CN103217924A - Dynamics modeling method of over-constrained heavy parallel machine tool applied to real-time control - Google Patents

Abstract

The invention discloses a dynamics modeling method of an over-constrained heavy parallel machine tool applied to real-time control, and belongs to the technical field of machine tool manufacturing. The method includes confirming positions, speed and accelerated speed of driving joints according to the structure of the machine tool, and confirming positions, speed and accelerated speed of component mass centers; building a speed relation between a movable platform and the driving joints; building force balance equations and torque equations of components of the over-constrained heavy parallel machine tool by means of a Newton-Euler method; according to structure parameters of the over-constrained heavy parallel machine tool and axial deformation situation of rod pieces of a branch chain, building a coordination equation of output errors between deformation and the movable platform; and finally combining the force balance equations, the torque equations and the coordination equation and building a dynamics model. The method takes the deformation of a connecting rod with poor rigidity of the heavy parallel machine tool into consideration and solves the problem that a traditional rigid body dynamics model is low in precision, and the built dynamics model meets real-time calculation requirements of a control system.

Description

The mistake of using towards real-time control retrains heavy parallel machine dynamic modeling method

Technical field

The invention belongs to the technology of Machine Tool field, related in particular to the heavy parallel machine Dynamic Modeling method of constraint.

Background technology

Parallel machine has in theory that the rigidity mass ratio is big, response speed is fast and advantage such as environmental suitability is strong, yet also has shortcomings such as little, the unusual position of work space shape is many.Driving redundancy is the effective ways that address this problem, and the driving redundancy can be carried out the demarcation certainly of parallel machine.In addition,, parallel machine was designed to constraint mechanism, can improves the rigidity of lathe by introducing extra connecting rod.Therefore, driving the redundant constraint parallel machine of crossing and have quiet preferably, dynamic property, is the lathe that a class has broad prospect of application.In order to realize high dynamic performance, in control system and structural design, need the high precision kinetic model, particularly require stricter based on precision and the real-time to kinetic model in the control system of the high-speed, high precision parallel machine of kinetic model.

General rigid dynamics modeling method commonly used, as newton-Euler's method method, Da Langbai-Lagrange's equation method and principle of virtual work method, can be used for setting up the rigid dynamics model of constraint and driving redundant parallel lathe, but it is not unique to drive separating of redundant parallel lathe rigid dynamics inverse problem.For example, Nahon and Angeles have summarized at four kinds of methods that drive redundant rigid dynamics inverse problem: quadrature decomposition method, pseudoinverse technique, Lagrangian method, direct method.For crossing constraint parallel institution kinetic model, its static equation number is less than the number of unknown force, can not directly find the solution.In order to find the solution all unknown force, in modeling, will cross the constraint parallel institution usually and be reduced to the non-constraint mechanism of crossing, at this moment, the number of static force equation equals the number of unknown force.Although these methods can be set up constraint and drive the kinetic model of redundant parallel lathe, but do not consider the distortion of the relatively poor rod member of rigidity in the modeling and will cross constraint mechanism to be reduced to the non-error that constraint mechanism brings of crossing, the kinetic model of being set up is inaccurate.Especially it is bigger to cross the distortion that retrains connecting rod in the heavy parallel machine, if do not consider rod deformation, the rigid dynamics model accuracy of being set up is lower.The structure that a kind of mistake retrains heavy parallel machine as shown in Figure 1, the parallel institution of this lathe part is made up of three kinematics side chains, wherein two side chains 1 are parallelogram sturcutre, respectively by slide block B ₁B ₂, B ₃B ₄, revolute pair (being called the joint again), fixed length connecting rod A ₁B ₁, A ₂B ₂, A ₃B ₃, A ₄B ₄And moving platform (A ₁A ₂A ₃A ₄A ₅) form, drive by the AC servo motor (not shown), and adopt the ball-screw transmission; The 3rd side chain 3 comprises member 51(B by revolute pair, shrinking connecting-rod ₅C) and member 52(CA ₅) and moving platform A ₁A ₂A ₃A ₄A ₅Form, by the AC servo machinery driving shrinking connecting-rod, its knife rest is fixed on moving platform.Though heavy parallel machine elastokinetics is more accurate, elastokinetics is too complicated, and the time that inverse dynamics calculates is long, can not be used for control system.Control system needs a large amount of complicated calculations in an interpolation cycle, the computing method difference, and the time of interpolation is also just different, and the interpolation cycle of this heavy machine tool generally is no more than 2ms.And when designing a calculating machine control system, usually require motion control computing time less than 10% of control cycle, thereby need the digital control system of heavy parallel machine to finish in the time of 0.2ms that interpolation calculating, kinematics are contraryly separated calculatings, contrary the separating of dynamics calculated and each controller calculating.In above-mentioned every calculation task, dynamics is contrary, and to separate calculating the most consuming time, should finish the contrary calculating of separating of dynamics with interior at 0.1ms.

