Graphics curve modeling

1. Curve
Modeling
Mrs.R.Shanthi Prabha M.Sc., M.Phil.,
Assistant Professor,
DEPARTMENT OF CS

2. Parametric
Polynomials

3. Parametric
Polynomials n
n-1

4. Parametric Polynomials
disadvantag
es
wiggles
degree high

flexibility
degree law
 No local

5. Solution:
Polynomial Splines

6. Spline
Curve

7. Spline
Curve In drafting terminology, a spline isa
flexible strip used to produce a
smooth curve through adesignated
set of points
Flexible Strip

8. Spline
Curve Several smallweights are distributed
along the length of the strip to hold
Weights

9. Spline Curve
 The termspline curve originally

10. Spline
Curve In modeling, the termspline curve
refers to a any composite curve
formed with polynomial section
satisfying specified continuity

11. Control
Points

control
points
shape

12. 
Control Points
defined
modified manipulated

13. Convex
Hull
convex hull

14. Convex
Hull
on
inside

15. Convex

Remind:
convex line
segment
within

16. Shape
of
the Spline Curve

17. Shape of the
Curve Control points
continuous parametric
two way
1. Interpolation Splines
2. Approximation Splines

18. Interpolation
Splines
passes
 When polynomial sections are fitted so
that the curvepasses
interpolate

19. Interpolation
Splines
digitize drawing
paths

20. Approximation
Splines
ppp
 When polynomial sections are fitted tothe
general control point pathwithout
necessarily passing through any control
point, the resulting curve is said to
approximate

21. Approximation
Splines design

22. Parametric
Continuity
Conditions

23. Parametric Continuity
Conditions
…Continuity!

24. Parametric Continuity Conditions


25. Parametric Continuity
Conditions

e
derivation

26. Parametric Continuity
Conditions Zero order parametric (C0):
 First order parametric (C1):

27. Parametric Continuity
Conditions Second order parametric (C2):

28. Parametric Continuity
Conditions


Simple Example:
[-1,0]
[0,1],
f(u ) = (u ,-u2, 0)
g(v) = (v ,v 2, 0 )
f'(u ) = ( 1, -2u, 0 ) ,f'(u) = ( 0, -2, 0 )
g'(u) = ( 1, 2v,0 ) ,g' (v) = ( 0, 2, 0)
C 1
 f' (0) = (0, -2, 0 ) not equal g' (v) = ( 0, 2, 0),
C 2c
Curvature
Curvature g(v ) = 2/(1 + 4v
)
f(u) = 2/(1 + 4u2)1.5
2 1.5
Two curve segments may beC 1continuous
2and evencurvature continuous ,but noCt
continuous

29. Problems
withParametric Continuity Conditions
Representation

30. Problems with Parametric
Continuity
Two line segments:

f'(u ) = B -A
g'(v ) = C-B
f'(u ) = B -A g'(v ) = C-B
Is it strange?
F(u) =A +u (B -A)/ | B -A|
f(u ) = A +u (B -A)
g(v) = B +v (C - B)
u is in the range of 0 and | B - A | and v is
in the range of 0 and | C – B|

31. Problems with Parametric
Continuity
Semi circle:
) C0)
f'(u ) = ( PIu sin(u 2 PI/2), PIu cos(u 2 PI/2), 0 )
f'(u) = ( PI2u 2 cos(u 2 PI/2), -PI2u 2 sin(u 2 PI/2), 0 )
f'(u ) × f'(u) = ( 0, 0, -PI3u 3 )
| f'(u )| = PIu
| f'(u ) × f'(u) | = PI3u 3
k (u) =1
g'(v) = ( PIv cos(v 2 PI/2), -PIv sin(v 2 PI/2), 0 )
g'(v) = ( -PI2v 2 cos(v 2 PI/2), -PI2v 2 cos(v 2 PI/2), 0 )
g'(v) × g' (v) = (0, 0, -PI3u 3 )
| g'(v) |= PIv
| g'(v ) × g'(v) | = PI3v 3

32. Problems with Parametric
Continuity
0 )
Semi circle:
u 2 =p f(u ) v 2 =q g(v).
f(p ) =( -cos(p PI/2), sin(p PI/2), 0 )
g(q ) =( sin(q PI/2), cos(q PI/2), 0)
Their derivatives are
f'(p ) = ( (PI/2) sin(p PI/2), (PI/2) cos(p PI/2),
f''(p ) = ( (PI/2)2 cos(p PI/2), -(PI/2)2 sin(p PI/2), 0 )
g'(q ) = ( (PI/2) cos(q PI/2), -(PI/2) sin(q PI/2), 0 )
g''(q ) = ( -(PI/2)2 sin(q PI/2), -(PI/2)2 cos(q PI/2), 0 )
C
f'(1) g'(0)
1c
g'(0) ( 0, -(PI/2)2, 0 )
( PI/2, 0, 0 )
f'(1)
C 2o

