Graphics Reference
In-Depth Information
3.5
3
2.5
1(1
−
t
) + 3
t
2
3
t
1.5
1
1(1
−
t
)
0.5
0
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
Fig. 8.2.
The top graph shows the result of linearly interpolating between values 1
and 3.
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
x
Fig. 8.3.
Interpolating between the points (1, 1) and (4, 3). Note that linear changes
in t give rise to equal spaces along the line.
between two locations (
x
1
,y
1
,z
1
)and(
x
2
,y
2
,z
2
). The interpolated position
is given by
x
=
x
1
(1
t
)+
x
2
t
y
=
y
1
(1
− t
)+
y
2
t
z
=
z
1
(1
−
−
t
)+
z
2
t
(8.5)
for 0
1 The parameter
t
could be generated from two frame values
within an animation. What is assured by this interpolant, is that equal steps
in
t
result in equal steps in
x
,
y
and
z
. Figure 8.3 illustrates this linear spacing
with a 2D example. We can write (8.2) in matrix form as follows:
≤
t
≤
n
1
n
2
n
=[(1
−
t
)
t
]
·
(8.6)
