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The hydraulic diameter is similarly defined as 4 times the cross-sectional area of a pipe A, divided by its "wetted" perimeter P. For a circular pipe of radius R, at full flow, this is = = as one would expect. This is equivalent to the above definition of the 2D mean diameter.
Pi is defined as the ratio of a circle's circumference to its diameter: [4] =. Or, equivalently, as the ratio of the circumference to twice the radius . The above formula can be rearranged to solve for the circumference: C = π ⋅ d = 2 π ⋅ r . {\displaystyle {C}=\pi \cdot {d}=2\pi \cdot {r}.\!}
In this context, a diameter is any chord which passes through the conic's centre. A diameter of an ellipse is any line passing through the centre of the ellipse. [7] Half of any such diameter may be called a semidiameter, although this term is most often a synonym for the radius of a circle or sphere. [8] The longest diameter is called the ...
For a fully filled duct or pipe whose cross-section is a convex regular polygon, the hydraulic diameter is equivalent to the diameter of a circle inscribed within the wetted perimeter. This can be seen as follows: The N {\displaystyle N} -sided regular polygon is a union of N {\displaystyle N} triangles, each of height D / 2 {\displaystyle D/2 ...
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.
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 ...
McDaniel shared the moment on Instagram, which has been viewed over 7 million times. In the video, McDaniel and Maverick stand before their family Christmas tree as they speak in whispers.
Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length. Using the fact that one part of one chord times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2r − x)x = (y / 2) 2. Solving for r, we find the required result.