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reverse 3D euclidean distance

Latest activity Commented on by Star Strider
on 25 Jul 2019
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');

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2 Answers

Answer by Star Strider
on 25 Jul 2019
 Accepted Answer

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.

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Works perfekt, thank you. Can you comment on the function you use?
Star Strider
on 25 Jul 2019
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).

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Answer by Akira Agata
on 25 Jul 2019

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

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