Graphics Reference
In-Depth Information
Using the second assumption (that all rays from patch
k
to patch
j
are essentially
the same) we see that the vector
C
j
)
, the unit
vector pointing from the center of patch
j
to the center of patch
k
. Since
u
jk
·
v
i
can be replaced by
u
jk
=
S
(
C
k
−
n
j
is
a constant, it can be factored out of the integral.
The form factor can then be written:
1
π
v
i
=
1
π
Ω
jk
v
i
·
n
j
d
u
jk
·
n
j
d
v
i
;
(31.42)
Ω
jk
=
1
π
1
d
v
i
u
jk
·
n
j
.
(31.43)
Ω
jk
The remaining integral is just the measure of the solid angle
Ω
jk
, which is the area
A
k
of patch
k
, divided by the square of the distance between the patches (i.e., by
2
), using the third assumption and scaled down by the cosine of the angle
between
n
k
and
u
jk
(by the Tilting principle). Thus, the form factor becomes
C
j
−
C
k
f
jk
=
1
π
A
k
2
|
u
jk
·
n
j
|·|
u
jk
·
n
k
|
.
(31.44)
C
j
−
C
k
Inline Exercise 31.4:
(a) The form of Equation 31.44 makes it evident that
f
jk
/
A
j
. Explain why, if
j
and
k
are mutually visible, exactly one of the
two dot products is negative.
(b) Suppose that patch
k
is enormous and occupies essentially all of the
hemisphere of visible directions from patch
j
. What will the value of
f
jk
be,
approximately?
A
k
=
f
kj
/
If we compute all the numbers
f
jk
and assemble them into a matrix, which
we multiply by a diagonal matrix
D
(
ρ
j
, and we
assemble the radiosity values
B
j
and emission values
E
j
into vectors
b
and
e
, then
the radiosity equation, under the assumptions listed above, becomes
ρ
)
whose
j
th diagonal entry is
b
=
e
+
D
(
ρ
)
Fb
.
(31.45)
This can be simplified (just like the integral form of the rendering equation) to
(
I
−
D
(
ρ
)
F
)
b
=
e
,
(31.46)
which is just a simple system of linear equations (albeit possibly with many
unknowns).
Standard techniques from linear algebra can be used to solve this equation
(see Exercise 31.2).
)
F
being “small” compared to the identity, that is, having all eigenvalues less than
one. This is a consequence of our assumption that all the reflectivities were less
than one (and your computation of the largest possible form factor). Note that
we do not suggest solving the equation by inverting the matrix; in general that's
O
(
n
3
)
, while approximation techniques like Gauss-Seidel work extremely well
(and much faster) in practice.
The existence of a solution depends on the matrix
D
(
ρ
