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| [[Image:3spheres.svg|500px|thumb|Figure 1. The plane ''z'' = 0, showing the three sphere centers, ''P''1, ''P''2, and ''P''3; their ''x'',''y''-coordinates; and the three sphere radii, ''r''1, ''r''2, and ''r''3. The two intersections of the three sphere surfaces are directly in front and directly behind the point designated intersections in the ''z'' = 0 plane.]]
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| In [[geometry]], '''trilateration''' is the process of determining absolute or relative locations of points by measurement of distances, using the geometry of [[circles]], [[sphere]]s or [[triangle]]s.<ref>[http://www.britannica.com/EBchecked/topic/605329/trilateration Encyclopædia Britannica]</ref><ref>[http://www.diracdelta.co.uk/science/source/t/r/trilateration/source.html diracdelta]</ref><ref>[http://www.globmaritime.com/technical-articles/marine-navigation/general-concepts/9622-trilateration-traverse-and-vertical-surveying.html Global Maritime]</ref><ref>[http://www.thefreedictionary.com/trilateration free dictionary]</ref> In addition to its interest as a geometric problem, trilateration does have practical applications in [[surveying]] and [[navigation]], including [[global positioning system]]s (GPS). In contrast to [[triangulation]], it does not involve the measurement of [[angle]]s.
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| In [[2D geometric model|two-dimensional geometry]], it is known that if a point lies on two circles, then the circle centers and the two radii provide sufficient information to narrow the possible locations down to two. Additional information may narrow the possibilities down to one unique location.
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| In three-dimensional geometry, when it is known that a point lies on the surfaces of three spheres, then the centers of the three spheres along with their radii provide sufficient information to narrow the possible locations down to no more than two (unless the centers lie on a straight line).
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| This article describes a method for determining the intersections of three sphere surfaces given the centers and radii of the three spheres.
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| == Derivation ==
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| The intersections of the surfaces of three spheres 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-colinear points lie on a unique plane. After finding the solution, it can be transformed back to the original three dimensional [[Cartesian coordinate system]].
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| We start with the equations for the three spheres:
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| :<math>r_1^2=x^2+y^2+z^2 \, </math>
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| :<math>r_2^2=(x-d)^2+y^2+z^2 \, </math>
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| :<math>r_3^2=(x-i)^2+(y-j)^2+z^2 \, </math>
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| We need to find a point located at (''x'', ''y'', ''z'') that satisfies all three equations.
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| First, we subtract the second equation from the first and solve for ''x'':
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| :<math>x=\frac{r_1^2-r_2^2+d^2}{2d}.</math>
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| We assume that the first two spheres intersect in more than one point, that is that
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| :<math>d-r_1 < r_2 < d+r_1. \, </math>
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| 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:
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| :<math>y^2+z^2=r_1^2-\frac{(r_1^2-r_2^2+d^2)^2}{4d^2}.</math>
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| <!-- Setting this formula equal to the formula for the third sphere finds: -->
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| Substituting <math>z^2=r_1^2-x^2-y^2</math> into the formula for the third sphere and solving for ''y'' there results:
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| :<math>y=\frac{r_1^2-r_3^2-x^2+(x-i)^2+j^2}{2j}=\frac{r_1^2-r_3^2+i^2+j^2}{2j}-\frac{i}{j}x.</math>
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| 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:
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| :<math>z=\pm \sqrt{r_1^2-x^2-y^2}.</math>
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| 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.
