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(b) Now use this to explain why every 3
×
3 skew-symmetric matrix has 0 as
an eigenvalue, hence
det
S
=
0.
Exercise 11.10:
In the description of the virtual-sphere controller, we used the
angle
Q
)
, which involves a dot product of
points
rather than vectors.
This worked only because we assumed that the center of our virtual sphere was
the origin.
(a) Suppose the center was some other point
B
; how would we have computed
θ
=cos
−
1
(
P
·
?
(b) Now suppose that the virtual sphere was no longer assumed to be a unit sphere.
How would we compute
θ
?
Exercise 11.11:
Use the 3D test bed to implement virtual-sphere control for
viewing.
Exercise 11.12:
Given a point
P
and a direction
v
, describe how to build a
transformation on
R
3
θ
about the line determined by
P
and
v
so
that the plane through
P
, orthogonal to
v
, rotates counterclockwise by
that rotates by
θ
θ
when
viewed from the point
P
+
v
.
Exercise 11.13:
We saw in Chapter 7 that if
f
is the test function for the plane
containing the point
Q
and with normal
n
, that is,
f
(
P
)=(
P
−
Q
)
·
n
, then we
could compute the intersection of the ray
t
A
+
t
w
with this plane by writing
g
(
t
)=
A
+
t
w
and solving
f
(
g
(
t
)) =
0. Suppose that
T
is a linear transformation
with 4
→
4matrix
M
—perhaps translation by some amount, or a rotation about
the
y
-axis by 30
◦
. We can imagine applying
T
to every point of the plane defined
by
f
to get a
new
plane, and then finding the intersection of our ray with the new
plane.
(a) George proposes that a test function
f
for the new plane can be defined by
f
(
P
)=
f
(
T
(
P
))
. Is he correct? If not, adjust his claim to a correct one.
(b) Show that
f
(
g
(
t
)) =
0 if and only if
f
(
g
(
t
)) =
0, where
g
(
t
)=
M
−
1
A
+
t
M
−
1
w
.
(c) Describe how you can use the idea of part (b) to compute a ray-object inter-
section between a ray and an object that has been transformed by some affine
transformation, assuming you know how to do ray-object intersection for some
standard form of the object. This allows you, for instance, to ray-trace a stretched,
rotated unit sphere (i.e., an ellipsoid) if you know only how to ray-trace the stan-
dard unit sphere.
(d) Often in ray tracing it's necessary to determine not only the intersec-
tion point, but also the normal vector at the intersection point. Suppose that
instead of intersecting a ray
R
with a transformed sphere, you intersect an inverse-
transformed version of
R
with the unit sphere at the origin, and find the inter-
section is at a point
P
=(
x
,
y
,
z
)
with normal vector
n
=
xyz
T
. How can
you find the normal vector to the transformed sphere? Note: This approach to ray
tracing is somewhat out of favor for complex scenes, as traversing the modeling
hierarchy for every ray is a big cost. Instead, flattening the hierarchy by converting
everything to triangles and using a spatial data structure to accelerate intersection
testing turns out to be faster in general, but not always: In a forest of 1 million
identical trees, each a copy of a single million-triangle tree, the spatial data struc-
ture approach fails. Instead, we make a spatial data structure whose leaves are
bounding boxes
for the individual trees, use the transforming-rays trick of this
exercise, and proceed to trace each ray in the archetype tree's modeling space,
possibly using a spatial data structure there to accelerate the computation. This is
another application of the Coordinate-System/Basis principle.
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