1
CS 537/CPE 537
537: Interacti
Interactive
e
Computer Graphics
Lecture XII
Instructor: Philippos Mordohai
Webpage: www.cs.stevens.edu/~mordohai
E
E-mail:
il Philippos.Mordohai@stevens.edu
Phili
M d h i@
d
Overview
• Recap of ANG Ch
Ch. 10
• ANG Ch. 12: Curves and Surfaces
2
Hierarchical Modeling (recap)
(ANG Ch. 10)
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3
Tree with Matrices
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4
C Definition of treenode
typedef struct treenode
{
GLfl t m[16];
GLfloat
[16]
void (*f)();
struct treenode *sibling;
struct treenode *child;
child;
} treenode;
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5
Display Function (figuretr
(figuretr.c)
c)
display(void)
{
glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_B
(
IT);
glLoadIdentity();
glColor3f(1.0,
lC l 3f(1 0 0
0.0,
0 0
0.0);
0)
traverse(&torso_node);
glutSwapBuffers();
}
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Traverse Function
void traverse(treenode* root)
{
if(root==NULL) return;
glPushMatrix();
glMultMatrixf(root->m);
root->f();
if(root->child!=NULL)
traverse(root->child);
glPopMatrix();
if(root->sibling!=NULL)
(
g
)
traverse(root->sibling);
}
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7
Graphical Objects and Scene
Graphs
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8
Imperative Programming Model
• Example:
p rotate a cube
cube data
glRotate
Application
results
• The rotation function must know how the
cube is represented
– Vertex list
– Edge list
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9
Object-Oriented Programming
M d l
Model
• In this model, the representation is stored with
the object
Application
message
Cube Object
• The application sends a message to the object
• The object contains functions (methods) which
allow it to transform itself
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10
Scene Graph
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11
Classes
Node
Camera
Light
TurnOff
DrawStyle
Color
Material
Transformation
Geometry
GLViewer
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Cube
Cylinder
S h
Sphere
Triangle
P l
Polygon
Line
12
Tree
Root
Camera
Light0
Color0
Transformation0
Object0
Transformation1
Object1
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Example
//Transformation Nodes for both Robot and Chair:
Transformation *Trans1=new Transformation;
Trans1->SetValue(TRANSLATION, -0.5, 0, 0, 2);
Trans1->AddChild(ChairColor);
Trans1->AddChild(RobotMat);
//Root Node:
Node *Root=new Node;
Root->AddChild(Trans1);
Root->AddChild(Camera1);
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Example
//Viewer:
GLViewer *MyViewer=new GLViewer;
MyViewer->Init(argc, argv);
MyViewer->SetValue(BACKCOLOR,
GREY);
MyViewer->SetValue(BUFFER,
DOUBLE);
MyViewer->CreateWin("Working
Hard", 500, 500);
GLViewer *MyViewer2=new GLViewer;
MyViewer2->Init(argc, argv);
MyViewer2->SetValue(BACKCOLOR,
MAGENTA);
MyViewer2->CreateWin("Working
y e e
eate
( o
g
Hard2", 200, 200);
MyViewer2->SetValue(BUFFER,
DOUBLE);
MyViewer->Show(Root);
MyViewer2->Show(Root);
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15
Curves and Surfaces
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16
Objectives
• Introduce types of curves and surfaces
– Explicit
– Implicit
– Parametric
– Strengths and weaknesses
eaknesses
• Discuss Modeling and Approximations
– Conditions
– Stability
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Escaping Flatland
• Until now we have worked with flat entities
such as lines and flat polygons
– Fit well with g
graphics
p
hardware
– Mathematically simple
• But the world is not composed
p
of flat
entities
– Need curves and curved surfaces
– May only have need at the application level
– Implementation can render them
approximately with flat primitives
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18
Modeling with Curves
interpolating data point
data points
approximating curve
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What Makes a Good Representation?
• There are many ways to represent curves
and surfaces
• Want a representation that is
– Stable
– Smooth
– Easy to evaluate
– Must we interpolate
p
or can we jjust come close
to data?
– Do we need derivatives?
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Explicit Representation
• Most familiar form of curve in 2D
y=f(x)
• Cannot represent
p
all curves
– Vertical lines
– Circles
• Extension to 3D
– y=f(x),
y=f(x) z=g(x)
– The form z = f(x,y) defines a surface
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Implicit Representation
• Two dimensional curve(s)
g(x,y)=0
• Much more robust
– All lines ax+by+c=0
– Circles x2+y2-r2=0
• Three dimensions g(x,y,z)=0 defines a surface
– Intersect two surface to get a curve
• In general, we cannot solve for points that satisfy
the equation
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22
Algebraic Surface
 x y z
i
i
j
j
k
0
k
• Quadric surface
2  i+j+k
• At most 10 terms
• Can solve intersection with a ray by
reducing problem to solving quadratic equation
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23
Parametric Curves
• Separate equation for each spatial variable
x=x(u)
T
p(u)=[x(u),
p(
)
[
(
),
y(
y(u),
),
z(u)]
(
)]
yy=y(u)
y(u)
z=z(u)
• For umax  u  umin we trace
t
outt a curve in
i two
t or
three dimensions
p(u)
p(umax)
p(umin)
p(
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Selecting Functions
• Usually we can select “good”
good functions
– not unique for a given spatial curve
– Approximate or interpolate known data
– Want functions which are easy to evaluate
– Want ffunctions
nctions which
hich are easy
eas to differentiate
• Computation of normals
• Connecting pieces (segments)
– Want functions which are smooth
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Parametric Lines
We can normalize u to be over the interval [0,1]
Line connecting two points p0 and p1
p(1)= p1
p(u)=(1-u)p
( ) ( ) 0+up1
p(0)) = p0
p(
p(1)= p0 +d
Ray from p0 in the direction d
d
p(u)=p0+ud
p(0) = p0
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Parametric Surfaces
• Surfaces require
q
2p
parameters
y
p(u,1)
x=x(u,v)
p(0,v)
yy=y(u,v)
y( , )
p(1 v)
p(1,v)
z=z(u,v)
x
p(u,v) = [x(u,v), y(u,v), z(u,v)]T
z p(u,0)
• Want same properties as curves:
– Smoothness
– Differentiability
Diff
i bili
– Ease of evaluation
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Normals
We can differentiate with respect to u and v to obtain
the normal at any point p
x (u , v) / u 
p(u , v) 
 y(u , v) / u 
u
 z(u , v) / u 
x (u , v) / v 
p(u , v) 
 y(u , v) / v 
v
 z(u , v) / v 
p(u , v) p(u , v)
n

