工程科学与技术   2021, Vol. 53 Issue (1): 146-154
2–UPR&2–RPU冗余并联机构动力学与功耗分析

Dynamics and Power Analysis of 2–UPR&2–RPU Redundant Parallel Mechanism
LIN Guangchun, LIAO Xunbao, ZHAO Rongkuan, CHEN Shichao
School of Mechanical Eng., Sichuan Univ., Chengdu 610065, China
Abstract: A novel 2–UPR&2–RPU parallel mechanism with redundance structure was proposed, which has two rotations and one movement (2R1T) degree of freedom. Based on the screw theory, the degrees of freedom and motion characteristics of the mechanism were analyzed, and the correctness of the analysis was verified by the modified G–K formula. By analyzing the constraints of each branch, the reverse solutions of position was solved by establishing the closed-loop vector equations of each branch. The velocity and acceleration of the mechanism were analyzed by the differential transformation method based on symbolic operation, then the velocity Jacobian matrix and the second order influence coefficient matrixes were obtained. On the basis of this, the Jacobian matrix of each branch was also figured out. The dynamics model of the mechanism was constructed by using the virtual work principle based on the kinematics. For the character that the distribution of the active joints’ drive forces was not unique due to the redundant actuation, the driving force and driving power equation were obtained by the method of 2-norm of driving force. Based on this, a calculation program was written for the constructed mathematical model by using MATLAB, and the constructed kinematics and dynamics model were simulated through specific numerical examples. Finally, in order to verify the correctness of the mathematical model constructed, a virtual prototype of the mechanism was established in ADAMS and simulated under the same conditions. The results were compared with those of MATLAB. The simulation results of both were basically consistent, indicating the correctness of the established mathematical model and showing the max errors of driving force of each rod with 0.091%, 1.83%, 1.04%, 1.40%, respectively. The results showed that under the given motion law, the mechanism does not produce rigid impact. At the beginning and the end of the motion, a certain flexible impact will occur. The driving force of each driving rod changes smoothly during the entire movement, and the total work by the method of 2-norm of the driving force is 1 143.2 J.
Key words: redundant parallel mechanism    kinematics    dynamics    power consumption

1 机构模型描述与自由度分析 1.1 机构描述

 图1 2–UPR&2–RPU 并联机构 Fig. 1 2–UPR&2–RPU parallel mechanism

2–UPR&2–RPU并联机构的定/动坐标系如图2所示，对图2中的参数做如下说明：

 图2 机构参数 Fig. 2 Parameters of mechanism

A1A3分别为两条UPR支链中的U副转动中心点，B1B3为其R副质心点，A2A4分别为两条RPU支链中R副的质心点，B2B4为其U副的转动中心点。

1.2 自由度分析

 $\;\;\;\;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\;\left\{ \begin{array}{l} {{{S}}_{11}} = [\begin{array}{*{20}{c}} 0&1&{0;\begin{array}{*{20}{c}} 0&0&0 \end{array}} \end{array}],\\ {{{S}}_{12}} = [\begin{array}{*{20}{c}} 1&0&0;\begin{array}{*{20}{c}} 0&0&{ - a} \end{array} \end{array}],\\ {{{S}}_{13}} = [\begin{array}{*{20}{c}} 0&0&0;\begin{array}{*{20}{c}} 0&{{M_{13}}}&{{N_{13}}} \end{array} \end{array}],\\ {{{S}}_{14}} = [\begin{array}{*{20}{c}} 1&0&{0;\begin{array}{*{20}{c}} 0&{\textit{z}}&{ - b} \end{array}} \end{array}] \end{array} \right.$ (1)

 $\;\;\;\;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\;\left\{ \begin{array}{l} {{S}}_{11}^r = [\begin{array}{*{20}{c}} 1&0&{0;\begin{array}{*{20}{c}} 0&0&0 \end{array}} \end{array}],\\ {{S}}_{12}^r = [\begin{array}{*{20}{c}} 0&0&{0;\begin{array}{*{20}{c}} 0&0&1 \end{array}} \end{array}] \end{array} \right.$ (2)

