Find out distance between two points based on distances to a moving object Is it possible to find the distance between points $a$ and $b$ based on their distances to a moving object $c$? For example, a train is moving and two observers, at arbitrary locations, measure their distances to the train over time. Now we want to determine the relative distance between observers based on their measurements. It feels like it is possible but I can not really wrap my head around this.  
 A: No, it is not possible with that information. Here is an easy counterexample:
Suppose that $d(A,C)=d(B,C)=r$. Then if we take the point $C$ as a center of some circle with radius $r$ then the points $A$ and $B$ can be anywhere on the circle so obviously their distance is not determined. You need additional information, for example, an angle between sides $AC$ and $BC$ and then you have a triangle which have the length of the third side $AB=d(A,B)$ determined by the length of the sides $AB$ and $BC$ and the angle between them.
With that angle you need only one measurement but without an angle the counterexample shows that it is not possible, even with many measurements, because $d(A,B)$ will be impossible to determine because the first measurement does not determine it and because there is no measurement that determine it, no matter the number of measurements.
But you can know something of course. suppose $d(A,C)=r_1$ and $d(B,C)=r_2$ and suppose, without losing generality that $r_2>r_1$. Then if we take $C$ as the center of the circles with radii $r_2$ and $r_1$ then you have $r_2-r_1\leq d(A,B)\leq r_2+r_1$ (draw picture).
A: Choose a coordinate system such that $a=(0,0)$ and such that $b=(d,0)$ for some $d>0$ and write $c(t) = (x(t),y(t))$.
From now on I'll omit the time dependencies for economy of notation.
Next we define $u=\|c-a\|=\sqrt{x^2+y^2}$ and $v=\|c-b\|=\sqrt{(x-d)^2+y^2}$ and we obtain $x=\frac{u^2-v^2+d^2}{2d}$ and $y=\sqrt{u^2-\frac{(u^2-v^2+d^2)^2}{4d^2}}$.
The expressions for $x$ and $y$ make sense whenever $u^2>\frac{(u^2-v^2+d^2)^2}{4d^2}$ or $$4u^2d^2>(u^2-v^2+d^2)^2=u^4+v^2+d^4-2u^2v^2+2u^2d^2-2v^2d^2,$$ which we conveniently rewrite to $$4u^2v^2>u^4+v^2+d^4+2u^2v^2-2u^2d^2-2v^2d^2=(u^2+v^2-d^2)^2.$$
Equivalently, we have $2uv>u^2+v^2-d^2>-2uv$ or $u^2+v^2+2uv>d^2>u^2+v^2-2uv$, which is more conveniently written as $(u+v)^2>d^2>(u-v)^2$, whence finally $u+v>d>|u-v|$.
By this cumbersome algebra we may thus conclude that whenever $u+v>d>|u-v|$, we have a point $c=(\frac{u^2-v^2+d^2}{2d},\sqrt{u^2-\frac{(u^2-v^2+d^2)^2}{4d^2}})$ such that $\|c-a\|=u$ and $\|c-b\|=v$.
Furthermore notice, that $c$ depends smoothly/continuously on $t$ whenever $u$ and $v$ do.
Thus, given functions $u(t)$ and $v(t)$, the only thing we can conclude about $d$ is that $\max_t|u(t)-v(t)|<d<\min_t u(t)+v(t)$, since for any such $d$ a path $c(t)$ exists such that $\|c(t)-a\|=u(t)$ and $\|c(t)-b\|=v(t)$ for all $t$.
As a nice corollary we have that a path $c(t)$ determines $d$ uniquely if and only if $u(t)+v(t)=d$ for some $t$ and $|u(t)-v(t)|=d$ for some $t$.
That is, if $c(t)$ crosses the line spanned by $a$ and $b$ somewhere in between these points and somewhere not in between these points.
A: In the coordinate system of the train we have
$$
d(p_1,p_2)=\sqrt{(u-\overline{u}\,)^2+(v-\overline{v}\,)^2}
$$
In the coordinate system of the reference point we have
$$
d(p_1,p_2)=\sqrt{(x-\overline{x}\,)^2+(y-\overline{y}\,)^2}
$$

