Construction of developable surfaces using generalized C-Bézier bases with shape parameters

Abstract

In this paper, we propose a novel method for constructing developable surfaces using generalized C-Bézier bases with shape parameters. Based on the duality between points and planes in 3D projective space, the generalized developable C-Bézier surfaces, whose shape can be adjusted by changing multiple shape parameters, are designed using control planes with extensional C-Bézier basis functions. With the shape parameters taking different values, a family of developable surfaces can be constructed, which keeps most of characteristics of classic developable Bézier surfaces. Furthermore, some interesting properties of the new developable surfaces, as well as the geometric continuity conditions between two adjacent generalized developable C-Bézier surfaces, are investigated. Finally, we illustrate the convenience and efficiency of the proposed methods by several convictive and representative numerical examples.

Appendix

1.1 A. The proof of the terminal properties for generalized C-Bézier basis functions

Proof

By induction on n. When n = 2, from (4) and (5), we obtain

$$ w_{0,2} (0;\,\alpha_{1} ) = 1,\;w_{2,2} (1;\,\alpha_{2} ) = \frac{{1 - \cos \alpha_{2} }}{{1 - \cos \alpha_{2} }} = 1, $$

$$ w_{1,2}^{(0)} (0;\,\alpha_{1} ,\alpha_{2} ) = w_{1,2} (0;\,\alpha_{1} ,\alpha_{2} ) = 0,\;w_{2,2}^{(0)} (0;\,\alpha_{2} ) = w_{2,2} (0;\,\alpha_{2} ) = \frac{1 - \cos 0}{{1 - \cos \alpha_{2} }} = 0, $$

$$ w_{{{0},{2}}}^{{(1)}} (0;\,\alpha_{{1}} ) = - \delta_{{{0},1}} w_{{{0},{1}}} (0;\alpha_{{1}} ) = - \delta_{{{0},1}} ,w_{{{2},{2}}}^{{(1)}} (0;\,\alpha_{{2}} ) = \delta_{1,1} w_{{1,{1}}} (0;\alpha_{2} ) = \delta_{1,1} \times 0 = 0, $$

$$ w_{0,2}^{(0)} (1;\,\alpha_{1} ) = w_{0,2} (1;\,\alpha_{1} ) = \frac{1 - \cos 0}{{1 - \cos \alpha_{1} }} = 0,\;w_{0,2}^{(1)} (1;\,\alpha_{1} ) = \left. {\frac{{\alpha_{1} \sin (\alpha_{1} - \alpha_{1} t)}}{{1 - \cos \alpha_{1} }}} \right|_{t = 1} = 0, $$

$$ w_{{{1},{2}}}^{{(0)}} ({1};\,\alpha_{{1}} ,\alpha_{{2}} ) = w_{{{1},{2}}} ({1};\,\alpha_{{1}} ,\alpha_{{2}} ) = \frac{{\cos (\alpha_{1} - \alpha_{1} ) - {\cos}\alpha_{1} }}{{1 - {\cos}\alpha_{1} }} - \frac{{1 - \cos (\alpha_{2} )}}{{1 - {\cos}\alpha_{2} }} = 0, $$

$$ w_{{{1},{2}}}^{{(1)}} ({0};\,\alpha_{{1}} ,\alpha_{{2}} ) = \left. {\left[ {\delta_{{0,1}} w_{{0,{1}}} (t;\alpha_{1} ) - \delta_{{1,1}} w_{{1,{1}}} (t;\alpha_{2} )} \right]} \right|_{t = 0} = \delta_{{0,1}} \times 1 - \delta_{{1,1}} \times 0 = \delta_{{0,1}} , $$

$$ w_{{{2},{2}}}^{{(2)}} ({0};\,\alpha_{{2}} ) = \delta_{{1,1}} w_{{{1},1}}^{{(1)}} (0;\,\alpha_{{2}} ) = \delta_{{1,1}} \delta (\alpha_{{2}} ),\,\;\delta (\alpha_{{2}} ) = \frac{{\alpha_{{2}} }}{{{\sin}\alpha_{2} }}. $$

Now assume the terminal properties in (9) hold for n = r. When n = r + 1, from formula (6) and the inductive hypothesis, for any $i = {0},{1}, \ldots ,r + 1$, we have

