guys!!
I'm a post doc in chemical engineering department.
During my experiment, I have achieved the distances among the dots. I can calculate the distance between the dots, and I want to convert that information to the coordinates of each dots.
Also, it is hard to figure out the dimension of coordinate system, yet.
Is there any possible ways to calculate the coordinates? Or is there any theoretical approaches for this kind of problem?
It would be great for me to get any clues for this problem!!
Thanks!!
If the representation space is 2 dimensional (see "remark about the dimensionality" at the end of this answer), here is a direction to attack the problem.
Fix 2 points $F_1$ and $F_2$ (sufficiently far apart, even at maximal distance in order to limitate the influence of measures' imprecision).
Let us consider that line $F_1F_2$ defines an $x$-axis and that the $y$ axis is defined by the perpendicular bissector line of $F_1F_2$. Divide all distances by $\tfrac12 F_1F_2$ making the normalized distance $F_1F_2$ equal to $2$ ; thus the coordiantes of $F_1,F_2$ can be taken resp. as $(-1,0)$ and $(1,0)$.
Consider (see graphics below) the following families of (orthogonal) curves that, each one, allows to cover the plane
ellipses $\frak{E}_u$: set of points M such that $MF_1+MF_2=u$.
hyperbolas $\frak{H}_v$ with 2 branches, $\begin{cases}\text{the set of points M such that} \ MF_1-MF_2=v, v > 0 \\ \text{the set of points M such that} \ MF_1-MF_2=v, v<0\end{cases}. \ \ \ $ The case $v=0$, characterizing the perpendicular bissector of $F_1F_2$, can be considered as a degenerated case of this family.
Ellipses are represented in red (from $u=2$ to to $u=3$ ; the case $u=2$ is line segment $F_1F_2$, i.e., a degenerate ellipse), hyperbolas in blue (from $v=-2,-1.8, -1.6, ... 2$ (the 2 cases $v=-2$ and $v=2$ being also degenerate cases reduced to half lines). An important feature of these two families of curves is that they share a same set of foci, i.e., points $F_1$ and $F_2$.
Being given a point $P$ of the given set of points, as you know distances $PF_1$ and $PF_2$, you are able to compute their sum and their difference, yielding a value of $u, u \geq 0$ and a value of $v, v\in \mathbb{R}$
This pair of values $(u,v)$ characterizes $2$ possible points, symmetrical with respect to $x$ axis (it is the fundamental "ambiguity" of your issue), thus $2$ possible pairs of cartesian coordinates (see computation below).
A second step is thus needed to "disambiguate" the $y$ coordinate. I have not done it but I have some tracks. One of these tracks is to operate in the same way with $2$ other points $F'_1$ and $F'_2$.
Computation of $(x,\pm y)$ coordinates out of a $(u,v)$ description:
$$\tag{1}\begin{cases} MF_1+MF_2=u \ \iff \ \sqrt{(x+1)^2+y^2} +\sqrt{(x-1)^2+y^2}=u & \ \ (a)\\ MF_1-MF_2=v \iff \ \sqrt{(x+1)^2+y^2} -\sqrt{(x-1)^2+y^2}=v& \ \ (b) \end{cases}.$$
Squaring and adding, one gets $4x^2+4y^2+4=u^2+v^2$, thus $x^2+y^2=r^2$ with $r:=\sqrt{\tfrac{u^2+v^2}{4}-1}$, giving
$$\tag{2}x=r \cos(\alpha), \ \ y=r \sin(\alpha)$$ for certain values of $\alpha$ (in general 2 values) that will be found by plugging the expressions given by (2) into (1)(a) for example.
Remark about the dimensionality of the representation space: have a look at "Cayley Menger determinant" (https://en.wikipedia.org/wiki/Distance_geometry_problem) which is a condition
$$det \left( \begin{array}{ccccc}0&AB^2&AC^2&AD^2&1\\ AB^2&0&BC^2&BD^2&1\\ AC^2&BC^2&0&CD^2&1\\ AD^2&BD^2&CD^2&0&1\\ 1&1&1&1&0\end{array}\right)=0,$$
that must be fullfilled by any set of 4 points $A,B,C,D$ if the representation space is $\mathbb{R}^2$. You can use it at first as a consistency test for your set of distances' data. If only few sets of 4 points give a LHS significantly different from $0$, you can incriminate errors in measurements, whereas, if a vast amoint of them differ from $0$, you can ask yourself if the real space of representation is a $\mathbb{R^n}$, with $n>2$ ; in this case, test with a "larger" Cayley-Menger determinant involving subsets of $n+2$ points if the representation space is $\mathbb{R}^n.$
Take a look at the nice figure(s) in (https://theinnerframe.wordpress.com/tag/triply-orthogonal-surfaces/).