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

1

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:

(i) they can approximate any curve, not just periodic

(ii) computations involved are high

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.

2

∑∑= =

≤≤≤≤=3

0

3

0

10,10,),(i j

ji

ij vuvuCvuP

•The parametric equation of Hermite bicubic surface patch is

[ ] [ ]

==

≤≤≤≤=

2323

bygiven is [C]matrix t coefficien The

1][,1][

10,10],][[][),(

curve. cubic Hermite similar to expanded becan equation The

CCCC

vvvVuuuU

vuVCUvuP

TT

T

3

=

00010203

10111213

20212223

30313233

][

CCCC

CCCC

CCCC

CCCC

C

≤≤≤≤=

][][

follows. as definedmatrix condition boundary or geometry theis ][

curve. cubic Hermitefor definedalready matrix Hermite theis][ Here

10,10],[]][][[][),(

obtained. ispatch surface Hermite

bicubic ofequation following thetscoefficien for the solving and

equation parametric theinto conditionsboundary theApplying

01000100 vv

H

T

HH

T

PPPPPP

PPPP

B

M

vuVMBMUvuP

4

=

=

=

st vectorscorner twictorstangent ve-ucorner

ctorstangent ve-corner vpointscorner

][][

][][][

11101110

01000100

11101110

uvu

v

uvuvuu

uvuvuu

vv

PP

PP

PPPP

PPPP

PPPPB

][]][[][][),(

][]][][[][),(

][]][[][][),(

bygiven are surface the

onpoint any at tors twist vecand ectors tangent vThe

=

=

=

vTuT

vT

HH

T

v

T

H

u

H

T

u

VMBMUvuP

VMBMUvuP

VMBMUvuP

5

curve. cubic Hermitefor already

definedmatrix Hermite aldifferenti The][][

][]][[][][),(

==

=

v

H

u

H

vT

H

u

H

T

uv

MM

Here

VMBMUvuP

Influencing factors

on position and

[ ]

[ ]

=

=

)(

)(

)(

)(

)(

][)()()()(),(

)(

)(

)(

)(

][)()()()(),(

4

3

2

1

4321

4

3

2

1

4321

vF

vF

vF

vF

vF

BuFuFuFuFvuP

vF

vF

vF

vF

BuFuFuFuFvuP

v

uuuu

u

6

on position and

tangent vectors[ ]

[ ]

=

=

)(

)(

)(

)(

][)()()()(),(

)(

)(

)(

)(

][)()()()(),(

4

3

2

1

4321

4

3

2

1

4321

vF

vF

vF

vF

BuFuFuFuFvuP

vF

vF

vF

vF

BuFuFuFuFvuP

v

v

v

v

uuuu

uv

v

v

v

v

sssF

ssssF

sssF

sssF

)(

2)(

32)(

132)(

necessary. as v"" and u""both represent

commonly chosen tobeen has below s"" variableThe

23

4

23

3

23

2

23

1

−=

+−=

+−=

+−=

7

sssF

sssF

sssF

sssF

s

s

s

s

23)(

143)(

66)(

66)(

:are functions basis theof sderivative The

2

4

2

3

2

2

2

1

4

−=

+−=

+−=

−=

[ ] [ ]

[ ][ ]

+++=

+++=

==

+++=

=

=

==

)()()()(),0(

,

)()()()(),1(

zero. are (u)F ofrest 1,(u)F 1,u with edge on the,

)()()()(

)()()()(

)()()()(][0001),0(

zero. are (u)F ofrest 1,(u)F 0,u with edge On the

114103112101

i2

014003012001

432101000100

4321

i1

PvFPvFPvFPvFvP

Further

PvFPvFPvFPvFvP

Similarly

PvFPvFPvFPvF

vFvFvFvFPPPP

vFvFvFvFBvP

vv

vv

T

vv

T

Proof

8

=

+++=

+++=

)(

)(

)(

)(

][

),1(

),0(

),1(

),0(

)()()()(),1(

)()()()(),0(

4

3

2

1

114103112101

014003012001

vF

vF

vF

vF

B

vP

vP

vP

vP

Hence

PvFPvFPvFPvFvP

PvFPvFPvFPvFvP

u

u

uvuvuuu

uvuvuuu

=

=][][

][][][

11101110

01000100

11101110

01000100

uvu

v

uvuvuu

uvuvuu

vv

vv

PP

PP

PPPP

PPPP

PPPP

PPPP

B

Proof

[ ]

[ ]

=

=

)(

)(

)(

)(

)(

][)1()1()1()1(),1(

)(

)(

)(

)(

][)0()0()0()0(),0(

1

4

3

2

1

4321

4

3

2

1

4321

vF

vF

vF

vF

vF

BFFFFvP

vF

vF

vF

vF

BFFFFvP

=

11101110

01000100

11101110

01000100

][

uvuvuu

uvuvuu

vv

vv

PPPP

PPPP

PPPP

PPPP

B

9

[ ]

