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A right frustum is a right pyramid or a right cone truncated perpendicularly to its axis; [3] otherwise, it is an oblique frustum. ... and the total surface area is
The surface of the spherical segment (excluding the bases) is called spherical zone. Geometric parameters for spherical segment. If the radius of the sphere is called R , the radii of the spherical segment bases are a and b , and the height of the segment (the distance from one parallel plane to the other) called h , then the volume of the ...
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.
The tenth problem of the Moscow Mathematical Papyrus asks for a calculation of the surface area of a hemisphere (Struve, Gillings) or possibly the area of a semi-cylinder (Peet). Below we assume that the problem refers to the area of a hemisphere. The text of problem 10 runs like this: "Example of calculating a basket.
The formula for the volume of a frustum of a paraboloid [23] [24] is: V = (π h/2)(r 1 2 + r 2 2), where h = height of the frustum, r 1 is the radius of the base of the frustum, and r 2 is the radius of the top of the frustum. This allows us to use a paraboloid frustum where that form appears more appropriate than a cone.
A bi-conic nose cone shape is simply a cone with length L 1 stacked on top of a frustum of a cone (commonly known as a conical transition section shape) with length L 2, where the base of the upper cone is equal in radius R 1 to the top radius of the smaller frustum with base radius R 2. = +
Regular polygons; Description Figure Second moment of area Comment A filled regular (equiliteral) triangle with a side length of a = = [6] The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.
Since the area of a triangle cannot be negative the spherical excess is always positive. It is not necessarily small, because the sum of the angles may attain 5 π (3 π for proper angles). For example, an octant of a sphere is a spherical triangle with three right angles, so that the excess is π /2.