[SOLVED] CS A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr

A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr1, through 012345
360 about an axis defined by x 200, z 0. Calculate the coordinates of P1 after it is rotated through 10 about the axis.
Figure Pr1
A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr2, through

Copyright By Assignmentchef assignmentchef

360 about an axis defined by x 200, z 0. Calculate the coordinates of P2 after it
is rotated through 20 about the axis.
300029 Engineering Visualization: Tutorial 9
Tutorial 9 Solid Modelling, Visible Surface Determination and Surface Details
Examples in lecture: Examples in tutorial: Homework:
Problems with two asterisks (**) Problems with one asterisk (*) All problems for this tutorial
P2 P2 = (160, 150, 0), P3 = (150, 200, 0),
P1 P0 x x = 200, z = 0.
P0 = (200, 50, 0), P1 = (50, 50, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
P1 P0 x x = 200, z = 0.
Figure Pr2
P0 = (200, 50, 0), P1 = (50, 50, 0),
P2 = (160, 150, 0), P3 = (150, 200, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
Dr. J.J. Zou, WSU School of Engineering

*Problem 4
360 about an axis defined by x 200, z 0. Calculate the coordinates of P4 after it
is rotated through 40 about the axis.
300029 Engineering Visualization: Tutorial 9
A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr3, through 012345
360 about an axis defined by x 200, z 0. Calculate the coordinates of P3 after it is rotated through 30 about the axis.
P2 P2 = (160, 150, 0), P3 = (150, 200, 0),
P1 P0 x x = 200, z = 0.
Figure Pr3
A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr4, through
P0 = (200, 50, 0), P1 = (50, 50, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
P1 P0 x x = 200, z = 0.
Figure Pr4
P0 = (200, 50, 0), P1 = (50, 50, 0),
P2 = (160, 150, 0), P3 = (150, 200, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
Dr. J.J. Zou, WSU School of Engineering

A viewer views two planes 1 and 2 along a projecting line L. The viewer is located at the origin (0, 0, 0). The planes are defined below:
Plane 1 : 3x2y5z40,
Plane 2 : 2x3y4z60.
The projecting line L is defined by the following parametric equations:
y 3, where is a real number. z 4 ,
(a) Find the intersections of the line with the planes.
(b) Which intersection is visible to the viewer? Give reasons to support your answer.
A viewer views two planes 1 and 2 along a projecting line L. The viewer is located at the origin (0, 0, 0). The planes are defined below:
Plane 1 : 5x4y2z30,
Plane 2 : 4x6y3z20.
The projecting line L is defined by the following parametric equations:
y 3, where is a real number. z 4 ,
(a) Find the intersections of the line with the planes.
(b) Which intersection is visible to the viewer? Give reasons to support your answer.
A viewer views two planes 1 and 2 along a projecting line L. The viewer is located at the origin (0, 0, 0). The planes are defined below:
Plane 1 : 2x5y3z40,
Plane 2 : 6x3y4z20.
The projecting line L is defined by the following parametric equations:
y 2, where is a real number. z3,
(a) Find the intersections of the line with the planes.
(b) Which intersection is visible to the viewer? Give reasons to support your answer.
300029 Engineering Visualization: Tutorial 9
Dr. J.J. Zou, WSU School of Engineering Page 3

