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Q: Geometry - shperes collision ( No Answer,   4 Comments )
Subject: Geometry - shperes collision
Category: Computers > Algorithms
Asked by: jonathontrimble-ga
List Price: $75.00
Posted: 14 Mar 2006 02:27 PST
Expires: 30 Mar 2006 03:45 PST
Question ID: 707081
I have three points (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) and I know the
distance from each point to another point d1, d2 and d3 respectively. 
I want to know the equation(s) that let me calculate the (x,y,z)
coordinate of the third point.

I need to be able to calculate this in a computer program so the
answer needs to be practical in that I want to simply supply the
coordniates not solve the equations.

I suspect (but don't know) that this can be solved by matrices and
involves the collision point of three spheres.

I'd also like to know how to expand this to include a fourth and a
fifth sphere into the equation.


Request for Question Clarification by hedgie-ga on 29 Mar 2006 01:45 PST
Hello jonathon

       When you say:  
         "I need to be able to calculate this  .. on a PC .."

 do you expect the answer to provide the code, or just the equations?
 The equations can be solved on ordinary computer, using standard
 libraries, but to write such program does requires programming skills.

 The last coment by ansel have some elements which are correct;
However the equations cannot be linearised. Three spheres will in
general intersect in two points (so you will have two solutions for
given d1 d2 d3) if the spheres are close enough.
Four and more spheres, in general case, will not intersect, so there will be 
no solution.

 General case means for most positions. An example of two speres 'just touching'
 is not a general case, it is 'special case' and the two spheres  will
define one point; One way to have N>3 spheres would be to discuss
special cases.
 Another way one can understand phrase "expand this .. to N spheres" would
be a LSQ solution - solution where d.i distances are aproximate, and one looks
for best fit to overdetermined situation:
Is that what you mean?

    are you still interested in the problem?
    and is correct set of equation a sufficient answer?

There is no answer at this time.

Subject: Re: Geometry - shperes collision
From: ansel001-ga on 14 Mar 2006 18:04 PST
I'm not quite sure what you are asking.  You go back and forth between
points and spheres.  You also mention a collision point.  A sphere
implies a non-zero radius.  If two spheres collide, the collision
point would be different than the center point of either of the two

If I look at only the first paragraph and assume you are only talking
about points, are you saying given the location of two points and the
three distances between those two points and a third point of unknown
location, what is the location of the third point?

You also mention a fourth and fifth sphere.  How do they fit in?

Please clarify.
Subject: Re: Geometry - shperes collision
From: ansel001-ga on 15 Mar 2006 01:47 PST
I think I finally figured out what you are talking about in the first paragraph.

You have three points


There is one additional point Q in an unknown location.

The distance from P1 to Q is d1
The distance from P2 to Q is d2
The distance from P3 to Q is d3

What are the coordinates of Q?
And this has nothing to do with spheres.

Is this understanding correct?

If so, there is in general, no unique solution.  In 3 space, it takes
four non-coplanar points to locate something uniquely.  For any three
points, there will always exist a plane that contains all of them. 
The fourth point Q, will usually not be in the plane.  If that is the
case, then when you find one possible location of Q above the plane,
there is another possible location of Q the same distance below the

As far as the spheres are concerned and the collision point of the
spheres, I still am not sure what you are referring to.
Subject: Re: Geometry - shperes collision
From: myoarin-ga on 15 Mar 2006 03:55 PST
Ansel, great restatement of the problem.

I would understand that the spheres are the 3d sweep of d1, d2, d3
from their relative points (but I was also confused about this before
your restatement).

There was a similar question some time ago, but I cannot find it.
Subject: Re: Geometry - shperes collision
From: ansel001-ga on 15 Mar 2006 13:25 PST

I think you are right about the spheres.  The intersection of the
spheres is the solution.

For a unique solution you need four non-coplanar points.


There is one additional point Q in an unknown location.

The distance from P1 to Q is d1
The distance from P2 to Q is d2
The distance from P3 to Q is d3
The distance from P4 to Q is d4

What are the coordinates of Q?

First set up the equations for the spheres of intersection: S1, S2, S3, S4

S1:  (x-x1)^2+(y-y1)^2+(z-z1)^2 = (d1)^2
S2:  (x-x2)^2+(y-y2)^2+(z-z2)^2 = (d2)^2
S3:  (x-x3)^2+(y-y3)^2+(z-z3)^2 = (d3)^2
S4:  (x-x4)^2+(y-y4)^2+(z-z4)^2 = (d4)^2

If you combine them correctly you can make all the squared terms drop
out.  Examples are:


This will leave you with three equations and three unknowns.  All of
the equations will be linear.  The solution to this is straight
forward and will give the coordinates of your point Q.

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