33. Geometric
Continuity
Conditions

34. Geometric Continuity
Conditions Two curve segments are said Gk
geometric continuous at thejoining
point if and only if there exists two
parameterizations, one for each
curve segment, such that all i-th
derivatives, i less than or equal to k,
computed with these new
parameterizations agree at the
joining point.

35. Geometric Continuity
Conditions TwoC 0 curve segments are saidG 1
geometric continuous at the joining point
if and only if vectors fu'( ) and gv'( ) are in
thseame direction at the joining point.
Note thautf'( ) andvg'( ) are evaluated at
taheejoniningpoint.

36. Geometric Continuity
Conditions
 Two C1 curve segments are said G2 geometric
continuous at the joining point if and only if
vector f''(u) - g''(v) is parallel to the tangent
vector at the joining point. Note that f''(u) and
g''(v) are evaluated at the joining point.

37. Geometric Continuity
Conditions


Example:
f(u) = ( -1 +u 2, 2u -u 2, 0 )
g(v ) =( 2u -u 2, 1 -u 2, 0 )
 f'(u ) = ( 2u,2 - 2u,0 )
f' (u)= ( 2, -2, 0 )
f'(u ) × f'(u) = ( 0, 0, -4 )
| f'(u ) | = 2SQRT(1 - 2u + 2u2 )
| f'(u ) × f'(u) | = 4
k (u) = 1/(2(1 - 2u + 2u2)1.5)
g'(v) = ( 2 - 2v,-2v,0)
g' (v) = ( -2, -2, 0)
g'(v) × g' (v) = ( 0, 0, -4)
| g'(v) | = 2SQRT(1 - 2v + 2v2 )
| g'(v) × g' (v) | =4
k (v) = 1/(2(1 - 2v + 2v2)1.5)

38. Geometric Continuity
Conditions

f'(1) = g'(0) = ( 2, 0, 0 ),
1C
not equal g''(0) = ( -2, -2, 0),
f''(1) =
not
C
( 2, -2, 0)
2
G2
parallel
C 1
f''(1) - g''(0) = ( 4, 0, 0 )
( 2, 0, 0 )
G 2o

39. Geometric Continuity
Conditions
 Geometric Continuity Conditions
derivatives
proportional

40. Geometric Continuity
Conditions

Zero order geometric (G0):
C )
First order geometric (G1):
C´ )
 Second order geometric (G2):
C˝ )

41. Parametric & Geometric
Continuity

differences curve
shape
geometric continuity
pulled
greater tangent vector

42. Spline
Interpolat
ion

43. Linear
Spline
Interpolation

44. Linear Spline
Interpolation linear
interpolation spline
linear polynomial
passes

45. Linear Spline
Interpolation
0 u 1p (u)  au  b ,

46. 

Linear Spline
Interpolation
e u a
b
0 1
u

47. Linear Spline
Interpolation

M M yU

48. Linear Spline
Interpolation matrix representation
 M ee
i y
 M h e
o
a
i
geometryspline
P (u)  U M M

49. Linear Spline
Interpolation
1
k0
k k
0 u 1P (u)   P b (u),
Blending Functions

50. Linear Spline
Interpolation
 The Linear spline blending function:
1
k0
k k
0 u 1P (u)   P b (u),

51. Linear Spline
Interpolation


linear
C0
linear 2 constraints
A polynomial of degree khas k+1
coefficients and thus requires
k+1 independent constraints to
uniquely determine it.

52. Cubic
Spline
Interpolation

53. 


Cubic Spline
Interpolation
Cubic splines are
used to:
Cubic splines are more flexible
for modeling arbitrary curve
shapes.