A texture image is mapped to a bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)],for 0u1and 0v1, definedbythefollowing16
control points:
p00 (12,8,20), p10 (13,12,10), p20 (27,8,10), p30 (28,12,20),
p01 (8,17,10), p11 (17,13,10), p21 (23,17,10), p31 (32,13,10),
p02 (12,23,10), p12 (13,27,10), p22 (27,23,10), p32 (28,27,10),
p03 (15,28,10), p13 (17,28,10), p23 (23,32,10), p33 (32,28,10).
The location of a point on the texture image is specified by its normalised coordinates (s, t), for 0 s 1 and 0 t 1, in the texture image coordinate system.
If (s, t) (0.1, 0.2), find the coordinates of the corresponding point on the bicubic Bezier surface patch in the world coordinate system after texture mapping.
A texture image is mapped to a bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)],for 0u1and 0v1, definedbythefollowing16
control points:
p00 (12,8,20), p10 (13,12,10), p20 (27,8,10), p30 (28,12,20),
p01 (8,17,10), p11 (17,13,10), p21 (23,17,10), p31 (32,13,10),
p02 (12,23,10), p12 (13,27,10), p22 (27,23,10), p32 (28,27,10),
p03 (15,28,10), p13 (17,28,10), p23 (23,32,10), p33 (32,28,10).
The location of a point on the texture image is specified by its normalised coordinates (s, t), for 0 s 1 and 0 t 1, in the texture image coordinate system.
If (s, t) (0, 1), find the coordinates of the corresponding point on the bicubic Bezier surface patch in the world coordinate system after texture mapping.
Problem 10
A texture image is mapped to a bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)],for 0u1and 0v1, definedbythefollowing16
control points:
p00 (12,8,20), p10 (13,12,10), p20 (27,8,10), p30 (28,12,20),
p01 (8,17,10), p11 (17,13,10), p21 (23,17,10), p31 (32,13,10),
p02 (12,23,10), p12 (13,27,10), p22 (27,23,10), p32 (28,27,10),
p03 (15,28,10), p13 (17,28,10), p23 (23,32,10), p33 (32,28,10).
The location of a point on the texture image is specified by its normalised coordinates (s, t), for 0 s 1 and 0 t 1, in the texture image coordinate system.
If (s, t) (1, 0), find the coordinates of the corresponding point on the bicubic Bezier surface patch in the world coordinate system after texture mapping.
300029 Engineering Visualization: Tutorial 9
Dr. J.J. Zou, WSU School of Engineering Page 4

A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr1, through 012345
360 about an axis defined by x 200, z 0. Calculate the coordinates of P1 after it is rotated through 10 about the axis.
Use homogeneous coordinates for 3D transformation.
Step 1: Select a reference point Pr (200, 0, 0) on the axis and then perform translation so that Pr is moved to the origin and P1 becomes P1 .
1 0 0 tx For 3D translation: PT(t ,t ,t )P, where T(t ,t ,t )0 1 0 ty.
50 1 0 0 200 Now, P 50, T 0 1 0 0 .
0 0 0 1 0 1 0001
0 0 1 tz 0 0 0 1
1 0 0 20050 15005000(200)1 150 PTP 0 1 0 0 50 0501500001 50 .
0 0 1 0 0 0500501001 0 0 0 0 1 1 0500500011 1
Step 2: Rotate P1 about the y-axis through an angle of 10 to become P1 .
300029 Engineering Visualization: Tutorial 9
Solutions to Problems in Tutorial 9
P2 P2 = (160, 150, 0), P3 = (150, 200, 0),
P1 P0 x x = 200, z = 0.
Figure Pr1
P0 = (200, 50, 0), P1 = (50, 50, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
Dr. J.J. Zou, WSU School of Engineering Page 5

For3Drotationabout y-axis: PR()P,whereR() 0
0 sin 0 1 0 0. 0 cos 0 0 0 1
150 cos10 0 Now,P 50 ,R 0 1
sin10 0 0 0.
1 0 0 0 1
0 sin10 0 cos10 0 sin10
0150 0 50 0 0
0 0 0 1 1
cos10(150)050sin10001
PRP 0 1 0
sin10 0 cos10
sin 0
147.7212 0(150)1500001 50 .
sin10(150)050cos10001 26.0472
300029 Engineering Visualization: Tutorial 9
0(150)0500011 1
Step 3: Perform translation so that Pr returns to its original location and P1
becomes the transformed point P1 required.
For 3D translation: PT(t ,t ,t )P, where T(t ,t ,t )0 1 0 ty.
1 xyz1 xyz
1 0 0 tx 0 0 1 tz
147.7212 1 0 0 200 Now,P 50 ,T0 1 0 0 .
1 0001
26.0472 0 0 1 0 1 0 0 200147.7212
PTP0 1 0 0 50
0 0 1 0 26.0472
0001 1
1 (147.7212) 0 50 0 26.0472 200 1
52.2788 0(147.7212)150026.047201 50 .
0(147.7212)050126.047201 26.0472 0(147.7212)050026.047211 1
Therefore, the transformed point is located at (52.2788, 50, 26.0472) .
Dr. J.J. Zou, WSU School of Engineering