• a. for t = 0, $w_{0,r + 1} (0{\kern 1pt} ;\,\alpha_{1} ) = 1 - \int_{0}^{{0}} {\delta_{0,r - 1} w_{0,r - 1} {(}x;\alpha_{1} {\text{)d}}x} = 1,$

  $$ w_{i,r + 1}^{(0)} (0;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = w_{i,r + 1} (0;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = 0,\;i = 1,2, \ldots ,r,\;w_{i,r + 1}^{(0)} (0;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = w_{i,r + 1} (0;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = 0,\;i = 1,2, \ldots ,r, $$
  $$ w_{i,r + 1}^{(0)} (0;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = w_{i,r + 1} (0;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = 0,\;i = 1,2, \ldots ,r,\;w_{r + 1,r + 1}^{(0)} (0;\,\alpha_{r + 1} ) = w_{r + 1,r + 1} (0;\,\alpha_{r + 1} ) = 0, $$
  $$ \begin{aligned} w_{i,r + 1}^{(j)} (0;\,\alpha_{i} ,\alpha_{i + 1} ) & = \left. {\frac{{{\text{d}}^{j} }}{{{\text{d}}t^{j} }}\left[ {\int_{0}^{t} {\left[ {\delta_{i - 1,r} \tilde{w}_{i - 1,r} (x;\alpha_{i} ) - \delta_{i,r} \tilde{w}_{i,r} (x;\alpha_{i + 1} )} \right]{\text{d}}x} } \right]} \right|_{t = 0} \\ & = \left. {\left[ {\delta_{i - 1,r} \tilde{w}_{i - 1,r}^{(j - 1)} (t;\alpha_{i} ) - \delta_{i,r} \tilde{w}_{i,r}^{(j - 1)} (t;\alpha_{i + 1} )} \right]} \right|_{t = 0} \\ & = \delta_{i - 1,r} w_{i - 1,r}^{(j - 1)} (0;\alpha_{i} ,\alpha_{i} ) - \delta_{i,r} w_{i,r}^{(j - 1)} (0;\alpha_{i + 1} ,\alpha_{i + 1} ) \\ & = 0,\;\,j = {1},{2}, \ldots ,i - 1,\,\;i = 2,3, \ldots ,r - 1, \\ \end{aligned} $$
  $$ \begin{aligned} w_{r,r + 1}^{(j)} (0;\,\alpha_{r} ,\alpha_{{r + {1}}} ) & = \left. {\frac{{{\text{d}}^{j} }}{{{\text{d}}t^{j} }}\left[ {\int_{0}^{t} {\left[ {\delta_{r - 1,r} \tilde{w}_{r - 1,r} (x;\alpha_{r} ) - \delta_{r,r} w_{r,r} (x;\alpha_{r + 1} )} \right]{\text{d}}x} } \right]} \right|_{t = 0} \\ & = \left. {\left[ {\delta_{r - 1,r} \tilde{w}_{r - 1,r}^{(j - 1)} (t;\alpha_{r} ) - \delta_{r,r} w_{r,r}^{(j - 1)} (t;\alpha_{r + 1} )} \right]} \right|_{t = 0} \\ & = \delta_{r - 1,r} w_{r - 1,r}^{(j - 1)} (0;\alpha_{r} ,\alpha_{r} ) - \delta_{r,r} w_{r,r}^{(j - 1)} (0;\alpha_{r + 1} ) \\ & = 0,\,\;j = {1},{2}, \ldots ,r - 1, \\ \end{aligned} $$
  $$ w_{r + 1,r + 1}^{(j)} (0;\,\alpha_{r + 1} ) = \left. {\frac{{{\text{d}}^{j} }}{{{\text{d}}t^{j} }}\left[ {\int_{0}^{t} {\delta_{r,r} w_{r,r} {(}x;\alpha_{r + 1} {\text{)d}}x} } \right]} \right|_{t = 0} = \delta_{r,r} w_{r,r}^{(j - 1)} {(}0;\alpha_{r + 1} {)} = 0,\,\;j = {1},{2}, \ldots ,r; $$
  $$ \begin{aligned} w_{{{0},r + {1}}}^{{(1)}} (0;\,\alpha_{{1}} ) & = \left. {\frac{{\text{d}}}{{{\text{d}}t}}\left[ {1 - \int_{0}^{t} {\delta_{0,r} w_{0,r} {(}x;\alpha_{1} {\text{)d}}x} } \right]} \right|_{t = 0} = \left. {\left[ { - \delta_{0,r} w_{0,r} {(}t;\alpha_{1} {)}} \right]} \right|_{t = 0} = - \delta_{0,(r + 1) - 1} w_{0,r} {(0}{\kern 1pt} ;\alpha_{1} {)} \\ & = - \delta_{0,(r + 1) - 1} \times {1} = - \delta_{{0,(r + {1}) - 1}} ; \\ \end{aligned} $$

