Can I Find an Analytical Solution for Spherical Trilateration?

In summary, this conversation is about finding the coordinates of a fourth point on a central sphere based on the angular distances between this point and three known points on the sphere. The usual method of transforming to Cartesian coordinates and then back to spherical coordinates results in a large error, so the goal is to find an analytic solution to avoid this issue. One possible approach is to treat the problem as the intersection of three planes, each representing a circle on the large sphere, and using vectorial methods to solve it in Cartesian coordinates. However, if staying in spherical coordinates is preferred, the cross-product in spherical coordinates would need to be used.
  • #1
strive
51
4
Hi

I am looking for an equation of intersection of 3 circles or 3 spheres, on the surface of the fourth (central) sphere, in a spherical coordinate circle. This should really be just a simple trilateration problem.

I know this is usually done by transforming the spherical coordinate system to a Cartesian coordinate system, then calculating the Intersection point, and then transforming it back to the spherical coordinate system. But due to large input errors my final error is then far too big to be useful.

Currently I am obtaining the result with a numerical calculation, but it takes too long (approx. 1 second for a single calculation), thus I would like to find an analytic solution. I have tried to derive it myself many times, but all I get is an unsolvable mix of sin and cosin functions (or tand and cotand)…

In short: I have the spherical coordinates of 3 points on the fourth ‘’center’’ sphere. On the center sphere I have a fourth point of which I do not know the coordinates, but I know all 3 angular distances between the first 3 points and the forth point.

How do I find the coordinates of the forth point, without having to transform everything into the Cartesian coordinate system?

I have scoured the internet and books for months, but failed to find a solution. Is it even possible? :nb)
I know it should be…
 
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  • #2
If I am envisioning this correctly, this can be convered to a problem of three planes that cut into a large sphere (in three circles) whose mutual intersection is non-empty and is already known to be on that large sphere. Is that correct?
 
  • #3
Yes that is correct.
I have tried to use equations for great circles and for small circles, but to no avail (although my math might be the reason, because the numerical solution works in this way).
I haven’t thought of trying with planes. But how do i define a straight plane in the spherical coordinate system?

Thank you for your reply.
 
  • #4
For simplicity, I think one should use vectorial methods as much as possible.

The normal vector to each of these three planes is radial [say, radially outward].
However, each plane is not tangent to the large sphere...
but parallel and slightly beneath that tangent plane by amount
determined by some circular geometry related to the versed sine [ http://en.wikipedia.org/wiki/Versine ].
Do you know the radii of the circles?

Now you have three planes (which do cut the sphere in circles.. but don't focus on that aspect).
As you say, there is a non-empty intersection of those three planes.
So, the problem becomes the intersection of three planes.
https://www.google.com/search?q="Intersection+of+Three+Planes"
..and you have some information that this intersection
lies on the large sphere (i.e. has a known distance from the center of the large sphere).

Does that help?
 
  • #5
Yes, I can calculate their radius, but initially I can only measure the angular separation between two antipodal points of the circle (the base of the angle is the large sphere).

The radius of the large sphere is not accurately known as it is virtual (imaginary is perhaps a better term). Of course I can calculate it, but with a considerable error.

So, the coordinates of each of the known 3 points constitute the normal of the belonging planes. And this really becomes just a vectorial problem and the solution would be obtained with a matrix?

I understand the solution in the Cartesian coordinate system, but I am unsure of how to represent it in spherical coordinates. I understand how this system works, but I don’t know how to represent anything that is straight in it.
I can transform from one system to another, but how do I represent a plane just by using the spherical coordinate system?
Can you give me the general equation of a straight line in coordinate system (let’s say a line that goes between point1 (ρ1, φ1, R) and point2 (ρ2, φ2, R))?
NumberedEquation1.gif

This is supposed to be it, but what does it actualy mean?
 