P2 P2 = (160, 150, 0), P3 = (150, 200, 0),
P1 P0 x x = 200, z = 0.
160 1 0 0 200 150 0 1 0 0
Now,P2 0 ,T0 0 1 0 . 1 0001
1 0 0 200160 1160015000(200)1 40 0 1 0 0 150 016011500001 150
P2TP2 0 0 1 0 0 016001501001 0 . 0 0 0 1 1 016001500011 1
Step 2: Rotate P2 about the y-axis through an angle of 20 to become P2 .
cos 0 sin 0 For3Drotationabout y-axis: P2Ry()P2,whereRy() 0 1 0 0.
300029 Engineering Visualization: Tutorial 9
A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr2, through 012345
360 about an axis defined by x 200, z 0. Calculate the coordinates of P2 after it is rotated through 20 about the axis.
Figure Pr2
Use homogeneous coordinates for 3D transformation.
Step 1: Select a reference point Pr (200, 0, 0) on the axis and then perform translation so that Pr is moved to the origin and P2 becomes P2 .
1 0 0 tx For 3D translation: P2T(tx,ty,tz)P2, where T(tx,ty,tz)0 1 0 ty.
P0 = (200, 50, 0), P1 = (50, 50, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
0 0 1 tz 0 0 0 1
sin 0 cos 0 0 0 0 1
Dr. J.J. Zou, WSU School of Engineering Page 7

cos20 0 sin20 040 P2 RP2 0 1 0 0150
sin20 0 cos20 0 0
0 0 0 11
cos20(40)0150sin20001
37.5877 0(40)11500001 150 .
sin20(40)0150cos20001 13.6808
0(40)01500011 1
Step 3: Perform translation so that Pr returns to its original location and P2
becomes the transformed point P2 required.
For 3D translation : P2 T(tx ,ty ,tz )P2 , where T(tx ,ty ,tz ) 0 1 0 ty .
37.5877 1 0 0 200 Now,P2 150 ,T0 1 0 0.
13.6808 0 0 1 0 1 0001
1 0 0 200 37.5877 P2TP20 1 0 0 150
0 0 1 0 13.6808
000 1 1
1 (37.5877) 0 150 0 13.6808 200 1
162.4123 0(37.5877)1150013.680801 150 .
0(37.5877)0150113.680801 13.6808 0(37.5877)0150013.680811 1
Therefore, the transformed point is located at (162.4123,150,13.6808).
300029 Engineering Visualization: Tutorial 9
40 cos20 0 sin20 0 Now,P2150, R 0 1 0 0.
0 sin20 0 cos20 0 1 0 0 0 1
1 0 0 tx 0 0 1 tz
Dr. J.J. Zou, WSU School of Engineering

A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr3, through 012345
360 about an axis defined by x 200, z 0. Calculate the coordinates of P3 after it is rotated through 30 about the axis.
Figure Pr3
Use homogeneous coordinates for 3D transformation.
Step 1: Select a reference point Pr (200, 0, 0) on the axis and then perform translation so that Pr is moved to the origin and P3 becomes P3 .
1 0 0 tx For 3D translation: P3T(tx,ty,tz)P3, where T(tx,ty,tz)0 1 0 ty.
150 1 0 0 200 Now,P3 200,T0 1 0 0 .
P1 P0 x x = 200, z = 0.
0 0 0 1 0 1 0001
1 0 0 200150 1150020000(200)1 50 P3T P3 0 1 0 0 200 015012000001 200.
0 0 1 0 0 015002001001 0 0 0 0 1 1 015002000011 1
Step 2: Rotate P3 about the y-axis through an angle of 30 to become P3 .
cos 0 sin 0
For3Drotationabout y-axis: P3Ry()P3,whereRy() 0 1 0 0. sin 0 cos 0
0 0 0 1 Dr. J.J. Zou, WSU School of Engineering Page 9
300029 Engineering Visualization: Tutorial 9
P2 P2 = (160, 150, 0), P3 = (150, 200, 0),
P0 = (200, 50, 0), P1 = (50, 50, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
0 0 1 tz 0 0 0 1

