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发布时间: 2018-12-16 |
计算机图形学 |
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收稿日期: 2018-05-21; 修回日期: 2018-07-17
基金项目: 国家自然科学基金项目(61861040);甘肃省科技基金项目(17YF1FA119);甘肃省教育厅科技成果转化基金项目(2017D-09)
第一作者简介:
汪凯, 1993年生, 男, 硕士研究生, 主要研究方向为计算机图形学、图形处理。E-mail:616688448@qq.com;
张贵仓, 男, 教授, 主要研究方向为计算机图形学、图形处理。E-mail:zhanggc@nwnu.edu.cn; 龚进慧, 女, 硕士研究生, 主要研究方向为图像处理。E-mail:321256487@qq.com.
中图法分类号: TP391.72
文献标识码: A
文章编号: 1006-8961(2018)12-1910-15
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摘要
目的
为了使扩展的曲线曲面保留传统Bézier方法以及B样条方法良好性质的同时,具备保形性、形状可调性、高阶连续性以及广泛的应用性,本文在拟扩展切比雪夫空间利用开花的性质构造了一组最优规范全正基,并利用该基进行曲线曲面构造。方法
首先构造一组最优规范全正基,并给出该基生成的拟三次TC-Bézier曲线的割角算法;接着利用最优规范全正基的线性组合构造拟三次均匀TC-B样条基,根据曲线的性质假设拟三次均匀B样条基函数具有规范性和
关键词
拟扩展切比雪夫空间; 最优规范全正基; 全正性; 高阶连续性; 保形性
Abstract
Objective
The Bézier and B-spline curves play an important role in traditional geometric design. With the development of the geometric industry over the recent years, the traditional Bézier and B-spline curves cannot meet people's needs due to defects. At the same time, many rational forms of Bézier curves are proposed, which solve the problems faced by traditional methods. However, rational methods have not only progressive problems, but also employ the improper use of weight factors, which can be destructived to the curve and surface design. In view of the abovementioned problems, a large number of Bernstein-like and B-spline-like basis functions with shape parameters are proposed. These methods are mainly constructed in trigonometric, hyperbolic and exponential function spaces, a combination of said spaces, and polynomial space. Although many improved methods are available, these methods are rarely applied in solving practical problems. In the final analysis, these methods increase the flexibility of the curve by adding shape parameters, compared with the traditional Bézier and B-spline methods. However, the method itself does not have the ability to replace the traditional method. Several aspects still need improvement. For example, the majority of these methods only discuss basic properties, such as non-negativity, partition of unity, symmetry, and linear independence. Shape preservation, total positivity, and variation diminishing are often overlooked, which are important properties for curve design. However, the basis function, which has total positivity, will ensure that the related curve contains variation diminishing and shape preservation. Therefore, possessing total positivity is highly important for basis function. In addition, constructing cubic curves and surfaces remains the main method among the improved methods. In general, these improved methods have
Key words
quasi extended Chebyshev space; optimal normalized totally positive basis; totally positivity; high-order continuity; shape preserving
0 引言
Bézier曲线以及B样条曲线在传统几何设计中具有举足轻重的作用。近年来,随着几何工业的发展,传统Bézier曲线以及B样条曲线因其本身的缺陷已经很难满足人们的需要。与此同时许多有理形式的Bézier曲线[1-2]被提出来,这解决了传统方法的问题,但有理化方法不仅存在渐进问题,而且权因子的使用不当会对曲线曲面设计产生一定的破坏性[3]。鉴于上述问题,大量带形状参数的类Bernstein基或类B样条基孕育而出,主要集中在三角函数空间[4-9]、双曲函数空间[10-11]、指数函数空间[12-13],以及该类空间与多项式空间的组合空间[14-18]等。
尽管改进的方法有很多,但是这些方法在实际问题的解决中却应用得很少。归根原因,与传统Bézier方法和B样条方法相比,这些方法虽然都通过添加形状参数增加了曲线的灵活性,在某些方面占有优势,但是其方法本身并不具备替代传统方法的能力,需要改进的地方还有很多。比如,这些方法大都只讨论了凸包性、规范性、几何不变性以及对称性等一些基本性质,而传统Bézier方法和B样条方法所拥有的像全正性、变差缩减性和保形性等重要性质往往被忽略。而函数空间基函数的全正性可保证生成曲线具有变差缩减性,也可保证大量的保形性质,并且其中的B基(最优规范全正基)具有最佳的保形效果[19-20]。因此是否具有全正性至关重要[21]。又比如,基于多项式空间构造的曲线曲面不能精确表示除抛物线以外的圆锥曲线;较高的基函数次数会增加曲线曲面的计算复杂度等。
此外,在改进的方法中以2次和3次曲线曲面为主,在通常情况下,这些方法构造的曲线曲面均可达到
由于拟扩展切比雪夫空间(QEC)具有适合构造B基的开花性质,成为适合几何设计的最大一类空间[22-27]。另外,基于三角函数空间的类Bézier基和类B样条基在保形设计中具有巨大的潜力[28-31],并且三角函数空间中的sin
1 B基的构造
文献[32]给出了当
$ \left\{ \begin{array}{l} {T_0}\left( t \right) = \left( {1 - \sin t} \right)\left( {1 - \alpha \sin t} \right)\\ {T_1}\left( t \right) = \left( {1 + \alpha } \right)\sin t\left( {1 - \sin t} \right)\\ {T_2}\left( t \right) = \left( {1 + \beta } \right)\cos t\left( {1 - \cos t} \right)\\ {T_3}\left( t \right) = \left( {1 - \cos t} \right)\left( {1 - \beta \cos t} \right) \end{array} \right. $ | (1) |
