Perspective Projection - Transformations and Projections in Computer Graphics - page 106

y

z

3

2

x

1

Figure 3.34: n -Point Perspective.

of vanishing points, depending on the existence of groups of straight, parallel lines on
the object (page 76).

Exercise 3.23: Calculate matrix (3.13) twice, first for the case where D =(0 , 0 , 1)
(viewer looking in the positive z direction) and then for D =(0 , 0 , 1) and B =(0 , 0 , − k )

(the standard position).

Exercise 3.24: Assuming a viewer at point B =(0 , 1 , 0) looking in direction D =
(0 , 1 , 1), calculate the projection of point P =(0 , 1 , 10).

Matrix T g of Equation (3.13) contains the expression 1 + f in the denominators
of certain elements, which may cause undefined values when f = 1. Since we assume

that vector D is normalized, d 2 + e 2 + f 2 must be equal to 1, so the case f =

−

− 1 implies

d = e = 0, which, in turn, implies D =(0 , 0 , 1) (i.e., a viewer looking in the negative

z direction). It turns out that T g can be used even in this case. When d = e =0,we
can write

−

T g [1 , 1] = e 2 + f + f 2
1+ f

= f (1 + f )
1+ f

= f = − 1 ,

T g [2 , 2] = d 2 + f + f 2
1+ f

= f (1 + f )
1+ f

= f = − 1 .

ae 2 af 2

T g [4 , 1] = cd + bde

− − af + cdf −

af 1+ f
1+ f

= − = a,

1+ f

bd 2 + ce + ade bf 2

T g [4 , 2] = −

− bf + cef −

bf 1+ f
1+ f

= − = b.

1+ f

Matrix elements T g [1 , 2] = T g [2 , 1] = − de/ (1 + f ) have the indefinite form 0 / 0, but we