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3 Model of electro-hydraulic proportional system
3.1 Dynamics of electro—hydraulic proportional valve
In this work， the electro-hydraulic proportional valve consists of proportional relief valves and SX-14 main valve．A transfer function from input current to the displacement of spool can be obtained as follows：
Xv(s)／Iv(s)=KI／(1+bs) (1) where Xv is the Laplace transform of xv，m；KI is the current gain of electro-hydraulic proportional valves，m／A；b is the time constant of the first order system，s：Iv=I(t)-Id，I(t)and Id are respectively the control current of proportional valve and the current to overcome dead band，A．
3.2 Flow equation of electro-hydraulic proportional valve
In this work，LUDV system was adopted in the experimental robotic excavator．According to the theory of LUDV system，the flow equation can be gotten：
Q1?cdwxv2cdwxv2?p?，I(t)?0
?p1? (2)
??cdwxv2?p1?pr?/?，I(t)?0 ?cdwxv2?p2?pr?/?，I(t)?0
?p2= (3)
Q2?cdwxv
2?cdwxv2?p?，I(t)?0
where ?pis the spring-setting pressure of load sense valve，MPa；cd is the flow coefficient m5／(N·s)；w is the area gradient of orifice，m2／m；ρ is the oil density, kg／m3;?p1and?p2 are the two orifices pressure，respectively, M Pa．When the flow of excavator is not saturated，?pis a nearly constant．In this work，the value was tested and gotten by experiment．In Fig.4，ps，p1s,and?prepresent the system pressure，the load sense valve pressure and the diference of pressure, respectively. The pressure experiment curves of the system show the variation of three kinds of pressures．Although Ps and pls change with load，their difference does not change with load，the value approximates to 2.0MPa.So，the effect of on the flow across the valve can be neglected．It is assumed that the flow across the valve is proportional to the size of orifice valve，and the flow is not influenced by
load．Then，Eqn．(2) can be simplified as
Q1=Kqxv(t),I(t)≥0 (4) where is the flow gain coefficient of valve, m2／s, and Kq?cdw2?p/?
Flg.4 Curves of pressure experiment under boom moving condition
3.3 Continuity equation of hydraulic cylinder
Generally speaking，construction machine does not permit external leakage．At
present，the external leakage can be controlled by sealing technology．On the other hand，it has been proven that the internal leakage of excavator is quite little by experiments．So, the influence of internal and external leakage of hydraulic system can be ignored．When the oil flows into head side of cylinder and discharges from rod side， the continuity equation can be written as
Q1?A1y?V1p1/?c Q2?A2y?V2p2/?c
????(5)
where V1 and V2 are the volumes of fluid flowing into and out the hydraulic cylinder, m3 ；?c is the effective bulk modulus(including liquid，air in oil and so on)，N/m2． 3.4 Force equilibrium equation of hydraulic cylinder
It is assumed that the mass of oil in hydraulic cylinder is negligible,and the load is rigid. Then the force equilibrium equation of hydraulic cylinder can be calculated from the Newton’s second law：
p1A1?p2A2?my?Bcy?Fc (6)
???where Bc is the viscous damping coefficient，N·s／m． 3.5 Simplified model of electro—hydraulic proportional system
After the Laplace transform of Eqns．(4)—(6)，the simplified model can be expressed as
Y?s??b1Xv?s??bfsFc?s?sa0s2?a1s?a2 (7)
where Y(s) is the Laplace transform of y;
2b1??cKq?A1V2?A2/A1;b1=V1V2;a0=V1V2m;a1=BcV1V2;a2??cV2A1?V1A2.
???????22?4 Parameters estimation
From the process of modeling and Eqn．(7)，it is clear that all parameters in the simplified model are related to the structure。the motional situation and the posture of excavator’s arm．Moreover，these parameters are time variable. So it is quite difficult to get accurate values and mathematic equations of these parameters. To solve this problem，those important parameters of model were estimated approximately by the estimation equation and method proposed in this work．
4.1 Equivalent mass estimation for load on hydraulic cylinder
The load of boom hydraulic cylinder(it is assumed there is no external load)consists of boom，dipper and bucket．In Fig.1，boom，dipper and bucket rotate around points O1，O2 and O3，respectively．So their motions are not straight line motions about the cylinders, that is to say, their motion directions are different from Y in Eqn.(5)．So，m in Eqn.(6)cannot be simply regarded as the sum mass of boom，dipper and bucket．
Considering O1 at an axis of manipulator, the torque and angular acceleration can begiven as follows：
M?Fc1lO1B?FclO1Bsin?
??ac1lOB?acsin?lOB
11(8)
where M and ? are the torque and angular acceleration of manipulator to O1，respectively;lQ1B is the length from point O1 to point B．According to the rotating law： M=J?,we get
FclO1Bsin??Jacsin?/lO1B
that is Fc?acJ/l2O1B （9） where J is the equivalent moment inertia of manipulator to point O1,kg·m2，and it can be written as follows：
J?J1?m1l2O1G1?J2?m2l2O1G2?J3?m3l2O1G3 (10) J1, J2 and J3 are the moment inertia of boom，dipper and bucket to their own bary center respectively．The values of them can be obtained by dynamic simulation based on the dynamic mode, J1=450.9N·m, J2=240.2N·m, J3=94.9N·m.
Comparing Eqn．(9)with Fc=mac，the equivalent mass at point B can be given：
m?J/l2O1B (11)
4.2 Estimation for load on hydraulic cylinder
The equivalent moment equation of manipulator to O1 is
??m1glOG??m2glOG??m3glOG? (12) FclO1Bsin111213where lOG?,lOG?and lOG? are the length from pointO1 to point G1′ ，G2′and G3′；，
111213respectively．Then，the counter force of load is
Fc?m1glOG??m2glOG??m3glOG?lO1Bsin? (13)
111213??4.3 Estimation for flow gain coefficient of valve
The flow of pump can be measured by flow transducer． The instrument used in this work was Multi—system 5050．The step response of flow of boom cylinder under the electro—hydraulic proportional valve controlled by the step curent is shown in Fig.5．At the same time，the curve verifies Eqn．[11]．Based on the experiment curve，the range of KqKl can be identified according to Eqns.(1)and(4)．And then，according to data in Fig.4，we can get：KqKl=2.825×10-4m3/(s·A)．