L11 Hermite Bicubic Surface Patch

Synthetic Surfaces

1) Hermite (Bicubic Surface) Patch

2) Bezier (Surface) Patch 3) B-Spline (Surface) Patch 4) Coons (Surface) Patch

5) Blending offset (Surface) Patch 6) Triangular (Surface) Patch

7) Sculptured (Surface) Patch

8) Rational surfaces (Surface) Patch

All these surfaces are based on polynomial forms.

Fourier series can also be used to approximate the surfaces instead. But they are not meant for general use. Because the facts are:

Hermite Bicubic Surface

•The parametric bicubic surface patch connects four corner data points and utilizes a bicubic equation.

•Therefore, 16 vectors or 16×3=48 scalars are required to determine the unknown coefficients in the equation. How?

•Corner points=4, corner tangent vectors=4×2=8, corner twist vectors=4.

∑∑

= =

≤

≤

≤

≤

=

3 0 3 0

1

0

,

1

0

,

)

,

(

i j j i ij

u

v

u

v

C

v

u

P

[

]

[

]





=

=

≤

≤

≤

≤

=

2 3 2 3

by

given

is

[C]

matrix

t

coefficien

The

1

]

[

,

1

]

[

1

0

,

1

0

],

][

[

]

[

)

,

(

curve.

cubic

Hermite

similar to

expanded

be

can

equation

The

C

C

C

C

v

v

v

V

u

u

u

U

v

u

V

C

U

v

u

P

T T T

























=

00 01 02 03 10 11 12 13 20 21 22 23 30 31 32 33

]

[

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

      ≤ ≤ ≤ ≤ = ] [ ] [ follows. as defined matrix condition boundary or geometry the is ] [ curve. cubic Hermite for defined already matrix Hermite the is ] [ Here 1 0 , 1 0 ], [ ] ][ ][ [ ] [ ) , ( obtained. is patch surface Hermite bicubic of equation following the ts coefficien for the solving and equation parametric the into conditions boundary the Applying 01 00 01 00 v v H T H H T P P P P P P P P P P B M v u V M B M U v u P       =       =           = st vectors corner twi ctors tangent ve -u corner ctors tangent ve -corner v points corner ] [ ] [ ] [ ] [ ] [ 11 10 11 10 01 00 01 00 11 10 11 10 uv u v uv uv u u uv uv u u v v P P P P P P P P P P P P P P P P B

]

[

]

][

[

]

[

]

[

)

,

(

]

[

]

][

][

[

]

[

)

,

(

]

[

]

][

[

]

[

]

[

)

,

(

by

given

are

surface

the

on

point

any

at

tors

twist vec

and

ectors

tangent v

The

=

=

=

vT u T vT H H T v T H u H T u

V

M

B

M

U

v

u

P

V

M

B

M

U

v

u

P

V

M

B

M

U

v

u

P

curve.

cubic

Hermite

for

already

defined

matrix

Hermite

al

differenti

The

]

[

]

[

]

[

]

][

[

]

[

]

[

)

,

(

=

=

=

v H u H vT H u H T uv

M

M

Here

V

M

B

M

U

v

u

P

Influencing factors on position and

[

]

[

]

                =               = ) ( ) ( ) ( ) ( ) ( ] [ ) ( ) ( ) ( ) ( ) , ( ) ( ) ( ) ( ) ( ] [ ) ( ) ( ) ( ) ( ) , ( 4 3 2 1 4 3 2 1 4 3 2 1 4 3 2 1 v F v F v F v F v F B u F u F u F u F v u P v F v F v F v F B u F u F u F u F v u P v u u u u u on position and tangent vectors

[

]

[

]

              =               = ) ( ) ( ) ( ) ( ] [ ) ( ) ( ) ( ) ( ) , ( ) ( ) ( ) ( ) ( ] [ ) ( ) ( ) ( ) ( ) , ( 4 3 2 1 4 3 2 1 4 3 2 1 4 3 2 1 v F v F v F v F B u F u F u F u F v u P v F v F v F v F B u F u F u F u F v u P v v v v u u u u uv v v v v

s

s

s

F

s

s

s

s

F

s

s

s

F

s

s

s

F

)

(

2

)

(

3

2

)

(

1

3

2

)