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| 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, ''z'' is equal to the [[Imaginary number|square root of a negative number]]: no real solution exists. If that circle touches the sphere on exactly one point, ''z'' is 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. <!-- If there is no solution, there is no solution. Furthermore one sphere surface is at a distance of r1 from (0,0) so there appears to be no basis for saying zero is the nearest solution. --> <!-- In the case of no solution, a not uncommon one when using [[noisy data]], the nearest solution is zero. One should be careful, though, to do a [[sanity check]] and assume zero only when the error is appropriately small. --> <!-- Trilateration is a mathematical method for finding the intersection of 3 sphere surfaces. It is purely mathematical and in no way depends on GPS. The comment below is inappropriate in an article on trilateration. --><!-- In the case of two solutions, some technique must be used to disambiguate between the two. This can be done mathematically, by using a fourth sphere with its center not being located on the same plane as the centers of the other three, and determining which point lies closest to the surface of this sphere. Or it can be done logically—for example, [[Global Positioning System|GPS receivers]] assume that the point that lies inside the orbit of the satellites is the correct one when faced with this ambiguity, because it is generally safe to assume that the user is never in space, outside the satellites' orbits. -->
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| == Preliminary and final computations == | |
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| The ''Derivation'' section 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, ''P''1, is at the origin, and (3) the sphere center, ''P''2, is on the ''x''-axis. In general the problem will not be given in a form such that these requirements are met.
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| This problem can be overcome as described below where the points, ''P''1, ''P''2, and ''P''3 are treated as vectors from the origin where indicated. ''P''1, ''P''2, and ''P''3 are of course expressed in the original coordinate system.
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| :<math>\hat e_x = \frac{ P2 - P1 }{ \| P2 - P1 \| }</math> is the unit vector in the direction from ''P''1 to ''P''2. | |
| :<math>i = \hat e_x \cdot ( P3 - P1 ) </math> is the signed magnitude of the ''x'' component, in the figure 1 coordinate system, of the vector from ''P''1 to ''P''3. | |
| :<math>\hat e_y = \frac{ P3 - P1 - i \; \hat e_x}{ \| P3 - P1 - i \; \hat e_x \| }</math> is the unit vector in the y direction. Note that the points ''P''1, ''P''2, and ''P''3 are all in the ''z'' = 0 plane of the figure 1 coordinate system.
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| The third basis unit vector is <math>\hat e_z = \hat e_x \times \hat e_y</math>. Therefore,
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| :<math>d = \| P2 - P1 \|</math> the distance between the centers ''P''1 and ''P''2 and
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| :<math>j = \hat e_y \cdot ( P3 - P1 ) </math> is the signed magnitude of the y component, in the figure 1 coordinate system, of the vector from ''P''1 to ''P''3.
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| Using <math>i, \; d</math> and <math>j</math> as computed above, solve for ''x'', ''y'' and ''z'' as described in the ''Derivation'' section. Then
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| :<math>\vec p_{1,2} = P1 + x \ \hat e_x + y \ \hat e_y \ \pm \ z \ \hat e_z </math>
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| gives the points in the original coordinate system since <math>\hat e_x, \; \hat e_y</math> and <math>\hat e_z</math>, the basis unit vectors, are expressed in the original coordinate system.
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| == See also ==
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| * [[Euclidean distance]]
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| * [[Multilateration]] – position estimation using measurements of time difference of arrival at (or from) three or more sites.
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| * [[Resection (orientation)]]
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| * [[Triangulation]]
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| * [[Global positioning system]]
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| == References ==
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| {{reflist}}
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| == External links ==
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| *[http://www.general-files.com/download/gs5bac8d73h32i0/IEEE-AES-96-Efficient%20Solution-and-Performance-Analysis-of-3-D%20Position-Estimation-by-Trilateration.pdf.html Efficient Solution and Performance Analysis of 3-D Position Estimation by Trilateration]
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| [[Category:Euclidean geometry]]
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| [[Category:Geodesy]]
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| [[Category:Wireless locating]]
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| [[ar:تثليث مساحي]]
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| [[bg:Трилатерация]]
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| [[de:Lateration]]
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| [[es:Trilateración]]
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| [[fr:Trilatération]]
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| [[it:Trilaterazione]]
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| [[ru:Трилатерация]]
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| [[sk:Trilaterácia]]
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Grateful to meet you! A few name is Eusebio Ledbetter. It's not a common concern but what I much like doing is bottle blouses collecting and now I do have time to have a look at on new things. Software developing is how I've support my family. My house is so in Vermont. I've been toiling on my website for some time now. Inspection it out here: http://prometeu.net
My weblog ... clash of clans hack no survey or password