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The red oblate spheroid (flattened sphere) corresponds to μ = 1, whereas the blue half-hyperboloid corresponds to ν = 45°. The azimuth φ = −60° measures the dihedral angle between the green xz half-plane and the yellow half-plane that includes the point P. The Cartesian coordinates of P are roughly (1.09, −1.89, 1.66).
For example, one sphere that is described in Cartesian coordinates with the equation x 2 + y 2 + z 2 = c 2 can be described in spherical coordinates by the simple equation r = c. (In this system—shown here in the mathematics convention—the sphere is adapted as a unit sphere, where the radius is set to unity and then can generally be ignored ...
More generally, one can consider the same problem on the d-dimensional sphere, or on a Riemannian manifold (in which case ||x i −x j || is replaced with the Riemannian distance between x i and x j). The problem originated in the paper by Michael Fekete who considered the one-dimensional, s = 0 case, answering a question of Issai Schur.
S 1: a 1-sphere is a circle of radius r; S 2: a 2-sphere is an ordinary sphere; S 3: a 3-sphere is a sphere in 4-dimensional Euclidean space. Spheres for n > 2 are sometimes called hyperspheres. The n-sphere of unit radius centered at the origin is denoted S n and is often referred to as "the" n-sphere. The ordinary sphere is a ...
Since spherical geometry violates the parallel postulate, there exists no such triangle on the surface of a sphere. The sum of the angles of a triangle on a sphere is 180°(1 + 4f), where f is the fraction of the sphere's surface that is enclosed by the triangle.
If the radius of the sphere is denoted by r and the height of the cap by h, the volume of the spherical sector is =. This may also be written as V = 2 π r 3 3 ( 1 − cos φ ) , {\displaystyle V={\frac {2\pi r^{3}}{3}}(1-\cos \varphi )\,,} where φ is half the cone aperture angle, i.e., φ is the angle between the rim of the cap and the ...
A simple example of a regular surface is given by the 2-sphere {(x, y, z) | x 2 + y 2 + z 2 = 1}; this surface can be covered by six Monge patches (two of each of the three types given above), taking h(u, v) = ± (1 − u 2 − v 2) 1/2. It can also be covered by two local parametrizations, using stereographic projection.
Solutions (not necessarily optimal) have been computed for every N ≤ 10,000. [2] Solutions up to N = 20 are shown below. [2] The obvious square packing is optimal for 1, 4, 9, 16, 25, and 36 circles (the six smallest square numbers), but ceases to be optimal for larger squares from 49 onwards.