 $\;\;\;\;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\;\;\;\;\left\{ \begin{array}{l} {{{S}}_{21}} = [\begin{array}{*{20}{c}} 0&1&{0\,;\begin{array}{*{20}{c}} 0&0&a \end{array}} \end{array}],\\ {{{S}}_{22}} = [\begin{array}{*{20}{c}} 0&0&{0\,;\begin{array}{*{20}{c}} {{L_{22}}}&0&{{N_{22}}} \end{array}} \end{array}],\\ {{{S}}_{23}} = [\begin{array}{*{20}{c}} 1&0&{0\,;\begin{array}{*{20}{c}} 0&{\textit{z}}&0 \end{array}} \end{array}],\\ {{{S}}_{24}} = [\begin{array}{*{20}{c}} 0&1&{0\,;\begin{array}{*{20}{c}} { - {\textit{z}}}&0&b \end{array}} \end{array}] \end{array} \right.$ (3)

 $\;\;\;\;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\;\;\;\;\left\{ \begin{array}{l} {{S}}_{21}^r = [\begin{array}{*{20}{c}} 0&1&{0\,;\begin{array}{*{20}{c}} { - {\textit{z}}}&0&0 \end{array}} \end{array}],\\ {{S}}_{22}^r = [\begin{array}{*{20}{c}} 0&0&{0\,;\begin{array}{*{20}{c}} 0&0&1 \end{array}} \end{array}] \end{array} \right.$ (4)

 $\;\;\;\;\;\; \;\;\; \;\;\; \;\;\; \;\;\; \;\;\;\;\;\;\;\;\;\left\{ \begin{array}{l} {{S}}_{r1}^{{\rm{pm}}} = [\begin{array}{*{20}{c}} 1&0&0\,;\;0&{\textit{z}}&0 \end{array}],\\ {{S}}_{r2}^{{\rm{pm}}} = [\begin{array}{*{20}{c}} 0&1&0\,;\;0&0&0 \end{array}],\\ {{S}}_{r3}^{{\rm{pm}}} = [\begin{array}{*{20}{c}} 0&0&0\,;\;0&0&1 \end{array}] \end{array} \right.$ (5)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;M = d(n - \rho - 1) + \sum\limits_{i = 1}^g {{f_i}} + \upsilon - \zeta$ (6)

2 机构运动学模型 2.1 位置逆解

2–UPR&2–RPU并联机构的位置逆解问题为已知机构的具体尺寸参数和动平台中心 $o$ 相对于定坐标系的位姿 $(x ,y ,{\textit{z}} ,\psi \, ,\theta \, ,\varphi )$ ，借助旋转矩阵确定机构各支链中驱动杆的杆长矢量 ${{{l}}_i}$

 \begin{aligned}[b] {{R}} =& {\bf{rot}}({\textit{z}}\,,\,\varphi ){\bf{rot}}(y ,\,\theta ){\bf{rot}}(x ,\,\psi ) = \\ &\left[ {\begin{array}{*{20}{c}} {\cos\; \theta \cos\; \varphi }&{\sin\; \psi \sin\; \theta \cos\; \varphi - \cos\; \psi \sin\; \varphi }&{\cos\; \psi \sin\; \theta \cos\; \varphi + \sin\; \psi \sin\; \varphi }\\ {\cos\; \theta \sin\; \varphi }&{\sin\; \psi \sin\; \theta \sin\; \varphi + \cos\; \psi \cos\; \varphi }&{\cos\; \psi \sin\; \theta \sin\; \varphi - \sin\; \psi \cos\; \varphi }\\ { - \sin\; \theta }&{\sin\; \psi \cos\; \theta }&{\cos\; \psi \cos\; \theta } \end{array}} \right] \end{aligned} (7)

