## 3 The Algebraic Sign of Virtual Power Flows
### Case \(x\) and \(y\) as Driven Links and \(z\) Driving Link  
Let us consider the case of a two dof gear train with \(x\) and \(y\) as driven links (i.e., \(P_{x}=T_{x}\omega_{x}<0\), \(P_{y}=T_{y}\omega_{y}<0\) and \(P_{z}=T_{z}\omega_{z}>0\)), as shown in Figure 2, 2 nd working mode. In this case the overall power balance equation is the following:  
\[\eta_{(z-x,y)}=\frac{\left|T_{x}\omega_{x}+T_{y}\omega_{y}\right|}{T_{z}\omega _{z}} \tag{15}\]  
**Case 1:**: \(\text{sign}\bigg{(}\dfrac{\omega_{x}}{\omega_{y}}\bigg{)}=-1\) (see Figure 2, 2 nd working mode)  
The following relations hold:  
\[T_{z}\omega_{zy}\eta_{y(z-x)}+T_{x}\omega_{xy} =0 \tag{16a}\] \[T_{z}\omega_{zx}\eta_{x(z-y)}+T_{y}\omega_{yx} =0 \tag{16b}\]  
Solving the system composed of the Equations (15) and (16), one obtains the following:  
\[\eta_{(z-x,y)}=\bigg{|}\dfrac{\eta_{x(z-y)}\omega_{y}\omega_{zx}-\eta_{y(z-x)} \omega_{x}\omega_{zy}}{\omega_{yx}\omega_{z}}\bigg{|} \tag{17}\]
**Subcase 2a:**: \(\dfrac{\omega_{y}}{\omega_{x}}<1\) and \(\text{sgn}\bigg{(}\dfrac{\omega_{x}}{\omega_{y}}\bigg{)}=+1\) (see Figure 3, 2 nd working mode)  
The following relations hold:  
\[T_{z}\omega_{zy}\eta_{y(z-x)}+T_{x}\omega_{xy} =0 \tag{18a}\] \[T_{z}\omega_{zx}+T_{y}\omega_{yx}\eta_{x(y-z)} =0 \tag{18b}\]  
Solving the system composed of the Equations (15) and (18), one obtains the following:  
\[\eta_{(z-x,y)}=\bigg{|}\dfrac{\omega_{y}\omega_{zx}-\eta_{x(y-z)}\eta_{y(z-x) }\omega_{x}\omega_{zy}}{\eta_{x(y-z)}\omega_{yx}\omega_{z}}\bigg{|} \tag{19}\]  
The analytical Expressions (18) and (19) can be combined into the following formula:  
\[\boxed{\eta_{(z-x,y)}=\bigg{|}\dfrac{B\omega_{y}\omega_{zx}-\eta_{y-(z-x)} \omega_{x}\omega_{zy}}{\omega_{yx}\omega_{z}}\bigg{|}} \tag{20}\]  
with  
\[B=\begin{cases}\eta_{x(z-y)}&\text{when sign}\bigg{(}\dfrac{\omega_{x}}{ \omega_{y}}\bigg{)}=-1\\ \dfrac{1}{\eta_{x(y-z)}}&\text{when }\bigg{(}\dfrac{\omega_{y}}{\omega_{x}} \bigg{)}<1\text{ and sign}\bigg{(}\dfrac{\omega_{x}}{\omega_{y}}\bigg{)}=+1 \end{cases} \tag{21}\]  
The proposed formula is more general than the one proposed by [PERSON] (see Equation (25) of [26]).  
The expressions of \(\eta_{x(y-z)}\) and \(\eta_{y(x-z)}\), for all possible combinations of \(x\), \(y\) and \(z\), are listed in Table 1 of reference [1] (see also Appendix A).  
**5. Numerical Example**  
We consider the case of the gear unit, shown in Figure 5, with \(x\) and \(y\) driving links and the \(z\)-driven link. In particular, \(\omega_{x}=8000\) rpm and \(\omega_{y}=7966.6\) rpm.  
For a planet gear ratio \(N_{j,i}=-1.5\),  
\[\tau_{1}=-\dfrac{1}{1-N_{j,i}}=-0.4\]  
and  
\[\tau_{2}=-\dfrac{N_{j,i}}{N_{j,i}-1}=-0.6\]  
Consequently, from [PERSON] equation, follows \(\omega_{z}=7980\) rpm.  
Since \(0<\dfrac{\omega_{x}}{\omega_{y}}<1\), the Formula (13) is applied. Considering the current value of \(N_{j,i}\), and assuming  
\[\eta_{z(x-y)}=0.900\]  
the entries 3a and 5a of Table 1 of the Appendix 1 give, respectively, the following:  
\[\eta_{y(x-z)}=\dfrac{N_{j,i}\eta_{z(x-y)}-1}{N_{j,i}-1}=0.940\] \[\eta_{x(y-z)}=\dfrac{N_{j,i}-\eta_{z(x-y)}}{N_{j,i}-1}=0.960\]  
For the prescribed numerical data, the Formula (13) yields the following:  
\[\eta_{(x,y-z)}=\left|\dfrac{\omega_{z}\omega_{yx}\eta_{y(x-z)}}{\omega_{x} \omega_{zy}-\dfrac{\eta_{y(x-z)}}{\eta_{x(y-z)}}\omega_{y}\omega_{zx}}\right|= 0.998\]