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I've been searching for a few days on a number of sites but I can't seem to find a good answer for this. I'm developing a collision detection program using Unscented Kalman Filter and predicting possible positions of different objects.

These objects have a certain area but my predictions are of course only propagating the objects center of mass and I get the respective variance for their positions.

That is, I get the position of their center of mass is given by the bivariate normal distributions $N_1$~$(\mu_1,\Sigma_1)$ for object 1 and $N_2$~$(\mu_1,\Sigma_2)$ for object 2 in Cartesian coordinates.

Now the objective is finding out the probability of them colliding with each other. If both the objects have area $A$, what is the probability of them colliding ?

I know how to assess this threat when only using the objects center of mass, but that distribution doesn't really contain the objects area. For example in cases when $\Sigma_1$ extremly small, I "know" that the objects center of mass is there and from that the object's edges (using the area, or if it's better, the objects $Width$ and $Length$)

So, long story short, is there a neat way to transform $N_1$ and $N_2$ to incorporate the area ?

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  • $\begingroup$ What is the shape of the objects? $\endgroup$
    – Tom Minka
    Commented Mar 8, 2014 at 15:22
  • $\begingroup$ @TomMinka they are vehicles, so they both have Length and Width. For example, L = 5m and W = 1.7m, but it is okay to assume that they are circular with radius L/2 to be able to discard the orientation. $\endgroup$
    – Dammi
    Commented Mar 10, 2014 at 7:24

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If the objects are circular, then the probability of them colliding is the probability that their distance is below a threshold. The difference between two independent normally distributed vectors is itself a normally distributed vector. The length of such a vector has a non-central chi-square distribution. So the problem boils down to computing the cumulative distribution of a non-central chi-squared variate. See the page on Probability of collision (two bivariate normal distributions) for more details.

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  • $\begingroup$ Oh wow, I read that question actually and didn't connect the dots (Have been thinking about this problem for a few weeks now). Thank you very much Tom! $\endgroup$
    – Dammi
    Commented Mar 10, 2014 at 15:06

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