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Point Triangulation in Fixed Space

  1. Dec 28, 2011 #1
    Hello,

    I'm an undergraduate electornics engineer.

    I'm trying to design a system where I am able to find my relative position in a room. Using electronics, I am able to find the angles between the corners of the room, as observed from the point. I also know the dimensions of the room. Using this information I should logically be able to find my position.

    I've been trying for the last 2 days to come up with an equation. However, everything I come up with is heavily linear dependant on each other; I'm going round in circles.

    The picture below better explains my problem:

    zl3y53.jpg

    Has anyone any ideas on how I might find my relative position using the angles phi[1-4] and the dimensions of the room?

    Thanks in advance,

    Chris Adams.
     
  2. jcsd
  3. Dec 28, 2011 #2
    Hello, matsalleh.

    The process you are describing is called resection, not triangulation.

    You need at least three known points (you have 4) to provide a 'fix' as surveyors speak or 'mark' as navigators speak.

    There are several methods of calculation available by searching the net.
     
  4. Dec 28, 2011 #3

    AlephZero

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  5. Dec 28, 2011 #4
    I googled it and came to this: http://www.map-reading.com/resect.php

    I don't understand how this helps me? I need an analytical way of calculating my position, using a computer/micro-controller.

    Haha, I had that idea just before I slept last night (sad I know). I didn't really know where to go with it?

    Can you nudge me in the right direction?

    Thanks,

    Chris.
     
  6. Dec 28, 2011 #5
  7. Dec 28, 2011 #6

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    Welcome to PF, matsalleh!


    Here's a different method.

    You can set up a series of 4 equations for each angle as a function of x and y.

    The 4 equations are:
    $$\phi_1(x,y) = \arctan{x \over y} + \arctan{W - x \over y}$$
    $$\phi_2(x,y) = \arctan{y \over W - x} + \arctan{L - y \over W - x}$$
    $$\phi_3(x,y) = \arctan{x \over L - y} + \arctan{W - x \over L - y}$$
    $$\phi_4(x,y) = \arctan{y \over x} + \arctan{L - y \over x}$$

    Typically you would minimize the squared errors of the angles, assuming you can measure all 4 angles with equal precision.
    That is, you would want to minimize the following function LSQ(x,y) by altering x and y:
    $$LSQ(x,y) = (\phi_1 - \phi_1(x,y))^2 + (\phi_2 - \phi_2(x,y))^2 + (\phi_3 - \phi_3(x,y))^2 + (\phi_4 - \phi_4(x,y))^2$$
    In this formula ##\phi_1, \phi_2, \phi_3, \phi_4## are the measured angles, and ##\phi_1(x,y), \phi_2(x,y), \phi_3(x,y), \phi_4(x,y)## are the formulas for the angles based on x and y.


    For a more accurate result you could use for instance the Levenberg-Marquardt algorithm to find an optimal solution.
     
    Last edited: Dec 28, 2011
  8. Dec 28, 2011 #7
    I don't know angles A,B and C :(

    It's been a few years since I've done maths like that. Alter x and y until phi[1-4] is satisfied?

    I'd like to implement this solution with C, is there kind of optimised algorithm to do this? I didn't really understand the last bit of your post.

    What about using arcs? I have a good feeling about using arcs.

    Thanks,

    Chris.
     
  9. Dec 28, 2011 #8

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    Yes.
    With phi[1-4] the system is over determined, meaning you will not be able to find a perfect solution.
    You can only minimize a predefined target function for which I suggest the sum of squares formula I gave (non-linear least squares method).



    In its simplest form you could start with an (x,y) in the middle of the room, take one step in each of the 4 directions and check in which direction the target function gets a lower value.
    Then repeat until you cannot find such a direction any more.

    There is a standard algorithm for problems like this, called the Levenberg-Marquardt algorithm:
    http://en.wikipedia.org/wiki/Levenberg–Marquardt_algorithm

    There are many implementations for it, which are referred to in the wikipedia article.
    If you are programming in C you may be interested in the levmar implementation which is freely downloadable:
    http://www.ics.forth.gr/~lourakis/levmar/
     
    Last edited: Dec 28, 2011
  10. Dec 28, 2011 #9

    AlephZero

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    It should be fairly obvious that [itex]\phi_1 + \phi_2 + \phi_3 + \phi_4 = 360[/itex] degrees, so you really only have 3 measurements not 4.

    The minimum amount of data is three points and two angles, but if you have three angles you can find the equations of the 3 circles (it's easy to find the center of each circle - see the website in my first link). Then you have 3 equations of the form

    [itex]x^2 + y^2 + a_1 x + b_1 y + c_1 = 0[/itex]
    [itex]x^2 + y^2 + a_2 x + b_2 y + c_2 = 0[/itex]
    [itex]x^2 + y^2 + a_3 x + b_3 y + c_3 = 0[/itex]

    You can subtract them in pairs to get rid of the [itex]x^2 + y^2[/itex] terms, then you have two simultaneous linear equations to solve for x and y.

