HQG TP H Ch Minh thi M HNH HA HNH HC

Trng H BCH KHOA Ngy thi: 07/01/2012

Khoa C Kh Thi gian: 60 pht

B mn THIT K MY c s dng ti liu.

1. Cho ng spline bc ba t nhin ni suy cc im P1(0, 0), P2(1, 1), P3(3, 0), P4(4, 2). Xc nh: a. Tip tuyn ng cong ti cc im trn

b. Cc phng trnh tham s tng on ng cong. c. To im nm trn phn on th hai khi t = 0,5.

2. Cho cc im iu khin V0(1/2, 3 / 2 ), V1(1, 0) v V2(0, 0).

a. Vit phng trnh ng cong NURBS vi cc gi tr trng s w0 = 1, w1 = 1 v w2 = 2 (hoc w0 = 2, w1 = 1 v w2 = 1) vi vct nt t = [0 0 0 1 1 1].

b. Tm P(0,25), P(0,75)

3. Cho 2 im A(1, 0, 0) B(1, 1, 3). Yu cu: a. Vit phng trnh tham s on thng qua 2 im A v B. b. Phng trnh tham s mt cong trn xoay c to ra bng php xoay on thng AB quanh

trc y v trc x.

p n

Bi 1.

1. Di vi spline t nhin :

)PP(3

)PP(3

)PP(3

)PP(3

P

P

P

P

2100

1410

0141

0012

34

24

13

12

'

3

'

2

'

1

'

0

63

39

09

33

P

P

P

P

2100

1410

0141

0012

'

3

'

2

'

1

'

0

T ay suy ra:

63

39

09

33

2100

1410

0141

0012

P

P

P

P1

'

3

'

2

'

1

'

0

15/443/2

15/23/5

15/73/5

15/263/2

P

P

P

P

'

3

'

2

'

1

'

0

2. Phng trnh tham s trn cc phn on:

15/73/5

15/263/2

11

00

0001

0100

1233

1122

1ttt)t(P 231

15/23/5

15/73/5

03

11

0001

0100

1233

1122

1ttt)t(P 232

15/443/2

15/23/5

24

03

0001

0100

1233

1122

1ttt)t(P 233

3. To im trn phn on 2 khi t = 0,5.

15/23/5

15/73/5

03

11

0001

0100

1233

1122

15,05,05,0)t(P 232

= [2 17/40]

= [2 0,425]

Bi 2.

1. Phng trnh ng cong NURBS c dng:

0,3 0 0 1,3 1 1 2,3 2 2

0,3 0 1,3 1 2,3 2

N (t)w V N (t)w V N (t)w V

P(t)

N (t)w N (t)w N (t)w

Ta thu c cc hm c s N0,3(t), N1,3(t) v N2,3(t) t phng trnh quy Cox - de Boor:

i i 1

i,1

1 vi t t tN (t)

0 ngc lai

v

i i k

i,k i,k 1 i k,k 1

i k 1 i i k i 1

(t t ) (t t)N (t) N (t) N (t)

(t t ) (t t )i+1,k-1(t)

trong , vct nt l [ti,..., ti+k].

Suy ra:

2 3

2,1

1 vi t 0 t t 1N (t)

0 ngc lai

3

1,2 2,1

3 2

(t t)N (t) N (t) 1 t

(t t )

2

2,2 2,1

3 2

(t t )N (t) N (t) t

(t t )

23

0,3 1,2

3 1

(t t)N (t) N (t) 1 t

(t t )

1 4

1,3 1,2 2,2

3 1 4 2

(t t ) (t t)N (t) N (t) N (t) 2t(1 t)

(t t ) (t t )

22

2,3 2,2

4 2

(t t )N (t) N (t) t

(t t )

Suy ra:

2 2

0 0 1 1 2 2

2 2

0 1 2

w V (1 t) w V 2t(1 t) w V tP(t)

w (1 t) w 2t(1 t) w t

22

22

t.2)t1(t2.1)t1(1

)0,0(t.2)0,1)(t1(t2.1)2/3,2/1()t1(1)t(P

Rt gn ta c:

)t1(2

)t1(3,

)t1(2

t3t21)t(P

2

2

2

2

2. To ti cc im t = 0,25 v t = 0,75:

)25,01(2

)25,01(3,

)25,01(2

25,0.325,0.21)25,0(P

2

2

2

2

34

39,

34

21)25,0(P

P(0,25)=[0,618, 0,458]

Tng t:

)75,01(2

)75,01(3,

)75,01(2

75,0.375,0.21)75,0(P

2

2

2

2

50

3,

50

13)75,0(P

P(0,25)=[0,26, 0,035]

Bi 3.

1. on thng AB vi A(1, 0, 0) B(1, 1, 3) c th biu din dng tham s L(t) = (x(t), y(t), z(t)) = A + (B A)t. T y, mi thnh phn ca on thng c vit li nh sau:

x(t) = 1 + (1 1)t = 1

y(t) = 0 + (1 0)t = t

z(t) = 0 + (3 0)t = 3t

2. Phng trnh tham s mt trn xoay quay quanh trc y:

0001

0cos0sin

0010

0sin0cos

1t3t1)t(P

Phng trnh tham s mt mt trn xoay quay quanh trc x:

0001

0cossin1

0sincos0

0101

1t3t1)t(P