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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.
The volume of a spherical cap with a curved base can be calculated by considering two spheres with radii and , separated by some distance , and for which their surfaces intersect at =. That is, the curvature of the base comes from sphere 2.
Strictly speaking, the laser diffraction equivalent diameter is the diameter of a sphere yielding, on the same detector geometry, the same diffraction pattern as the particle. In the size regimen where the Fraunhofer approximation is valid, this diameter corresponds to the projected area diameter of the particle in random orientation. For ...
where S n − 1 (r) is an (n − 1)-sphere of radius r (being the surface of an n-ball of radius r) and dA is the area element (equivalently, the (n − 1)-dimensional volume element). The surface area of the sphere satisfies a proportionality equation similar to the one for the volume of a ball: If A n − 1 ( r ) is the surface area of an ( n ...
Volume-based particle size equals the diameter of the sphere that has the same volume as a given particle. Typically used in sieve analysis , as shape hypothesis ( sieve's mesh size as the sphere diameter).
As the local density of a packing in an infinite space can vary depending on the volume over which it is measured, the problem is usually to maximise the average or asymptotic density, measured over a large enough volume. For equal spheres in three dimensions, the densest packing uses approximately 74% of the volume.
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.
Thus, the segment volume equals the sum of three volumes: two right circular cylinders one of radius a and the second of radius b (both of height /) and a sphere of radius /. The curved surface area of the spherical zone—which excludes the top and bottom bases—is given by =.