Curves&Surfaces Lec Ppt

7/31/2019 Curves&Surfaces Lec Ppt

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UNIT II

CURVES & SURFACES


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Contents

Bresenhams line drawing algorithm

Circle drawing algorithms

A simple technique

The mid-point circle algorithm

Curves & Surfaces

Polygon Mesh

Hermite curve

Bezier Curve

B-Spline Curve


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The Bresenham Line Algorithm

Meet JACK BRESENHAM

The big advantage of this algorithm is thatit uses only integer calculations

Jack Bresenham workedfor 27 years at IBM

b e f o r e e n t e r i n g

academia. Bresenham

developed his famous

algorithms at IBM in the

e a r l y 1 9 6 0 s


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The Big Idea

Move across thex axis in unit intervals and ateach step choose between two differentycoordinates

2 3 4 5

2

4

3

5 For example, fromposition (2, 3) we

have to choose

between (3, 3) and

(3, 4)

We would like the

point that is closer to

the original line

(xk

,yk

)

(xk+1,yk)

(xk+1,yk+1)


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They coordinate on the mathematical line at

xk+1 is:

Deriving The Bresenham Line Algorithm

At sample positionxk+1 the

vertical separations from

the mathematical line are

labelled dupperand dlower

bxmy k )1(

y

yk

yk+1

xk+1

dlower

dupper


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So, dupperand dlowerare given as follows:

and:

We can use these to make a simple decision

about which pixel is closer to the mathematical

line

Deriving The Bresenham Line Algorithm (cont)

klower yyd

kk ybxm

)1(

yyd kupper )1(

bxmy kk )1(1


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This simple decision is based on the differencebetween the two pixel positions:

Lets substitute m with y/xwhere x and

y are the differences between the end-points:


122)1(2 byxmdd kkupperlower

)122)1(2()(

byxx

yxddx kkupperlower

)12(222 bxyyxxy kk

cyxxy kk 22


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So, a decision parameterpkfor the kth stepalong a line is given by:

The sign of the decision parameterpkis thesame as that ofdlowerdupper

Ifpkis negative, then we choose the lowerpixel, otherwise we choose the upper pixel


cyxxy

ddxp

kk

upperlowerk

22

)(


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Remember coordinate changes occur along

thex axis in unit steps so we can doeverything with integer calculations

At step k+1 the decision parameter is given as:

Subtractingpkfrom this we get:


cyxxyp kkk 111 22

)(2)(2 111 kkkkkk yyxxxypp


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But,xk+1 is the same asxk+1 so:

whereyk+1 - ykis either 0 or 1 depending onthe sign ofpk

The first decision parameter p0 is evaluated at

(x0, y0) is given as:


)(22 11 kkkk yyxypp

xyp 20


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The Bresenham Line Algorithm

BRESENHAMS LINE DRAWING ALGORITHM(for |m| < 1.0)

1. Input the two line end-points, storing the left end-point in (x0, y0)

2. Plot the point (x0, y0)

3. Calculate the constants x, y, 2y, and (2y - 2x) and get the firstvalue for the decision parameter as:

4. At eachxkalong the line, starting at k = 0, perform the following test. Ifpk

< 0, the next point to plot is

(xk+1, yk) and:

xyp 20

ypp kk 21


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The Bresenham Line Algorithm (cont)

warning! The algorithm and derivation above

assumes slopes are less than 1. for other

slopes we need to adjust the algorithm slightly

Otherwise, the next point to plot is (xk+1, yk+1) and:

5. Repeat step 4 (x 1) times

xypp kk 221


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Bresenham Example

Lets have a go at this

Lets plot the line from (20, 10) to (30, 18)

First off calculate all of the constants:

x: 10

y: 8

2y: 16

2y - 2x: -4

Calculate the initial decision parameterp0:

p0= 2yx = 6


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Bresenham Example (cont)

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

k pk (xk+1,yk+1)

0

1

2

3

4

56

7

8

9


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Bresenham Exercise

Go through the steps of the Bresenham line

drawing algorithm for a line going from

(21,12) to (29,16)


