Perspective Projection - Transformations and Projections in Computer Graphics - page 102

P ∗ = PT . We illustrate this for two points.
1 :Point P =(1 , 1 , 1) is projected to P ∗ =(0 , 0 , 0) because

⎛
⎝

⎞
⎠

s 00 sr
0100
−

(1 , 1 , 1 , 1) =(0 , 0 , 0 , 0) .

s 00 sr

0 − 10 − 2 rs

1 / √ 2(2 2 :Point P =(2 k, 0 , 2 k ) is projected to P ∗ =(0 ,

− − r ) , 0) because

⎛ ⎞

s 00 sr
0100
−s

⎝ ⎠

(2 k, 0 , 2 k, 1) =(0 , − 1 , 0 , 2 s (2 − r )) .

00 sr

0 − 10 − 2 rs

Exercise 3.19: The product

⎛ ⎞

s 00 sr
0100
−

⎝ ⎠

(0 , 0 , 0 , 1)

s 00 sr

0 − 10 − 2 rs

equals (0 , − 1 , 0 , − 2 sr ) , which suggests that the origin (0 , 0 , 0) is projected on the screen

at point P ∗ =(0 ,k/ √ 2 , 0). This, however, does not make sense since point (0 , 0 , 0) was
originally “behind” the viewer and should remain behind it after all the transformations.
What's the explanation?

Mighty is geometry; joined with art, resistless.
—Euripides

Note . Notice the rightmost column of matrix T [Equation (3.8)]. The first and
third elements of that column are nonzero, which indicates that the projection plane
intercepts the x and z axes. This is discussed in detail on page 110.
Example 2 . The viewer is located at B =( kβ )

and is looking in direction D =( α, 0 ,β ) (i.e., toward the origin). Matrices T 1 , T 2 , T 3 ,
and T p below are similar to the ones from the previous example. The result is

− k sin θ, 0 , − k cos θ )=( − kα, 0 , −

T = T 1 T 2 T 3 T p

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

1000
0100
0010
kα

β 0 α 0
0 100
−

1000
0100
0010
00

1000
0100
000 r
0001

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

α 0 β 0
0 001 0 kβ 1 − k 1