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Derivation of Volume of Cube
The amount of space occupied by a three dimensional object is called as volume. volume of room is greater than volume of an match box. cubic unit are used to measure the volume of solid objects.
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From Wikipedia
Volume rendering is a technique used to display a 2D projection of a 3D discretely sampleddata set.
A typical 3D data set is a group of 2D slice images acquired by a CT, MRI, or MicroCTscanner. Usually these are acquired in a regular pattern (e.g., one slice every millimeter) and usually have a regular number of image pixels in a regular pattern. This is an example of a regular volumetric grid, with each volume element, or voxel represented by a single value that is obtained by sampling the immediate area surrounding the voxel.
To render a 2D projection of the 3D data set, one first needs to define a camera in space relative to the volume. Also, one needs to define the opacity and color of every voxel. This is usually defined using an RGBA (for red, green, blue, alpha) transfer function that defines the RGBA value for every possible voxel value.
A volume may be viewed by extracting surfaces of equal values from the volume and rendering them as polygonal meshes or by rendering the volume directly as a block of data. The marching cubes algorithm is a common technique for extracting a surface from volume data. Direct volume rendering is a computationally intensive task that may be performed in several ways.
Direct volume rendering
A direct volume renderer requires every sample value to be mapped to opacity and a color. This is done with a "transfer function" which can be a simple ramp, a piecewise linear function or an arbitrary table. Once converted to an RGBA (for red, green, blue, alpha) value, the composed RGBA result is projected on correspondent pixel of the frame buffer. The way this is done depends on the rendering technique.
A combination of these techniques is possible. For instance, a shear warp implementation could use texturing hardware to draw the aligned slices in the offscreen buffer.
Volume ray casting
The technique of volume ray casting can be derived directly from the rendering equation. It provides results of very high quality, usually considered to provide the best image quality. Volume ray casting is classified as image based volume rendering technique, as the computation emanates from the output image, not the input volume data as is the case with object based techniques. In this technique, a ray is generated for each desired image pixel. Using a simple camera model, the ray starts at the center of projection of the camera (usually the eye point) and passes through the image pixel on the imaginary image plane floating in between the camera and the volume to be rendered. The ray is clipped by the boundaries of the volume in order to save time. Then the ray is sampled at regular or adaptive intervals throughout the volume. The data is interpolated at each sample point, the transfer function applied to form an RGBA sample, the sample is composited onto the accumulated RGBA of the ray, and the process repeated until the ray exits the volume. The RGBA color is converted to an RGB color and deposited in the corresponding image pixel. The process is repeated for every pixel on the screen to form the completed image.
Splatting
This is a technique which trades quality for speed. Here, every volume element is splatted, as Lee Westover said, like a snow ball, on to the viewing surface in back to front order. These splats are rendered as disks whose properties (color and transparency) vary diametrically in normal (Gaussian) manner. Flat disks and those with other kinds of property distribution are also used depending on the application.
Shear warp
The shear warp approach to volume rendering was developed by Cameron and Undrill, popularized by Philippe Lacroute and Marc Levoy. In this technique, the viewing transformation is transformed such that the nearest face of the volume becomes axis aligned with an offscreen image buffer with a fixed scale of voxels to pixels. The volume is then rendered into this buffer using the far more favorable memory alignment and fixed scaling and blending factors. Once all slices of the volume have been rendered, the buffer is then warped into the desired orientation and scaled in the displayed image.
This technique is relatively fast in software at the cost of less accurate sampling and potentially worse image quality compared to ray casting. There is memory overhead for storing multiple copies of the volume, for the ability to have near axis aligned volumes. This overhead can be mitigated using run length encoding.
Texture mapping
Many 3D graphics systems use texture mapping to apply images, or textures, to geometric objects. Commodity PC graphics cards are fast at texturing and can efficiently render slices of a 3D volume, with real time interaction capabilities. WorkstationGPUs are even faster, and are the basis for much of the production volume visualization used in medical imaging, oil and gas, and other markets (2007). In earlier years, dedicated 3D texture mapping systems were used on graphics systems such as Silicon GraphicsInfiniteReality, HPVisualize FXgraphics accelerator, and others. This technique was first described by Bill Hibbard and Dave Santek.
These slices can either be aligned with the volume and rendered at an angle to the viewer, or aligned with the viewing plane and sampled from unaligned slices through the volume. Graphics hardware support for 3D textures is needed for the second technique.
Volume aligned texturing produces images of reasonable quality, though there is often a noticeable transition when the volume is rotated.
Hardwareaccelerated volume rendering
Due to the extremely parallel nature of direct volume rendering, special purpose volume rendering hardware was a rich research topic before GPU volume rendering became fast enough. The m
In arithmetic and algebra, the cube of a number n is its third power— the result of the number multiplying by itself three times:
 n^{3} = n× n× n.
This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the onethird power.
A perfect cube (also called a cube number, or sometimes just a cube) is a number which is the cube of an integer.
The sequence of nonnegative perfect cubes starts :
 0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507, 85184, 91125, 97736, 103823, 110592, 117649, 125000, 132651, 140608, 148877, 157464, 166375, 175616, 185193, 195112, 205379, 216000, 226981, 238328...
Geometrically speaking, a positive number m is a perfect cube if and only if one can arrange m solid unit cubes into a larger, solid cube. For example, 27 small cubes can be arranged into one larger one with the appearance of a Rubik's Cube, since 3 × 3 × 3 = 27.
