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A brief guide to texture mapping in Processing

Color interpolation

As seen in Graphic Composition in Processing , one can obtain surfaces as collections of polygons by means of the definition of a vertexwithin the couple beginShape() - endShape() . It is possible to assign a color to one or more vertices, in order to make the color variations continuous ( gradient ). For example, you can try to run the code size(200,200,P3D);beginShape(TRIANGLE_STRIP); fill(240, 0, 0); vertex(20,31, 33);fill(240, 150, 0); vertex(80, 40, 38); fill(250, 250, 0); vertex(75, 88, 50);vertex(49, 85, 74); endShape(); in order to obtain a continuous nuance from red to yellow in the strip of two triangles.

Bilinear interpolation

The graphical system performs an interpolation of color values assigned to the vertices. This type of bilinear interpolation is defined in the following way:

  • For each polygon of the collection
    • For each side of the polygon one assigns to each point on the segment the color obtained by means of linear interpolation of the colors of the vertices i e j that define the polygon: C ij α (1 - α) C i α C j
    • A scan line scans the polygon (or, better, its projection on the image window) intersecting at each step two sides in two points l ed r whose colors have already been identified as C l e C r . In each point of the scan line the color is determined by linear interpolation C lr β (1 - β) C l β C r

A significative example of interpolation of colors associated to the vertices of a cube can be found in examples of Processing , in the code RGB Cube .


When modeling a complex scene by means of a composition of simple graphical elements one cannot go beyond a certainthreshold of complexity. Let us think about the example of a modelization of a natural scene, where one has to representeach single vegetal element, including the grass of a meadow. It is unconceivable to do this manually. It would bepossible to set and control the grass elements by means of some algorithms. This is an approach taken, for example, inrendering the hair and skin of characters of the most sophisticated animation movies (see for example, the Incredibles ). Otherwise, especially in case of interactive graphics, one has to resort to using textures . In other words, one employs images that represent the visual texture of the surfaces and map them onthe polygons that model the objects of the scene. In order to have a qualitative rendering of the surfaces it is necessaryto limit the detail level to fragments not smaller than one pixel and, thus, the texture mapping is inserted in the rendering chain at the rastering level of the graphic primitives, i.e. where one passes from a 3Dgeometric description to the illumination of the pixels on the display. It is at this level that the removal of thehidden surfaces takes place, since we are interested only in the visible fragments.

In Processing, a texture is defined within a block beginShape() - endShape() by means of the function texture() that has as unique parameter a variable of type PImage . The following calls to vertex() can contain, as last couple of parameters, the point of the texture correspondingto the vertex. In fact, each texture image is parameterized by means of two variables u and v , that can be referred directly to the line and column of a texel (pixel of a texture) or, alternatively, normalized between 0 and 1 , in such a way that one can ignore the dimension as well as the width and height of the texture itself. Themeaning of the parameters u and v is established by the command textureMode() with parameter IMAGE or NORMALIZED .

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Media processing in processing. OpenStax CNX. Nov 10, 2010 Download for free at http://cnx.org/content/col10268/1.14
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