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Bresenham Exercise (cont)

17

16

15

14

13

12

11

10

18

292726252423222120 28 30

k pk (xk+1,yk+1)

0

1

2

3

4

56

7

8


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Bresenham Line Algorithm Summary

The Bresenham line algorithm has the following advantages:

An fast incremental algorithm

Uses only integer calculations

Comparing this to the DDA algorithm, DDA has the followingproblems:

Accumulation of round-off errors can make the pixelated

line drift away from what was intended

The rounding operations and floating point arithmeticinvolved are time consuming


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A Simple Circle Drawing Algorithm

The equation for a circle is:

where ris the radius of the circle So, we can write a simple circle drawing

algorithm by solving the equation fory at unit

x intervals using:

222ryx

22 xry


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A Simple Circle Drawing Algorithm (cont)

20020 220 y

20120 221 y

20220 222 y

61920 2219 y

02020 2220 y


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A Simple Circle Drawing Algorithm (cont)

However, unsurprisingly this is not a brilliantsolution!

Firstly, the resulting circle has large gaps where

the slope approaches the vertical Secondly, the calculations are not very efficient

The square (multiply) operations

The square root operation try really hard to avoid

these!

We need a more efficient, more accurate solution


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Eight-Way Symmetry

The first thing we can notice to make our circledrawing algorithm more efficient is that circles

centred at (0, 0) have eight-way symmetry

(x, y)

(y, x)

(y, -x)

(x, -y)(-x, -y)

(-y, -x)

(-y, x)

(-x, y)

2

R


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Mid-Point Circle Algorithm

Similarly to the case with lines, there

is an incremental algorithm for

drawing circles the mid-point circle

algorithm In the mid-point circle algorithm we

use eight-way symmetry so only ever

calculate the points for the top right

eighth of a circle, and then usesymmetry to get the rest of the

points

The mid-point circle

a l g o r i t h m w a sd e v e l o p e d b y J a c k

B r e s e n h a m .


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Mid-Point Circle Algorithm (cont)

(xk+1, yk)

(xk+1, yk-1)

(xk, yk)

Assume that we havejust plotted point (xk, yk)

The next point is a

choice between (xk+1, yk)and (xk+1, yk-1)

We would like to choose

the point that is nearest tothe actual circle

So how do we make this choice?


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6

2 3 41

5

4

3


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M

6

2 3 41

5

4

3


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M

6

2 3 41

5

4

3


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Lets re-jig the equation of the circle slightly togive us:

The equation evaluates as follows:

By evaluating this function at the midpointbetween the candidate pixels we can make ourdecision

222),( ryxyxfcirc

,0

,0

,0

),( yxfcirc

boundarycircletheinsideis),(if yx

boundarycircleon theis),(if yx

boundarycircletheoutsideis),(if yx


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Assuming we have just plotted the pixel at(xk,yk) so we need to choose between(xk+1,yk) and (xk+1,yk-1)

Our decision variable can be defined as:

Ifpk< 0 the midpoint is inside the circle andand the pixel atykis closer to the circle

Otherwise the midpoint is outside andyk-1 is

closer

222 )2

1()1(

)2

1,1(

ryx

yxfp

kk

kkcirck


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To ensure things are as efficient as possible we cando all of our calculations incrementally

First consider:

or:

whereyk+1 is eitherykoryk-1 depending on the signofpk

2212111

21]1)1[(

21,1

ryx

yxfp

kk

kkcirck

1)()()1(2 122

11 kkkkkkk yyyyxpp


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The first decision variable is given as:

Then ifpk< 0 then the next decision variable

is given as:

Ifpk> 0 then the decision variable is:

r

rr

rfp circ

45

)

2

1(1

)2

1,1(

22

0

12 11 kkk xpp

1212 11 kkkk yxpp


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The Mid-Point Circle Algorithm

1. Input radius rand circle centre (xc, yc), then set the

coordinates for the first point on the circumference of a

circle centred on the origin as:

