基于向量投影响应面的数控机床几何误差分配方法

doi:10.13229/j.cnki.jdxbgxb20211089

Abstract

Abstract:

Taking the geometric errors as the research object， this paper presents an approach for the error allocation of CNC machine tool based on the vector projection response surface. Based on the multi-body system theory， a comprehensive error model of machine tool was established to reveal the quantitative relationship between the machining accuracy and geometric errors of the machine tool； by applying the vector projection response surface method， a machining accuracy reliability model was developed to predict and evaluate the machining ability of machine tools， then a machining accuracy reliability sensitivity model was proposed to sequence the influence of distribution parameters of geometric errors on the machining accuracy reliability， and therefore it conducted the identification and optimization of the geometric errors which have larger affect to the machining accuracy reliability. Hence， a general approach for the error allocation of machine tools was formed to realize the reasonable optimization of the geometric error parameters and overall improvement of machining ability of machine tool. The proposed approach was implemented to a four-axis CNC machining center， and the results verified the effectiveness and competitiveness of the approach.

Key words: CNC machine tool, machining accuracy reliability, sensitivity, vector projection response surface method

CLC Number:

TH161

Zi-ling ZHANG,Xiong HU,Yin QI,Wei WANG,Zhi-qiang TAO,Zhi-feng LIU. An approach for error allocation of machine tool based on vector projection response surface method[J].Journal of Jilin University(Engineering and Technology Edition), 2022, 52(2): 384-391.

Figures/Tables 6

Fig.1

Table 1

Ideal and error homogenous transformation matrices of machine tool"

相邻体	体间理想静止、理想运动特征矩阵	体间实际静止、运动误差特征矩阵
0-1 X轴	$C 01 p = I 4 × 4$	$Δ C 01 p = I 4 × 4$
0-1 X轴	$C 01 s = 100 x 010000100001$	$Δ C 01 s = 1 - Δ γ x Δ β x Δ x x Δ γ x 1 - Δ α x Δ y x - Δ β x Δ α x 1 Δ z x 0001$
1-2 Y轴	$C 12 p = I 4 × 4$	$Δ C 12 p = 1 - Δ γ x y 00 Δ γ x y 100 00100001$
1-2 Y轴	$C 12 s = 1000 010 y 00100001$	$Δ C 12 s = 1 - Δ γ y Δ β y Δ x y Δ γ y 1 - Δ α y Δ y y - Δ β y Δ α y 1 Δ z y 0001$
0-4 Z轴	$C 04 p = I 4 × 4$	$Δ C 04 p = 10 Δ β x z 0 01 - Δ α y z 0 - Δ β x z Δ α y z 10 0001$
0-4 Z轴	$C 04 s = 10000100 001 z 0001$	$Δ C 04 s = 1 - Δ γ z Δ β z Δ x z Δ γ z 1 - Δ α z Δ y z - Δ β z Δ α z 1 Δ z z 0001$
4-5 B轴	$C 45 p = I 4 × 4$	$Δ C 45 p = 1 - Δ γ y b 00 Δ γ y b 1 - Δ α y b 0 0 Δ α y b 1 Δ z y b 0001$
4-5 B轴	$C 45 s = c o s B 0 - s i n B 0 0100 s i n B 0 c o s B 0 0001$	$Δ C 45 s = 1 - Δ γ B Δ β B Δ x B Δ γ B 1 - Δ α B Δ y B - Δ β B Δ α B 1 Δ z B 0001$

Table 1

Fig.2

Fig.3

Fig.4

Fig.5

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