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Orientation Dependence - Modeling and Rendering
The outcomes of interpolated-shading models are dependent of the projected polygon's orientation. Because values are interpolated among vertices and across horizontal scan lines, the results may be different when the polygon is rotated. This consequence is mainly obvious when the orientation modifies slowly among successive frames of an animation. A same problem cab also happens in visible-surface determination while the z value at each point is interpolated by the z values allocated to each vertex. Both issues can be resolved through decomposing polygons into triangles. Instead, the solution is rotation- independent, although expensive, interpolation methods which solve problem without the requirement for decomposition.
What is the use of Projection reference point? In Perspective projection, the object positions are transformed to the view plane with these converged projection line and the p
Boundary Fill Algorithm Boundary fill algorithm is suitable when the boundary has single color while flood fill algorithm is more suitable for filling regions which are defined
Anti- aliasing: Most aliasing artifacts, when appear in a static image at a moderate resolution, are often tolerable, and in many cases, negligible. However, they can have a signi
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Simulating Accelerations - Computer Animation In previous block, we have seen the dominance and function of mathematics in computer graphics and now, we will undertake the invo
Question) Compute the following: a) Size of 420 × 300 image at 240 pixels per inch. b) Resolution (per square inch) of 3 × 2 inch image that has 768×512 pixels. c) H
Overview of Hardware Primitives With the advancement of computer technology, computer graphics has become a practical tool used in real life applications in diverse areas such
To prove ‾P (1) = p n Solution : since in the above case we determine each term excluding B n,n (u) will have numerous of (1 - u) i (i = 0 to n) consequently by using u = 1
uniform scaling and differential scaling with the help of diagram
To prove: P (u = 0) = p0 Solution : = p 0 B n,0 (u) + p 1 B n, 1 (u) +...... + p n B n, n (u)...............(1) B n,i (u) = n c i u i (1 - u) n-i B n,0
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