u
v
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Parametric Planes
n
point-vector form
r
p(u,v)=p
p(u,v)
p0+uq+vr
q
n=qxr
p0
three-point form
n
p2
q = p1 – p0
r = p2 – p0
p1
p0
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Parametric Sphere
x(u,v) = r cos  sin 
y(u,v) = r sin sin 
z(u,v) = r cos 
360    0
180    0
 constant:
t t circles
i l off constant
t t longitude
l it d
 constant: circles of constant latitude
differentiate to show n = p
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Curve Segments
• After normalizing u, each curve is written
p(u)=[x(u), y(u), z(u)]T, 1  u  0
• In classical numerical methods, we design a single
global curve
• In computer
p
g
graphics
p
and CAD,, it is better to
design small connected curve segments
p(u)
p(0)
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join point p(1) = q(0)
q(u)
q(1)
31
Parametric Polynomial Curves
N
M
L
i 0
j 0
k 0
x(u )   c xi u i y (u )   c yj u j z (u )   c zk u k
•If N=M=K, we need to determine 3(N+1) coefficients
•Equivalently
q
y we need 3(N+1)
(
) independent
p
conditions
•Noting that the curves for x, y and z are independent,
we can define each independently in an identical manner
•We will use the form p(u ) 
where p can be any of x, y, z
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L
c
k
u
k
k 0
32
Why Polynomials
• Easy to evaluate
• Continuous and differentiable everywhere
– Must
M t worry about
b t continuity
ti it att join
j i points
i t
including continuity of derivatives
p(u)
q(u)
join point p(1) = q(0)
but p’(1)  q’(0)
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Cubic Parametric Polynomials
• N=M=L=3, g
gives balance between ease of
evaluation and flexibility
in
design
3
p(u )   c k u k
k 0
• Four coefficients to determine for each of x, y and
z
• Seek four independent conditions for various
values of u resulting in 4 equations in 4 unknowns
for each of x,, y and z
– Conditions are a mixture of continuity
requirements at the join points and conditions
for fitting the data
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Cubic Polynomial
y
Surfaces
p(u,v)=[x(u,v), y(u,v), z(u,v)]T
where
3
3
p(u , v)   cij u i v j
i 0 j 0
p is any of x, y or z
Need 48 coefficients ( 3 independent sets of 16) to
determine a surface patch
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Designing Parametric Cubic
Curves
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Objectives
• Introduce the types of curves
– Interpolating
– Hermite
– Bezier
– B-spline
B spline
• Analyze their performance
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Matrix-Vector
Matrix
Vector Form
3
p(u )   c k u k
k 0
c 0 
 