(

necessary.

as

v"

"

and

u"

"

both

represent

commonly

chosen to

been

has

below

s"

"

variable

The

2 3 4 2 3 3 2 3 2 2 3 1

−

=

+

−

=

+

−

=

+

−

=

s

s

s

F

s

s

s

F

s

s

s

F

s s s

1

4

3

)

(

6

6

)

(

6

6

)

(

:

are

functions

basis

the

of

s

derivative

The

2 2 2 2 1 4

+

−

=

+

−

=

−

=

[

] [

]

[

]

[

]

+ + + = + + + = = = + + + = = = = = ) ( ) ( ) ( ) ( ) , 0 ( , ) ( ) ( ) ( ) ( ) , 1 ( zero. are (u) F of rest 1, (u) F 1, u with edge on the , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ] [ 0 0 0 1 ) , 0 ( zero. are (u) F of rest 1, (u) F 0, u with edge On the 11 4 10 3 11 2 10 1 i 2 01 4 00 3 01 2 00 1 4 3 2 1 01 00 01 00 4 3 2 1 i 1 P v F P v F P v F P v F v P Further P v F P v F P v F P v F v P Similarly P v F P v F P v F P v F v F v F v F v F P P P P v F v F v F v F B v P v v v v T v v T Proof               =               + + + = + + + = ) ( ) ( ) ( ) ( ] [ ) , 1 ( ) , 0 ( ) , 1 ( ) , 0 ( ) ( ) ( ) ( ) ( ) , 1 ( ) ( ) ( ) ( ) ( ) , 0 ( 4 3 2 1 11 4 10 3 11 2 10 1 01 4 00 3 01 2 00 1 v F v F v F v F B v P v P v P v P Hence P v F P v F P v F P v F v P P v F P v F P v F P v F v P u u uv uv u u u uv uv u u u       =             = ] [ ] [ ] [ ] [ ] [ 11 10 11 10 01 00 01 00 11 10 11 10 01 00 01 00 uv u v uv uv u u uv uv u u v v v v P P P P P P P P P P P P P P P P P P P P B Proof

[

]

[

]

                  =               = ) ( ) ( ) ( ) ( ) ( ] [ ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) , 1 ( ) ( ) ( ) ( ) ( ] [ ) 0 ( ) 0 ( ) 0 ( ) 0 ( ) , 0 ( 1 4 3 2 1 4 3 2 1 4 3 2 1 4 3 2 1 v F v F v F v F v F B F F F F v P v F v F v F v F B F F F F v P             = 11 10 11 10 01 00 01 00 11 10 11 10 01 00 01 00 ] [ uv uv u u uv uv u u v v v v P P P P P P P P P P P P P P P P B

[

]

[

]

      =             = ) ( ) ( ) ( ] [ ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) , 1 ( ) ( ) ( ) ( ] [ ) 0 ( ) 0 ( ) 0 ( ) 0 ( ) , 0 ( 2 1 4 3 2 1 4 3 2 1 4 3 2 1 v F v F v F B F F F F v P v F v F v F B F F F F v P u u u u u u u u u u

[

] [

]

[

]

[

]

) ( ) ( ) ( ) ( ) 0 , ( , ) ( ) ( ) ( ) ( ) 1 , ( zero. are (u) F of rest 1, (v) F 1, with v edge on the , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 0 0 0 1 ] [ ) ( ) ( ) ( ) ( ) 0 , ( zero. are (u) F of rest 1, (v) F 0, with v edge On the 11 4 01 3 11 2 01 1 i 2 10 4 00 3 10 2 00 1 10 00 10 00 4 3 2 1 4 3 2 1 i 1 P u F P u F P u F P u F u P Further P u F P u F P u F P u F u P Similarly P u F P u F P u F P u F P P P P u F u F u F u F B u F u F u F u F u P u u u u T u u T + + + = + + + = = = + + + = = = = = ] [ ) ( ) ( ) ( ) ( ) 1 , ( ) 0 , ( ) 1 , ( ) 0 , ( ) ( ) ( ) ( ) ( ) 1 , ( ) ( ) ( ) ( ) ( ) 0 , ( 4 3 2 1 11 4 01 3 11 2 01 1 10 4 00 3 10 2 00 1 B u F u F u F u F u P u P u P u P Hence P u F P u F P u F P u F u P P u F P u F P u F P u F u P T T v v uv uv v v v uv uv v v v               =               + + + = + + + =

Continuity of Blent Hermite Bicubic Patches

• Similar to the Hermite cubic splines, the Bicubic

Hermite patches give maximum C

1

continuity from

one patch to the next, though they give C

2

continuity

inside each of the patches.

