Beta-Splines: A Flexible Model - Bezier and Splines in Image Processing and Machine Vision - page 135

P 0 ( t ) = { w − 2 ( t )+ w − 1 ( t )+ w 0 ( t ) } V 0 + w 1 ( t ) V 1 ,

P m +1 ( t )= w − 2 ( t ) V m− 1 + { w − 1 ( t )+ w 0 ( t )+ w 1 ( t ) } V m .

Upon substitution of these expressions, the basis functions become

2 t 3 /δ 0 ( t ) 2 t 3 /δ 0 ( t ) P 0 ( t ) = { 1 − } V 0 + { } V 1 ,

t )) 3 /δ m +1 ( t ) P m +1 ( t )= { 2( β 1 ,m +1 ( t )(1 − } V m− 1

t )) 3 /δ m +1 }

+ { 1 − 2( β 1 ,m +1 ( t )(1 − V m .

As t variesfrom0to1, P 0 ( t ) traces a straight line segment starting at V 0
and ending at a point distant 2 /γ 0 along the vector V 0 V 1 . Similarly, P m +1 ( t )
also traces a straight line segment from a point

2 α 1 ,m /γ m along the vector
V m − 1 V m to the terminal point V m . Use of triple vertices interpolates the
end points.

−

Phantom Vertices

Phantom vertices are auxiliary vertices that are generally created for the pur-
pose of additional pieces of curves. As these vertices are inaccessible to the
users and are not displayed, they are named phantom vertices . Normally, they
are defined in terms of the original control polygon vertices, and at each end
point, the curve interpolates a specified point. This means P 1 (0) = P 0 and
P m (1) = P m .
From equation (6.1), solving for the phantom vertices we get,

2 α 1 , 0 − ) / 2 α 1 , 0 ,

V 1 =( γ 0 P 0 −{ ( γ 0 − 2) V 0 +2 V 1 }

(6.41)

2 α 1 ,m − 2) V m +2 α 1 ,m V m− 1 }

V m +1 =( γ m P m −{ ( γ m − ) / 2 .

First derivatives are then
P 1 (0) = 6 α 1 , 0 ( α 1 , 0 V 1 +( α 1 , 0 −

− 1) V 0 + V 1 ) /γ 0 ,

P m (1) = 6( α 1 ,m V m− 1 +( α 1 ,m −

− 1) V m + V m +1 ) /γ m .

Substituting the expressions of phantom vertices in equation (6.41), the above
expressions become

P 1 (0) = 3( { 2( α 1 , 0 +1) V 1 +( γ 0 − 2 α 1 , 0 − 2) V 0 } /γ 0 − P 0 ) ,

P m (1) = 3( (2 α 1 ,m ( α 1 ,m +1) 2 α 1 ,m ( α 1 ,m +1) V m− 1 }

{ − γ m ) V m − /γ 0 + P m ) .

Similarly, the second derivative vector at each end point of the curve is

P 1 (0) = 6( (2 α 1 , 0 + α 2 , 0 − ( γ 0 +2 α 1 , 0 + α 2 , 0 −

{ 2) V 1 − 2) V 0 } /γ 0

+ P 0 ) ,
P m (1) = 6(

(6.42)

2 α 1 ,m +2 α 1 ,m + α 2 ,m ) V m− 1 − 2 α 1 ,m +2 α 1 ,m

{ ( − ( γ m −

+ α 2 ,m ) V m } /γ m + P m ) .

The first and second derivative vectors at each end point of the curve are,
in general, linearly independent. Thus, the curvature is non-zero at each end
point of the curve.