 ${}^O{{{R}}_i} = \left[ {\begin{array}{*{20}{c}} {\cos\; {\beta _i}}&{\sin\; {\alpha _i}\sin\; {\beta _i}}&{\cos\; {\alpha _i}\sin\; {\beta _i}} \\ 0&{\cos\; {\alpha _i}}&{ - \sin\; {\alpha _i}} \\ { - \sin\; {\beta _i}}&{\sin\; {\alpha _i}\cos\; {\beta _i}}&{\cos\; {\alpha _i}\cos\; {\beta _i}} \end{array}} \right]$ (8)

${}^O{{{R}}_i}$ 的第3列即为分支方向的单位向量 ${{{e}}_i}$ ，即 ${{{e}}_i} =$ ${[{\rm{cos}}\;{\alpha _i}\;\sin \;{\beta _i} \;- \;\sin \;{\alpha _i}\;\cos \;{\alpha _i}\;\cos \;{\beta _i}]^{\rm{T}}}$ ，对于RPU支链，则 ${\alpha _i} = 0$

 ${{{l}}_i} = {l_i}{{{e}}_i} = {{p}} + {{{b}}_i} - {{{a}}_i}$ (9)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ \begin{array}{l} {{c}}_1^{\rm{T}}({{p}} - {{{a}}_1}) = {\rm{0}}\,\,,\;{l_{2y}} = 0\\ {{c}}_3^{\rm{T}}({{p}} - {{{a}}_3}) = {\rm{0}}\,\,\,,\;{l_{4y}} = 0 \end{array} \right.$ (10)

 $\left\{ \begin{array}{l} x\cos\; \theta \cos\; \varphi - (a - y)\cos\; \theta \sin\; \varphi - {\textit{z}}\sin\; \theta = 0, \\ x\cos\; \theta \cos\; \varphi + (y + a)\cos\; \theta \sin\; \varphi - {\textit{z}}\sin\; \theta = 0, \\ b\cos\; \theta \sin\; \varphi = 0, \\ - b\cos\; \theta \sin\; \varphi = 0 \\ \end{array} \right.$ (11)

 $x\cos\; \theta \cos\; \varphi - {\textit{z}}\sin\; \theta = 0$ (12)

 $\left\{ {\begin{array}{*{20}{l}} {{l_1} = \sqrt {\begin{array}{*{20}{l}} {{{({\textit{z}}\tan\; \theta + b\sin\; \psi \sin\; \theta )}^2} + {{( - a + b\cos\; \psi )}^2} + } {{{({\textit{z}} + b\sin\; \psi \cos\; \theta )}^2}}, \end{array}} }\\ {{l_2} = \sqrt {{{({\textit{z}}\tan\; \theta - a + b\cos\; \theta )}^2} + {{({\textit{z}} - b\sin\; \theta )}^2}} },\\ {{l_3} = \sqrt {\begin{array}{*{20}{l}} {{{({\textit{z}}\tan\; \theta - b\sin\; \psi \sin\; \theta )}^2} + {{(a - b\cos\; \psi )}^2} + } {{{({\textit{z}} - b\sin\; \psi \cos\; \theta )}^2}} \end{array}} },\\ {{l_4} = \sqrt {\sqrt {{{({\textit{z}}\tan\; \theta + a - b\cos\; \theta )}^2} + {{({\textit{z}} + b\sin\; \theta )}^2}} } } \end{array}} \right.$ (13)
2.2 速度分析

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{l_i} = \sqrt {{{({l_i}{{{e}}_i})}^{\rm{T}}}({l_i}{{{e}}_i})} = {f_i}({\textit{z}}\,,\psi \,,\theta )$ (14)

