﻿ 锥柱面螺旋线短程线与网格缠绕成型技术
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Spiral Curve and Geodesic Line of Column and ConeWith Texture Winding Formation of Grid Structure
TI Yafeng, JI Baofeng, HUANG Cheng, WANG Shixun, LI Xiongkui
Beijing Institute of Aerospace Systems Engineering, Beijing 100076
Abstract: According to the demand of the gird structure skin and rib texture winding formantion, this article discuss the spiral curve and geodesic line of column and cone by means of the differential geometry method. The problem of the composite grid structure skin and rib texture winding formation was solved. The results show that the arc length and curvature calculation method of the spiral curve and geodesic line of column and cone can be used for the design and manufacture of the mold forming device of the composite grid structure and fiber winding.The calculation method of arc length and curvature of logarithmic spiral and archimedes spiral can be used for the design and production of the soft mold forming device of the cone composite grid structure and fiber winding.The condition for the automation manufacture of the composite trellis structure has created.At the same time, for engineering needs, different kinds of grid structure are selected for different processing and forming methods, which can greatly reduce the processing difficulty and realize low cost and fast mass production.
Key words: Column and cone     Spiral curve     Geodesic line     Grid structure     Texture winding formation
0 引言

1 圆柱锥侧面上的螺旋线与复合材料网格结构的筋条缠绕成型

 $\left\{ \begin{array}{l} x = r\cos \theta \\ y = r\sin \theta \\ z = r\cot \alpha \cdot \theta \end{array} \right.$ (1)

 $\left\{ \begin{array}{l} \frac{{{\rm{d}}x}}{{{\rm{d}}\theta }} = - r\sin \theta \\ \frac{{{\rm{d}}y}}{{{\rm{d}}\theta }} = r\cos \theta \\ \frac{{{\rm{d}}z}}{{{\rm{d}}\theta }} = r\cot \alpha \end{array} \right.$ (2)

 $\begin{array}{*{20}{c}} {\cos \omega = \frac{{\frac{{{\rm{d}}z}}{{{\rm{d}}\theta }}}}{{\sqrt {{{\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}z}}{{{\rm{d}}\theta }}} \right)}^2}} }}}\\ { = \frac{{r\cot \alpha }}{{\sqrt {{{\left( { - r\sin \theta } \right)}^2} + {{\left( {r\cos \theta } \right)}^2} + {{\left( {r\cot \alpha } \right)}^2}} }} = \cos \alpha } \end{array}$ (3)

 $\left\{ \begin{array}{l} x = \rho \sin \beta cos\theta \\ y = \rho \sin \beta \sin \theta \\ z = \rho \cos \beta \\ \rho = {\rho _0}{{\rm{e}}^{\left( {\frac{{\sin \beta }}{{\tan \alpha }}\theta } \right)}} \end{array} \right.$ (4)

 $\left\{ \begin{array}{l} x = a\theta \sin \beta \cos \theta \\ y = a\theta \sin \beta \sin \theta \\ z = a\theta \cos \beta \end{array} \right.$ (5)

 $\cos \omega = \frac{1}{{\sqrt {1 + {\theta ^2}{{\sin }^2}\beta } }}$

 $\begin{array}{l} S = \int_{{\theta _1}}^{{\theta _2}} {\sqrt {{{\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}z}}{{{\rm{d}}\theta }}} \right)}^2}} {\rm{d}}\theta } \\ \;\;\;\; = \int_{{\theta _1}}^{{\theta _2}} {r\csc \alpha {\rm{d}}\theta } = r\csc \alpha \left( {{\theta _2} - {\theta _1}} \right) \end{array}$ (6)

