Introduction to CAD/CAM - home.iitk.ac.inhome.iitk.ac.in/~jrkumar/download/ME761_Lecture-6 CAD.pdf · computers. CAD/CAM in addition to its particular emphasis on the use of computer

1

Introduction to CAD/CAM

IITK ME 761A Dr. J. Ramkumar

2

• CAD can be defined as the use of computer systems to perform

certain functions in the design process.

• CAM is the use of computer systems to plan, manage and control

the operations of manufacturing plant through either direct or

indirect computer interface with the plant’s production resources.


3

From CAM definition, the application of CAM falls into

two broad categories:

1. Computer monitoring and control .

Computer ProcessProcess data

Control signals

Computer ProcessProcess data


4

2. Manufacturing support application .

Control signalsComputer Mfg

operations

Process data


5

The Product Cycle and CAD/CAM

In order to establish the scope and definition of CAD/CAM

in an engineering environment and identify existing and

future related tools, a study of a typical product cycle is

necessary. The following Figure shows a flowchart of such a

cycle.


6

The Manufacturing Process

The Design Process

Synthesis Analysis The CAD Process

The CAM Process

Design

needs

Design definitions,

specifications, and

requirements

Collecting relevant

design information

and feasibility study

Design conceptualization

Design modeling

and simulationDesign

analysis

Design

optimization

Design

evaluation

Design documentation and

communication

Process planning

Order materials

Design and

procurement of new

tools

Production planning

NC, CNC, DNC

programming

ProductionQuality control Packaging

Marketing

Shipping

Typical Product Life Cycle


7

• The product begins with a need which is identified based on customers'

and markets' demands.

• The product goes through two main processes from the idea

conceptualization to the finished product:

1. The design process.

2. The manufacturing process.

The main sub-processes that constitute the design process are:

1. Synthesis.

2. Analysis.


8

Implementation of a Typical CAD Process on a

CAD/CAM system

Delineation of

geometric model

Definition translator

Geometric model

Design and

Analysis algorithms

Drafting and

detailing

Documentation

To CAM Process

Interface algorithms

Design changes


9

CAD Tools Required to Support the Design Process

Design phase Required CAD tools

Design conceptualization Geometric modeling techniques; Graphics aids;

manipulations; and visualization

Design modeling and simulation Same as above; animation; assemblies; special

modeling packages.

Design analysis Analysis packages; customized programs and

packages.

Design optimization Customized applications; structural optimization.

Design evaluation Dimensioning; tolerances; BOM; NC.

Design communication and documentation Drafting and detailing…


10

Implementation of a Typical CAM Process on a CAD/CAM

system

Geometric model

Interface algorithms

Process planning

Inspection

Assembly

Packaging

To shipping and marketing

NC programs


11

Manufacturing phase Required CAM tools

Process planning CAPP techniques; cost analysis;

material and tooling specification.

Part programming NC programming

Inspection CAQ; and Inspection software

Assembly Robotics simulation and

programming

CAM Tools Required to Support the Design Process


12

Definitions of CAD Tools Based on Their Constituents

Computer graphics

concepts

Design tools

Geometric modeling

CAD

tools


13

Definition of CAD Tools Based on Their Implementation in a Design

Environment

Design tools + Computer

Hardware

(control unit; display terminals;

I/O devices

Software (graphics;

modeling; applications

programs

= CAD tools


14

Definitions of CAM Tools Based on Their Constituents

Networking

concepts

Mfg toolsCAD

CAM

tools


15

Definition of CAM Tools Based on Their Implementation in a

Manufacturing Environment

Mfg tools + Computer

Hardware

(control unit; display terminals;

I/O devices

Software (CAD; NC;

MRP; CAPP…)= CAM tools

Networking


16

Definitions of CAD/CAM Tools Based on Their

Constituents

Mfg tools

Networking

Design tools

Geometric

modeling

Computer

graphics

concepts

CAD/CAM

tools


17

Definition of CAD/CAM Tools Based on Their Implementation in an

Engineering Environment

Design and

Mfg tools

Hardware

Software = CAD/CAM tools

Networking

+ Computer


18

Geometric modeling

of conceptual design

Is design evaluation

Possible with available

Standard software?

Design testing

And evaluation

Is final design

Applicable?

Drafting

Documentation

Process planning

Are there manufacturing

discrepancies in CAD

databases?

NC programming

Machining

Inspection

Assembly

Develop customized

programs and packages

No

Yes

Yes

Yes

Geometric modeling and graphics package

Design

package

Programming

package

No

No

CAPP package

NC

package

Inspection

And Robotics

package

Typical Utilization of CAD/CAM Systems in an Industrial Environment


19

Automation and CAD/CAM

Automation can be defined as the technology concerned

with the application of complex mechanical, electronic,

and computer-based systems in the operation and control

of manufacturing systems.


20

Types of Manufacturing Systems

1. Continuous-flow processes. Continuous dedicated production of large amount of bulk

product. Continuous manufacturing is represented by chemicals, plastics, petroleum, and

food industries.

2. Mass production of discrete products. Dedicated production of large quantities of one

product (with perhaps limited model variations). Examples include automobiles,

appliances and engine blocks.

