The solid cut from an infinite circular cylinder by two planes is a cylindrical segment or a truncated cylinder. The simplest case is when one of the cutting planes is perpendicular to the axis of the cylinder. Then the cylindrical segment has a circular base. The main interest of this page is to see how a truncated cylinder can be developed into a plane.
This in another example: The volume of a cylindrical segment it is easy to obtain if we notice that two copies of the cylindrical segment one of them turned upsidedown, together form a cylinder. If the cutting plane is not perpendicular to the axis, the section is an ellipse. An ellipse is commonly defined as the locus of points P such that the sum of the distances from P to two fixed points F1, F2 (called foci) are constant. We are going to follow Hilbert and CohnVossen's book 'Geometry and the Imagination' to see a wonderful demonstration of this fact: "A circular cylinder intersects every plane at right angles to its axis in a circle. A plane not at right angles to the axis nor parallel to it intersects the cylinder in a curve that looks like an ellipse. We shall prove this curve really is an ellipse. To this end, we take a sphere that just fits into the cylinder, and move it within the cylinder until it touches the intersecting plane (Fig. 9)." Hilbert and CohnVossen. Geometry and the Imagination. Chelsea Publishing Company. pag.7.
"We then take another such sphere and do the same thing with it ont the other side of the plane. The spheres touch the cylinder in two circles and touch the intersecting plane at two points, F1 and F2. Let B be any point on the curve of intersection of the plane with the cylinder. Consider the straight line through B lying on the cylinder (i.e. parallel to the axis). It meets the circle of contact of the spheres at two points P1 and P2. BF1 and BP1 are tangents to a fixed sphere through a fixed point B, and all such tangents must be equal, because of the rotational symmetry of the sphere. Thus BF1=BP1; and similarly BF2=BP2. It follows that But by the rotational symmetry of our figure, the distance P1P2 is independent of the point B on the curve. Therefore BF1+BF2 is constant for all points B of the section; i.e. the curve is an ellipse with foci at F1 and F2." "The fact that we have just proved can also be formulated in terms of the theory of projections as follows: The shadow that a circle throws onto an oblique plane is an ellipse if the light rays are perpendicular to the plane of the circle." (Hilbert and CohnVossen. Geometry and the Imagination)
REFERENCES
Hilbert and CohnVossen. Geometry and the Imagination. Chelsea Publishing Company. pag.7.
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We study different prisms and we can see how they develop into a plane net. Then we explain how to calculate the lateral surface area.
Every ellipse has two foci and if we add the distance between a point on the ellipse and these two foci we get a constant.
The first drawing of a plane net of a regular dodecahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 .
Transforming a circle we can get an ellipse (as Archimedes did to calculate its area). From the equation of a circle we can deduce the equation of an ellipse.
In his book 'On Conoids and Spheroids', Archimedes calculated the area of an ellipse. We can see an intuitive approach to Archimedes' ideas.
In his book 'On Conoids and Spheroids', Archimedes calculated the area of an ellipse. It si a good example of a rigorous proof using a double reductio ad absurdum.
We can cut in half a cube by a plane and get a section that is a regular hexagon. Using eight of this pieces we can made a truncated octahedron.
Using eight half cubes we can make a truncated octahedron. The cube tesselate the space an so do the truncated octahedron. We can calculate the volume of a truncated octahedron.
Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the truncated octahedron.
The truncated octahedron is an Archimedean solid. It has 8 regular hexagonal faces and 6 square faces. Its volume can be calculated knowing the volume of an octahedron.
The volume of an octahedron is four times the volume of a tetrahedron. It is easy to calculate and then we can get the volume of a tetrahedron.
You can chamfer a cube and then you get a polyhedron similar (but not equal) to a truncated octahedron. You can get also a rhombic dodecahedron.
