Graphics Reference
In-Depth Information
As shown in Fig. 2.39, the interpolated vector should be
2
2
2
2
=
+
w
i
j
Using Eq. (2.31), we find that
sin 45
sin 9 0
sin 45
si n 90
w
=
i
+
j
2
2
2
2
w
=
i
+
j
Although Eq. (2.31) assumes that
u and
ˆ
v are unit vectors, it will work with non-unit vectors.
ˆ
For example, if u
90 .
Now let's find the interpolated vector when
=
2 i and v
=
4 j , then
=
=
05.
Therefore,
2
2
2
2
= 2 i
2 2 j
w
=
2 i
+
4 j
+
The magnitude of w is
2 2
2 2 2
10
w
=
+
=
=
3162
So, even though the rotated angle is 50% of the separating angle, the vector's length exceeds the
half-way point between 2 and 4, which is because the interpolation is spherical rather than linear.
If the length of the interpolated vector is required to be linearly related to , then we must
normalize u and v and scale the interpolated vector as follows:
sin1
v
sin
sin
sin ˆ
w
=
1
u
+
v
u
ˆ
+
For instance, using the above example, where u
=
2 i , v
=
4 j , and
=
05
then
4 sin 45
sin 90
j
sin 45
sin 90
w
=
05
×
2
+
05
×
i
+
and
3 2
2
j
2
2
w
=
i
+
which are correct.
Needless to say, the interpolation works in 3D as well as 2D.
 
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