Vrftjkry

My problem setup is as shown below:

I know the location (x,y) of fixed points p1+, p1-, p2+, and p2-, and want to find the position marked "o" by trilateration. However, I do not know the individual distances r1+, r1-, r2+ and r2-. Rather, I have the following system of equations available:

$begin{align}
frac{1}{r_{1+}}-frac{1}{r_{1-}}&=V_1\
frac{1}{r_{2+}}-frac{1}{r_{2-}}&=V_2\
frac{1}{r_{3+}}-frac{1}{r_{3-}}&=V_3\
frac{1}{r_{4+}}-frac{1}{r_{4-}}&=V_4\
frac{1}{r_{5+}}-frac{1}{r_{5-}}&=V_5
end{align}$

I couldn't figure out an analytical solution to this system of equations, even though there are really only 3 unknowns -- (x,y) of the the point marked o, and thought the solution needs to be found numerically...

Thus my question is, is my conclusion correct in that analytical solution isn't possible?

I realize now that I haven't taken into account the fact that the values $r1+, r1-$, etc. are only known through the system of equations you give. But nevertheless, I am going to show that even if we know these four values, we are practically unable to find an exact solution.

I recall first that an ellipse with focii $F$ and $F'$ is the locus of points $M$ such that the sum of distances $MF+MF'$ is a constant $k$.

Here, point $O$ belongs in particular to

• the ellipse with focii $p1-$ and $p2-$ and constant $k=r1-+r2-$, and to




• the ellipse with focii $p1+$ and $p2+$ and constant $k=r1++r2+$.

Otherwise said, point $O$ is an intersection point of these two ellipses. Knowing that an ellipse has a second degree equation, the coordinates of $O$ involve a $2^2=4$th degree equation (a rapid check : indeed, two ellipses can have up to 4 intersection points).

Thus, theoretically, you could get an (awfully complicated) analytical answer, but practically speaking, you have to look for a numerical method.

Remarks :

1) One could have chosen other ellipses such as the one with focii $p2+$ and $p1-$. There is no guarantee that the $binom{4}{2}=6$ ellipses and the $binom{6}{2}=15$ intersection points (at least) you can get in this way will intersect in a very same point due to measuring errors. Therefore, the numerical method you will use needs a "least squares" step.

2) one could have used hyperbolas instead of ellipses.

3) the fact that the 4 points $p1+, ...$ belong to a same circle hasn't been used.