3. Batch production. Production of medium lot sizes of the same product. The lot may be

produced once or repeated periodically. Examples: books, clothing and certain industrial

machinery.

4. Job-shop production. Production of low quantities, often one of a kind, of specialized

products. The products are often customized and technologically complex. Examples:

prototypes, aircraft, machine tools and other equipment.


21

Production

quantity

Continuous-flow

production

Mass production

Batch production

Job shop

production

Product variety


22

Category Automation achievements

Continuous-flow process •Flow process from beginning to end

•Sensors technology available to measure important process

variables

•Use of sophisticated control and optimization strategies

•Fully computer automated lines

Mass production of discrete products •Automated transfer machines

•Dial indexing machines

•Partially and fully automated assembly lines

•Industrial robots for spot welding, part handling, machine

loading, spray painting, etc.

•Automated material handling systems

•Computer production monitoring

Batch production •Numerical control (NC), direct numerical control (DNC),

computer numerical control (CNC).

•Adaptive control machining

•Robots for arc welding, parts handling, etc.

•CIM systems.

Job shop production •Numerical control, computer numerical control


23

Most of the automated production systems implemented today make use of

computers. CAD/CAM in addition to its particular emphasis on the use of

computer technology, is also distinguished by the fact that it includes not only the

manufacturing operations but also the design and planning functions that precede

manufacturing.

To emphasize the differences in scope between automation and CAD/CAM,

consider the following mathematical model:

Computer Technology in Automation


24

321 TBTBQTTTlc ++=

BQ

T

Q

TT

Tlc

321 cycle life its during

product ofunit each on spent timeaverage The

++=

=

.product different each for once edaccomplish are that activities

other theallfor andproduct thedesigningfor required timeThe3 =T

.production ofbatch each

for up setting and planning with associated timeThe2 =T

cycle. lifeproduct t the throughouproduced batches ofnumber The =B

product. ofunit one produce torequired timeThe1 =T

batch.each in produced units ofnumber The=Q

cycle lifeproduct theduring time total=clTT


25

Automation technology is concerned with reducing

the with emphasis on the unit production time

CAD/CAM concerned with reducing all three terms, but is perhaps

focused on terms. The emphasis in CAD/CAM includes

the design and planning function of the product life cycle.

1T

32 &TT

The most important term in mass production and batch

production

become very important in job shop

manufacturing.

1T

32 &TT

21 &TT


26

Advantages of CAD/CAM systems

• Greater flexibility.

• Reduced lead times.

• Reduced inventories.

• Increased Productivity.

• Improved customer service.

• Improved quality.

• Improved communications with suppliers.

• Better product design.

• Greater manufacturing control.

• Supported integration.

• Reduced costs.

• Increased utilization.

• Reduction of machine tools.

• Less floor space.


27

Introduction to Geometric Transformation

• Computer graphics plays an important role in the product

development process by generating, presenting, and

manipulating geometric models of objects.

• During the product development process, for proper

understanding of designs, it is necessary not only to generate

geometric models of objects but also to perform such

manipulations on these objects as rotation, translation and

scaling.


28

Introduction to Geometric transformation

• Essentially, computer graphics is concerned with generating, presenting and

manipulating models of an object and its different views using computer hardware,

software and graphic devices.

• Usually the numerical data generated by a computer at very high speeds is hard to

interpret unless one represents the data in graphic format and it is even better if the

graphic can be manipulated to be viewed from different sides, enlarged or reduced

in size.

• Geometric transformation is one of the basic techniques that is used to accomplish

these graphic functions involving scale change, translation to another location or

rotating it by a certain angle to get a better view of it.


29

2D Transformations


30

Rotation About the OriginTo rotate a line or polygon, we must rotate each of

its vertices.

To rotate point (x1,y1) to point (x2,y2) we observe:

From the illustration we know that:

sin (A + B) = y2/r cos (A + B) = x2/r

sin A = y1/r cos A = x1/r

x-axis

(x1,y1)

(x2,y2)

A

Br

(0,0)

y-axis


31

Rotation About the Origin

From the double angle formulas: sin (A + B) = sinAcosB + cosAsinB

cos (A + B)= cosAcosB - sinAsinB

Substituting: y2/r = (y1/r)cosB + (x1/r)sinB

Therefore: y2 = y1cosB + x1sinB

We have x2 = x1cosB - y1sinB

y2 = x1sinB + y1cosB

P2 = R P1.

(x1)

(y1)

(cosB -sinB)

(sinB cosB)

(x2)

(y2) =


32

TranslationsMoving an object is called a translation. We translate a point by adding to the x and y coordinates,

respectively, the amount the point should be shifted in the x and y directions. We translate an

object by translating each vertex in the object.

P2 = P1 + T

T = ( tx )

( ty )

P1 = ( x1 )

( y1 )

P2 = (x1 + tx)

(y1 + ty)

Ty

Tx


33

Scaling

Changing the size of an object is called a scale. We scale an object by scaling the x and y coordinates

of each vertex in the object.

P2 = S P1

.

S = (sx 0)

(0 sy)P1 = ( x1 )

( y1 )

P2 = (sxx1)

(syy1)


34

Homogeneous CoordinatesAlthough the formulas we have shown are usually the most efficient way to implement programs to

do scales, rotations and translations, it is easier to use matrix transformations to represent and

manipulate them.