54. Cubic Spline
Interpolation
 Cubic interpolation spline
cubic polynomial
passes

55. Cubic Spline
Interpolation
p (u)  au3
 bu2
 cu  d , 0  u 1

56. Linear Spline
Interpolation


a
b c d
conditions
n curve section.
boundary
“joints”
coefficients.
p (u)  au3
 bu2
 cu  d , 0  u 1

57. 1.
2.
3.
4.
Cubic Spline Interpolation
Methods for setting the boundaryconditions
for cubic interpolation splines:




58. Natural
Cubic
Splines

59. Natural Cubic
Splines

 C2

60. Natural Cubic
Splines


n+1
n
4n
4×2

61. 
Natural Cubic
Splines
n-1
4
first second
each curvederivatives
pass control point
1
2
3
4. C 1

62. Natural Cubic
Splines


5
6

63. Natural Cubic Splines
We still needtwo

64. Natural Cubic
SplinesMethod 1:
p o
7
8

65. Natural Cubic
SplinesMethod 2:
Add P n n
1.
7
8
P1 P
3

66. Natural Cubic
Splines
Disadvantage:
 No “Local Control

67. Hermit
Splines

68. Hermit
Splines
i
 Hermit splines is aninterpolating
piecewise cubic polynomial with
sapecified tangent

69. Hermit
Splines Hermit splines a adjusted
locally
dependent
endpoint constraints unlike the
natural cubic splines

70. HERMIT SPLINES
 Example: Change inMagnitude of
T 0

71. Hermit
Splines Example: Change inDirection of T

72. Hermit
Splines
k1
k
p (0)  Dp
p (1)  Dp
 P(u)
p +
 Theboundary conditions
p (0)  p
k
p (1)  p
k1
Dp c
e o

73. Hermit
Splines
p k  0
p
 1
 k1   
Dp  0

k
 
Dp k1  3
0 0 1 a
1
 b

   
0 c 
  
0 d
constraints 1 1
matrix 0 1
2 1
k1
k
k1
p (0)  Dp
p (1)  Dp
p (1)  p
p (0)  p
k
Boundary conditions

74. Hermit
Splines
  
   
 
   
Dp

Dp 0 d 
0 c  0
1
 b
p
 1
 1 a0 0
1 1
0 1
3 2 1k1 
k
k1
p k 0
Dp
M e
i t i x

75. Blending Functions
H(u)k 0 k1 3k1 1 k 2
P(u) p H (u)p H(u) Dp H(u)Dp
   
 
  


Dp

Dp
p
0 d 
0 c
  
 0
1
 b
 1
1 a0 0 0
1 1
0 1
3 2 1k1 
k
k1
p k

76. Hermit Splines
Hermit Blending
Functions

77. Hermit Splines
 useful
not
approximate
without
values
in modeling
spline
input
slopes
difficult specify
slopes
But…

78. Cardin
al
Spline

79. Cardinal Splines
 Cardinal splines
specified
tangents
 do not have
give values endpoint tangents
For a cardinal spline, the value for the
slope at a control point is
calculated from the coordinatesof
the two adjacent controlpoints.

80. Cardinal
Splines
 A cardinal splinesection
four
points
 middle control points
section endpoints other two
points used calculation
endpoint slopes

81. Cardinal
Splines
 P(0) 
1
(1 t)(p  p ) 2
 P(1) 
1
(1 t)(p  p ) 2
 The boundary conditions:
k
k
P (0)  p
P (1)  p
k2
k1k1
k1
slopes
p 1

82. Cardinal
Splines
 P(0) 
1
(1 t)(p  p ) 2
 P(1) 
1
(1 t)(p  p ) 2
 The boundary conditions:
k
k
P (0)  p
P (1)  p
k2
k1k1
k1

83. Cardinal
Splines
 t
tension parameter
 Tension parameter
loosely
tightly
k
k
P (0) 
1
(1 t)( p  p )
2
P (1) 
1
(1 t)( p  p )
2
P (0)  p
P (1) p
k2
k1k1
k1

84. Cardinal Splines
 t=0
Catmull-Rom splines
Overhauser splines.

85. Cardinal
Splines
k
k
P (0) 
1
(1 t)( p  p )
2
P (1) 
1
(1 t)( p  p )
2
P (0)  p
P (1) p
k2
k1k1
k1

86. Cardinal
Splines
k
k
P (0) 
1
(1 t)( p  p )
2
P (1) 
1
(1 t)( p  p )
2
P (0)  p
P (1) p
k2
k1k1
k1
32
CAR (u)0 k 1
k
k2k1k1
k2k1
k1
CAR (u)  p p CAR (u)  p CAR (u)  p
 p [( s 2)u3
 (3  2s)u2
 su]  p ( su3
su2
)
(su3
 2su2
su)  p [( 2 s)u3
 (s3)u2
1]P (u) p
 CAR r

87. Cardinal
Splines
Cardinal
functions