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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 ...
In computational geometry, the smallest enclosing box problem is that of finding the oriented minimum bounding box enclosing a set of points. It is a type of bounding volume. "Smallest" may refer to volume, area, perimeter, etc. of the box. It is sufficient to find the smallest enclosing box for the convex hull of the objects in question. It is ...
The ratio of the volume of a sphere to the volume of its circumscribed cylinder is 2:3, as was determined by Archimedes. The principal formulae derived in On the Sphere and Cylinder are those mentioned above: the surface area of the sphere, the volume of the contained ball, and surface area and volume of the cylinder.
Area#Area formulas – Size of a two-dimensional surface; Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities
A sphere enclosed by its axis-aligned minimum bounding box (in 3 dimensions) In geometry, the minimum bounding box or smallest bounding box (also known as the minimum enclosing box or smallest enclosing box) for a point set S in N dimensions is the box with the smallest measure (area, volume, or hypervolume in higher dimensions) within which all the points lie.
Consider the linear subspace of the n-dimensional Euclidean space R n that is spanned by a collection of linearly independent vectors , …,. To find the volume element of the subspace, it is useful to know the fact from linear algebra that the volume of the parallelepiped spanned by the is the square root of the determinant of the Gramian matrix of the : (), = ….
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An a × b rectangle can be packed with 1 × n strips if and only if n divides a or n divides b. [15] [16] de Bruijn's theorem: A box can be packed with a harmonic brick a × a b × a b c if the box has dimensions a p × a b q × a b c r for some natural numbers p, q, r (i.e., the box is a multiple of the brick.) [15]