Calculate Position  Angle Difference of Arrivalby murrdpirate0 Tags: angle, arrival, difference, position 

#1
May912, 04:07 PM

P: 29

I'm trying to work out the mathematics of this for a school project.
Say you have 4 beacons arranged in a square (I think you only need 3, actually) at known positions relative to one another. Now say you are in a random position within this square, at a random height, and at a random orientation (yaw, pitch, and roll). If you know both the zenith and azimuth angle to each beacon, can you completely solve your 3D position and orientation? I believe that you can. It's setting up the equations that's taking me a while. If anyone can point me in the right direction, so to speak, I'd greatly appreciate it. 



#2
Jul212, 02:03 AM

P: 65

and why don't you try to give a diagrammatic representation ? 



#3
Jul212, 04:25 AM

P: 4,570

I'm assuming that you are triangulating from (three of four) of the beacons in one plane and not using any kind of curved geometry for simplicity. If this is the case you should be able to find a position on the plane corresponding to the position, by solving some simultaneous equations. Essentially the intuition behind it is that each angle traces a line from each beacon and the intersection of these lines will dictate your position. If all beacons have the exact angle information, you will get a unique answer. You can factor in uncertainty in the readings by getting a 'best guess', but this is a little more complicated. If you assume that you have an exact reading for all angle information at the same point in time, you solve the simultaneous equations to get your final position that is in the plane. Have you had any experience doing this? 



#4
Jul1112, 12:53 PM

P: 29

Calculate Position  Angle Difference of Arrival
vrmuth, the position needs to be calculated in reference to the position of any one of the beacons, each of which will have a 3D coordinate in reference to a global coordinate system. See the link below for a diagram.
chiro, thanks for the welcome. I actually found that while 3 beacons is enough the vast majority of the time, I need 4 beacons to be absolutely sure I have a single solution. I don't want to have to constrain the 4 beacons to a single plane, however, as I think that would cause construction challenges. I do have a solution using simultaneous equations, but it's iterative and not quite as fast as I'd like. See this thread I started on that (but note that I've since added a 4th beacon): http://www.physicsforums.com/showthread.php?t=608591 One challenge is that due to inexact angle measurements, the angles to each of the beacons do not intersect at a point, they intersect in a series of points or not at all. So I have to assume an intersection point and iterate to find what lengths to each beacon fit both the intersection point and the measured angles. Thanks for the responses. 



#5
Jul1112, 07:25 PM

P: 4,570

You may be able to use the redundancy of a fourth beacon to simplify the equations a little bit, even if they going to use some kind of numeric scheme.
The other thing also to consider so that you could place the beacons in a configuration that allows you to get a specific kind of expression which could be factorized. For example you can create a triangulation orientation that allows you to get a polynomial (after you do all the algebra) and then depending on the angles and lengths, you end up getting simpler polynomials like say a quadratic times a quartic if you have an 8th degree polynomial and both of those can be solved analytically. So this means that if you are allowed to place the beacons in many possible configurations, you choose the best one that allows simplicity of doing calculations to triangulate position. 


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