2. Calculate the initial value of the decision parameter as:

3. Starting with k = 0 at each positionxk, perform thefollowing test. Ifpk< 0, the next point along the circle

centred on (0, 0) is (xk+1, yk) and:

),0(),( 00 ryx

rp 4

50

1211

kkk

xpp


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The Mid-Point Circle Algorithm (cont)

4. Otherwise the next point along the circle is (xk+1, yk-1) and:

5. Determine symmetry points in the other seven octants

6. Move each calculated pixel position (x, y) onto the circular

path centred at (xc, yc) to plot the coordinate values:

7. Repeat steps 3 to 5 untilx >= y

111 212 kkkk yxpp

cxxx cyyy


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Mid-Point Circle Algorithm Example

To see the mid-point circle algorithm in action

lets use it to draw a circle centred at (0,0) with

radius 10


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Mid-Point Circle Algorithm Example (cont)

9

7

6

5

4

3

2

1

0

8

976543210 8 10

10k pk (xk+1,yk+1) 2xk+1 2yk+1

0

1

2

3

4

5

6


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Mid-Point Circle Algorithm Exercise

Use the mid-point circle algorithm to draw the

circle centred at (0,0) with radius 15


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Mid-Point Circle Algorithm Example (cont)

k pk (xk+1,yk+1) 2xk+1 2yk+1

0

1

2

3

4

5

6

78

9

10

11

12

9

7

6

54

3

2

1

0

8

976543210 8 10

10

131211 14

15

13

12

14

11

16

15 16


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Mid-Point Circle Algorithm Summary

The key insights in the mid-point circle

algorithm are:

Eight-way symmetry can hugely reduce the work

in drawing a circle

Moving in unit steps along the x axis at each point

along the circles edge we need to choose

between two possible y coordinates


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CURVES & SURFACES


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The need to represent curves and surfaces arises in two cases:

In modeling existing objects (a car , a face, a mountain)

In modeling from scratch where no preexisting physical object is being represented

In first case mathematical object may be unavailable.

One can use as a model the coordinates of the infinitely many points of the object but this notfeasible for computer with finite storage.

We merely approximate the object with pieces of planes, spheres or other shapes the t are easy todescribe mathematically and require that points on our model be close to corresponding points onthe object.

In second case, the user creates the object in the modeling process.

To create the object the user may sculpt the object interactively, describe it mathematically of givean approximate description to be filled in by some program.

In CAD the computer representation is used later to generate physical realizations of the abstractlydesigned object.


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Curves and Surfaces

Often we are required to represent surfaces that are notplanar in nature.

To do so the parametric representation of 2-D curves and3-D surfaces may be employed.

In general for any genus of surface or curve, there is both aparametric and an implicit representation.

In computer graphics it is often more convenient to adoptthe parametric form.


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Three most common representations for 3D surfaces are polygon mesh surfaces

parametric surfaces

quadric surfaces

How do we draw surfaces? Approximate with polygons

Draw polygons

How do we specify a surface? Explicit, implicit, parametric

How do we approximate a surface? Interpolation (use only points)

Hermite (use points and tangents)

Bezier (use points, and more points for tangents)


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Polygon mesh

A polygon mesh or unstructured grid is a collection ofvertices, edges and faces that defines the shape of apolyhedral object in 3D computer graphics and solidmodeling.

Thefaces usually consist of triangles, quadrilaterals or othersimple convex polygons, since this simplifies rendering, butmay also be composed of more general concave polygons, orpolygons with holes.

