal projection were used to plot points 5 0 , 7 0 and 3 0 .

Points 6 0 , 8 0 and 4 0 are constructed as symmetrical. The constructed curve of intersection of the surface of the torus by the plane is expressed by a 4th order algebraic equation.

The intersection curves of a torus with a plane parallel to the axis are shown in Fig. 12 below. They have a common name - Perseus curves. (Perseus- Geometer of Ancient Greece). These are curves of the fourth order. The shape of the curves depends on the distance from the secant plane to the axis of the torus.