LEC_15_Projections.pdf

Nihar Ranjan Roy

What is Projection?

It can be defined as mapping of an

object which is in 3D domain into 2D

domain.

Types of Projections

Projection

Parallel

Perspective

3

Parallel Projection Taxonomy

Parallel

Orthographic

Multiview Axonometric

Isometric Dimetric Trimetric

Oblique

Cavalier Cabinet

Taxonomy of Perspective Projection

Perspective

Projection

One point

Two Point

Three point

5

Perspective Projection (

PP

)

Its generalization of the principles

used by an artist in preparing

perspective drawing on 3D objects and

scenes.

One point Perspective

• Depends on how many principal axes

intersect with view plane.

• Parallel lines not parallel to view plane

have the same vanishing point.

Two Point Perspective

Three point perspective

Perspective projection

10

N

Center of projection (eye)

View Plane

View plane Normal

Characteristics of PP

• Perspective foreshortening:-illusion that length of the objects appear smaller as their distance from the center of projection changes.

COP

• Principal Vanishing points:- apparent intersection of the lines parallel to principal axes with axes.

• View confusion

:-12

Center of projection

Objects behind the center of projection are

• Topological Distortion:

13

P1

p2

Transformation for Perspective

ΔAOC and ΔA’O’C are similar Δ (x’/x)=(d/(z+d)) similarly

(y’/y)=(d/(z+d))

(z’/z)=(d/(z+d)) where z`=0

Variations of Projection Matrix

P

=

P

=

where r=1/d

Problem Session

Q. A line from A(10,-10,10) and

B(10,-10,0) is being viewed from the point

P(0,0,20). Find the projection of this

line on xy plane.

Solution

COP



C(0,0,20) and the projection plane is z’=0

COP



C(0,0,-d)

COP



C(0,0,-(-20))

P

=

R=-1/20=-0.05

A’=(20,-20,0) and B’=(10,-10,0).

17 X Y Z 1 = 10 10 -10 -10 10 0 1 1

=

problem

Using origin as center of projection,

derive the perspective projection on to

the plane passing through the point

R

₀

(x

₀

,y

₀

,z

₀

) and having the normal

vector N=n

1

i+n

2

j+n

3

k

Help

According to property of vector

OP’=α.OP

x`i+y`j+z`k= α(xi+yj+zk)

X`= αx

Y`= αy

Z`= αz

Equation of the plane

n

x+n

y+n

z+d=0

Since point P`(x`,y`,z`) line on

the plane it must satisfy the

equation

20

P(x,y,z)

P’(x’,y’,z’)

R₀(x₀,y₀,z₀) N=n₁i+n₂j+n₃k

o

x

21

P(x,y,z)

P’(x’,y’,z’)

R₀(x₀,y₀,z₀) N=n₁i+n₂j+n₃k

o

x

Problem

Find the projection matrix for the

previous case if the center of projection

is at C(x

c

,y

c

,z

c

)

Solution

• Steps involved

– Translate the center of projection to origin – Perform the projection as in previous case – Translate back the center of projection

Here R₀=(x₀-x_c,y₀-y_c,z₀-z_c)

So the the new composite transformation matrix is

=T_c. P_per,origin .T_-c

Where P_per,origin=Perspective transformation matrix with COP at origin.

Parallel projection taxonomy

Parallel

Orthographic

Multiview Axonometric

Isometric Dimetric Trimetric

Oblique

Cavalier Cabinet

Oblique Projetion

The projection vector is not perpendicular to projection plane.

Cavalier The lines perpendicular to the projection plane are preserved in length.

L L/2

Cabinet The lines perpendicular to projection plane are ½ their true length.

26

Parallel

Oblique

Orthographic Projections

The projection vector is perpendicular to the

projection plane

Multiview Parallel projections



Direction of

projection is parallel to principal axis.

TOP view,BOTTOM view,SIDE views..ie u read it in

mechanical drawing.

Axonometric parallel projections



projection vector

is not parallel to any of the principal axes.

Axonometric Projections

• Isometric projection vector makes equal angle with all the principal axes.

• Dimetric makes equal angle with two principal axes.

• Trimetric  makes unequal angles with three principal axes.

problem

Derive the matrix for parallel projection

onto the xy plane in the direction of

projection vector v=ai+bj+ck.

Parallel projection

V=ai +bj +ck Z’=0

PP’ has the direction of V PP`=λV

X`-x= λ a Y`-y= λ b Z`-z= λ c

But z`=0 λ =-(z/c)

x’=x-(az/c) y’=y-(bz/c) X’ Y’ Z’ 1 =

1 0 -(a/c) 0 0 1 -(b/c) 0 0 0 0 0 0 0 0 1

Orthographic Projection on XY

V=ai+bj+ck

a=b=0



V=ck

P

=

P

=

31

1 0 -(a/c) 0 0 1 -(b/c) 0 0 0 0 0 0 0 0 1

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

0 0 0 1 _X

Y Z

P(x,y,z)

Oblique Projection

f

OA=x’=f cos Ө

AP’=y’=f sin Ө

PP’ and V have same direction V= α PP’

a= α(x`-x)= α f cos Ө

b= α(y`-y)= α fsin Ө

c= α (z’-1)= - α

z’=0  c=- α on xy plane

Oblique projection is special case of Parallel projection

Poblique=

1 0 -(a/c) 0 0 1 -(b/c) 0 0 0 0 0 0 0 0 1

1 0 f cos Ө 0 0 1 f sin Ө 0 0 0 0 0 0 0 0 1

=

L=1

Cavalier



there is no foreshortening ie

f=1

Cabinet



foreshortening is ½ ie

f=

/

2

33

1 0 f cos Ө 0 0 1 f sin Ө 0 0 0 0 0 0 0 0 1

1 0 (½) cos Ө 0 0 1 (1_/

2)sin Ө 0

0 0 0 0 0 0 0 1