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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 ...
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.
Packing circles in a square - closely related to spreading points in a unit square with the objective of finding the greatest minimal separation, d n, between points. To convert between these two formulations of the problem, the square side for unit circles will be L = 2 + 2 / d n {\displaystyle L=2+2/d_{n}} .
For example, the volume of a rectangular box is found by measuring and multiplying its length, width, and height (often labeled x, y, and z). This concept of ordinary space is called Euclidean space because it corresponds to Euclid's geometry , which was originally abstracted from the spatial experiences of everyday life.
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.
If the sides of the cube were multiplied by 2, its surface area would be multiplied by the square of 2 and become 24 m 2. Its volume would be multiplied by the cube of 2 and become 8 m 3. The original cube (1 m sides) has a surface area to volume ratio of 6:1. The larger (2 m sides) cube has a surface area to volume ratio of (24/8) 3:1.
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The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m-1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus