Express distance between two points in given proximity of other points I have two movable receivers, both receivers get straight line signals from fixed transmitters, where the strength of the signal is given in DBm. Assuming the signal strength as a measure for distance, (not a pure measure, since walls/objects alter the signal strength, but let's not dive into physics) how would I express the distance between the two receivers, provided there are enough transmitters in range?
My guess is that this is a case of trilateration, and that by using circle intersections, we can narrow down the location of the receivers in a simple plane. Next we can express the distance between the two receivers within the plane. 


I think there is no solution unless more information is provided.
Suppose the distance information is exact. The measurements are translation and rotation invariant, so we can assume that one receiver is at the origin (in $ \mathbb{R}^2$) and the other is located at $(x,0)$ (and $x \neq 0$). The goal is to determine $|x|$.
Let $\phi:\mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}^2$ be given by $\phi(x,t) = (\|t\|, \|t-(x,0)\|)$. If a transmitter is at $t \in \mathbb{R}^2$, and the other receiver is at $(x,0)$, then the distance measurements for that receiver are given by $\phi(x,t)$.
We note that if $R$ is a reflection through the $x$-axis, then $\phi(x,Pt) = \phi(x,t)$, so we might as well assume that $t_2 \ge 0$.
We want to determine if the measurements $d_k = \phi(x,t_k)$ uniquely determine $|x|$.
Suppose $\phi(x,t) = d$. Then $d_1^2 = t_1^2+t_2^2$, $d_2^2 = (t_1-x)^2 + t_2^2$. Subtracting gives $x^2-2 x t_1 = d_2^2-d_1^2$, from which we get $t_1 = \frac{1}{2}(x + \frac{d_1^2-d_2^2}{x})$, $t_2 = \sqrt{d_1^2-t_1^2}$. 
Let $X = \{ (x,d) | x \neq 0, d_2 >0, d_1 > \frac{1}{2} | x + \frac{d_1^2-d_2^2}{x} | \}$. Note that $X$ is open and non-empty (for example, $(1,(\sqrt{2},1)) \in X$).
Define $t:X  \to \mathbb{R}^2$ by $t_1(x,d) = \frac{1}{2}(x + \frac{d_1^2-d_2^2}{x})$ and $t_2(x,d) = \sqrt{d_1^2-t_1(x,d)^2}$.
Then we have $\phi(x,t(x,d)) = d$ for all $x$ such that $(x,d) \in X$.
In particular, if for some $\hat{x}$ we have $(\hat{x}, d_k) \in X$ for all $k$, then we have $\phi(x,t(x,d_k)) = d_k$ for $x$ in some neighbourhood of $\hat{x}$, hence the measurements do not uniquely determine $|x|$.