• b. for t = 1,

  $$ w_{r + 1,r + 1} (1;\,\alpha_{r + 1} ) = \delta_{r,r} \int_{0}^{1} {w_{r,r} {(}x;\alpha_{r + 1} {\text{)d}}x} = \left[ {\int_{0}^{1} {w_{r,r} {(}x;\alpha_{r + 1} {\text{)d}}x} } \right]^{ - 1} \int_{0}^{1} {w_{r,r} {(}x;\alpha_{r + 1} {\text{)d}}x} = 1\;, $$
  $$ \begin{gathered} w_{0,r + 1}^{{({0})}} (1;\,\alpha_{1} ) = w_{0,r + 1} (1;\,\alpha_{1} ) = 1 - \int_{0}^{{1}} {\delta_{0,r} w_{0,r} {(}x;\alpha_{1} {\text{)d}}x} \\ = {1} - \frac{{\int_{0}^{{1}} {w_{0,r} {(}x;\alpha_{1} {\text{)d}}x} }}{{\delta_{0,r} }} = {1} - \frac{{\int_{0}^{{1}} {w_{0,r} {(}x;\alpha_{1} {\text{)d}}x} }}{{\int_{0}^{{1}} {w_{0,r} {(}t;\alpha_{1} {\text{)d}}t} }} = 1 - 1 = 0, \\ \end{gathered} $$
  $$ w_{i,r + 1}^{(0)} (1;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = w_{i,r + 1} (1;\,\alpha_{i} ,\alpha_{{i + {1}}} ) = 0,\,\;i = 1,2, \ldots ,r, $$
  $$ w_{0,r + 1}^{(k)} (1;\,\alpha_{1} ) = \left. {\frac{{{\text{d}}^{k} }}{{{\text{d}}{\kern 1pt} t^{k} }}\left[ {1 - \int_{0}^{t} {\delta_{0,r} w_{0,r} {(}x;\alpha_{1} {\text{)d}}x} } \right]} \right|_{t = 1} = - \delta_{0,r} w_{0,r}^{(k - 1)} {(}0;\alpha_{1} {)} = 0,\,\;k = {1},{2}, \ldots ,r, $$
  $$ \begin{gathered} w_{1,r + 1}^{(k)} (1;\,\alpha_{1} ,\alpha_{2} ) = \left. {\frac{{{\text{d}}^{k} }}{{{\text{d}}{\kern 1pt} t^{k} }}\left[ {\int_{0}^{t} {\left[ {\delta_{0,r} w_{0,r} (x;\alpha_{1} ) - \delta_{1,r} \tilde{w}_{1,r} (x;\alpha_{2} )} \right]{\text{d}}x} } \right]} \right|_{t = 1} \\ = \left. {\left[ {\delta_{0,r} w_{0,r}^{(k - 1)} (t;\alpha_{1} ) - \delta_{1,r} \tilde{w}_{1,r}^{(k - 1)} (t;\alpha_{2} )} \right]} \right|_{t = 1} \\ = \delta_{0,r} w_{0,r}^{(k - 1)} (1;\alpha_{1} ) - \delta_{1,r} w_{1,r}^{(k - 1)} (1;\alpha_{2} ,\alpha_{2} ) \\ = 0,\,\;k = 1,2, \ldots ,r - 1{,} \\ \end{gathered} $$
  $$ \begin{gathered} w_{i,r + 1}^{(k)} (1;\,\alpha_{i} ,\alpha_{i + 1} ) = \left. {\frac{{{\text{d}}^{k} }}{{{\text{d}}{\kern 1pt} t^{k} }}\left[ {\int_{0}^{t} {\left[ {\delta_{i - 1,r} \tilde{w}_{i - 1,r} (x;\alpha_{i} ) - \delta_{i,r} \tilde{w}_{i,r} (x;\alpha_{i + 1} )} \right]{\text{d}}x} } \right]} \right|_{t = 1} \\ = \left. {\left[ {\delta_{i - 1,r} \tilde{w}_{i - 1,r}^{(k - 1)} (t;\alpha_{i} ) - \delta_{i,r} \tilde{w}_{i,r}^{(k - 1)} (t;\alpha_{i + 1} )} \right]} \right|_{t = 1} \\ = \delta_{i - 1,r} w_{i - 1,r}^{(k - 1)} (1;\alpha_{i} ,\alpha_{i} ) - \delta_{i,r} w_{i,r}^{(k - 1)} (1;\alpha_{i + 1} ,\alpha_{i + 1} ) \\ = 0,\,\;k = {1},{2}, \ldots ,r - i,\,\;i = {2},{3}, \ldots ,r - 1; \\ \end{gathered} $$