cos30 0 sin30 050 P3 RP3 0 1 0 0200
sin30 0 cos30 0 0
0 0 0 11
cos30(50)0200sin30001
43.30137 0(50)12000001 200 .
sin30(50)0200cos30001 25
0(50)02000011 1
Step 3: Perform translation so that Pr returns to its original location and P3
becomes the transformed point P3 required.
For 3D translation: P3T(tx,ty,tz)P3, where T(tx,ty,tz)0 1 0 ty.
43.3013 1 0 0 200 Now,P3 200 ,T0 1 0 0.
25 0 0 1 0 1 0001
1 0 0 200 43.3013 P3TP30 1 0 0 200
001 0 25 000 1 1
1(43.3013)02000252001 156.6987
200 . 25
0(43.3013)120002501 0(43.3013)020012501 0(43.3013)020002511
1 Therefore, the transformed point is located at (156.6987, 200, 25) .
300029 Engineering Visualization: Tutorial 9
50 cos30 0 sin30 0 Now,P3200, R 0 1 0 0.
0 sin30 0 cos30 0 1 0 0 0 1
1 0 0 tx 0 0 1 tz
Dr. J.J. Zou, WSU School of Engineering

Use homogeneous coordinates for 3D transformation.
Step 1: Select a reference point Pr (200, 0, 0) on the axis and then perform translation so that Pr is moved to the origin and P4 becomes P4 .
1 0 0 tx For 3D translation: P4T(tx,ty,tz)P4, where T(tx,ty,tz)0 1 0 ty.
60 1 0 0 200 Now,P4 300,T0 1 0 0 .
0 0 0 1 0 1 0001
1 0 0 20060 160030000(200)1 140 P4TP4 0 1 0 0 300 06013000001 300.
0 0 1 0 0 06003001001 0 0 0 0 1 1 06003000011 1
Step 2: Rotate P4 about the y-axis through an angle of 40 to become P4 .
300029 Engineering Visualization: Tutorial 9
*Problem 4
A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr4, through 012345
360 about an axis defined by x 200, z 0. Calculate the coordinates of P4 after it is rotated through 40 about the axis.
P2 P2 = (160, 150, 0), P3 = (150, 200, 0),
P1 P0 x x = 200, z = 0.
Figure Pr4
P0 = (200, 50, 0), P1 = (50, 50, 0), P4 = (60, 300, 0), P5 = (200, 300, 0).
0 0 1 tz 0 0 0 1
Dr. J.J. Zou, WSU School of Engineering Page 11

140 cos40 0 sin40 0 Now,P4300, R 0 1 0 0.
0 sin40 0 cos40 0 1 0 0 0 1
cos40 0 sin40 0140 P4RP4 0 1 0 0 300
sin40 0 cos40 0 0
0 0 0 1 1
cos40(140)0300sin40001
107.2462 0(140)13000001 300 .
sin40(140)0300cos40001 89.9903
0(140)03000011 1
Step 3: Perform translation so that Pr returns to its original location and P4
becomes the transformed point P4 required.
For 3D translation : P4 T(tx ,ty ,tz )P4 , where T(tx ,ty ,tz ) 0 1 0 ty .
107.2462 1 0 0 200 Now,P4 300 ,T0 1 0 0.
89.9903 0 0 1 0 1 0001
1 0 0 200107.2462 P4TP40 1 0 0 300
0 0 1 0 89.9903
0001 1 1(107.2462)0300089.99032001 92.7538
0(107.2462)1300089.990301 300 . 0(107.2462)0300189.990301 89.9903 0(107.2462)0300089.990311 1
Therefore, the transformed point is located at (92.7538, 300, 89.9903) .
300029 Engineering Visualization: Tutorial 9
cos 0 sin 0 For3Drotationabout y-axis: P4Ry()P4,whereRy() 0 1 0 0.
sin 0 cos 0 0 0 0 1
1 0 0 tx 0 0 1 tz
Dr. J.J. Zou, WSU School of Engineering