并分析了其具有非负性、权性、拟对称性、单峰性以及端点性质。但实际上在
对任意的
$ \mathit{\Phi }\left( t \right) = \left( \begin{array}{l} {\sin ^2}t,\left( {1 - \sin t} \right)\left( {1 - \alpha \sin t} \right),\\ \left( {1 - \cos t} \right)\left( {1 - \beta \cos t} \right) \end{array} \right) $ |
由文献[22]的定理3.1,只需证明函数空间
$ {\rm{D}}{\mathit{\boldsymbol{T}}_{\alpha ,\beta }} = \left\{ \begin{array}{l} 2\sin t\cos t,2\alpha \sin t\cos t - \left( {\alpha + 1} \right)\cos t,\\ \left( {\beta + 1} \right)\sin t - 2\beta \sin t\cos t \end{array} \right\} $ |
为3维QEC空间。
定理1 对于任意的
证明对任意的
$ \begin{array}{*{20}{c}} {{\xi _0}\left[ {2\sin t\cos t} \right] + }\\ {{\xi _1}\left[ {2\alpha \sin t\cos t - \cos t - \alpha \cos t} \right] + }\\ {{\xi _2}\left[ {\left( {\beta + 1} \right)\cos t - 2\beta \sin t\cos t} \right] = 0} \end{array} $ | (2) |
当
当
最后可得
下面先证明
$ \begin{array}{*{20}{c}} {u\left( t \right) = }\\ {{{\left[ {\frac{{2\alpha \sin t\cos t - \left( {\alpha + 1} \right)\cos t}}{{\sin t\cos t}}} \right]}^\prime } = \left( {\alpha + 1} \right)\frac{{\cos t}}{{{{\sin }^2}t}}} \end{array} $ |
$ \begin{array}{*{20}{c}} {v\left( t \right) = }\\ {{{\left[ {\frac{{\left( {\beta + 1} \right)\sin t - 2\beta \sin t\cos t}}{{\sin t\cos t}}} \right]}^\prime } = \left( {\beta + 1} \right)\frac{{\sin t}}{{{{\cos }^2}t}}} \end{array} $ |
直接计算得
$ u'\left( t \right) = \left( {\alpha + 1} \right)\frac{{ - 1 - {{\cos }^2}t}}{{{{\sin }^3}t}} $ |
$ v'\left( t \right) = \left( {\beta + 1} \right)\frac{{1 + {{\sin }^2}t}}{{{{\cos }^3}t}} $ |
因此,关于函数
$ \begin{array}{*{20}{c}} {W\left( {u,v} \right)\left( t \right) = u\left( t \right)v'\left( t \right) - u'\left( t \right)v\left( t \right) = }\\ {\left( {\alpha + 1} \right)\left( {\beta + 1} \right)\frac{3}{{{{\sin }^2}t \cdot {{\cos }^2}t}} > 0,\forall t \in \left( {0,{\rm{ \mathsf{ π} }}/2} \right)} \end{array} $ |
对
$ {w_0}\left( t \right) = 2\sin t\cos t $ |
$ {w_1}\left( t \right) = Au\left( t \right) + Bv\left( t \right) $ |
$ {w_2}\left( t \right) = C\frac{{W\left( {u,v} \right)\left( t \right)}}{{{{\left[ {Au\left( t \right) + Bv\left( t \right)} \right]}^2}}} $ |
式中,
$ {\mathit{u}_0}\left( t \right) = {w_0}\left( t \right) $ |
$ {u_1}\left( t \right) = {w_0}\left( t \right)\int_a^t {{w_1}\left( {{t_1}} \right){\rm{d}}{t_1}} $ |
$ {u_2}\left( t \right) = {w_0}\left( t \right)\int_a^t {{w_1}\left( {{t_1}} \right)\int_a^{{t_1}} {{w_2}\left( {{t_2}} \right){\rm{d}}{t_2}{\rm{d}}{t_1}} } $ |
可以验证,函数
接着,我们证明
$ \begin{array}{*{20}{c}} {F\left( t \right) = {C_1}\left[ {2\sin t\cos t} \right] + }\\ {{C_2}\left[ {2\alpha \sin t\cos t - \left( {\alpha + 1} \right)\cos t} \right] + }\\ {{C_3}\left[ {\left( {\beta + 1} \right)\sin t - 2\beta \sin t\cos t} \right]} \end{array} $ |
式中,
假设
在此情况下,如果
如果
如果
$ \begin{array}{*{20}{c}} {F\left( t \right) = {C_1}\left[ {2\sin t\cos t} \right] + }\\ {{C_3}\left[ {\left( {\beta + 1} \right)\sin t - 2\beta \sin t\cos t} \right] = }\\ {\sin t\left[ {2{C_1}\cos t + {C_3}\left( {1 + \beta - 2\beta \cos t} \right)} \right]} \end{array} $ |
很明显
$ f\left( x \right) = 1 + \beta - 2\beta \cos \;t > 1 + \beta - 2\beta = 1 - \beta \ge 0 $ |
所以
如果
$ \begin{array}{*{20}{c}} {g\left( t \right) = 2{C_1}\cos t + {C_3}\left( {1 + \beta - 2\beta \cos t} \right)}\\ {g'\left( t \right) = 2\sin t\left( {\beta {C_3} - {C_1}} \right)} \end{array} $ |
可知当
可知当
可见
由于函数空间
定理2 对于任意的
$ \left\{ \begin{array}{l} {T_0}\left( t \right) = \left( {1 - \sin t} \right)\left( {1 - \alpha \sin t} \right)\\ {T_1}\left( t \right) = \left( {1 + \alpha } \right)\sin t\left( {1 - \sin t} \right)\\ {T_2}\left( t \right) = \left( {1 + \beta } \right)\cos t\left( {1 - \cos t} \right)\\ {T_3}\left( t \right) = \left( {1 - \cos t} \right)\left( {1 - \beta \cos t} \right) \end{array} \right. $ | (3) |
证明对于任意的
$ \begin{array}{*{20}{c}} {\mathit{\Phi }\left( 0 \right) = \left( {0,1,0} \right)}&{\mathit{\Phi }\left( {{\rm{ \mathsf{ π} }}/2} \right) = \left( {1,0,1} \right)} \end{array} $ |