[ ]

=

=

)(

)(

)(

)(

][)1()1()1()1(),1(

)(

)(

)(][)0()0()0()0(),0(

4

3

2

1

4321

4

3

2

1

4321

vF

vF

vF

vF

BFFFFvP

vF

vF

vFBFFFFvP

uuuu

u

uuuu

u

[ ] [ ]

[ ][ ]

)()()()()0,(

,

)()()()()1,(

zero. are (u)F ofrest 1,(v)F 1, with vedge on the,

)()()()(

)()()()(

0001][)()()()()0,(

zero. are (u)F ofrest 1,(v)F 0, with vedge On the

114013112011

i2

104003102001

100010004321

4321

i1

PuFPuFPuFPuFuP

Further

PuFPuFPuFPuFuP

Similarly

PuFPuFPuFPuF

PPPPuFuFuFuF

BuFuFuFuFuP

uu

uu

T

uu

T

+++=

+++=

==

+++=

=

=

==

10

][

)(

)(

)(

)(

)1,(

)0,(

)1,(

)0,(

)()()()()1,(

)()()()()0,(

4

3

2

1

114013112011

104003102001

B

uF

uF

uF

uF

uP

uP

uP

uP

Hence

PuFPuFPuFPuFuP

PuFPuFPuFPuFuP

TT

v

v

uvuvvvv

uvuvvvv

=

+++=

+++=

Continuity of Blent Hermite Bicubic Patches

• Similar to the Hermite cubic splines, the Bicubic

Hermite patches give maximum C1 continuity from

one patch to the next, though they give C2 continuity

inside each of the patches.

• While blending two Bicubic Hermite patches, the

necessary conditions are:

11

– 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, C0 continuity

• [Pu(0,v)]patch2=K[Pu(1,v)]patch 1, C1 continuity

The B matrix for the

adjacent patches are

shown. Only these

elements need to be

constrained, rest could

be arbitrary for C1

12

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.

13121110

03

30

02

20

01

10

00

00

3

0

3

0

),(

10,10,),(

vuCvuCvuCvuC

vuCvuCvuCvuCvuP

vuvuCvuPi j

ji

ij

++++

++++=

≤≤≤≤=∑∑= =

13

3

3

2

210

33

33

32

23

31

13

30

03

23

32

22

22

21

12

20

02

13

31

12

21

11

11

10

01

),(

HCC. toreducesit cases, 1 vand 0 vofeach For

uCuCuCCvuP

vuCvuCvuCvuC

vuCvuCvuCvuC

vuCvuCvuCvuC

+++=

==

+++

++++

++++

• The second way to prove is:

2) Let u edges coincide. P00 coincides with P10, and P01 coincideswith P11. Pv00=Pv10 and Pv01=Pv11. All four twist vectors will bezero. Pu00=Pu10= Pu01=Pu11=0.

[ ]

4

3

2

1

01000100

01000100

)(

)(

)(

)(

0000

00000001),0(

,0

vv

vv

vF

vF

vF

vF

PPPP

PPPP

vP

uFor

=

=

14

[ ]

014003012001

4

3

2

1

01000100

01000100

014003012001

4

)()()()()(),1(

)(

)(

)(

)(

0000

00000010),1(

,1

)()()()()(),0(

)(0000

vv

vv

vv

vv

PvFPvFPvFPvFvPvP

vF

vF

vF

vF

PPPP

PPPP

vP

uFor

PvFPvFPvFPvFvPvP

vF

+++==

=

=

+++==

===

++=+++

++++

++++

++++

++==∑∑= =

matrix, [B]resultant The zero. toequal are allrest

ˆ;ˆ;

ˆˆ

ˆˆ),(

0110000

0

33

33

32

23

31

13

30

03

23

32

22

22

21

12

20

02

13

31

12

21

11

11

10

01

03

30

02

20

01

10

00

00

0

3

0

3

0

C

sLCrLCPC

Hence

svLruLPvuCvuCvuCvuC

vuCvuCvuCvuC

vuCvuCvuCvuC

vuCvuCvuCvuC

svLruLPvuCvuP

ij

vu

vu

vu

i j

ji

ij By equivalence, find the bicubic

planar surface patch.

15

+++

+

=

=

00ˆˆ

00ˆˆ

ˆˆˆˆˆ

ˆˆˆ

[B]

be togoing is

][]][][[][),(

thebuildyou when

matrix, [B]resultant The zero. toequal are allrest

00

00

rLrL

rLrL

sLsLsLrLPrLP

sLsLsLPP

VMBMUvuP

C

uu

uu

vvvuu

vvv

T

HH

T

ij

Can you prove it?