Perspective Projections
Perspective projections are more commonly used
and require some additional effort to derive.
The first step is choosing the form of the matrix. A simple version which
projects onto the plane given by z=-d was given earlier by
The more complex
transformation we wish to perform will in addition need the ability to scale and
translate each of x, y, and z. The following matrix has all the correct elements
in place.
The choice of the coefficient of -1, which makes h' = -z, is arbitrary.
Scaling this number can be matched by scaling each of the other parameters.
Let's first look at how y gets mapped to y'. For y = -z*top/near, we would
like y'/h' = 1. Thus,
(-F*z*top/near + B*z)/(-z) = 1
Similarly, for y = -z*bottom/near, we would like y'/h' = -1. Thus,
(-F*z*bottom/near + B*z)/(-z) = -1
Solving these two equations for F and B gives
F = 2 near/(top-bottom)
B = (top+bottom)/(top-bottom)
The mapping for x is determined in an analogous way, giving
E = 2 near/(right-left)
A = (right+left)/(right-left)
Lastly, let's look at the mapping for z. For z=-near, we would like z'/h' =
-1. Thus
C(-near)/near + D/near = -1 .
For z=-far, we would like z'/h' = 1. Thus
C(-far)/far + D/far = 1
Solving these equations for C and D gives:
C = -(far+near)/(far-near)
D = -2*far*near/(far-near)
OpenGL calls for Perspective Projections
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
followed by one of: glFrustum(left, right, bottom, top, near, far)
gluPerspective(fovy, aspect, near, far)
The glFrustum() call uses the parameters as described above. An
alternative specification is to use field-of-view and an aspect
ratio to specify the image plane parameters. In gluPerspective(),
fovy gives the field-of-view in the y-direction, measured in degrees
and centred about y=0. The aspect ratio gives the relative horizontal size of
the image, centred about x=0.
Non-linearity in Perspective Transformations
The perspective
transformations produce a z-coordinate for use in visibility calculations. It
is, however, a non-linear function of the original z coordinate. To illustrate
this, consider a railroad viewed in perspective as follows. tracks
left: x= -1, y= -1
right: x= 1, y= -1
view volume
left = -1, right = 1
bot = -1, top = 1
near = 1, far = 4
In this scene, what
happens to z in VCS and NDCS as we move along the track? The following
expressions tell us what we wish to know.
It can be directly seen that z_NDCS is a non-linear function of z_VCS.
What does this look like in the image plane? Let's determine z_VCS as a
function of x_NDCS. With this we can point at the image and ask What is the
real distance of this point?
Lastly, what does the train track look like in NDCS?
Straight lines in VCS correspond to straight lines in NDCS, although the
'speed' at which one moves along them is not the same in the two coordinate
systems.
