## 2 Modeling of the Milling Process with Active Control  
In case of a milling tool with \(n\) evenly spaced teeth which is almost flexible to the rigid workpiece, a generalized 2-degrees of freedom mathematical model is [40; 41]:  
\[M_{m}\ddot{\mathbf{x}}_{c}(t)+C_{m}\dot{\mathbf{x}}_{c}(t)+K_{m}\mathbf{x}_{c }(t)=\mathbf{F}_{m}(\mathbf{t}) \tag{1}\]  
the equivalent mass, damping and stiffness matrices are illustrated by the terms \(M_{m},C_{m}\) and \(K_{m}\), respectively.  
\[M_{m}=\left[\begin{array}{cc}m_{mx}&0\\ 0&m_{my}\end{array}\right]\in\Re^{2\times 2},C_{m}=\left[\begin{array}{cc}c_{mx}&0\\ 0&c_{my}\end{array}\right]\in\Re^{2\times 2},K_{m}=\left[\begin{array}{cc}k_{mx}&0\\ 0&k_{my}\end{array}\right]\in\Re^{2\times 2} \tag{2}\]  
Again, \(\mathbf{x}_{c}(t)=[x\ y]^{T}\) illustrates the displacement of the tool along \(x\) and \(y\) components. \(\mathbf{F}_{m}(t)=[F_{fx}\ F_{fy}]^{T}\) illustrates \(x\) and \(y\) component cutting forces. Along the \(x\) and \(y\) axes, the new form of Equation (1) is:  
\[\left[\begin{array}{cc}m_{mx}&0\\ 0&m_{my}\end{array}\right]\left[\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right]+\left[\begin{array}{cc}c_{mx}&0\\ 0&c_{my}\end{array}\right]\left[\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right]+\left[\begin{array}{cc}k_{mx}&0\\ 0&k_{my}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}F_{fx}\\ F_{fy}\end{array}\right] \tag{3}\]  
which generates:  
\[\begin{array}{c}m_{mx}\dot{x}+c_{mx}\dot{x}+k_{mx}x=F_{fx}\\ m_{my}\dot{y}+c_{my}\dot{y}+k_{my}y=F_{fy}\end{array} \tag{4}\]  
It is utter necessary to elaborate the dynamics of the cutting forces along the \(x\) and \(y\) components. Figure 1 illustrates the dynamics of the milling process [42; 43]. The closed form equations representing the nonlinear cutting forces along \(x\) and \(y\) components are illustrated as [20]:  
\[\begin{array}{c}F_{x}=+\frac{N}{2\pi}\left\{\begin{array}{c}\xi_{1}\Delta x ^{3}+\eta_{1}\Delta y^{3}+\zeta_{2}\Delta x^{2}+\eta_{2}\Delta y^{2}+\zeta_{3 }\Delta x+\eta_{3}\Delta y\\ +3\gamma_{1}\Delta x^{2}\Delta y+3\gamma_{2}\Delta x\Delta y^{2}+2\gamma_{3} \Delta x\Delta y+\gamma_{4}\end{array}\right\}\\ F_{y}=-\frac{N}{2\pi}\left\{\begin{array}{c}\zeta_{1}^{*}\Delta x^{3}+\eta _{1}^{*}\Delta y^{3}+\zeta_{2}^{*}\Delta x^{2}+\eta_{2}^{*}\Delta y^{2}+\zeta _{3}^{*}\Delta x+\eta_{3}^{*}\Delta y\\ +3\gamma_{1}^{*}\Delta x^{2}\Delta y+3\gamma_{2}^{*}\Delta x\Delta y^{2}+2 \gamma_{3}^{*}\Delta x\Delta y+\gamma_{4}^{*}\end{array}\right\}\end{array} \tag{5}\]  
where \(\Delta x+x(t-\tau)=x(t)\) and \(\Delta y+y(t-\tau)=y(t)\). The time delay is illustrated as \(\tau=\frac{2\pi}{n\Omega}\), \(\Omega=\) spindle speed (\(rad/s\)). If we consider the present and previous tool period instances, then it is represented by \(\left[x(t)\ y(t)\right]\) and \(\left[x(t-\tau)\ y(t-\tau)\right]\), respectively. Considering the start immersion angle as 0 and the exit angle as \(\frac{\pi}{2}\), The calculation of the coefficients for half-immersion up-milling is:  
\[\begin{array}{c}\zeta_{1}=\frac{1}{4}\left[\zeta_{1}+\frac{3}{4}\pi\delta_{ 1}\right],\eta_{1}=\frac{1}{4}\left[\delta_{1}+\frac{3}{4}\pi\zeta_{1}\right], \zeta_{2}=\frac{1}{4}\left[\zeta_{2}+2\delta_{2}\right],\\ \eta_{2}=\frac{1}{4}\left[\delta_{2}+2\zeta_{2}\right],\zeta_{3}=\frac{1}{2} \left[\zeta_{3}+\frac{1}{2}\pi\delta