reverse 3D euclidean distance

Hi there,

I am standing at an unknown point U(x,y,z) in the room. I can measure 3 (euclidean) distances D to 3 known points P in the room. I try to find the point, where I am at. My equation system looks like this:

(x-3)²+(y-1)²+(z-4)²=D1²=81 
(x-12)²+(y-1)²+(z-4)²=D2²=36
(x-34)²+(y-2)²+(z-4)²=D3²=601

I can put the known points into a matrix P, the measured distance in a vector D:

P=[3 1 4;12 1 4; 34 2 4]
P =
     3     1     4
    12     1     4
    34     2     4
    
D=[81 36 601]

Do you know how I can find U(x,y,z) ?

I am not sure if I can use

D = pdist(X,'euclidean');

Answer 1

[Star Strider]

Try this:

P=[3 1 4;12 1 4; 34 2 4];
D=[81 36 601];

fcn = @(b,x) (b(1)-x(:,1)).^2 + (b(2)-x(:,2)).^2 + (b(3)-x(:,3)).^2;
B = fminsearch(@(b) norm(D(:) - fcn(b,P)), [1; 1; 1])

producing:

B =
   10.0000
    5.0000
   -0.0000

that are the (x,y,z) coordinates, as best fminsearch can calculate them.

[Ferdinand Grosse-Dunker]

Works perfekt, thank you. Can you comment on the function you use?

[Star Strider]

As always, my pleasure.

The documentation for the fminsearch function is at the link. It is an unconstrained optimiser that uses a derivative-free method to find the minimum.

The code I use for my objective function ‘fcn’ and as an argument to fminsearch are Anonymous Functions. They are quite useful for coding short functions, although they have their limitations.

I use the norm function so that the fminsearch function finds the minimum value that satisfies the sum-of-squares criterion (since this is essentially a curve-fiting problem).

Answer 2

[Akira Agata]

There should be 2 answers.

Here is my try.

P = [3 1 4;12 1 4; 34 2 4];
D = [81 36 601];

func = @(x) (vecnorm(x - P(1,:))-sqrt(D(1)))^2+...
  (vecnorm(x - P(2,:))-sqrt(D(2)))^2+...
  (vecnorm(x - P(3,:))-sqrt(D(3)))^2;

x1 = fminsearch(func,[1 1 1]);
x2 = fminsearch(func,[10 10 10]);

>> x0

x0 =

10.0000 5.0000 -0.0000

>> x1

x1 =

10.0000 5.0000 8.0000