Re: Triangulation Problem

• To: mathgroup at smc.vnet.net
• Subject: [mg3058] Re: [mg2968] Triangulation Problem
• From: Preston Nichols <nichols at godel.math.cmu.edu>
• Date: Tue, 30 Jan 1996 02:21:59 -0500
• Sender: owner-wri-mathgroup at wri.com

```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

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,

Sqrt[(a-b).(a-b)]

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
equations:

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
unknowns;

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
together.

---------------------------------------------------
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

---------------------------------------------------

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

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