Beta-Splines: A Flexible Model - Bezier and Splines in Image Processing and Machine Vision - page 130

We obtain the coecient functions, c mn , in equation(6.2) for m =0 , 1 , 2 , 3and
n = 1 , 0 , 1 once we differentiate the basis functions and get their values

at t =0and t = 1. These provide us a system of 15 linear equations in 16
unknowns. Hence, we need one more constraint to determine the coecients
uniquely. The adequate constraint is chosen to satisfy the convex hull property
to normalize the basis functions at t =0.

− 2 , −

c 3 ,− 2 + c 2 ,− 2 + c 1 ,− 2 + c 0 ,− 2 =0

c 3 ,r + c 2 ,r + c 1 ,r + c 0 ,r = c 0 ,r− 1 r = − 1 , 0 , 1

c 0 , 1 =0

β 1 ( c 3 , −

2+2 c 2 ,

−

2+ c 1 ,− 2

=0

β 1 (3 c 3 ,r +2 c 2 ,r + c 1 ,r

= c 1 ,r− 1 r =

−

1 , 0 , 1 (6.18)

c 1 , 1 =0
3(2 β 1 + β 2 ) c 3 ,− 2 +2( β 1 + β 2 ) c 2 ,− 2 + β 2 c 1 ,− 2 =0
3(2 β 1 + β 2 ) c 3 ,r +2( β 1 + β 2 ) c 2 ,r + β 2 c 1 ,r =2 c 2 ,r− 1 r = − 1 , 0 , 1

c 2 , 1 =0 .

The convex hull property to normalize the basis function at t =0is

c 0 ,− 2 + c 0 ,− 1 + c 0 , 0 + c 0 , 1 =1 . (6.19)

Note that, c 0 , 1 , c 1 , 1 ,and c 2 , 1 are zero. Hence we have effectively thirteen
equations in thirteen unknowns. The unknowns are coecient functions of β 1
and β 2 . Barsky [15] used a computer algebra system “REDUCE” to determine
the coecients as

c 0 ,− 2 =2 β 1 /δ
c 1 ,− 2 =

6 β 1 /δ
c 2 ,− 2 =6 β 1 /δ
c 3 ,− 2 =

−

2 β 1 /δ
c 0 ,− 1 =4 β 1 +4 β 1 + β 2 /δ
c 1 ,− 1 =6 β 1 ( β 1 −

−

1) /δ

2 β 1 − 2 β 1 −

c 2 ,− 1 =3( β 2 ) /δ

c 3 ,− 1 =2( β 1 + β 1 + β 1 + β 2 ) /δ
c 0 , 0 =2 /δ
c 1 , 0 =6 β 1 /δ
c 2 , 0 = 3(2 β 1 + β 2 ) /δ
c 3 , 0 =

− (6.20)

2( β 1 + β 1 + β 2 +1) /δ

−

c 3 , 1 =2 /δ,

with δ =2 β 1 +4 β 1 +4 β 1 + β 2 +2 .
β 1 and β 2 are the shape parameters because they are defined through
the unit tangent vector and curvature vector. Hence, one can use these two
parameters to control the shape of a curve at the time of design. Given the
two pieces of curve segments, say P 1 ( t )and P 2 ( t ), β 1 and β 2 are visualized
through