Appunti di Grafica Computazionale Avanzata

!

6 2 ,##

• 3 7 2 @

• 3

3

•

•

A 6

$ B C

! !

# 7 #

Poligonale di controllo

Control Points Curva di Bezier

+ 3

n

(t) = p · (t)

B b

n k n,k

k=0

7

p R ≥ 2

d

ε d

k D 2 #

n

(t) = (1 − t ∈ [0, 1]

k n−k

t t)

b n,h k

- + , #

P

i

E - 3 2 #7

7

n (t) = 1

b

n,k

k=0

!" #

. '

(1 − = 1

n

nk=0 k n−k

t t)

k

n n

(a + =

n k n−k

b) a b

k

k=0

F 2F

n n

(t + 1 − = (1 −

k n−k

t) t t)

k

k=0

+ # B C 9

$

6 G

OK NO

- 8 (t) ≥ 0

b

n,k

.

k

(t) = (1 − ≥ 0 ∈ [0, 1]

k n−k

b t t) t

n,k t

" # ! :;

#

• 3

•

( 9 # +

, E 6

2

prima curva seconda curva

0, =

0 0, =

E k k n

(0) = (1) =

b b

n,k n,k

1, =0 1, =

k k n

!" %

* , # 7

2 3 (t) = (1 −

b b t)

n,k n,n−k

& # 7

!7 7 !

(t) = 1

b

n,k

7

# ! , (t)k =

b nt

n,k

) B C 9

&!

? 7

#

$ B

!" !'

7 ! - Disegno iniziale

Disegno modificato

9 3 @ 3

: 7 7 ;7

3

# B C +

!" !

3 9

# ,

! !( )

E 2 : #

;7 !

* +

! "#$ %"&

+, # $ % 7

2

b 1

b 0 b 2

2

2 2−k

(t) = (1 −

k

B t t) b

2 k

k

k=0

# 8 $

1

=

t 2

#

b b

0 1 b 1 10 = (1 − +

b t)b tb

0 1

b 0 b 2

b b

0 1

b 1 11 = (1 − +

b t)b tb

1 2

b 0 b 2

A

10 11

b b

b 1 20 11 11

= (1 − +

b t)b tb

b 0 b 2

! !( ,

6 12

20 =

b t .

+ - -

20 = (1 − − + ] + − + ]

b t)[/1 t)b tb t[(1 t)b tb

0 1 1 2

= 1

2 2 )

+ 2t(1 − + = (

(1 − b t)b t b B

t) 0 0 2 2 2

b 2

b 1 b

3 3

3 3−k

(t) = (1 −

k

B t t) b

b 3 k

0 k=0 k

#

12

=

t

# #

b 2

b 1 1

b 1 10 = (1 − +

1 b b b t)b tb

3

2 0 1

11 = (1 − +

b t)b tb

b 1 2

0 1

b 12

0 = (1−t)b +tb

b 2 3

7 7 #

10 11 11 12

b b b b

b 2

1

b 1

1 b

b 2

1 2

b b 20 10 11

1 3 = (1 − +

1 b t)b tb

b 2

0 b 21 11 12

= (1 − +

0

b b t)b tb

0

$ # 7

20 21

b b

! !( *

b 2

1

b 1

1 b

b 2

1 2

b b

1 3

1

b b 3

2

0 b 30 20 21

= (1 − +

0

0 b t)b tb

b 0

b 2

b 1 b 3

4

4 4−k

(t) = (1 −

k

B t t) b

b 4 k

0 k=0

b k

4

#

13

=

t b 2 10 = (1 − +

b t)b tb

0 1

11 = (1 − +

b t)b tb

1 2

b 1 1

b 12 = (1 − +

1 b t)b tb

b 2 2 3

b

1 3 13 = (1 − +

b t)b tb

3 4

b 0 b

1

b 4

1 3

b 0

# # 7 #

1

=

t 3 b 2 1

b

1 2

b 20 10 11

1 = (1 − +

b b t)b tb

1 21 11 12

= (1 − +

2 b t)b tb

b 2 b

b

1 3 22 12 13

= (1 − +

2 b t)b tb

1

b 0

b 2

0 b b

1

b 4

0 3

-.! !(/ 0

9 b 2 1

2 b

b

1 2

1

b 1 30 20 21

= (1 − +

b 2 b t)b tb

b

1 2 31 21 22

= (1 − +

b t)b tb

b

2 3

b 3

b 0 1

1

b 3

b

0

b 0 b

0 1 4

b 3

' - b 2 1

2 b

b

1 2

1

b 1

b 2

b

1 40 30 31

= (1 − +

2 b t)b tb

3

b 3

0 b

b

2 3

1

b 0

1

b 0 b

b 04

0 b

1 4

b 3

' ("#$ %"&) *

7

0

b

i

= (1 − r−i

ri

b t)b

i

+ !