$ \mathit{\Phi '}\left( 0 \right) = \left( {0, - \left( {\alpha + 1} \right),0} \right) $ |
$ \mathit{\Phi '}\left( {{\rm{ \mathsf{ π} }}/2} \right) = \left( {0,0,1 + \beta } \right) $ |
$ \mathit{\Phi ''}\left( 0 \right) = \left( {2,2\alpha ,1 - \beta } \right) $ |
$ \mathit{\Phi ''}\left( {{\rm{ \mathsf{ π} }}/2} \right) = \left( { - 2,1 - \alpha ,2\beta } \right) $ |
参考文献[34]中关于开花性质有(关于开花性质的定义可参见文献[34]的基础知识部分, 更加具体细节可参考[35-37],限于篇幅,这里将不再赘述)
$ {\mathit{\Pi }_0} = \mathit{\Phi }\left( 0 \right) = \left( {0,1,0} \right) $ |
$ {\mathit{\Pi }_3} = \mathit{\Phi }\left( {{\rm{ \mathsf{ π} }}/2} \right) = \left( {1,0,1} \right) $ |
$ \left\{ {{\mathit{\Pi }_1}} \right\} = Os{c_1}\mathit{\Phi }\left( 0 \right) \cap Os{c_2}\mathit{\Phi }\left( {{\rm{ \mathsf{ π} }}/2} \right) = \left( {0,0,0} \right) $ |
$ \left\{ {{\mathit{\Pi }_2}} \right\} = Os{c_2}\mathit{\Phi }\left( 0 \right) \cap Os{c_1}\mathit{\Phi }\left( {{\rm{ \mathsf{ π} }}/2} \right) = \left( {1,0,0} \right) $ |
对
$ \left\{ \begin{array}{l} {T_2}\left( t \right) + {T_3}\left( t \right) = {\sin ^2}t\\ {T_0}\left( t \right) = \left( {1 - \sin t} \right)\left( {1 - \alpha \sin t} \right)\\ {T_3}\left( t \right) = \left( {1 - \cos t} \right)\left( {1 - \beta \cos t} \right) \end{array} \right. $ |
由上式连同
首先,证明
$ \sum\limits_{i = 0}^3 {{\xi _i}{T_i}\left( t \right)} = 0 $ | (4) |
两边对
$ \sum\limits_{i = 0}^3 {{\xi _i}{{T'}_i}\left( t \right)} = 0 $ | (5) |
把
$ \left\{ \begin{array}{l} {\xi _0} = 0\\ \left( {\alpha + 1} \right)\left( {{\xi _0} - {\xi _1}} \right) = 0 \end{array} \right. $ |
由此可得
其次可以很容易验证
因为
对于
因此文献[22]的定理2.18可知,在
2 拟三次TC-Bézier曲线
定义1 对于给定控制顶点
$ \begin{array}{*{20}{c}} {Q\left( t \right) = \sum\limits_{i = 0}^3 {{T_i}\left( t \right){P_i}} }\\ {t \in \left[ {0,{\rm{ \mathsf{ π} }}/2} \right];\;\;\;\alpha ,\beta \in \left[ {0,1} \right]} \end{array} $ | (6) |
为带两个形状参数
文献[32]给出了在
割角算法是生成曲线的一种稳定和高效的算法。为此,将拟三次TC-Bézier曲线写成式(7)的形式。图 1给出了割角算法的例子。
$ \begin{array}{*{20}{c}} {Q\left( t \right) = \left( {\begin{array}{*{20}{c}} {1 - {{\sin }^2}t}&{1 - {{\cos }^2}t} \end{array}} \right) \times }\\ {\left( {\begin{array}{*{20}{c}} {1 - \sin t}&{\sin t}&0\\ 0&{\cos t}&{1 - \cos t} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} {\frac{{1 - \alpha \sin t}}{{{{\cos }^2}t}}}&{\frac{{\sin t\left( {\alpha - \sin t} \right)}}{{{{\cos }^2}t}}}&0&0\\ 0&{\frac{{\cos t}}{{\sin t + \cos t}}}&{\frac{{\sin t}}{{\sin t + \cos t}}}&0\\ 0&0&{\frac{{\cos t\left( {\beta - \cos t} \right)}}{{{{\sin }^2}t}}}&{\frac{{1 - \beta \cos t}}{{{{\sin }^2}t}}} \end{array}} \right)\\\times \left( {\begin{array}{*{20}{c}} {{P_0}}\\ {{P_1}}\\ {{P_2}}\\ {{P_3}} \end{array}} \right)} \end{array} $ | (7) |
3 拟三次TC-B样条曲线曲面及应用
3.1 拟三次均匀TC-B样条基的构造
在式(3)的基础上,令欲构造的拟三次均匀TC-B样条基为
$ \left\{ \begin{array}{l} {N_0}\left( t \right) = {x_1}{T_3}\left( t \right)\\ {N_1}\left( t \right) = {x_2}{T_0}\left( t \right) + {x_3}{T_1}\left( t \right) + {x_4}{T_2}\left( t \right) + {x_5}{T_3}\left( t \right)\\ {N_2}\left( t \right) = {x_6}{T_0}\left( t \right) + {x_7}{T_1}\left( t \right) + {x_8}{T_2}\left( t \right) + {x_9}{T_3}\left( t \right)\\ {N_3}\left( t \right) = {x_{10}}{T_0}\left( t \right) \end{array} \right. $ | (8) |
式中,
为了确定待定系数的值,首先预设由式(8)定义的结构具有规范性、
$ \psi = 4 + 3\alpha + 3\beta + 2\alpha \beta $ |
$ {x_1} = {x_2} = \frac{{1 + \alpha }}{\psi } $ |
$ {x_3} = {x_8} = \frac{{2 + \alpha + \beta }}{\psi } $ |
$ {x_9} = {x_{10}} = \frac{{1 + \beta }}{\psi } $ |
$ {x_4} = {x_5} = {x_6} = {x_7} = \frac{{2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)}}{\psi } $ |
定义2 给定节点向量
传统三次B样条基函数为分段函数,这里定义的拟三次均匀TC-B样条基也可用分段形式给出。设
$ \begin{array}{*{20}{c}} {{B_i}\left( u \right) = }\\ {\left\{ \begin{array}{l} {N_{i,0}}\left( {\frac{{\rm{ \mathsf{ π} }}}{2} \cdot \frac{{u - {u_i}}}{h}} \right)\;\;\;\;\;\;\;u \in \left[ {{u_i},{u_{i + 1}}} \right)\\ {N_{i,1}}\left( {\frac{{\rm{ \mathsf{ π} }}}{2} \cdot \frac{{u - {u_{i + 1}}}}{h}} \right)\;\;\;\;\;u \in \left[ {{u_{i + 1}},{u_{i + 2}}} \right)\\ {N_{i,2}}\left( {\frac{{\rm{ \mathsf{ π} }}}{2} \cdot \frac{{u - {u_{i + 2}}}}{h}} \right)\;\;\;\;\;u \in \left[ {{u_{i + 2}},{u_{i + 3}}} \right)\\ {N_{i,3}}\left( {\frac{{\rm{ \mathsf{ π} }}}{2} \cdot \frac{{u - {u_{i + 3}}}}{h}} \right)\;\;\;\;\;u \in \left[ {{u_{i + 3}},{u_{i + 4}}} \right)\\ 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;u \notin \left[ {{u_i},{u_{i + 4}}} \right) \end{array} \right.} \end{array} $ | (9) |