c1 

define c 
c 2 
 
 c3 
then
1
u 
u   2
u 
 3
u 
p(u )  u c  c u
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T
T
38
Interpolating Curve
p1
p0
p3
p2
Given four data (control) points p0 , p1 ,p2 , p3
determine cubic p(u) which passes through them
Must find c0 ,c1 ,c2 , c3
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Interpolation Equations
Apply the interpolating conditions at u=0,
u 0 1/3,
1/3 2/3,
2/3 1
p0=p(0)=c0
p1=p(1/3)=c
p( ) 0+(1/3)c
( ) 1+(1/3)
( )2c2+(1/3)
( )3c3
p2=p(2/3)=c0+(2/3)c1+(2/3)2c2+(2/3)3c3
p3=p(1)=c0+c1+c2+c3
or in matrix form with p = [p0 p1 p2 p3]T
p=Ac
1

1
A

1

1
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0
1
 
 3
 2
 
 3
1
0
2
1
 
 
 3
2
 2
 
 3
1
0 
3
1
  
  
 3 
3
 2 
  
 3 
1 
40
Interpolation Matrix
Solving for c we find the interpolation matrix
0
0
0 
 1
  5.5

9

4
.
5
1
1
M I  A   9  22.5 18  4.5



4
.
5
13
.
5

13
.
5
4
.
5


c=MIp
Note that MI does not depend on input data and
can be used for each segment in x, y, and z
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Interpolating Multiple Segments
use p =
[p0 p1 p2 p3]T
use p = [p3 p4 p5 p6]T
Get continuity at join points but not
continuity of derivatives
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Blending Functions
Rewriting the equation for p(u)
p(u)=uTc=uTMIp = b(u)Tp
where b(u) = [b0(u) b1(u) b2(u) b3(u)]T is
an array of blending polynomials such that
p(u)
( ) = b0(u)p
( ) 0+ b1(u)p
( ) 1+ b2(u)p
( ) 2+ b3(u)p
( ) 3
b0(u) = -4.5(u-1/3)(u-2/3)(u-1)
b1(u) = 13.5u (u-2/3)(u-1)
b2(u) = -13.5u (u-1/3)(u-1)
b3(u)
( ) = 4.5u
4 5 (u-1/3)(u-2/3)
( 1/3)( 2/3)
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43
Blending Functions
• These functions are not smooth
– Hence the interpolation polynomial is not
smooth
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Interpolating Patch
3
p (u , v)  
i o
3
j
u
 cij v
i
j 0
Need 16 conditions to determine the 16 coefficients cij
Choose at u,v = 0, 1/3, 2/3, 1
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Matrix Form
Define
fi v = [1 v v2 v3]T
C = [cij]
P = [pij]
p(u,v) = uTCv
If we observe that for constant u (v),
( ) we obtain
interpolating curve in v (u), we can show
C=M
C
MIPMI
p(u,v) = uTMIPMITv
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Blending Patches
3
p (u , v)  
i o
3
 b (u ) b
i
j 0
j
(v ) pij
Each bi(u)bj(v) is a blending patch
Shows that we can build and analyze surfaces
from our knowledge
g of curves
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Other Types of Curves and
S f
Surfaces
• How can we get around the limitations of
the interpolating form
– Lack of smoothness
– Discontinuous derivatives at join points
• W
We have
h
four
f
conditions
diti
(for
(f cubics)
bi ) that
th t
we can apply to each segment
– Use them not only for
f interpolation
– Need only come close to the data
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Hermite Form
pp’(0)
(0)
p(0)
(0)
pp’(1)
( )
p(1)
(1)
Use two interpolating conditions and
t derivative
two
d i ti conditions
diti
per segmentt
Ensures continuity and first derivative
continuity between segments
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Equations