= 1, = 0, −

r . . . , n i . . . , n r

r

b

0 = (t)

r r

b b

0

+, # $ % 6 2

+ 2 ! 6 @ 2 +

1 − ≥ 0

t t

! , #

> 6 2 #

!( 3

b

0 2

2

b b

0 1

1 1

1

b b b

0 2

1

b b b b 3

0 2

1

# ! ! +

+ "#$ # "

+, # # : ;

H 2 ! ## # 9

#

# + 1

n n

E j j

1 ·

= + 1 ·

b

b b

j−1 j

+1 − 1

j n n

= 1, · · ·

j , n

$ # # +, # I

b b

0 n+1

, - 7 : #

; ! G +, # 6 2

7 J # -

, "#$ %%

+, # 6 ? -

9 7 =

r

7 - 7 (s)

0, n

C

. . . , n

, 3 9

# - # $ %

- .

7 7 -

= (t)P (t)

n

C(t) b c

n,i i

i=0

n−1

d

= (t) = (t) · [n(P − )]

c(t) c b P

n−1,i i+1 i

dt i=0

,? 7 I

6 7 # G 2

n−1 (t)Q

b

n−1, i i

i=0

>

(0) = − ) (1) = − )

C n(P P C n(P P

1 0 n n−1

, !!"

- 7 6 7

1

/

3 G E

6 2 # !

$ ! 9 #

! !

7 7 #

7 6 - + !

# ! 9

i

C i

7

0 1

C C

6

#

D # $ 1

C

+ ! # 7 ? J

1

G

# 7 ? 2

0 1

C C

+

Knots 9 : ;

2!

0 1

+ 3 2 ,

E 3

,

3

9 : ; 6

+ 1

M m

+, ,

≤ ≤ ≤ [m )

m m . . . m , m

0 1 n i i+1

=

2!

7

# = =

m m m m

i i i+1 i+k−1

# # 9

? 6 3

! !!"

m i

= 3

2

# =

m m

i i+1

# ?

!! !!

m i

9- !

1 !! !

= 0 = 1

m m

0 m

+ 3 2 # -

(u)

N p

i,p

-

> 6 " E 2 #

p #

$ : ! #

= 0 0

p

? 3 ! ;

1, ≤

m m < m

i i+1

=

N

i,0 0, altrimenti

9 7 -# 7 3 #

> 7 $K

− −

u u u u

i i+p+1

(u) = · (u) + · (u)

N N N

i,p i,p−1 i+1,p−1

− −

u u u u

i+p i i+p+1 i+1

+, 7 3 # 7 I

7

$ , [0, 3] = 0

p

9 = 7 7 7

= 0 = 1 = 2 = 3

m m m m

0 1 2 3

$ 7

= 0 = 0 [0, 1)

i p

1,

=

N

0,0 0, f uori

0 1 2 3

$ 7

= 1 = 0 [1, 2)

i p

1,

=

N

1,0 0, f uori

0 1 2 3

' - 7

= 2 = 0 [2, 3]

i p

#

"

1,

=

N

2,0 0, f uori

0 1 2 3

9 # - 3 # 7

# !

9 6 !

p

− −

u u u u

0 2

(u) = · (u) + · (u) =

N N N

0,1 0,0 1,0

− −

u u u u

1 0 2 1

· (u) + (2 − · (u)

u N u) N

0,0 1,0

E

[0, 1)

u ε J

· 1 + (2 − · 0 =

u u) u

[1, 2)

u ε

· 0 + (2 − · 1 = 2 −

u u) u

,

[2, 3] 0 0 1 2 3

N

1,1

− −

u u u u

1 3

(u) = · (u) + · (u) =

N N N

1,1 1,0 2,1

− −

u u u u

2 1 3 2

(u) + 3 − · (u)

− 1 · u N

u N 1,0 2,0

E

[1, 2)

u ε J

− 1 · 1 + (3 − · 0 = − 1

u u) u

[2, 3)

u ε

− 1 · 0 + (3 − · 1 = 3 −

u u) u

,

[1, 2] 0 0 1 2 3

0 7 ? @!

? 9

1

= 2

p p

− −

u u u u

0 3

(u) = · (u) + =

N N

0,2 0,1

− −

u u u u

2 0 3 1

(u) + 0, 5(3 − (u).

0, 5u · u)N

N 0,1 1,1

3 %

E

[0, 1)

u ε

2

0, 5 · [1, 2)

u u ε

2

0, 5 · (3 − u)

,

[2, 3]

2

0, 5 · (−3 + 6u − 2u )

$

N N

1,2 2,3

:6 2 , #

22

; -#

! .* 01"

$

n

= (u)P

C(u) N

i,p i

i=0

!! "" 0%"

22 ! # ! 7

# ! !

3

# E #

N

i,p

- 9 !

2

+ #

(u) ≥ 0

N

i,p :3 , ; #

" [u

N , u

i,p i i+p+1

E 6 2 2 6 (u)

N

i,p

#

( ! E

(u) = 1

N

i,p

, #

* :? ! 3 # ;

= + + 1.

m n p + =

& (u) p−k

N ε c

i,p