91Ó°ÊÓ

A parametric vector form of a B-spline curveis given as\({\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\;,\,\;0 \le t \le 1\).

It is to be shown that for \(0 \le t \le 1\), \({\bf{x}}\left( t \right)\) in the convex hull of the control points \({{\rm{p}}_1}\), \({{\rm{p}}_2}\), \({{\rm{p}}_3}\), \({{\rm{p}}_4}\) and for the translation \({\rm{x}}\left( t \right) + {\rm{b}}\), the parametric curve is also a B-spline.

For \({\rm{x}}\left( t \right)\) to be the convex combination of the control points, the constant of the its parametric curve must sum up to 1. To check, expand \({\rm{x}}\left( t \right)\) as shown below:

\(\begin{array}{c}{\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\;\\ = \frac{1}{6}\left[ {\left( {1 - {t^3} + 3{t^2} - 3t} \right){{\bf{p}}_o} + \left( {3{t^3} - 6{t^2} + 4} \right){{\bf{p}}_1} + \left( {3{t^2} - 3{t^3} + 3t + 1} \right){{\bf{p}}^2} + {t^3}{{\bf{p}}_3}} \right]\end{array}\)

The sum of the constant is \(\frac{1}{6} + \frac{4}{6} + \frac{1}{6}\) which is 1 only.

So, \({\rm{x}}\left( t \right)\) is the convex combination of the control points.

Consider \({\rm{x}}\left( t \right)\) as shown below:

\(\begin{array}{c}{\bf{x}}\left( t \right) = \frac{1}{6}\left[ {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\;\\ = \frac{1}{6}\left[ {\left( {1 - {t^3} + 3{t^2} - 3t} \right){{\bf{p}}_o} + \left( {3{t^3} - 6{t^2} + 4} \right){{\bf{p}}_1} + \left( {3{t^2} - 3{t^3} + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\end{array}\)

Simplify \({\bf{x}}\left( t \right)\) by adding \(\frac{1}{6}\left( {6{\bf{b}}} \right)\) to it, as shown below:

\(\begin{array}{c}{\bf{x}}\left( t \right) + {\bf{b}} = \frac{1}{6}\left[ {\left( {1 - {t^3} + 3{t^2} - 3t} \right){{\bf{p}}_o} + \left( {3{t^3} - 6{t^2} + 4} \right){{\bf{p}}_1} + \left( {3{t^2} - 3{t^3} + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right] + {\bf{b}}\\ = \frac{1}{6}\left[ {\left( {1 - {t^3} + 3{t^2} - 3t} \right){{\bf{p}}_o} + \left( {3{t^3} - 6{t^2} + 4} \right){{\bf{p}}_1} + \left( {3{t^2} - 3{t^3} + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right]\\ + \frac{1}{6}\left( {6{\bf{b}}} \right)\\ = \frac{1}{6}\left[ \begin{array}{l}\left( {1 - {t^3} + 3{t^2} - 3t} \right)\left( {{{\bf{p}}_o} + {\bf{b}}} \right) + \left( {3{t^3} - 6{t^2} + 4} \right)\left( {{{\bf{p}}_1} + {\bf{b}}} \right) + \left( {3{t^2} - 3{t^3} + 3t + 1} \right)\\\left( {{{\bf{p}}_2} + {\bf{b}}} \right) + {t^3}\left( {{{\bf{p}}_3} + {\bf{b}}} \right)\end{array} \right]\end{array}\)

As \({\bf{x}}\left( t \right) + {\bf{b}}\) is translated using points \(\left( {{{\bf{p}}_0} + {\bf{b}},\,{{\bf{p}}_1} + {\bf{b}},{{\bf{p}}_2} + {\bf{b}}} \right)\). This shows that \({\bf{x}}\left( t \right) + {\bf{b}}\)is a cubic B-spline with control point \(\left( {{{\bf{p}}_0} + {\bf{b}},\,{{\bf{p}}_1} + {\bf{b}},{{\bf{p}}_2} + {\bf{b}}} \right)\).