The motion control effect of heavy parallel machine mainly depends on model accuracy, therefore sets up heavy parallel machine high precision kinetic model, and can be applied to real-time control, and to improving heavy parallel machine precision, it is significant to improve its practicability level.

Summary of the invention

The objective of the invention is to disclose a kind of mistake of using towards real-time control and retrain heavy parallel machine dynamic modeling method, overcome the defective that present rigid dynamics model accuracy is low, the elastokinetics model can not be applied to real-time control system.

Dynamic modeling method of the present invention may further comprise the steps:

1), on the basis of kinematics analysis, obtains the position in each active joint of lathe, speed and acceleration according to crossing the structure that retrains heavy parallel machine; And each member barycenter position in the lathe different coordinates, speed and acceleration are also set up initiatively length velocity relation in joint of moving platform and each according to Jacobi matrix simultaneously;

2) according to crossing the situation that each member of heavy parallel machine carries out force analysis that retrains, adopt Niu Dun Euler's method set up equilibrium equation and the momental equation that retrains each member in the heavy parallel machine;

3), set up the equation of comptability of the output error of the axial deformation that retrains in the heavy parallel machine side chain and moving platform according to the rod member axial deformation situation of crossing structural parameters and the relatively poor side chain of rigidity retrain heavy parallel machine;

4) last, equilibrium equation, momental equation and error equation of comptability simultaneous are set up kinetic model, and obtain Model parameter.

The invention provides a kind of initial design stage at heavy parallel machine, towards the heavy parallel machine high precision Dynamic Modeling method method that real-time control is used, its characteristics and beneficial effect are:

1. compare with existing parallel machine dynamic modeling method, the kinetic model that this method is set up has been considered the distortion of the relatively poor rod member of rigidity, compares with the rigid dynamics model, has improved model accuracy.In addition, this method is fit to drive redundant and crosses the Dynamic Modeling that retrains parallel machine.

2. the kinetic model counting yield height set up of this patent satisfies the requirement of control system real-time, can be applied in the control method of parallel machine based on model, thereby improve machine tool accuracy.

Description of drawings

Below in conjunction with the drawings and specific embodiments the present invention is described in further detail.

Fig. 1 is that the present invention is to driving the redundant analysis chart that retrains axial parallel machine tool structure of crossing;

Fig. 2 is a moving platform error synoptic diagram of the present invention.

Fig. 3 is the site error and the angular error synoptic diagram of moving platform of the present invention, and wherein (a) is the site error of x and y direction, (b) is rotation error.

Embodiment

Method embodiment of the present invention may further comprise the steps as follows:

1) according to crossing the structure that retrains heavy parallel machine, by the kinematics analysis that lathe carries out, obtains the position in each joint of lathe, speed and acceleration; And each member barycenter position in the lathe different coordinates, speed and acceleration are also set up the moving platform of this lathe and the length velocity relation in each active joint according to Jacobi matrix simultaneously; Specifically comprise:

11) set up the fixed coordinate system O-XY that is fixed on the lathe frame, as shown in Figure 1, the O point is the contact point C on two columns of lathe and ground ₁, C ₃Mid point; The X-axis level to the right, Y-axis is vertically upward; Be based upon the fixed coordinate system O '-xy on the moving platform, the center of moving platform is designated as O ', the x axle horizontal to the right, the y axle is vertically upward.