 $\left\{ \begin{array}{l} \frac{{{\rm{d}}x}}{{{\rm{d}}\theta }} = {\rho _0}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\frac{{{{\sin }^2}\beta }}{{\tan \alpha }}\cos \theta - {\rho _0}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\sin \beta \sin \theta \\ \frac{{{\rm{d}}y}}{{{\rm{d}}\theta }} = {\rho _0}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\frac{{{{\sin }^2}\beta }}{{\tan \alpha }}\sin \theta - {\rho _0}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\sin \beta \sin \theta \\ \frac{{{\rm{d}}z}}{{{\rm{d}}\theta }} = {\rho _0}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\frac{{\sin \beta }}{{\tan \alpha }}\cos \beta \end{array} \right.$ (7)
 $\begin{array}{l} S = \int_{{\theta _1}}^{{\theta _2}} {{\rho _0}\frac{{\sin \beta }}{{\sin \alpha }}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}{\rm{d}}\theta } = {\rho _0}\frac{{\sin \beta }}{{\sin \alpha }} \cdot \frac{{\tan \alpha }}{{\sin \beta }}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}/_{{\theta _1}}^{{\theta _2}}\\ = \frac{{{\rho _0}}}{{\cos \alpha }}\left[ {{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}{\theta _2}}} - {{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}{\theta _1}}}} \right] \end{array}$ (8)

 $\left\{ \begin{array}{l} \frac{{{\rm{d}}x}}{{{\rm{d}}\theta }} = a\sin \beta \cos \theta - a\theta \sin \beta \sin \theta \\ \frac{{{\rm{d}}y}}{{{\rm{d}}\theta }} = a\sin \beta \sin \theta + a\theta \sin \beta \cos \theta \\ \frac{{{\rm{d}}z}}{{{\rm{d}}\theta }} = a\cos \beta \end{array} \right.$ (9)
 $S = \int_{{\theta _1}}^{{\theta _2}} {a\sqrt {1 + {\theta ^2}{{\sin }^2}\beta } {\rm{d}}\theta }$ (10)

 $\begin{array}{*{20}{c}} {S = \frac{a}{{\sin \beta }}\int_{{\rm{arsh}}\left( {{\theta _1}\sin \beta } \right)}^{{\rm{arsh}}\left( {{\theta _2}\sin \beta } \right)} {{\rm{c}}{{\rm{h}}^2}t{\rm{d}}t} }\\ { = \frac{a}{{\sin \beta }}\int_{{\rm{arsh}}\left( {{\theta _1}\sin \beta } \right)}^{{\rm{arsh}}\left( {{\theta _2}\sin \beta } \right)} {\frac{{{\rm{ch2}}t + 1}}{2}{\rm{d}}t} }\\ { = \frac{a}{{2\sin \beta }}\left( {\frac{{{\rm{sh}}2t}}{2} + t} \right)/_{{\rm{arsh}}\left( {{\theta _1}\sin \beta } \right)}^{{\rm{arsh}}\left( {{\theta _2}\sin \beta } \right)}}\\ { = \frac{a}{{2\sin \beta }}\left( {{\rm{sh}}t{\rm{ch}}t + t} \right)/_{{\rm{arsh}}\left( {{\theta _1}\sin \beta } \right)}^{{\rm{arsh}}\left( {{\theta _2}\sin \beta } \right)}}\\ { = \frac{a}{{2\sin \beta }}\left( {{\theta _2}\sin \beta \sqrt {\theta _2^2{{\sin }^2}\beta + 1} - {\theta _1}\sin \beta \sqrt {\theta _1^2{{\sin }^2}\beta + 1} + } \right.}\\ {\left. {{\rm{arsh}}\left( {{\theta _2}\sin \beta } \right) - {\rm{arsh}}\left( {{\theta _1}\sin \beta } \right)} \right)} \end{array}$ (11)

 $\left\{ \begin{array}{l} \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\theta ^2}}} = - r\cos \theta \\ \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\theta ^2}}} = - r\sin \theta \\ \frac{{{{\rm{d}}^2}z}}{{{\rm{d}}{\theta ^2}}} = 0 \end{array} \right.$ (12)

 $\begin{array}{l} K = \sqrt {\frac{{\left[ {{{\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}z}}{{{\rm{d}}\theta }}} \right)}^2}} \right]\left[ {{{\left( {\frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\theta ^2}}}} \right)}^2} + {{\left( {\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\theta ^2}}}} \right)}^2} + {{\left( {\frac{{{{\rm{d}}^2}z}}{{{\rm{d}}{\theta ^2}}}} \right)}^2}} \right] - \left( {\frac{{{\rm{d}}x}}{{{\rm{d}}\theta }} \cdot \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\theta ^2}}} + \frac{{{\rm{d}}y}}{{{\rm{d}}\theta }} \cdot \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\theta ^2}}} + \frac{{{\rm{d}}z}}{{{\rm{d}}\theta }} \cdot \frac{{{{\rm{d}}^2}z}}{{{\rm{d}}{\theta ^2}}}} \right)}}{{{{\left[ {{{\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}\theta }}} \right)}^2} + {{\left( {\frac{{{\rm{d}}z}}{{{\rm{d}}\theta }}} \right)}^2}} \right]}^3}}}} \\ = \frac{{{{\sin }^2}\alpha }}{r} \end{array}$ (13)

