## 3 Control Strategy
### Optimization of Pressure Differential \(\Delta p\) between the HA and the Maximum Load Pressure  
One case should be considered: as long as the EM starts to run, the joystick returns to its maximum position in an instant and the EM speed cannot reach to its target speed as quickly as possible. Hence, the flow rate of the pump will be too low to drive the load in time and the velocity of the load cannot be guaranteed. Meanwhile, when the HE works in a crane mode and pilot-oil discloses the multi-way valve, the output pressure of the pump is at a rather low level and the pressure in the hydraulic cylinder will drop suddenly with a big oscillation, especially in the case that the load is heavy and the load had not been lowered to the ground. As a matter of fact, one of the disadvantage of the HA is that only when the pressure of the HA is larger than the load pressure can it release the oil to drive the load. If the pressure differential \(\Delta p\) between the HA and the maximum load pressure is too high, though the actuator will start immediately, excess loss will be produced across the hydraulic control valve. While if the pressure differential \(\Delta p\) between the HA and the maximum load pressure is too low, it cannot build up the pressure quickly to drive the load. Therefore, the pressure differential \(\Delta p\) between the HA and the maximum load pressure should be optimized to ensure the control performance and the energy saving. The maximum load pressure \(p_{\mathrm{Lmax}}\) is determined from the expression  
\[p_{\mathrm{Lmax}}=\mathrm{max}(p_{\mathrm{Lb}},p_{\mathrm{Ls}}) \tag{1}\]  
where, \(p_{\mathrm{Lb}}\) is the non-rod side chamber of the hydraulic cylinder, MPa. \(p_{\mathrm{Ls}}\) is the rod side chamber of the hydraulic cylinder, MPa. The target velocity of the actuator is given by  
\[v_{\mathrm{t}}=k\cdot(p_{\mathrm{Lb}}-p_{\mathrm{Ls}}) \tag{2}\]  
where, \(k\) is the proportionality coefficient between the target velocity and the joystick pressure. The actual velocity of the load is calculated as  
\[v=\frac{q_{\mathrm{P}}\cdot n_{\mathrm{P}}-(C_{\mathrm{ip}}+C_{\mathrm{ep}})p _{\mathrm{P}}-Q_{\mathrm{bp}}+Q_{\mathrm{ac}}}{A_{\mathrm{L}}} \tag{3}\]  
where, \(q_{\mathrm{P}}\) is the displacement of the pump, m\({}^{3}\)/rad. \(n_{\mathrm{P}}\) is the speed of the pump, rad/s. \(C_{\mathrm{ip}}\) and \(C_{\mathrm{ep}}\) are the internal and external leakage coefficient of the pump, respectively, m\({}^{3}\)/(Pa\(\cdot\)s). \(p_{\mathrm{P}}\) is the output port pressure of the pump, Pa. \(Q_{\mathrm{bp}}\) is the bypass flow rate of the pump, m\({}^{3}\)/s. \(Q_{\mathrm{ac}}\) is the flow rate of the HA, and m\({}^{3}\)/s. \(A_{\mathrm{L}}\) is the effective area of driving chamber of the cylinder, m\({}^{2}\). The [PERSON]'s law expression for this case is  
\[p_{a0}{V_{0}}^{n}=p_{\rm ax}(V_{0}\pm\Delta V)^{n} \tag{4}\]  
where, \(p_{a0}\) is the pre-charge pressure of the HA, MPa. \(V_{0}\) is the gas volume under the pressure \(p_{a0}\), m\({}^{3}\). \(p_{\rm ax}\) is the pressure of the HA, MPa. \(\Delta V\) is the volume change of the HA, m\({}^{3}\). \(n\) is the polytrophic exponent, in this research, and \(n\) can be set at 1.4 because the time for releasing the flow to the multi-way valve is short, which is less than 5 s. When the HA is charged, the minus sign is chosen in Equation (4). The flow rate of the HA can be achieved by differentiating the Equation (4) with respect to time, as follows:  
\[Q_{\rm ac}=\frac{d\Delta V}{dt}=\pm\frac{V_{0}p_{a0}^{\frac{1}{n}}}{n}(p_{\rm ax })^{\frac{-1-n}{n}}\frac{dp_{\rm ax}}{dt} \tag{5}\]  
When the idle mode is cancelled, the HA is the main power device to drive the load in the initial stage, namely,  
\[Q_{\rm ac}\approx vA_{\rm L} \tag{6}\]  
When considering that the pressures of HA satisfies,  
\[0.25p_{a2}<p_{a0}<0.9p_{a1} \tag{7}\]  
where \(p_{a1}\) and \(p_{a2}\) are the minimum and maximum working pressure of HA, respectively.