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Re: Triangulation Problem

David Hoare wrote:
I am looking for the trigonometric / algebreic  formulae one would  
use to triangulate an unknown position, knowing 3 fixed points and  
the respective distances to the unknown point. (make sense?) Oh - in  
3 dimensional space.


It's for a program I'm writing, and it's been a while since I've done
this sort of thing - - my math has left me!
Any Help would be GREATLY appreciated!!

I have seen essentially two responses to this, and I have a few  
comments on each.

1.  Anthony D. Gleckler's approach [mg3007], slightly paraphrased  
(and the approach I used):

If a and b are points (given as lists of coordinates), then the  
distance between them is, in Mathematica,


Let the three given points be
a = {a1,a2,a3};
b = {b1,b2,b3};
c = {c1,c2,c3};

let the associated distances be A,B,C, and let the unknown point be

X = {x,y,z};

Then the desired point is the simultaneous solution of these three  

spherea := ( (X-a).(X-a) == A^2 );
sphereb := ( (X-b).(X-b) == B^2 );
spherec := ( (X-c).(X-c) == C^2 );

each of which just states the desired equality of distances, but  
with both sides squared.  This is three quadratic equations in three  

sols = Solve[{eqa,eqb,eqc}, X]

will return the solution as replacement rules, and

soughtpoints = X/.sols

will give the unknown point as a list of coordinates -- but not  
really!  First of all, the Mathematica commands given above work  
fine for numerical values for a, b, c, A, B, and C.  If you try to  
get a general symbolic solution, however, the expressions get  
unmanageably big.

But there's another issue where writing the necessary code is not  
hard, but deciding just what you want the code to do requires some  
reflection.  Unless the measurements of the given data (a, b, c, A,  
B, and C) are absolutely perfect, the system of equations will have  
TWO DISTINCT SOLUTIONS, possibly complex (having equal and opposite  
imaginary parts), and the soughtpoints as computed above will  
actually be a list of TWO DISTINCT POINTS, possibly with complex  
coordinates.  Each of the equations shperea, sphereb, and spherec  
define a sphere; two of the spheres generically intersect in a  
circle, which will generically intersect the third sphere in two  
points.  These two points (if real) determine a line perpendicular  
to the plane containing the given points a,b,c, and the two points  
are equidistant from that plane.

If the measurements of a, b, c, A, B, and C are only slightly off,  
these two points will be very close together, and simply averaging  
them will give a reasonable, usable result:

thepoint = Apply[Plus, soughtpoints]/2

(Conveniently, this will give a real result even if complex  
solutions are involved!)

You might, however, wish to use the distance between the two  
solutions as an indicator of the quality of your data (i.e. the  
given points and distances).  At the simplest level, a relatively  
"large" distance between the two solution points indicates that your  
values for A, B, and C *can't* be accepted as distances of a, b,  
and c from a common fourth point; you could use that "largeness"  
simply as a tipoff that something's wrong with the data.  With some  
more effort, you could also convert the size of that distance into a  
numerical measurement of how well all your pieces are fitting  

2.  Dave Rusin's approach [mg3015]

I really enjoyed Dave's method, because it so closely follows what  
we might do if we could directly manipulate abstract geometrical  
objects in 3-space with our hands.  He suggests locating and  
explicitly parametrizing the circle where the first two spheres  
intersect.  (I also think parametrizing curves is underappreciated  
in general.)  This can be done in Mathematica by using some tricks  
with NullSpace and LinearAlgebra`Orthogonalization`GramSchmidt.

Dave's method will lead to the same two solutions for less than  
perfect data, but this time they appear in the (ususally) multiple  
solutions of the trigonometric equation

> D3^2 = (r cos(t)-a)^2 + (r sin(t)-b)^2 + (x-c)^2

the solving of which involves taking multi-valued arcsines.  (In 1,  
we might have been alerted to multiple solutions by the fact that  
we were solving quadratic equations.)


Preston Nichols
Visiting Assistant Professor
Department of Mathematics
Carnegie Mellon University
(seeking employment for Fall 1996)

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