A prediction method of mechanical product assembly precision based on the fusion of measured samples and assembly feature fidelity samples

Abstract

Customized mechanical products based on orders have characteristics such as small batch sizes and multi-variety in production, which results in less availability of sample data related to assembly. Therefore, a problem of prediction of assembly precision exists due to the small number of samples. This paper studies the prediction method of mechanical product assembly precision based on the fusion of measured samples and feature fidelity samples. First, an assembly-feature fidelity sample (AFFS) generation method based on measured data is proposed, which expands the amount of mechanical product assembly feature samples through the meta-model concept. Then, a fusion method of measured samples and AFFS of mechanical products based on a double-layer learning network is proposed, which improves the under-fitting of a few samples and enhances the generalization ability of the model. Finally, case studies of gear-shaft assembly structure that are common in mechanical products, such as engines, gearboxes, and traction machines, were examined to verify the method we proposed and compare them with a tolerance analysis method and a machine learning method. The average errors of the assembly precision predicted by the method in this paper are 1.2% and 5.0%, respectively, which are better than the outcomes of SVR and NNs. The average errors of SVR and NNs are 3.2% and 2.8% in case 1 and 19% and 17% in case 2, respectively. The results show that the method in this paper has advantages of smaller error fluctuations and the best accuracy stability.

Appendices

Appendix 1

The corresponding Jacobian matrix J_FE0 of FE₀ is

$$ {J_{FE0}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right]. $$

(A1.1)

The corresponding Jacobian matrixes J_FE1 − J_FE4 of FE₁ − FE₄ are

$$ {J_{FE1 - FE4}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{35}&0\\ 0&1&0&{ - 35}&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right]. $$

(A1.2)

The corresponding Jacobian matrix ${J^{\prime }_{FE0}}$ of $FE^{\prime }_0$ is

$$ {J^{\prime}_{FE0}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right]. $$

(A1.3)

The corresponding Jacobian matrixes ${J^{\prime }_{FE1}} - {J^{\prime }_{FE4}}$ of $FE^{\prime }_1 - FE^{\prime }_4$ are

$$ {J^{\prime}_{FE1 - FE4}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{35}&{ - 60}\\ 0&1&0&{ - 35}&0&0\\ 0&0&1&{60}&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right]. $$

(A1.4)

The corresponding Jacobian matrix J_FE5 of FE₅ is

$$ {J_{FE5}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{35}&{ - 75}\\ 0&1&0&{ - 35}&0&0\\ 0&0&1&{75}&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right]. $$

(A1.5)

T₁ − T₄ are

$$ {T_{1}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{2}} - 32}}{2}}&0&{\frac{{{x^{2}} - 32}}{2} + {x^{1}} - 35}&{\frac{{{x^{2}} - 32}}{{10}}}&0&{\frac{{{x^{2}} - 32}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A1.6)

$$ {T_{2}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{3}} - 32}}{2}}&0&{\frac{{{x^{3}} - 32}}{2}}&{\frac{{{x^{3}} - 32}}{{10}}}&0&{\frac{{{x^{3}} - 32}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A1.7)

$$ {T_{3}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{4}} - 12}}{2}}&0&{\frac{{{x^{4}} - 12}}{2}}&{\frac{{{x^{4}} - 12}}{{10}}}&0&{\frac{{{x^{4}} - 12}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A1.8)

$$ {T_{4}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{5}} - 12}}{2}}&0&{\frac{{{x^{5}} - 12}}{2}}&{\frac{{{x^{5}} - 12}}{{{x^{11}}}}}&0&{\frac{{{x^{5}} - 12}}{{{x^{11}}}}} \end{array}} \right]^{\text{T}}}; $$

(A1.9)