In order to represent a translation as a matrix operation we use 3 x 3 matrices and pad our points to

become 1 x 3 matrices.

cos ø -sin ø 0

Rø = sin ø cos ø 0

0 0 1

Sx 0 0

S = 0 Sy 0

0 0 1

1 0 Tx

T = 0 1 Ty

0 0 1

Point P =

(x)

(y)

(1)


35

Composite Transformations - Scaling

Given our three basic transformations we can create other transformations.

Scaling with a fixed point

A problem with the scale transformation is that it also moves the object being scaled.

Scale a line between (2, 1) (4,1) to twice its length.

(2 0 0) =

(0 1 0)

(0 0 1)

(2 0 0) =

(0 1 0)

(0 0 1)

0 1 2 3 4 5 6 7 8 9 10

Before

After

(2)

(1)

(1)

(4)

(1)

(1)

(4)

(1)

(1)

(8)

(1)

(1)


36

Composite Transforms - Scaling (cont.)

If we scale a line between (0, 1) (2,1) to twice its length, the left-hand endpoint does not move.

(2 0 0) =

(0 1 0)

(0 0 1)

(2 0 0) =

(0 1 0)

(0 0 1)

(0,0) is known as a fixed point for the basic scaling transformation. We can used composite

transformations to create a scale transformation with different fixed points.

0 1 2 3 4 5 6 7 8 9 10

Before

After

(0)

(1)

(1)

(0)

(1)

(1)

(2)

(1)

(1)

(0)

(1)

(1)


37

Fixed Point Scaling

Scale by 2 with fixed point = (2,1)

• Translate the point (2,1) to the origin

• Scale by 2

• Translate origin to point (2,1)

(1 0 2) (2 0 0) (1 0 -2) = (2 0 -2)

(0 1 1) (0 1 0) (0 1 -1) (0 1 0)

(0 0 1) (0 0 1) (0 0 1) (0 0 1)

(2 0 -2) =

(0 1 0)

(0 0 1)

(2 0 -2) =

(0 1 0)

(0 0 1) 0 1 2 3 4 5 6 7 8 9 10

Before

After

(2)

(1)

(1)

(2)

(1)

(1)

(4)

(1)

(1)

(6)

(1)

(1)


38

More Fixed Point ScalingScale by 2 with fixed point = (3,1)

• Translate the point (3,1) to the origin

• Scale by 2

• Translate origin to point (3,1)

(1 0 3) (2 0 0) (1 0 -3) = (2 0 -3)

(0 1 1) (0 1 0) (0 1 -1) (0 1 0)

(0 0 1) (0 0 1) (0 0 1) (0 0 1)

(2 0 -3) =

(0 1 0)

(0 0 1)

(2 0 -3) =

(0 1 0)

(0 0 1)

0 1 2 3 4 5 6 7 8 9 10

Before

After

(2)

(1)

(1)

(1)

(1)

(1)

(4)

(1)

(1)

(5)

(1)

(1)


39

Rotation about a Fixed Point

Rotation of ø Degrees About Point (x,y)

• Translate (x,y) to origin

• Rotate

• Translate origin to (x,y)

(1 0 x) (cosø -sinø 0)(1 0 -x) (cosø -sinø -xcosø + ysinø + x )

(0 1 y) (sinø cosø 0)(0 1 -y) = (sinø cosø -xsinø - ycosø + y )

(0 0 1) (0 0 1)(0 0 1) (0 0 1 )

You rotate the box by rotating each vertex.

(x,y) (x,y)


40

Shears

Original Data y Shear x Shear

1 0 0 1 b 0

a 1 0 0 1 0

0 0 1 0 0 1

GRAPHICS --> x shear --> GRAPHICS


41

Reflections

Reflection about the y-axis Reflection about the x-axis

-1 0 0 1 0 0

0 1 0 0 -1 0

0 0 1 0 0 1


42

More Reflections

Reflection about the origin Reflection about the line y=x

-1 0 0 0 1 0

0 -1 0 1 0 0

0 0 1 0 0 1


IITK ME 761A Dr. J. Ramkumar43

Concatenation of Transformations

• Concatenating

– affine transformations by multiplying together

– sequences of the basic transformations

➔ define an arbitrary transformation directly

– ex) three successive transformations

( )( )( ) CBApApBCq ==

A B Cp q

CBAM =

Mpq = Mp q

CBAp1 = Ap

p2 = Bp1

q = Cp2

q = CBp1

q = CBAp

p1 p2

IITK ME 761A Dr. J. Ramkumar 44

Matrix Concatenation Properties

• Associative properties

• Transformation is not commutative

– Order of transformation may affect transformation position

321321321 M)MM()MM(MMMM ==


Two-dimensional composite transformation (1)

• Composite transformation

– A sequence of transformations

– Calculate composite transformation matrix rather than applying individual transformations

• Composite two-dimensional translations

– Apply two successive translations, T1 and T2

– Composite transformation matrix in coordinate form

PM'P

PMM'P 12

=

=

PTTPTTP

ttTT

ttTT

yx

yx

==

=

=

)()('