  • #6
Suppose you are given position vectors [possibly in terms of some coordinate system] C1, C2, C3 with origin at O.
If these position vectors lie on a fourth sphere centered at O, then magnitudes of C1, C2, C3 should be equal to each other and the radius of the large sphere.

So, you are trying to avoid cartesian coordinates... and stay in spherical coordinates as much as possible?
http://mathworld.wolfram.com/SphericalCoordinates.html

(Just saw this... maybe useful... but not immediately for my proposed method
http://math.stackexchange.com/quest...-intersection-of-three-spheres-full-solutions
)

According to http://jwooten37830.com/blog/wp-content/uploads/2014/01/ConeAndPlane.pdf , Eq. 12 is the equation of a plane in spherical coordinates. (I don't have time to check that.)

In my method, the key idea is that each circle on the large sphere [who center point and radius is known] is encoded by a unique plane.
Circle 1 has normal-vector along O-C1, etc...
Given the angular separation (given as a central angle with vertex at O) of the antipodal points of a circle on the big sphere,
you can calculate [using the versed sine of half-of-that-angular-separation] how far down to move the plane [essentially changing only rho].
So, by now, one has three planes corresponding to the three circles.

Then you can use the methods of http://mathworld.wolfram.com/Plane-PlaneIntersection.html repeatedly (or https://www.google.com/search?q="Intersection+of+Three+Planes" ).
However, if you insist on staying in spherical coordinates, you'll have to get expressions for the cross-product in spherical coordinates ( http://math.stackexchange.com/quest...lculating-dot-and-cross-products-in-spherical )

(This person https://rbrundritt.wordpress.com/2008/10/14/points-of-intersection-of-two-planes/ doing the intersection of two planes through the center... suggests working in cartesian coordinates.)

In my comment earlier, about "vectorial methods",
I was thinking about using normal-vectors to planes-that-cut-the-sphere and using linear algebra [in cartesian coordinates for computational simplicity],
as opposed to writing analytic functions describing circles [in some coordinate system] and seeking solutions somehow.
 
  • #7
Beautiful!
By using the equations from ’The Intersection of a Cone and a Plane' paper and the plane intersection method, I think I can now make this work.

The problem with using the transformations to and from the Cartesian coordinate system is in my initial measurement errors, as they are significantly increased by transformations. The error can be increased up to 50 times in the entire trilateration procedure if I use the Cartesian coordinate system, and I cannot afford such a high error. The numerical solution increases the error by an almost unnoticeable amount, but it gives 3 points of intersections, as the circles overlap due to the measurement errors. Thus I have to average the 3 obtained results to obtain the final position, and this again increases the error.
A simplified 2-D version of this problem has shown that an analytical solution partially covers the measuring error thus the final error of the calculated position is actually slightly lower than the input error. On top of this is the calculation time improvement, thus an analytical solution will really help in many ways.

Thank you for your time and help, I am much obliged.
 

What is spherical trilateration?

Spherical trilateration is a mathematical technique used to determine the location of a point in three-dimensional space using the distances from that point to three known reference points.

How does spherical trilateration work?

Spherical trilateration works by intersecting three spheres, each with a radius equal to the distance from the reference point to the unknown point. The intersection of these spheres results in two possible points, and the correct one can be determined by using a fourth reference point.

What are the applications of spherical trilateration?

Spherical trilateration has various applications, including GPS navigation, surveying and mapping, and determining the location of objects in space. It is also used in land surveying, geodesy, and wireless communication systems.

What are the limitations of spherical trilateration?

Spherical trilateration is limited by the accuracy of the reference points and the measurements. If the reference points are not accurately known, the calculated location of the unknown point will also be inaccurate. Additionally, any measurement errors can also affect the accuracy of the results.

Are there any alternatives to spherical trilateration?

Yes, there are alternative methods for determining the location of a point in space, such as triangulation, multilateration, and the global positioning system (GPS). These methods may be more suitable depending on the specific application and the available technology.

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