    [itex](a_2 - a_1) x + (b_2 - b_1) y + (c_2 - c_1) = 0[/itex]
    [itex](a_3 - a_1) x + (b_3 - b_1) y + (c_3 - c_1) = 0[/itex]
     
  11. Dec 28, 2011 #10

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    For reference I have edited post #6 with the proper equations.

    Note that the formulas guarantee that the sum of the angles is 360 degrees.
    It remains useful to measure the 4th angle to minimize the effects of measurement errors and to maximize the precision of the result.
     
  12. Dec 28, 2011 #11
    What three circles? Would you be able to draw a diagram?

    That algorithm goes way beyond my understanding of mathematics.

    And I really have no clue on how to use levmar. All the examples talk about curve fitting and stuff.

    *straight over my head*

    I get how you have derived derived the first 4 equations; simple trig.

    However, I've never heard of this minimising square thing? (function LSQ) How is that derived?

    Is there a systematic way in altering x and y to minimise the function LSQ?

    Thanks,

    Chris.
     
  13. Dec 28, 2011 #12

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    It is the method of the least squares:
    http://en.wikipedia.org/wiki/Least_squares


    Here's a C code fragment (untested) that should do the trick without Levenberg-Marquardt.

    This code will find your x and y coordinates with an accuracy of dx and dy that you can arbitrarily choose.

    If a proper solution is found the function returns 0.
    *pstdev_angle gives the average mismatch of the angles (as a standard deviation).
    If it is too big, you have a misfit, which would occur if your angles are wrong.
    *px and *py receive the solution for x and y.

    If no solution was found, the function returns 1.

    For this code to work, you need to implement another function LSQ(x,y) that calculates the formulas I gave.

    Code (Text):

    int fit_angles(double W, double H, const double phi[4], double dx, double dy, double *px, double *py, double *pstdev_angle)
    {
       double x = W/2;
       double y = H/2;
       double minsq = LSQ(x, y);
       double stepsq;
       int keepgoing = 1;

       while (keepgoing && (x >= 0) && (y >= 0) && (x <= W) && (y <= H))
       {
          keepgoing = 0;
          stepsq = LSQ(x+dx, y);
          if (stepsq < minsq)
          {
             minsq = stepsq;
             x += dx;
             keepgoing = 1;
          }
          stepsq = LSQ(x-dx, y);
          if (stepsq < minsq)
          {
             minsq = stepsq;
             x -= dx;
             keepgoing = 1;
          }
          stepsq = LSQ(x, y+dy);
          if (stepsq < minsq)
          {
             minsq = stepsq;
             y += dy;
             keepgoing = 1;
          }
          stepsq = LSQ(x, y-dy);
          if (stepsq < minsq)
          {
             minsq = lsq1;
             y -= dy;
             keepgoing = 1;
          }
       }

       *px = x;
       *py = y;
       *pstdev_angle = sqrt(minsq / 4);

       return keepgoing;
    }
     
     
  14. Dec 28, 2011 #13
    I can't imagine why not, some are right angles in your case, since they are the angles at the corners of your room. The others can easily be obtained from the coordinates and are fixed for any room.

    You should also look up the 'danger circle' since it is impossible to solve certain points within the room, with certain of the corners.

    http://www.surveyequipment.com/PDFs/QuickGuide_TPS800_Resection.pdf

    This is also the reason a cotangent formula was used not a tangent formula - to avoid division by zero.
     
    Last edited: Dec 28, 2011
  15. Dec 29, 2011 #14
    I actually designed a bit of code using 2 for loops before I saw your example. It's actually surprisingly quick:

    Code (Text):

    #include <stdio.h>
    #include <math.h>

    #define STEP 1
    #define THRESHOLD 0.0005
    #define ANGLES 4

    float lsq(int w, int l, float x, float y, float p[]);

    int main(void){
       
        float W, L;
       
        float p[ANGLES];
       
        int i;
       
        float x, y;
       
        float test;
       
        printf("Room width:\n");
        scanf("%f", &W);
        printf("Room length:\n");
        scanf("%f", &L);
       
        for(i = 0; i <= (ANGLES - 1); ++i){
            printf("Angle %d:\n", (i+1));
            scanf("%f", &p[i]);
        }

        for(i = 0; i <= (ANGLES -1); ++i)
            p[i] = ((p[i]*M_PI)/180);
       
        for(x = 0; x <= W; x += STEP){
            for(y = 0; y <= L; y += STEP){
                printf("Trying x: %f and y: %f", x, y);
                printf("\r");
                test = lsq(W, L, x, y, p);
                if(test <= THRESHOLD)
                    break;
            }
            if(test <= THRESHOLD)
                break;
        }
       
