graphics & visualization: principles &...

Graphics & Visualization

Chapter 4

Projections and Viewing

Transformations

Graphics & Visualization: Principles & Algorithms Chapter 4

Graphics & Visualization: Principles & Algorithms Chapter 4 2

• In computer graphics there are:

3D models

2D output devices (displays and printers)

• Projective mapping (projection):

Must take place at some point in the graphics pipeline

Usually after the culling stages and before the rendering stage

• Viewing transformation:

Defines the transition from the world coordinate

system (WCS) to canonical screen system (CSS)

via the eye coordinate system (ECS)

Specifies the clipping bounds (for frustum culling)

in ECS

Introduction


• All objects are initially defined in their own local coordinate system

• These objects are unified in WCS

• The WCS is essentially used to define the model of a 3D synthetic

world

• The transition from WCS to ECS:

Simplifies a number of operations including culling (e.g. the specification of

the clipping bounds by the user) and projection

• The transition from ECS to CSS:

Ensures that all objects that survived culling will be defined in a canonical

space (usually [-1,1])

• Objects defined in a canonical space:

Can easily be scaled to the actual coordinates of any display device or monitor

Maintain high floating-point accuracy

Coordinate Systems


• Projection: techniques for the creation of the image of an object

onto another simpler object (e.g. line, plane, surface)

• Projection lines are defined by:

Center of projection

Points on the image being projected

• The intersection of a projector with the simpler object (e.g. the plane

of projection) forms the image of a point of the original object.

• Projections can be defined in spaces of arbitrary dimension.

• In Computer Graphics & Visualization:

Projections are from 3D space onto 2D space

The 2D space is referred to as the plane of projection & models the output

device

Projections


• Two projections are of interest:

Perspective: the distance of the center of projection from the plane of

projection is finite

Parallel: the distance of the center of projection from the plane of

projection is infinite

• Projective mappings are not affine transformations, so they cannot

be described by affine transformation matrices.

Differences between affine transformations and projective mappings:

Projections(2)


• Parallel lines: are not projected onto parallel lines

their projections seem to meet at a vanishing point

• Parallel lines are projected onto parallel lines only when their plane

is parallel to the plane of projection

• A straight line will map to a straight line

Example:

• Ratios of distances are not preserved:

• Cross ratios are preserved:

• To fully describe the projective image of a line we need the image of

3 points on the line (affine transformations need only 2)

Projections(3)

ab a b

bd b d

ac a c

cd c dab a b

bd b d


• Models the viewing system of our eyes

• Can be abstracted by a pinhole camera:

Center of projection: the pinhole

Plane of projection: the image plane (where the image is formed)

Creates an inverted image

Derive an upright image by placing the image ‘in front’ of the pinhole

Perspective Projection


• Suppose that:

Center of projection is the origin

Plane of projection is perpendicular to the

negative z-axis at distance d from the center

• A point P = [x, y, z]T is projected onto P΄ = [x΄, y΄, d]T

• P1 and P1΄ are the projections of P and P΄ onto the yz-plane

• are similar so:

• Similarly :

• The above are not linear equations (division by z)

Perspective Projection (2)

and 1 2 1 2OP P OP P ·y y d y

yd z z

1 2 1 2

22

P P P P

OPOP·d xx

z


• Trick to express the perspective-projection equations in matrix form:

Use matrix

which alters the homogeneous coordinate and

maps the coordinates of a point [x, y, z]T as follows:

• Divide by the homogeneous coordinate:


0 0 0

0 0 0

0 0 0

0 0 1 0

d

d

d

PERP

·

··

·

1

x x d

y y d

z z d

z

PERP

·

·

· ·/

·

1

x d

x d z

y d y dz

zz d

dz


• Perspective shortening:

The size of the projection of an object is inversely proportional to its distance

from the center of projection

• Known in ancient times, then forgotten

• Leonardo da Vinci studied the laws of perspective

• Older paintings had no perspective (symbolic criteria prevailed)



• Center of projection is at infinite distance from the projection plane

• Projection lines are parallel to each other

• To describe parallel projection one must specify:

The direction of projection (a vector)

The plane of projection

• Two types of parallel projections:

