[SOLVED] 计算机代考 A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is def

A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is defined by four control points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1) and p3(0,0,1). Calculate the coordinates of a point on the curve segment when u  0.3.
*Problem 2
A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is defined by four control points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1) and p3(0,0,1). Calculate the tangent vector of a point on the curve segment when u  0.
A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is defined by four control points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1) and p3(0,0,1). Calculate the tangent vector of a point on the curve segment when u  1.

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There are seven points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1), p3(0,0,1), p4(x4,y4,z4), p5(0,1,0) and p6(1,1,0). A cubic Bezier curve segment P(u)[x(u) y(u) z(u)], for 0u1 isdefinedby p0, p1, p2 and p3. curve segment Q(v)[x(v) y(v) z(v)], for 0v1 is defined by p3, p4, p5 and p6. Two segments join at p3. It is required to have C1 continuity at the joined point p3. Calculate the coordinates of p4.
There are seven points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1), p3(0,0,1), p4 (0, 1, 2), p5 (0, 1, 0) and p6 (1, 1, 0). A cubic Bezier curve segment P(u)[x(u) y(u) z(u)], for 0u1 isdefinedby p0, p1, p2 and p3. curve segment Q(v)[x(v) y(v) z(v)], for 0v1 is defined by p3, p4, p5 and p6. Two segments join at p3. Do they have G1 continuity at the joined point p3 ? Give reasons to support your answer.
300029 Engineering Visualization: Tutorial 8
Examples in lecture: Examples in tutorial: Homework:
Tutorial 8 Curves and Surfaces
Problems with two asterisks (**) Problems with one asterisk (*) All problems for this tutorial
Dr. J.J. Zou, WSU School of Engineering Page 1

A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
Calculate the coordinates of a point on the surface patch when u  0.1 and v  0.2. Problem 7
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
Calculate the surface normal at a point on the surface patch when u  0 and v  1. Problem 8
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
Calculate the surface normal at a point on the surface patch when u  1 and v  0. Problem 9
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
Another bicubic Bezier surface patch Q(s,t)[x(s,t) y(s,t) z(s,t)],for 0s1 and 0  t  1 is defined by the following 16 control points:
q00 (12,8,20), q01 (x01,y01,z01), q02 (12,7,30), q03 (9,12,30),
q10 (13,12,10), q11 (x11,y11,z11), q12 (13,3,30), q13 (9,4,10), q20 (27,8,10), q21 (x21,y21,z21), q22 (27,7,30), q23 (31,16,10),
300029 Engineering Visualization: Tutorial 8
Dr. J.J. Zou, WSU School of Engineering Page 2

Another bicubic Bezier surface patch Q(s,t)[x(s,t) y(s,t) z(s,t)],for 0s1 and 0  t  1 is defined by the following 16 control points:
q00 (12,8,20), q01 (18,1,30), q02 (12,7,30), q03 (9,12,30),
q10 (13,12,10), q11 (9,12,30), q12 (13,3,30), q13 (9,4,10), q20 (27,8,10), q21 (31,1,32), q22 (27,7,30), q23 (31,16,10), q30 (28,12,20), q31 (22,11,30), q32 (28,3,30), q33 (24,4,30).
Two surface patches join along an edge defined by their common control points, namely,p q ,p q ,p q andp q .DotheyhaveG1 continuityon
00 00 10 10 20 20 30 30
the joined edge? Give reasons to support your answer.
300029 Engineering Visualization: Tutorial 8
q30 (28,12,20), q31 (x31,y31,z31), q32 (28,3,30), q33 (24,4,30).
Two surface patches join along an edge defined by their common control points,
namely,p q,p q,p q andp q.ItisrequiredtohaveC1 00 00 10 10 20 20 30 30
continuity on the joined edge. Calculate the coordinates of q01, q11, q21 and q31. Problem 10
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
Dr. J.J. Zou, WSU School of Engineering Page 3

