# A Cooperative Control Method for Excavation Support Robot with Desired Position/Posture
## 2 ESR System Modeling and Description
### Hydraulic System Model with AHC for ESR  
Considering the workload and other on-site environmental factors, a digital AHC cannot be used effectively for excavation operations. An ESR with a cylindrical AHC working stroke is regulated by controlling the opening degree of the Electro-Hydraulic Servo (EHS) valve and its opening time. According to the mechanical structure and movement form of the ESR, the two cylindrical AHCs on one side are connected to each other. The lateral cylindrical AHC works alone. A model of the hydraulic system with AHC for ESR is illustrated in Figure 5. As mentioned above, the lateral AHCs belong to the single-stage hydraulic cylinder. The EHS reversing valve is identified as an ideal spool valve structure, which makes the return oil pressure of the spool valve zero; the oil supply pressure of the whole system is stable, and is not affected by the oil in the spool valve cavity. The hydraulic system with the lateral AHC for ESR is indicated in Figure 6.  
In Figure 6, three sets of balance equations are listed, such as the EHS diverter valve flow equation, the lateral AHC flow continuity equation, and the force balance equation with external load. (1) The EHS diverter valve flow equation [29] is expressed as  
\[Q_{L}=K_{q}x_{p}-K_{c}p_{L} \tag{1}\]  
where \(K_{q}\) is expressed as the flow gain of the EHS valve (directional valve) \((\mathrm{m}^{-3}\cdot\mathrm{s}^{-1}/\mathrm{m})\), \(x_{p}\) is the stroke (displacement) of the EHS valve spool (mm), \(K_{c}\) indicates the flow pressure coefficient of the EHS valve \(\left(\mathrm{m}^{3}/(\mathrm{s}/\mathrm{Pa})\right)\), and \(p_{L}\) represents the lateral AHC pressure (Pa). (2) The fluid continuity equations [30] for lateral AHC  
The fluid flowing into the lateral AHC rodless cavity of the ESR is \(Q_{i}\), which is expressed as:  
\[Q_{i}=A_{p}\frac{dx_{p}}{dt}+\frac{V_{i}}{\beta_{e}}\frac{dp_{i}}{dt}+C_{pi}( p_{i}-p_{0})+C_{pe}p_{i} \tag{2}\]  
where \(A_{p}\) represents the effective area of the AHC rod-end piston \((\mathrm{m}^{2})\), \(x_{p}\) is the travel stroke (displacement) of the AHC piston \((\mathrm{m})\), \(p_{i}\) is the oil pressure in the AHC rodless cavity \((\mathrm{Pa})\), \(p_{0}\) is the oil pressure of the AHC rod cavity \((\mathrm{Pa})\), \(C_{pi}\) is the leakage coefficient in the laterally adjusted EHS cylinder, \(C_{pe}\) is the external leakage coefficient of the laterally adjusted EHS cylinder, and \(V_{i}\) is the volume in the rodless cavity of the AHC \((\mathrm{m}^{3})\); herein, \(V_{i}=V_{i0}+A_{p}x_{p}\), \(\beta_{e}\) is the elastic modulus of the effective volume \((\mathrm{Pa})\). Figure 4: The geometrical dimensions of the core structure of the ESR. The flow rate in the rod cavity of the lateral AHC of ESR is \(Q_{0}\), which is expressed as:  
\[Q_{i}=A_{p}^{\prime}\frac{dx_{p}}{dt}-\frac{V_{0}}{\beta_{e}}\frac{dp_{0}}{dt}+C_{ pi}(p_{i}-p_{0})-C_{pe}p_{i} \tag{3}\]  
where \(A_{p}^{\prime}\) is the effective area of the AHC rod end piston (\(\text{m}^{2}\)), \(A_{p}^{\prime}=A_{p}-\pi r^{2}\), and \(V_{0}\) is the volume inside the AHC rodless cavity (\(\text{m}^{3}\)); herein, \(V_{0}=V_{0}^{\prime}+A_{p}^{\prime}x_{p}\). Set \(Q_{L}=\frac{\text{Q}_{i}+\text{Q}_{0}}{2}\), \(p_{L}=p_{i}-p_{0}\), combining Equations (2) and (3), leading to  
\[Q_{L}=(A_{p}-\pi r^{2})\frac{dx_{p}}{dt}+(C_{pi}+\frac{C_{pe}}{2})p_{L}+0.5 \beta_{e}\bigg{[}\bigg{(}V_{0}\frac{dp_{i}}{dt}-V_{0}^{\prime}\frac{dp_{0}}{ dt}\bigg{)}+\bigg{(}A_{p}x_{p}\frac{dp_{i}}{dt}+A_{p}^{\prime}x_{p}\frac{dp_{0}}{ dt}\bigg{)}\bigg{]} \tag{4}\]  
Then, set to equalize the compression flow in the hydraulic cylinder, that is, \(V_{0}=V_{0}^{\prime}=V_{0}\). Assuming that \(V_{t}=\frac{V_{0}}{2}=\frac{V_{0}^{\prime}}{2}\) holds, with this in mind, \(A_{p}x_{p}\ll V_{0}\),\(A_{p}^{\prime}x_{p}\ll V_{0}\), with \(\frac{dx_{p}}{dt}\approx 0\),\(\frac{dp_{0}}{dt}\approx 0\).