[SOLVED] CS代考 300029 Engineering Visualization: Tutorial 10

300029 Engineering Visualization: Tutorial 10
Examples in lecture: Examples in tutorial: Homework:
Tutorial 10 Illumination Models
Problems with two asterisks (**) Problems with one asterisk (*) All problems for this tutorial

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Using the diffuse reflection model on a Lambertian surface (Figure P1), the amount of light observed by a viewer is proportional to cos, where  is the angle between the direction L (L 1) from a point on the surface to the light source and the
surface normal N ( N  1) at the point. Given L  (0.3785, 0.018, 0.9254) and N  (0.4888, 0.1835, 0.8529) , calculate cos and find the angle  in degrees.
Using the diffuse reflection model on a Lambertian surface, the amount of light observed by a viewer is proportional to cos, where  is the angle between the direction L (L 1) from a point on the surface to the light source and the surface
normal N ( N  1) at the point. Given L  (0.7408, 0.105, 0.6634) and N  (0.9102, 0.2611, 0.3214) , calculate cos and find the angle  in degrees.
*Problem 3
In the Phong illumination model (Figure P3), L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point,
R ( R  1) is the direction of reflection from the point and V ( V  1) is the direction
from the point to the viewer. Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is
the angle between R and V . Given L  (0.3785, 0.018, 0.9254) N  (0.4888, 0.1835, 0.8529) , calculate R.
Dr. J.J. Zou, WSU School of Engineering

In the Phong illumination model, L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R ( R  1)
is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is proportional to
cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . Given L  (0.7408, 0.105, 0.6634) and
N  (0.9102, 0.2611, 0.3214) , calculate R. Problem 5
In the Phong illumination model (Figure P5), L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point,
R ( R  1) is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is
proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . It is given that L  (0.3785, 0.018, 0.9254), N  (0.4888, 0.1835, 0.8529) and V  (0.2565, 0.4548, 0.8529). Calculate cos and
find  in degrees.
In the Phong illumination model (Figure P6), L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point,
R ( R  1) is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is
proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . It is given that L  (0.7408, 0.105, 0.6634), N  (0.9102, 0.2611, 0.3214) and V  (0.8511, 0.4152, 0.3214). Calculate cos and
find  in degrees.
300029 Engineering Visualization: Tutorial 10
Dr. J.J. Zou, WSU School of Engineering

300029 Engineering Visualization: Tutorial 10
In the simplified Phong illumination model (Figure P7), L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H (H 1) is the halfway vector between
L and V . Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H. It is given that L  (0.3785, 0.018, 0.9254) and
V  (0.2565, 0.4548, 0.8529). Calculate H.
In the simplified Phong illumination model, L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H ( H  1) is the halfway vector between L and V .
Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H . It is given that L  (0.7408, 0.105, 0.6634) and V  (0.8511, 0.4152, 0.3214). Calculate
Dr. J.J. Zou, WSU School of Engineering Page 3

300029 Engineering Visualization: Tutorial 10
In the simplified Phong illumination model (Figure P9), L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H (H 1) is the halfway vector between
L and V . Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H . It is given that L  (0.3785, 0.018, 0.9254), N  (0.4888, 0.1835, 0.8529)
and V  (0.2565, 0.4548, 0.8529). Calculate cos and find  in degrees.
Problem 10
In the simplified Phong illumination model (Figure P10), L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H (H 1) is the halfway vector between
L and V . Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H . It is given that L  (0.7408, 0.105, 0.6634), N  (0.9102, 0.2611, 0.3214)
and V  (0.8511, 0.4152, 0.3214). Calculate cos and find  in degrees.
Figure P10
Dr. J.J. Zou, WSU School of Engineering

Using the diffuse reflection model on a Lambertian surface (Figure P1), the amount of light observed by a viewer is proportional to cos, where  is the angle between the direction L (L 1) from a point on the surface to the light source and the
surface normal N ( N  1) at the point. Given L  (0.3785, 0.018, 0.9254) N  (0.4888, 0.1835, 0.8529) , calculate cos and find the angle  in degrees.
(0.3785, 0.018, 0.9254)(0.4888, 0.1835, 0.8529)  0.3785  0.4888  0.018  0.1835  0.9254  0.8529  0.9776.
 cos1(0.9776)12.1509.
300029 Engineering Visualization: Tutorial 10
Solutions to Problems in Tutorial 10
Dr. J.J. Zou, WSU School of Engineering

normal N ( N  1) at the point. Given L  (0.7408, 0.105, 0.6634) N  (0.9102, 0.2611, 0.3214) , calculate cos and find the angle  in degrees.
(0.7408, 0.105, 0.6634)(0.9102, 0.2611, 0.3214)  0.7408  0.9102  0.105  0.2611  0.6634  0.3214  0.915.
  cos 1 (0.915)  23.7961.
300029 Engineering Visualization: Tutorial 10
Using the diffuse reflection model on a Lambertian surface, the amount of light observed by a viewer is proportional to cos, where  is the angle between the direction L (L 1) from a point on the surface to the light source and the surface
Dr. J.J. Zou, WSU School of Engineering

