Angel and Shreiner: Interactive Computer Graphics, Seventh
Edition
Chapter 6 Solutions
6.1 Point sources produce a very harsh lighting. Such images are
characterized by abrupt transitions between light and dark. The ambient
light in a real scene is dependent on both the lights on the scene and the
6.2 With smooth shading, the colors at the vertices are interpolated across
the polygon. If the polygon is large, the viewer can be close to points in
6.3 If we were to take into account a light source being obscured by an
object, we would have to have all polygons available so as to test for this
6.4 A viewer sees a point on an illuminated surface as a source of light.
Hence, the amount of light received by the viewer from each point on the
6.5 Materials absorb light from sources. Thus, a surface that appears red
under white light appears so because the surface absorbs all wavelengths of
6.6 See problem 1.2. We also can argue from symmetry as follows.
Suppose that we place one of the vertices at (0,1,0). We can place the
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6.7 Let ψbe the angle between the normal and the halfway vector, φbe
the angle between the viewer and the reflection angle, and θbe the angle
between the normal and the light source. If all the vectors lie in the same
6.8 If the surface is curved, we have to compute the normal at each point;
if it is flat the normal is constant. If both the viewer and the light source
6.13 Without loss of generality, we can consider the problem in two
dimensions. Suppose that the first material has a velocity of light of v1and
the second material has a light velocity of v2. Furthermore, assume that
the axis y= 0 separates the two materials.
Place a point light source at (0, h) where h > 0 and a viewer at (x, y)
where y < 0. Light will travel in a straight line from the source to a point
(t, 0) where it will leave the first material and enter the second. It will
6.14 If we orient the surface so it points in the direction of the halfway
vector, then we have equal angles between the halfway vector and the
6.19 Shading requires that when we transform normals and points, we
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6.21 Probably the easiest approach to this problem is to rotate the given
plane to plane z= 0 and rotate the light source and objects in the same
6.23 A global rendering approach would generate all shadows correctly. In
a global renderer, as each point is shaded, a calculation is done to see
which light sources shine on it. The projection approach assumes that we
can project each polygon onto all other polygons. If the shadow of a given
6.26 The normal at each point on the sphere points from the origin to that
point. Given this normal nand the location of the viewer p, we can
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