Question

The “B” in B-spline refers to the fact that a segment \({\bf{x}}\left( t \right)\)may be written in terms of a basis matrix, \(\,{M_S}\) , in a form similar to a Bézier curve. That is,

\({\bf{x}}\left( t \right) = G{M_S}{\bf{u}}\left( t \right)\)for \(\,0 \le t \le 1\)

where \(G\) is the geometry matrix \(\,\left( {{{\bf{p}}_{\bf{0}}}\,\,\,\,{{\bf{p}}_{{\bf{1}}\,\,\,}}\,{{\bf{p}}_{\bf{2}}}\,\,\,{{\bf{p}}_{\bf{3}}}} \right)\)and \({\bf{u}}\left( {\bf{t}} \right)\) is the column vector \(\left( {1,\,\,t,\,\,{t^2},\,{t^3}} \right)\) . In a uniform B-spline, each segment uses the same basis matrix \(\,{M_S}\), but the geometry matrix changes. Construct the basis matrix \(\,{M_S}\) for \({\bf{x}}\left( t \right)\).

Short answer

The basis matrix is \({M_S} = \left( {\begin{array}{{}}1&{ - 3}&3&{ - 1}\\4&0&{ - 6}&3\\1&3&3&{ - 3}\\0&0&0&1\end{array}} \right)\).

Step-by-step solution

Write matrix \({\bf{x}}\left( t \right) = G{M_S}{\bf{u}}\left( t \right)\)for control points\(\,\left( {{{\bf{p}}_{\bf{0}}}\,\,\,\,{{\bf{p}}_{{\bf{1}}\,\,\,}}\,{{\bf{p}}_{\bf{2}}}\,\,\,{{\bf{p}}_{\bf{3}}}} \right)\)

The matrix \({\rm{x}}\left( t \right) = G{M_S}{\rm{u}}\left( t \right)\)is given as \({\bf{x}}\left( t \right) = \frac{1}{6}\left( {\left( {1 - 3t + 3{t^2} - {t^3}} \right){{\bf{p}}_o} + \left( {4 - 6{t^2} + 3{t^3}} \right){{\bf{p}}_1} + \left( {1 + 3t + 3{t^2} - 3{t^3}} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right)\) for \(0 \le t \le 1\) .

Step 2:Write vector of weights of \({\bf{x}}\left( t \right)\)

The weighted vector of \({\rm{x}}\left( t \right)\)is\(\frac{1}{6}\left( {\begin{array}{{}}{1 - 3t + 3{t^2} - {t^3}}\\{4 - 6{t^2} + 3{t^3}}\\{1 + 3t + 3{t^2} - 3{t^3}}\\{\,\,\,\,\,\,{t^3}}\end{array}} \right)\).

The weighted matrix

Factor the weighted vector as\({M_S}{\bf{u}}\left( t \right)\), where\({\bf{u}}\left( t \right)\)is the column vector involving ascending powers of t as shown below:

\({M_S}{\bf{u}}\left( t \right) = \left( {\begin{array}{{}}1&{ - 3}&3&{ - 1}\\4&0&{ - 6}&3\\1&3&3&{ - 3}\\0&0&0&1\end{array}} \right)\left( \begin{array}{l}1\\t\\{t^2}\\{t^3}\end{array} \right)\)

Compare and write \({M_S}\)

The basis matrix is \({M_S} = \left( {\begin{array}{{}}1&{ - 3}&3&{ - 1}\\4&0&{ - 6}&3\\1&3&3&{ - 3}\\0&0&0&1\end{array}} \right)\).