L  N  (0.3785, 0.018, 0.9254)  (0.4888, 0.1835, 0.8529)  0.3785  0.4888  0.018  0.1835  0.9254  0.8529  0.9776.
R  2(L  N)N  L
 2(0.9776)(0.4888, 0.1835, 0.8529)  (0.3785, 0.018, 0.9254)
 ((2)(0.9776)(0.4888)  0.3785, (2)(0.9776)(0.1835)  0.018, (2)(0.9776)(0.8529)  0.9254)  (0.5772, 0.3407, 0.7421).
R  L  (R  N)N  (L  N)N  2(L  N)N  R  2(L  N)N  L
300029 Engineering Visualization: Tutorial 10
*Problem 3
In the Phong illumination model (Figure P3), L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point,
R ( R  1) is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is
proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . Given L  (0.3785, 0.018, 0.9254) and
N  (0.4888, 0.1835, 0.8529) , calculate R.
Dr. J.J. Zou, WSU School of Engineering Page 7

In the Phong illumination model, L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R ( R  1)
is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is proportional to
cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . Given L  (0.7408, 0.105, 0.6634) and
N  (0.9102, 0.2611, 0.3214) , calculate R. Solution:
R  L  (R  N)N  (L  N)N  2(L  N)N  R  2(L  N)N  L
L  N  (0.7408, 0.105, 0.6634)  (0.9102, 0.2611, 0.3214)  0.7408  0.9102  0.105  0.2611  0.6634  0.3214  0.915.
R  2(L  N)N  L
 2(0.915)(0.9102, 0.2611, 0.3214)  (0.7408, 0.105, 0.6634)
 ((2)(0.915)(0.9102)  0.7408, (2)(0.915)(0.2611)  0.105, (2)(0.915)(0.3214)  0.6634)  (0.9249, 0.3728,  0.0753).
300029 Engineering Visualization: Tutorial 10
Dr. J.J. Zou, WSU School of Engineering Page 8

In the Phong illumination model (Figure P5), L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point,
R ( R  1) is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is
proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . It is given that L  (0.3785, 0.018, 0.9254), N  (0.4888, 0.1835, 0.8529) and V  (0.2565, 0.4548, 0.8529). Calculate cos and
find  in degrees.
L  N  (0.3785, 0.018, 0.9254)  (0.4888, 0.1835, 0.8529)  0.3785  0.4888  0.018  0.1835  0.9254  0.8529  0.9776.
R  2(L  N)N  L
 2(0.9776)(0.4888, 0.1835, 0.8529)  (0.3785, 0.018, 0.9254)
 ((2)(0.9776)(0.4888)  0.3785, (2)(0.9776)(0.1835)  0.018, (2)(0.9776)(0.8529)  0.9254)  (0.5772, 0.3407, 0.7421).
cos  R  V
 (0.5772, 0.3407, 0.7421)  (0.2565, 0.4548, 0.8529)  0.5772  0.2565  0.3407  0.4548  0.7421  0.8529  0.9359.
 cos1(0.9359)  20.6199.
R  L  (R  N)N  (L  N)N  2(L  N)N  R  2(L  N)N  L
300029 Engineering Visualization: Tutorial 10
Dr. J.J. Zou, WSU School of Engineering

In the Phong illumination model (Figure P6), L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point,
R ( R  1) is the direction of reflection from the point and V ( V  1) is the direction from the point to the viewer. Then, the specular reflection from the surface is
proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between R and V . It is given that L  (0.7408, 0.105, 0.6634), N  (0.9102, 0.2611, 0.3214) and V  (0.8511, 0.4152, 0.3214). Calculate cos and
find  in degrees.
L  N  (0.7408, 0.105, 0.6634)  (0.9102, 0.2611, 0.3214)  0.7408  0.9102  0.105  0.2611  0.6634  0.3214  0.915.
R  2(L  N)N  L
 2(0.915)(0.9102, 0.2611, 0.3214)  (0.7408, 0.105, 0.6634)
 ((2)(0.915)(0.9102)  0.7408, (2)(0.915)(0.2611)  0.105, (2)(0.915)(0.3214)  0.6634)  (0.9249, 0.3728,  0.0753).
cos  R  V
(0.9249, 0.3728, 0.0753)(0.8511, 0.4152, 0.3214)  0.9249  0.8511  0.3728  0.4152  (0.0753)  0.3214  0.9177.
 cos1(0.9359)  23.4072.
R  L  (R  N)N  (L  N)N  2(L  N)N  R  2(L  N)N  L
300029 Engineering Visualization: Tutorial 10
Dr. J.J. Zou, WSU School of Engineering

