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A node contained determined distances from itself to other three nodes s1, s2 and s3. The other nodes are defined as S1 = (ax, ay). S2= (bx, by). S3= (Cx, Cy). The distances from node N to each of the three nodes is r1 and r2. The Trilateration problem is to find the coordinates of the unknown node N= (nx, ny) from the given information.
A complicating factor is that the unknown node coordinates are moving and the measurements contain measurement features.
The intersection of surfaces in a sphere can be found by formulating equations of the three spheres and then solving for the unknowns in the x, y and z planes. A simplified model of the equations is found by assuming the centers of the spheres to be at the Z = 0 PLANE. By placing one center at the origin, and another at the x-axis, it will be possible to come up with equations since any non-collinear points always lie on the x-axis. Once we have found the solution, we will transform it into the original three dimensional Cartesian coordinate system.
Example.
A sailor is located a point A in the Indian Ocean. Due to the malfunctioned GPS system on the sail ship, he cannot know his exact location. Upon asking the location form passing fishermen, he is told that he is located 625 mile from the Bole. However he cannot figure out his exact coordinates and location. He is certain that he is located anywhere on a circle 625 miles from Bole. A second group of fishermen informed him that he is located 700 miles from Addis.
This second information brings some insight on his probable location. Combining this information with the one from the first group of fishermen will give us the two circles that intersect at some point. Ther points of intersection give the exact location of the sailor. That is 625 miles from Bole AND 700 miles from Addis.
If a third group of sailors informs him that he is located 500 miles from Tuson, it leads to the elimination of one of the possibilities. A third circle is probable to intersect with one of the two points thus giving as definite location.
The same concept can be applied in three dimensional spaces dealing with spheres instead of circles.
We will draw a circle of radius of radius R1around Bole represented by point A, a circle of radius R2 around Addis represented by point B and a circle of radius R3 around Tuson represented by point C. The circle will overlap in a region rather than intersecting at a single point. The unknown node is somewhere in this region. Each pair of circles yields two intersecting points and the three circles yields the intersecting points. Three of the points are however, placed closely to each other while the rest are situated apart. The unknown point is mostly likely to be situated in the middle of the cluster. A computation of the distance between the three points will certainly yield points in the intersecting circles.
First we compute the distance between the circle centers.
r12x-r1x=dx
r12y-r2 y=dy
d2=dx2+dy2
∆2= ∆
Next we find the point C.
s2+u2=a2
t2+u2=b2 Subtracting the two equations we get,
s2-t2=a2-b2
s-ts+t=a2-b2
Since s+t=∆ and t=∆-s then it gives the expression for S as follows.
s-∆-s∆=a2=b2
2s∆-∆2=a2-b2
Thus s= ∆2+a2-b22
The basic equation of a circle is as follows
We need to find the location of a point on the (x, y, z) plane that satisfies all the three equations.
In order to do so, we subtract the second equation from the first.
() – () to give We know that s2+u2=a21 Brinker, R. C. (2005). The Surveying Handbook. Springer.
A basic assumption is that the two spheres intersect in more than one point thus, d-r1 < r2
Substituting into equation one above yields y=r2- x2
A rearrangement of the equation gives the equation of the z-coordinate. . Thus we have the solution of the three points, x, y and z. Z cab be expressed as the positive or negative value. Also it is possible to have a zero, or more than one solution to the problem.
The z equation is translated as an equation of the circle found to intersect the first and the second spheres. If the circle falls outside or inside of the sphere, then z is equal to a square root of the negative number. It therefore, implies that there is no real solution found. Likewise, if the circle intersects the sphere on only one location, z is equal to zero. The last scenario possible is if the circle touches the surface at exactly two points then z is equal to plus or minus the square root of a real number.
The sphere centers designated above were placed such that all lie in the designated z plane, z=0. The sphere center S1 is at the origin and the sphere S2 is on the X-axis.
One way of finding the coordinates of the unknown point is through computing the centroid of the cluster. An average of the centroids X coordinates gives the coordinates of the X coordinates. The centroids Y coordinates is the coordinates of the Y points. If the circle intersection of the close coordinates is found, adding this to the cluster and re-computing the centre of the centroid will enable us to zero in the most probable points. A repeat of this procedure is performed until n intersection points are added to the cluster. K in this case denotes the number of circles. The final result of the cluster gives the location of the unknown node N.References
Chandra, A. (2005). Surveying Problem Solution With Theory And Objective Type Questions. New Age International.
Comfort, L. K. (2010). Designing Resilience. University of Pittsburgh Pre.
D. M. Akbar Hussain, A. Q. (2008). Wireless Networks, Information Processing and Systems:. Springer,.
Engineering, A. S. (2004). The Glossary of the Mapping Sciences. ASCE Publications.
Field, H. L. (2011). Landscape Surveying. Cengage Learning.