        printf("\n\nX is %f\n", x);
        printf("Y is %f", y);
       
        return 0;
    }

    float lsq(int W, int L, float x, float y, float p[]){

        float pTest[4];
       
        int i;
       
        float output = 0.0;
           
        pTest[0] = atan(x/y) + atan((W-x)/y);
        pTest[1] = atan(y/(W-x)) + atan((L-y)/(W-x));
        pTest[2] = atan(x/(L-y)) + atan((W-x)/(L-y));
        pTest[3] = atan(y/x) + atan((L-y)/x);
       
        for(i = 0; i <= (ANGLES - 1); ++i){
            output = (output + pow((p[i] - pTest[i]), 2.0));
        }
       
        return output;
       
    }
     
    My next question is, how do I calculate a good value for my threshold? (my LSQ)

    • My device's angle measurement has a resolution of 1.8 degrees.
    • My grid size is 1cm. My room size is up to 10 meter squared, but typically only around 5 sqm.

    Also it's possible to minimise using only 2 angles. What gives the best results, using 2, 3 or 4 angles?

    Ahh yeah I see now. I could, except I'd have a really big dead zone using only 3 corners...

    I suppose I could algorithmically decide between corner C and D?

    Thanks,

    Chris.
     
  16. Dec 29, 2011 #15

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    :smile:

    I recommend not using that threshold at all but to find the actual minimum.

    It only speeds up your code by a factor of 2, but it loses accuracy unnecessarily and it gives your algorithm a preference for low x and y coordinates which is asymmetric.


    I would include an additional check that the resulting average mismatch of the angles is less than say ##3 \times 1.8^\circ \approx 5^\circ## to pick up on measurements that are totally wrong.

    With a grid size of 1 cm and area 10 m² your algorithm needs ##\propto 10^5## iterations, which is just doable.

    Note that the algorithm I gave needs ##\propto 300## iterations.


    More measurements means better results with the additional bonus of being able to detect measurements that are totally wrong.

    If you want to minimize on equipment 2 angles will suffice.
    This should give you an accuracy of about 10 cm.
     
    Last edited: Dec 29, 2011
  17. Dec 29, 2011 #16
    With plenty of computing power available one simple way is to calculate all possibilities and take an average. This is much simpler than Least Squares and leads to a simpler answer.

    Least Squares is not necessarily the best answer as some of the triangles in the fix will contain greater uncertainty than others and this varies with position. A weighting scheme helps, but where do you stop correcting it?

    Another (surveyors) solution is to use three angles. In general this gives three position lines that enclose the real fix within a small triangle.
    The real fix is then at the centroid of this triangle.
     
  18. Dec 29, 2011 #17

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    That depends on the most significant source of measurement errors in the angles.
    Is it a limitation of the device used to measure the angles?
    Is it dictated by the distance to the corners?
    Or is there some other dependency on position?

    If it is the first, then the LSQ method I proposed gives the optimal result, since it assumes equal standard deviations in the measurement errors of the angles.

    Otherwise a weighted LSQ could be used if you have more information on the accuracy of the angle measurements.
    In my opinion, a (possibly weighted) LSQ gives the best results since it makes the best statistical fit.


    EDIT: Alternatively, you could create a lookup table.
    You would have to calibrate that.
    So you would measure the angles from a number of given positions and put those in a lookup table.
    Then you could determine your position by simply looking up the closest entry, possibly interpolating with nearby entries.
     
    Last edited: Dec 29, 2011
  19. Dec 29, 2011 #18
    Can you just confirm how your algorithm works? It starts at half way and then tests each direction?

    I don't see how it's possible to get zero without using a really really small step size (infinitely small) ?

    And yes, it's a device limitation, the stepper motor I'm using only moves in 1.8degrees.
     
  20. Dec 29, 2011 #19

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    Yes.
    You can compare it to the situation that you stand in the middle of a landscape with a valley.
    To reach the lowest point you follow the path with the steepest descent.


    You won't get zero.

    You will reach some minimum (greater than zero) while you're still at most dx or dy away from the best minimum (also greater than zero).
    To get even closer to the real minimum you could reduce the step sizes dx and dy and retry.

    So with a step size of 1 cm you get within 1 cm of the best position.
    Closer is probably not useful for you, since you are limited to an accuracy of 1.8 degrees in your angles.
     
  21. Dec 29, 2011 #20
    Okay, so my question is: If I can't get zero, what's my threshold for minimum?

    Are you suggesting also, that I gradually decrease my step size until a solution is found?

    So try with dx=1, no solution found, try dx = 0.1, no solution found, ...

    ??

    Thanks,

    Chris
     
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