In the simplified Phong illumination model (Figure P7), L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H (H 1) is the halfway vector between
L and V . Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between
N and H. It is given that L  (0.3785, 0.018, 0.9254) V  (0.2565, 0.4548, 0.8529). Calculate H.
HLV. LV
300029 Engineering Visualization: Tutorial 10
L  V  (0.3785, 0.018, 0.9254)  (0.2565, 0.4548, 0.8529)  (0.635, 0.4728, 1.7783).
L  V  (0.635)2  (0.4728)2  (1.7783)2  1.9466.
H  (0.635, 0.4728, 1.7783) 1.9466
 (0.3262, 0.2429, 0.9136).
Dr. J.J. Zou, WSU School of Engineering

300029 Engineering Visualization: Tutorial 10
In the simplified Phong illumination model, L ( L  1) is the direction from a point on the surface to the light source, N (N 1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H ( H  1) is the halfway vector between L and V .
Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H . It is given that L  (0.7408, 0.105, 0.6634) and V  (0.8511, 0.4152, 0.3214). Calculate
H. Solution:
HLV. LV
L  V  (0.7408, 0.105, 0.6634)  (0.8511, 0.4152, 0.3214)  (1.5919, 0.5202, 0.9848).
L  V  (1.5919)2  (0.5202)2  (0.9848)2  1.9429.
H  (1.5919, 0.5202, 0.9848) 1.9429
 (0.8194, 0.2678, 0.5069).
Dr. J.J. Zou, WSU School of Engineering

In the simplified Phong illumination model (Figure P9), L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H (H 1) is the halfway vector between
L and V . Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H . It is given that L  (0.3785, 0.018, 0.9254), N  (0.4888, 0.1835, 0.8529)
and V  (0.2565, 0.4548, 0.8529). Calculate cos and find  in degrees.
Find H. HLV.
300029 Engineering Visualization: Tutorial 10
L  V  (0.3785, 0.018, 0.9254)  (0.2565, 0.4548, 0.8529)
 (0.635, 0.4728, 1.7783).
L  V  (0.635)2  (0.4728)2  (1.7783)2  1.9466.
H  (0.635, 0.4728, 1.7783) 1.9466
 (0.3262, 0.2429, 0.9136).
Find cos and . cos NH
 (0.4888, 0.1835, 0.8529)  (0.3262, 0.2429, 0.9136)  0.4888  0.3262  0.1835  0.2429  0.8529  0.9136  0.9832.
  cos 1 (0.9832)  10.5243.
Dr. J.J. Zou, WSU School of Engineering

Problem 10
In the simplified Phong illumination model (Figure P10), L (L 1) is the direction from a point on the surface to the light source, N ( N  1) is the surface normal at the point, R (R 1) is the direction of reflection from the point, V (V 1) is the direction from the point to the viewer and H (H 1) is the halfway vector between
L and V . Then, the specular reflection from the surface is proportional to cosn  where n is the surface’s specular reflection exponent and  is the angle between N and H . It is given that L  (0.7408, 0.105, 0.6634), N  (0.9102, 0.2611, 0.3214)
and V  (0.8511, 0.4152, 0.3214). Calculate cos and find  in degrees.
Find H. HLV.
300029 Engineering Visualization: Tutorial 10
Figure P10
L  V  (0.7408, 0.105, 0.6634)  (0.8511, 0.4152, 0.3214)
 (1.5919, 0.5202, 0.9848).
L  V  (1.5919)2  (0.5202)2  (0.9848)2  1.9429.
H  (1.5919, 0.5202, 0.9848) 1.9429
 (0.8194, 0.2678, 0.5069).
Find cos and . cos NH
 (0.9102, 0.2611, 0.3214)  (0.8194, 0.2678, 0.5069)  0.9102  0.8194  0.2611  0.2678  0.3214  0.5069  0.9786.
  cos 1 (0.9832)  11.8621.
Dr. J.J. Zou, WSU School of Engineering

程序代写 CS代考加微信: assignmentchef QQ: 1823890830 Email: [email protected]