The pattern between every perfect cube from negative infinity to positive infinity is as follows,
n^{3} = (n− 1)^{3} + (3n− 3)n + 1.
Cubes in number theory
There is no smallest perfect cube, since negative integers are included. For example, (−4) × (−4) × (−4) = −64. For any n, (−n)^{3} = −(n^{3}).
Base ten
Unlike perfect squares, perfect cubes do not have a small number of possibilities for the last two digits. Except for cubes divisible by 5, where only 25, 75 and 00 can be the last two digits, any pair of digits with the last digit odd can be a perfect cube. With even cubes, there is considerable restriction, for only 00, o2, e4, o6 and e8 can be the last two digits of a perfect cube (where o stands for any odd digit and e for any even digit). Some cube numbers are also square numbers, for example 64 is a square number (8 × 8) and a cube number (4 × 4 × 4); this happens if and only if the number is a perfect sixth power.
It is, however, easy to show that most numbers are not perfect cubes because all perfect cubes must have digital root1, 8 or 9. Moreover, the digital root of any number's cube can be determined by the remainder the number gives when divided by 3:
 If the number is divisible by 3, its cube has digital root 9;
 If it has a remainder of 1 when divided by 3, its cube has digital root 1;
 If it has a remainder of 2 when divided by 3, its cube has digital root 8.
Waring's problem for cubes
Every positive integer can be written as the sum of nine (or fewer) positive cubes. This upper limit of nine cubes cannot be reduced because, for example, 23 cannot be written as the sum of fewer than nine positive cubes:
 23 = 2^{3} + 2^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3}.
Fermat's last theorem for cubes
The equation x^{3} + y^{3} = z^{3} has no nontrivial (i.e. xyz≠ 0) solutions in integers. In fact, it has none in Eisenstein integers.
Both of these statements are also true for the equation x^{3} + y^{3} = 3z^{3}.
Sums of rational cubes
Every positive rational number is the sum of three positive rational cubes, and there are rationals that are not the sum of two rational cubes.
Sum of first ''n'' cubes
The sum of the first n cubes is the n^{th}triangle number squared:
 1^3+2^3+\dots+n^3 = (1+2+\dots+n)^2=\left(\frac{n(n+1)}{2}\right)^2.
For example, the sum of the first 5 cubes is the square of the 5th triangular number,
 1^3+2^3+3^3+4^3+5^3 = 15^2 \,
A similar result can be given for the sum of the first yodd cubes,
 1^3+3^3+\dots+(2y1)^3 = (xy)^2
but {x,y} must satisfy the negative Pell equation x^22y^2 = 1. For example, for y = 5 and 29, then,
 1^3+3^3+\dots+9^3 = (7*5)^2 \,
 1^3+3^3+\dots+57^3 = (41*29)^2
and so on. Also, every evenperfect number, except the first one, is the sum of the first 2^{(pâˆ’1)/2}odd cubes,
 28 = 2^2(2^31) = 1^3+3^3
 496 = 2^4(2^51) = 1^3+3^3+5^3+7^3
 8128 = 2^6(2^71) = 1^3+3^3+5^3+7^3+9^3+11^3+13^3+15^3
Sum of cubes in arithmetic progression
There are examples of cubes in arithmetic progression whose sum is a cube,
 3^3+4^3+5^3 = 6^3
 11^3+12^3+13^3+14^3 = 20^3
 31^3+33^3+35^3+37^3+39^3+41^3 = 66^3
with the first one also known as Plato's number. The formula F for finding the sum of an n number of cubes in arithmetic progression with common difference d and initial cube a^{3},
 F(d,a,n) = a^3+(a+d)^3+(a+2d)^3+...+(a+dnd)^3
is given by,
 F(d,a,n) = (n/4)(2ad+dn)(2a^22ad+2adnd^2n+d^2n^2)
A parametric solution to,
 F(d,a,n) = y^3
is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d> 1, such as d = {2,3,5,7,11,13,37,39}, etc.
History
Determination of the Cube of large numbers was very common in many ancient civilizations. From Encyclopedia
cube in geometry, regular solid bounded by six equal squares. All adjacent faces of a cube are perpendicular to each other; any one face of a cube may be its base. The dimensions of a cube are the lengths of the three edges which meet at any vertex. The volume of a cube is equal to the product of its dimensions, and since its dimensions are equal, the volume is equal to the third power, or cube, of any one of its dimensions. Hence, in arithmetic and algebra, the cube of a number or letter is that number or letter raised to the third power. For example, the cube of 4 is 4 3 =4Ã—4Ã—4=64. The problem of constructing a cube with a volume equal to twice that of a given cube using only a compass and a straightedge is known as the problem of the duplication of the cube and is one of the famous geometric problems of antiquity . The cube, or hexahedron, is one of only five regular polyhedra (see polyhedron ).
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Answers:i denote pi with P. volume area cube a^3 6a^2 cuboid abc 2(ab+bc+ca) cylinder Pr^2h 2Pr^2+h(2Pr) cone (1/3)bh Pr^2+Prs sphere (4/3)Pr^3 4Pr^2
Answers:A derived unit is obtained by combining base units by multiplication, division or both of these operations. It's units is derived from a similar combination of base units. Hence, volume = length x length x length = metre x metre x metre = m^3 See, multiplication of the same base units, metre gives you volume.
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Answers:a square of 16 m means 4 m long and 4 meter wide and that cube must be also 4 m high. so 4 x 4 x 4 = 64 m volume
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