# Brake Torque Sensor Calibration Device
## 5 Calculation of the Calibration Device Error
The calculation example for a weight of 2.5 is as follows:  
\[m_{2.5+\mu}=m_{2.5}+\mu_{2.5}=0.25488019+0.00005=0.25488519\text{ kg}\]  
\[m_{2.5-\mu}=m_{2.5}-\mu_{2.5}=0.25488019-0.00005=0.25487519\text{ kg}\]  
In the next step, the values of the maximum and minimum forces induced by the weight with the given tolerance were calculated according to relation (12), as follows:  
\[F_{max/min}=m_{i\pm\mu}\cdot g\quad i=\{2.5;\ 5;\ 10;\ 50\} \tag{12}\]  
\[F_{2.5 max}=m_{2.5+\mu}\cdot g=0.25488519\cdot g.80853=2.50004903\text{ N}\]  
\[F_{2.5 max}=m_{2.5-\mu}\cdot g=0.25487519\cdot g.80853=2.49995095\text{ N}\]  
The values for other calibration weights were calculated in the same way and the results are shown in Table 5.  
Since the actual diameter of the calibration wheel detected by the 3D scanner, which measures with an uncertainty of \(U=\pm 1.2\) um, is \(d_{s}=400.1334\) mm, it was necessary to add or subtract this uncertainty from the actual value of the diameter of the calibration wheel according to relation (13):  
\[d_{s\pm U}=d_{s}\pm U \tag{13}\]  
where  
\(d_{s\pm U}\) is the actual diameter of the calibration wheel enlarged or reduced by the uncertainty of the 3D scanner [m]. \(d_{s}\) is the actual diameter of the calibration wheel [m]. \(U\) is the uncertainty of the 3D scanner [m]. \[d_{s+U}=d_{s}+U=0.4001334+0.0000012=0.4001346\text{ m}\]  
\[d_{s-U}=d_{s}-U=0.4001334-0.0000012=0.4001322\text{ m}\]  
Subsequently, the actual maximum and minimum values of the radius of the calibration wheel could be calculated, where half the thickness of the steel strip with a thickness of \(h_{p}=0.07\) mm was added. This radius was calculated using relation (14), as follows:  
\[m_{2.5+\mu}=m_{2.5}+\mu_{2.5}=0.25488019+0.00005=0.25488519\text{ kg}\]0  
where  
\(r_{max/min}\) is the maximum or minimum radius of the calibration wheel [m]. \(d_{s\pm U}\) is the actual diameter of the calibration wheel enlarged or reduced by the uncertainty of the 3D scanner [m]. \(h_{p/2}\) is half the thickness of the steel strip [m]. \(r_{max}=\frac{d_{s+U}}{2}+h_{p/2}=\frac{0.4001346}{2}+0.000035=0.2001023\text{ m}\) \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Weight Value [N]** & **Max. Mass Value [kg]** & **Max. Weight Value [N]** & **Min. Mass Value [kg]** & **Min. Weight Value [N]** \\ \hline
2.5 & 0.25488519 & 2.50004903 & 0.25487519 & 2.49995095 \\
5 & 0.50976538 & 5.00004902 & 0.50975538 & 4.99995094 \\
10 & 1.01953076 & 10.00009805 & 1.01951076 & 9.99990187 \\
50 & 5.09765382 & 50.00049042 & 5.09755382 & 49.99950957 \\ \hline \hline \end{tabular}
\end{table} 
Table 5: Table of calculated values. \[r_{max}=\frac{d_{s-U}}{2}+h_{p/2}=\frac{0.4001322}{2}+0.000035=0.2001011\ \text{m}\]  
After finding the actual maximum and minimum radii of the calibration wheel, the maximum and minimum values of the torque induced by the force of the weight on the given arm could be calculated according to relations (15) and (16):  
\[M_{i_{max}}=F_{i_{max}}.r_{max}\quad i=\{2.5;\ 5;\ 10;\ 50\} \tag{15}\]  
\[M_{i_{min}}=F_{i_{min}}.r_{min}\quad i=\{2.5;\ 5;\ 10;\ 50\} \tag{16}\]  
where  
\(M_{i_{max/min}}\) is the maximum or minimum value of the torque induced by the given weight [Nm]. \(F_{i_{max/min}}\) is the maximum or minimum value of the calibration weight [N]. \(r_{max/min}\) is the maximum or minimum value of the radius of the calibration wheel [m]. \[M_{2.5_{max}}=F_{2.5_{max}}.r_{max}=2.50004903\cdot 0.2001023=0.50026556\ \text{m}\]  
\[M_{2.5_{min}}=F_{2.5_{min}}.r_{max}=2.49995095\cdot 0.2001011=0.50024294\ \text{m}\]  
The calculation procedure for other calibration weights is the same as in the previous calculation and the results are shown in Table 6.  
During calibration, the required torque values were achieved by combining calibration weights. Therefore, it was necessary to calculate the tolerance for each torque value (MM = 0.5, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15) depending on the weight combination used.