# Calibration of a 3-Component External Balance for MAVs Wind Tunnel Research
## 4 Balance Calibration
### Residuals  
Table 7 shows the non-dimensional mean residual values \(\overline{R_{F_{i}}}\left(\%FS\right)\) defined as follows:  
\[\overline{R_{F_{i}}}\left(\%FS\right)=\frac{R_{F_{i}}}{FS_{F_{i}}}\cdot 100 \tag{26}\] \begin{table}
\begin{tabular}{c c c c c} \hline \multicolumn{6}{c}{**First-Order Normalized Contributions**} \\ \hline i & \(A_{ij}\cdot U_{j}/A_{ii}\cdot U_{i}\) & j = 1 & j = 2 & j = 3 \\ \hline
1 & \(A_{1j}\cdot U_{j}/A_{11}\cdot U_{1}\) & 1 & \(-1.16\cdot 10^{-2}\) & \(-7.85\cdot 10^{-3}\) \\ \hline
2 & \(A_{2j}\cdot U_{j}/A_{22}\cdot U_{2}\) & \(-1.60\cdot 10^{-2}\) & 1 & \(-3.48\cdot 10^{-2}\) \\ \hline
3 & \(A_{3j}\cdot U_{j}/A_{33}\cdot U_{3}\) & \(3.01\cdot 10^{-3}\) & \(2.92\cdot 10^{-2}\) & 1 \\ \hline \end{tabular}
\end{table} 
Table 5: Contributions to the total load cells for first-order interactions. where \(R_{F_{i}}\) is the residual for each component measured with the balance (\(F_{i}\)) and \(FS\) is the full scale corresponding to each of them, which will be different for each component. Then, for the lift component, the full scale will be equal to \(FS_{L}=600\) g, while for the drag component, it will be \(FS_{D}=200\) g and for the pitching moment, it will be \(FS_{M_{y}}=8\) g\(\cdot\)m. Moreover, Table 7 presents the non-dimensional standard deviation \(s_{F_{i}}(\%FS)\) of the residuals for each component measured with the balance (\(F_{i}\)), which represents the mean square error of the fit model as follows:  
\[s_{F_{i}}^{2}=\frac{1}{n-p}\Big{[}[R]\;[R]^{T}\Big{]}_{ii}\;\;\text{with}\;i=1,2,3;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;\;\;s_{F_{i}}=\sqrt{s_{F_{i}}^{2}} \tag{27}\]  
where \(n-p\) are the freedom degrees with \(p=9\). Several sources of error can affect the performance of the balance, but the uncertainty can be determined by considering that a generic force measurement by the balance equals the real force (applied load) acting on it, minus the corresponding residual. \[\hat{F}_{i}=F_{i_{real-load}}-R_{i}=f\Big{(}F_{i_{real-load}}\;,\;R_{i}\Big{)} \tag{28}\]  
Then, the combined uncertainty can be computed as follows:  
\[u_{\hat{F}_{i}}^{2}=\Bigg{(}\frac{\partial f}{\partial F_{i_{real-load}}} \Bigg{)}^{2}u_{F_{real-load}}^{2}+\Bigg{(}\frac{\partial f}{\partial R_{i}} \Bigg{)}^{2}u_{R_{i}}^{2}=u_{F_{real-load}}^{2}+u_{R_{i}}^{2} \tag{29}\]  
The uncertainty in the value of the real load arises because the input of the applied load in the calibration bench does not exactly match the nominal value. This discrepancy is due to the weights used during calibration, which were measured using an analytical balance with a resolution of 1 mg. Additionally, friction in the pulleys, misalignments, and deformations in the mechanical structure of the bench can introduce uncertainties estimated to be on the order of \(10^{-2}\) of the load, as determined during the balance's calibration in the mid-range. The uncertainty of the residual can be estimated using the standard deviation, calculated from the variance of the residuals. The expanded uncertainty \(U_{\hat{F}_{i}}\) is given by the following equation:  
\[U_{\hat{F}_{i}}=k\cdot u_{\hat{F}_{i}} \tag{30}\] \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline  & & & & & & & & \\ \hline i & \(j=1\), \(k=1\) & \(j=2\), \(k=2\) & \(j=3\), \(k=3\) & \(j=2\), \(k=1\) & \(j=3\), \(k=1\) & \(j=3\), \(k=2\) & Total \\ \hline
1 & 8.81\(\cdot\)\(10^{-3}\) & 6.43\(\cdot\)\(10^{-3}\) & 2.90\(\cdot\)\(10^{-2}\) & \(-\)1.20\(\cdot\)\(10^{-2}\) & \(-\)8.45\(\cdot\)\(10^{-5}\) & \(-\)2.23\(\cdot\)\(10^{-4}\) & 3.19\(\cdot\)\(10^{-2}\) \\ \hline
2 & \(-\)2.14\(\cdot\)\(10^{-3}\) & 2.65\(\cdot\)\(10^{-3}\) & \(-\)9.96\(\cdot\)\(10^{-3}\) & 4.76\(\cdot\)\(10^{-4}\) & \(-\)3.08\(\cdot\)\(10^{-4}\) & \(-\)1.27\(\cdot\)\(10^{-3}\) & \(-\)1.05\(\cdot\)\(10^{-2}\) \\ \hline
3 & \(-\)1.77\(\cdot\)\(10^{-3}\) & 1.32\(\cdot\)\(10^{-3}\) & \(-\)8.58\(\cdot\)\(10^{-3}\) & 1.56\(\cdot\)\(10^{-3}\) & 2.69\(\cdot\)\(10^{-4}\) & \(-\)3.77\(\cdot\)\(10^{-3}\) & \(-\)1.09\(\cdot\)\(10^{-2}\) \\ \hline \hline \end{tabular}
\end{table}