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P
2
(0) =
β
1
P
1
(1)
(6.21)
and
P
2
(0) =
β
1
P
1
(1) +
β
2
P
1
(1)
.
(6.22)
We can observe that
β
1
= 1 provides the continuity of the parametric
first derivative vector and,
β
1
=1and
β
2
= 0 provides the continuity of
the parametric first and second derivative vectors. For
β
1
>
0and
β
2
≥
0,
they form a basis, i.e., they are linearly independent. With the coecients
so determined, one can compute the four
w
n
values in equation (6.2). These
values after simplification can be written as
w
−
2
(
β
1
,β
2
,t
)=2
β
1
(1
t
)
3
/δ
−
(6.23)
w
−
1
(
β
1
,β
2
,t
)=[2
β
1
t
[
t
2
3
t
+3]+2
β
1
[
t
3
3
t
2
+2]
−
−
(6.24)
+2
β
1
[
t
3
3
t
+2]+
β
2
[2
t
3
3
t
2
+ 1]]
/δ
−
−
w
0
(
β
1
,β
2
,t
)=[2
β
1
t
2
[
−
t
+3]+2
β
1
t
[
−
t
2
+3]
(6.25)
+
β
2
t
2
[
−
2
t
+3]+2[
−
t
3
+ 1]]
/δ
w
1
(
β
1
,β
2
,t
)=2
t
3
/δ.
(6.26)
6.3 Design Criteria for a Curve
In order to design a curve with two pieces of curve segments, say
P
1
(
t
)and
P
2
(
t
), we need to maintain position continuity, first order continuity, and
curvature continuity. The
i
th curve segment in terms of
β
1
and
β
2
can be
written as
P
i
(
t
)=(2
β
1
(1
t
)
3
/δ
)
V
i−
2
−
+ ([2
β
1
t
[
t
2
3
t
+3]+2
β
1
[
t
3
3
t
2
+2]
−
−
+2
β
1
[
t
3
3
t
+2]+
β
2
[2
t
3
3
t
2
+ 1]]
/δ
)
V
i−
1
−
−
+ ([2
β
1
t
[
t
2
3
t
+3]+2
β
1
[
t
3
3
t
2
+2]
−
−
(6.27)
+2
β
1
[
t
3
3
t
+2]+
β
2
[2
t
3
3
t
2
+ 1]]
/δ
)
V
i−
0
−
−
+ ([2
β
1
t
2
[
t
2
+3]
−
t
+3]+2
β
1
t
[
−
+
β
2
t
2
[
t
3
+ 1]]
/δ
)
V
i
+1
.
−
2
t
+3]+2[
−
The first derivative of the curve
P
i
(
t
) can be computed through
w
−
6
β
1
(1
t
2
)
/δ
2
(
β
1
,β
2
,t
)=
−
−
(6.28)
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