Blending Surfaces

IntroductionBlending n. 1. The act of mingling. 1913 Webster

2. (Paint.) The method of laying on different tints so that they may mingle together while wet, and shade into

each other insensibly. --Weale. 1913 Webster

Introduction

The process of mixing several base objects to form a new object.

The process of providing smooth transition between intersecting surfaces or smooth connection between disjoint surfaces.

A General Blending model

We have seen a Belnding method before !

(where ?)

Lets presents a simple scheme for point blending:

1 2 1 2

1 2

, ,..., , ,...,, ,...,

1 2( , ,..., )

n n

n

i i i i i ii i i

n

F u b u P

u u u u

A General Blending Model

Bezier and Bspline representation is exactly of this form.

Q. Why use Points as the Base objects?

A. There is no reason

A General Blending Model Let Q be an arbitrary parametrically defined

objects. The general parametric equation is we receive is:

Q – base objectsb – blending functions

1 2 1 2

1 2

, ,..., , ,...,, ,...,

1 2( , ,..., )

n n

n

i i i i i ii i i

n

G u b u Q

u u u u

Blending example Blending a set of curves for example: We use a continues function b which

satisfy the following conditions:

Then blending and , two parametric curves on the same domain is:

0 0 0 b s 1 b 1 1b

1c t 2c t

1 2, 1S s t b s C t b s C t

Blending example We can immediately see that: S is a surface. S(0,t) is a curve. (which one ?) S(1,t) is a curve. (which one ?)

Q. Can we blend in this way surfaces ? A. Yes

1 2, 1S s t b s C t b s C t 0 0 0 b s 1 b 1 1b

Blending Function

We will use the Bernstein functions to create a smooth blending function.

Let be the i-th Bernstein basis function of degree n.

lets define :

,i nB s

1 0,3 1,3

2 1

1 0

0 1

0 1

1

s

b s B s B s s

s

b s b s

Blending functions

General Equation

Let S1 and S2 be two smooth surfaces then we can define:

1 1 2 2, , ,

[0,1]

S s t b s S s t b s S s t

s

Rail curves

S is a blending surface smoothly connecting S1 and S2 along the rail curves S1(0,t) , S2(1,t)

The intersection problem

Finding the intersection curve between two surfaces is a Hard problem.

Algebraic solutions – complex , good for low dimensionality.

Numerically solutions – not accurate, loose parameterization.

The intersection problem

Solution: Numerically find points on the

intersection curve. Construct a curve C that

interpolate the points. Locally change the surfaces so

they pass through C.

Curve/Surface Blending Model Let c(t) be a smooth curve on [c,d] S1(s,t) a smooth surface on [a,b]X[c,d] We define:

1 0,2

2 1

1 0

0 1

0 1

1

s

b s B s s

s

b s b s

Blending function

Curve / Surface Blending Model

The new parametric surface we get is:

1 2 1

1

, ,

, [0,1] [ , ]

0,

, , S 1

S s t b s C t b s S s t

s t X c d

S t C t

S s t S s t

1 0,2

2 1

1 0

0 1

0 1

1

s

b s B s s

s

b s b s

Curve / Surface Blending

We can easily see that the interpolated curve pass through the new Surface.

To finish the algorithm we will use the model presented earlier on our problem.

Curve / Surface Blending

C(t) is a curve defined on [a,b] S1(s,t) is a surface defined on

[a,b]x[c,d] C1=S1(h(v)) a curve on S1 h(v) is a function from [0,1] to

[a,b]X[c,d]

For simplicity:

3

0 ,30

0j 0 0

v [0,1]

P , [ , ] [ , ]

j ji

j j

h v P B v

X Y a b X c d

Curve / Surface Blending

We need to create a blending erea. This is done by sweeping h(v) to

the right.

And the blending area is:

1 0 ( ,0)j jP P e

1 3

1 30 0

,

, [0,1] [0,1]

ij i ji j

f u v P B u B v

u v X

Curve / Surface Blending Thus the blending surface is:

1 2 1, ,

, [0,1] [0,1]

1

S u v b u C g v b u S f u v

u v X

g v v a vb

3 surfaces – 2 curves

Can we use a similar approach for more variables ?

Yes we can …

Surface/Surface – Corner Blending

Surface/Surface – Corner Blending

Blending is done in the parameter space.

Intersection curve can be approximated !

Blending functions

21

1 ,,

0 ,i

ii

i i i

s t Hb s t

s t O

O R H M M G

Constructing b1 definitions Bernstein of degree 5

f- mapping (rotation / translation)

3

0

, ,5i

b s t B i

1 2

1

:

:

:

i i

i i i

i i i

f P O

f PP e

f PR e

Bernstein triangular

C(s,t) = Bernstein triangular Edges are bizier curves. Fits our parameters (c1)

Blending function

1

1

1 11

3 3

1 ,

0 ,

, ,,

1 , ,

, ,

s t H

s t O

b f s t s t Mb s t

b f s t s t M

c s t s t G

Blend by pointwise interpolation

Given two surfaces P(u,v) , Q(s,t) Let A(w) , B(w) two respective contact

curves: A(w)=P(u(w),v(w)) B(w)=Q(s(w),t(w))

We pick two vectors in the tangent plane.

pointwise interpolation

A general form of the vectors:

2 2

, ,

, ,

a u v

b s t

T l P u w v w k P u w v w

T l Q s w t w k Q s w v w

pointwise interpolation

Using global functions M0 and M1 :

0 1 0 1

0 0 0 1

0 0 0 1

1 0 1 1

1 0 1 1

0 0 0 1

0 0 0 1

0 0 0 1

0 0 0 1

1 M 0

' 0 M ' 0

0 M 1

' 0 M ' 0

0 N 0

' 1 N ' 0

0 N 0

' 0 N ' 1

a bf h M h A M h B N h T N h T

M h h

M h h

M h h

M h h

N h h

N h h

N h h

N h h

Blend by pointwise interpolation

0 1 0 1, a bF h w M h A w M h B w N h T w N h T w

And the new surface is:

Choices of functions

There are many choices for M and N.

Tangent vectors T are more application driven.

Example:aT N t

Geometric correspondence

Hard problem There is No good solution.

Fanout surface technique

Using intrinsic properties of the curves !

( ), ,

,, ,

, ,

u vn

u v

f n

P u v P u vP u v

P u v P u v

P P u v p P u v

Fanout surface technique If P is a point on A. (the contact

curve)

And the curve becomes:

, ,P u v P u a v a A a

f nP A a p A a

The fanout surface Using a and p as parameters gives us

the fanout surface:

And in a similar way:

, ,f n nP A a p A a P u a v a p P u a v a

, ,f n nQ B b q B b Q s b t b q Q s b t b

Funout surfaces intersection The intersection of the fanout surfaces

gives us the needed correspondence. 3 equations , 4 unknowns, one

parameter

Q. Where are the 3 equations?A. Next page…

, ,f fP a p Q b q

Correspondence solution

, ,

, [ , , , , , ]

, [ , , , , , ]

1) , ,

2) , ,

3) , ,

f f

f f f fx y z

f f f fx y z

f fx x

f fy y

f fz z

P a p Q b q

P a p P a p P a p P a p

Q b q Q b q Q b q Q b q

P a p Q b q

P a p Q b q

P a p Q b q

Correspondence solution

a=a(w) , p=p(w) , b=b(w) , q=q(w) We have a parametric solution

from degree 1 = curve !

THE END