Graphics Reference
In-Depth Information
Fig. 3.6
Vector
v
is derived
from
a
part of
v
1
and
b
part
of
v
2
Furthermore,
m
=
a
cos
tθ
n
=
b
cos
(
1
−
t) θ
where
m
+
n
=
1
.
(3.9)
From (
3.8
)
a
sin
tθ
sin
(
1
b
=
−
t) θ
and from (
3.9
) we get
a
sin
tθ
cos
(
1
−
t) θ
a
cos
tθ
+
=
1
.
sin
(
1
−
t) θ
Solving for
a
we find that
sin
(
1
−
t) θ
a
=
sin
θ
sin
tθ
sin
θ
b
=
.
Therefore, the final spherical interpolant is
sin
(
1
−
t
)
θ
sin
tθ
sin
θ
v
=
v
1
+
v
2
.
(3.10)
sin
θ
To see how this operates, let's consider
a
simpl
e
exercise of interpolating be-
tween two unit vectors
1
/
√
21
/
√
2
T
. The angle
θ
between the
vectors is 135°. Equation (
3.10
) is used to interpolate individually the
x
- and the
y
-components individually:
T
[
10
]
and
[−
]
−
sin
(
1
t)
135°
sin 135°
−
sin
t
135°
sin 135°
1
√
2
v
x
=
×
(
1
)
+
×
.
sin
(
1
t)
135°
sin 135°
−
sin
t
135°
sin 135°
×
1
√
2
v
y
=
×
(
0
)
+