 $\;\;\;\;\;\;\;\;\;\;\;{\dot l_i} = \frac{{\partial {f_i}({\textit{z}}\,,\psi \,,\theta )}}{{\partial {\textit{z}}}}\dot {\textit{z}} + \frac{{\partial {f_i}({\textit{z}}\,,\psi \,,\theta )}}{{\partial \psi }}\dot \psi + \frac{{\partial {f_i}({\textit{z}}\,,\psi ,\theta )}}{{\partial \theta }}\dot \theta$ (15)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ \begin{array}{l} {{\dot l}_1} \\ {{\dot l}_2} \\ {{\dot l}_3} \\ {{\dot l}_4} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{{\partial {f_1}}}{{\partial {\textit{z}}}}}&{\dfrac{{\partial {f_1}}}{{\partial \psi }}}&{\dfrac{{\partial {f_1}}}{{\partial \theta }}} \\ {\dfrac{{\partial {f_2}}}{{\partial {\textit{z}}}}}&{\dfrac{{\partial {f_2}}}{{\partial \psi }}}&{\dfrac{{\partial {f_2}}}{{\partial \theta }}} \\ {\dfrac{{\partial {f_3}}}{{\partial {\textit{z}}}}}&{\dfrac{{\partial {f_3}}}{{\partial \psi }}}&{\dfrac{{\partial {f_3}}}{{\partial \theta }}} \\ {\dfrac{{\partial {f_4}}}{{\partial {\textit{z}}}}}&{\dfrac{{\partial {f_4}}}{{\partial \psi }}}&{\dfrac{{\partial {f_4}}}{{\partial \theta }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\dot {\textit{z}}} \\ {\dot \psi } \\ {\dot \theta } \end{array}} \right]$ (16)

$\dot {{l}} = {[\begin{array}{*{20}{c}} {{{\dot l}_1}}&{{{\dot l}_2}}&{{{\dot l}_3}}&{{{\dot l}_4}} \end{array}]^{\rm{T}}}$ ${{{V}}_o} = {[\begin{array}{*{20}{c}} {\dot {\textit{z}}}&{\dot \psi }&{\dot \theta } \end{array}]^{\rm{T}}}$ ，则有：

 $\dot {{l}} = {{J}}{{{V}}_o}$ (17)

2.3 加速度分析

 $\ddot {{l}} = {{{V}}^{\rm{T}}_o}{{h}}{{{V}}_o} + {{J}}{\dot {{V}}_o}$ (18)

2.4 支链雅克比求解

 ${{{v}}_{Bi}} = {{{\omega}} _{li}} \times {l_i}{{{e}}_i} + {\dot l_i}{{{e}}_i}$ (19)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{{e}}_i} \times {{{v}}_{Bi}} = {l_i}{{{\omega}} _{li}} - {l_i}{{{e}}_i}({{{\omega}} _{li}} \cdot {{{e}}_i})$ (20)

 ${{{\omega}} _{li}} = \frac{{{{{e}}_i} \times {{{v}}_{Bi}}}}{{{l_i}}}$ (21)

${{{v}}_{Bi}} = {{{\omega}} _o} \times {{{b}}_i} + {{{v}}_o}$ 代入式（21）可得：

 \begin{aligned}[b] {{{\omega}} _{li}} = \frac{1}{{{l_i}}}[{{{e}}_i} \times ({{{v}}_o} + {{{\omega}} _o} \times & {{{b}}_i})] = \frac{1}{{{l_i}}}({\hat {{e}}_i}{{{v}}_o} - {\hat {{e}}_i}{\hat {{b}}_i}{{{\omega}} _o}) = \\ &{{{J}}_{\omega i}}{\left[ {\begin{array}{*{20}{c}} {{{{v}}_o}}&{{{{\omega}} _o}} \end{array}} \right]^{\rm{T}}} \end{aligned} (22)

${{{J}}_{\omega i}} =$ $[\begin{array}{*{20}{c}} {{{\hat {{e}}}_i}}&{ - {{\hat {{e}}}_i}{{\hat {{b}}}_i}} \end{array}]/ {l_i}$