Orthographic: the direction of projection is normal to the plane

of projection

Oblique: the direction of projection is not necessarily normal to

the plane of projection

Parallel Projection


• Usually employs one of the main planes as the plane of projection

• Suppose the xy-plane is used

• A point P = [x, y, z]T will be projected onto P΄ = [x΄,y΄,z΄]T =[x,y,0] T

as follows:

where

Orthographic Projection

· ORTHO

P P P

1 0 0 0

0 1 0 0

0 0 0 0

0 0 0 1

ORTHOP


• Let the direction of projection be:

and the plane of projection be the xy-plane

• The projection P΄ = [x΄,y΄,z΄]T of a point P = [x, y, z]T will be

(1) for some scalar λ

• The z-coordinate of P΄ is 0, so (1) becomes:

• The other 2 coordinates of P΄ are:

Oblique Projection [ , , ]T

x y zDOP DOP DOPDOP

· P P DOP

0 · or z

z

zz DOP

DOP

· · and ·yx

x

z z

DOPDOPx x DOP x z y y z

DOP DOP


• Matrix form:

•

Oblique Projection (2)

1 0 0

0 1 0( )

0 0 0 0

0 0 0 1

x

z

y

z

DOP

DOP

DOP

DOP

OBLIQUEP DOP

( )· OBLIQUE

P P DOP P


EXAMPLE 1: Perspective Projection of a Cube

Determine the perspective projections of a cube

of side 1 when

(a)the plane of projection is z=-1 and

(b)the plane of projection is z=-10.

The cube is placed on the plane of projection.

SOLUTION (a)

• Represent the vertices of the cube as a 4×8 matrix:

Projection Example 1

0 1 1 0 0 1 1 0

0 0 1 1 0 0 1 1

1 1 1 1 2 2 2 2

1 1 1 1 1 1 1 1

C


• Multiply the perspective projection matrix (d=-1) by C:

• Normalize by the homogeneous coordinates:

Projection Example 1 (2)

1 0 0 0 0 1 1 0 0 1 1 0

0 1 0 0 0 0 1 1 0 0 1 1· ·

0 0 1 0 1 1 1 1 2 2 2 2

0 0 1 0 1 1 1 1 2 2 2 2

PERP C C

1 10 1 1 0 0 0

2 2

1 10 0 1 1 0 0

2 2

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1


(b) The original cube is:

• Multiply the perspective projection matrix (d=-10) by C’:

• Normalize by the homogeneous coordinates:


0 1 1 0 0 1 1 0

0 0 1 1 0 0 1 1

10 10 10 10 11 11 11 11

1 1 1 1 1 1 1 1

C

10 0 0 0 0 10 10 0 0 10 10 0

0 10 0 0 0 0 10 10 0 0 10 10·

0 0 10 0 100 100 100 100 110 110 110 110

0 0 1 0 10 10 10 10 11 11 11 11

C

10 100 1 1 0 0 0

11 11

10 100 0 1 1 0 0

11 11

10 10 10 10 10 10 10 10

1 1 1 1 1 1 1 1


EXAMPLE 2: Perspective Projection onto an Arbitrary Plane

Compute the perspective projection of a point P=[x, y, z]T onto the

arbitrary plane Π which is specified by the point R0=[x0,y0,z0]T and a

normal vector . The center of projection is O.

SOLUTION

• P΄=[x ΄, y ΄, z ΄]T is the projection of P=[x, y, z]T

• are collinear so

for some scalar a

• In the projection equations:

(1)

scalar a must be determined.


[ , , ]T

x y zn n nN

and OP OP ·a OP OP

' , ' , 'x ax y ay z az


• Vector is on the plane of projection, so :

• Substitute the values of x, y, z from (1), set c = nx x0 + ny y0 + nz z0 and

solve for a:

• The projection equations include a division by a combination of x,y,z

• Put in matrix form by altering the homogeneous coordinate:


0R P

0 0 0

0 0 0

· 0

( ) ( ) ( ) 0

x y z

x y z x y z

N

or n x x n y y n z z

or n x n y n z n x n y n z

0R P

x y z

ca

n x n y n z

,

0 0 0

0 0 0

0 0 0

0x y z

c

c

c

n n n

PERP


EXAMPLE 3: Oblique projection with Azimuth & Elevation Angles

Sometimes (particularly in architectural design), oblique projections are

specified in terms of the azimuth and elevation angles φ and θ that define

the relation of the direction to the plane of projection. Determine the

projection matrix in this case.

SOLUTION

• Let xy be the plane of projection

• The direction of the projection vector is:

• So:


[cos cos ,cos sin ,sin ]T DOP

cos1 0 0

tan

sin0 1 0( , )

tan

0 0 0 0

0 0 0 1

OBLIQUEP


EXAMPLE 4: Oblique projection onto an Arbitrary Plane

Determine the oblique projection mapping onto an arbitrary plane Π that

is specified by a point R0=[x0,y0,z0]T and a normal vector .