• While blending two Bicubic Hermite patches, the

necessary conditions are:

– Same curves (C0 continuity) at the common edge

– Same direction of tangent vectors (C1 continuity) at the common edge

– The magnitudes of the tangent vectors do not have to be the same

Blending Two Hermite Patches along u edges

• [P(0,v)]

patch2

=[P(1,v)]

patch1

,

C

0

continuity

• [P

u

(0,v)]

patch2

=K[P

u

(1,v)]

patch 1

, C

1

continuity

The B matrix for the adjacent patches are shown. Only these elements need to be constrained, rest could be arbitrary for C1 be arbitrary for C1 continuity

HERMITE BICUBIC PATCH IS A “SIMPLE EXTENSION” OF THE HERMITE CUBIC CURVE

• There are two ways to prove it.

1) Substitute u=1 or v=1 in the parametric equation of the Hermite patch, it degenerates to that of HCC.

1 3 1 2 1 1 1 0 0 3 30 0 2 20 0 1 10 0 0 00 3 0 3 0

)

,

(

1

0

,

1

0

,

)

,

(

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

P

v

u

v

u

C

v

u

P

i j j i ij

+

+

+

+

+

+

+

+

=

≤

≤

≤

≤

=

∑∑

= = 3 3 33 3 2 23 3 1 13 3 0 03 2 3 32 2 2 22 2 1 12 2 0 02 1 3 31 1 2 21 1 1 11 1 0 01

HCC.

to

reduces

it

cases,

1

v

and

0

v

of

each

For

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

v

u

C

=

=

+

+

+

+

+

+

+

+

+

+

+

• The second way to prove is:

2) Let u edges coincide. P₀₀ coincides with P₁₀, and P₀₁ coincides with P₁₁. Pv₀₀=Pv₁₀ and Pv₀₁=Pv₁₁. All four twist vectors will be zero. P_u00=P_u10= P_u01=P_u11=0.

[

]

4 3 2 1 01 00 01 00 01 00 01 00 ) ( ) ( ) ( ) ( 0 0 0 0 0 0 0 0 0 0 0 1 ) , 0 ( , 0 v v v v v F v F v F v F P P P P P P P P v P u For                           = =

[

]

01 4 00 3 01 2 00 1 4 3 2 1 01 00 01 00 01 00 01 00 01 4 00 3 01 2 00 1 4 ) ( ) ( ) ( ) ( ) ( ) , 1 ( ) ( ) ( ) ( ) ( 0 0 0 0 0 0 0 0 0 0 1 0 ) , 1 ( , 1 ) ( ) ( ) ( ) ( ) ( ) , 0 ( ) ( 0 0 0 0 v v v v v v v v P v F P v F P v F P v F v P v P v F v F v F v F P P P P P P P P v P u For P v F P v F P v F P v F v P v P v F + + + = =                           = = + + + = =    

= = = + + = + + + + + + + + + + + + + + + + + = =

∑∑

= = matrix, [B] resultant The zero. to equal are all rest ˆ ; ˆ ; ˆ ˆ ˆ ˆ ) , ( 01 10 0 00 0 3 3 33 3 2 23 3 1 13 3 0 03 2 3 32 2 2 22 2 1 12 2 0 02 1 3 31 1 2 21 1 1 11 1 0 01 0 3 30 0 2 20 0 1 10 0 0 00 0 3 0 3 0 C s L C r L C P C Hence s vL r uL P v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C v u C s vL r uL P v u C v u P ij v u v u v u i j j i

ij By equivalence, find the bicubic

planar surface patch.

      + + + + = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ be to going is ] [ ] ][ ][ [ ] [ ) , ( the build you when matrix, [B] resultant The zero. to equal are all rest 0 0 s L s L s L r L P r L P s L s L s L P P V M B M U v u P C v v v T H H T ij