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\dot {{\omega}} _{li}} = \frac{{{{\dot {{e}}}_i} \times {{{v}}_{Bi}} + {{{e}}_i} \times {{\dot {{v}}}_{Bi}} - {{{\omega}} _{li}}{{\dot l}_i}}}{{{l_i}}}$ (23)

 \begin{aligned}[b] {{{v}}_{mi}} =& {{{\omega }}_{li}} \times {l_{mi}}{{{e}}_i} + {\dot l_i}{{{e}}_i} = {l_{mi}}{\hat {{e}}_i}{{{J}}_{ \omega i}}{{{v}}_o} + \\ & ({{{e}}_i}{{e}}_i^{\rm{T}} - {{{e}}_i}{{e}}_i^{\rm{T}}{\hat {{b}}_i}){{{\omega}} _o} = {{{J}}_{vi}}{\left[ {\begin{array}{*{20}{c}} {{{{v}}_o}}&{{{{\omega}} _o}} \end{array}} \right]^{\rm{T}}} \end{aligned} (24)

${{{J}}_{vi}} = {l_{mi}}{\hat {{e}}_i}{{{J}}_{\omega i}} + ({{{e}}_i}{{e}}_i^{\rm{T}} - {{{e}}_i}{{e}}_i^{\rm{T}}{\hat {{b}}_i})$

 \begin{aligned}[b] {{{a}}_{mi}} =& {\dot {{\omega}} _{li}} \times {l_{mi}}{{{e}}_i} + {{{\omega}} _{li}} \times ({{{\omega}} _{li}} \times {{{e}}_i}){l_{mi}} + \\ &{{{e}}_i}{\ddot l_i} + 2({{{\omega}} _{li}} \times {{{e}}_i}){\dot l_i} \end{aligned} (25)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left[ {\begin{array}{*{20}{c}} {{{{v}}_{mi}}}&{{{{\omega }}_{mi}}} \end{array}} \right]^{\rm{T}}} = {{{J}}_{li}}{\left[ {\begin{array}{*{20}{c}} {{{{v}}_o}}&{{{{\omega}} _o}} \end{array}} \right]^{\rm{T}}}$ (26)

3 动力学及驱动功率分析

3.1 动力学分析

${{{I}}_o}$ 表示动平台的惯性矩阵在动坐标系，则动平台的惯性矩阵在定坐标系中表示为 ${{{I}}_O} = {{R}}{{{I}}_o}{{{R}}^{\rm{T}}}$

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{{F}}_o} = \left[ {\begin{array}{*{20}{c}} {{{{f}}_O} + {m_o}{{g}} - {m_o}{{\dot {{v}}}_o}} \\ {{{{n}}_O} - {{{I}}_O}{{{\omega}} _o} - {{{\omega}} _o} \times ({{{I}}_O}{{{\omega}} _o})} \end{array}} \right]$ (27)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{{F}}_{li}} = \left[ {\begin{array}{*{20}{c}} {{m_{li}}{{g}} - {m_{li}}{{{a}}_{mi}}} \\ { - {}^O{{{I}}_{li}}{{\dot {{\omega}} }_{li}} - {{{\omega}} _{li}} \times ({}^O{{{I}}_{li}}{{{\omega}} _{li}})} \end{array}} \right]$ (28)

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\delta {{x}}_o^{\rm{T}}{{{F}}_o} + \delta {{{q}}^{\rm{T}}}{{F}} + \sum\limits_{i = 1}^4 {\delta {{x}}_{li}^{\rm{T}}{{{F}}_{li}}} = 0$ (29)

 ${\;\;\;\;\;\;\;\;\;\;\;\delta} {{x}}_o^{\rm{T}}({{{F}}_o} + {{{J}}^{\rm{T}}}{{F}} + \sum\limits_{i = 1}^4 {{{J}}_{li}^{\rm{T}}{{{F}}_{li}}} ) = 0$ (30)