At articulation point B _iFoundation is with B _iA _iDirection is the moving coordinate system B of x ' axle _i-x ' y ' and the moving coordinate system B parallel with fixed coordinate system O-XY _i-xy.Obtain according to the geometry site among the figure:

r+a _i＝b _i+q _ie ₂+ln _i，i＝1,2,3,4 (1)

Wherein, r=[x y] ^TBe O ' some position vector in the O-XY coordinate system, a _iBe A _iThe position vector of point in O '-xy coordinate system, a _i=[a _Ixa _Iy] ^T, b _iExpression from an O to C _iVector, q _iExpression articulation point B _iY coordinate in coordinate system O-XY, e ₂=[0 1] ^T, l and n _iRepresent connecting rod A respectively _iB _iLength and unit vector.

Connecting rod A ₅B ₅Constrain equation can be expressed as

r+a ₅＝b ₅+(l ₅₁+l ₅₂)n ₅ (2)

Wherein, a ₅Be A ₅At the position vector of O '-xy coordinate system, b ₅For from an O to a B ₅Vector, l ₅₁And l ₅₂Be respectively the length of member 51 and member 52, n ₅Be connecting rod A ₅B ₅Unit vector.

12) first order derivative is asked to the time simultaneously in the both sides of expression formula (1), and the speed that draws each slide block is:

${\stackrel{\·}{q}}_i=\frac{{n_i}^T\stackrel{\·}{r}}{n_{\mathrm{iy}}},(i=\mathrm{1,2,3,4}),{\stackrel{\·}{q}}_5=0---\left(3\right)$

Wherein,

Be O ' some velocity in the O-XY coordinate system, connecting rod A _iB _iAnd A ₅B ₅Angular velocity be respectively: ${\mathrm{\ω}}_i=\frac{(\stackrel{\·}{r}-{\stackrel{\·}{q}}_i{e}_2)}{l}\·\left[\begin{array}{c}{-n}_{\mathrm{iy}}\\ n_{\mathrm{ix}}\end{array}\right],$ ${\mathrm{\ω}}_5=\frac{\stackrel{\·}{r}}{l_{51}+l_{52}}\·\left[\begin{array}{c}{-n}_{5y}\\ n_{5x}\end{array}\right],$ Wherein, n _IxAnd n _IyBe divided into n _iComponent on X and Y-axis (i=1,2,3,4,5).

The speed of the barycenter of member 51 and member 52 in the O-XY coordinate system is expressed as respectively:

${\stackrel{\·}{r}}_{g51}=\frac{l_{51}{\mathrm{\ω}}_5}{2}\left[\begin{array}{c}-n_{5y}\\ n_{5x}\end{array}\right]---\left(4\right)$

${\stackrel{\·}{r}}_{g52}=\frac{{\stackrel{\·}{l}}_{52}}{2}{n}_5+(l_{51}+\frac{l_{52}}{2}){\mathrm{\ω}}_5\left[\begin{array}{c}{-n}_{5y}\\ n_{5x}\end{array}\right]---\left(5\right)$

13) second derivative is asked to the time simultaneously in the both sides of expression formula (1), and the acceleration that draws each slide block is:

${\stackrel{\·\·}{q}}_i=\frac{n_i\stackrel{\·\·}{r}+{\mathrm{\ω}}_i^2{l}_i}{n_{\mathrm{iy}}},(i=\mathrm{1,2,3,4}),{\stackrel{\·\·}{q}}_5=0---\left(6\right)$

Wherein,

Be O ' some acceleration in the O-XY coordinate system, connecting rod A _iB _iAnd A ₅B ₅Angular acceleration be respectively: ${\mathrm{\ϵ}}_i=\frac{\stackrel{\·\·}{r}-{\stackrel{\·\·}{q}}_i{e}_2}{l}\·\left[\begin{array}{c}-n_{\mathrm{iy}}\\ n_{\mathrm{ix}}\end{array}\right],$ ${\mathrm{\ϵ}}_5=\frac{\stackrel{\·\·}{r}\·{\left[\begin{array}{cc}-n_{5y}& n_{5x}\end{array}\right]}^T-2{\stackrel{\·}{l}}_{52}{\mathrm{\ω}}_5}{l_{51}+l_{52}}.$ The acceleration of the barycenter of member 51 and member 52 in coordinate system O-XY is respectively:

${\stackrel{\·\·}{r}}_{g51}=\frac{l_{51}{\mathrm{\ϵ}}_5}{2}\left[\begin{array}{c}-n_{5y}\\ n_{5x}\end{array}\right]-\frac{l_{51}{\mathrm{\ω}}_5^2{n}_5}{2}---\left(7\right)$

${\stackrel{\·\·}{r}}_{g52}=\frac{l_{52}}{2}{n}_5+{\stackrel{\·}{l}}_{52}{\mathrm{\ω}}_5\left[\begin{array}{c}-n_{5y}\\ n_{5x}\end{array}\right]+(l_{51}+\frac{l_{52}}{2}){\mathrm{\ϵ}}_5\left[\begin{array}{c}-n_{5y}\\ n_{5x}\end{array}\right]-(l_{51}+\frac{l_{52}}{2}){\mathrm{\ω}}_5^2{n}_5---\left(8\right)$

2) according to retraining the situation that each member of heavy parallel machine carries out force analysis, utilize newton-Euler's method to write out the Newton's equation and the Eulerian equation of each member respectively, specifically comprise crossing:

21) connecting rod A _iB _iEquilibrium equation as shown in Equation (9):

$F_{\mathrm{Bi}}+F_{\mathrm{Bi}}+m_{i}g-m_i{\stackrel{\·\·}{r}}_{\mathrm{gi}},(i=\mathrm{1,2,3,4})---\left(9\right)$

Wherein, m _iBe A _iB _iThe quality of bar, F _AiBe connecting rod A _iB _iAct on the power on the moving platform, and F _Ai=[F _{Aix '}F _{Aiy '}] ^T, It is rod piece A _iB _iAcceleration vector of center of mass, g=[0 g] ^T, and g is acceleration of gravity, F _B1And F _B2It is slide block B ₁B ₂Act on connecting rod A ₁B ₁And A ₂B ₂Constraining force, F _B3And F _B4It is slide block B ₃B ₄Act on connecting rod A ₃B ₃And A ₄B ₄Constraining force, and F _Bi=[F _{Bix '}F _{Biy '}] ^TConnecting rod A _iB _iMomental equation as shown in Equation (10):

$F_{\mathrm{Ai}}.l\left[\begin{array}{c}-n_{\mathrm{iy}}\\ n_{\mathrm{ix}}\end{array}\right]-m_i(g+{\stackrel{\·\·}{q}}_i).r_{\mathrm{cxi}}-m_i{\stackrel{\·\·}{r}}_{\mathrm{ci}}\·\left[\begin{array}{c}-r_{\mathrm{cyi}}\\ r_{\mathrm{cxi}}\end{array}\right]-J_i{\mathrm{\ϵ}}_i=0,(i=\mathrm{1,2,3,4})---\left(10\right)$

Wherein, J _iBe connecting rod A _iB _iBased on B _iThe moment of inertia of-xy coordinate system, r _CiBe connecting rod A _iB _iAt B _iPosition vector under the-xy coordinate system, r _CxiAnd r _CyiIt is respectively the component of x axle and y axle.

Connecting rod A ₅B ₅Equilibrium equation and momental equation respectively shown in formula (11) and (12):

$F_{A5}+F_{B5}+m_{51}(g-{\stackrel{\·\·}{r}}_{g51})+m_{52}(g-{\stackrel{\·\·}{r}}_{g52})=0---\left(11\right)$

$F_{A5}.(l_{51}+l_{52})\left[\begin{array}{c}-n_{5y}\\ n_{5x}\end{array}\right]-m_{51}g.r_{\mathrm{cx}51}-m_{51}{\stackrel{\·\·}{r}}_{g51}\left[\begin{array}{c}-r_{\mathrm{cy}51}\\ r_{\mathrm{cx}51}\end{array}\right]-m_{52}g{.r}_{\mathrm{cx}52}-m_{52}{\stackrel{\·\·}{r}}_{g52}\left[\begin{array}{c}-r_{\mathrm{cy}52}\\ r_{\mathrm{cx}52}\end{array}\right]-(J_{51}+J_{52}){\mathrm{\ϵ}}_5=0---\left(12\right)$

