Re: Triangulation Problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg3007] Re: Triangulation Problem*From*: in2optix at aol.com (In2Optix)*Date*: Mon, 22 Jan 1996 03:06:02 -0500

In article <4diafm$84a at dragonfly.wri.com>, techie at io.org (david hoare) writes: >I am looking for the trigonometric / algebreic formulae one would use >to triangulate an unknown position, knowing 3 fixed points and the >respective distances to the unknown point. (make sence?) >Oh - in 3 dimensional space. > >And actually, the reverse operations would be handy too...ie: given a >known central point, AND 3 other fixed points, how to find the lengths >of the lines required to meet at that first point, in 3D. > >It's for a program I'm writing, and it's been a while since I've done >this sort of thing - - my math has left me! >Any Help would be GREATLY appreciated!! > >Thanks, >David. >techie at io.org > > If you know the distance to the unknown point, but not the direction (which is the opposite of what would happen in a typical surveying scenario) then the unknown point lies on a sphere. Each of your three known points would be surrounded by spheres of radius equal to their respective distance to the unknown point. Solving the equations of the three spheres will result in the intersection point(s). If you think about the problem in these terms, one will realize that in the majority of cases, two solutions will be found. Equations of the spheres: sphere1=(x-x1)^2+(y-y1)^2+(z-z1)^2==r1^2; sphere2=(x-x2)^2+(y-y2)^2+(z-z2)^2==r2^2; sphere3=(x-x3)^2+(y-y3)^2+(z-z3)^2==r3^2; where (xi, yi, zi) is the position of the three known points and ri is the distance from point i to the unknown point. Now solve for x, y, and z. Solve[{sphere1,sphere2,sphere3},{x,y,z}] As for the reverse problem, you only need to reverse the sphere equations given above (i.e., solve for ri). For example: r1= Sqrt[(x-x1)^2+(y-y1)^2+(z-z1)^2] Good luck, Anthony D. Gleckler Kaman Aerospace Corporation ==== [MESSAGE SEPARATOR] ====