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SCIENTIA SINICA Informationis, Volume 47 , Issue 12 : 1694-1708(2017) https://doi.org/10.1360/N112016-00240

Construction of a class of quintic curves with rational offsets

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  • ReceivedOct 9, 2016
  • AcceptedDec 11, 2016
  • PublishedMay 18, 2017

Abstract


Funded by

国家自然科学基金(61272300,61100084)

浙江省一流学科A类 (浙江财经大学统计学)资助 和浙江省教育厅科研基金(Y201223321)


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  • Figure 1

    (Color online) A necessary and sufficient condition on the control polygon of a class I quintic OR curve: there are points $\boldsymbol{Q}_i$, $i=0,1,2$, such that $\Delta\boldsymbol{P}_0~:~4(\boldsymbol{Q}_0~-~\boldsymbol{P}_1)~=~2(\boldsymbol{P}_2~-~\boldsymbol{Q}_0)~:~3(\boldsymbol{Q}_1~-~\boldsymbol{P}_2)~=~3(\boldsymbol{P}_3~-~\boldsymbol{Q}_1)~:~2(\boldsymbol{Q}_2~-~\boldsymbol{P}_3)~=~4(\boldsymbol{P}_4~-~\boldsymbol{Q}_2)~:~\Delta\boldsymbol{P}_4$, where the ratio determines the parameter value of the cusp

  • Figure 2

    (Color online) $C^1$ Hermite interpolation using class I quintic curves with rational offsets. Given $\boldsymbol{P}_0~=~0,~\boldsymbol{P}_1~=~-1~+~2{\rm{i}},~\boldsymbol{P}_4~=~10~+~4{\rm{i}},~\boldsymbol{P}_5~=~8~+~{\rm{i}}$, and $a_0~=~2$, we get four class I quintic OR curves: (a) $\boldsymbol{P}_2\approx~1.187~+~5.650{\rm{i}}$, $\boldsymbol{P}_3~\approx~7.125~+~6.050{\rm{i}}$; (b) $\boldsymbol{P}_2~\approx~2.858~+~8.064{\rm{i}}$, $\boldsymbol{P}_3~\approx~8.079~+~7.429{\rm{i}}$; (c) $\boldsymbol{P}_2~\approx~10.193~-~3.358{\rm{i}}$, $\boldsymbol{P}_3~\approx~12.271~+~0.903{\rm{i}}$; (d) $\boldsymbol{P}_2~\approx~-~2.470~-~3.361{\rm{i}}$, $\boldsymbol{P}_3~\approx~5.035~+~0.901{\rm{i}}$

  • Figure 3

    (Color online) Selection of the parameter $a_0$ can be used to adjust the shape of the resultant class I quintic OR curve

  • Figure 4

    (Color online) $C^1$ Hermite interpolation using class II quintic OR curves. Given $\boldsymbol{P}_i$, $i=0,1,4,5$, and a real parameter $a_0$, there is a unique quintic OR curve meeting the condition

  •   

    Algorithm 1 Construct class I

    Require: 复数表示的两端控制顶点 $\boldsymbol{P}_i$, $i=0,1,4,5$; 一个实参数 $a_0$.

    Output: 复数表示的4组控制顶点$\boldsymbol{P}_2^{(i)}$, $\boldsymbol{P}_3^{(i)}$, $i=0,\ldots,3$.

    计算$\boldsymbol{A}~=~\frac{5}{a_0}\Delta\boldsymbol{P}_0$, $\boldsymbol{D}~=~5\Delta\boldsymbol{P}_4$;

    构造关于$\boldsymbol{B}$和$\boldsymbol{C}$的二元四次方程(10);

    构造关于$\boldsymbol{B}$和$\boldsymbol{C}$的二元一次方程(9);

    用数值求解方法求解(9)和(10)构成的非线性方程组, 求解得到$\boldsymbol{B}^{(i)}$和$\boldsymbol{C}^{(i)}$, $i=0,\ldots,3$;

    代入(8)求得$\boldsymbol{P}_2^{(i)}$, $\boldsymbol{P}_3^{(i)}$, $i=0,\ldots,3$.

  •   

    Algorithm 2 Construct class II

    Require: 复数表示的两端控制顶点 $\boldsymbol{P}_i$, $i=0,1,4,5$, 一个实参数 $a_0$.

    Output: 复数表示的控制顶点$\boldsymbol{P}_2$, $\boldsymbol{P}_3$.

    计算$\boldsymbol{z}_0~=~\frac{5\Delta\boldsymbol{P}_0}{a_0}$, $\boldsymbol{z}_1~=~5\Delta\boldsymbol{P}_4$;

    构造二元一次复方程(12);

    将方程(12)的实部和虚部分解, 得二元一次实方程组并求解$a_1$, $a_2$;

    代入(11)求得控制顶点$\boldsymbol{P}_2$, $\boldsymbol{P}_3$.