Wherein, J ₅₁And J ₅₂Be respectively connecting rod B ₅C and CA ₅With respect to coordinate system B ₅The moment of inertia of-xy, m ₅₁And m ₅₂Be respectively the quality of member 51 and member 52, F _A5And F _B5Expression moving platform and beam effect are at connecting rod A ₅B ₅Last acting force, r _Cx51And r _Cy51Represent member B respectively ₅The C barycenter is at coordinate system B ₅Position vector is along the component of x and y among-the xy.

21) the equilibrium equation public affairs of moving platform are as shown in Equation (13):

$F_e+M(g-\stackrel{\·\·}{r})-\underset{i=1}{\overset{5}{\mathrm{\Σ}}}{F}_{\mathrm{Ai}}=0---\left(13\right)$

Wherein, M is the quality of moving platform, F _eFor being applied to the external force on the moving platform.

If the barycenter of moving platform is R with respect to the position vector of coordinate system O '-xy _c=[R _CxR _Cy] ^T, then the Eulerian equation of moving platform is as shown in Equation (14):

$-M\·g\·R_{\mathrm{cx}}-M\stackrel{\·\·}{r}\left[\begin{array}{c}-R_{\mathrm{cy}}\\ R_{\mathrm{cx}}\end{array}\right]+M_e-\underset{i=1}{\overset{5}{\mathrm{\Σ}}}(F_{\mathrm{Ai}}\·\left[\begin{array}{c}-A_{\mathrm{iy}}\\ A_{\mathrm{ix}}\end{array}\right])=0---\left(14\right)$

Wherein, M _eBe the moment that is applied on the platform, A _IxAnd A _IyBe that an O ' is to A _iVector along F _Aix'And F _{Aiy '}The component of direction.

3) according to the rod member axial deformation situation of crossing structural parameters and the side chain retrain heavy parallel machine, the equation of comptability of output error of setting up the axial deformation that retrains in the heavy parallel machine side chain and moving platform is as follows:

(when actual implementation, because each side chain in the lathe is all very long, so the power that is applied on them produces distortion.The distortion of rod member is bigger, can produce very big influence to kinetic model.)

Connecting rod A _iB _iAxial deformation such as formula (15):

${\mathrm{\δ}}_i={\∫}_0^l{F}_u\mathrm{du}=\frac{-n_i}{E{S}_i}(F_{\mathrm{Bi}}+m_i\mathrm{gl}/2-m_i{\stackrel{\·\·}{r}}_{\mathrm{gi}}l/2),(i=\mathrm{1,2,3,4})---\left(15\right)$

Wherein, E represents connecting rod A _iB _iElastic modulus, δ _iExpression connecting rod A _iB _iAxial deformation, S _iBe connecting rod A _iB _iCross-sectional area, F _uExpression connecting rod A _iB _iGo up B _iDistance is the suffered power of u point,

Be connecting rod A _iB _iThe acceleration of barycenter.

Can calculate connecting rod A equally ₅B ₅Axial deformation such as formula (16):

${\mathrm{\δ}}_5=\frac{1}{2E\·S_{52}}(2F_{{A5x}^{\′}}+m_{52}{l}_{52}{n}_5(g-{\stackrel{\·\·}{r}}_{g52}))+\frac{1}{2E\·S_{51}}{m}_{51}{l}_{51}{n}_5(g-{\stackrel{\·\·}{r}}_{g51})---\left(16\right)$

Wherein, S ₅₁And S ₅₂It is the cross-sectional area of member 51 and member 52.

