# A New Method for Haptic Shape Discriminability Detection
## 2 Angle Sorting
### Estimation of Standard Deviation
And the probability of each angle being selected during the process of this category is \(P_{2,n}\) (\(n=22,\,24,\,26,\,28,\,30\)), as show in Table 1. For instance, the angle of the second position is \(26^{\circ}\), and the probability of its occurrence is  
\[P_{L2}=\frac{f(x=26;\mu=22;\sigma=1.4)}{\sum_{n=22}^{30}P_{2,n}}=\frac{0.0048} {0.2850+0.1027+0.0048+0+0}=0.0123 \tag{2}\]  
\begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \multirow{2}{*}{**Angle**} & \multicolumn{2}{c}{**Second**} & \multicolumn{2}{c}{**Third**} & \multicolumn{2}{c}{**Fourth**} & \multicolumn{2}{c}{**Fifth**} & \multicolumn{2}{c}{**Sixth**} \\ \cline{2-10}  & \(P_{2,n}\) & \(P_{L2}(26^{\circ})\) & \(P_{3,n}\) & \(P_{L3}(22^{\circ})\) & \(P_{4,n}\) & \(P_{L4}(24^{\circ})\) & \(P_{5,n}\) & \(P_{L4}(28^{\circ})\) & \(P_{6,n}\) & \(P_{L6}(30^{\circ})\) \\ \hline \(22^{\circ}\) & 0.2850 & & 0.2850 &  
In Equation (2), \(P_{L2}\) is the probability of choosing this angle (\(26^{\circ}\)) among the remaining angle stimuli (\(26^{\circ}\), \(22^{\circ}\), \(24^{\circ}\), \(28^{\circ}\), \(30^{\circ}\)) at the second position. At the third position, the rest angles are \(22^{\circ}\), \(24^{\circ}\), \(28^{\circ}\) and \(30^{\circ}\), so \(\mu=min(22,\ 24,\ 28,\ 30)=22\). \(26\) has been placed in the second position, and according to the assumption that once each stimulus was selected, it could not be selected again, its probability of being selected is 0. However, the rest of angles still obey the normal distribution. Therefore, the probability of each category of angle being selected is \(P_{3,n}\) (\(n=22,24,26,28,30\)), as show in Table 1. The angle of the third position is \(22^{\circ}\), and the probability of its occurrence is  
\[P_{L3}=\frac{f(x=22;\mu=22;\sigma=1.4)}{\sum_{n=22}^{30}P_{2,n}}=\frac{0.2850 }{0.2850+0.1027+0+0+0}=0.7350 \tag{3}\]  
In Equation (3), \(P_{L3}\) is the probability of choosing this angle (\(22^{\circ}\)) among the remaining angle stimuli (\(22^{\circ}\), \(24^{\circ}\), \(28^{\circ}\), \(30^{\circ}\)). Similarly, we have \(P_{L4}=0.9833\), \(P_{L5}=0.7350\), \(P_{L6}=1\). Therefore, when \(\sigma=1.4\) and \(AS_{1}=(26^{\circ},22^{\circ},24^{\circ},28^{\circ},30^{\circ})\), the probability is \(P(\sigma,AS_{1})=P_{L2}P_{L3}P_{L4}P_{L5}P_{L6}=0.0065\), as shown in Figure 2.0  
Here, we know that this result has nothing to do with the mean \(\mu\) and \(x\) range of the normal distribution and is based on the angle sorting \(AS_{1}=(26^{\circ},22^{\circ},24^{\circ},28^{\circ},30^{\circ})\) and the assumption of \(\sigma=1.4\). The result, \(P(\sigma,AS_{1})\), changes with each different value chosen for \(\sigma\). According to the above data analysis model, when \(AS_{1}\) is determined, we can construct the relationship function of \(P(\sigma,AS_{1})=f(\sigma)\) with \(\sigma\) as the independent variable. For example, when \(AS_{1}=(26^{\circ},22^{\circ},24^{\circ},28^{\circ},30^{\circ})\), we can obtain the relationship between \(\sigma\) and \(P(\sigma,AS_{1})\), as shown in Figure 2(a). We can see from Figure 2(a) that when the angle sorting is \(AS_{1}=(26^{\circ},22^{\circ},24^{\circ},28^{\circ},30^{\circ})\), \(P(\sigma,AS_{1})\) will change with the change of \(\sigma\); there is a peak value in the change of \(P(\sigma,\ AS_{1})\), as shown in point A (2.6, 0.0327) in Figure 2(a). The probability of \(P(\sigma,\ AS_{1})\) corresponding to the peak is the maximum probability in this angle sorting. Therefore, we can obtain the participant's \(\sigma\) corresponding to the peak value, \(\sigma=2.6\). Two special situations need to be considered in the standard deviation estimation. First, when the angle sorting is completely correct, \(AS_{2}=(22^{\circ},24^{\circ},26^{\circ},28^{\circ},30^{\circ})\), the relationship between \(P(\sigma,AS_{2})\) and \(\sigma\) is shown in Figure 2(b).