),(

),(

1212

222

111

)tt,tt(T)t,t(T)t,t(T y2y1x2x1y1x1y2x2 ++=



• Composite two-dimensional rotations

– Two successive rotations, R1 and R2 into a point P

– Multiply two rotation matrices to get composite transformation matrix

• Composite two-dimensional scaling

P)}(R)(R{'P

}P)(R{)(R'P

21

21

=

=

P)ss,ss(S'P

)ss,ss(S)s,s(S)s,s(S

y2y1x2x1

y2y1x2x1y2x2y1x1

=

=

P)(R'P

)(R)(R)(R

21

2121

+=

+=



• General two-dimensional Pivot-point rotation

– Graphics package provide only origin rotation

– Perform a translate-rotate-translate sequence• Translate the object to move pivot-point position to origin

• Rotate the object

• Translate the object back to the original position

– Composite matrix in coordinates form ),y,x(R)y,x(T)(R)y,x(T rrrrrr =−−



• Example of pivot-point rotation


Pivot-Point Rotation

−−

+−−

=

−

−

−

100

sin)cos1(cossin

sin)cos1(sincos

100

10

01

100

0cossin

0sincos

100

10

01

rr

rr

r

r

r

r

xy

yx

y

x

y

x

( ) ( ) ( ) ( ) ,,,, rrrrrr yxRyxTRyxT =−−

Translate Rotate Translate

(xr,yr) (xr,yr) (xr,yr)(xr,yr)


General Fixed-Point Scaling

−

−

=

−

−

100

)1(0

)1(0

100

10

01

100

00

00

100

10

01

yfy

xfx

f

fx

f

f

sys

sxs

y

x

s

s

y

x

y

Translate Scale Translate

(xr,yr) (xr,yr) (xr,yr) (xr,yr)

( ) ( ) ( ) ( )yxffffyxff ssyxSyxTssSyxT ,,,, ,, =−−



• General two-dimensional scaling directions

– Perform a rotate-scaling-rotate sequence

– Composite matrix in coordinates form

+−

−+

=

−

100

0cosssinssincos)ss(

0sincos)ss(sinscoss

)(R)s,s(S)(R

22

2112

122

22

1

211

s1

s2


General Scaling Directions

• Converted to a parallelogram

+−

−+

=−

100

0cossinsincos)(

0sincos)(sincos

)(),()( 2

2

2

112

12

2

2

2

1

21

1

ssss

ssss

RssSR

x

y

(0,0)

(3/2,1/2)

(1/2,3/2) (2,2)

x

y

(0,0) (1,0)

(0,1) (1,1)

x

y s2

s1

Scale

IITK ME 761A Dr. J. Ramkumar53

General Rotation (1/2)

• Three successive rotations about the three axes

rotation of a cube about the z axis rotation of a cube about the y axis

rotation of a cube about the x axis

?


General Rotation (2/2)

zyxRRR=R

−

−

−

1000

0100

00cossin

00sincos

1000

0cos0sin

0010

0sin0cos

1000

0cossin0

0sincos0

0001

Examples




3D Transformation

◼ A 3D point (x,y,z) – x,y, and z coordinates

◼ We will still use column vectors to represent points

◼ Homogeneous coordinates of a 3D point

(x,y,z,1)

◼ Transformation will be performed using 4x4 matrix

x

y

z

58

Right-handed Coordinate System

x

y

z

;zyx = ;xzy = yxz =

Left hand coordinate system

Not used in this class and

Not in OpenGL

59/94

3D Transformation

◼ Very similar to 2D transformation

◼ Translation transformation

=

11000

100

010

001

1

z

y

x

t

t

t

z

y

x

x

y

x

Homogenous coordinates

60

3D Transformation

◼ Very similar to 2D transformation

◼ Scaling transformation

=

11000

000

000

000

1

z

y

x

s

s

s

z

y

x

z

y

x

Homogenous coordinates

61

3D Transformation

◼ 3D rotation is done around a rotation axis

◼ Fundamental rotations – rotate about x, y, or z axes

◼ Counter-clockwise rotation is referred to as positive rotation

(when you look down negative axis)

x

y

z

+

62

3D Transformation

◼ Rotation about z – similar to 2D rotation

x

y

z

+

−

=

11000

0100

00cossin

00sincos

1

z

y

x

z

y

x

zz

yxy

yxx

=

+=

−=

'

cossin'

sincos'

63

3D Transformation

◼ Rotation about y: z -> y, y -> x, x->z

−=

11000

0cos0sin

0010

0sin0cos

1

z

y

x

z

y

x

yy

xzx

xzz

=

+=

−=

'

cossin'

sincos'

y

z

x

x

y

z

64

3D Transformation

◼ Rotation about x (z -> x, y -> z, x->y)

x

y

z

−=

11000

0cossin0

0sincos0

0001

1

z

y

x

z

y

x

xx

zyz

zyy

=

+=

−=

'

cossin'

sincos'

z

x

y

65

Inverse of 3D Transformations

◼ Invert the transformation

◼ In general, X= AX’->x’=A-1X

=

11000

100

010

001

1

z

y

x

t

t

t

z

y

x

x

y

x

−

−

−

=

1

'

'

'

1000

100

010

001

1

z

y

x

t

t

t

z

y

x

x

y

x

−

=

11000

0100

00cossin

00sincos

1

z

y

x

z

y

x

−=

1

'

'

'

1000

0100

00cossin

00sincos

1

z

y

x

z

y

x

)(zR)( −zR

T(tx,ty,tz)T(-tx,-ty,-tz)

3D Rotation about Arbitrary Axes

◼ Rotate p about the by the angle

x

y

zr

),,( zyxp =

r

67

3-D Rotation

◼ General rotations in 3-D require rotating about an arbitrary axis of rotation

◼ Deriving the rotation matrix for such a rotation directly is a good exercise in linear algebra

◼ The general rotation matrix is a combination of coordinate-axis rotations and translations!