式中,
为此,可以将在每个节点区间
由拟三次均匀TC-B样条基
引理1 以下等式成立
$ \left\{ \begin{array}{l} {x_1} + {x_5} + {x_9} = 1\\ {x_2} + {x_6} + {x_{10}} = 1\\ {x_3} + {x_7} = 1\\ {x_1} = {x_2}\\ {x_5} = {x_6}\\ {x_9} = {x_{10}}\\ \left( {1 + \beta } \right){x_1} = - \left( {1 + \alpha } \right){x_2} + \left( {1 + \alpha } \right){x_3}\\ - \left( {1 + \beta } \right){x_4} + \left( {1 + \beta } \right){x_5} = - \left( {1 + \alpha } \right){x_6} + \left( {1 + \alpha } \right){x_7}\\ - \left( {1 + \beta } \right){x_8} + \left( {1 + \beta } \right){x_9} = - \left( {1 + \alpha } \right){x_{10}}\\ 2\beta {x_1} = 2\alpha {x_2} - 2\left( {1 + \alpha } \right){x_3} + \left( {1 + \beta } \right){x_4} + \left( {1 - \beta } \right){x_5}\\ \left( {1 - \alpha } \right){x_2} + \left( {1 + \alpha } \right){x_3} - 2\left( {\beta + 1} \right){x_4} + 2\beta {x_5} = \\ 2\alpha {x_6} - 2\left( {1 + \alpha } \right){x_7} + \left( {1 + \beta } \right){x_8} + \left( {1 - \beta } \right){x_9}\\ \left( {1 + \alpha } \right){x_6} + \left( {1 + \alpha } \right){x_7} - 2\left( {1 + \beta } \right){x_8} + 2\beta {x_9} = 2\alpha {x_{10}} \end{array} \right. $ |
3.2 拟三次均匀TC-B样条基的性质
从拟三次均匀TC-B样条基函数的构造方法和B基的性质,容易推出拟三次均匀TC-B样条基函数具有非负性、规范性、拟对称性以及端点性质。下面证明拟三次均匀TC-B样条基函数具有线性无关性、全正性以及
定理3线性无关性。即对任意
证明:对任意
$ {\xi _0}{N_0}\left( t \right) + {\xi _1}{N_1}\left( t \right) + {\xi _2}{N_2}\left( t \right) + {\xi _3}{N_3}\left( t \right) = 0 $ |
整理有
$ \begin{array}{*{20}{c}} {\left( {{x_2}{\xi _1} + {x_6}{\xi _2} + {x_{10}}{\xi _3}} \right){T_0}\left( t \right) + }\\ {\left( {{x_3}{\xi _1} + {x_7}{\xi _2}} \right){T_1}\left( t \right) + }\\ {\left( {{x_4}{\xi _1} + {x_8}{\xi _2}} \right){T_2}\left( t \right) + }\\ {\left( {{x_1}{\xi _0} + {x_5}{\xi _1} + {x_9}{\xi _2}} \right){T_3}\left( t \right) = 0} \end{array} $ |
因为{
$ \left\{ \begin{array}{l} {x_2}{\xi _1} + {x_6}{\xi _2} + {x_{10}}{\xi _3} = 0\\ {x_3}{\xi _1} + {x_7}{\xi _2} = 0\\ {x_4}{\xi _1} + {x_8}{\xi _2} = 0\\ {x_1}{\xi _0} + {x_5}{\xi _1} + {x_9}{\xi _2} = 0 \end{array} \right. $ |
相应系数行列式有
$ \begin{array}{*{20}{c}} {\left| \mathit{\boldsymbol{D}} \right| = \left| {\begin{array}{*{20}{c}} 0&{{x_2}}&{{x_6}}&{{x_{10}}}\\ 0&{{x_3}}&{{x_7}}&0\\ 0&{{x_4}}&{{x_8}}&0\\ {{x_1}}&{{x_5}}&{{x_9}}&0 \end{array}} \right| = - {x_1}{x_{10}}\left( {{x_3}{x_8} - {x_4}{x_7}} \right) = }\\ {{x_1}{x_{10}}\frac{{4{{\left( {1 + \alpha } \right)}^2}{{\left( {1 + \beta } \right)}^2} - {{\left( {2 + \alpha + \beta } \right)}^2}}}{{{\psi ^2}}} = {x_1}{x_{10}} \cdot }\\ {\frac{{\left( {2\alpha \beta + \alpha + \beta } \right) \cdot \left[ {2\left( {1 + \alpha } \right)\left( {1 + \beta } \right) + 2 + \alpha + \beta } \right]}}{{{\psi ^2}}} > 0} \end{array} $ |
因此{
定理4 全正性。对任意
证明:对任意的
$ \left( {{N_0},{N_1},{N_2},{N_3}} \right) = \left( {{T_3},{T_2},{T_1},{T_0}} \right)\mathit{\boldsymbol{H}} $ |
其中转换矩阵为
$ \mathit{\boldsymbol{H}} = \left[ {\begin{array}{*{20}{c}} {{x_1}}&{{x_5}}&{{x_9}}&0\\ 0&{{x_4}}&{{x_8}}&0\\ 0&{{x_3}}&{{x_7}}&0\\ 0&{{x_2}}&{{x_6}}&{{x_{10}}} \end{array}} \right] $ |
由定理2可知,(
$ \begin{array}{l} \left| {\begin{array}{*{20}{c}} {{x_5}}&{{x_9}}\\ {{x_4}}&{{x_8}} \end{array}} \right| = {x_8} \cdot {x_5} - {x_9} \cdot {x_4} = \\ \frac{{2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)}}{\psi }\frac{{1 + \alpha }}{\psi } > 0 \end{array} $ |
$ \begin{array}{l} \left| {\begin{array}{*{20}{c}} {{x_5}}&{{x_9}}\\ {{x_3}}&{{x_7}} \end{array}} \right| = {x_7} \cdot {x_5} - {x_9} \cdot {x_3} = \frac{{{{\left[ {2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)} \right]}^2}}}{{{\psi ^2}}} - \\ \frac{{1 + \beta }}{\psi } \cdot \frac{{2 + \alpha + \beta }}{\psi } = \\ \left[ \begin{array}{l} \frac{{4{\alpha ^2}{\beta ^2} + 8{\alpha ^2}\beta + 4{\alpha ^2} + 8\alpha {\beta ^2}}}{{{\psi ^2}}} + \\ \frac{{15\alpha \beta + 7\alpha + 3{\beta ^2} + 5\beta + 2}}{{{\psi ^2}}} \end{array} \right] > 0 \end{array} $ |
