Graphics Reference
In-Depth Information
The task is to find the values of the constants associated with the polynomials
V
1
and
V
2
. The requirements are:
1. A cubic function
V
2
must grow from 0 to 1 for 0
1.
2. The slope at a point
t
, must equal the slope at the point (1
≤
t
≤
−
t
). This
ensures slope symmetry over the range of the function.
3. The value
V
2
at any point
t
must also produce (1
−
V
2
)at(1
−
t
). This
ensures curve symmetry.
•
To satisfy the first requirement:
V
2
=
at
3
+
bt
2
+
ct
+
d
(8.11)
therefore,
t
=0
,d
=0for
V
2
=0,andwhen
t
=1
,V
2
=
a
+
b
+
c
.
•
To satisfy the second requirement, we differentiate
V
2
to obtain the slope
d
V
2
d
t
=3
at
2
+2
bt
+
c
=3
a
(1
− t
)
2
+2
b
(1
− t
)+
c
(8.12)
and equating constants we discover
c
=0and0=3
a
+2
b
•
To satisfy the third requirement,
at
3
+
bt
2
=1
t
)
3
+
b
(1
t
)
2
]
−
[
a
(1
−
−
(8.13)
where we discover 1 =
a
+
b
.But0=3
a
+2
b
, therefore
a
=
−
2and
b
=3.
Therefore
2
t
3
+3
t
2
(8.14)
To find the curve's mirror curve, which starts at 1 and collapses to 0 as
t
moves from 0 to 1, we subtract (8.14) from 1:
V
2
=
−
V
1
=2
t
3
3
t
2
+ 1
−
(8.15)
Therefore, the two polynomials are
2
t
3
+3
t
2
V
1
=2
t
3
V
2
=
−
3
t
2
+ 1
−
(8.16)
and are shown in Figure 8.7. They can be used as interpolants as follows:
n
=
n
1
V
1
+
n
2
V
2
(8.17)
which in matrix form is
n
=
2
t
3
n
1
n
2
2
t
3
+3
t
2
3
t
2
+1
−
−
·
(8.18)
⎡
⎤
2
−
2
n
1
n
2
n
=
t
3
1
·
⎣
⎦
·
−
33
00
10
t
2
t
1
(8.19)