 $\left\{ \begin{array}{l} \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\theta ^2}}} = {\rho _0}\sin \beta {{\rm{e}}^{\frac{{\sin \beta }}{{{{\tan }^\alpha }}}\theta }}\left( {\frac{{{{\sin }^2}\beta }}{{{{\tan }^2}\alpha }}\cos \theta - 2\frac{{\sin \beta }}{{\tan \alpha }}\sin \theta - \cos \theta } \right)\\ \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\theta ^2}}} = {\rho _0}\sin \beta {{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\left( {\frac{{{{\sin }^2}\beta }}{{{{\tan }^2}\alpha }}\sin \theta + 2\frac{{\sin \beta }}{{\tan \alpha }}\cos \theta - \sin \theta } \right)\\ \frac{{{{\rm{d}}^2}z}}{{{\rm{d}}{\theta ^2}}} = {\rho _0}{{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}\theta }}\frac{{{{\sin }^2}\beta }}{{{{\tan }^2}\alpha }}\cos \beta \end{array} \right.$ (14)

 $K = \frac{{\cos \alpha }}{{{\rho _0}\tan \alpha \sin \beta {{\rm{e}}^{\frac{{\sin \beta }}{{\tan \alpha }}}}}}\sqrt {{{\sin }^2}\alpha {{\sin }^2}\beta + 2{{\sin }^2}\alpha {{\tan }^4}\alpha {{\sin }^2}\beta + {{\sin }^2}\alpha {{\tan }^4}\alpha - {{\tan }^2}\alpha {{\sin }^2}\beta }$ (15)

 $\left\{ \begin{array}{l} \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\theta ^2}}} = - 2a\sin \beta \sin \theta - a\theta \sin \beta \cos \theta \\ \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\theta ^2}}} = 2a\sin \beta \cos \theta - a\theta \sin \beta \sin \theta \\ \frac{{{{\rm{d}}^2}z}}{{{\rm{d}}{\theta ^2}}} = 0 \end{array} \right.$ (16)

 $K = \frac{{\sin \beta }}{{a\left( {1 + {\theta ^2}{{\sin }^2}\beta } \right)}}\sqrt {\frac{{4 + 3{\theta ^2}{{\sin }^2}\beta + {\theta ^2} + {\theta ^4}{{\sin }^2}\beta }}{{1 + {\theta ^2}{{\sin }^2}\beta }}}$ (17)
2 圆柱圆锥侧面上的短程线与复合材料网格结构的蒙皮缠绕成型

2.1 圆柱侧面上的短程线

 图 1 圆柱面上的短程线 Figure 1 Geodesicline of column

 $\left\{ \begin{array}{l} y = y\left( x \right),\\ z = z\left( x \right), \end{array} \right.\left( {{x_0} \le x \le {x_1}} \right)$

$\left\{ \begin{array}{l} y\left( {{x_0}} \right) = {y_0},y\left( {{x_1}} \right) = {y_1}\\ z({x_0}) = {z_0},z({x_1}) = {z_1} \end{array} \right.$ 下，求泛函J[y, z]取极小值的解y(x)和z(x)。

 $I\left[ {y,z} \right] = \int_{{x_0}}^{{x_1}} {\left[ {\sqrt {1 + {{y'}^2} + {{z'}^2}} + \lambda \left( x \right)\left( {z - \sqrt {1 - {x^2}} } \right)} \right]{\rm{d}}x}$ (18)

 $H = \sqrt {1 + {{y'}^2} + {{z'}^2}} + \lambda \left( x \right)\left( {z - \sqrt {1 - {x^2}} } \right)$ (19)