A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is defined by four control points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1) and p3(0,0,1). Calculate the coordinates of a point on the curve segment when u  0.3.
P(u)  [x(u) y(u) z(u)]
(1u)3p 3u(1u)2p 3u2(1u)p u3p .
P(0.3)  (1 0.3)3 p  3(0.3)(1 0.3)2 p  3(0.3)2 (1 0.3)p  (0.3)3 p 0123
0.343p0 0.441p1 0.189p2 0.027p3.
x(0.3)  (0.343)(1)  (0.441)(0)  (0.189)(0)  (0.027)(0)  0.343. y(0.3)  (0.343)(0)  (0.441)(1)  (0.189)(1)  (0.027)(0)  0.63. z(0.3)  (0.343)(0)  (0.441)(0)  (0.189)(1)  (0.027)(1)  0.216.
300029 Engineering Visualization: Tutorial 8
Solutions to Problems in Tutorial 8
Dr. J.J. Zou, WSU School of Engineering

*Problem 2
A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is defined by four control points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1) and p3(0,0,1). Calculate the tangent vector of a point on the curve segment when u  0.
P(u)  [x(u) y(u) z(u)]
(1u)3p 3u(1u)2p 3u2(1u)p u3p .
The tangent vector is :
dP(u)  dx(u) dy(u) dz(u)
300029 Engineering Visualization: Tutorial 8
 du du du 
3(1u)2(1)p 3(1u)2 3u(2)(1u)(1)p 3(2u)(1u)3u2(1)p 3u2p
3(1u)2p (1u)3(1u)3u(2)p 3u2(1u)up 3u2p 0123
3(1u)2p (1u)39up 3u23up 3u2p. 0123
dx(u)  (3)(1)  3(0)  3. du u0
dy(u)  (3)(0)  3(1)  3. du u0
dz(u)  (3)(0)  3(0)  0. du u0
 3(10)2p (10)39(0)p 3(0)23(0)p 3(0)2p 0123
3p0 3p1.
Dr. J.J. Zou, WSU School of Engineering

A cubic Bezier curve segment P(u)  [x(u) y(u) z(u)], for 0  u  1 is defined by four control points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1) and p3(0,0,1). Calculate the tangent vector of a point on the curve segment when u  1.
P(u)  [x(u) y(u) z(u)]
(1u)3p 3u(1u)2p 3u2(1u)p u3p .
The tangent vector is :
dP(u)  dx(u) dy(u) dz(u)
300029 Engineering Visualization: Tutorial 8
 du du du 
3(1u)2(1)p 3(1u)2 3u(2)(1u)(1)p 3(2u)(1u)3u2(1)p 3u2p
3(1u)2p (1u)3(1u)3u(2)p 3u2(1u)up 3u2p 0123
3(1u)2p (1u)39up 3u23up 3u2p. 0123
dx(u)  (3)(0)  3(0)  0. du u1
dy(u)  (3)(1)  3(0)  3. du u1
dz(u)  (3)(1)  3(1)  0. du u1
 3(11)2p (11)39(1)p 3(1)23(1)p 3(1)2p 0123
3p2 3p3.
Dr. J.J. Zou, WSU School of Engineering

There are seven points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1), p3(0,0,1), p4(x4,y4,z4), p5(0,1,0) and p6(1,1,0). A cubic Bezier curve segment P(u)[x(u) y(u) z(u)], for 0u1 isdefinedby p0, p1, p2 and p3. curve segment Q(v)[x(v) y(v) z(v)], for 0v1 is defined by p3, p4, p5 and p6. Two segments join at p3. It is required to have C1 continuity at the joined point p3. Calculate the coordinates of p4.
Similarly, the tangent vector of Q(v) at p3 is:
dQ(v)3(p4 p3). dv
To have C1 continuity at p3 , the following condition must be met : dP(u)  dQ(v) , i.e.,
3(p3 p2)3(p4 p3).
p4 2p3 p2 2(0,0,1)(0,1,1)(0,1,1).
According to the definition of a Bezier curve, the tangent vector of P(u) at p3 is : dP(u)3(p3 p2).
300029 Engineering Visualization: Tutorial 8
Dr. J.J. Zou, WSU School of Engineering

