Question

Question: 6. A B-spline is built out of B-spline segments, described in Exercise 2. Let \({{\bf{p}}_{\bf{o}}}........{{\bf{p}}_{\bf{4}}}\)be control points. For \(0 \le t \le 1\), let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\)be determined by the geometry matrices \(\left( {{{\bf{p}}_{\bf{o}}}\,{{\bf{p}}_{\bf{1}}}\,{{\bf{p}}_{\bf{2}}}\,{{\bf{p}}_{\bf{3}}}} \right)\)and \(\left( {\,{{\bf{p}}_{\bf{1}}}\,\,{{\bf{p}}_{\bf{2}}}\,\,{{\bf{p}}_{\bf{3}}}\,{{\bf{p}}_{\bf{4}}}} \right)\)respectively. Notice how the two segments share three control points. The two segments do not overlap, however—they join at a common endpoint, close to \({{\bf{p}}_2}\) .

a. Show that the combined curve has \({G_0}\) continuity—that is,\({\bf{x}}\left( 1 \right) = {\bf{y}}\left( 0 \right)\).

b. Show that the curve has \({C^1}\) continuity at the join point, \({\bf{x}}\left( 1 \right)\). That is, show that \({\bf{x}}'\left( 1 \right) = {\bf{y}}'\left( 0 \right)\).

Short answer

a) Both \({\bf{x}}\left( 1 \right) = {\bf{y}}\left( 0 \right) = \left( {{{\bf{p}}_1} + 4{{\bf{p}}_2} + {{\bf{p}}_3}} \right)/6\).

b) It is shown that the B-spline is \({C^1}\) continuous at mid-point.

Step-by-step solution

Step 1:Use the parametric form of the B-spline curve with control points\(\left\{ {{{\bf{p}}_{\bf{0}}}{\bf{,}}{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\)

The parametric vector form of a B-spline curve is defined as \({\bf{x}}\left( t \right) = \frac{1}{6}\left( {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t\left( {1 - {t^2}} \right) - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right)\) for \(0 \le t \le 1\).

Step 2:Find the value \(x\left( 0 \right)\) and \(x\left( 1 \right)\)

As \({\bf{x}}\left( t \right) = \frac{1}{6}\left( {{{\left( {1 - t} \right)}^3}{{\bf{p}}_o} + \left( {3t\left( {1 - {t^2}} \right) - 3t + 4} \right){{\bf{p}}_1} + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}} \right)\).

So, \({\bf{x}}\left( 0 \right) = \left( {{{\bf{p}}_0} + 4{{\bf{p}}_1} + {{\bf{p}}_2}} \right)/6\)and \({\bf{x}}\left( 1 \right) = \left( {{{\bf{p}}_1} + 4{{\bf{p}}_2} + {{\bf{p}}_3}} \right)/6\).

Use \({\bf{x}}\left( 0 \right)\) to find \({\bf{y}}\left( 0 \right)\) in accordance with its control points \(\left\{ {{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}{\bf{,}}{{\bf{p}}_{\bf{4}}}} \right\}\)

Replace \({{\bf{p}}_0}\) by \({{\bf{p}}_1}\), \({{\bf{p}}_1}\) by \({{\bf{p}}_2}\), and \({{\bf{p}}_2}\) by \({{\bf{p}}_3}\).

So, \({\bf{y}}\left( 0 \right) = \left( {{{\bf{p}}_1} + 4{{\bf{p}}_2} + {{\bf{p}}_3}} \right)/6\).

Draw a conclusion

Both \({\bf{x}}\left( 1 \right) = {\bf{y}}\left( 0 \right) = \left( {{{\bf{p}}_1} + 4{{\bf{p}}_2} + {{\bf{p}}_3}} \right)/6\). This implies that B-spline is \({G^o}\) continuous at mid-point, as stated in part (a).

use the result of exercise 4 part (a)

If \({\bf{x'}}\left( 1 \right) = \left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)/2\)and \({\bf{x'}}\left( 0 \right) = \left( {{{\bf{p}}_2} - {{\bf{p}}_0}} \right)/2\), then \({\bf{y'}}\left( 0 \right) = {\bf{x'}}\left( 1 \right) = \left( {{{\bf{p}}_3} - {{\bf{p}}_1}} \right)/2\).

This concludes that B-spline is \({C^1}\) continuous at mid-point, as stated in part (b).