# Experimental Research into the Efficiency of an Electromagnetic Mill
## 4 Analysis of Results of the Experiment
The regression formula was expressed as a Taylor series with four squared variables. \[ \begin{array}{l}Y=c_{1}+c_{2}x_{1}+c_{3}x_{2}+c_{4}x_{3}+c_{5}x_{4}+c_{6} \frac{x_{7}^{2}}{2\pi}+c_{7}x_{1}x_{2}+c_{8}x_{1}x_{3}+c_{9}x_{1}x_{4}+\\ +c_{10}\frac{x_{7}^{2}}{2\pi}+c_{11}x_{2}x_{3}+c_{12}x_{2}x_{4}+c_{13}\frac{x_{ 7}^{2}}{2\pi}+c_{14}x_{3}x_{4}+c_{15}\frac{x_{7}^{2}}{2\pi}.\end{array}  \tag{3}\]  
\begin{table}
\begin{tabular}{c c c c c c c c c c} \hline \hline \multirow{2}{*}{ \begin{tabular}{c} **No. of** \\ **Experiment** \\ \end{tabular}  } & \multirow{2}{*}{ \begin{tabular}{c} **O** \\ **Millstone** \\ **[mm]** \\ \end{tabular}  } & \multirow{2}{*}{ \begin{tabular}{c} **Mass of** \\ **Millstones** \\ **[kg]** \\ \end{tabular}  } & \multirow{2}{*}{ \begin{tabular}{c} **WS** \\ **Mass** \\ **[kg]** \\ \end{tabular}  } & \multirow{2}{*}{\ \begin{tabular}{c} **Grinding** \\ **Mars** \\ **[s]** \\ \end{tabular}  } & \multirow{2}{*}{
\begin{tabular}{c} **WS** \\ **Mars** \\ **[s]  
The least squares method was used to calculate the regression parameters. The sum of squares of deviations of a dependent variable value (3) from the experimental values in a set of plan points is defined by the formula:  
\[\Delta=\sum_{n=1}^{S}\Bigg{(}Y_{Tn}-\Bigg{(}c_{1}+c_{2}x_{1n}+c_{3}x_{2n}+\ldots +c_{14}x_{3n}x_{4n}+c_{15}\frac{x_{4n}^{2}}{2!}\Bigg{)}\Bigg{)}^{2}. \tag{4}\]  
The value \(\Delta\) is a function of regression parameters \(c_{1},\ldots,c_{15}\). By comparing partial derivatives of this function to zero, respectively, through variables \(c_{1},\ldots,c_{15}\), an algebraic system of equations (ASE) was obtained in the form:  
\[ \begin{array}{l}\sum\limits_{n=1}^{S}2\bigg{(}Y_{Tn}-\Bigg{(}c_{1}+c_{2}x_{1 n}+c_{3}x_{2n}+\ldots+c_{14}x_{3n}x_{4n}+c_{15}\frac{x_{4n}^{2}}{2!}\Bigg{)} \Bigg{)}.1=0;\\ \sum\limits_{n=1}^{S}2\bigg{(}Y_{Tn}-\Bigg{(}c_{1}+c_{2}x_{1n}+c_{3}x_{2n}+ \ldots+c_{14}x_{3n}x_{4n}+c_{15}\frac{x_{4n}^{2}}{2!}\Bigg{)}\Bigg{)}.x_{1n}=0 ;\\ \sum\limits_{n=1}^{S}2\bigg{(}Y_{Tn}-\Bigg{(}c_{1}+c_{2}x_{1n}+c_{3}x_{2n}+ \ldots+c_{14}x_{3n}x_{4n}+c_{15}\frac{x_{4n}^{2}}{2!}\Bigg{)}\Bigg{)}.\frac{x_{ 4n}^{2}}{2!}=0.\end{array}  \tag{5}\]  
Regression Equation (3) in vector form is:  
\[Y=\overset{\rightarrow}{T}(x_{1},x_{2},x_{3},x_{4})\overset{\rightarrow}{c}, \tag{6}\]  
where  
\[\overset{\rightarrow}{T}(x_{1},\ldots,x_{4})=\left[1\ x_{1}\ \ldots\ x_{4}\ \frac{x_{1}^{2}}{2!}\ \ldots\ \frac{x_{4}^{2}}{2!}\right] \tag{7}\]  
denotes row of Taylor series of degree II with 4 independent variables;  
\[\overset{\rightarrow}{c}=\big{[}c_{1}\ \ \ldots\ \ c_{15}\big{]}^{T} \tag{8}\]  
is regression parameter column. ASE (5) utilizing designations (7), (8) can be expressed as  
\[\sum\limits_{n=1}^{S}\overset{\rightarrow^{T}}{T}(x_{1n},\ldots,x_{4n}) \overset{\rightarrow}{T}(x_{1n},\ldots,x_{4n})\overset{\rightarrow}{c}= \sum\limits_{n=1}^{S}Y_{Tn}\overset{\rightarrow^{T}}{T}(x_{1n},\ldots,x_{4n}) \tag{9}\]  
Figure 3: Separation of WS into fractions in test no. 10.  
or  
\[\sum_{n=1}^{S}\overset{\rightarrow}{T}_{D}(x_{1n},\ldots,x_{4n})\overset{ \rightarrow}{c}=\sum_{n=1}^{S}Y_{Tn}\overset{\rightarrow}{T}^{T}(x_{1n},\ldots, x_{4n}), \tag{10}\]  
where  
\[\overset{\rightarrow}{T}_{D}(x_{1n},\ldots,x_{4n})=\overset{\rightarrow}{T}^ {T}(x_{1n},\ldots,x_{4n})\overset{\rightarrow}{T}(x_{1n},\ldots,x_{4n}), \tag{11}\]  
denotes value of a Taylor pair in the \(n\)-th plan point. To calculate regression parameters \(c_{1},\ldots,c_{15}\) (6) of the TF it is necessary to define a linear ASE (10), based on the data in Table 1 and to solve it numerically. The value of function \(Y_{Tn}\) in this TF is the total mass of the ground WS \(m_{z\Sigma}\). For the purposes of the quantitative analysis of the impact of factors and their mutual effect on the course of the function, a linear transformation of the factor space was conducted.