Interpolating conditions are the same at ends
p(0) = p0 = c0
p(1) = p3 = c0+c1+c2+c3
Differentiating we find p’(u) = c1+2uc2+3u2c3
Evaluating at end points
p (0) = p’
p’(0)
p 0 = c1
p’(1) = p’3 = c1+2c2+3c3
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Matrix Form
 p 0  1
 p  1
q   3  
pp'0  0
  
 p'3 0
0 0 0
1 1 1
c
1 0 0

1 2 3
Solving, we find cc=M
MHq where MH is the Hermite matrix
M
H
0
0
0
1
0

0
1
0


 3 3  2  1


1
 2 2 1
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Blending Polynomials
p(u) = b(u)Tq
2 u 3  3 u 2  1

3
2 

2

3
u
u 

b(u )  3
 u  2 u2  u 


3
2
 u u

Although these functions are smooth, the Hermite form
i nott usedd directly
is
di tl in
i Computer
C
t Graphics
G hi andd CAD
because we usually have control points but not derivatives
However the Hermite form is the basis of the Bezier form
However,
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Parametric and Geometric
C i i
Continuity
• We can require the derivatives of x, y,and z to
each be continuous at join points (parametric
continuity)
• Alternately, we can only require that the
tangents of the resulting curve be continuous
(geometric continuity)
• The latter gives more flexibility as we need to
satisfy
f only two conditions rather than three at
each join point
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Example
• Here the p and q have the same tangents
at the ends of the segment but different
derivatives
• Generate different
H
Hermite
i curves
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Higher Dimensional Approximations
• The techniques
q
for both interpolating
p
g and
Hermite curves can be used with higher
dimensional parametric polynomials
• For
F the
h interpolating
i
l i form,
f
the
h resulting
l i matrix
i
becomes increasingly more ill-conditioned and
the resulting curves less smooth and more prone
to numerical errors
• In both cases, there is more work in rendering
the resulting polynomial curves and surfaces
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Bezier and Spline Curves and
Surfaces
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Objectives
• Introduce the Bezier curves and surfaces
• Derive the required matrices
• Introduce
I
d
the
h B
B-spline
li and
d compare iit to
the standard cubic Bezier
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Bezier’ss Idea
Bezier
• In graphics and CAD,
CAD we do not usually
have derivative data
• Bezier suggested using the same 4 data
points as with the cubic interpolating curve
to approximate the derivatives in the
Hermite form
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Approximating Derivatives
p2
p1
p2 located at u=2/3
p1 located at u=1/3
p1  p0
p' (0) 
1/ 3
p3  p 2
p' (1) 
1/ 3
slope p’(1)
slope p’(0)
p0
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u
p3
59
Equations
Interpolating conditions are the same
p(0) = p0 = c0
p(1)) = p3 = c0+c1+c2+c3
p(
Approximating derivative conditions
p (0) = 3(p1- p0) = c0
p’(0)
p’(1) = 3(p3- p2) = c1+2c2+3c3
Solve four linear equations for c=MBp
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Bezier Matrix
0
0
1
 3 3
0

