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Lines, L. (1965), Solid geometry: With Chapters on Space-lattices, Sphere-packs and Crystals, Dover. Reprint of 1935 edition. A problem on page 101 describes the shape formed by a sphere with a cylinder removed as a "napkin ring" and asks for a proof that the volume is the same as that of a sphere with diameter equal to the length of the hole.
Many problems in the chemical and physical sciences can be related to packing problems where more than one size of sphere is available. Here there is a choice between separating the spheres into regions of close-packed equal spheres, or combining the multiple sizes of spheres into a compound or interstitial packing.
In what is called the napkin ring problem, one shows by Cavalieri's principle that when a hole is drilled straight through the centre of a sphere where the remaining band has height , the volume of the remaining material surprisingly does not depend on the size of the sphere. The cross-section of the remaining ring is a plane annulus, whose ...
On the Sphere and Cylinder (Greek: Περὶ σφαίρας καὶ κυλίνδρου) is a treatise that was published by Archimedes in two volumes c. 225 BCE. [1] It most notably details how to find the surface area of a sphere and the volume of the contained ball and the analogous values for a cylinder, and was the first to do so. [2]
In the case of the finite sphere packing problem, these objects are restricted to equally-sized spheres. Such a packing of spheres determines a specific volume known as the convex hull of the packing, defined as the smallest convex set that includes all the spheres.
An example of a spherical cap in blue (and another in red) In geometry, a spherical cap or spherical dome is a portion of a sphere or of a ball cut off by a plane.It is also a spherical segment of one base, i.e., bounded by a single plane.
The condition of balance ensures that the volume of the cone plus the volume of the sphere is equal to the volume of the cylinder. The volume of the cylinder is the cross section area, times the height, which is 2, or . Archimedes could also find the volume of the cone using the mechanical method, since, in modern terms, the integral involved ...
For most practical purposes, the volume inside a sphere inscribed in a cube can be approximated as 52.4% of the volume of the cube, since V = π / 6 d 3, where d is the diameter of the sphere and also the length of a side of the cube and π / 6 ≈ 0.5236.
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