Graphics Reference
In-Depth Information
subtended by the small sphere is therefore
r
2
π
π
m
(Ω) =
4
(26.48)
4
π
R
2
=
π
r
2
R
2
.
(26.49)
Substituting this in Equations 26.47 and 26.43, we get
r
2
R
2
R
2
L
π
4
π
=Φ
, so
(26.50)
Φ
L
=
r
2
)
.
(26.51)
4
π
(
π
P
We've made two approximations in this computation: that the dot product is
always 1, and that the area occluded by the emitting sphere is
Q
r
2
. If you instead
evaluate the integral exactly, you'll see that the two approximations exactly cancel
each other.
π
R
H
r
We'll now consider a similar example and analysis. This time we have a
small disk-shaped emitter of radius
r
, that emits light only on one side (see
Figure 26.28).
We enclose it in a hemisphere
H
of radius
R
, and first compute the power
density at the North Pole
P
just as before; once again the power density is
Figure 26.28: A Lambertian emit-
ting disk that radiates on one side
only.
r
2
R
2
D
P
=
Lm
(Ω) =
L
π
.
(26.52)
At a point like
Q
that's off-axis by the angle
φ
, the Tilting principle applies, and the
power density arriving at
Q
is only
cos
times that arriving at
P
. Thus, the total
power arriving at all points of the hemisphere (which must be the total emitted
power
Φ
)is
φ
Φ=
H
)
L
π
r
2
R
2
(cos
φ
(26.53)
R
2
=
L
π
r
2
cos
φ
(26.54)
H
by pulling the constant out of the integral. Further simplifying,
R
2
S
+
r
2
R
2
Φ=
L
π
cos
φ
(26.55)
r
2
=
L
π
cos
φ
(26.56)
S
+
because the area of
H
is
R
2
times that of
S
+
. Finally, by the Average height prin-
ciple, we get
Φ=
L
π
r
2
π
(26.57)
so that
Φ
L
=
r
2
)
.
(26.58)
π
(
π
