Solution Manual for Interactive Computer Graphics, 8th Edition

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Angel and Shreiner: Interactive Computer Graphics, EighthEditionChapter 1 Solutions1.1 The main advantage of the pipeline is that each primitive can beprocessed independently. Not only does this architecture lead to fastperformance, it reduces memory requirements because we need not keep allobjects available. The main disadvantage is that we cannot handle mostglobal effects such as shadows, reflections, and blending in a physicallycorrect manner.1.2 If we use lines of constant longitude and lines of constant latitude, theirintersections define a set of quadrilaterals. If we draw diagonals for eachquadrilateral, we get a set of triangles. However, at the poles all the curvesof constant longitude meet and we get triangles rather than quadrilaterals.1.3 We derive this algorithm later in Chapter 6. First, we can form thetetrahedron by finding four equally spaced points on a unit sphere centeredat the origin. One approach is to start with one point on thezaxis(0,0,1). We then can place the other three points in a plane of constantz.One of these three points can be placed on theyaxis. To satisfy therequirement that the points be equidistant, the point must be at(0,22/3,1/3). The other two can be found by symmetry to be at(6/3,2/3,1/3) and (6/3,2/3,1/3).We can subdivide each face of the tetrahedron into four equilateraltriangles by bisecting the sides and connecting the bisectors. However, thebisectors of the sides are not on the unit circle so we must push thesepoints out to the unit circle by scaling the values. We can continue thisprocess recursively on each of the triangles created by the bisection process.1.4 Suppose that the line segment is between the points (x1, y1) and(x2, y2) We can use the endpoints of the line segment to determine theslope and y intercept of a line of which the segment is part, i.e.y=mx+h=y2y1x2x1x+y1y2y1x2x1x1.Note that we can deal with horizontal and vertical line segments as specialcases. We can find the intersections with the sides of the window bysubstitutingy=ymax,y=ymin,x=xmax, andx=xmin(the equationsfor the sides of the window) into the above equation. We can check the1

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