The direction of projection is given by the vector

SOLUTION

Transform plane Π to coincide with the xy-plane, use the oblique

projection matrix, undo the first transform:

• Step 1: Translate R0 to the origin,

• Step 2: Align with the positive z-axis (by using from[Ex 3.12])

• Step 3: Use the oblique projection matrix with the direction of

projection transformed according to previous steps:


[ , , ]T

x y zDOP DOP DOPDOP

[ , , ]T

x y zn n nN

( )0

T R

N ( )A N

( )· ( )·

0DOP A N T R DOP


• Step 4: Undo the alignment

• Step 5: undo the translation

• Finally:

Projection Example 4 (2) ( )1A N

( )0

T R

, ( ) ( )· ( ) · ( )· ( )· ( )

1

OBLIQUE 0 OBLIQUE 0P DOP T R A N P DOP A N T R


• Defines:

the process of coordinate conversion:

The clipping boundaries (for frustum culling) in ECS

(Right-handed coordinate systems)

• Viewing Transformation: a. WCS-to-ECS conversion

b. ECS-to-CSS conversion

orthographic projections

perspective projections

• z-coordinate is maintained by the ECS-to-CSS conversion

Stages following VT (e.g. hidden surface elimination) require 3D

information

Viewing Transformation (VT)


• ECS can be defined within the WCS by:

The ECS origin E

The direction of view

The up direction

• The origin E represents the point of view (location of an observer)

• The vector

defines the up direction

need not be perpendicular to

• WCS to ECS conversion steps:

1. Define the ECS axes xe, ye and ze

2. Perform the WCS-to-ECS conversion for all objects

WCS to ECS

g

up

up

g


1. Define the ECS axes

• Sufficient information (E, , ) since the coordinate system is

right-handed

• Align xe- and ye-axes with the WCS axes with the convention:

xe is the horizontal axis & increases to the right

ye is the vertical axis & increases upwards

ze-axis points towards the observer (right-handed ECS)

Compute xe- and ye-axes as cross products:

WCS to ECS (2)

gup

e e

e e e

x up z

y z x

ez g


2. WCS-to-ECS conversion

i. Establish matrix

ii. Pre-multiply all object vertices by it

• From [Ex. 3.16] this conversion requires 2 transformations:

Translation by

Change of basis (rotational transformation)

• Let the WCS coordinates of the ECS unit axis vectors be:

• Then:

WCS to ECS (3)

WCS ECSM

[ , , ]T

x y zE E E E

[ , , ] , [ , , ] and [ , , ]T T T

x y z x y z x y za a a b b b c c c e e ex y z

0 1 0 0

0 0 1 0 ·

0 0 0 1

0 0 0 1 0 0 0 1

x y z x

x y z y

x y z z

a a a E

b b b E

c c c E

WCS ECSM


• Orthographic projection onto the xy-plane:

Select a region of space (view volume) that will be mapped to CSS

rectangular parallelepiped

can be defined by 2 opposite vertices

xe = l, the left clip plane

xe = r, the right clip plane, (r > l)

ye = b, the bottom clip plane

ye = t, the top clip plane, (t > b)

ze = n, the near clip plane

ze = f , the far clip plane, ( f < n, since

the ze axis points toward the observer)

ECS to CSS: Orthographic projection


• To maintain the z-coordinate use the ortho. proj. MORTHO = ID

• Converting the view volume into CSS:

Translation & Scaling

Map the (l, b, n) values to -1 and (r, t, f) values to 1

• Thus a WCS point Xw = [xw, yw, zw]T is mapped into CSS by:

ECS to CSS: Orthographic projection(2)

2 2 2( , , )· ( , , )·

2 2 2

2 20 0 0 0 01 0 0

2

2 20 0 0 0 00 1 0

· 2

2 20 0 0 0 00 0 1

2

0 0 0 10 0 0 1 0 0 0 1

r l t b n f

r l t b f n

r lr l

r l r l r l

t bt b

t b t b t b

n fn f

f n f n f n

ORTHO

ECS CSSM S T ID

· · ORTHO

s ECS CSS WCS ECS wX M M X


• View volume is a truncated pyramid, symmetrical about the

-ze axis

• This view volume is specified by:

θ , the angle of the field of view in the y-direction

aspect, the width to height ratio of a cross section of the pyramid

ze = n, the near clipping plane

ze = f , the far clipping plane ( f < n)