 $\left\{ \begin{array}{l} {H_y} - \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{H_{y'}}} \right) = 0\\ {H_z} - \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{H_{z'}}} \right) = 0 \end{array} \right.$ (20)

 $\left\{ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{y'}}{{\sqrt {1 + {{y'}^2} + {{z'}^2}} }}} \right) = 0\\ \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{z'}}{{\sqrt {1 + {{y'}^2} + {{z'}^2}} }}} \right) = \lambda \left( x \right) \end{array} \right.$ (21)

 ${\rm{d}}s = \sqrt {1 + {{y'}^2} + {{z'}^2}} {\rm{d}}x$ (22)

 $\left\{ \begin{array}{l} \frac{{{\rm{d}}y}}{{{\rm{d}}s}} = {c_1}\\ \frac{{{\rm{d}}z}}{{{\rm{d}}s}} = \omega \left( x \right) \end{array} \right.$ (23)

 ${\rm{d}}x = - \frac{{\sqrt {1 - {x^2}} }}{x}\omega \left( x \right){\rm{d}}s$ (24)

 $\begin{array}{*{20}{c}} {{{\left( {{\rm{d}}s} \right)}^2} = {{\left( {{\rm{d}}x} \right)}^2} + {{\left( {{\rm{d}}y} \right)}^2} + {{\left( {{\rm{d}}z} \right)}^2}}\\ { = \left[ {\frac{{1 - {x^2}}}{{{x^2}}}\omega \left( x \right) + C_1^2 + {\omega ^2}\left( x \right)} \right]{{\left( {{\rm{d}}s} \right)}^2}} \end{array}$ (25)

 $\omega \left( x \right) = \sqrt {1 - C_1^2} x$ (26)

 $- \frac{{{\rm{d}}x}}{{\sqrt {1 - {x^2}} }} = \sqrt {1 - C_1^2} {\rm{d}}s$ (27)

 $x = \cos \left( {\sqrt {1 - C_1^2} s + {C_2}} \right)$ (28)

 $z = \sqrt {1 - {x^2}} = \sin \left( {\sqrt {1 - C_1^2} s + {C_2}} \right)$ (29)

 $y = {C_1}s + {C_3}$ (30)

 $\left\{ \begin{array}{l} x = \cos \left( {\sqrt {1 - C_1^2} s + {C_2}} \right)\\ y = {C_1}s + {C_2}\\ z = \sin \left( {\sqrt {1 - C_1^2} s + {C_2}} \right) \end{array} \right.$ (31)

$\left\{ \begin{array}{l} s = {\rm{csc}}\alpha \cdot \theta \\ {c_1} = {\rm{cos}}\alpha \\ {c_2} = 0 \end{array} \right.$ 代入式(31)，

$\left\{ \begin{array}{l} x = {\rm{cos}}({\rm{sin}}\alpha \cdot {\rm{csc}}\alpha \cdot \theta + 0)\\ y = {\rm{cos}}\alpha \cdot {\rm{csc}}\alpha \cdot \theta + 0\\ z = {\rm{sin}}({\rm{sin}}\alpha \cdot {\rm{csc}}\alpha \cdot \theta + 0) \end{array} \right.$

$\left\{ \begin{array}{l} x = {\rm{cos}}\theta \\ y = {\rm{cot}}\alpha \cdot \theta \\ z = {\rm{sin}}\theta \end{array} \right.$

2.2 圆锥侧面上的短程线及螺旋线的展开

 $\left\{ \begin{array}{l} y = y\left( x \right),\\ z = z\left( x \right), \end{array} \right.\left( {{x_0} \le x \le {x_1}} \right)$
 图 2 圆锥面上的短程线 Figure 2 Geodesicline of cone

$\left\{ \begin{array}{l} y\left( {{x_0}} \right) = {y_0},y\left( {{x_1}} \right) = {y_1}\\ z\left( {{x_0}} \right) = {z_0},z\left( {{x_1}} \right) = {z_1} \end{array} \right.$ 下，求泛函J [y, z]取极小值的解y(x)和z(x)。

 $\begin{array}{*{20}{c}} {I\left( {y,z} \right) = \int_{{x_0}}^{{x_1}} {\left[ {\sqrt {1 + {{y'}^2} + {{z'}^2}} + \lambda \left( x \right)\left( z \right.} \right.} }\\ {\left. {\left. { - \sqrt {{y^2}{{\tan }^2}\theta - {x^2}} } \right)} \right]{\rm{d}}x} \end{array}$ (32)

