## 2 Optimization Problem
### Determination of Gearbox Efficiency  
The efficiency of the gearbox (%), \(\eta_{gb}\) is calculated by:  
\[\eta_{gb}=100-\frac{100\cdot P_{l}}{P_{m}} \tag{13}\]  
Here, \(P_{l}\) is the total power loss, which can be determined by [52],  
\[P_{l}=P_{lg}+P_{lb}+P_{ls}+P_{Z0} \tag{14}\]  
where \(P_{lg}\), \(P_{lb}\), \(P_{ls}\), and \(P_{zo}\) are the power loss in the gears, in the bearings, in the seals, and in the idle motion (kW), which can be found as follows:  
Figure 1: Two-stage helical gearbox with second-stage double gear sets.  
(+) Determination of \(P_{lg}\):  
\[\text{P}_{lg}=\sum_{i=1}^{2}\text{P}_{\text{lgi}} \tag{15}\]  
where  
\[P_{lgi}=P_{gi^{*}}(1-\eta_{gi}) \tag{16}\]  
Here, \(\eta_{gi}\) is the efficiency of the \(i\) stage, which is found by [53]:  
\[\eta_{gi}=1-\left(\frac{1+1/\text{u}_{\text{i}}}{\beta_{\text{ai}}+\beta_{\text {ri}}}\right)\cdot\frac{\text{f}_{\text{i}}}{2}\cdot\left(\beta_{\text{ai}}^{2} +\beta_{\text{ri}}^{2}\right) \tag{17}\]  
In the above equation, \(\beta_{\text{ai}}\) and \(\beta_{\text{ri}}\) are the arc of approach and recess on \(i\) stage, which can be calculated by [53]:  
\[\beta_{\text{ai}}=\frac{\left(\text{R}_{\text{e2i}}^{2}-\text{R}_{02i}^{2} \right)^{1/2}-\text{R}_{\text{2i}}\cdot\text{sin}\alpha}{\text{R}_{01i}} \tag{18}\]  
\[\beta_{\text{ri}}=\frac{\left(\text{R}_{\text{ei1}}^{2}-\text{R}_{01i}^{2} \right)^{1/2}-\text{R}_{\text{1i}}\cdot\text{sin}\alpha}{\text{R}_{01i}} \tag{19}\]  
where \(\alpha\) is the pressure angle (rad). It is assumed that the gear has an involute gear profile with a pressure angle of 20 degrees; f is the friction coefficient, which can be determined by [11]:  
- if v \(\leq\) 0.424 (m/s):  
\[\text{f}=-0.0877\cdot\text{v}+0.0525 \tag{20}\]  
- if \(v>\) 0.424 (m/s):  
\[\text{f}=0.0028\cdot\text{v}+0.0104 \tag{21}\]  
where v is the sliding velocity of the gear (m/s).  
(+) Determination of \(P_{lb}\)[52]:  
\[P_{lb}=\sum_{i=1}^{6}f_{b}\cdot F_{i}\cdot v_{bi} \tag{22}\]  
where \(i=1\div 6\), and \(f_{b}=0.0011\) is the friction coefficient of the bearing (for the radical ball bearings with angular contact) [52]; Fi is the load of the bearing (N); and \(v_{bi}\) is the peripheral velocity of the bearing (m/s).  
(+) Determination of \(P_{ls}\)[52]:  
\[P_{\text{ls}}=\sum_{i=1}^{2}\text{P}_{\text{si}} \tag{23}\]  
Here, \(i=1\div 2\), is the ordinal number of the seal; and \(P_{si}\) is the power loss in seal \(i\) (kW), which can be determined by:  
\[P_{si}=\left[145-1.6\cdot t_{oil}+350\cdot loglog(VG_{40}+0.8)\right]\cdot d_ {s}^{2}\cdot n\cdot 10^{-7} \tag{24}\]  
where \(d_{s}\) and \(n\) are the diameter (mm) and the revolution of the shaft; and VG\({}_{40}\) is the ISO Viscosity Grades number.  
(+) Determination of \(P_{20}\)[52]:  
\[P_{Z0}=\sum_{i=1}^{k}T_{Hi}\cdot\frac{\pi\cdot n_{i}}{30} \tag{25}\]In (25), k = 2, is the total gear pair number; n is the driven gear revolution; and \(T_{Hi}\) is the hydraulic torque of power loss (Nm), which is found by [52]:  
\[T_{Hi}=C_{Sp}\cdot C_{1}\cdot e^{\frac{C_{Sp}\cdot e}{\Psi_{0}}} \tag{26}\]  
Here, \(C_{Sp}=1\), for stage 1 in the case of the involved oil having to pass until the mesh, and for stage 2, \(C_{Sp}\) can be found by (Figure 2):  
\[C_{Sp}=\left(\frac{4\cdot e_{max}}{3\cdot h_{C}}\right)^{1.5}\cdot\frac{2\cdot h _{C}}{l_{hi}} \tag{27}\]  
With \(\mathrm{l_{hi}}\), which is calculated by [52]:  
\[\mathrm{l_{hi}}=(1.2\div 2.0)\cdot\mathrm{d_{a2i}} \tag{28}\]  
In (26), \(C_{1}\) and \(C_{2}\) are calculated by [52]:  
\[C_{1}=0.063\cdot\left(\frac{\mathrm{e_{1}}+\mathrm{e_{2}}}{\mathrm{e_{0}}} \right)+0.0128\cdot\left(\frac{\mathrm{b}}{\mathrm{b_{0}}}\right) \tag{29}\]  
\[C_{2}=\frac{\mathrm{e_{1}}+\mathrm{e_{2}}}{80\cdot\mathrm{e_{0}}}+0.2 \tag{30}\]  
With \(\mathrm{e_{0}}=\mathrm{b_{0}}=10\) (mm).