There are seven points, namely, p0(1,0,0), p1(0,1,0), p2(0,1,1), p3(0,0,1), p4 (0, 1, 2), p5 (0, 1, 0) and p6 (1, 1, 0). A cubic Bezier curve segment P(u)[x(u) y(u) z(u)], for 0u1 isdefinedby p0, p1, p2 and p3. curve segment Q(v)[x(v) y(v) z(v)], for 0v1 is defined by p3, p4, p5 and p6. Two segments join at p3. Do they have G1 continuity at the joined point p3 ? Give reasons to support your answer.
Similarly, the tangent vector of Q(v) at p3 is:
dQ(v)3(p4 p3). dv
To have G1 continuity at p3 , the following condition must be met : dP(u)  k dQ(v) , where k is a constant and k  0.
In other words,
3(p3 p2)k3(p4 p3)  p3 p2 k(p4 p3)
 (0,0,1)(0,1,1)k(0,1,2)(0,0,1)  (0,1,0)k(0,1,1).
However, there is no positive value of k that meets the above requirement. Therefore, the segments do not have G1 continuity at the joined point p3.
According to the definition of a Bezier curve, the tangent vector of P(u) at p3 is: dP(u)3(p3 p2).
300029 Engineering Visualization: Tutorial 8
Dr. J.J. Zou, WSU School of Engineering

Calculate the coordinates of a point on the surface patch when u  0.1 and v  0.2. Solution:
P(u,v)  [x(u,v) y(u,v) z(u,v)]
(1u)3(p (1v)3 3p (1v)2v3p (1v)v2 p v3)
00 01 02 03
3(1u)2u(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 10 11 12 13
3(1u)u2(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 20 21 22 23
u3(p (1v)3 3p (1v)2v3p (1v)v2 p v3). 30 31 32 33
P(0.1, 0.2)(10.1)3(p (10.2)3 3p (10.2)2(0.2)3p (10.2)(0.2)2 p (0.2)3) 00 01 02 03
3(10.1)2(0.1)(p (10.2)3 3p (10.2)2(0.2)3p (10.2)(0.2)2 p (0.2)3) 10 11 12 13
3(10.1)(0.1)2(p (10.2)3 3p (10.2)2(0.2)3p (10.2)(0.2)2 p (0.2)3) 20 21 22 23
(0.1)3(p (10.2)3 3p (10.2)2(0.2)3p (10.2)(0.2)2 p (0.2)3) 30 31 32 33
 0.729(0.512p00  0.384p01  0.096p02  0.008p03)
 0.243(0.512p10  0.384p11  0.096p12  0.008p13)  0.027(0.512p20  0.384p21  0.096p22  0.008p23)  0.001(0.512p30  0.384p31  0.096p32  0.008p33).
x(0.1, 0.2)  0.729((0.512)(12)  (0.384)(8)  (0.096)(12)  (0.008)(15))
 0.243((0.512)(13)  (0.384)(17)  (0.096)(13)  (0.008)(17))  0.027((0.512)(27)  (0.384)(23)  (0.096)(27)  (0.008)(23))  0.001((0.512)(28)  (0.384)(32)  (0.096)(28)  (0.008)(32))
y(0.1, 0.2)  0.729((0.512)(8)  (0.384)(17)  (0.096)(23)  (0.008)(28))
 0.243((0.512)(12)  (0.384)(13)  (0.096)(27)  (0.008)(28))  0.027((0.512)(8)  (0.384)(17)  (0.096)(23)  (0.008)(32))  0.001((0.512)(12)  (0.384)(13)  (0.096)(27)  (0.008)(28))
 13.2755.
z(0.1, 0.2)  0.729((0.512)(20)  (0.384)(10)  (0.096)(10)  (0.008)(10))
 0.243((0.512)(10)  (0.384)(10)  (0.096)(10)  (0.008)(10))  0.027((0.512)(10)  (0.384)(10)  (0.096)(10)  (0.008)(10))  0.001((0.512)(20)  (0.384)(10)  (0.096)(10)  (0.008)(10))
 10.9296.
Dr. J.J. Zou, WSU School of Engineering
300029 Engineering Visualization: Tutorial 8
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).