 $H = \sqrt {1 + {{y'}^2} + {{z'}^2}} + \lambda \left( x \right)\left( {z - \sqrt {{y^2}{{\tan }^2}\theta - {x^2}} } \right)$ (33)

 $\left\{ \begin{array}{l} {H_y} - \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{H_{y'}}} \right) = 0\\ {H_z} - \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{H_{z'}}} \right) = 0 \end{array} \right.$ (34)
 $\left\{ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{y'}}{{\sqrt {1 + {{y'}^2} + {{z'}^2}} }}} \right) = - \frac{{y{{\tan }^2}\theta }}{{\sqrt {{y^2}{{\tan }^2}\theta - {x^2}} }}\lambda \left( x \right)\\ \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{z'}}{{\sqrt {1 + {{y'}^2} + {{z'}^2}} }}} \right) = \lambda \left( x \right) \end{array} \right.$ (35)

 图 3 圆柱螺旋线的展开 Figure 3 Expansion of the column spiral curve

 图 4 圆锥侧面展开图 Figure 4 Spread out side of the cone

 $r = r\left( \mathit{\Theta } \right)$ (36)

y=rsinΘ则方程组：x=r(Θ)·cosΘy=r(Θ)·sinΘ是曲线的参数方程，其中参数为极角Θ，曲线的切线斜率是:

 $\begin{array}{*{20}{c}} {y' = \frac{{\frac{{{\rm{d}}y}}{{{\rm{d}}\mathit{\Theta }}}}}{{\frac{{{\rm{d}}x}}{{{\rm{d}}\mathit{\Theta }}}}} = \frac{{r'\left( \mathit{\Theta } \right)\sin \mathit{\Theta } + r\left( \mathit{\Theta } \right)\cos \mathit{\Theta }}}{{r'\left( \mathit{\Theta } \right)\cos \mathit{\Theta } - r\left( \mathit{\Theta } \right)\sin \mathit{\Theta }}}}\\ { = \frac{{r'\left( \mathit{\Theta } \right)\tan \mathit{\Theta } + r\left( \mathit{\Theta } \right)}}{{r'\left( \mathit{\Theta } \right) - r\left( \mathit{\Theta } \right)\tan \mathit{\Theta }}}} \end{array}$ (37)

 $\alpha = \gamma - \mathit{\Theta }$ (38)
 $\tan \alpha = \tan \left( {\gamma - \mathit{\Theta }} \right) = \frac{{y' - \tan \mathit{\Theta }}}{{1 + y'\tan \mathit{\Theta }}}$ (39)

y′的表达式(37)代入(39)：

 $\begin{array}{l} \tan \alpha = \frac{{\frac{{r'\left( \mathit{\Theta } \right)\tan \mathit{\Theta } + r\left( \mathit{\Theta } \right)}}{{r'\left( \mathit{\Theta } \right) - r\left( \mathit{\Theta } \right)\tan \mathit{\Theta }}} - \tan \mathit{\Theta }}}{{1 + \frac{{r'\left( \mathit{\Theta } \right)\tan \mathit{\Theta } + r\left( \mathit{\Theta } \right)}}{{r'\left( \mathit{\Theta } \right) - r\left( \mathit{\Theta } \right)\tan \mathit{\Theta }}}\tan \mathit{\Theta }}}\\ = \frac{{r'\left( \mathit{\Theta } \right)\tan \mathit{\Theta } + r\left( \mathit{\Theta } \right) - r'\left( \mathit{\Theta } \right)\tan \mathit{\Theta } + r\left( \mathit{\Theta } \right){{\tan }^2}\mathit{\Theta }}}{{r'\left( \mathit{\Theta } \right) - r\left( \mathit{\Theta } \right)\tan \mathit{\Theta + }r'\left( \mathit{\Theta } \right){{\tan }^2}\mathit{\Theta } + r\left( \mathit{\Theta } \right)\tan \mathit{\Theta }}}\\ = \frac{{r\left( \mathit{\Theta } \right)}}{{r'\left( \mathit{\Theta } \right)}} \end{array}$ (40)

