Graphics Reference
In-Depth Information
method of rendering. Through this, we hope you'll gain insight into the ways
we represent the real world in computer graphics. We start with a discussion of
visibility.
The matter of selecting only visible points might not even have occurred to Dürer,
but it is important for us. One way for Dürer to determine whether a point
P
is
visible is to place the pointer on
P
and see whether the string runs in a straight
line to the screw eye, or has to bend around some other part of the lute or other
object to get there. In making our simplified model of the process, we'll dis-
regard the issue of visibility determination, for the time being, not because it's
unimportant—Chapter 36 is entirely about data structures for improving visibility
computations—but because it is both tricky in general and unnecessary for the
model we'll be drawing in this simple renderer.
To mimic the woodcut's algorithm, we'll use algebra and geometry, and a very
simple description of a very simple shape: We'll describe a cube by giving the
locations of its six corners (or
vertices),
and by noting which pairs of vertices are
connected by an edge. Thus, our model of a cube can be considered a
wire-frame
model,
in which an object is created by connecting pieces of wire together.
To make the geometry simple, we'll establish a particularly nice way of
measuring coordinates in the room. The origin of our coordinate system (see
Figure 3.3) will be at the screw eye on the wall, and will be denoted
E
(for “eye”).
The drawing frame will be in the
z
=
1 plane (i.e., we measure the distance from
the screw eye to the plane of the drawing frame, and choose a system of units in
which this distance is 1). The point on this plane that's closest to the screw eye
will be called
T
; its coordinates are
(
0, 0, 1
)
. We'll make the
y
-axis vertical and
the
x
-axis horizontal, as shown.
The frame, within the
z
=
1 plane, will extend from the point
(
x
min
,
y
min
,1
)
to
the point
(
x
max
,
y
max
,1
)
, where
x
min
<
y
max
, as the names suggest,
and where, for simplicity and consistency with the drawing (in which the frame
appears fairly square
1
), we'll assume that the width,
x
max
−
x
max
and
y
min
<
x
min
, is the same as
the height,
y
max
−
y
min
. To be precise, two opposite corners of the frame are points
whose 3D coordinates are
(
x
min
,
y
min
,1
)
and
(
x
max
,
y
max
,1
)
.
Inline Exercise 3.3:
What are the coordinates of the other two corners of the
frame?
We'll also imagine that the paper that fills the frame (when the shutter is
closed) is actually graph paper, whose lower-left corner is labeled
(
x
min
,
y
min
)
and whose upper-right corner is labeled
(
x
max
,
y
max
)
, therefore providing us with
1. The panel that fills the frame does
not
appear square, however; we'll stick with square-
ness for simplicity.