$ \begin{array}{l} \left| {\begin{array}{*{20}{c}} {{x_5}}&{{x_9}}\\ {{x_2}}&{{x_6}} \end{array}} \right| = {x_6} \cdot {x_5} - {x_9} \cdot {x_2} = \frac{{{{\left[ {2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)} \right]}^2}}}{{{\psi ^2}}} - \\ \frac{{1 + \alpha }}{\psi } \cdot \frac{{1 + \beta }}{\psi } = \\ \frac{{\left( {1 + \alpha } \right)\left( {1 + \beta } \right)\left[ {4\left( {1 + \alpha } \right)\left( {1 + \beta } \right) - 1} \right]}}{{{\psi ^2}}} > 0 \end{array} $ |
$ \begin{array}{l} \left| {\begin{array}{*{20}{c}} {{x_4}}&{{x_8}}\\ {{x_3}}&{{x_7}} \end{array}} \right| = {x_4} \cdot {x_7} - {x_3} \cdot {x_8} = \\ \frac{{{{\left[ {2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)} \right]}^2}}}{{{\psi ^2}}} - \frac{{{{\left[ {2 + \alpha + \beta } \right]}^2}}}{{{\psi ^2}}} = \\ \frac{{\left( {2\alpha \beta + \alpha + \beta } \right)\left[ {2\left( {1 + \alpha } \right)\left( {1 + \beta } \right) + 2 + \alpha + \beta } \right]}}{{{\psi ^2}}} > 0 \end{array} $ |
$ \begin{array}{l} \left| {\begin{array}{*{20}{c}} {{x_4}}&{{x_8}}\\ {{x_2}}&{{x_6}} \end{array}} \right| = {x_4} \cdot {x_6} - {x_2} \cdot {x_8} = \\ \frac{{{{\left[ {2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)} \right]}^2}}}{{{\psi ^2}}} - \frac{{\left( {2 + \alpha + \beta } \right)\left( {1 + \alpha } \right)}}{{{\psi ^2}}} = \\ \left[ \begin{array}{l} \frac{{4{\alpha ^2}{\beta ^2} + 8{\alpha ^2}\beta + 4{\alpha ^2} + 8\alpha {\beta ^2}}}{{{\psi ^2}}} + \\ \frac{{16\alpha \beta + 8\alpha + 4{\beta ^2} + 8\beta + 2}}{{{\psi ^2}}} \end{array} \right] > 0 \end{array} $ |
$ \begin{array}{l} \left| {\begin{array}{*{20}{c}} {{x_3}}&{{x_7}}\\ {{x_2}}&{{x_6}} \end{array}} \right| = {x_3} \cdot {x_6} - {x_2} \cdot {x_7} = \frac{{2\left( {1 + \alpha } \right)\left( {1 + \beta } \right)}}{\psi } \cdot \\ \left( {\frac{{2 + \alpha + \beta }}{\psi } - \frac{{1 + \alpha }}{\psi }} \right) = \frac{{2\left( {1 + \alpha } \right){{\left( {1 + \beta } \right)}^2}}}{{{\psi ^2}}} > 0 \end{array} $ |
据此容易验证
定理5 连续性。对于均匀节点向量
证明:不失一般性,考虑拟三次均匀TC-B样条基
$ \left\{ \begin{array}{l} T_0^{\left( {2n - 1} \right)} = {\left( { - 1} \right)^n}\left[ {\left( {1 + \alpha } \right)\cos t + {2^{2n - 2}}\alpha \sin 2t} \right]\\ T_1^{\left( {2n - 1} \right)} = {\left( { - 1} \right)^n}\left( {1 + \alpha } \right)\left( { - \cos t + {2^{2n - 2}}\sin 2t} \right)\\ T_2^{\left( {2n - 1} \right)} = {\left( { - 1} \right)^n}\left( {1 + \beta } \right)\left( {\sin t - {2^{2n - 2}}\sin 2t} \right)\\ T_3^{\left( {2n - 1} \right)} = {\left( { - 1} \right)^n}\left[ { - \left( {\beta + 1} \right)\sin t + {2^{2n - 2}}\beta \sin 2t} \right] \end{array} \right. $ |
1) 当
$ \left\{ \begin{array}{l} {{T'}_0} = - \left( {1 + \alpha } \right)\cos t - \alpha \sin 2t\\ {{T'}_1} = \left( {1 + \alpha } \right)\left( {\cos t - \sin 2t} \right)\\ {{T'}_2} = \left( {1 + \beta } \right)\left( { - \sin t + \sin 2t} \right)\\ {{T'}_3} = \left( {1 + \beta } \right)\sin t - \beta \sin 2t \end{array} \right. $ |
显然成立。
2) 假设当
$ \left\{ \begin{array}{l} T_0^{\left( {2k - 1} \right)} = {\left( { - 1} \right)^k}\left[ {\left( {1 + \alpha } \right)\cos t + {2^{2k - 2}}\alpha \sin 2t} \right]\\ T_1^{\left( {2k - 1} \right)} = {\left( { - 1} \right)^k}\left( {1 + \alpha } \right)\left( { - \cos t + {2^{2k - 2}}\sin 2t} \right)\\ T_2^{\left( {2k - 1} \right)} = {\left( { - 1} \right)^k}\left( {1 + \beta } \right)\left( {\sin t - {2^{2k - 2}}\sin 2t} \right)\\ T_3^{\left( {2k - 1} \right)} = {\left( { - 1} \right)^k}\left[ { - \left( {\beta + 1} \right)\sin t + {2^{2k - 2}}\beta \sin 2t} \right] \end{array} \right. $ |
对式(3)对应B基的
