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cmathmathematical-optimization

Is there a way to adjust this system of non linear equation

I've been to solve this problem for the last 2 days with no success. say we have this system : system

the has an infinite number of solutions when :

condition

to get value of the angle Theta, I use this formula :

solution

The value that I get is correct but only in some points, because the system isn't solvable. and here is how it looks like :

anlge

the red curve represents the "must be" value and the blue one represents that I actually get.

after ploting the value of the sum :

condition

here is what I get : error as you can see it's sinus curve, that seems to influence the "must be" value of the angle.

to simulate this I've been using this program : 

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


int main (){
  float xin =0;
  float yin =0;
  float zin =0;
  float A =0;
  float B =0;
  float C =0;
  float angle =0;
  float outangle =0 ;
  float angledeg =0;
  while (angledeg <=(360*3)) {

    angle = angledeg * M_PI/180;
    xin = 0.11;
    yin = (sin(angle) / sqrt(3))  + xin;
    zin = (xin - (sin(angle +(120*M_PI/180))));
    A  = yin - xin;
    B  = xin - zin;
    C  = zin - yin;
    outangle = atan2((A*sqrt(3)) , (B -C) )   * 180/M_PI; // 100% correct 
    printf ("%lf \n" , outangle);
    angledeg++;
  }

  return 0;
}

So my QUESTION is : how could I use the value of sum to adjust the resulted value of "must be" value of the angle ( the red curve)

UPDATE

I don't know if this make a sense but it works : Removing sqrt(3) :

yin = (sin(angle) / sqrt(3))  + xin;

I really don't get it, but it works fine ? any idea why ?

Solution

  • You can dramatically simplify the situation by considering that

    sin(t)+sin(t+120°)+sin(t+240°)=0

    which just means that the (weight) center point of any equilateral triangle on the unit circle is the origin.

    Thus

    sin(t)/sqrt(3)+sin(t+120°)+sin(t+240°)=sin(t)*(1/sqrt(3)-1)

    So the only values of t=theta that make the system solvable are t=0° and t=180°.

    In computing 

    y=x+sin(t)/sqrt(3); z=x-sin(t+120°)

    you produce as value for C

    z-y = - sin(t+120°) - sin(t)/sqrt(3)

    and thus

    B-C = 2*sin(t+120°) + sin(t)/sqrt(3)
    = -sin(t) + sqrt(3)*cos(t) + sin(t)/sqrt(3)

    where in some kind of expectation you only expect the middle term, which is true for sin(t)=0, again for t=0° or t=180°.