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While the circle has a relatively low maximum packing density, it does not have the lowest possible, even among centrally-symmetric convex shapes: the smoothed octagon has a packing density of about 0.902414, the smallest known for centrally-symmetric convex shapes and conjectured to be the smallest possible. [3]
Of these, solutions for n = 2, 3, 4, 7, 19, and 37 achieve a packing density greater than any smaller number > 1. (Higher density records all have rattles.) [ 10 ] See also
The related circle packing problem deals with packing circles, possibly of different sizes, on a surface, for instance the plane or a sphere. The counterparts of a circle in other dimensions can never be packed with complete efficiency in dimensions larger than one (in a one-dimensional universe, the circle analogue is just two points). That is ...
The optimal packing density or packing constant associated with a supply collection is the supremum of upper densities obtained by packings that are subcollections of the supply collection. If the supply collection consists of convex bodies of bounded diameter, there exists a packing whose packing density is equal to the packing constant, and ...
A circle packing for a five-vertex planar graph. The circle packing theorem (also known as the Koebe–Andreev–Thurston theorem) describes the possible tangency relations between circles in the plane whose interiors are disjoint. A circle packing is a connected collection of circles (in general, on any Riemann surface) whose interiors are ...
Circle packing in a square is a packing problem in recreational mathematics, where the aim is to pack n unit circles into the smallest possible square. Equivalently, the problem is to arrange n points in a unit square aiming to get the greatest minimal separation, d n , between points. [ 1 ]
Number of inner spheres Maximum radius of inner spheres [1] Packing density Optimality Arrangement Diagram Exact form Approximate 1 1.0000 1 Trivially optimal.
Using the number density as a function of spatial coordinates, the total number of objects N in the entire volume V can be calculated as = (,,), where dV = dx dy dz is a volume element. If each object possesses the same mass m 0 , the total mass m of all the objects in the volume V can be expressed as m = ∭ V m 0 n ( x , y , z ) d V ...