Trilateration
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Trilateration
This article describes a method for
determining the intersections of three
spheresurfaces given the centers and
radii of the three spheres.
More generally, trilateration methods
involve the determination of absolute
or relative locations of points by
measurement of distances, using the
geometryofspheres ortriangle
In contrast totriangulationit
does not involve the measurement of
angles.
Trilateration is mainly used in
surveyingandnavigation, including
global positioning systems (GPS).
Figure 1. The plane, z=0, showing the 3 sphere centers, P1, P2, and P3; their x,y
In a2D planeusing two reference coordinates; and the 3 sphere radii, r1, r2, and r3. The two intersections of the three
sphere surfaces are directly in front and directly behind the point designated intersections
points is normally sufficient to leave
in the z=0 plane.
only two possibilities for the location
determined, and the tie must be broken
by including a third reference point or other information.
In 3D space, using three reference points similarly leaves only two possibilities, and the tie is broken by including a
fourth reference point or other information.
Derivation
The solution is found by formulating the equations for the three sphere surfaces and then solving the three equations.
for the three unknowns, x, y, and z. To simplify the calculations, the equations are formulated so that the centers of
the spheres are on the z=0 plane. Also the formulation is such that one center is at the origin, and one other is on the
x-axis. It is possible to formulate the equations in this manner since any three non-collinear points lie on a plane.
After finding the solution it can be transformed back to the original three-dimensional.Cartesian coordinate system.
We start with the equations for the three spheres:
We need to find a point located at (x,y,z) that satisfies all three equations.
First we subtract the second equation from the first and solve for x:
We assume that the first two spheres intersect in more than one point, that is that d-r1 < r2 < d+r1. In this case
substituting the equation for x back into the equation for the first sphere produces the equation for a circle, the
solution to the intersection of the first two spheres:
Trilateration 2
Substituting: into the formula for the third sphere and solving for y there results:
Now that we have the x- and y-coordinates of the solution point, we can simply rearrange the formula for the first
sphere to find the z-coordinate:
Now we have the solution to all three points x, y, and z. Because z is expressed as the positive or negative square root,
It is possible for there to be zero, one or two solutions to the problem.
This last part can be visualized as taking the circle found from intersecting the first and second sphere and
intersecting that with the third sphere. If that circle falls entirely outside or inside of the sphere, zis equal to the
square root of a negative numberno real solution exists. If that circle touches the sphere at exactly one point.
equal to zero. If that circle touches the surface of the sphere at two points, then z is equal to plus or minus the square
root of a positive number.
Preliminary and final computations
The section Derivation pointed out that the coordinate system in which the sphere centers are designated must be
such that (1) all three centers are in the plane, Z = 0, (2) the sphere center, P1, is at the origin, and (3) the sphere
center, P2, is on the X axis. In general the problem will not be given in a form such that these requirements are met.
This problem can be overcome as described below where the points, P1, P2, and P3 are treated as vectors from the
origin where indicated. P1, P2, and P3 are of course expressed in the original coordinate system.
is the unit vector in the direction from P1 to P2.
is the signed magnitude of the x component, in the figure 1 coordinate system, of the
vector from P1 to P3.
is the unitized y component, in the figure 1 coordinate system, of the vector
from P1 to P3. Note that: has been defined in such a manner that the points P1, P2, and P3 are all in the
z=0 plane of the figure 1 coordinate system as required.
The third basis unit vector is . Therefore,
the distance between the centers P1 and P2 and
Is the signed magnitude of the y component, in the figure 1 coordinate system, of the
vector from P1 to P3.
Using as computed above, solve for x, y and z as described in the Derivation section. Then
gives the points in the original coordinate system since , the basis unit vectors, are expressed in
the original coordinate system.
Trilateration 3
Application
Trilateration is mainly used insurveyingandnavigationincludingglobal positioning systems (GPS).
See also
• Euclidean distance
• Multilaterationposition estimation using measurements of time difference of arrival at (or from) three or more
sites.
• Resection
• Triangulation
• Global positioning system
References
Encyclopedia Britannicahttp://www.britannica.com/EBchecked/topic/605329/trilateration
dirac delta(http://www.diracdelta.co.uk/science/source/t/r/trilateration/source.html)
global maritimeThe provided text is a URL and does not contain translatable content.
free dictionary(http://www.thefreedictionary.com/trilateration)
Article Sources and Contributors 4
Article Sources and Contributors
TrilaterationSourceThe provided text is a URL, not translatable content. Contributors: 4johnny, AMag, Adrian.benko, Ahseaton, Antandrus, Ashawley, Barrylb, Bepa, Braindrain0000,
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Image Sources, Licenses and Contributors
No translatable text provided in the input.http://en.wikipedia.org/w/index.php?title=File:3spheres.jpg Free Art License
License
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