$ \left\{ \begin{array}{l} {\left[ {T_0^{\left( {2k - 1} \right)}} \right]^{\prime \prime }} = {\left( { - 1} \right)^{k + 1}}\left[ {\left( {1 + \alpha } \right)\cos t + } \right.\\ \left. {{2^{2\left( {k + 1} \right) - 2}}\alpha \sin 2t} \right] = T_0^{\left( {2\left( {k + 1} \right) - 1} \right)}\\ {\left[ {T_1^{\left( {2k - 1} \right)}} \right]^{\prime \prime }} = {\left( { - 1} \right)^{k + 1}}\left( {1 + \alpha } \right) \cdot \\ \left( { - \cos t + {2^{2\left( {k + 1} \right) - 2}}\sin 2t} \right) = T_1^{\left( {2\left( {k + 1} \right) - 1} \right)}\\ {\left[ {T_2^{\left( {2k - 1} \right)}} \right]^{\prime \prime }} = {\left( { - 1} \right)^{k + 1}}\left( {1 + \beta } \right) \cdot \\ \left( {\sin t - {2^{2\left( {k + 1} \right) - 2}}\sin 2t} \right) = T_2^{\left( {2\left( {k + 1} \right) - 1} \right)}\\ {\left[ {T_3^{\left( {2k - 1} \right)}} \right]^{\prime \prime }} = {\left( { - 1} \right)^{k + 1}}\left[ { - \left( {\beta + 1} \right)\sin t + } \right.\\ \left. {{2^{2\left( {k + 1} \right) - 2}}\beta \sin 2t} \right] = T_3^{\left( {2\left( {k + 1} \right) - 1} \right)} \end{array} \right. $ |
显然对
$ \begin{array}{l} N_0^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = {x_1}T_3^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = \\ {\left( { - 1} \right)^n}\left[ { - \left( {\beta + 1} \right)} \right]\frac{{1 + \alpha }}{\psi } \end{array} $ |
$ \begin{array}{l} N_1^{\left( {2n - 1} \right)}\left( 0 \right) = {x_2}T_0^{\left( {2n - 1} \right)}\left( 0 \right) + {x_3}T_1^{\left( {2n - 1} \right)}\left( 0 \right) + \\ {x_4}T_2^{\left( {2n - 1} \right)}\left( 0 \right) + {x_5}T_3^{\left( {2n - 1} \right)}\left( 0 \right) = \\ {\left( { - 1} \right)^n}\left[ { - \left( {1 + \beta } \right)} \right]\frac{{1 + \alpha }}{\psi } \end{array} $ |
$ \begin{array}{l} N_1^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = {x_2}T_0^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) + {x_3}T_1^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) + \\ {x_4}T_2^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) + {x_5}T_3^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = 0 \end{array} $ |
$ \begin{array}{l} N_2^{\left( {2n - 1} \right)}\left( 0 \right) = {x_6}T_0^{\left( {2n - 1} \right)}\left( 0 \right) + {x_7}T_1^{\left( {2n - 1} \right)}\left( 0 \right) + \\ {x_8}T_2^{\left( {2n - 1} \right)}\left( 0 \right) + {x_9}T_3^{\left( {2n - 1} \right)}\left( 0 \right) = 0 \end{array} $ |
$ \begin{array}{l} N_2^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = \\ {x_6}T_0^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) + {x_7}T_1^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) + \\ {x_8}T_2^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) + {x_9}T_3^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = \\ {\left( { - 1} \right)^n}\left( {1 + \alpha } \right)\frac{{1 + \beta }}{\psi } \end{array} $ |
$ \begin{array}{l} N_3^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = {x_{10}}T_0^{\left( {2n - 1} \right)}\left( {{\rm{ \mathsf{ π} }}/2} \right) = 0\\ N_0^{\left( {2n - 1} \right)}\left( 0 \right) = {x_1}T_3^{\left( {2n - 1} \right)}\left( 0 \right) = 0 \end{array} $ |
显然有
3.3 拟三次均匀B样条曲线
3.3.1 曲线的定义与性质
定义3 给定控制顶点
$ Q\left( u \right) = \sum\limits_{i = 0}^n {{P_i}{B_i}\left( u \right)} ,u \in \left[ {{u_3},{u_{n + 1}}} \right] $ |
式中,
$ {R_i}\left( t \right) = Q\left( {u\left( t \right)} \right) = \sum\limits_{k = 0}^3 {{P_{i - k}}{N_k}\left( t \right)} $ | (10) |
式中,
由拟三次均匀TC-B样条基的性质,可以得到拟三次均匀TC-B样条曲线具有几何不变性与仿射不变性、拟对称性、端点性质以及凸包性,由拟三次均匀TC-B样条基的全正基又可推出拟三次均匀TC-B样条曲线具有变差缩减性,进而又可推出曲线具有保凸性。这说明曲线
与传统B样条方法相同,拟三次均匀TC-B样条曲线也具有局部调整性质,在不改变控制顶点的情况下,可以通过改变参数值对拟三次均匀TC-B样条曲线进行局部调整。为了方便讨论将式(10)写成
$ \begin{array}{*{20}{c}} {{R_i}\left( t \right) = }\\ {\left( {{x_2}{P_{i - 1}} + {x_6}{P_{i - 2}} + {x_{10}}{P_{i - 3}}} \right){T_0}\left( t \right) + }\\ {\left( {{x_3}{P_{i - 1}} + {x_7}{P_{i - 2}}} \right){T_1}\left( t \right) + }\\ {\left( {{x_4}{P_{i - 1}} + {x_8}{P_{i - 8}}} \right){T_2}\left( t \right) + }\\ {\left( {{x_1}{P_i} + {x_5}{P_{i - 1}} + {x_9}{P_{i - 2}}} \right){T_3}\left( t \right)} \end{array} $ | (11) |
故可知,若改变节点区间
相较于传统B样条曲线,拟三次均匀TC-B样条曲线具有更好的连续性,根据定理5可推出以下的连续性质。
定理6 对均匀节点向量