In theory, moving platform can not rotate, but considers that rod member axial deformation, moving platform have small rotation d γ, as shown in Figure 2.Connecting rod A _iB _iAxial deformation and the moving platform site error between relation as formula (17):

δ _i＝n _i(a′ _i-a _i)＝n _i((T-I)a _i+[dx dy] ^T)(i=1,2,3,4,5) (17)

In the formula, dx is the directions X error, and dy is the Y deflection error, $T=\left[\begin{array}{cc}\cos \left(\mathrm{d\γ}\right)& \sin \left(\mathrm{d\γ}\right)\\ -\sin \left(\mathrm{d\γ}\right)& \cos \left(\mathrm{d\γ}\right)\end{array}\right],$ A ' _iPut A after the expression moving platform generation minor rotation _iPosition vector in coordinate system O '-xy, I are the unit square formation of two dimension.

Equation (17) can be write as

$\left[\begin{array}{c}{\mathrm{\δ}}_1\\ {\mathrm{\δ}}_2\\ {\mathrm{\δ}}_3\\ {\mathrm{\δ}}_4\\ {\mathrm{\δ}}_5\end{array}\right]=Q\left[\begin{array}{c}\mathrm{dx}\\ \mathrm{dy}\\ \mathrm{d\γ}\end{array}\right]---\left(18\right)$

In the formula, $Q=\left[\begin{array}{ccc}{n}_{1x}& n_{1y}& (n_{1x}{a}_{1y}-n_{1y}{a}_{1x})\\ n_{2x}& n_{2y}& (n_{2x}{a}_{2y}-n_{2y}{a}_{2x})\\ n_{3x}& n_{3y}& (n_{3x}{a}_{3y}-n_{3y}{a}_{3x})\\ n_{4x}& n_{4y}& (n_{4x}{a}_{4y}-n_{4y}{a}_{4x})\\ n_{5x}& n_{5y}& (n_{5x}{a}_{5y}-n_{5y}{a}_{5x})\end{array}\right].$

System of equations is set up as shown in Equation (19) in simultaneous equations (15), (16) and (18):

$Q\left[\begin{array}{c}\mathrm{dx}\\ \mathrm{dy}\\ \mathrm{d\γ}\end{array}\right]=\left[\begin{array}{c}\frac{1}{2E{S}_1}(2F_{A1x^{\′}}+m_{1}l{n}_1(g-{\stackrel{\·\·}{r}}_{g1}))\\ \frac{1}{2E{S}_2}(2F_{A2x^{\′}}+m_{2}l{n}_2(g-{\stackrel{\·\·}{r}}_{g2}))\\ \frac{1}{2E{S}_3}(2F_{A3x^{\′}}+m_{3}l{n}_3(g-{\stackrel{\·\·}{r}}_{g3}))\\ \frac{1}{2E{S}_4}(2F_{A4x^{\′}}+m_{4}l{n}_4(g-{\stackrel{\·\·}{r}}_{g4}))\\ \frac{1}{2E{S}_{52}}(2F_{A5x^{\′}}{l}_{52}+m_{52}{l}_{52}{n}_5(g-{\stackrel{\·\·}{r}}_{g52}))+\frac{m_{51}{l}_{51}{n}_5(g-{\stackrel{\·\·}{r}}_{g51})}{2E{S}_{51}}\end{array}\right]---\left(19\right)$

4) last, the equation of comptability (19) of the equilibrium equation of simultaneous member (9), (11), (13) and momental equation (10), (12), (14) and connecting rod axial deformation and moving platform output error obtains kinetic model, and obtains Model parameter; Described parameter comprises: connecting rod A _iB _iGive the moving platform acting force along bar length direction component F _Aix', moving platform site error dx, dy and rotation error d γ, slide block acts on the power F on the connecting rod _BiAnd driving force, be specially at first simultaneous formula (13), (14) and (19) obtain F _Aix'(i=1,2,3,4,5), dx, dy and d γ obtain F according to formula (10) and (12) then _Aiy', further obtain F according to formula (9) and (11) _Bi, at last to slide block B ₁B ₂And B ₃B ₄Carry out force analysis, according to slide block and member B ₅The equilibrium equation of C can calculate driving force.

Carry out Computer Simulation according to the driving force method for solving, compare with the model of not considering rod deformation, the error of the X that obtains, the position of Y-axis and corner, see Fig. 3 (a), wherein, circular curve and normalized curve are respectively X and Y-axis site error, curve is represented angular error among Fig. 3 (b), can find out that thus the model that the present invention sets up is more accurate, improve the precision of kinetic model, and then improved the precision of heavy machine tool.

