[SOLVED] CS 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 surfaces 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 surfaces 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 surfaces 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 surfaces 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 surfaces 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 surfaces 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 surfaces 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 surfaces 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.
cos1(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 surfaces 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 surfaces 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 surfaces 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.
cos1(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 surfaces 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.
cos1(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 surfaces 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.
HLV. LV
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 surfaces 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:
HLV. LV
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 surfaces 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. HLV.
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 NH
(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 surfaces 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. HLV.
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 NH
(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

CS: assignmentchef QQ: 1823890830 Email: [email protected]