Question

Let \({\bf{x}}\left( t \right)\) and \(y\left( t \right)\) be segments of a B-spline as in Exercise 6. Show that the curve has \({C^2}\) continuity (as well as \({C^1}\) continuity) at \({\bf{x}}\left( 1 \right)\). That is, show that \({\bf{x}}''\left( 1 \right) = {\bf{y}}''\left( 0 \right)\). This higher-order continuity is desirable in CAD applications such as automotive body design, since the curves and surfaces appear much smoother. However, B-splines require three times the computation of Bézier curves, for curves of comparable length. For surfaces, B-splines require nine times the computation of Bézier surfaces. Programmers often choose Bézier surfaces for applications (such as an airplane cockpit simulator) that require real-time rendering.

Short answer

As \({\bf{y}}''(0) = {\bf{x}}''\left( 1 \right)\), the curves have\({C^2}\) continuity.

Step-by-step solution

Step 1:Describe the given information

It is given that\({\bf{x}}\left( t \right)\)and\({\bf{y}}\left( t \right)\)are the B-Spline curves their combined curve has\({C^2}\)continuity including\({C^1}\)continuityat\({\bf{x}}\left( 1 \right)\).

It has already been shown that \({\bf{x}}''(0) = {{\bf{p}}_o} - 2{{\bf{p}}_1} + {{\bf{p}}_2}\), \({\bf{x}}''(1) = {{\bf{p}}_1} - 2{{\bf{p}}_2} + {{\bf{p}}_3}\)and if \({\bf{x}}''\left( 0 \right)\) is the control point of \({\bf{y}}\left( t \right)\), then \({\bf{y}}''(0) = {{\bf{p}}_1} - 2{{\bf{p}}_2} + {{\bf{p}}_3}\)which is also equal to \({\bf{x}}''(1)\).

Draw a conclusion

As \({\bf{y}}''(0) = {\bf{x}}''\left( 1 \right)\), it can be concluded that the curves have\({C^2}\) continuity.