# Method for Volume of Irregular Shape Pellets Estimation Using 2D Imaging Measurement
## 3 Results
### Estimation of Pellet Volume by Dividing Them into Elliptical Layers
 } & \multirow{2}{*}{13.2} & \multirow{2}{*}{13.27} & \multirow{2}{*}{0.07} & \multirow{2}{*}{13.10} & \multirow{2}{*}{0.10} & \multirow{2}{*}{15.79} & \multirow{2}{*}{2.59} & \multirow{2}{*}{15.61} & \multirow{2}{*}{2.41} \\  & & & & & & & & & & \\ \hline \multirow{3}{*}{8} & \multirow{2}{*}{ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.0} & \multirow{2}{*}{14.27} & \multirow{2}{*}{0.27} & \multirow{2}{*}{14.14} & \multirow{2}{*}{0.14} & \multirow{2}{*}{16.65} & \multirow{2}{*}{16.46} & \multirow{2}{*}{2.46} & \multirow{2}{*}{16.46} \\  & & & & & & & & & & \\ \hline \multirow{3}{*}{9} & \multirow{2}{*}{ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.4} & \multirow{2}{*}{13.70} & \multirow{2}{*}{0.70} & \multirow{2}{*}{13.70} & \multirow{2}{*}{0.70} & \multirow{2}{*}{13.79} & \multirow{2}{*}{0.61} & \multirow{2}{*}{13.63} & \multirow{2}{*}{0.77} \\  & & & & & & & & & & \\ \hline \multirow{3}{*}{10} & \multirow{2}{*}{ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.6} & \multirow{2}{*}{14.28} & \multirow{2}{*}{0.32} & \multirow{2}{*}{14.20} & \multirow{2}{*}{0.40} & \multirow{2}{*}{16.04} & \multirow{2}{*}{1.44} & \multirow{2}{*}{15.82} & \multirow{2}{*}{1.22} \\  & & & & & & & & & \\ \hline \multirow{3}{*}{ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.6} & \multirow{2}{*}{14.28} & \multirow{2}{*}{0.32} & \multirow{2}{*}{14.20} & \multirow{2}{*}{0.40} & \multirow{2}{*}{16.04} & \multirow{2}{*}{1.44} & \multirow{2}{*}{15.82} & \multirow{2}{*}{1.22} \\  & & & & & & & & & \\ \hline \multirow{3}{*}{ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.7} & \multirow{2}{*}{0.735} & \multirow{2}{*}{0.654} & \multirow{2}{*}{13.63} & \multirow{2}{*}{0.97} & \multirow{2}{*}{14.7} & \multirow{2}{*}{0.1} \\  & & & & & & & & & & \\ \hline \multirow{3}{*}{\ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.7} & \multirow{2}{*}{0.735} & \multirow{2}{*}{0.654} & \multirow{2}{*}{1.191} & \multirow{2}{*}{0.314} \\  & & & & & & & & & & \\ \hline \multirow{3}{*}{ \begin{tabular}{c} Largest area 1 \\ Rotated by 90\({}^{\circ}\) \\ \end{tabular}  } & \multirow{2}{*}{14.7} & \multirow{2}{*}{0.73} & \multirow{2}{*}{0.932} & \multirow{2}{*}{0.811} & \multirow{2}{*}{1.380} & \multirow{2}{*}{0.532}  
The evaluation of the received results showed evident matching of the results with those received from the control equipment. Evaluation of the results of separate samples showed that absolute error does not exceed 0.15 mm of the average size of the pellet (D\({}_{50}\)). The relative error is less than 5%, when the absolute error of approximation of the pellet using the ellipse is up to 0.37 and relative error is \(>\) 12%. Therefore, it can be stated that the proposed method for evaluating separate pellets allows for achieving better matching between the measuring results and control equipment results. Figure 15: Measuring of volume of formed pellet models based on method of image processing. All graphs present comparison of results to results received using water displacement volume evaluation method: (**a**) cross-sections of pellet model layers approximated using circles; (**b**) cross-sections of pellet model layers approximated using ellipses. In the (**a**,**b**) graphs, particle volume calculation was made by evaluating 2 images (largest area and rotated by 90\({}^{\circ}\)). In (**c**,**d**)—cross-sections of pellet model layers approximated using circles from 1 image (largest area or rotated by 90\({}^{\circ}\)). In (**e**,**f**)—cross-sections of pellet model layers approximated using ellipses from 1 image (largest area or rotated by 90\({}^{\circ}\)).