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Consider a circle in with center at the origin and radius . Gauss's circle problem asks how many points there are inside this circle of the form ( m , n ) {\displaystyle (m,n)} where m {\displaystyle m} and n {\displaystyle n} are both integers.
where C is the circumference of a circle, d is the diameter, and r is the radius.More generally, = where L and w are, respectively, the perimeter and the width of any curve of constant width.
[1] [4] That is, if a given munitions design has a CEP of 100 m, when 100 munitions are targeted at the same point, an average of 50 will fall within a circle with a radius of 100 m about that point. There are associated concepts, such as the DRMS (distance root mean square), which is the square root of the average squared distance error, a ...
The circle with center and radius () intersects circle orthogonal. Angle between two circles If the radius ρ {\displaystyle \rho } of the circle centered at P {\displaystyle P} is different from Π ( P ) {\displaystyle {\sqrt {\Pi (P)}}} one gets the angle of intersection φ {\displaystyle \varphi } between the two circles applying the Law of ...
The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and because the sequence tends to a circle, the corresponding formula–that the area is half the circumference times the radius–namely, A = 1 / 2 × 2πr × r, holds for a circle.
The area within a circle is equal to the radius multiplied by half the circumference, or A = r x C /2 = r x r x π.. Liu Hui argued: "Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the ...
The algorithm selects one point p randomly and uniformly from P, and recursively finds the minimal circle containing P – {p}, i.e. all of the other points in P except p. If the returned circle also encloses p, it is the minimal circle for the whole of P and is returned. Otherwise, point p must lie on the boundary of the result circle.
Machin-like formulas for π can be constructed by finding a set of integers , =, where all the prime factorisations of + , taken together, use a number of distinct primes , and then using either linear algebra or the LLL basis-reduction algorithm to construct linear combinations of arctangents of . For example, in the Størmer formula ...