Graphics Programs Reference
In-Depth Information
Next, we calculate the local coordinates of this point on the screen. Vector c is
first obtained by c = α ( p
2 , 2). The local
axes on the screen are computed next from Equation (3.17). They are u =(2 , 0 , 0)
and w =(0 , 4 ,
b )
a =(2 / 5)(0 , 0 , 10)
(0 , 2 , 2) = (0 ,
4). We normalize th e m by dividing each by its magnitude, obtaining
u =(1 , 0 , 0) and w =(0 , 1 / 2 ,
1 / 2). (Note that u is in the x direction and w is in
the yz plane.)
T hu s, the sc reen coordinates of c are u c =(1 , 0 , 0)
(0 ,
2 , 2) = 0 a nd w c =
(0 , 1 / 2 ,
8. The projected point is therefore 8 units away
from the center of the screen C . Note that this equals the absolute value of vector c .
As an added bonus, we compute the plane equation of the screen. Let ( x, y, z )
be a general point on the screen. The vector from the center (point C )to( x, y, z )is
( x
1 / 2)
(0 ,
2 , 2) =
2). This vector must be perpendicular to the normal to the screen
(vector a ), which implies
0 ,y
3 ,z
0= a
( x, y
3 ,z
2) = (0 , 2 , 2)
( x, y
3 ,z
2) ,
or
y + z =5 .
This equation relates the y and z coordinates of all the points on the screen. Any point
with coordinates ( x, y, 5
y ) is therefore on the screen regardless of the value of x . Note
that the projected point P also satisfies this relation.
Exercise 3.26: Generalize the previous example to the case of a general point P =
( x, y, z ).
Example 2 . Again we give a simple example, illustrated in Figure 3 .3 7. The s cr een
is centered on the origin at a 45 angle, and the viewer is at point (
k/ 2), a
distance o f k units from the screen. To simplify the notation, we introduce the quantity
k/ 2 , 0 ,
ψ = k/ 2. From Figure 3.36a it is clear that a =( ψ, 0 )and b =
a =(
ψ, 0 ,
ψ ).
The center of the screen is, as always, at a + b , which is point (0 , 0 , 0).
x
P
Screen
45 0
z
Viewer
Figure 3.37: Viewer Rotated About the y Axis.
Thefirststepistodetermine α :
2
2 ψ 2
|
a
|
2 ψ
x + z +2 ψ .
α =
b ) =
( x + ψ, y, z + ψ ) =
a
( P
( ψ, 0 )
 
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