300029 Engineering Visualization: Tutorial 9
A viewer views two planes 1 and 2 along a projecting line L. The viewer is located at the origin (0, 0, 0). The planes are defined below:
Plane 1 : 3x2y5z40,
Plane 2 : 2x3y4z60.
The projecting line L is defined by the following parametric equations:
y 3, where is a real number. z 4 ,
(a) Find the intersections of the line with the planes.
(b) Which intersection is visible to the viewer? Give reasons to support your answer.
Find the intersection P (x , y , z ) of L and . 1111 1
3x 2 y 5z 4 0
3(2)2(3)5(4)40 (6620)40 4/200.2. x1 22(0.2)0.4.
y1 33(0.2)0.6. z1 44(0.2)0.8.
Find the intersection P (x , y , z ) of L and . 2222 2
2x3y4z60
2(2)3(3)4(4)60 (4916)60 6/110.5455. x2 22(0.5455)1.0909.
y2 33(0.5455)1.6364. z2 44(0.5455)2.1818.
Find the distance from P to the viewer. 1
dist1 (x 0)2 (y 0)2 (z 0)2 (0.4)2 (0.6)2 (0.8)2 1.077. 111
Find the distance from P to the viewer. 2
dist2 (x 0)2 (y 0)2 (z 0)2 (1.0909)2 (1.6364)2 (2.1818)2 2.9374. 222
Because dist1 is less than dist2, the intersection of L with 1 is visible to the viewer.
Dr. J.J. Zou, WSU School of Engineering Page 13

A viewer views two planes 1 and 2 along a projecting line L. The viewer is located at the origin (0, 0, 0). The planes are defined below:
Plane 1 : 5x4y2z30,
Plane 2 : 4x6y3z20.
The projecting line L is defined by the following parametric equations:
y 3, where is a real number. z 4 ,
(a) Find the intersections of the line with the planes.
(b) Which intersection is visible to the viewer? Give reasons to support your answer.
Find the intersection P (x , y , z ) of L and . 1111 1
5x4y2z30
5(2)4(3)2(4)30 (10128)30 3/100.3. x1 2 2(0.3) 0.6.
y1 33(0.3)0.9.
z1 4 4(0.3) 1.2.
Find the intersection P (x , y , z ) of L and . 2222 2
4x 6y 3z 2 0
4(2)6(3)3(4)20 x2 22(0.0909)0.1818.
y2 33(0.0909)0.2727.
z2 44(0.0909)0.3636.
(81812)20
2/220.0909.
Find the distance from P to the viewer. 1
300029 Engineering Visualization: Tutorial 9
dist1 (x 0)2 (y 0)2 (z 0)2 (0.6)2 (0.9)2 (1.2)2 1.6155. 111
Find the distance from P to the viewer. 2
dist2 (x 0)2 (y 0)2 (z 0)2 (0.1818)2 (0.2727)2 (0.3636)2 0.4896. 222
Because dist2 is less than dist1, the intersection of L with 2 is visible to the viewer.
Dr. J.J. Zou, WSU School of Engineering Page 14

A viewer views two planes 1 and 2 along a projecting line L. The viewer is located at the origin (0, 0, 0). The planes are defined below:
Plane 1 : 2x5y3z40,
Plane 2 : 6x3y4z20.
The projecting line L is defined by the following parametric equations:
y 2, where is a real number. z3,
(a) Find the intersections of the line with the planes.
(b) Which intersection is visible to the viewer? Give reasons to support your answer.
Find the intersection P (x , y , z ) of L and . 1111 1
2x 5y 3z 4 0
2(4)5(2)3(3)40 (8109)40 x1 4 4(0.5714) 2.2857.
y1 2 2(0.5714) 1.1429. z1 33(0.5714)1.7143.
Find the intersection P (x , y , z ) of L and . 2222 2
4/70.5714.
2/300.0667.
6x 3y 4z 2 0
6(4)3(2)4(3)20 x2 44(0.0667)0.2667.
y2 22(0.0667)0.1333. z2 33(0.0667)0.2.
(24612)20
Find the distance from P to the viewer. 1
300029 Engineering Visualization: Tutorial 9
dist1 (x 0)2 (y 0)2 (z 0)2 (2.2857)2 (1.1429)2 (1.7143)2 3.0772. 111
Find the distance from P to the viewer. 2
dist2 (x 0)2 (y 0)2 (z 0)2 (0.2667)2 (0.1333)2 (0.2)2 0.359. 222
Because dist2 is less than dist1, the intersection of L with 2 is visible to the viewer.
Dr. J.J. Zou, WSU School of Engineering Page 15

A texture image is mapped to a bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)],for 0u1and 0v1, definedbythefollowing16
control points:
p00 (12,8,20), p10 (13,12,10), p20 (27,8,10), p30 (28,12,20),
p01 (8,17,10), p11 (17,13,10), p21 (23,17,10), p31 (32,13,10),
p02 (12,23,10), p12 (13,27,10), p22 (27,23,10), p32 (28,27,10),
p03 (15,28,10), p13 (17,2

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