68



x

y

zr

),,( zyxp =

r

69

3-D Rotation

◼ General rotations in 3-D require rotating about an arbitrary axis of rotation

◼ Deriving the rotation matrix for such a rotation directly is a good exercise in linear algebra

◼ Standard approach: express general rotation as composition of canonical rotations

◆Rotations about x, y, z

70

Composing Canonical Rotations

◼ Goal: rotate about arbitrary vector r by θ

◆ Idea: we know how to rotate about x,y,z

⧫ Set up a transformation that superimposes rotation axis onto

one coordinate axis

⧫ Rotate about the coordinate axis

⧫ Translate and rotate object back via inverse of the

transformation matrix

71

Composing Canonical Rotations

◼ Goal: rotate about arbitrary vector r by θ

◆ Idea: we know how to rotate about x,y,z

⧫ So, rotate about z by - until r lies in the xz plane

⧫ Then rotate about y by -β until r coincides with +z

⧫ Then rotate about z by θ

⧫ Then reverse the rotation about y (by β )

⧫ Then reverse the rotation about z (by )

72



x

y

zr

),,( zyxp =

r

73


◼ Translate so that rotation axis passes through the origin

x

y

zr

TRRRRRT zyzyz )()()()()(1 −−−

74


◼ Rotation by about z-axis

x

y

z

−

r

TRRRRRT zyzyy )()()()()(1 −−−

75


◼ Rotation by about y-axis

x

y

z

−

r

x


76



x

y

z

r

x


77


◼ Rotation by about y-axis

x

y

z

r

x


78



x

y

z

r


79


◼ Translate the object back to original point

x

y

z

r


80


◼ Final transformation matrix for rotating about an arbitrary

axis

x

y

z

r


81



axis

x

y

z

r


82



axis


=

1000

0

0

0

333231

232221

131211

rrr

rrr

rrr



axis


A 3 by 3 Rotation matrix—orthogonal matrix

=

1000

0

0

0

333231

232221

131211

rrr

rrr

rrr

Rotation Matrices

◼ Orthonormal matrix:

◆orthogonal (columns/rows linearly independent)

◆normalized (columns/rows length of 1)

85

Rotation Matrices




◼ The inverse of an orthogonal matrix is just its transpose:

86

T

rrr

rrr

rrr

rrr

rrr

rrr

=

−

333231

232221

131211

1

333231

232221

131211

Rotation Matrices




◼ The inverse of an orthogonal matrix is just its transpose:

87

T

rrr

rrr

rrr

rrr

rrr

rrr

=

−

333231

232221

131211

1

333231

232221

131211

93IITK ME 761A Dr. J. Ramkumar

Examples




Examples














Splines and Bezier Curves


• |Representing Curves

– A number of small line-segments joined

• Interpolation

• Parametric equations

• Types of curve we study

– Natural Cubic Splines and Bezier Curves

• Derivation

• Implementation


“Computers can’t draw curves.”

The more points/line segments

that are used, the smoother the

curve.


Why have curves ?

• Representation of “irregular surfaces”

• Example: Auto industry (car body design)

– Artist’s representation

– Clay / wood models

– Digitizing

– Surface modeling (“body in white”)

– Scaling and smoothening

– Tool and die Manufacturing


Curve representation

• Problem: How to represent a curve easily and efficiently

• “Brute force and ignorance” approaches:

• storing a curve as many small straight line segments– doesn’t work well when scaled

– inconvenient to have to specify so many points

– need lots of points to make the curve look smooth

• working out the equation that represents the curve– difficult for complex curves

– moving an individual point requires re-calculation of the entire curve


Solution - Interpolation

•Define a small number of points

•Use a technique called “interpolation” toinvent the extra points for us.

•Join the points with a series of (short)

straight lines


The need for smoothness

• So far, mostly polygons

• Can approximate any geometry, but

– Only approximate

– Need lots of polygons to hide discontinuities

• Storage problems

– Math problems

• Normal direction

• Texture coordinates

– Not very convenient as modeling tool

• Gets even worse in animation

– Almost always need smooth motion


Geometric modeling

• Representing world geometry in the computer

– Discrete vs. continuous again

• At least some part has to be done by a human

– Will see some automatic methods soon

• Reality: humans are discrete in their actions

• Specify a few points, have computer create something which

“makes sense”

– Examples: goes through points, goes “near” points


Requirements

• Want mathematical smoothness

– Some number of continuous derivatives of P

• Local control

– Local data changes have local effect

• Continuous with respect to the data

– No wiggling if data changes slightly

• Low computational effort


A solution

• Use SEVERAL polynomials

• Complete curve consists of several pieces

• All pieces are of low order

– Third order is the most common

• Pieces join smoothly

• This is the idea of spline curves

– or just “splines”


Parametric Equations - linear

x0 ,y0

x1,y1

dy

dx

(13,8)

(2,3)

P0

P1

(7.5,4)

Walk from P0 to P1 at a constant

speed.