A cross section of curved object and its polygon representation


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polygon mesh representation

Pointers to a vertex list

V=(v1,v2,v3,v4)={(x1,y1,z1)(x4,y4,z4)}

P1={1,2,4}

P2={4,2,3}

Each vertex stored just once, considerable space is saved

It is still difficult to find polygons that share an edge

Shared edge are drawn twice

Pointers to an edge list

V=(v1,v2,v3,v4)={(x1,y1,z1)(x4,y4,z4)}

E1={v1,v2,p1,*}

E2={v2,v3,p2,*}E3={v3,v4,p2,*}

E4={v4,v2,p1,p2}

E5={v4,v1,p1,*}

P1={E1,E4,E5}

P2={E2,E3,E4}

E=(v1,v2,p1,p2,pn)

Avoid

redundant clipping

Transformation

Scan conversion

If Edge is shared by n

no. of polygon

Explicit representation

P={(x1,y1,z1),(x2,y2,z2),(x3,y3,z3),..(xn,yn,zn)}

v1

v2v3

v4

v1

v2v3

v4

p1 p2

p1 p2

E1

E2

E3E5

E4


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Explicit Representation

Curve in 2D: y = f(x)

Curve in 3D: y = f(x), z = g(x)

Surface in 3D: z = f(x,y)

Problems:

How about a vertical line x = c as y = f(x)?

Circle y = (r2 x2)1/2 two or zero values for x

Too dependent on coordinate system

Rarely used in computer graphics


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Implicit Representation

Curve in 2D: f(x,y) = 0

Line: ax + by + c = 0

Circle: x2

+ y2

r2

= 0

Surface in 3d: f(x,y,z) = 0

Plane: ax + by + cz + d = 0

Sphere: x2 + y2 + z2 r2 = 0


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Sphere definition

For example, the implicit form of a sphere is:

In general all implicit representations of a 3-d surface are ofthe form:

The parametric form of a sphere is:

x y z r2 2 2 2 0

f x y z( , , ) 0

f:( , ) (cos cos ,sin ,cos sin )


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Parametric form

The general form of a 3-d surface in its parametric (orexplicit) representation is:

When rendering such curves and surfaces using polygons,the parametric representation is more convenient as thesurface is defined in terms of a parametric variable orvariables.

This allows the exact determination of the value of thesurface at regular intervals, thus allowing an approximationto the surface by taking a number of such samples.

f u v f u v f u v f u vx y z( , ) ( ( , ), ( , ), ( , ))


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Example - 1

determine the representation of the functionf() = sin().

This is a parametric description of a curve in 2 dimensions

with parameter .

This is an example of an unbounded curve (in that we cantake values of from -...+. Well limit our curve to the

domain (0...2). This gives the following curve:


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Now we must determine how fine or coarse a representation we need to use

in order to faithfully capture this curve.

We will sample the curve at regular intervals of along the length of the

curve. In this example, the curve will be sampled at regular points a unit

distance apart (i.e. at = 0, 1, 2...).

This yields the following sample points which we will join by straight lineswhich is the way the curve will be finally displayed on the raster:


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Representing Curves

There are many different methods of representing general

curves. The most common are:

Cubic Splines

Bezier Curves

B-splines

NURBS and -splines


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Interpolation vs. Approximation

Given a set ofnpoints, to create a curve we either

interpolate the points (curve passes through all points)

approximate the points (points describe convex hullof curve)

Points on curve = knot points

Points on convex hull (off curve) = control points

Interpolate Approximate

Control points

Convex hull

knot points


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Splines

To interpolate we can use a simple polynomial spline.

With n points we require a polynomial of degree n-1

(order n polynomial).

Letf(u) be the parameterised polynomial where 0 u 1

bauuf cbuauuf 2 dcubuauuf 23

Linear Quadratic Cubic


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Splines

These polynomials are plots off(u) with respect to u

for each u, there is one and only onef(u)

curve cannot turn back on itself

Use polynomials for each axis (in 2D we have 2 polys):

As before we limit u to [0,1], although the polynomial isdefined for all values ofu.

yyyy

xxxx

ducubuauy

ducubuaux

23

23


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yyyy

xxxx

ducubuauyducubuaux

23

23

S li


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Cup .12323

23

23

u

ddd

ccc

bbb

aaa

uuuuzuyux

ducubuauz

ducubuauy

ducubuaux

zyx

zyx

zyx

zyx

zzzz

yyyy

xxxx

Splines

For a 3D spline, we have 3 polynomials:

up

Defines the variation in x withdistance u along the curve

12 unknowns4 3D points required

S li


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If we have more than 4 points werequire a polynomial of higher degree

higher degree polynomials are more

difficult to control

they exhibit unwanted wiggles

(oscillations)

