Bezier Blossoms CS 418 Interactive Computer Graphics John C. Hart - - PowerPoint PPT Presentation

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Dec 01, 2022 352 likes •543 views

Bezier Blossoms CS 418 Interactive Computer Graphics John C. Hart de Casteljau de Casteljau algorithm p 1 evaluates a point on a p 12 Bezier curve by p 2 p 012 1- t scaffolding lerps p 123 p 0123 Blossoming renames the p 01 control

SLIDE 1

Bezier Blossoms

CS 418 Interactive Computer Graphics John C. Hart

SLIDE 2

de Casteljau

de Casteljau algorithm

evaluates a point on a Bezier curve by scaffolding lerps

Blossoming renames the

control and intermediate points, like p12, using a polar form, like p(0,t,1) p0 p1 p3 p2 p01 p12 p23 p012 p123 p0123 t 1-t

SLIDE 3

de Casteljau

de Casteljau algorithm

evaluates a point on a Bezier curve by scaffolding lerps

Blossoming renames the

control and intermediate points, like p12, using a polar form, like p(0,t,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p012 p123 p0123 t 1-t p01 p12 p23

SLIDE 4

de Casteljau

de Casteljau algorithm

evaluates a point on a Bezier curve by scaffolding lerps

Blossoming renames the

control and intermediate points, like p12, using a polar form, like p(0,t,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p012 p123 p0123 t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 5

de Casteljau

de Casteljau algorithm

evaluates a point on a Bezier curve by scaffolding lerps

Blossoming renames the

control and intermediate points, like p12, using a polar form, like p(0,t,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t) p0123

SLIDE 6

de Casteljau

de Casteljau algorithm

evaluates a point on a Bezier curve by scaffolding lerps

Blossoming renames the

control and intermediate points, like p12, using a polar form, like p(0,t,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 7

Blossoming Rules

1. # of parameters = degree

Cubic: p(_,_,_) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 8

Blossoming Rules

1. # of parameters = degree

Cubic: p(_,_,_)

2. Order doesn’t matter

p(a,b,c) = p(b,a,c) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 9

Blossoming Rules

1. # of parameters = degree

Cubic: p(_,_,_)

2. Order doesn’t matter

p(a,b,c) = p(b,a,c)

3. Setting up the board

p(0,0,0), p(0,0,1), p(0,1,1), p(1,1,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 10

Blossoming Rules

1. # of parameters = degree

Cubic: p(_,_,_)

2. Order doesn’t matter

p(a,b,c) = p(b,a,c)

3. Setting up the board

p(0,0,0), p(0,0,1), p(0,1,1), p(1,1,1)

4. Winning the game

p(t,t,t) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 11

Placing Blossoms

SLIDE 12

Evaluation

p(t) = p(t,t,t) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 13

Evaluation

p(t) = p(t,t,t) = (1-t) p(t,t,0) + t p(t,t,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 14

Evaluation

p(t) = p(t,t,t) = (1-t) p(t,t,0) + t p(t,t,1) = (1-t)[(1-t) p(t,0,0) + t p(t,0,1)] + t [(1-t) p(t,0,1) + t p(t,1,1)] p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 15

Evaluation

p(t) = p(t,t,t) = (1-t) p(t,t,0) + t p(t,t,1) = (1-t)[(1-t) p(t,0,0) + t p(t,0,1)] + t [(1-t) p(t,0,1) + t p(t,1,1)] = (1-t)2 p(t,0,0) + 2 (1-t) t p(t,0,1) + t2 p(t,1,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 16

Evaluation

p(t) = p(t,t,t) = (1-t) p(t,t,0) + t p(t,t,1) = (1-t)[(1-t) p(t,0,0) + t p(t,0,1)] + t [(1-t) p(t,0,1) + t p(t,1,1)] = (1-t)2 p(t,0,0) + 2 (1-t) t p(t,0,1) + t2 p(t,1,1) = (1-t)2[(1-t)p(0,0,0)+tp(1,0,0)]+2(1-t)t[(1-t)p(0,0,1)+tp(1,0,1)]+t2[(1-t)p(0,1,1)+tp(1,1,1)] p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)

SLIDE 17

Evaluation

p(t) = p(t,t,t) = (1-t) p(t,t,0) + t p(t,t,1) = (1-t)[(1-t) p(t,0,0) + t p(t,0,1)] + t [(1-t) p(t,0,1) + t p(t,1,1)] = (1-t)2 p(t,0,0) + 2 (1-t) t p(t,0,1) + t2 p(t,1,1) = (1-t)2[(1-t)p(0,0,0)+tp(1,0,0)]+2(1-t)t[(1-t)p(0,0,1)+tp(1,0,1)]+t2[(1-t)p(0,1,1)+tp(1,1,1)] = (1-t)3 p(0,0,0) + 3 (1-t)2 t p(0,0,1) + 3 (1-t) t2 p(0,1,1) + t3 p(1,1,1) p(0,0,0) p(1,0,0) p(1,1,1) p(1,1,0) p(t,t,0) p(1,t,t) p(t,t,t) t 1-t p(t,0,0) p(1,t,0) p(1,1,t)