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For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016 [7]) is 2π ⋅ 6371 2 | sin 90° − sin 66.56° | = 21.04 million km 2 (8.12 million sq mi), or 0.5 ⋅ | sin 90° − sin 66.56° | = 4.125% of the total surface area of the Earth.
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 2-sphere, because it is a 2-dimensional surface which is embedded in 3-dimensional space.
Canonically, and originally, in 1958 when Smoot was a Lambda Chi Alpha pledge at MIT (class of 1962), the bridge was measured to be 364.4 Smoots, plus or minus one ear, using Mr. Smoot himself as a ruler. [17] At the time, Smoot was 5 feet, 7 inches, or 170 cm, tall. [18] Google Earth and Google Calculator include the smoot as a unit of ...
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 ...
(The Sun's diameter is 400 times as large and its distance also; the Sun is 200,000 to 500,000 times as bright as the full Moon (figures vary), corresponding to an angular diameter ratio of 450 to 700, so a celestial body with a diameter of 2.5–4″ and the same brightness per unit solid angle would have the same brightness as the full Moon.)
Thus, the segment volume equals the sum of three volumes: two right circular cylinders one of radius a and the second of radius b (both of height /) and a sphere of radius /. The curved surface area of the spherical zone—which excludes the top and bottom bases—is given by =.
Lines, L. (1965), Solid geometry: With Chapters on Space-lattices, Sphere-packs and Crystals, Dover. Reprint of 1935 edition. A problem on page 101 describes the shape formed by a sphere with a cylinder removed as a "napkin ring" and asks for a proof that the volume is the same as that of a sphere with diameter equal to the length of the hole.
The formula for the surface area of a sphere is more difficult to derive: because a sphere has nonzero Gaussian curvature, it cannot be flattened out. The formula for the surface area of a sphere was first obtained by Archimedes in his work On the Sphere and Cylinder. The formula is: [6] A = 4πr 2 (sphere), where r is the radius