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Archimedes also devised a way to calculate the volume of an irregular object, by submerging it underwater and measure the difference between the initial and final water volume. The water volume difference is the volume of the object.
Measurement of volume by displacement, (a) before and (b) after an object has been submerged; the amount by which the liquid rises in the cylinder (∆V) is equal to the volume of the object. The most widely known anecdote about Archimedes tells of how he invented a method for determining the volume of an object with an irregular shape.
The magnitude of buoyancy force may be appreciated a bit more from the following argument. Consider any object of arbitrary shape and volume V surrounded by a liquid. The force the liquid exerts on an object within the liquid is equal to the weight of the liquid with a volume equal to that of the object. This force is applied in a direction ...
Their intercepts with the dashed lines show that when the volume increases 8 (2³) times, the surface area increases 4 (2²) times. The surface-area-to-volume ratio or surface-to-volume ratio (denoted as SA:V, SA/V, or sa/vol) is the ratio between surface area and volume of an object or collection of objects.
Many well-known objects are regularly used as casual units of volume. They include: Double-decker bus The approximate volume of a double-decker bus, abbreviated to DDB, has been used informally to describe the size of hole created by a major sewer collapse. For example, a report might refer to "a 4 DDB hole".
An object immersed in a liquid displaces an amount of fluid equal to the object's volume. Thus, buoyancy is expressed through Archimedes' principle, which states that the weight of the object is reduced by its volume multiplied by the density of the fluid. If the weight of the object is less than this displaced quantity, the object floats; if ...
An irregular volume in space can be approximated by an irregular triangulated surface, and irregular tetrahedral volume elements. In numerical analysis , complicated three-dimensional shapes are commonly broken down into, or approximated by, a polygonal mesh of irregular tetrahedra in the process of setting up the equations for finite element ...
The volume of a rhombicuboctahedron can be determined by slicing it into two square cupolas and one octagonal prism. Given that the edge length a {\displaystyle a} , its surface area and volume is: [ 7 ] A = ( 18 + 2 3 ) a 2 ≈ 21.464 a 2 , V = 12 + 10 2 3 a 3 ≈ 8.714 a 3 . {\displaystyle {\begin{aligned}A&=\left(18+2{\sqrt {3}}\right)a^{2 ...