Graphics Reference
In-Depth Information
y
we want to compute
v
·
m
.Howis
m
related to the surface normal
n
of the original
model?
The original surface normal
n
was defined by the property that it was orthog-
onal to every vector
u
that was tangent to the surface. The new normal vector
m
must be orthogonal to all the transformed tangent vectors, which are tangent to
the transformed surface. In other words, we need to have
x
n
m
·
Mu
=
0
(10.102)
for every tangent vector
u
to the surface. In fact, we can go further. For
any
vector
u
,we'dliketohave
y
·
Mu
=
n
·
m
u
,
(10.103)
that is, we'd like to be sure that the angle between an untransformed vector and
n
is the same as the angle between a transformed vector and
m
.
x
Before working through this, let's look at a couple of examples. In the case
of the transformation
T
1
, the vector perpendicular to the bottom side of the house
(we'll use this as our vector
n
) should be transformed so that it's still perpendicular
to the bottom of the transformed house. This is achieved by rotating it by 30
◦
(see
Figure 10.20).
If we just
translate
the house, then
n
again should be transformed just the way
we transform ordinary vectors, that is, not at all.
But what about when we shear the house, as with example transformation
T
3
?
The associated vector transformation is still a shearing transformation; it takes a
vertical vector and tilts it! But the vector
n
, if it's to remain perpendicular to the
bottom of the house, must not be changed at all (see Figure 10.21). So, in this
case, we see the necessity of transforming covectors differently from vectors.
m
Figure 10.20: For a rotation, the
normal vector rotates the same
way as all other vectors.
y
Let's write down, once again, what we want. We're looking for a vector
m
that
satisfies
x
m
·
(
Mu
)=
n
·
u
(10.104)
n
for every possible vector
v
. To make the algebra more obvious, let's swap the order
of the vectors and say that we want
y
(
Mu
)
·
m
=
u
·
n
.
(10.105)
b
can be written
a
T
b
, we can rewrite this as
Recalling that
a
·
(
Mu
)
T
m
=
u
T
n
.
x
(10.106)
Remembering that
(
AB
)
T
=
A
T
B
T
, and then simplifying, we get
m
(
Mu
)
T
m
=
u
T
n
(10.107)
(
u
T
M
T
)
m
=
u
T
n
(10.108)
Figure 10.21: While the vertical
sides of the house are sheared,
the normal vector to the house's
bottom remains unchanged.
u
T
(
M
T
m
)=
u
T
n
,
(10.109)
where the last step follows from the associativity of matrix multiplication. This
last equality is of the form
u
·
a
=
u
·
b
for all
u
. Such an equality holds if and
only if
a
=
b
, that is, if and only if
M
T
m
=
n
,
(10.110)