Calculate the surface normal at a point on the surface patch when u  0 and v  1. Solution:
The surface normal is equal to the cross-product of vector P u and vector P v, where P(u,v)  [x(u,v) y(u,v) z(u,v)]
(1u)3(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 00 01 02 03
3(1u)2u(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 10 11 12 13
3(1u)u2(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 20 21 22 23
u3(p (1v)3 3p (1v)2v3p (1v)v2 p v3). 30 31 32 33
Step1: Find P u.
P(u,v)3(1u)2(1)(p (1v)3 3p (1v)2v3p (1v)v2 p v3)
P(u, v) u
 3(1 0)2 (1)(p (11)3  3p (11)2 (1)  3p (11)(12 )  p 13 ) 00 01 02 03
00 01 02 03
300029 Engineering Visualization: Tutorial 8
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
32(1u)(1)u(1u)2(p (1v)3 3p (1v)2v3p (1v)v2 p v3)
10 11 12 13
3(1)u2 (1u)(2u)(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 20 21 22 23
3u2(p (1v)3 3p (1v)2v3p (1v)v2 p v3). 30 31 32 33
32(10)(1)(0)(10)2(p (11)3 3p (11)2(1)3p (11)(12)p 13)
Step2: Find P v.
P(u, v)(1v)3(p (1u)3 3p (1u)2u3p (1u)u2 p u3)
00 10 20 30
3(1v)2v(p (1u)3 3p (1u)2u3p (1u)u2 p u3) 01 11 21 31
3(1v)v2(p (1u)3 3p (1u)2u3p (1u)v2 p u3) 02 12 22 32
v3(p (1u)3 3p (1u)2u3p (1u)u2 p u3). 03 13 23 33
10 11 12 13
 3(1)(02 )  (1 0)(2(0))(p (11)3  3p (11)2 (1)  3p (11)(12 )  p 13 ) 20 21 22 23
3(02)(p (11)3 3p (11)2(1)3p (11)(12)p 13) 30 31 32 33
3p03 3p13
 (3)(15, 28,10)  (3)(17, 28, 10)  (6, 0,  60).
Dr. J.J. Zou, WSU School of Engineering

P(u, v) 3(1v)2(1)(p (1u)3 3p (1u)2u3p (1u)u2 p u3)
P(u, v) v
3(11)2(1)(p (10)3 3p (10)2(0)3p (10)(02)p (03)) 00 10 20 30
01 11 21 31
3(1)v2 (1v)(2v)(p (1u)3 3p (1u)2u3p (1u)u2 p u3) 02 12 22 32
3v2(p (1u)3 3p (1u)2u3p (1u)u2 p u3). 03 13 23 33
300029 Engineering Visualization: Tutorial 8
00 10 20 30
32(1v)(1)v(1v)2(p (1u)3 3p (1u)2u3p (1u)u2 p u3)
 32(11)(1)(1)  (11)2 (p (1 0)3  3p (1 0)2 (0)  3p (1 0)(02 )  p (03 ))
01 11 21 31
 3(1)(12 )  (11)(2(1))(p
3(12)(p (10)3 3p (10)2(0)3p (10)(02)p (03))
03 13 23 33 3p02 3p03
 (3)(12, 23,10)  (3)(15, 28,10)  (9,15,0).
Step3: Find surface normal N for u0 and v1.
Let i, j and k be the unit vectors along the positive x – axis, positive y – axis and positivez-axis, respectively.
 i j k 6 0 60
u0,v1 9 15 0
 i(0)(0)  (15)(60) j(6)(0)  (9)(60) k(6)(15)  (9)(0)  900i  540j  90k
 (900,  540, 90).
(1 0)3  3p (1 0)2 (0)  3p (1 0)(02 )  p (03 )) 02 12 22 32
P(u, v) N u
 P(u, v)  v
u0,v1 
Dr. J.J. Zou, WSU School of Engineering

A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 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).
Calculate the surface normal at a point on the surface patch when u  1 and v  0. Solution:
The surface normal is equal to the cross-product of vector P u and vector P v, where P(u,v)  [x(u,v) y(u,v) z(u,v)]
(1u)3(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 00 01 02 03
3(1u)2u(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 10 11 12 13
3(1u)u2(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 20 21 22 23
u3(p (1v)3 3p (1v)2v3p (1v)v2 p v3). 30 31 32 33
Step1: Find P u.
P(u,v)3(1u)2(1)(p (1v)3 3p (1v)2v3p (1v)v2 p v3)
P(u, v) u
3(11)2(1)(p (10)3 3p (10)2(0)3p (10)(02)p (03)) 00 01 02 03
00 01 02 03
300029 Engineering Visualization: Tutorial 8
32(1u)(1)u(1u)2(p (1v)3 3p (1v)2v3p (1v)v2 p v3)
10 11 12 13
3(1)u2 (1u)(2u)(p (1v)3 3p (1v)2v3p (1v)v2 p v3) 20 21 22 23
3u2(p (1v)3 3p (1v)2v3p (1v)v2 p v3). 30 31 32 33
 32(11)(1)(1)  (11)2 (p (1 0)3  3p (1 0)2 (0)  3p (1 0)(02 )  p (03 ))
 (3)(27,8,10)(3)(28,12,20)  (3,12,90).
Step2: Find P v.
P(u, v)(1v)3(p (1u)3 3p (1u)2u3p (1u)u2 p u3)
00 10 20 30
3(1v)2v(p (1u)3 3p (1u)2u3p (1u)u2 p u3) 01 11 21 31
3(1v)v2(p (1u)3 3p (1u)2u3p (1u)v2 p u3) 02 12 22 32
v3(p (1u)3 3p (1u)2u3p (1u)u2 p u3). 03 13 23 33
10 11 12 13
 3(1)(12 )  (11)(2(1))(p
3(12)(p (10)3 3p (10)2(0)3p (10)(02)p (03))
30 31 32 33 3p20 3p30
(1 0)3  3p (1 0)2 (0)  3p (1 0)(02 )  p (03 )) 20 21 22 23
Dr. J.J. Zou, WSU School of Engineering

