Triangulation is the process of determining the location of
a point using angles formed between it and points with known
locations
Trialateration is the process of determining the location of
a point using the distances between it and points with known
locations
For Our Purposes:
We won't distinguish between the two
Using Distances in One Dimension
The Problem:
Given one or more points with known locations,
determine the location of a point using the distances
between it and those points
Notation:
p \in \mathbb{R} denotes a point
\cal{C(p, \ell)} denotes the set of points
that are exactly distance \ell from p
Using Distances in One Dimension (cont.)
With One Known Location:
Given the location of the point p and the
distance from p to a, determine the
location of the point a
A Specific Example:
p is located at 4
The distance from p to a is 2
Visualization:
Using Distances in One Dimension (cont.)
With One Known Location (cont.):
You can see from the example that the problem is
under-identified when there is only one known location
A Specific Example with Two Known Locations:
p is located at 4
The distance from p to a is 2
q is located at 9
The distance from q to a is 3
Visualization:
Using Distances in Two Dimensions
Generalizing the Notation:
p \in \mathbb{R}^2 now denotes a point
on the plane (i.e., in two dimensions)
Recall:
A circle is the set of points in a plane that
are a given distance (called the radius) from
a given point (called the center)
The Implication:
We need to find the intersections of circles
Using Distances in Two Dimensions (cont.)
An Example:
p = (2.0, 3.0) and \ell(p) = 5.0
q = (4.5, 10.0) and \ell(q) = 3.0
r = (7.5, 6.5) and \ell(r) = 2.5
Visualization:
Using Distances in Two Dimensions (cont.)
Recall:
Letting (h, k) denote the center of the circle,
and r denote its radius, the circle is defined as the
set of points (x, y) that satisfy:
\sqrt{(x - h)^2 + (y - k)^2} = r
Squaring Both Sides:
(x - h)^2 + (y - k)^2 = r^2
Using Distances in Two Dimensions (cont.)
Intersection of Two Circles:
They may not intersect
They may intersect at one point
They may intersect at two points
Finding the Intersection:
Requires a little algebra, but isn't difficult (as long as
you account for the different possibilities)