I'm looking for some assistance with solving the below equations in Java
(a-x1)^2 + (b-y1)^2 = r1^2 + r^2
(a-x2)^2 + (b-y2)^2 = r2^2 + r^2
(a-x3)^2 + (b-y3)^2 = r3^2 + r^2
Values of x1, y1, r1, x2, y2, r2 & x3, y3, r3 are known.
I need to solve for a, b, r
How to go about doing this in Java? I checked the Commons Maths library but didn't find how I could achieve this. It helps with linear equations though.
I think you need linear equations for Gaussian elimination.
If a, b, and r are what you need to solve for, it's obvious that these are non-linear equations.
You'll need a non-linear solver, like Newton-Raphson.
You'll have to linearize your equations. Calculate the Jacobean for the differentials da, db, and dr.
You'll start with an initial guess
a = a(old)
b = b(old)
r = r(old)
use a linearized version of the equations to calculate an increment
2*(a(old)-x1)*da + 2*(b(old)-y1)*db = 2*r(old)*dr
2*(a(old)-x2)*da + 2*(b(old)-y2)*db = 2*r(old)*dr
2*(a(old)-x3)*da + 2*(b(old)-y3)*db = 2*r(old)*dr
update your guess
a(new) = a(old) + da
b(new) = b(old) + db
r(new) = r(old) + dr
and repeat until it converges (if it converges).
You should never solve linear equations using Gaussian elimination: it suffers from a number of problems. A better idea is to do LU decomposition and forward-back substitution.
If my linearized equations are correct, they take the form A(dx) = 0. What should the boundary condition be?
(a, b) are the coordinates for the center of the circle; r is the radius.
Do you really have three points (x1, y1), (x2, y2), and (x3, y3)? Or do you have lots more points? If it's the latter, you'll need a least squares fit.
hope this method can give you some ideas:
public int[] getCoordinates(float XR_1, float YR_1, float XR_2, float YR_2,
float XR_3, float YR_3, int R1, int R2, int R3) {
//define the positions
int XU_1 = 0, YU_1 = 0, XU_2 = 0, YU_2 = 0, XU, YU;
//define variables and arrays that needed
float D0[][] = new float[17][50];
float D1[][] = new float[17][50];
float f[][] = new float[17][50];
float fmin_1 = 0;
float fmin_2 = 0;
//define columns and rows
int i, j;
//Y goes from 0 to 49
for(j=0; j<=49; j++){
//X goes from 0 to 16
for(i=0; i<=16; i++){
D0[i][j] = (float) (Math.pow((i-XR_1),2) + Math.pow((j-YR_1),2) - Math.pow(R1,2));
D1[i][j] = (float) (Math.pow((i-XR_2),2) + Math.pow((j-YR_2),2) - Math.pow(R2,2));
f[i][j] = (float) Math.sqrt(Math.pow(D0[i][j], 2) + Math.pow(D1[i][j], 2));
//get two position where f[i][j] are the minimum
//initialise the minimum two positions
if(i==0 & j==0){
fmin_1 = f[i][j];
XU_1 = i;
YU_1 = j;
}
else if(j==0 & i==1){
if(f[i][j] < fmin_1){
fmin_2 = fmin_1;
fmin_1 = f[i][j];
XU_2 = XU_1;
XU_1 = i;
YU_2 = YU_1;
YU_1 = j;
}
else {
fmin_2 = f[i][j];
XU_2 = i;
YU_2 = j;
}
}
else{
if(f[i][j] < fmin_1){
fmin_2 = fmin_1;
fmin_1 = f[i][j];
XU_2 = XU_1;
XU_1 = i;
YU_2 = YU_1;
YU_1 = j;
}
else if(f[i][j] < fmin_2){
fmin_2 = f[i][j];
XU_2 = i;
YU_2 = j;
}
}
}
}
this method gives two closest points in the coordinate system, you can use the similar way to get the most ideal one.
