Perspective Projection - Transformations and Projections in Computer Graphics - page 112

(Try to convince yourself that α is positive in the gray area above and to the right of
the screen because x + z +2 ψ is positive in this area.)
Thenextstepistocomputevector d

d = b + α ( P − b )

2 ψ
x + z +2 ψ ( x + ψ, y, z + ψ )

=( − ψ, 0 , − ψ )+

ψ
x + z +2 ψ ( x

= − z, y, z − x ) .

Notice that P =(0 , 0 , 0) is transformed to P ∗ =(0 , 0 , 0).

Also, every point P =

x ) is transformed to P ∗ =(0 , 0 , 0).
Since the screen is centered at the origin, we have c = α ( P

( x, 0 , −

b )+ b =
d . The next step is to calculate the local screen vectors u and w from Equation (3.17).
This is straightforward and results in u =(0 ,

− b ) − a = α ( P −

− ψ, 0) and w = ( ψ, 0 , − ψ ). After normal-

1 , 0) and w =(1 / √ 2 , 0 , 1 / √ 2). Notice that u is the y

ization, these become u =(0 , − −

axis and w is in the xz plane.
The screen equation is obtained from a •

( x, y, z ) = 0, which implies ψ ( x + z )=0

or x = − z . The last step is to derive the transformation matrix. From

X
H

z )
x + z +2 ψ ,

ψ ( x − Y

H

ψy
x + z +2 ψ ,

Z
H

x )
x + z +2 ψ ,

ψ ( z −

x ∗ = ∗ = ∗ = = = =

we get

⎛
⎝

⎞
⎠

ψ 0 − ψ 1

0 ψ 00

( X, Y, Z, H )=( x, y, z, 1) .

− ψ 0 ψ 1

000 ψ

(Notice the two 1's in the last column. They indicate that the projection plane intercepts
the x and z axes but not the y axis. This is a two-point perspective.)

3.9 A Coordinate-Free Approach: II

This approach to the problem of perspective projection also uses vectors instead of
coordinates, but we assume that the following are given (Figure 3.38):
1. the position of the viewer (vector b );
2. the direction and distance from the viewer to the projection plane (vector a );
3. an “up” vector Z , which determines the direction of the local screen vector w ;
4. two viewing half-angles h and v , an approach that is handy when we want to
limit the projected image to certain viewing angles, as in Figure 3.13.