MB   3  6 3

 1 3  3
0

0
0

1
p(u)) = uTMBp = b(u)
p(
( )Tp
blending functions
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Blending Functions
 (1  u )3 

2
3u (1  u ) 

b(u ) 
2 u 2 (1  u )


3


u
Note that all zeros are at 0 and 1 which forces
the functions to be smooth over (0,1)
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Bernstein Polynomials
• The blending functions are a special case
of the Bernstein polynomials
d!
d k
k
bkd (u ) 
u (1  u )
k!(d  k )!
• These polynomials give the blending
polynomials for any degree Bezier form
– All zeros at 0 and 1
– For any degree they all sum to 1
– They are all between 0 and 1 inside (0,1)
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Convex Hull Property
• The properties of the Bernstein polynomials
ensure that all Bezier curves lie in the convex hull
of their control points
• Hence, even though we do not interpolate all the
data, we cannot be too far away
p1
p2
convex hull
Bezier curve
p0
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p3
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Bezier Patches
U i same d
Using
data
t array P=[pij] as with
ith iinterpolating
t
l ti fform
3
3
p (u, v)   bi (u ) b j (v) pij  uT M B P MTB v
i 0 j 0
Patch lies in
convex hull
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Analysis
• Although the Bezier form is much better than the
interpolating form, the derivatives are not
continuous at join points
• Can we do better?
– Go to higher order Bezier
• More work
• Derivative continuity still only approximate
• Supported by OpenGL
– Apply different conditions
• Tricky without letting order increase
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B Splines
B-Splines
T
• Basis splines:
p
use the data at p
p=[p
[pi-2
i 2 pi-1
i 1 pi pi+1] to
define curve only between pi-1 and pi
• Allows us to apply more continuity conditions to
each
h segmentt
• For cubics, we can have continuity of function, first
and second derivatives at join points
• Cost is 3 times as much work for curves
– Add one new p
point each time rather than
three
• For surfaces, we do 9 times as much work
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Cubic B-spline
B spline
p(u) = uTMSp = b(u)Tp
4
1
1
 3 0
3

MS   3  6 3

 1 3  3
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0
0
0

1
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Blending Functions
3


(1  u )

2
3 
1  4  6 u  3u 
b(u ) 
6 1  3u  3 u 2  3 u 2 


3

u

convex hull property
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B-Spline
B
Spline Patches
3
3
p (u , v)   bi (u ) b j (v) pij  u M S P M v
T
T
S
i 0 j 0
defined over only 1/9 of region
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Splines and Basis
• If we examine the cubic B-spline
B spline from the
perspective of each control (data) point,
point contributes ((through
g the
each interior p
blending functions) to four segments
• We can rewrite p(
p(u)) in terms of the data
points as
p(u)  Bi(u) pi
defining the basis functions {Bi(u)}
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Basis Functions
In terms of the blending
gp
polynomials
y
0

 (u  2)
b0
 b1 (u  1)
Bi (u )  
(u )
 b2
 b3 (u  1)

