CS代写 A solid is generated by rotating a polygon P P P P P P , shown in Figure Pr

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 : 3x2y5z40,
Plane 2 : 2x3y4z60.
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 : 5x4y2z30,
Plane 2 : 4x6y3z20.
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 : 2x5y3z40,
Plane 2 : 6x3y4z20.
The projecting line L is defined by the following parametric equations:
y  2, where  is a real number.  z3,
(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 0u1and 0v1, 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 0u1and 0v1, 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 0u1and 0v1, 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: P’T(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 20050 15005000(200)1 150 P’TP 0 1 0 0 50 0501500001  50 .
0 0 1 0 0  0500501001   0  0 0 0 1 1  0500500011   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: P”R()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
0150 0 50  0 0 
 0 0 0 1 1 
 cos10(150)050sin10001 
P”RP’ 0 1 0
 sin10 0 cos10
sin  0
147.7212   0(150)1500001    50 .
sin10(150)050cos10001  26.0472 
300029 Engineering Visualization: Tutorial 9
 0(150)0500011   1 
 Step 3: Perform translation so that Pr returns to its original location and P1 ‘ ‘
becomes the transformed point P1”’ required.
For 3D translation: P”’T(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 ,T0 1 0 0 .
 1  0001
26.0472 0 0 1 0 1 0 0 200147.7212
P”’TP”0 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)150026.047201    50 .
 0(147.7212)050126.047201  26.0472  0(147.7212)050026.047211   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 ,T0 0 1 0 . 1 0001
1 0 0 200160 1160015000(200)1 40 0 1 0 0 150  016011500001  150
P2’TP2 0 0 1 0  0  016001501001  0 . 0 0 0 1 1  016001500011   1 
        Step 2: Rotate P2 ‘ about the y-axis through an angle of 20 to become P2 ‘ ‘ .
cos 0 sin 0 For3Drotationabout y-axis: P2”Ry()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: P2’T(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 040 P2” RP2’  0 1 0 0150
sin20 0 cos20 0 0 
 0 0 0 11
 cos20(40)0150sin20001 
37.5877   0(40)11500001    150 .
sin20(40)0150cos20001  13.6808 
 0(40)01500011   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 ,T0 1 0 0.
13.6808 0 0 1 0 1 0001
1 0 0 200 37.5877 P2”’TP2”0 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)1150013.680801    150 .
 0(37.5877)0150113.680801  13.6808  0(37.5877)0150013.680811   1 
Therefore, the transformed point is located at (162.4123,150,13.6808).
300029 Engineering Visualization: Tutorial 9
40 cos20 0 sin20 0 Now,P2’150, 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: P3’T(tx,ty,tz)P3, where T(tx,ty,tz)0 1 0 ty.
150 1 0 0  200 Now,P3 200,T0 1 0 0 .
P1 P0 x x = 200, z = 0.
 0  0 0 1 0  1 0001
1 0 0 200150 1150020000(200)1 50 P3’T P3  0 1 0 0 200   015012000001   200.
0 0 1 0  0   015002001001   0  0 0 0 1 1  015002000011  1
 Step 2: Rotate P3 ‘ about the y-axis through an angle of 30 to become P3 ‘ ‘ .
cos 0 sin 0
For3Drotationabout y-axis: P3”Ry()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 050 P3” RP3’  0 1 0 0200
sin30 0 cos30 0 0 
 0 0 0 11
 cos30(50)0200sin30001 
43.30137   0(50)12000001    200 .
sin30(50)0200cos30001  25 
 0(50)02000011   1 
 Step 3: Perform translation so that Pr returns to its original location and P3 ‘ ‘
becomes the transformed point P3 ‘ ‘ ‘ required.
For 3D translation: P3”’T(tx,ty,tz)P3”, where T(tx,ty,tz)0 1 0 ty.
 43.3013 1 0 0 200 Now,P3” 200 ,T0 1 0 0.
 25  0 0 1 0  1 0001
1 0 0 200 43.3013 P3”’TP3”0 1 0 0  200 
001 0 25  000 1 1 
1(43.3013)02000252001 156.6987
200 . 25 
  0(43.3013)120002501     0(43.3013)020012501    0(43.3013)020002511  
1  Therefore, the transformed point is located at (156.6987, 200, 25) .
300029 Engineering Visualization: Tutorial 9
50  cos30 0 sin30 0 Now,P3’200, 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: P4’T(tx,ty,tz)P4, where T(tx,ty,tz)0 1 0 ty.
60 1 0 0 200 Now,P4 300,T0 1 0 0 .
 0  0 0 1 0  1 0001
1 0 0 20060 160030000(200)1 140 P4’TP4 0 1 0 0 300 06013000001 300.
0 0 1 0  0   06003001001   0  0 0 0 1 1  06003000011   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,P4’300, R 0 1 0 0.
 0  sin40 0 cos40 0  1   0 0 0 1
 cos40 0 sin40 0140 P4”RP4’ 0 1 0 0 300 
sin40 0 cos40 0 0 
 0 0 0 1 1 
 cos40(140)0300sin40001 
107.2462   0(140)13000001    300 .
sin40(140)0300cos40001  89.9903 
 0(140)03000011   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 ,T0 1 0 0.
89.9903 0 0 1 0  1  0001
1 0 0 200107.2462 P4”’TP4”0 1 0 0  300 
0 0 1 0  89.9903 
0001 1  1(107.2462)0300089.99032001 92.7538
  0(107.2462)1300089.990301    300 .  0(107.2462)0300189.990301  89.9903  0(107.2462)0300089.990311   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: P4”Ry()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 : 3x2y5z40,
Plane 2 : 2x3y4z60.
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)40  (6620)40   4/200.2. x1 22(0.2)0.4.
y1 33(0.2)0.6. z1 44(0.2)0.8.
Find the intersection P (x , y , z ) of L and  . 2222 2
2x3y4z60
 2(2)3(3)4(4)60  (4916)60  6/110.5455. x2 22(0.5455)1.0909.
y2 33(0.5455)1.6364. z2 44(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 : 5x4y2z30,
Plane 2 : 4x6y3z20.
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
5x4y2z30
 5(2)4(3)2(4)30  (10128)30  3/100.3. x1  2  2(0.3)  0.6.
y1 33(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)20 x2 22(0.0909)0.1818.
y2 33(0.0909)0.2727.
z2 44(0.0909)0.3636.
 (81812)20
  2/220.0909.
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 : 2x5y3z40,
Plane 2 : 6x3y4z20.
The projecting line L is defined by the following parametric equations:
y  2, where  is a real number.  z3,
(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)40  (8109)40 x1  4  4(0.5714)  2.2857.
y1  2  2(0.5714)  1.1429. z1 33(0.5714)1.7143.
Find the intersection P (x , y , z ) of L and  . 2222 2
  4/70.5714.
  2/300.0667.
6x  3y  4z  2  0
 6(4)3(2)4(3)20 x2 44(0.0667)0.2667.
y2 22(0.0667)0.1333. z2 33(0.0667)0.2.
 (24612)20
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

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