An inverse reconstruction approach considering uncertainty and correlation for vehicle-vehicle collision accidents

Abstract

Due to the limitation of measurements and the complexity of vehicle collision, uncertain factors inevitably exist in traffic accidents, and they usually are not independent. This makes traffic accident reconstruction difficult to implement using traditional deterministic inverse methods. This paper presents an uncertain inverse reconstruction approach to effectively and stably reconstruct vehicle-vehicle collision considering uncertainty and correlation. A complex finite element (FE) model is first established to simulate a vehicle-vehicle collision. Then, using the optimal Latin hypercube sampling, the kriging approximate model as the forward problem model is built to replace the time-consuming FE model with side impact. Thus, the model of the inverse problem can be represented according to the fitted surrogate model. Subsequently, based on the point estimation method and Nataf transformation, the uncertain inverse problem with correlation influence is transformed into several deterministic inverse problems with independence. It not only realizes vehicle-vehicle collision accident reconstruction under uncertainty and correlation, but largely improves the computational efficiency of the inverse solution process. Also, according to statistical moments of the identified parameters, we calculate their probability density functions to comprehensively assess these parameters and obtain more information by applying the maximum entropy principle. Finally, compared with the Monte Carlo simulation, the presented method is effective and reliable for vehicle collision accident reconstruction under uncertainty and correlation.

Appendices

Appendix A

Table 6 Thirty training samples generated using the Latin hypercube sampling method

Table 7 The comparisons of the output responses for seven testing samples on the simulation and calculation results

Appendix B

In this paper, three-point estimation method is used, namely q = 3, to calculate the position coefficients and the corresponding concentrated probabilities. Equation (15) can be written as follows:

$$ \Big\{{\displaystyle \begin{array}{l}{p}_{h,1}+{p}_{h,2}+{p}_{h,3}=1/n\\ {}{p}_{h,1}{\xi}_{h,1}^1+{p}_{h,2}{\xi}_{h,2}^1+{p}_{h,3}{\xi}_{h,3}^1={\gamma}_{h,1}\\ {}{p}_{h,1}{\xi}_{h,1}^2+{p}_{h,2}{\xi}_{h,2}^2+{p}_{h,3}{\xi}_{h,3}^2={\gamma}_{h,2}\\ {}{p}_{h,1}{\xi}_{h,1}^3+{p}_{h,2}{\xi}_{h,2}^3+{p}_{h,3}{\xi}_{h,3}^3={\gamma}_{h,3}\\ {}{p}_{h,1}{\xi}_{h,1}^4+{p}_{h,2}{\xi}_{h,2}^4+{p}_{h,3}{\xi}_{h,3}^4={\gamma}_{h,4}\\ {}{p}_{h,1}{\xi}_{h,1}^5+{p}_{h,2}{\xi}_{h,2}^5+{p}_{h,3}{\xi}_{h,3}^5={\gamma}_{h,5}\end{array}} $$

(B.1)

Among the three location points, one is the mean. So, there is a point coefficient of 0. We take the third point coefficient ${\xi}_{h,3}^1=0$ and rewrite the Eq. (B.1) by

$$ \Big\{{\displaystyle \begin{array}{l}{p}_{h,1}+{p}_{h,2}+{p}_{h,3}=1/n\\ {}{\xi}_{h,3}^1=0\\ {}{p}_{h,1}{\xi}_{h,1}^1+{p}_{h,2}{\xi}_{h,2}^1={\gamma}_{h,1}\\ {}{p}_{h,1}{\xi}_{h,1}^2+{p}_{h,2}{\xi}_{h,2}^2={\gamma}_{h,2}\\ {}{p}_{h,1}{\xi}_{h,1}^3+{p}_{h,2}{\xi}_{h,2}^3={\gamma}_{h,3}\\ {}{p}_{h,1}{\xi}_{h,1}^4+{p}_{h,2}{\xi}_{h,2}^4={\gamma}_{h,4}\end{array}} $$

(B.2)

In Eq. (5), ${\gamma}_{X_h,j}$ takes 0 and 1 when j = 1 and j = 2, respectively. Equation (B.2) is written as follows:

(B.3)

According to the formulas ① and ② from Eq. (B.3), we obtain

$$ {p}_{h,1}=\frac{1}{\xi_{h,1}\left({\xi}_{h,1}-{\xi}_{h,2}\right)},{p}_{h,2}=\frac{-1}{\xi_{h,2}\left({\xi}_{h,1}-{\xi}_{h,2}\right)} $$

(B.4)

Substituting Eq. (B.4) into Eq. (B.3), the parameters p_{h, 3}, ξ_{h, 1}, and ξ_{h, 2} are obtained as follows:

$$ {\displaystyle \begin{array}{l}{p}_{h,3}=\frac{1}{n}-{p}_{h,1}-{p}_{h,2}=\frac{1}{n}-\frac{1}{\gamma_{h,4}-{\gamma}_{h,3}^2}\\ {}{\xi}_{h,1}=\frac{\gamma_{h,3}}{2}+\sqrt{\gamma_{h,4}-\frac{3{\gamma}_{h,3}^2}{4}},{\xi}_{h,2}=\frac{\gamma_{h,3}}{2}-\sqrt{\gamma_{h,4}-\frac{3{\gamma}_{h,3}^2}{4}}\end{array}} $$

(B.5)

Combining Eqs. (B.4) and (B.5) and ${\xi}_{h,3}^1=0$, we can obtain the analytic solutions of Eq. (16) as follows:

$$ \Big\{{\displaystyle \begin{array}{l}{\xi}_{h,k}=\frac{\gamma_{h,3}}{2}+{\left(-1\right)}^{3-k}\sqrt{\gamma_{h,4}-\frac{3{\gamma}_{h,3}^2}{4}}\\ {}{\xi}_{h,3}=0\\ {}{p}_{h,k}=\frac{{\left(-1\right)}^{3-k}}{\xi_{h,k}\left({\xi}_{h,1}-{\xi}_{h,2}\right)}\\ {}{p}_{h,3}=\frac{1}{n}-{p}_{h,1}-{p}_{h,2}=\frac{1}{n}-\frac{1}{\gamma_{h,4}-{\gamma}_{h,3}^2}\end{array}},k=1,2;h=1,2,\cdots, n $$

(B.6)