# Reverse Engineering Used to Profile a Gerotor Pump Rotor
## 3 Results
### Calculation of the Enwrapping Profile  
In order to determine the coordinates of the \(C_{1}\) point on the \(Z_{4}\) rotor, which is in enwrapping with the \(C_{2}\) point on the \(Z_{5}\) rotor, the \"virtual pole\" method is used. Using the enwrapping condition, the value of the \(\varphi_{1}\) angle for which the normal to the \(\Sigma_{1}\) profile, taken through the \(C_{1}\) point, passes through the gearing pole is established. For \(u=42.1^{\circ}\), it results \(\varphi_{1}=\left[\arcsin\left(\frac{32.26\cdot\sin 42.1^{\circ}}{22} \right)\right]-42.1^{\circ}\), so \(\varphi_{1}=37.348^{\circ}\). When the \(X_{1}O_{1}Y_{1}\) system rotates by the \(\varphi_{1}\) angle, the normal passes through the gearing pole, and the coordinates of the \(C_{1}\) point in the \(x_{1}O_{1}y_{1}\) system will be given by the coordinate transformation:  
\[x_{1}=\omega_{3}^{T}(\varphi_{1})\cdot X_{1}; \tag{26}\]  
\[ \begin{pmatrix}x_{1}\\ y_{1}\end{pmatrix} = \begin{pmatrix}\cos\varphi_{1}&-\sin\varphi_{1}\\ \sin\varphi_{1}&\cos\varphi_{1}\end{pmatrix} \cdot \begin{pmatrix}X_{1}\\ Y_{1}\end{pmatrix} = \begin{pmatrix}X_{1}\cdot\cos\varphi_{1}-Y_{1}\cdot\sin \varphi_{1}\\ X_{1}\cdot\sin\varphi_{1}+Y_{1}\cdot\cos\varphi_{1}\end{pmatrix} . \tag{27}\]  
For \(u=42.1^{\circ}\), it results in  
\[\left\{ \begin{array}{l}X_{1}=-R_{0}+r\cdot\cos u=-20.388;\\ Y_{1}=r\cdot\sin u=10.72;\end{array} \right.\]  
\[\left\{ \begin{array}{l}x_{1}=X_{1}\cdot\cos\varphi_{1}-Y_{1}\cdot\sin \varphi_{1}=-30.294;\\ y_{1}=X_{1}\cdot\sin\varphi_{1}+Y_{1}\cdot\cos\varphi_{1}=-3.846.\end{array} \right.\]  
In the \(x_{2}O_{2}y_{2}\) system, the coordinate point \(x_{1}y_{1}\), previously calculated, has the coordinates:  
\[x_{2}=x_{1}+A; \tag{28}\]  
\[ \begin{pmatrix}x_{2}\\ y_{2}\end{pmatrix} = \begin{pmatrix}x_{1}\\ y_{1}\end{pmatrix} + \begin{pmatrix}A_{12}\\ 0\end{pmatrix} . \tag{29}\]  
For \(u=42.1^{\circ}\), it results in \( \begin{pmatrix}x_{2}\\ y_{2}\end{pmatrix} = \begin{pmatrix}-30.294+4.4\\ -3.846\end{pmatrix} = \begin{pmatrix}-25.894\\ -3.846\end{pmatrix} \). This point corresponds to a \(\varphi_{2}\) rotation of the \(X_{2}O_{2}Y_{2}\) system, with the value of \(\varphi_{2}\) being given by  
\[R_{r_{1}}\cdot\varphi_{1}=R_{r_{2}}\cdot\varphi_{2}. \tag{30}\]  
It results in \(\varphi_{2}=\frac{R_{r_{1}}}{R_{r_{2}}}\cdot\varphi_{1}=\frac{5}{4}\cdot \varphi_{1}=46.685^{\circ}\). Therefore, the coordinates of this point in the \(X_{2}O_{2}Y_{2}\) system will be  
\[X_{2}=\omega_{3}(\varphi_{2})\cdot x_{2}; \tag{31}\]  
\[ \begin{pmatrix}X_{2}\\ Y_{2}\end{pmatrix} = \begin{pmatrix}\cos\varphi_{2}&\sin\varphi_{2}\\ -\sin\varphi_{2}&\cos\varphi_{2}\end{pmatrix} \cdot \begin{pmatrix}x_{2}\\ y_{2}\end{pmatrix} = \begin{pmatrix}x_{2}\cdot\cos\varphi_{2}+y_{2}\cdot\sin \varphi_{2}\\ -x_{2}\cdot\sin\varphi_{2}+y_{2}\cdot\cos\varphi_{2}\end{pmatrix} . \tag{32}\]  
In these conditions, \( \begin{pmatrix}X_{2}\\ Y_{2}\end{pmatrix} = \begin{pmatrix}-20.562\\ 16.202\end{pmatrix} \). For \(u=42.1^{\circ}\) and \(A_{12}=4.4\), results:  
\[i=\frac{5}{4};R_{r_{1}}-R_{r_{2}}=A_{12}=R_{r_{1}}-\frac{R_{r_{1}}}{i};R_{r_{1} }=A_{12}\cdot\left(\frac{i}{i-1}\right  
By developing,  
\[\left( \begin{array}{c}X_{2}\\ Y_{2}\end{array} \right)=\left( \begin{array}{c}X_{1}\cdot\cos\varphi_{1}\cdot \cos\varphi_{2}-Y_{1}\cdot\sin\varphi_{2}\cdot\cos\varphi_{2}+\\ -X_{1}\cdot\cos\varphi_{1}\cdot\sin\varphi_{2}+Y_{1}\cdot\sin\varphi_{1}\cdot \sin\varphi_{2}-\\ +A_{12}\cdot\cos\varphi_{2}+X_{1}\cdot\sin\varphi_{1}\cdot\sin\varphi_{2}+Y_{1 }\cdot\cos\varphi_{1}\cdot\sin\varphi_{2}\\ -A_{12}\cdot\sin\varphi_{2}+X_{1}\cdot\sin\varphi_{1}\cdot\cos\varphi_{2}+Y_{1 }\cdot\cos\varphi_{1}\cdot\cos\varphi_{2}\end{array} \right)= \tag{34}\] \[=\left(