ECS to CSS: Perspective projection


• Projection takes place onto the near clipping plane ze = n

• Compute the other clipping boundaries:

• Use a modified version of the

perspective projection matrix:

• z-coordinate :

Preserve it for hidden surface & other computations in screen space

Simply keeping ze will deform objects

Use a mapping that preserves lines & planes

zs = A + B / ze , where A,B are constants

ECS to CSS: Perspective projection (2)

top: | |·tan( )2

bottom: -

right: ·

left: -

t n

b t

r t aspect

l r

0 0 0

0 0 0

0 0 0

0 0 1 0

d

d

d

PERP


• Require that (ze = n) (zs = n) and (ze = f ) (zs = f ) so

A = ( n + f ) and B = -nf

• This mapping will not alter the boundaries ze = n , ze = f

but this will not be true for ze values between them

• Thus the perspective projection matrix is:

Makes the w-coordinate equal to ze

Must be followed by a division by ze

(perspective division)

Clipping boundaries are not affected

Transforms the truncated pyramid into

a rectangular parallelepiped :


0 0 0

0 0 0

0 0

0 0 1 0

n

n

n f nf

VTP


• Situation similar to the setting before the orthographic projection

• This view volume is already symmetrical about the –ze axis

• ECS-to-CSS conversion steps: 1. PVT

2. translation along ze

3. scaling


2 2 2( , , )· (0,0, )·

2

20 0 0

1 0 0 00 0 0

2 0 1 0 00 0 0 0 0 0

· · 0 00 0 12 20 0 0 0 0 1 0

0 0 0 1

0 0 0 1

n f

r l t b f n

r ln

nt b n f

n f nf

f n

PERSP

ECS CSS VTM S T P

20 0 0

20 0 0

20 0

0 0 1 0

n

r l

n

t b

n f nf

f n f n


• Thus a WCS point Xw = [xw, yw, zw]T is converted into CSS by:

followed by the perspective division by w (= ze)

• Frustum culling is performed before the perspective division

ensuring that the coordinates of every point on every object are

within the clipping bounds:

-w ≤ x, y, z ≤ w

• The coordinates of every point are now in the range [-1,1]


· ·

x

y

z

w

PERSP

ECS CSS WCS ECS wM M X

/

x

yw

z

w

sX


Example:

• Take the boundary points with ECS coordinates

[l, b, n, 1]T and [0, 0, f, 1]T

• Apply the PVT matrix:

• The homogeneous coordinate is no longer 1

• Apply the scaling & translation matrices:


2 2

0 0

0 0· ·

1 1

l ln

b bn

n n f f

n f

VT VTP P

2 2

0 0

0 0· · · ·

ln n

bn n

n n f f

n n f f

S T S T


Example(continued):

• Due to the symmetry of the truncated pyramid about –ze :

r = -l, t = -b so r - l = -2l, t - b = -2b

• Perspective division:

which are the CSS values of the points


1 0

0

1 0 0/ /

1 1

1 1

n

nn f

n f

n f


A. Truncated Pyramid Not Symmetrical about ze-Axis

e.g. stereo viewing where two viewpoints are slightly offset on the

xe –axis

Extended Viewing Transformation



• Parameters of the clipping planes:

ze = n0, the near clipping plane (as before)

ze = f0, the far clipping plane, f0 < n0 (as before)

ye = b0, the ye-coordinate of the bottom clipping plane at its

intersection with the near clipping plane

ye = t0, the ye-coordinate of the top clipping plane at its intersection

with the near clipping plane

xe = l0, the xe-coordinate of the left clipping plane at its intersection


xe = r0, the xe-coordinate of the right clipping plane at its intersection


Extended Viewing Transformation(2)


A. Truncated Pyramid Not Symmetrical about ze-Axis (steps)

• Convert the non-symmetrical pyramid to symmetrical about ze with a shear

transformation on the xy-plane

(1)

• Determine the A, B parameters

Take the shear on the y-coordinate: map the midpoint of t0b0 to 0

Similarly for the xe shear factor:

• So (1) becomes:


1 0 0

0 1 0

0 0 1 0

0 0 0 1

A

B

xySH

0 0 0 00

0

· 0 2 2

b t b tB n B

n

0 0

02

l rA

n

0 0

0

0 0

0

1 0 02

0 1 02

0 0 1 0

0 0 0 1

l r

n

b t

n

NON SYMSH


A. Truncated Pyramid Not Symmetrical about ze-Axis (steps)

• Change the clipping boundaries, to reflect the symmetrical shape of

the new pyramid:

• Substitute the above into

which is equivalent to the original with the clipping bounds

replaced by the initial clipping bounds.