${\rm{cot}}\alpha = \frac{{r'\left( \mathit{\Theta } \right)}}{{r\left( \mathit{\Theta } \right)}}$ ，两边对Θ积分，解为：

 $\cot \alpha = \ln r\left( \mathit{\Theta } \right) + c = \ln \sqrt {{x^2} + {y^2}} + c$ (41)

y=0为初始点(图中B点)，此时Θ=0，代入式(41)则：

 $c = - \ln {x_0}$ (42)

 $\cot \alpha \cdot \mathit{\Theta } = \ln \frac{{r\left( \mathit{\Theta } \right)}}{{{x_0}}}$ (43)

 图 5 阿基米德螺线 Figure 5 Archimedes spiral

 $\begin{array}{*{20}{c}} {S = \int_{{\mathit{\Theta }_1}}^{{\mathit{\Theta }_2}} {\sqrt {{r^2} + {{r'}^2}} {\rm{d}}\mathit{\Theta }} = {x_0}\sqrt {1 + {{\cot }^2}\alpha } \int_{{\mathit{\Theta }_1}}^{{\mathit{\Theta }_2}} {{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}{\rm{d}}\mathit{\Theta }} }\\ { = \frac{{{x_0}\sqrt {1 + {{\cot }^2}\alpha } }}{{\cot \alpha }}\left( {{{\rm{e}}^{\cot \alpha \cdot {\mathit{\Theta }_2}}} - {{\rm{e}}^{\cot \alpha \cdot {\mathit{\Theta }_1}}}} \right)}\\ { = \frac{{{x_0}}}{{\cos \alpha }}\left( {{{\rm{e}}^{\frac{{{\mathit{\Theta }_2}}}{{\tan \alpha }}}} - {{\rm{e}}^{\frac{{{\mathit{\Theta }_1}}}{{\tan \alpha }}}}} \right)} \end{array}$ (44)

 $S = \int_{{\mathit{\Theta }_1}}^{{\mathit{\Theta }_2}} {\sqrt {{r^2} + {{r'}^2}} {\rm{d}}\mathit{\Theta }} = \int_{{\mathit{\Theta }_1}}^{{\mathit{\Theta }_2}} {\sqrt {{{\left( {{x_0}\mathit{\Theta }} \right)}^2} + x_0^2} {\rm{d}}\mathit{\Theta }}$

 $\begin{array}{*{20}{c}} {S = {x_0}\int_{{\rm{arsh}}{\mathit{\Theta }_1}}^{{\rm{arsh}}{\mathit{\Theta }_2}} {{\rm{c}}{{\rm{h}}^2}t{\rm{d}}t} = {x_0}\int_{{\rm{arsh}}{\mathit{\Theta }_1}}^{{\rm{arsh}}{\mathit{\Theta }_2}} {\frac{{{\rm{ch}}2t + 1}}{2}{\rm{d}}t} }\\ { = \frac{{{x_0}}}{2}\left( {\frac{{{\rm{sh}}2t}}{2} + t} \right)/_{{\rm{arsh}}{\mathit{\Theta }_1}}^{{\rm{arsh}}{\mathit{\Theta }_2}} = \frac{{{x_0}}}{2}\left( {{\rm{sh}}t{\rm{ch}}t + t} \right)/_{{\rm{arsh}}{\mathit{\Theta }_1}}^{{\rm{arsh}}{\mathit{\Theta }_2}}}\\ { = \frac{{{x_0}}}{2}\left( {{\mathit{\Theta }_2}\sqrt {\mathit{\Theta }_2^2 + 1} - {\mathit{\Theta }_1}\sqrt {\mathit{\Theta }_1^2 + 1} + {\rm{arsh}}{\mathit{\Theta }_2} - {\rm{arsh}}{\mathit{\Theta }_1}} \right)} \end{array}$ (45)

 ${x_0} = \frac{a}{{\sin \beta }},螺距 z= 2\pi a\cos \beta ,a = \frac{z}{{2\pi \cos \beta }}$
 ${x_0} = \frac{z}{{2\pi \cos \beta {\rm{sin}}\beta }} = \frac{z}{{\pi \sin 2\beta }}$ (46)

