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Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be cubic B茅zier 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

Expert verified

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

01

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

02

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 \(x'\left( 0 \right) = - 3{p_0} + 3{p_1}\).

03

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

04

Apply \({C^1}\) continuity rule

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

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

05

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).

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Most popular questions from this chapter

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the 鈥渓eft鈥 side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.

Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d so that \(\left( {f:d} \right)\) the hyperplane H described in the exercise.

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Question: 23. Let \({{\bf{v}}_1} = \left( \begin{array}{l}1\\1\end{array} \right)\), \({{\bf{v}}_2} = \left( \begin{array}{l}3\\0\end{array} \right)\), \({{\bf{v}}_3} = \left( \begin{array}{l}5\\3\end{array} \right)\) and \({\bf{p}} = \left( \begin{array}{l}4\\1\end{array} \right)\). Find a hyperplane \(f:d\) (in this case, a line) that strictly separates \({\bf{p}}\) from \({\rm{conv}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\).

The parametric vector form of a B-spline curve was defined in the Practice Problems as

\({\bf{x}}\left( t \right) = \frac{1}{6}\left[ \begin{array}{l}{\left( {1 - t} \right)^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}\end{array} \right]\;\), for \(0 \le t \le 1\) where \({{\bf{p}}_o}\) , \({{\bf{p}}_1}\), \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\) are the control points.

a. Show that for \(0 \le t \le 1\), \({\bf{x}}\left( t \right)\) is in the convex hull of the control points.

b. Suppose that a B-spline curve \({\bf{x}}\left( t \right)\)is translated to \({\bf{x}}\left( t \right) + {\bf{b}}\) (as in Exercise 1). Show that this new curve is again a B-spline.

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