0 0

0 0 0 00 0

0 0 0 00 0

2 2

2 2

n n f f

l r l rl l r r

b t b tb b t t

PERSP

ECS CSSM

0

0 0

0

0 0

0 0 0 0

0 0 0 0

20 0 0

20 0 0

20 0

0 0 1 0

n

r l

n

t b

n f n f

f n f n

PERSP

ECS CSSM

PERSP

ECS CSSM



• Thus it is not necessary to have initial clipping bounds; can name

them n, f, l, r, b, t from the start.

• The ECSCSS mapping in the case of non-symmetrical

perspective projection is thus:


20 0 0

1 0 02

20 0 0

0 1 0· · 2

20 0 0 0 1 0

0 0 0 10 0 1 0

nl r

r ln

nb t

t bn

n f nf

f n f n

PERSP NON SYM PERSP

ECS CSS ECS CSS NON SYMM M SH

20 0

20 0

20 0

0 0 1 0

n l r

r l r l

n b t

t b t b

n f nf

f n f n


B. Oblique Projection

• Useful, e.g., in the computation of oblique views for 3D displays:

is not sufficient

Direction of projection must be taken into account

• View volume: a six-sided parallelepiped specified by:

The 6 parameters used for the non-symmetrical pyramid (n0, f0, l0, r0, b0, t0)

The direction of projection vector

• Steps:

1. Translate the view volume so that (l0, b0, n0)-point moves to the ECS origin

2. Perform a shear in the xy-plane


ORTHO

ECS CSSM

DOP


B. Oblique Projection (steps)

• Take the point defined by the origin and

Shear the (DOPy) coordinate to 0:

Similarly for the x-shear factor:

• The required transformation is:

• Alter the clipping bounds to reflect the new rectangular parallelepiped

n = 0, f = f0 - n0, l = 0, r = r0 - l0, b = 0, t = t0 - b0

• ECSCSS mapping for a general parallel projection is thus:


[ , , ]T

x y zDOP DOP DOPDOP

· 0y

y z

z

DOPDOP B DOP B

DOP

x

z

DOPA

DOP

0

0

0

1 0 01 0 0

0 1 00 1 0· ·

0 0 1

0 0 0 10 0 1 0

0 0 0 1

x

z

y

z

DOP

DOP l

DOP b

DOP n

PARALLEL PARALLELSH T

· · PARALLEL ORTHO

ECS CSS ECS CSS PARALLEL PARALLELM M SH T


• Frustum culling:

Eliminates primitives (parts or whole) that lie outside the view volume(frustum)

Is implemented by 3D clipping algorithms

• The viewing transformation defines the 3D clipping boundaries.

• Clipping takes place in CSS, after the application of or

but before the division by w.

• Clipping boundaries for: perspective projection: -w ≤ x, y, z ≤ w

orthographic or parallel projection: -1 ≤ x, y, z ≤ 1

• Why clip in 3D and not in 2D, after throwing away z?

In perspective projection, objects behind the center of projection E would

appear upside-down (no sufficient information in clipping)

Avoids possible perspective division by 0

The near and far clipping planes limit the depth range & enable the optimal

allocation of the bits of the depth buffer

Frustum Culling & the Viewing Transformation

PERSP

ECS CSSM

ORTHO

ECS CSSM


• Viewport: rectangular part of the screen where the contents

of the view volume are displayed.

• Defined by its bottom-left and top-right corners

[xmin, ymin]T and [xmax, ymax]

T in pixel coordinates

[xmin, ymin, zmin]T and [xmax, ymax, zmax]

T to maintain the z-coordinate

• Viewport transformation converts objects from CSS into the

viewport coordinate system (VCS) with a scaling & a translation:

• This is a generalization of the 2D window-to-viewport transformation [Ex. 3.8].

• The size of the viewport defines the final size of the objects on screen

The Viewport Transformation

1 0 0 0 0 0 0 02 2 2 2

0 1 0 0 0 0 0 0·2 2 2 2

0 0 1 0 0 0 0 02 2

0 0 0 1 0 0 0 1

min max max min max min min max

min max max min max min min max

min max max min max min

x x x x x x x x

y y y y y y y y

z z z z z z

VIEWPORT

CSS VCSM

2 2

0 0 0 1

min maxz z

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