Start from P0 at t=0

Arrive at P1 at t=1

dx = x1 - x0

dy = y1 - y0

Where are you at a general time t?

x(t) = x0 + t.dx

y(t) = y0 + t.dyEquation(s) of a straight line as afunction of an arbitrary parametert.


Why use parametric equations?

• One, two, three or n-dimensional representation is

possible

• Can handle “infinite slope” of tangents

• Can represent multi-valued functions


• Natural Cubic Spline

• Bezier Curves


Splines

• Define the “knots”

– (x0,y0) - (x3,y3)

• Calculate all other points

spline

knots


Continuity

• Parametric continuity Cx

– Only P is continuous: C0

• Positional continuity

– P and first derivative dP/du are continuous: C1

• Tangential continuity

– P + first + second: C2

• Curvature continuity

• Geometric continuity Gx

– Only directions have to match


)0(G

)1(G

(a)

(b)

(c)

continuity order th0

continuity order st1

continuity order nd2)2(G

)0(C

)1(C

)2(C

Order of continuity


Wiggling effect

• Example:

– Four data points

– Third degree polynomial

– Might look something like:

• This the ONLY third degree polynomial which fits the data

• Wiggling gets much worse with higher degree


Polynomial parametric equations

• To represent a straight line -

linear parametric equation (i.e.

one where the highest power of t

was 1).

• For curves, we need polynomial

equations. Why? Because the

graphs of polynomial equations

wiggle!


Natural Cubic Spline

• spline, n. 1 A long narrow and relatively thin piece or strip of wood, metal, etc. 2 A flexible strip of wood or rubber used by draftsmen in laying out broad sweeping curves, as in railroad work.

• A natural spline defines the curve that minimizes the potential energy of an idealized elastic strip.


• A natural cubic spline defines a curve, in which the points P(u) that define each segment of the spline are represented as a cubic P(u) = a0 + a1u + a2u

2 + a3u3

• Where u is between 0 and 1 and a0, a1, a2 and a3 are (as

yet) undetermined parameters.

• We assume that the positions of n+1 control points Pk,

where k = 0, 1,…, n are given, and the 1st and 2nd

derivatives of P(u) are continuous at each interior control

point.


• Undesirable wiggles and oscillations for higher orders

• The lowest order polynomials to satisfy the following conditions

Why cubic?

Caution! Heavy math ahead!


minu

maxu

u

0u

1u

2u

nu1−nu

)(uQ

A parametrically defined curve

=

)(

)(

)(

)(

uz

uy

ux

uQ

sequenceknot uniform then , If

vectorknot

sequenceknot

iu

uuuu

uuuuuu

i

nn

nn

=

==

−

−

),,,,( 110

max110min

Parameterization


0u 1u 2u 3u iu 1+iu mu

)(uY)(1 uY

)(0 uY)(2 uY

)(uYm

i 1+i

10

,,2,1,0),()(Y

),(

1

i

1

=−

−=

+

+

u

niuu

uuYu

uuuuY

ii

i

ii

non-uniform

“Re-parameterization”


0V

1V

2V

3V

4V

5V6V

Interpolation

0V

1V 2V

3V 4V

5V

6V

Approximation

Interpolation versus Approximation


0y1y

2y

3y

4y

5y6y

)(0 uY )(uY

)(5 uY

u

6=u

0=u

++=

=

+++=

=

+

+

1

)1(

)1(

1

≡32)1(

≡)0(

≡)1(

≡)0(

iiiii

iii

iiiiii

iii

DdcbY

DbY

ydcbaY

yaY

5,...,010

)( 32

=

+++=

iu

uducubau iiii

Yi

11

11

)(2

2)(3

++

++

++−=

−−−=

=

=

iiiii

iiiii

ii

ii

DDyyd

DDyyc

Db

ya

1 2 3 4 5

Hermite Interpolation


Hermite Spline• Say the user provides

• A cubic spline has degree 3, and is of the form:

– For some constants a, b, c and d derived from the control points, but how?

• We have constraints:

– The curve must pass through x0 when t=0

– The derivative must be x’0 when t=0

– The curve must pass through x1 when t=1

– The derivative must be x’1 when t=1

dctbtatx +++= 23

1010 ,,, xxxx


10

))(1() 11

−−+− +−

avrage hted weig

(y (2) i iii yyy

directly specifying (3)

1,, +ii yyy and through parabola a fitting e.g.

llyautomatica D Computing

(1)

-1i

i

1−iy

iy

1+iy1−iP

iP

1+iP

How to specify slope Di


Hermite Spline

A Hermite spline is a curve for which the user provides:

– The endpoints of the curve

– The parametric derivatives of the curve at the endpoints

• The parametric derivatives are dx/dt, dy/dt, dz/dt

– That is enough to define a cubic Hermite spline, more derivatives are required for higher order curves


Hermite Spline

• Solving for the unknowns gives:

• Rearranging gives:

0

0

0101

0101

233

22

xd

xc

xxxxb

xxxxa

=

=

−−−=

+++−=

)2(

)(x

)132(

)32(

23

0

23

1

23

0

23

1

tttx

tt

ttx

ttxx

+−+

−+

+−+

+−=

−

−

−

−

=

10121

0011

1032

00322

3

0101t

t

t

xxxxx

or


Hermite curves in 2D and 3D

• We have defined only 1D splines:

x = f(t:x0,x1,x’0,x’1)

• For higher dimensions, define the control points in higher

dimensions (that is, as vectors)

−

−

−

−

=

10121

0011

1032

00322

3

0101

0101

0101

t

t

t

zzzz

yyyy

xxxx

z

y

x


Basis Functions• A point on a Hermite curve is obtained by multiplying each control

point by some function and summing

• The functions are called basis functions

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

x1

x0

x'1

x'0


0V

1V

2V

3V

4V

5V6V

An interpolating cubic spline

),(

))(),(()(

iii

iii

yxV

uyuxuQ

=

=

Hermite and Cubic Spline Interpolation


(knots) jointsat continuity )2(C

0y1y

2y

iy

1+iy

my)(0 uY )(uY

)(1 uYm−

)(1 uY

mm

ii

ii

ii

ii

yY

yY

miYY

YY

yY

yY

=

=

−=

=

=

=

−

−

−

−

)1(

)0(

11 )0()1(

)0()1(

)0(

)1(

1

00

)2()2(

1

)1()1(

1

1

44 −m

2

need two more Eq’s to determine all coefficients

10)( 32 −+++= miuducubauY iiiii : note

Spline Interpolation

0)1(

0)0(

)2(

1

)2(

0

=

=

−mY

Y

Natural Spline


11

)(34

)(2

2)(3

262

1111

11

11

11

−

−=++

++−=

−−−=

=

=

=+

−++−

++

++

−−

mi

yyDDD

DDyyd

DDyyc

Db

ya

cdc

iiiii

iiiii

iiiii

ii

ii

iii

tion,simplifica and onsubstitutiBy

(4)

0)1(

0)0(

)1(

)0(

)0()1(

)0()1(

)0(

)1(

)2(

1

)2(

0

1

00

)2()2(

1

)1()1(

1

1

=

=

=

=

=

=

=

=

−

−

−

−

−

m

mm

ii

ii

ii

ii

Y

Y

yY

yY

YY

YY

yY

yY

(8)

(7)

(6)

(5)

(4)

(3)

(2)

(1)

11 − mi

32 tdtctbay iiiii +++=

Deriving Natural Cubic Splines

)(32

(8) from Similarly,

)(32

tion,simplifica and onsubstitutiBy

02

(7), From

conditions-end Applying

11

0110

0

−− −=+

−=+

=

mmmm yyDD

yyDD

c

145


−

−

−

−

=

−

−

)(3

)(3

)(3

)(3

21

141

....

141

141

141

12

1

2

02

01

1

0

mm

mm

m yy

yy

yy

yy

D

D

D

Tri-Diagonal Matrix for Natural Cubic Spline


iP

mP0P

2P

1P

)(34 1111 −++− −=++ iiiii yyDDD

−

−

−

−

=

−

−

)(3

)(3

)(3

)(3

411

141

....

141

141

141

114

10

2

02

1

1

0

m

mm

m

m yy

yy

yy

yy

D

D

D

Closed Curves



Examples


149


Exercise

Bezier Curves

• An alternative to splines

• M. Bezier was a French mathematician who worked for the Renault

motor car company.

• He invented his curves to allow his firm’s computers to describe the

shape of car bodies.


? and provide to How 41 R R

1P

4P

1R4R

mation Approxi

and

!!! points control additional two provide , and providing of Instead

32

41

PP

RR

1P

4P

2P3P

Bezier Approximation


Bezier curves

• Typically, cubic polynomials

• Similar to Hermite interpolation

– Special way of specifying end tangents

• Requires special set of coefficients

• Need four points

– Two at the ends of a segment

– Two control tangent vectors


Bezier curves

• Control polygon: control points connected to each other

• Easy to generalize to higher order

– Insert more control points

– Hermite – third order only


De Casteljau algorithm

• Can compute any point on the curve in a few iterations

• No polynomials, pure geometry

• Repeated linear interpolation

– Repeated order of the curve times

• The algorithm can be used as definition of the curve



Third order, u=0.75








De Casteljau Algorithm

3P

1P2P

t

t−1

t−1 t−1

t−1t−1

t

tt

t

0P0P

1P

2P

3P

t−1

t

t−1

t

t−1

t

10)1( tPPt +−

21)1( tPPt +−

32)1( tPPt +−

t−1

t

t−1

t

2

2

10

2 )1(2)1(

Pt

tPtPt

+

−+−

3

2

21

2 )1(2)1(

Pt

tPtPt

+

−+−

t−1

t

3

3

2

2

1

2

0

3

)1(3

)1(3

)1(

Pt

Ptt

tPt

Pt

+

−+

−+

−

ii

i

tti

−

=

−

3

3

0

)1(3


0

1

1

3)1( t−

2)1(3 -tt

3t

)1(3 2 -tt

Now,

1)1(3

33

0

=−

−

=

ii

i

tti

What does it mean ?

