Air France Flight 447 Disaster Summary
Application
For the Air France Flight 447 disaster, Stone et al. [17] described this problem and analyzed the results of this analysis. The analysis shows that all impact points are contained within a 20 nautical mile radius circle from the point at which the emergency situation began. The results of this analysis are represented by a second distribution which is circular normal with center at the last known position. One of the important advantages of circular normal distribution is that sensitive to shifts in the centering of the searching process. On 6th June, an oil slick was found in known area of the Atlantic Ocean. This area is away from the Brazilian coast about 640 kilometers and it is located approximately halfway between …show more content…
Without loss of generality, we let the target be losted in one of two regions (i.e., n=2) and let ₁=₂=(1/2). Using the above Kuhn--Tucker conditions, we have,
(((e²-1))/(A₁A₂))[(1+z₃)(∫₀¹e^{(x₁²(t_{k})+y₁²(t_{k}))}dt_{k}+∫₀¹e^{(x₂²(t_{k})+y₂²(t_{k}))}dt_{k})] -2z₁[x₁(t_{k})(x₁(t_{k})-₁)] -2z₂[y₁(t_{k})(y₁(t_{k})-₁)]=0, (16) z₁[ r₁-(x₁(t_{k})-₁)²-(y₁(t_{k})-₁)²]=0, (17) z₂[ r₂-(x₂(t_{k})-₂)²-(y₂(t_{k})-₂)²]=0, (18) z₃[(1/(A₁A₂))∫₀¹e^{(x₁²(t_{k})+y₁²(t_{k}))}dt_{k}∫₀¹e^{(x₂²(t_{k})+y₂²(t_{k}))}dt_{k}-1]=0. (19)
By using (15) we get, x_{i}^{∗}(t_{k})=√(r_{i}-(y_{i}^{∗}(t_{k})-_{i})²)+_{i}, i=1,2. Suppose that the first trace of debris floating on one of two circles in the two regions, respectively. In addition, let the center of the two circles and the radius of them be considered as in the following Table 1.
Table 1. The values of (_{i},_{i}) and r_{i},
For the Air France Flight 447 disaster, Stone et al. [17] described this problem and analyzed the results of this analysis. The analysis shows that all impact points are contained within a 20 nautical mile radius circle from the point at which the emergency situation began. The results of this analysis are represented by a second distribution which is circular normal with center at the last known position. One of the important advantages of circular normal distribution is that sensitive to shifts in the centering of the searching process. On 6th June, an oil slick was found in known area of the Atlantic Ocean. This area is away from the Brazilian coast about 640 kilometers and it is located approximately halfway between …show more content…
Without loss of generality, we let the target be losted in one of two regions (i.e., n=2) and let ₁=₂=(1/2). Using the above Kuhn--Tucker conditions, we have,
(((e²-1))/(A₁A₂))[(1+z₃)(∫₀¹e^{(x₁²(t_{k})+y₁²(t_{k}))}dt_{k}+∫₀¹e^{(x₂²(t_{k})+y₂²(t_{k}))}dt_{k})] -2z₁[x₁(t_{k})(x₁(t_{k})-₁)] -2z₂[y₁(t_{k})(y₁(t_{k})-₁)]=0, (16) z₁[ r₁-(x₁(t_{k})-₁)²-(y₁(t_{k})-₁)²]=0, (17) z₂[ r₂-(x₂(t_{k})-₂)²-(y₂(t_{k})-₂)²]=0, (18) z₃[(1/(A₁A₂))∫₀¹e^{(x₁²(t_{k})+y₁²(t_{k}))}dt_{k}∫₀¹e^{(x₂²(t_{k})+y₂²(t_{k}))}dt_{k}-1]=0. (19)
By using (15) we get, x_{i}^{∗}(t_{k})=√(r_{i}-(y_{i}^{∗}(t_{k})-_{i})²)+_{i}, i=1,2. Suppose that the first trace of debris floating on one of two circles in the two regions, respectively. In addition, let the center of the two circles and the radius of them be considered as in the following Table 1.
Table 1. The values of (_{i},_{i}) and r_{i},