$T^{\prime }_1-T^{\prime }_4$ are

$$ {T^{\prime}_{1}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{7}} - 32}}{2}}&0&{\frac{{{x^{7}} - 32}}{2} + {x^{6}} - 35}&{\frac{{{x^{7}} - 32}}{{10}}}&0&{\frac{{{x^{7}} - 32}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A1.10)

$$ {T^{\prime}_{2}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{8}} - 32}}{2}}&0&{\frac{{{x^{8}} - 32}}{2}}&{\frac{{{x^{8}} - 32}}{{10}}}&0&{\frac{{{x^{8}} - 32}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A1.11)

$$ {T_{3'}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{9}} - 12}}{2}}&0&{\frac{{{x^{9}} - 12}}{2}}&{\frac{{{x^{9}} - 12}}{{10}}}&0&{\frac{{{x^{9}} - 12}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A1.12)

$$ {T^{\prime}_{4}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{10}} - 12}}{2}}&0&{\frac{{{x^{10}} - 12}}{2}}&{\frac{{{x^{10}} - 12}}{{{x^{12}} - 49}}}&0&{\frac{{{x^{10}} - 12}}{{{x^{12}} - 49}}} \end{array}} \right]^{\text{T}}}; $$

(A1.13)

T₅ is

$$ {T_{5}} = {\left[ {\begin{array}{*{20}{c}} 0&{{x^{12}} - 80}&0&{\frac{{{x^{12}} - 80}}{{{x^{10}}}}}&0&{\frac{{{x^{12}} - 80}}{{{x^{10}}}}} \end{array}} \right]^{\text{T}}}. $$

(A1.14)

Appendix : 2

The corresponding Jacobian matrixes J_FE0 − J_FE2 of FE₀ − FE₂ are

$$ {J_{FE0}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{35}&0\\ 0&1&0&{ - 35}&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right], $$

(A2.1)

$$ {J_{FE1}} = \left[ {\begin{array}{*{20}{c}} {\text{1}}&{\text{0}}&{\text{0}}&{\text{0}}&{{\text{35}}}&{{\text{ - 37}}{\text{.5}}}\\ {\text{0}}&{\text{1}}&{\text{0}}&{{\text{ - 35}}}&{\text{0}}&{\text{0}}\\ {\text{0}}&{\text{0}}&{\text{1}}&{{\text{37}}{\text{.5}}}&{\text{0}}&{\text{0}}\\ {\text{0}}&{\text{0}}&{\text{0}}&{\text{1}}&{\text{0}}&{\text{0}}\\ {\text{0}}&{\text{0}}&{\text{0}}&{\text{0}}&{\text{1}}&{\text{0}}\\ {\text{0}}&{\text{0}}&{\text{0}}&{\text{0}}&{\text{0}}&{\text{1}} \end{array}} \right], $$

(A2.2)

$$ {J_{FE2}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{35}&{ - 45}\\ 0&1&0&{ - 35}&0&0\\ 0&0&1&{45}&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}} \right]. $$

(A2.3)

T₀ − T₂ are

$$ {T_{0}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{1}} - 12}}{2}}&0&{\frac{{{x^{1}} - 12}}{2}}&{\frac{{{x^{1}} - 12}}{{10}}}&0&{\frac{{{x^{1}} - 12}}{{10}}} \end{array}} \right]^{\text{T}}}, $$

(A2.4)

$$ \begin{array}{@{}rcl@{}} {T_{1}} &= &{\bigg[ {\begin{array}{*{20}{c}} {\frac{{{x^{3}} - 15}}{2}}&{{x^{4}} - 35}&{\frac{{{x^{3}} - 15}}{2}}&{\min \left( {\frac{{{x^{4}} - 35}}{{20}},\frac{{{x^{3}} - 15}}{{14}}} \right)}&0 \end{array}}}\\ &&~~{ {\begin{array}{*{20}{c}} {\min \left( {\frac{{{x^{4}} - 35}}{{20}},\frac{{{x^{3}} - 15}}{{14}}} \right)} \end{array}} \bigg]^{\text{T}}}, \end{array} $$

(A2.5)

$$ {T_{2}} = {\left[ {\begin{array}{*{20}{c}} {\frac{{{x^{3}} - 15}}{2}}&0&{\frac{{{x^{3}} - 15}}{2}}&{\frac{{{x^{1}} - 15}}{{15}}}&0&{\frac{{{x^{3}} - 15}}{{15}}} \end{array}} \right]^{\text{T}}}. $$

(A2.6)