0

u i2
i  2  u  i 1
i 1  u  i
i  u  i 1
i 1  u  i  2
u i2
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Piecewise Polynomial Curve
Linear combination of basis functions
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Generalizing Splines
• We can extend to splines of any degree
• Data and conditions do not have to given
at equally spaced values (the knots)
– Nonuniform and uniform splines
– Can
C h
have repeated
d kknots
• Can force spline to interpolate points
• C
Cox-de
d B
Boor recursion
i gives
i
method
h d off
evaluation
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NURBS
• Nonuniform Rational B
B-Spline
Spline curves and
surfaces add a fourth variable w to x,y,z
– Can interpret
p as weight
g to g
give more
importance to some control data
– Can also interpret as moving to homogeneous
coordinate
di
• Requires a perspective division
– NURBS act correctly for perspective viewing
• Quadrics are a special case of NURBS
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Rendering Curves and Surfaces
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Objectives
• Introduce methods to draw curves
– Approximate with lines
– Finite Differences
• Derive the recursive method for evaluation
off Bezier
B i curves and
d surfaces
f
• Learn how to convert all polynomial data to
data for Bezier polynomials
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Evaluating Polynomials
• Simplest method to render a polynomial
curve is to evaluate the polynomial at many
points
po
sa
and
d form
o a
an app
approximating
o
a g po
polyline
y e
• For surfaces we can form an approximating
mesh of triangles or quadrilaterals
• Use Horner’s method to evaluate
polynomials
p(u)=c0+u(c1+u(c2+uc3))
– 3 multiplications/evaluation for cubic
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Finite Differences
For equally spaced {uk} we define finite differences
 p(u k )  p (u k )
(0)
 p(u k )  p (u k  1)  p (u k )
(1)
( m 1)

p(u k )  
(m)
p(u k  1)  
(m)
p(u k )
For a polynomial of degree n,
the nth finite difference is constant
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Building a Finite Difference Table
p(u)=1+3u+2u
p(u)
1+3u+2u2+u3
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Finding the Next Values
Starting at the bottom, we can work up generating new
values for the polynomial

(m)
( m 1)
p (u k  1)  
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p (u k )  
(m)
p (u k )
81
de Casteljau Recursion
• We can use the convex hull property of
Bezier curves to obtain an efficient
q
any
y
recursive method that does not require
function evaluations
– Uses only the values at the control points
• Based on the idea that “any polynomial
and any part of a polynomial is a Bezier
polynomial for properly chosen control
data”
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Splitting a Cubic Bezier
p0, p1 , p2 , p3 determine a cubic Bezier polynomial
and its convex hull
Consider left half l(u) and right half r(u)
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l(u) and r(u)
Since l(u)
( ) and r(u)
( ) are Bezier curves,, we should be able to
find two sets of control points {l0, l1, l2, l3} and {r0, r1, r2, r3}
that determine them
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Convex Hulls
{l0, l1, l2, l3} and {r0, r1, r2, r3}each have a convex hull
th t is
that
i closer
l
tto p(u)
( ) than
th the
th convex hull
h ll off {p
{ 0, p1, p2, p3}
This is known as the variation diminishing property.
The polyline from l0 to l3 (= r0) to r3 is an approximation
to p(u). Repeating recursively we get better approximations.
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Equations
Start with Bezier equations p(u)
p(u)=uTMBp
l(u) must interpolate p(0) and p(1/2)
l(0) = l0 = p0
l(1) = l3 = p(1/2) = 1/8( p0 +3 p1 +3 p2 + p3 )
Matching slopes
slopes, taking into account that l(u)
l( ) and r(u)
( )
only go over half the distance as p(u)
l (0) = 3(l1 - l0) = pp’(0)
l’(0)
(0) = 3/2(p1 - p0 )
l’(1) = 3(l3 – l2) = p’(1/2) = 3/8(- p0 - p1+ p2 + p3)
S
Symmetric
t i equations
ti
h ld ffor r(u)
hold
( )
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Efficient Form
l0 = p0
r3 = p3
l1 = ½(p0 + p1)
r2 = ½(p2 + p3)
l2 = ½(l1 + ½( p1 + p2))
r1 = ½(r2 + ½( p1 + p2))
l3 = r0 = ½(l2 + r1)
Requires only shifts and adds!
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Every Curve is a Bezier Curve
• We can render a given polynomial using the
recursive method if we find control points for its
representation as a Bezier curve
• Suppose that p(u) is given as an interpolating
curve with control points q
p(u)=uTMIq
• There exist Bezier control points p such that
p(u)=uTMBp
• Equating and solving, we find p=MB-1MIq
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Matrices
Interpolating to Bezier
B S li tto B
B-Spline
Bezier
i
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 1
 5