Quadratic Cubic Quartic Quintic

Splines


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In general we use cubic polynomials for curves in CG:

minimal ups & downs and faster to compute than high

degree polynomials lowest degree which allows non-planar curves

(quadratics require 3 points, 3 points always lie in the

same plane)

Splines


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Defining the Cubic Spline

Normally we supply 4 points we wish the spline to pass through. These are 2 endpoints

2 derivatives of the end points

If we have more than 4 points we must employ more than 1 spline use apiecewise cubic polynomial

for n points, we have n1/3 individual cubic segments

without further constraints these will not join smoothly

smoothnon-smooth


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Curve Continuity Piecewise Curve Segments

To ensure a smooth connection between curve segments we enforcefurther continuityconstraints

2 types of continuity:

parametric continuity, denoted Cn where n = degree of continuity

geometric continuity, denoted Gn

Given a curve such that at pointp, 2 segments ci(u) and ci+1(u) meetthen:

0

1

1

u

n

in

u

n

in

n

du

ucd

du

ucdC

01 1 ii ccp

0

1

1

u

n

in

u

n

in

n

du

ucd

du

ucdG

differentials are equal differentials are proportional

G t i ti it


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Geometric continuity

In this case we require only parametric derivative of two curves to be

proportional to each other at their intersection point

If two curve segments joint together there curve has G0 continuity

If the direction of two segments tangent vectors are equal at the

joint point the curve has G1 continuity

In CAD G1 is often required

Mathematically.

2

2

1

2

2

2

11

0

1

Gdu

Qd

du

Pd

Gdu

dQ

du

dP

GQP

n

n

n

Parametric Continuity


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Parametric Continuity 0th order

Here curve simply meets

1st order

If the tangent vectors of two curves segment are equal (indirection as well as in magnitude) at the joint point

2nd

order If both the first and second parametric derivatives of the

two curve section are same at the their intersection

Nth order

If the nthe derivative are equal at the joint point.


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since we want these curves to fit together

reasonably ...

Zero order parametric continuity

First order parametric continuity

Second order parametric continuity


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Examples of Continuity

c0

c1

c2


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Cubic Parametric Curves

A curve segment p(u) is defined by constraints on end-points, tangent vectors, and continuity between curvesegments.

Each cubic polynomial has 4 co-efficients, so fourconstraints will be needed.

Remember:

This allows us to formulate 4 equations in the 4unknowns, and then solve for the unknowns.

Cup .12323

23

23

u

ddd

ccc

bbb

aaa

uuuuzuyux

ducubuauz

ducubuauy

ducubuaux

zyx

zyx

zyx

zyx

zzzz

yyyy

xxxx


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Geometry Matrix

To see how the co-efficients can depend on 4 constraints, recall that aparametric cubic curve is defined by

Rewrite the co-efficient matrix as where M is a 4x4 basis matrix,

G is a 4-element matrix ofgeometric contraints, called the geometry matrix.

The geometric contraints are just the conditions, such as endpoints, ortangent vectors, that define the curve.

Gxrefers to the column vector of just the x components; Gyand Gz aresimilarly defined

G or M, or both G and M, differ for each type of curve.

Cu )p(uM.GC


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Geometry Matrix

The elements ofG and M are constants so the product

G.M.u is just three cubic polynomials in u.

Expanding:

4

3

2

1

44434241

34333231

24232221

14131211

231

G

G

G

G

mmmm

mmmm

mmmm

mmmm

uuuuzuyuxup

zyx

zyx

zyx

zyx

ggg

ggg

ggg

ggg

mmmm

mmmm

mmmm

mmmm

uuu

444

333

222

111

44434241

34333231

24232221

14131211

231


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Blending Functions

We can read this equation in the following way:

The point p(u) is a weighted sum of the columns of the

geometry matrix G, each of which represents a point or a

vector in 3-space

Multiplying out justx(u) gives:

x

x

x

x

gmummumu

gmummumu

gmummumu

gmummumuux

4443424

2

14

3

3433323

2

13

3

24232222

123

1413121

2

11

3

)(

)(

)(

)()(

Blending functions

Blending Functions


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Blending Functions

This emphasizes that the curve is a weighted sum of the

elements of the geometry matrix.

