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I am trying to work out if there is a way to calculate some coordinates relative to each other simply by knowing $3$ or more distances from some unknown points.

I do not have a distance matrix, I simply have distances from myself to $3$ or more points. These distances are calculated from signal strength, not geometry.

For instance, if I am given \begin{align*} d_1 & = 7.0711\\ d_2 & = 2.8284\\ d_3 & = 3.1623 \end{align*} I assign my location as $(0,0)$.

Then I move somewhere, recalculate and find my new distances are \begin{align*} d_1 & = 7.6158\\ d_2 & = 5.6569\\ d_3 & = 3.1623 \end{align*}

These are real values and I know from plotting some points and calculating distances that the second set means my new location is $(2,2)$. Is there some way to calculate this from knowing only the two sets of distances and assigning myself a point of origin?

Note : I have searched already and most questions were related to distances and known points or distance matrices so I don't think I'm asking a duplicate question.

Thanks

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  • $\begingroup$ Welcome to MathSE. Please see this tutorial on how to format mathematics on this site. $\endgroup$ Commented Jun 12, 2015 at 11:47
  • $\begingroup$ Get some more distances from different positions, then have a look at Multilateration or Trilateration $\endgroup$ Commented Jun 14, 2015 at 15:48
  • $\begingroup$ Yes I'm aware of trilateration but the issue is the same. I'll have 3 positions (with unknown coordinates) with 3 sets of distances to 3 points with unknown coordinates. What I'm trying to do is essentially work out the distance between two sets of distances if this is possible. $\endgroup$ Commented Jun 15, 2015 at 14:25

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Two measures do not suffice. Consider that the observer takes three successive positions $(0,0),(a_1,0),(a_2,b_2)$ ($b_1=0$ to prevent rotations). The data are the $9$ associated distances to the vertices of the unknown triangle $ABC$. There are also $9$ unknowns: $a_1,a_2,b_2$ and the coordinates of $A,B,C$. The solution is not unique because the whole figure is invariant by symmetry with respect with $O$ and by symmetry with respect with $Ox$.

EDIT 1. In fact, there are $8$ figure-solutions and $4$ solutions for $a_1$ -of the form $(\alpha,-\alpha,\beta,-\beta)$-. In particular, the distance between the first and second position of the observer is not uniquely defined!! Moreover, the minimal polynomial of $a_1$ is not solvable; in particular, the figure is not constructible with ruler and compass.

EDIT 2. Answer to Matt. Consider that the observer takes only two successive positions $(0,0),(a_1,0)$. The data are the $6$ associated distances to the vertices of the unknown triangle $ABC$. Unfortunately, there are $7$ unknowns: $a_1$ and the coordinates of $A,B,C$. Thus there are an infinity of solutions (in particular for $a_1$).

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    $\begingroup$ The solutions are all loosely equal to each other though aren't they? If A, B and C are fixed then the two sets of distances refer to two unique points. Isn't all that changes the coordinates of A, B and C? For my problem, if I could find a solution just for a1 (i.e. the distance between the two sets of distances) then that would help me greatly. Is this possible? $\endgroup$ Commented Jun 14, 2015 at 14:25

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