To spend time of inverse dynamics of the redundant heavy parallel machine of constraint be 0.05ms with being configured to Duo2 2.2GHZ CPU of Intel and 1.96GB internal memory Microcomputer Calculation to use executive system of the present invention.Counting yield can satisfy the requirement of real-time control system.

Claims (5)

1. a mistake of using towards real-time control retrains heavy parallel machine dynamic modeling method, it is characterized in that this method may further comprise the steps:

1),, obtains the position in each active joint of lathe, speed and acceleration by kinematics analysis according to crossing the structure that retrains heavy parallel machine; And each member barycenter position in the lathe different coordinates, speed and acceleration are also set up initiatively length velocity relation in joint of moving platform and each according to Jacobi matrix simultaneously;

2) according to crossing the situation that each member of heavy parallel machine carries out force analysis that retrains, adopt newton-Euler's method to set up equilibrium equation and the momental equation that retrains each member in the heavy parallel machine;

3), set up the equation of comptability of the output error of the axial deformation that retrains in the heavy parallel machine side chain and moving platform according to the rod member axial deformation situation of crossing structural parameters and the side chain retrain heavy parallel machine;

4) last, equilibrium equation, momental equation and error equation of comptability simultaneous are set up kinetic model, and obtain Model parameter.

2. method according to claim 1 is characterized in that described step 1) specifically comprises:

11) set up the fixed coordinate system O-XY that is fixed on the lathe frame, the O point is the contact point C on two columns of lathe and ground ₁, C ₃Mid point; The X-axis level to the right, Y-axis is vertically upward; Be based upon the fixed coordinate system O '-xy on the moving platform, the center of moving platform is designated as O ', the x axle horizontal to the right, the y axle is vertically upward.

At articulation point B _iFoundation is with B _iA _iDirection is the moving coordinate system B of x ' axle _i-x ' y ' and the moving coordinate system B parallel with fixed coordinate system O-XY _i-xy.Obtain according to the geometry site among the figure:

r+a _i＝b _i+q _ie ₂+ln _i，i＝1,2,3,4 (1)

Wherein, r=[x y] ^TBe O ' some position vector in the O-XY coordinate system, a _iBe A _iThe position vector of point in O '-xy coordinate system, a _i=[a _Ixa _Iy] ^T, b _iExpression from an O to C _iVector, q _iExpression articulation point B _iY coordinate in coordinate system O-XY, e ₂=[0 1] ^T, l and n _iRepresent connecting rod A respectively _iB _iLength and unit vector.

Connecting rod A ₅B ₅Constrain equation be expressed as

r+a ₅＝b ₅+(l ₅₁+l ₅₂)n ₅ (2)

Wherein, a ₅Be A ₅At the position vector of O '-xy coordinate system, b ₅For from an O to a B ₅Vector, l ₅₁And l ₅₂Be respectively the length of member (51) and member (52), n ₅Be connecting rod A ₅B ₅Unit vector;

12) first order derivative is asked to the time simultaneously in the both sides of expression formula (1), and the speed that draws each slide block is:

Wherein,

Be O ' some velocity in the O-XY coordinate system, connecting rod A _iB _iAnd A ₅B ₅Angular velocity be respectively:

Wherein, n _IxAnd n _IyBe divided into n _iComponent on X and Y-axis, i=1,2,3,4,5;

The speed of barycenter in the O-XY coordinate system of member (51) and member (52) is expressed as respectively:

13) second derivative is asked to the time simultaneously in the both sides of expression formula (1), and the acceleration that draws each slide block is:

Wherein,

Be O ' some acceleration in the O-XY coordinate system, connecting rod A _iB _iAnd A ₅B ₅Angular acceleration be respectively:

The acceleration of barycenter in coordinate system O-XY of member (51) and member (52) is respectively:

3. as method as described in the claim 2, it is characterized in that described step 2) specifically comprise:

21) connecting rod A _iB _iEquilibrium equation as shown in Equation (9):