The weights are each cubic polynomials of the parameter u,and are called the blending functions.

The blending functions B are given by

This is similar to piecewise linear approximation, for whichonly two geometric constraints (i.e. the endpoints of the

line) are needed.

So each curve segment is a straight line defined by theendpoints G1 and G2 :

Mu

Linear Interpolation (straight line)


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Linear Interpolation (straight line)

We can represent its equation in three ways

1. As weight average of control pointsX (u)=(1-u) g1x+ u g2x

2. As polynomial in t

X (u)=(g2x-g1x) u + g1x

3. As matrix form

G1

G2

101

11]gg[)( 1x2x

uux

B0(t) p0 +B1(t)p1 Blending function

When t=0 p0 , t=1p1 , t=.5midpoint

Curve is based at p0 & a vector(p1-p0)is added

which is scaled by t

Geometry Matrix, Geometry Basis, Polynomial

Basis


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The key to defining a parametric cubic curvetherefore lies in the basis matrix M.

Depending on the nature of this matrix, specific

forms of curves may be created.

HERMITE CURVE

BEZIER CURVE

UNIFORM NONRATIONAL B-SPLINE NONUNIFORM, NONRATIONAL B-SPLINE

OTHER SPLINE FORMS


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Hermite Curves

The Hermite form of a cubic polynomial curve

segment is determined by constraints on the

endpoints P1 and P4, and tangent vectors atthe endpoints R1 and R4.

NOTE: LATER P2, P3 WILL BE USED INSTEAD OF TANGENT VECTORS TO DEFINE THE CURVE


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Hermite Curves - Examples

P1P4

R1

R4

Only the direction of R1 varies

Only the magnitude of R1 varies


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Hermite Geometry Vector

The Hermite Geometry vector GH is

GHxis thexcomponent of GH so:

4

1

4

1

R

R

P

P

GH

x

x

x

x

Hx

R

R

P

P

G

4

1

4

1


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Hermite Curves

The Hermite basis matrix, MH, relates the Hermite Geometryvector GH to the polynomial co-efficients.

Therefore:

where

HxHxxxx GMTdtctbtatx 23

)(

123

tttT


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Hermite Curves

The constraints on x(0) and x(1) are found by directsubstitution into the previous equation:

xHxH

xHxH

PGMx

PGMx

4

1

1111)1(

1000)0(


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Hermite Curves

The tangent vector constraints on x(0) and x(1) are foundby differentiation, i.e:

So:

and

HxH GMtttx 0123)('2

HxHx GMRx 0100)0(' 1

HxHx GMRx 0123)1(' 4


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Hermite Curves The four constraints can be written in matrix form as:

The only way that this equation can be satisfied is if MH is

the inverse of the given 4x4 matrix, so:

HxHHx

x

x

x

x

GMG

R

R

P

P

0123

0100

1111

1000

4

1

4

1

0001

01001233

1122

0123

01001111

10001

HM

Matrix Inverse


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Matrix Inverse

H i Bl di F i


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Hermite Blending Functions We know that:

The Hermite blending functions BH are given by , since these

weight the geometry vector GH.

Therefore:

HH GMTtp )(

HMT

HHHH GBGMTtp .)(

4

23

1

23

423

1

23

)(

)2(

32

132

Rtt

Rttt

Ptt

Ptt


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Hermite Curves - Blending Fuctions

P1 P4

R1

R4

t

f(t)

1

1

Labels show

which geometry

element isweighted.

T H i j i d 4


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Two Hermite curves joined at p4

p1

p4p7

Y(t)

t

0,

7

4

7

4

4

1

4

1

withk

R

kR

P

P

and

R

R

P

P

x

x

x

x

Both curves share a

common end point with

G1 continuity


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Bzier Curves

The drawback of the Hermite form is the need to explicitly specify the

tangent vectors.

