Question

Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be cubic Bézier curves with control points \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\)and \(\left\{ {{{\bf{p}}_{\bf{3}}}{\bf{,}}{{\bf{p}}_{\bf{4}}}{\bf{,}}{{\bf{p}}_{\bf{5}}}{\bf{,}}{{\bf{p}}_{\bf{6}}}} \right\}\) respectively, so that \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) are joined at \({{\bf{p}}_3}\) . The following questions refer to the curve consisting of \({\bf{x}}\left( t \right)\) followed by \(y\left( t \right)\). For simplicity, assume that the curve is in \({\mathbb{R}^2}\).

a. What condition on the control points will guarantee that the curve has \({C^1}\) continuity at \({{\bf{p}}_3}\) ? Justify your answer.

b. What happens when \({\bf{x'}}\left( 1 \right)\) and \({\bf{y'}}\left( 1 \right)\) are both the zero vector?

Short answer

a) The \({C^1}\) continuity is \({{\bf{p}}_3} = \frac{{\left( {{{\bf{p}}_4} + {{\bf{p}}_2}} \right)}}{2}\), and\({{\bf{p}}_3}\)is the mid-point of the line segment\({{\bf{p}}_2}\)to\({{\bf{p}}_4}\).

b) The line joining \({{\bf{p}}_4}\) and \({{\bf{p}}_2}\) is point \({{\bf{p}}_3}\) only.

Step-by-step solution

Apply the standard parameterization on \(\left\{ {{{\bf{p}}_{\bf{0}}}{\bf{,}}{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\)

For the points \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}\,{{\bf{p}}_{\bf{1}}}{\bf{,}}\,{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\), \({\rm{x}}\left( t \right)\) has the standard parameterization as \({\bf{x}}\left( t \right) = \left( {1 - 3t + 3{t^2} - {t^3}} \right){{\bf{p}}_o} + \left( {3t - 6{t^2} + 3{t^3}} \right){{\bf{p}}_1} + \left( {3{t^2} - 3{t^3}} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_{\bf{3}}}\).

Find the value of \({\bf{x'}}\left( 1 \right)\)

\({\bf{x'}}\left( t \right) = \left( { - 3t + 6t - 3{t^2}} \right){{\bf{p}}_o} + \left( {3 - 12t + 9{t^2}} \right){{\bf{p}}_1} + \left( {6t - 9{t^2}} \right){{\bf{p}}_2} + 3{t^2}{{\bf{p}}_3}\).

So, \(x'\left( 1 \right) = - 3{p_2} + 3{p_3}\) and \(x'\left( 0 \right) = - 3{p_0} + 3{p_1}\).

Use \({\bf{x'}}\left( 0 \right)\) to find \({\bf{y'}}\left( 0 \right)\) in accordance with its control points

\(\left\{ {{{\bf{p}}_{\bf{3}}}{\bf{,}}{{\bf{p}}_{\bf{4}}}{\bf{,}}{{\bf{p}}_{\bf{5}}}{\bf{,}}{{\bf{p}}_{\bf{6}}}} \right\}\)

Replace \({{\bf{p}}_0}\) by \({{\bf{p}}_3}\) and \({{\bf{p}}_1}\) by \({{\bf{p}}_4}\).

So, \({\bf{y}}'\left( 0 \right) = - 3{{\bf{p}}_3} + 3{{\bf{p}}_4}\).

Apply \({C^1}\) continuity rule

According to \({C^1}\) continuity, \({\bf{x}}'\left( 0 \right) = {\bf{y}}'\left( 0 \right)\), that is shown below:

\(\begin{array}{c} - 3{{\bf{p}}_3} + 3{{\bf{p}}_4} = - 3{{\bf{p}}_0} + 3{{\bf{p}}_1}\\{{\bf{p}}_3} = \frac{{\left( {{{\bf{p}}_4} + {{\bf{p}}_2}} \right)}}{2}\end{array}\)

Draw a conclusion

It is concluded that\({{\bf{p}}_3}\)is the midpoint of the line joining\({{\bf{p}}_4}\)and\({{\bf{p}}_2}\)as\({{\bf{p}}_3} = \frac{{\left( {{{\bf{p}}_4} + {{\bf{p}}_2}} \right)}}{2}\), which is the required condition of part (a).

If\({\bf{x}}'\left( 1 \right) = {\bf{y}}'\left( 0 \right) = 0\), \({{\bf{p}}_2} = {{\bf{p}}_3} = {{\bf{p}}_4}\), as\({\bf{x'}}\left( 1 \right) = - 3{{\bf{p}}_2} + 3{{\bf{p}}_3}\), and \({\bf{y'}}\left( 0 \right) = - 3{{\bf{p}}_3} + 3{{\bf{p}}_4}\).

This concludes that the line joining \({{\bf{p}}_4}\) and \({{\bf{p}}_2}\) is \({{\bf{p}}_3}\) only, which is the required condition of part (b).