Graphics Reference
In-Depth Information
For
T
, the (unsigned) area is the length of the vector
a
=
A
yz
,
A
zx
,
A
xy
. Because
T
is tilted only in the
yz
-plane,
A
yz
=
0. And because the
x
- and
z
-coordinates for
T
and
T
agree, we have
A
zx
=
A
zx
.
(7.131)
That leaves only
A
xy
to consider:
A
xy
=(
a
x
b
y
−
a
y
b
x
)+(
b
x
c
y
−
c
x
b
y
)+(
c
x
a
y
−
a
x
c
y
)
.
(7.132)
Remembering that the normal to the plane of
T
is
0
cos
θ
T
, we know
θ
sin
the plane equation: Every point
(
x
,
y
,
z
)
on this plane must satisfy
0
x
+cos(
θ
)
y
+sin(
θ
)
z
=
K
(7.133)
where
K
is some constant. (You should make sure that you understand why this is
true.) This means that for any point
(
x
,
y
,
z
)
on the plane,
K
cos(
y
=
−
tan(
θ
)
z
+
)
.
(7.134)
θ
Applying this to
a
y
,
b
y
, and
c
y
in Equation 7.132 and canceling many terms, we
get
A
xy
=(
a
x
tan(
θ
)
b
z
−
tan(
θ
)
a
z
b
x
)+(
b
x
tan(
θ
)
c
z
−
c
x
tan(
θ
)
b
z
)
+(
c
x
tan(
θ
)
a
z
−
a
x
tan(
θ
)
c
z
)
(7.135)
=
−
tan(
θ
)
A
zx
.
(7.136)
The length of
a
(hence the area of
T
)isthus
area
=
(
A
yz
)
2
+(
A
zx
)
2
+(
A
xy
)
2
(7.137)
=
0
2
+
A
zx
+(
−
tan(
θ
)
A
zx
)
2
(7.138)
=
(
1
+tan
2
(
θ
)
A
zx
(7.139)
=
±
sec(
θ
)
|
A
zx
|
,
(7.140)
. Thus, the area of
T
is
while the area of
T
is
|
A
zx
|
|
cos
θ|
times that of
T
.
There remains the question of the sign. If
cos
θ>
0, and the signed area of
A
B
C
is positive, then the vertices
A
,
B
, and
C
are organized counterclockwise
as viewed from the tip of
n
, and hence the signed area of
ABC
is also positive.
A
B
C
is positive, then
the vertices
A
,
B
, and
C
as viewed from the tip of
n
are organized clockwise, and
hence the signed area of
On the other hand, if
cos
θ<
0 and the signed area of
ABC
is negative. If we reverse the order of
A
,
B
, and
C
, in both cases, both signed areas change sign. So, in all four possible cases, the
ratio of unsigned areas is
|
cos
θ|
, and the sign of the ratio of signed areas is the
n
.
sign of
cos
θ
, hence the ratio of signed areas is exactly
cos
θ
=
n
·
Inline Exercise 7.18:
Suppose that instead of projecting
T
onto the
xy
-plane
by projecting in the
y
-direction, we projected in the
n
direction. What would
be the relationship between the signed area of the projected triangle
T
and
that of
T
?