The Bzier form of the cubic polynomial curve segment indirectly specifiesthe endpoint tangent vector.

Such curves are constrained by their endpoints: P1 and P4, and also byintermediate points that are not on the curve: P2 and P3.

The starting and ending tangent vectors are determined by the vectorsP1P2 and P3P4 and are related to the Hermite R1 and R4 by:

)(3)1('

)(3)0('

344

121

PPpR

PPpR


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Examples of some Bzier Curves

B i C


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Bzier Curves

The reason for using the constant 3 is apparent from the following:

Consider the Bezier curve defined by the 4 equally spaced points:(0,0), (0,1), (0,2), (0,3).

It's obvious that this curve has the definition:

Therefore:

Now we can see that if velocity is to be constant everywhere onthe line:

)()( 141 PPtPtp

14)(' PPtp

)(3)0(' 12141 PPPPpR

)(3)1(' 34144 PPPPpR

P1 P2 P3 P4


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Bzier Geometry Vector and Change of Basis

The Bzier geometry vector is:

A change of basis matrix MHB defines the relationship

between the Hermite geometry vector GH and the Bzier

geometry vector GB as follows:

4

3

2

1

P

P

P

P

GB

BHBH GM

P

PP

P

R

RP

P

G

4

3

2

1

4

1

4

1

3300

00331000

0001

)(3)0(' 12141 PPPPpR

)(3)1('34144

PPPPpR

Bzier Basis Matrix


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Bzier Basis Matrix

To find the Bezier basis matrix, MB, consider:

Therefore, we simply calculate:

BB

BHBH

BHBH

HH

GMT

GMMT

GMMT

GMTtp

)(

)(

0001

0033

0363

1331

HBHB MMM

B t i P l i l


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Bernstein Polynomials

We now have:

The four weights are known as the Bernstein Polynomials

BB GMTtp )(

4

3

32

2

2

1

3

)1(3

)1(3

)1(

Pt

Ptt

Ptt

Pt


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Bernstein Polynomials

P1

P2 P3

P4

Labels show

which geometry

element is

weighted.

G l B t i F f B i C


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General Bernstein Form for Bzier Curves

The Bezier curve p(t) based on the (L+1) points P0,P1,,PL isgiven by:

where are the Bernstein polynomials, and the k-

th Bernstein polynomial is defined as:

and

)()p( 0 tBPt

L

kk

L

k

)(tBL

k

kkLL

k ttk

LtB

)1()( kL

kLk

L

k

L

for

)!(!

!

J i i S t


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Joining Segments

Consider the following two Bezier curve segments, joinedat P4:

P1

P2

P3

P4

P5P

6

P7

Points P3, P4 and P5 are collinear

Curve Continuity


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Curve Continuity

G1 continuity is provided at the endpoint when

i.e. the 3 points P3, P4 and P5 must be distinct and collinear.

In the more restrictive case when k=1, there is C1 continuity in

adition to G1 continuity.

0),( 5443 kPPkPP

Convex Hull Property


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The Bernstein blending polynomials are everywhere non-

negative.

In addition, their sum is everywhere unity (i.e. 1).

Thus, each curve segment, which is just the sum of four control

points weighted by the polynomials, is completely containedwithin the convex hull of the four control points.

Convex Hull

C H ll P t


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The convex hull for 2D curves is the convex polygonformed by the 4 control points (e.g. like a rubber bandaround them)

For 3D curves, the convex hull is the convex polyhedron

formed by the control points (e.g. like cling-filmstretched around them.)

The convex hull property holds for all cubics defined byweighted sums of control points if the blendingfunctions are nonnegative and sum to one.


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One advantageous result of the fact that the blending

polynomials sum to 1, is that the value of the fourth

polynomial can be found by subtracting the first three

from 1. The convex hull property is useful for clipping and

collision detection, where we can perform tests on the

convex hull of a curve before having to perform

expensive intersection tests.

Curves&Surfaces Lec Ppt

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