10)1(3

)1(3)1(3)1()(

0001

0033

0363

1331

1)(

33

0

4

3

3

2

2

2

1

3

4

3

2

1

23

−

=

+++−=

=

−

−

−

=

−

=

t, Ptti

PtP-ttP-ttPttQ

GMT

P

P

P

P

ttttQ

i

ii

i

bb

3))1(( tt −+



Examples



Bezier Polynomial Function

ini

in ttini

ntB −−

−= )1.(.

)!(!

!)(


P(t)= σ𝑖=0𝑛 𝐶𝑖𝐵𝑖𝑛 (𝑡)

Ai is the control points (xi, yi)

Parametric equations for x(t),y(t)

)(.)(0

tBxtx in

n

i

i=

=

)(.)(0

tByty in

n

i

i=

=


Numerical problem


Question: Generate a Bezier curve using the following control points: (2,0), (4,3),

(5,2), (4,-2), (5,-3), (6,-2)

Solution: Control points =6 ,Therefore n=5

= σ𝑖=05 𝐴𝑖 𝑛 𝐶𝑖 𝑡

𝑖 (1-t) n-i

X(t) =σ𝑖=05 𝑥𝑖 5 𝐶𝑖 𝑡

𝑖 (1-t) 5-i

Y(t)= σ𝑖=05 𝑦𝑖 5 𝐶𝑖 𝑡

𝑖 (1-t) 5-i

P(t)= σ𝑖=05 𝐴𝑖 𝐵𝑖𝑛(𝑡)


Some Bezier Curves


Bezier Curves - properties

• Not all of the control points are on the line

– Some just attract it towards themselves

• Points have “influence” over the course of the line

• “Influence” (attraction) is calculated from a polynomial expression

• (show demo applet)


1P

3P

2P

4P

Convex Hull property


Bezier Curve Properties• The first and last control points are interpolated

• The tangent to the curve at the first control point is along the line joining the first and second control points

• The tangent at the last control point is along the line joining the second last and last control points

• The curve lies entirely within the convex hull of its control points– The Bernstein polynomials (the basis functions) sum to 1 and are everywhere positive

• They can be rendered in many ways– E.g.: Convert to line segments with a subdivision algorithm


Disadvantages

• The degree of the Bezier curve depends on the

number of control points.

• The Bezier curve lacks local control. Changing

the position of one control point affects the entire

curve.


Invariance• Translational invariance means that translating the control points

and then evaluating the curve is the same as evaluating and then translating the curve

• Rotational invariance means that rotating the control points and then evaluating the curve is the same as evaluating and then rotating the curve

• These properties are essential for parametric curves used in graphics

• It is easy to prove that Bezier curves, Hermite curves and everything else we will study are translation and rotation invariant


Longer Curves• A single cubic Bezier or Hermite curve can only capture a small class of

curves

– At most 2 inflection points

• One solution is to raise the degree

– Allows more control, at the expense of more control points and higher degree polynomials

– Control is not local, one control point influences entire curve

• Alternate, most common solution is to join pieces of cubic curve together into piecewise cubic curves

– Total curve can be broken into pieces, each of which is cubic

– Local control: Each control point only influences a limited part of the curve

– Interaction and design is much easier


Piecewise Bezier Curve

“knot”P0,0

P0,1 P0,2

P0,3

P1,0

P1,1

P1,2

P1,3


Continuity• When two curves are joined, we typically want some degree of continuity

across the boundary (the knot)– C0, “C-zero”, point-wise continuous, curves share the same point where they join

– C1, “C-one”, continuous derivatives, curves share the same parametric derivatives where they join

– C2, “C-two”, continuous second derivatives, curves share the same parametric second derivatives where they join

– Higher orders possible

• Question: How do we ensure that two Hermite curves are C1 across a knot?

• Question: How do we ensure that two Bezier curves are C0, or C1, or C2

across a knot?


Achieving Continuity• For Hermite curves, the user specifies the derivatives, so C1 is achieved

simply by sharing points and derivatives across the knot

• For Bezier curves:

– They interpolate their endpoints, so C0 is achieved by sharing control points

– The parametric derivative is a constant multiple of the vector joining the first/last 2 control points

– So C1 is achieved by setting P0,3=P1,0=J, and making P0,2 and J and P1,1 collinear, with J-P0,2=P1,1-J

– C2 comes from further constraints on P0,1 and P1,2


Bezier Continuity

P0,0

P0,1 P0,2

J

P1,1

P1,2

P1,3

Disclaimer: PowerPoint curves are not Bezier curves, they are

interpolating piecewise quadratic curves! This diagram is an

approximation.


Problem with Bezier Curves

• To make a long continuous curve with Bezier segments requires

using many segments

• Maintaining continuity requires constraints on the control point

positions

– The user cannot arbitrarily move control vertices and automatically

maintain continuity

– The constraints must be explicitly maintained

– It is not intuitive to have control points that are not free


Bezier Basis Functions for d=3

0

0.2

0.4

0.6

0.8

1

1.2

B0

B1

B2

B3


Documents

Introduction to CAD/CAM - home.iitk.ac.inhome.iitk.ac.in/~jrkumar/download/ME761_Lecture-6 CAD.pdf · computers. CAD/CAM in addition to its particular emphasis on the use of computer