From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Triangulation using sphere intersects
Date: Wed, 28 Jan 2009 11:03:04 +0000 (UTC)
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"Roger Stafford" <> wrote in message <glohe8$ai1$>...
> "gregoire " <> wrote in message <glmsf6$7b6$>...
> > Thanks David for your response, some interesting reads there, especially as I may have to look into errors and noise variation in my data at a later stage.
> > 
> > I dont wish to find a new method (though the Newton-Rapshon one seems interesting) for now Im just trying to work out how to 
> > 
> > a) use plot3 to plot many circles together in 3d space, which are of the form:-
> > 
> >  Q = P + R*u*cos(s) + R*v*sin(s)
> > where s is an (angular) parameter and Q an arbitrary point on the circle. u,v are necessary orthogonal vectors, 
> > 
> > as that is the way I have already coded and successfully calculated the necessary parts, from my data
> > 
> > and b) attempting to find potential intersections betweens these circles 
> > 
> > Thanks again
>   I don't think I agree with you, Gregoire, that finding the intersections of the parametric circles is the best way to find the intersection points of triplets of spheres.  The way I outlined to you involving three intersecting planes and then a certain distance in either direction along a line orthogonal to the plane of the three centers is a far better method which does not involve iterative methods like Newton-Raphson.  You get the two points all in one clean shot without convergence worries or iterations.  The fact that the circles will run through these points does not help you in my opinion.  I think you will find that things are easier that way.
> Roger Stafford

i think a more general solution is the newton-raphson approach, especially if he expects to be considering uncertainty in the future.  if the spheres are not known exactly, or there are introduced errors where maybe two of the spheres only get very close but don't actually intersect, then an iterative approach may be the only option to find the best possible answer.