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Given a state vector given by $ x = {[r, v, a]}^{T} $ (Range, Velocity, Acceleration) the Time to Hit is the the time which holds the following: $$ r + v {T}_{tth} + \frac{a {T}_{tth}^{2}}{2} = 0 $$ Now, given the State Vector covariance $ P $ what would the the Time to Hit Covariance?

If the Time to Hit was given by a regular function it would be easy to do given its Jacobian, yet how can it be for this kind of calculation (Implicit Function)?

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  • $\begingroup$ Couldn't you just use the quadratic formula to solve for $T_{tth}$ in terms of $r$, $v$, and $a$, then use the Jacobian method that you're already familiar with? $\endgroup$ – Jason R Jun 11 '14 at 11:28
  • $\begingroup$ @JasonR, how would you handle the plus or minus? Go ahead and write the answer. I will mark it as right. $\endgroup$ – Royi Jun 11 '14 at 12:09
  • $\begingroup$ I would have to work it out, which I don't have time to do right now, but I think that the variance should be the same for both solutions (plus and minus). $\endgroup$ – Jason R Jun 11 '14 at 13:36
  • $\begingroup$ @JasonR, I don't think so. Since you calculating the derivative of the solution and it will be completely different for the + / -. $\endgroup$ – Royi Jun 11 '14 at 18:56
  • $\begingroup$ You could be right. If you know some conditions on the shape of the parabola (i.e. whether $a$ is constrained to have a certain sign), then you might be able to discard one of the solutions. Based on the physical interpretation of the problem, there should be a unique solution for $t \ge 0$. But, it looks like you may have a real answer below anyway. $\endgroup$ – Jason R Jun 11 '14 at 19:15
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I think I have the solution.
I'd be happy to hear others' thought.

Defining $ F \left(r, v, a, {T}_{tth} \right) = r + v {T}_{tth} + \frac{a {{T}_{tth}}^{2}}{2} $ which is the implicit function which connects all variables. Since we're dealing with non linear function the variance is given by:

$$ var \left( {T}_{tth} \right) = J P {J}^{T} $$

Where $ P $ is the state vector covariance at a given time and $ J = \left[ \frac{\partial {T}_{tth} }{\partial r} \frac{\partial {T}_{tth} }{\partial v} \frac{\partial {T}_{tth} }{\partial a} \right] $

Using Total Derivative Law on the Implicit Function: $$ \frac{\partial {T}_{tth} }{\partial r} = -\frac{{F}_{r}}{{F}_{{T}_{tth}}}, \frac{\partial {T}_{tth} }{\partial v} = -\frac{{F}_{v}}{{F}_{{T}_{tth}}}, \frac{\partial {T}_{tth} }{\partial a} = -\frac{{F}_{a}}{{F}_{{T}_{tth}}} $$

Where $ {F}_{x} $ is the partial derivative of $ F $ with respect to $ x $.

Each individual component would be:

$$ \frac{\partial {T}_{tth} }{\partial r}=-\frac{1}{v+aT_{tth}}, \frac{\partial {T}_{tth} }{\partial v}=-\frac{T_{tth}}{v+aT_{tth}}, \frac{\partial {T}_{tth} }{\partial a}=-\frac{T^2_{tth}}{2\left(v+aT_{tth}\right)} $$

Transform the state covariance matrix with $J P {J}^{T}$ and you'd have the right solution.

Given a diagonal state covariance matrix, we should have:

$$ P_{tth}=\frac{\sigma^2_r}{\left(v+aT_{tth}\right)^2}+\frac{T^2_{tth}\sigma^2_v}{\left(v+aT_{tth}\right)^2}+\frac{T^4_{tth}\sigma^2_a}{4\left(v+aT_{tth}\right)^2} $$

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  • $\begingroup$ I don't know much about the topic, but your math looks correct. $\endgroup$ – Phonon Jun 12 '14 at 18:37
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    $\begingroup$ It's great to have someone asking a question and doing the right thing by sharing the solution once he found it. $\endgroup$ – David Aug 11 at 19:22

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