P(u, v) v
 3(1 0)2 (1)(p (11)3  3p (11)2 (1)  3p (11)(12 )  p 00 10 20 30
3(1)v2 (1v)(2v)(p (1u)3 3p (1u)2u3p (1u)u2 p u3) 02 12 22 32
3v2(p (1u)3 3p (1u)2u3p (1u)u2 p u3). 03 13 23 33
03 13 23 33 3p30 3p31
 (3)(28,12, 20)  (3)(32,13,10)  (12, 3,  30).
Step3: Find surface normal N for u0 and v1.
Let i, j and k be the unit vectors along the positive x – axis, positive y – axis and positivez-axis, respectively.
 i j k  3 12 90
u0,v1 12 3 30
 i(12)(30)  (3)(90) j(3)(30)  (12)(90) k(3)(3)  (12)(12)  630i1170j135k
 (630, 1170, 135).
300029 Engineering Visualization: Tutorial 8
P(u, v) 3(1v)2(1)(p (1u)3 3p (1u)2u3p (1u)u2 p u3)
00 10 20 30
32(1v)(1)v(1v)2(p (1u)3 3p (1u)2u3p (1u)u2 p u3)
01 11 21 31
 32(1 0)(1)(0)  (1 0)2 (p (11)3  3p (11)2 (1)  3p
 3(1)(02 )  (1 0)(2(0))(p
3(02)(p (11)3 3p (11)2(1)3p (11)(12)p (13))
(11)(12 )  p 01 11 21 31
(11)3  3p (11)2 (1)  3p (11)(12 )  p (13 )) 02 12 22 32
P(u, v) N u
 P(u, v)  v
u0,v1 
Dr. J.J. Zou, WSU School of Engineering

Another bicubic Bezier surface patch Q(s,t)[x(s,t) y(s,t) z(s,t)],for 0s1 and 0  t  1 is defined by the following 16 control points:
q00 (12,8,20), q01 (x01,y01,z01), q02 (12,7,30), q03 (9,12,30),
q10 (13,12,10), q11 (x11,y11,z11), q12 (13,3,30), q13 (9,4,10), q20 (27,8,10), q21 (x21,y21,z21), q22 (27,7,30), q23 (31,16,10), q30 (28,12,20), q31 (x31,y31,z31), q32 (28,3,30), q33 (24,4,30).
To have C1 continuity on the joined edge, the following conditions must be met : (1) p01 p00 q00 q01;
(2) p11 p10 q10 q11;
(3) p21 p20 q20 q21;
(4) p31 p30 q30 q31.
Thus, we can find :
q01 q00 p01 p00 (12,8,20)(8,17,10)(12,8,20)(16,1,30);
q11 q10 p11 p10 (13,12,10)(17,13,10)(13,12,10)(9,11,30); q21 q20 p21 p20 (27,8,10)(23,17,10)(27,8,10)(31,1,30); q31  q30  p31  p30  (28,12, 20)  (32,13,10)  (28,12, 20)  (24,11, 30);
Two surface patches join along an edge defined by their common control points, namely,p q,p q,p q andp q.ItisrequiredtohaveC1
00 00 10 10 20 20 30 30
continuity on the joined edge. Calculate the coordinates of q01, q11, q21 and q31.
300029 Engineering Visualization: Tutorial 8
A bicubic Bezier surface patch P(u,v)[x(u,v) y(u,v) z(u,v)], for 0u1 and 0  v  1 is defined by the following 16 control points:
p00  (12,8,20), p10 (13,

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