• c. for t = 0 and $i = j$,

  $$ \begin{gathered} w_{{{1},r + 1}}^{{({1})}} (0;\,\alpha_{{1}} ,\alpha_{{2}} ) = \left. {\frac{{{\text{d}}^{1} }}{{{\text{d}}t^{1} }}\left[ {\int_{0}^{t} {\left[ {\delta_{0,r} w_{0,r} (x;\alpha_{1} ) - \delta_{1,r} \tilde{w}_{1,r} (x;\alpha_{2} )} \right]{\text{d}}x} } \right]} \right|_{t = 0} \\ = \delta_{0,r} w_{0,r} (0;\alpha_{1} ) - \delta_{1,r} w_{1,r} (0;\alpha_{2} ,\alpha_{2} ) = \delta_{0,r} \times 1 - \delta_{1,r} \times 0 = \delta_{0,r} , \\ \end{gathered} $$
  $$ \begin{gathered} w_{i,r + 1}^{(i)} (0;\,\alpha_{i} ,\alpha_{i + 1} ) = \left. {\frac{{{\text{d}}^{i} }}{{{\text{d}}t^{i} }}\left[ {\int_{0}^{t} {\left[ {\delta_{i - 1,r} \tilde{w}_{i - 1,r} (x;\alpha_{i} ) - \delta_{i,r} \tilde{w}_{i,r} (x;\alpha_{i + 1} )} \right]{\text{d}}x} } \right]} \right|_{t = 0} \\ = \left. {\left[ {\delta_{i - 1,r} \tilde{w}_{i - 1,r}^{(i - 1)} (t;\alpha_{i} ) - \delta_{i,r} \tilde{w}_{i,r}^{(i - 1)} (t;\alpha_{i + 1} )} \right]} \right|_{t = 0} \\ = \delta_{i - 1,r} w_{i - 1,r}^{(i - 1)} (0;\alpha_{i} ,\alpha_{i} ) - \delta_{i,r} w_{i,r}^{(i - 1)} (0;\alpha_{i + 1} ,\alpha_{i + 1} ) \\ = \delta_{i - 1,r} \times \left[ {\delta_{{(i - {1}) - 1,r - 1}} \delta_{(i - 1) - 2,r - 2} \ldots \delta_{0,r - (i - 1)} } \right] - \delta_{i,r} \times {0} \\ = \delta_{i - 1,(r + 1) - 1} \delta_{i - 2,(r + 1) - 2} \ldots \delta_{0,(r + 1) - i} ,\,\;i = {2},{3}, \ldots ,r - 1, \\ \end{gathered} $$
  $$ \begin{gathered} w_{r,r + 1}^{(r)} (0;\,\alpha_{r} ,\alpha_{r + 1} ) = \left. {\frac{{{\text{d}}^{r} }}{{{\text{d}}t^{r} }}\left[ {\int_{0}^{t} {\left[ {\delta_{r - 1,r} \tilde{w}_{r - 1,r} (x;\alpha_{r} ) - \delta_{r,r} w_{r,r} (x;\alpha_{r + 1} )} \right]{\text{d}}x} } \right]} \right|_{t = 0} \\ = \left. {\left[ {\delta_{r - 1,r} \tilde{w}_{r - 1,r}^{(r - 1)} (t;\alpha_{r} ) - \delta_{r,r} w_{r,r}^{(r - 1)} (t;\alpha_{r + 1} )} \right]} \right|_{t = 0} \\ = \delta_{r - 1,r} w_{r - 1,r}^{(r - 1)} (0;\alpha_{r} ,\alpha_{r} ) - \delta_{r,r} w_{r,r}^{(r - 1)} (0;\alpha_{r + 1} ) \\ = \delta_{r - 1,r} \times \left[ {\delta_{{(r - {1}) - 1,r - 1}} \delta_{(r - 1) - 2,r - 2} \ldots \delta_{0,r - (r - 1)} } \right] - \delta_{r,r} \times 0 \\ = \delta_{r - 1,(r + 1) - 1} \delta_{r - 2,(r + 1) - 2} \cdots \delta_{0,(r + 1) - r} ; \\ \end{gathered} $$

