Graphics Reference
In-Depth Information
so
m
=(
M
T
)
−
1
n
,
(10.111)
where we are assuming that
M
is invertible.
We can therefore conclude that the covector
φ
n
transforms to the covector
φ
(
M
T
)
−
1
n
. For this reason, the inverse transpose is sometimes referred to as the
covector transformation
or (because of its frequent application to normal vec-
tors) the
normal transform.
Note that if we choose to write covectors as row
vectors, then th
e t
ranspose is not needed, but we have to multiply the row vector
on the
right
by
M
−
1
.
The normal transform, in its natural mathematical setting, goes in the opposite
direction: It takes a normal vector in the codomain of
T
M
and produces one in the
domain; the matrix for this
adjoint transformation
is
M
T
. Because we need to
use it in the other direction, we end up with an inverse as well.
Taking our shearing transformation,
T
3
, as an example, when written in
xyw
-
space the matrix
M
for the transformation is
⎡
⎤
120
010
001
⎣
⎦
,
(10.112)
and hence
M
is
12
01
,
(10.113)
while the normal transform is
(
M
−
1
)
T
=
1
T
=
10
−
.
2
01
−
(10.114)
21
φ
n
, where
n
=
2
1
, for example, becomes the covector
Hence the covector
φ
m
,
where
m
=
10
−
n
=
2
−
.
21
3
Inline Exercise 10.24:
(a) Find an equation (in coordinates, not vector form)
for a line passing through the point
P
=(
1, 1
)
, with normal vector
n
=
2
1
.
(b) Find a second point
Q
on this line. (c) Find
P
=
T
3
(
P
)
and
Q
=
T
3
(
Q
)
,
and a coordinate equation of the line joining
P
and
Q
. (d) Verify that the
normal to this second line is in fact proportional to
m
=
2
−
, confirming
that the normal transform really did properly transform the normal vector to
this line.
3
Inline Exercise 10.25:
We assumed that the matrix
M
was invertible when we
computed the normal transform. Give an intuitive explanation of why, if
M
is
degenerate (i.e., not invertible), it's impossible to define a normal transform.
Hint: Suppose that
u
, in the discussion above, is sent to
0
by
M
, but that
u
·
n
is nonzero.