证明:不失一般性,只考虑拟三次均匀TC-B样条曲线在节点
$ \begin{array}{*{20}{c}} {Q\left( u \right) = {R_{i + 1}}\left( t \right) = {P_{i + 1}}{N_0}\left( t \right) + {P_i}{N_1}\left( t \right) + }\\ {{P_{i - 1}}{N_2}\left( t \right) + {P_{i - 2}}{N_3}\left( t \right),u \in \left[ {{u_{i + 1}},{u_{i + 2}}} \right]}\\ {Q\left( u \right) = {R_i}\left( t \right) = {P_i}{N_0}\left( t \right) + {P_{i - 1}}{N_1}\left( t \right) + }\\ {{P_{i - 2}}{N_2}\left( t \right) + {P_{i - 3}}{N_3}\left( t \right),u \in \left[ {{u_i},{u_{i + 1}}} \right]} \end{array} $ | (12) |
所以
$ \begin{array}{*{20}{c}} {Q\left( {u_{i + 1}^ - } \right) = {R_{i + 1}}\left( 0 \right) = }\\ {{P_{i + 1}}{N_0}\left( 0 \right) + {P_i}{N_1}\left( 0 \right) + {P_{i - 1}}{N_2}\left( 0 \right) + {P_{i - 2}}{N_3}\left( 0 \right)}\\ {Q\left( {u_{i + 1}^ + } \right) = {R_i}\left( {{\rm{ \mathsf{ π} }}/2} \right) = }\\ {{P_i}{N_0}\left( {{\rm{ \mathsf{ π} }}/2} \right) + {P_{i - 1}}{N_1}\left( {{\rm{ \mathsf{ π} }}/2} \right) + }\\ {{P_{i - 2}}{N_2}\left( {{\rm{ \mathsf{ π} }}/2} \right) + {P_{i - 3}}{N_3}\left( {{\rm{ \mathsf{ π} }}/2} \right)} \end{array} $ |
由定理5知道
3.3.2 曲线的应用
案例1 整圆的精确表示
拟三次均匀TC-B样条曲线也可精确地表示圆锥曲线,但与文献[32]中方法不同的是,它不需通过图像拼接或对称变换就可以一次生成整圆。当
案例2 旋转体的生成
图 5(a)(c)分别给出了在给定控制多边形的情况下,不同参数生成的花瓶旋转曲面的母线,其中图 5(a)的参数为
3.4 拟三次均匀TC-B样条曲面
3.4.1 曲面的定义与性质
定义4 给定控制顶点
$ \begin{array}{*{20}{c}} {Q\left( {u,v} \right) = }\\ {\sum\limits_{i = 0}^m {\sum\limits_{j = 0}^n {{P_{i,j}}{B_i}\left( {u;{\alpha _1},{\beta _1}} \right){B_j}\left( {v;{\alpha _2},{\beta _2}} \right)} } } \end{array} $ | (13) |
式中,参数
$ q\left( {u,v} \right) = \sum\limits_{k = i - 3}^i {\sum\limits_{l = j - 3}^j {{P_{k,l}}{B_k}\left( {u,{\alpha _1},{\beta _1}} \right){B_l}\left( {v,{\alpha _2},{\beta _2}} \right)} } $ |
也可以用局部参数表示成
$ \begin{array}{*{20}{c}} {{R_{ij}}\left( {s,t} \right) = q\left( {u\left( s \right),v\left( t \right)} \right) = }\\ {\sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k,j - l}}{N_k}\left( {s;{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } } \end{array} $ |
式中
定理7 拟三次均匀TC-B样条曲面片的基{
证明:对于曲面片非负性可由
对于规范性可由式(14)得到证明
$ \begin{array}{*{20}{c}} {\sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{N_k}\left( {s;{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } = }\\ {\sum\limits_{k = 0}^3 {{N_k}\left( {s;{\alpha _1},{\beta _1}} \right)} \sum\limits_{l = 0}^3 {{N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} = }\\ {\sum\limits_{k = 0}^3 {{N_k}\left( {s;{\alpha _1},{\beta _1}} \right)} = 1} \end{array} $ | (14) |
对于线性无关性,考虑对任意
$ \sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{\xi _{kl}}{N_k}\left( {s;{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } = 0 $ |
整理有
$ \sum\limits_{k = 0}^3 {{N_k}\left( {s;{\alpha _1},{\beta _1}} \right)} \sum\limits_{l = 3}^3 {{\xi _{kl}}{N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} = 0 $ |
而
$ \sum\limits_{l = 0}^3 {{\xi _{kl}}{N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} = 0\left( {k = 0,1,2,3} \right) $ |
所以
除了变差缩减性以外,拟三次均匀TC-B样条曲线的性质均可以推广到拟三次均匀TC-B样条曲面上,如仿射不变性、规范性、凸包性以及局部控制性等,传统B样条曲面也含有该类性质,且这些性质容易证明,限于篇幅这里就不再赘述,下面证明拟三次均匀TC-B样条曲面特有的高阶连续性,即具有
定理8 对均匀节点向量
证明:首先证明
$ \begin{array}{*{20}{c}} {{R_{ij}}\left( {s,t} \right) = q\left( {u\left( s \right),v\left( t \right)} \right) = }\\ {\sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k,j - l}}{N_k}\left( {s;{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } }\\ {\left( {u,v} \right) \subset \left[ {{u_i},{u_{i + 1}}} \right] \times \left[ {{v_i},{v_{i + 1}}} \right]} \end{array} $ |
$ \begin{array}{*{20}{c}} {{R_{i + 1,j}}\left( {s,t} \right) = q\left( {u\left( s \right),v\left( t \right)} \right) = }\\ {\sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k + 1,j - l}}{N_k}\left( {s;{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } }\\ {\left( {u,v} \right) \subset \left[ {{u_{i + 1}},{u_{i + 2}}} \right] \times \left[ {{v_i},{v_{i + 1}}} \right]} \end{array} $ |
所以