 $\left\{ \begin{array}{l} x = {x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}\cos \mathit{\Theta }\\ y = {x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}\sin \mathit{\Theta } \end{array} \right.$ (47)
 $\left( \begin{array}{l} \frac{{{\rm{d}}x}}{{{\rm{d}}\mathit{\Theta }}} = {x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}\left( {\cot \alpha \cos \mathit{\Theta } - \sin \mathit{\Theta }} \right)\\ \frac{{{\rm{d}}y}}{{{\rm{d}}\mathit{\Theta }}} = {x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}\left( {\cot \alpha \sin \mathit{\Theta } + \cos \mathit{\Theta }} \right) \end{array} \right.$ (48)
 $\left( \begin{array}{l} \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\mathit{\Theta }^2}}} = {x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}\left( {{{\cot }^2}\alpha \cos \mathit{\Theta } - 2\cot \alpha \sin \mathit{\Theta } - \cos \mathit{\Theta }} \right)\\ \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\mathit{\Theta }^2}}} = {x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}\left( {{{\cot }^2}\alpha \sin \mathit{\Theta } + 2\cot \alpha \cos \mathit{\Theta } - \sin \mathit{\Theta }} \right) \end{array} \right.$ (49)

 $K = \frac{{\left| {\frac{{{\rm{d}}x}}{{{\rm{d}}\mathit{\Theta }}} \cdot \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\mathit{\Theta }^2}}} - \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\mathit{\Theta }^2}}} \cdot \frac{{{\rm{d}}y}}{{{\rm{d}}\mathit{\Theta }}}} \right|}}{{{{\left[ {{{\left( {\frac{{{\rm{d}}x}}{{{\rm{d}}\mathit{\Theta }}}} \right)}^2} + {{\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}\mathit{\Theta }}}} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\sin \alpha }}{{{x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}}}$ (50)

 $R = \frac{{{x_0}{{\rm{e}}^{\cot \alpha \cdot \mathit{\Theta }}}}}{{\sin \alpha }}$ (51)

 $\left\{ \begin{array}{l} x = {x_0}\mathit{\Theta }\cos \mathit{\Theta }\\ y = {x_0}\mathit{\Theta }\sin \mathit{\Theta } \end{array} \right.$ (52)
 $\left\{ \begin{array}{l} \frac{{{\rm{d}}x}}{{{\rm{d}}\mathit{\Theta }}} = {x_0}\cos \mathit{\Theta } - {x_0}\mathit{\Theta }\sin \mathit{\Theta }\\ \frac{{{\rm{d}}y}}{{{\rm{d}}\mathit{\Theta }}} = {x_0}\sin \mathit{\Theta } + {x_0}\mathit{\Theta }\cos \mathit{\Theta } \end{array} \right.$ (53)
 $\left\{ \begin{array}{l} \frac{{{{\rm{d}}^2}x}}{{{\rm{d}}{\mathit{\Theta }^2}}} = - 2{x_0}\sin \mathit{\Theta } - {x_0}\mathit{\Theta }\cos \mathit{\Theta }\\ \frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{\mathit{\Theta }^2}}} = 2{x_0}\cos \mathit{\Theta } - {x_0}\mathit{\Theta }\sin \mathit{\Theta } \end{array} \right.$ (54)

 $K = \frac{{2 + {\mathit{\Theta }^2}}}{{{x_0}{{\left( {1 + {\mathit{\Theta }^2}} \right)}^{\frac{3}{2}}}}}$ (55)

 $R = \frac{{{x_0}{{\left( {1 + {\mathit{\Theta }^2}} \right)}^{\frac{3}{2}}}}}{{2 + {\mathit{\Theta }^2}}}$ (56)

x2+y2=(zh)2tan2βr=htanβA(r, 0, 0)，B(x0, 0, z0)，0＜z0h。设BxOz的坐标为(xB, zB)，求xBzB

 图 6 圆锥面短程线的展开 Figure 6 Expansion of the cone geodesic line
 $\overrightarrow {OB} = \left( {\frac{{h - {z_0}}}{{\cos \beta }}\sin \left( {\beta + {\rm{ \mathit{ π} sin}}\beta } \right),h - \frac{{h - {z_0}}}{{\cos \beta }}\cos \left( {\beta + {\rm{ \mathit{ π} sin}}\beta } \right)} \right)$ (57)
 ${x_{\rm{B}}} = \frac{{h - {z_0}}}{{\cos \beta }}\sin \left( {\beta + {\rm{ \mathit{ π} sin}}\beta } \right)$ (58)
 ${z_{\rm{B}}} = h - \frac{{h - {z_0}}}{{\cos \beta }}\cos \left( {\beta + {\rm{ \mathit{ π} sin}}\beta } \right)$ (59)

