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Q: Triangulation (Lateration) ( No Answer,   2 Comments )
Question  
Subject: Triangulation (Lateration)
Category: Science
Asked by: flattire2006-ga
List Price: $2.00
Posted: 28 Oct 2004 13:19 PDT
Expires: 29 Oct 2004 15:24 PDT
Question ID: 421339
Triangulatio (Lateration)

I am working on a project in which I have 3 known points (A B C), one
unknown point(F), and all the distances.

Where can I find an algorithm, equation, or formula to get the point F
with the given information?  I have developed a formula, but it is
very lengthy.

Also, can this be done with one two known point and all of the needed distances?

Thanks.

Request for Question Clarification by mathtalk-ga on 28 Oct 2004 15:20 PDT
Hi, flattire2006-ga:

If by your last question you meant can one triangulate using two known
points and the necessary distances from them to an unknown point, the
difficulty is that you would obtain (assuming the values involved are
feasible, satisfying the "triangle inequality) two solutions
(symmetric with respect to the line drawn through the two known
points).  This is because unsigned distances can't distinguish between
these two "mirror image" solutions.

Using three known points breaks the possible ambiguity of a solution,
and also allows for measurement uncertainties to be handled in a
stable fashion, ie. giving a best least squares fit to the data.

regards, mathtalk-ga
Answer  
There is no answer at this time.

Comments  
Subject: Re: Triangulation (Lateration)
From: hfshaw-ga on 28 Oct 2004 15:46 PDT
 
Assuming your question deals with locating the position of a point in
2 dimensions (i.e., on a plane), Mathtalk is correct; in general, you
need the distances to three known points to uniquely determine the
coordinates of the fourth point.  If you need to locate a point in
three dimensions, you actually need to know the distance to a fourth
known point.  (This is how GPS receivers work to find the latitude,
longitude, and elevation of the receiver.)

For the 2-dimensional case, you can visualize why this is so by
considering that if you know you are a given distance (R1) from a
known point at (x1, y2), you know you are somewhere on the circle with
radius R1 surrounding that point.  If, in addition, you know you are a
distance R2 from a second point (x2,y2), you know that you must also
be on a circle of radius R2 from that point.  If (R1 + R2) <
sqrt[(x1-x2)^2 + (y1-y2)^2] (this is what Mathtalk means by the
triangle inequality), the two circles do not intersect, and there is
no physically meaningful solution to the location problem.  If R1 + R2
> sqrt[(x1-x2)^2 + (y1-y2)^2], the two circles intersect at two
points, and you need the distance to a third point to determine which
of these points is the actual location.  If (R1 + R2) = sqrt[(x1-x2)^2
+ (y1-y2)^2], one has the special case that all three points lie on a
single line, and in this case only, you only need the distances to two
known points to determine the coordinates of a third point.

See for http://www.restena.lu/convict/Jeunes/beacon/Positioning.htm
for a nice illustration of these concepts, as well as the equations
you are looking for to determine the coordinates of your unknown
point.
Subject: Re: Triangulation (Lateration)
From: flattire2006-ga on 29 Oct 2004 15:24 PDT
 
http://www.restena.lu/convict/Jeunes/beacon/Positioning.htm

That url contains the exact information I was looking for.  Thank you.

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