 ${{{F}}_o} + {{{J}}^{\rm{T}}}{{F}} + \sum\limits_{i = 1}^4 {{{J}}_{li}^{\rm{T}}{{{F}}_{li}}} = 0$ (31)

 ${\;\;\;\;\;\;\;\;\;\;\;\;{{F}}} = - {({{{J}}^{\rm{T}}})^ + }({{{F}}_o} + \displaystyle\sum\limits_{i = 1}^4 {{{J}}_{li}^{\rm{T}}{{{F}}_{li}}} ) + {{{F}}_0}{{K}}$ (32)

 $\begin{array}{c} {({{{J}}^{\rm{T}}})^ + } = {{J}}{({{{J}}^{\rm{T}}}{{J}})^{ - 1}},\\ {{{F}}_0} = ({{E}} - {({{{J}}^{\rm{T}}})^ + }{{{J}}^{\rm{T}}})\text{。} \end{array}$

3.2 驱动功率分析

 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{F}} = - {({{{J}}^{\rm{T}}})^ + }\bigg({{{F}}_o} + \sum\limits_{i = 1}^4 {{{J}}_{li}^{\rm{T}}{{{F}}_{li}}} \bigg)$ (33)

 ${{P}} = {\rm{diag}}(\dot {{l}}){{F}}$ (34)

 ${\;\;\;\;\;\;\;\;W} = \int_{{{\rm{t_s}}}}^{{\rm{{t_e}}}} {{{{P}}_{\rm{s}}}} {\rm{d}}t \approx \sum {\frac{{{t_{\rm{e}}} - {t_{\rm{s}}}}}{{{n_t} - 1}}} \cdot {P_{\rm{s}}}\bigg(\frac{{{t_{k + 1}} - {t_k}}}{2}\bigg)$ (35)

4 数值模拟与仿真分析

 ${{{I}}_o}/({\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}) = {\rm{diag}}(\begin{array}{*{20}{c}} {0.000\;36}&{0.000\;22}&{0.000\;56} \end{array})\text{,}$
 ${{{I}}_{li}}/({\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}) = {\rm{diag}}(\begin{array}{*{20}{c}} {0.050\;99}&{0.050\;99}&{0.001\;48} \end{array})\text{。}$

 $\left\{\begin{array}{l}{{{f}}_o}/{\rm{N}} = {[\begin{array}{*{20}{c}} 0&0&{40} \end{array}]^{\rm{T}}},\\ {{{n}}_o}/({\rm{ N}} \cdot {\rm{m}}) = {[\begin{array}{*{20}{c}} {200}&{200}&0 \end{array}]^{\rm{T}}}\text{。}\end{array}\right.$

 $\left\{ \begin{array}{l} {\textit{z}}(t) = 0.25 - 0.05\cos (\text{π} t{\rm{/}}5), \\ \psi (t) = {\rm{0}}{\rm{.1sin(}}\text{π} t{\rm{/}}5{\rm{)}} - (\text{π} {\rm{/50)cos(}}\text{π} t{\rm{/}}5{\rm{) + }}\text{π} {\rm{/50}}, \\ \theta (t) = 0.1\cos (\text{π} t{\rm{/}}5) - 0.1 \text{。}\\ \end{array} \right.$

 图3 杆长变化曲线 Fig. 3 Length curves of rods

 图4 驱动杆速度曲线 Fig. 4 Velocity curves of driving rods

 图5 驱动杆加速度曲线 Fig. 5 Acceleration curves of driving rods

 图6 驱动力仿真曲线 Fig. 6 Simulation curves of driving forces

 图7 驱动功率仿真曲线 Fig. 7 Simulation curves of driving power

 图8 总功率仿真曲线 Fig. 8 Simulation curves of sum of driving power

 图9 虚拟样机模型 Fig. 9 Virtual prototype model