Wherein, m _iBe A _iB _iThe quality of bar, F _AiBe connecting rod A _iB _iAct on the power on the moving platform, and F _Ai=[F _{Aix '}F _{Aiy '}] ^T,

It is rod piece A _iB _iAcceleration vector of center of mass, g=[0 g] ^T, and g is acceleration of gravity, F _B1And F _B2It is slide block B ₁B ₂Act on connecting rod A ₁B ₁And A ₂B ₂Constraining force, F _B3And F _B4It is slide block B ₃B ₄Act on connecting rod A ₃B ₃And A ₄B ₄Constraining force, and F _Bi=[F _{Bix '}F _{Biy '}] ^TConnecting rod A _iB _iMomental equation as shown in Equation (10):

Wherein, J _iBe connecting rod A _iB _iBased on B _iThe moment of inertia of-xy coordinate system, r _CiBe connecting rod A _iB _iAt B _iPosition vector under the-xy coordinate system, r _CxiAnd r _CyiIt is respectively the component of x axle and y axle;

Connecting rod A ₅B ₅Equilibrium equation and momental equation respectively shown in formula (11) and (12):

Wherein, J ₅₁And J ₅₂Be respectively connecting rod B ₅C and CA ₅With respect to coordinate system B ₅The moment of inertia of-xy, m ₅₁And m ₅₂Be respectively the quality of member 51 and member 52, F _A5And F _B5Expression moving platform and beam effect are at connecting rod A ₅B ₅Last acting force, r _Cx51And r _Cy51Represent member B respectively ₅The C barycenter is at coordinate system B ₅Position vector is along the component of x and y among-the xy;

21) the equilibrium equation public affairs of moving platform are as shown in Equation (13):

Wherein, M is the quality of moving platform, F _eFor being applied to the external force on the moving platform.

If the barycenter of moving platform is R with respect to the position vector of coordinate system O '-xy _c=[R _CxR _Cy] ^T, then the Eulerian equation of moving platform is as shown in Equation (14):

Wherein, M _eBe the moment that is applied on the platform, A _IxAnd A _IyBe that an O ' is to A _iVector along F _Aix'And F _{Aiy '}The component of direction.

4. as method as described in the claim 2, it is characterized in that it is as follows that described step 3) was set up the equation of comptability of output error of the axial deformation that retrains in the heavy parallel machine side chain and moving platform:

Connecting rod A _iB _iAxial deformation such as formula (15):

Wherein, E represents connecting rod A _iB _iElastic modulus, δ _iExpression connecting rod A _iB _iAxial deformation, S _iBe connecting rod A _iB _iCross-sectional area, F _uExpression connecting rod A _iB _iGo up B _iDistance is the suffered power of u point,

Be connecting rod A _iB _iThe acceleration of barycenter.

Can calculate connecting rod A equally ₅B ₅Axial deformation such as formula (16):

Wherein, S ₅₁And S ₅₂It is the cross-sectional area of member (51) and member (52);

Consider the rod member axial deformation, moving platform has small rotation d γ, connecting rod A _iB _iAxial deformation and the moving platform site error between relation as formula (17):

δ _i＝n _i(a′ _i-a _i)＝n _i((T-I)a _i+[dx dy] ^T)(i=1,2,3,4,5) (17)

In the formula, dx is the directions X error, and dy is the Y deflection error,

A ' _iPut A after the expression moving platform generation minor rotation _iPosition vector in coordinate system O '-xy, I are the unit square formation of two dimension.

Equation (17) can be write as

In the formula,

The equation of comptability group that connecting rod axial deformation and moving platform output error are set up in simultaneous equations (15), (16) and (18) is as shown in Equation (19):

5. as method as described in the claim 4, it is characterized in that described step 4) specifically comprises:

Simultaneous formula (13) at first, (14) and (19) obtain F _Aix'(i=1,2,3,4,5), dx, dy and d γ obtain F according to formula (10) and (12) then _Aiy', further obtain F according to formula (9) and (11) _Bi, at last to slide block B ₁B ₂And B ₃B ₄Carry out force analysis, according to slide block and member B ₅The equilibrium equation of C can calculate driving force.

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