• d. for $i = j = r + 1$,

  $$ \begin{gathered} w_{r + 1,r + 1}^{(r + 1)} (0;\,\alpha_{r + 1} ) = \left. {\frac{{{\text{d}}^{r} }}{{{\text{d}}{\kern 1pt} t^{r} }}\left[ {\delta_{r,r} w_{r,r} (t;\,\alpha_{r + 1} )} \right]} \right|_{t = 0} = \delta_{r,r} w_{r,r}^{(r)} (0;\,\alpha_{r + 1} ) \\ = \delta_{r,r} \left[ {\delta_{r - 1,r - 1} \delta_{r - 2,r - 2} \cdots \delta_{{{1},{1}}} \delta (\alpha_{r + 1} )} \right] \\ = \delta_{r,r} \delta_{r - 1,r - 1} \cdots \delta_{{{1},{1}}} \delta (\alpha_{r + 1} ),\;\,\delta (\alpha_{r + 1} ) = \alpha_{r + 1} {/}\sin \alpha_{r + 1} . \\ \end{gathered} $$

These indicate that the terminal properties listed in (9) also hold for n = r + 1. This ends the proof.

1.2 The proof of linear independence for generalized C-Bézier basis functions

Proof

To check the linear independence of $\{ w_{{{0},n}} (t)\;,\;w_{{{1},n}} (t), \ldots ,w_{n,n} (t)\}$, we consider a linear combination as follows:

$$ l_{{0}} w_{{{0},n}} (t;\,\alpha_{1} ) + \sum\limits_{i = 1}^{n - 1} {l_{i} w_{i,n} (t;\,\alpha_{i} ,\alpha_{{i + {1}}} )} + l_{n} w_{n,n} (t;\,\alpha_{n} ) = 0. $$

(56)

By taking the k-order derivative of Eq. (56) about t on both sides, we get

$$ l_{{0}} w_{{{0},n}}^{(k)} (t;\,\alpha_{1} ) + \sum\limits_{i = 1}^{n - 1} {l_{i} w_{i,n}^{(k)} (t;\,\alpha_{i} ,\alpha_{{i + {1}}} )} + l_{n} w_{n,n}^{(k)} (t;\,\alpha_{n} ) = 0,\quad k = {0},{1}, \ldots ,n. $$

(57)

For t = 0, according to (57) and combining with (9), we can obtain the following system of linear equations with respect to $l_{i} ,\,\;i = 0,1, \ldots ,n$

$$ \left\{ {\begin{array}{*{20}l} {l_{0} w_{0,n}^{(0)} (0;\,\alpha_{1} ) = 0,} \hfill \\ {l_{0} w_{0,n}^{(1)} (0\,;\,\alpha_{1} ) + l_{1} w_{1,n}^{(1)} (0;\,\alpha_{1} ,\alpha_{2} ) = 0,} \hfill \\ {l_{0} w_{0,n}^{(2)} (0\,;\,\alpha_{1} ) + l_{1} w_{1,n}^{(2)} (0;\,\alpha_{1} ,\alpha_{2} ) + l_{2} w_{2,n}^{(2)} (0;\,\alpha_{2} ,\alpha_{3} ) = 0,} \hfill \\ \cdots \hfill \\ {l_{0} w_{0,n}^{(n)} (0;\,\alpha_{1} ) + l_{1} w_{1,n}^{(n)} (0;\,\alpha_{1} ,\alpha_{2} ) + \cdots + l_{n} w_{n - 1,n}^{(n)} (0;\,\alpha_{n - 1} ,\alpha_{n} ) + l_{n} w_{n,n}^{(n)} (0;\,\alpha_{n} ) = 0,} \hfill \\ \end{array} } \right. $$

(58)

where $w_{0,n}^{(0)} (0;\,\alpha_{1} ) = w_{0,n} (0;\,\alpha_{1} ) = {1}$, $w_{i,n}^{(i)} (0;\,\alpha_{i} ,\alpha_{i + 1} ) \ne 0,\,\;i = 1,2, \ldots ,n - 1$ and $w_{n,n}^{(n)} (0;\,\alpha_{n} ) \ne 0$.

Thus, it is obvious that $l_{i} = 0,\;i = 0,1, \ldots ,n$, meaning that $w_{i,n} (t),\,\;i = 0,1, \ldots ,n$ are linearly independent.