6
1

MB MI  1

 3
 0
0
0
3

2
3
3
2
0

1
0
1
MB MS  
0

0
3
0
4
4
2
1
1
2
4
4
0 
1 
3 
5
 
6
1 
0
0
0

1
89
Example
These three curves were all generated from the same
original data using Bezier recursion by converting all
control point data to Bezier control points
Bezier
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Interpolating
B Spline
p
90
Surfaces
• Can apply the recursive method to surfaces if we
recall that for a Bezier patch curves of constant u
(or v) are Bezier curves in u (or v)
• First subdivide in u
– Process creates new points
– Some of the original points are discarded
original and discarded
original and kept
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new
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Second Subdivision
16 final points for
1 of 4 patches created
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Surface Subdivision Limitations
• The test for whether the new convex hull is
flat enough is more difficult than for curves
• Renderers use fixed number of
subdivisions or let user pick the number
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Utah Teapot
• Most famous data set in computer graphics
• Widely available as a list of 306 3D vertices and
p
the indices that define 32 Bezier patches
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Rendering Quadrics
• Anyy q
quadric can be written as the q
quadratic form
pTAp+bTp+c=0 where p=[x, y, z]T
with A,
A b and c giving the coefficients
• Render by ray casting
– Intersect with parametric ray p()
p()=p
p0+d
d that
passes through a pixel
– Yields a scalar quadratic equation
• No solution: ray misses quadric
• One solution: ray tangent to quadric
• Two solutions: entry and exit points
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Curves and Surfaces in OpenGL
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Objectives
• Introduce OpenGL evaluators
• Learn to render polynomial curves and
surfaces
• Discuss quadrics in OpenGL
– GLUT Quadrics
– GLU Quadrics
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What Does OpenGL
p
Support?
pp
• Evaluators: a g
general mechanism for working
g with
the Bernstein polynomials
– Can use any degree polynomials
– Can use in 1-4 dimensions
– Automatic generation of normals and texture
coordinates
– NURBS supported in GLU
• Quadrics
– GLU and GLUT contain polynomial
approximations
app
o
at o s o
of quad
quadrics
cs
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One-Dimensional
O
e
e s o a Evaluators
a ua o s
• Evaluate a Bernstein polynomial of any
degree at a set of specified values
• Can evaluate a variety of variables
– Points along a 2, 3 or 4 dimensional curve
– Colors
– Normals
– Texture Coordinates
• We can set up multiple evaluators that are
all evaluated for the same value
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Setting Up an Evaluator
what we want to evaluate
max and min of u
glMap1f(type,u_min,u_max,stride,
1
order, pointer_to_array)
separation bet
between
een
1+degree of polynomial
data points
pointer to control data
Each type must be enabled by glEnable(type)
Types include 3D and 4D points, RGBA colors, normals,
t t
texture
coordinates
di t etc.
t
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Example
C
Consider
id an evaluator
l t ffor a cubic
bi B
Bezier
i curve over (0
(0,1)
1)
point data[ ]={
]={…………..};
}; * /3d data /*
/
glMap1f(GL_MAP_VERTEX_3,0.0,1.0,3,4,data);
cubic
data are 3D vertices
d t are arranged
data
d as x,y,z,x,y,z……
three floats between data points in array
glEnable(GL_MAP_VERTEX_3);
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Evaluating
• The function glEvalCoord1f(u) causes all
enabled evaluators to be evaluated for the
specified u
– Can replace glVertex, glNormal,
glTexCoord
• The values of u need not be equally
spaced
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Example
• Consider the previous evaluator that was set up
for a cubic Bezier over (0,1)
• Suppose that we want to approximate the curve
with a 100 point polyline
glBegin(GL_LINE_STRIP)