 $\left\{ \begin{array}{l} x = \left( {{x_{\rm{B}}} - r} \right)t + r,\\ z = {z_{\rm{B}}}t, \end{array} \right.t \in \left[ {0,1} \right]$ (60)

 $\left| {MN} \right| = \left| {MN'} \right|,\angle AMN' = \varphi = \theta \cdot \sin \beta$
 图 7 圆锥面短程线 Figure 7 Geodesic line of cone

[与前述(46)式Θ=sinβ·θ结论相同]

 $\therefore \left\{ \begin{array}{l} x = \left| {MN} \right|\sin \beta \cos \theta \\ y = \left| {MN} \right|\sin \beta \sin \theta \\ z = h - \left| {MN} \right|\cos \beta \end{array} \right.$ (61)
 $\begin{array}{*{20}{c}} {\because \overrightarrow {ON'} = \overrightarrow {OM} + \overrightarrow {MN'} = \left( {0,h} \right) + \left| {MN'} \right|\left( {\sin \left( {\varphi + \beta } \right),} \right.} \\ {\left. { - \cos \left( {\varphi + \beta } \right)} \right)} \end{array}$ (62)
 $\therefore \overrightarrow {ON'} = \left( {\left| {MN'} \right|\sin \left( {\varphi + \beta } \right),h - \left| {MN'} \right|\cos \left( {\varphi + \beta } \right)} \right)$ (63)

N′在AB上，

 $\therefore \frac{{h - \left| {MN'} \right|\cos \left( {\varphi + \beta } \right)}}{{{z_{\rm{B}}}}} = \frac{{\left| {MN'} \right|\sin \left( {\varphi + \beta } \right) - r}}{{{x_{\rm{B}}} - r}}$ (64)
 $\therefore \left| {MN'} \right| = \frac{{h\left( {{x_{\rm{B}}} - r} \right) + {z_{\rm{B}}}r}}{{{z_{\rm{B}}}\sin \left( {\varphi + \beta } \right) + \left( {{x_{\rm{B}}} - r} \right)\cos \left( {\varphi + \beta } \right)}}$ (65)
 $\therefore \left\{ \begin{array}{l} x = \frac{{h\left( {{x_{\rm{B}}} - r} \right) + {z_{\rm{B}}}r}}{{{z_{\rm{B}}}\sin \left( {\varphi + \beta } \right) + \left( {{x_{\rm{B}}} - r} \right)\cos \left( {\varphi + \beta } \right)}}\sin \beta \cos \theta \\ y = \frac{{h\left( {{x_{\rm{B}}} - r} \right) + {z_{\rm{B}}}r}}{{{z_{\rm{B}}}\sin \left( {\varphi + \beta } \right) + \left( {{x_{\rm{B}}} - r} \right)\cos \left( {\varphi + \beta } \right)}}\sin \beta \sin \theta \\ z = h - \frac{{h\left( {{x_{\rm{B}}} - r} \right) + {z_{\rm{B}}}r}}{{{z_{\rm{B}}}\sin \left( {\varphi + \beta } \right) + \left( {{x_{\rm{B}}} - r} \right)\cos \left( {\varphi + \beta } \right)}}\cos \beta \end{array} \right.$ (66)

3 结论

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http://dx.doi.org/10.12044/j.issn.1007-2330.2018.01.005

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#### 文章信息

TI Yafeng, JI Baofeng, HUANG Cheng, WANG Shixun, LI Xiongkui

Spiral Curve and Geodesic Line of Column and ConeWith Texture Winding Formation of Grid Structure

Aerospace Materials & Technology, 2018, 48(1): 22-29.
http://dx.doi.org/10.12044/j.issn.1007-2330.2018.01.005