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Chapter 8: Q8E (page 437)

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 B茅zier curves, for curves of comparable length. For surfaces, B-splines require nine times the computation of B茅zier surfaces. Programmers often choose B茅zier surfaces for applications (such as an airplane cockpit simulator) that require real-time rendering.

Short Answer

Expert verified

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

Step by step solution

01

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 \({\bf{x}}''(1)\).

02

Draw a conclusion

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

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Most popular questions from this chapter

In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{aligned}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{aligned}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each of the given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{8}}\\{\bf{4}}\end{aligned}} \right)\)

b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{6}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right)\)

c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{ - {\bf{1}}}\\{ - {\bf{5}}}\end{aligned}} \right)\)

Explain why a cubic Bezier curve is completely determined by \({\mathop{\rm x}\nolimits} \left( 0 \right)\), \(x'\left( 0 \right)\), \({\mathop{\rm x}\nolimits} \left( 1 \right)\), and \(x'\left( 1 \right)\).

Question: 29. Prove that the open ball \(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x:}}\left\| {{\rm{x - p}}} \right\| < \delta } \right\}\)is a convex set. (Hint: Use the Triangle Inequality).

Repeat Exercise 25 with\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\{\bf{2}}\\{ - {\bf{4}}}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{8}}\\{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{{\bf{10}}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{8}}\end{array}} \right]\), and \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{.{\bf{9}}}\\{{\bf{2}}.{\bf{0}}}\\{ - {\bf{3}}.{\bf{7}}}\end{array}} \right]\).

Suppose that\(\left\{ {{p_1},{p_2},{p_3}} \right\}\)is an affinely independent set in\({\mathbb{R}^{\bf{n}}}\)and q is an arbitrary point in\({\mathbb{R}^{\bf{n}}}\). Show that the translated set\(\left\{ {{p_1} + q,{p_2} + q,{p_3} + {\bf{q}}} \right\}\)is also affinely independent.
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