for(i=0; i<100; i++)
glEvalCoord1f(
l
l
d1f( (float)
(fl
) i/100.0);
i/100 0)
glEnd();
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Equally Spaced Points
Rather than use a loop, we can set up an equally
spaced mesh (grid) and then evaluate it with
one function call
glMapGrid(100, 0.0, 1.0);
sets up 100 equally-spaced points on (0,1)
glEvalMesh1(GL_LINE,
lE lM h1(GL LINE 0,
0 99);
99)
renders lines between adjacent evaluated
points
i t ffrom point
i t 0 tto point
i t 99
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Bezier Surfaces
• Similar
S
a p
procedure
ocedu e to 1D but use 2D e
evaluators
a uato s in u
and v
• Set up
p with
glMap2f(type, u_min,
_
umax, u_stride,
_
u_order, v_min, v_max, v_stride,
v_order, pointer_to_data)
• Evaluate with glEvalCoord2f(u,v)
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Example
bicubic over (0,1) x (0,1)
point data[4][4]={………};
glMap2f(GL_MAP_VERTEX_3, 0.0, 1.0, 3, 4,
0.0, 1.0, 12, 4, data);
Note that in v direction data points
are separated by 12 floats since array
data is stored by rows
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Rendering with Lines
must draw in both directions
for(j=0;j<100;j++) {
glBegin(GL_LINE_STRIP);
for(i=0;i<100;i++)
glEvalCoord2f((float)
2
i/100.0,
/100 0 (float) j/100.0);
/100 0
glEnd();
glBegin(GL_LINE_STRIP);
for(i=0;i<100;i++)
glEvalCoord2f((float) j/100.0, (float) i/100.0);
glEnd();
}
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Rendering with Quadrilaterals
We can form a quad mesh and render with lines
for(j=0; j<99; j++) {
glBegin(GL_QUAD_STRIP);
l
i (G Q
S
)
for(i=0; i<100; i++) {
glEvalCoord2f ((float) i/100.0,
(float) j/100
j/100.0);
0);
glEvalCoord2f ((float)(i+1)/100.0,
(float)j/100.0);
}
glEnd():
}
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Uniform Meshes
• We can form a 2D mesh (grid) in a similar
manner to 1D for uniform spacing
glMapGrid2(u
g
p
( _num,
, u_min,
, u_max,
,
v_num, v_min, v_max)
• Can evaluate as before with lines or if
want filled polygons
glEvalMesh2( GL_FILL, u_start,
u_num, v_start, v_num)
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Rendering with Lighting
• If we use filled p
polygons,
yg
we have to shade
or we will see solid color uniform rendering
• Can specify lights and materials but we
need normals
– Let OpenGL find them
glEnable(GL_AUTO_NORMAL);
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NURBS
• OpenGL supports NURBS surfaces
through the GLU library
• Why GLU?
– Can use evaluators in 4D with standard
OpenGL library
– However, there are manyy complexities
p
with
NURBS that need a lot of code
– There are five NURBS surface functions plus
f nctions for trimming curves
functions
c r es that can remo
remove
e
pieces of a NURBS surface
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Quadrics
• Quadrics
Q
are in both the GLU and GLUT
libraries
– Both use polygonal approximations where the
application specifies the resolution
– Sphere: lines of longitude and lattitude
• GLU: disks
disks, cylinders
cylinders, spheres
– Can apply transformations to scale, orient,
and position
• GLUT: Platonic solids, torus, Utah teapot,
cone
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Quadric Objects in GLU
• GLU can automaticallyy generate
g
normals and
texture coordinates
• Quadrics are objects that include properties
such
h as h
how we would
ld lik
like th
the object
bj t tto b
be
rendered
disk
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partial
ti l di
disk
k
sphere
113
Defining a Cylinder
GLUquadricOBJ
q
*p;
p
P = gluNewQuadric(); /*set up object */
gluQuadricDrawStyle(GLU_LINE);/*render
style*/
gluCylinder(p, BASE_RADIUS, TOP_RADIUS,
BASE_HEIGHT, sections, slices);
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GLUT Objects
Each has a wire and a solid form
glutWireCone()
glutWireTorus()
glutWireTeapot()
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GLUT Platonic Solids
glutWireTetrahedron()
glutWireDodecahedron()
glutWireOctahedron()
glutWireIcosahedron()
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