$ \begin{array}{l} q\left( {u_{i + 1}^ + ,v\left( t \right)} \right) = {R_{ij}}\left( {{\rm{ \mathsf{ π} /2}},t} \right) = \\ \sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k,j - l}}{N_k}\left( {{\rm{ \mathsf{ π} /2}};{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } \end{array} $ |
$ \begin{array}{l} q\left( {u_{i + 1}^ - ,v\left( t \right)} \right) = {R_{i + 1,j}}\left( {{\rm{0}},t} \right) = \\ \sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k + 1,j - l}}{N_k}\left( {s;{\alpha _1},{\beta _1}} \right){N_l}\left( {t;{\alpha _2},{\beta _2}} \right)} } \end{array} $ |
$ \begin{array}{l} \frac{{{\partial ^{2n - 1}}}}{{\partial {s^{2n - 1}}}}{R_{ij}}\left( {{\rm{ \mathsf{ π} }}/2,t} \right) = \\ \sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k,j - l}}{N_k}\left( {{\rm{ \mathsf{ π} }}/2;{\alpha _1},{\beta _1}} \right)N_l^{\left( {2n - 1} \right)}\left( {t;{\alpha _2},{\beta _2}} \right)} } \end{array} $ |
$ \begin{array}{l} \frac{{{\partial ^{2n - 1}}}}{{\partial {s^{2n - 1}}}}{R_{i + 1,,j}}\left( {0,t} \right) = \\ \sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k + 1,j - l}}{N_k}\left( {0;{\alpha _1},{\beta _1}} \right)N_l^{\left( {2n - 1} \right)}\left( {t;{\alpha _2},{\beta _2}} \right)} } \end{array} $ |
而由定理5的结论
$ \begin{array}{*{20}{c}} {\frac{{{\partial ^2}}}{{\partial {t^2}}}{R_{ij}}\left( {{\rm{ \mathsf{ π} }}/2,t} \right) = }\\ {\sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k,j - l}}{N_k}\left( {{\rm{ \mathsf{ π} }}/2;{\alpha _1},{\beta _1}} \right){{N''}_i}\left( {{\rm{ \mathsf{ π} }}/2;{\alpha _2},{\beta _2}} \right)} } } \end{array} $ |
$ \begin{array}{*{20}{c}} {\frac{{{\partial ^2}}}{{\partial {t^2}}}{R_{i + 1,,j}}\left( {0,t} \right) = }\\ {\sum\limits_{k = 0}^3 {\sum\limits_{l = 0}^3 {{P_{i - k + 1,j - l}}{N_k}\left( {0;{\alpha _1},{\beta _1}} \right){{N''}_i}\left( {0;{\alpha _2},{\beta _2}} \right)} } } \end{array} $ |
根据引理1有
3.4.2 曲面的应用
案例3 椭球面和球面的精确表示
拟三次均匀TC-B样条曲面可精确表示椭球面和球面。
设
$ \left\{ \begin{array}{l} x\left( {s,t} \right) = \sum\limits_{k = 0}^3 {\sum\limits_{l = 3}^3 {{N_k}\left( {{s_{k,l}};{\alpha _1},{\beta _1}} \right){N_l}\left( {{t_{k,l}};{\alpha _2},{\beta _2}} \right){x_{k,l}}} } \\ y\left( {s,t} \right) = \sum\limits_{k = 0}^3 {\sum\limits_{l = 3}^3 {{N_k}\left( {{s_{k,l}};{\alpha _1},{\beta _1}} \right){N_l}\left( {{t_{k,l}};{\alpha _2},{\beta _2}} \right){y_{k,l}}} } \\ z\left( {s,t} \right) = \sum\limits_{k = 0}^3 {\sum\limits_{l = 3}^3 {{N_k}\left( {{s_{k,l}};{\alpha _1},{\beta _1}} \right){N_l}\left( {{t_{k,l}};{\alpha _2},{\beta _2}} \right){z_{k,l}}} } \end{array} \right. $ |
而椭球面的参数方程为
$ \left\{ \begin{array}{l} x\left( {s,t} \right) = X + a\sin \left( s \right)\sin \left( t \right)\\ y\left( {s,t} \right) = Y + b\sin \left( s \right)\cos \left( t \right)\\ z\left( {s,t} \right) = Z + c\cos \left( s \right) \end{array} \right. $ |
当给定控制顶点
$ \left( {\begin{array}{*{20}{c}} {{P_1}}&{{P_2}}&{{P_3}}&{{P_4}}\\ {{P_5}}&{{P_5}}&{{P_5}}&{{P_5}}\\ {{P_3}}&{{P_4}}&{{P_1}}&{{P_2}}\\ {{P_6}}&{{P_6}}&{{P_6}}&{{P_6}} \end{array}} \right) $ |
式中,
$ \left( {\begin{array}{*{20}{c}} {{P_1}}&{{P_2}}&{{P_3}}&{{P_4}}&{{P_1}}&{{P_2}}&{{P_3}}\\ {{P_5}}&{{P_5}}&{{P_5}}&{{P_5}}&{{P_5}}&{{P_5}}&{{P_5}}\\ {{P_3}}&{{P_4}}&{{P_1}}&{{P_2}}&{{P_3}}&{{P_4}}&{{P_1}}\\ {{P_6}}&{{P_6}}&{{P_6}}&{{P_6}}&{{P_6}}&{{P_6}}&{{P_6}}\\ {{P_1}}&{{P_2}}&{{P_3}}&{{P_4}}&{{P_1}}&{{P_2}}&{{P_3}} \end{array}} \right) $ |
可以表示一个完整的椭球面;当
案例4 旋转曲面的直接生成
设
$ \begin{array}{*{20}{c}} {{P_{i0}}\left( {{x_i},0,z} \right),{P_{i1}}\left( {0, - {x_i},{z_i}} \right)}\\ {{P_{i2}}\left( { - {x_i},0,{z_i}} \right),{P_{i3}}\left( {0,{x_i},{z_i}} \right)}\\ {{P_{i4}} = {P_{i0}},{P_{i5}} = {P_{i1}},{P_{i6}} = {P_{i2}}}\\ {\left( {i = 0,1,2, \cdots ,6} \right)} \end{array} $ |
式中,
4 结论
传统文献对Bézier方法和B样条方法改进时,只关注于能否增加曲线灵活度以及是否能够逼近或精确表示某一类曲线、曲面,因而构造的曲线、曲面只保留了Bézier方法和B样条方法的一些基本性质,如凸包性、仿射不变性、对称性等,像变差缩减性、全正性、保形性等重要性质往往被忽略。鉴于传统改进方法的该类问题,本文从适合造型设计的保形性上出发,构造了一组最优规范全正基,并设计出具有高阶连续性的曲线曲面。另外还给出了大量的曲线曲面应用案例,这进一步说明了本文构造方法的适用性。为了设计出更加符合实际需求的造型,需要对曲线曲面的形状进行详细讨论,如尖点、拐点、重结点、凸性等,限于篇幅将另文叙述。
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