Can't get the solution on this nonlinear system of equations using vpasolve

Hi, so i'm trying to find solutions for E1 and E2, but sadly it doesn't show me an answer

clc
clear all

Px=[6.56044 -4.12471 -19.6017 -18.2934 -1.8923 21.1196 17.6706];
Py=[0.995643 0.00147957 1.14159 2.07962 6.30745 11.3302 4.55826];

L=82.4;
l1=35.7;
L2=39.7;

for j=1:7
    PX(j)=-Px(j);
    PY(j)=L-Py(j);

end


j=1

syms X1 X2

    
E1=l1*cos(X1)+L2*cos(X2)-PX(1); 
E2=l1*sin(X1)+L2*sin(X2)-PY(1);


Y=vpasolve([E1,E2],[X1,X2]),[1.5,1.7]

A=Y.X1
B=Y.X2

Answer 1

[Torsten]

For that your two equations can have a solution, one can show that

abs((PX^2+PY^2-l1^2-L2^2)/(2*l1*L2))

must be <= 1.

This is only the case for i=6, and for this case, vpasolve gives a solution.

Px=[6.56044 -4.12471 -19.6017 -18.2934 -1.8923 21.1196 17.6706];
Py=[0.995643 0.00147957 1.14159 2.07962 6.30745 11.3302 4.55826];
L=82.4;
l1=35.7;
L2=39.7;
syms X1 X2
x1 = zeros(7,1);
x2 = zeros(7,1);
for i=1:7
    PX(i)=-Px(i);
    PY(i)=L-Py(i);
    value = abs((PX(i).^2+PY(i).^2-l1^2-L2^2)/(2*l1*L2))
    if value > 1
      continue  
    end
    E1 = l1*cos(X1)+L2*cos(X2)-PX(i) == 0;
    E2 = l1*sin(X1)+L2*sin(X2)-PY(i) == 0;
    Y = vpasolve([E1,E2],[X1,X2]);
    Y.X1
    Y.X2
    x1(i) = Y.X1;
    x2(i) = Y.X2;
end

value = 1.3473

value = 1.3956

value = 1.4593

value = 1.3884

value = 1.0383

value = 0.9336

ans = 

1.6665945097644129747979749283503

ans = 

2.0330490977381021058751812658667

value = 1.2422

[José David Castillo Blanco]

may I ask what does

abs((PX^2+PY^2-l1^2-L2^2)/(2*l1*L2))

must be <= 1.

mean?

[Torsten]

It means that the absolute value of the number

(PX^2+PY^2-l1^2-L2^2)/(2*l1*L2)

must be less or equal 1 in order that your system of equations can have a solution.

[José David Castillo Blanco]

Is there any way to change the tolerance? i need a solution for the other points, no matter if they are not that close

[Torsten]

Here is a least-squares solution. The array "error" indicates how good the two equations could be solved.

Px=[6.56044 -4.12471 -19.6017 -18.2934 -1.8923 21.1196 17.6706];
Py=[0.995643 0.00147957 1.14159 2.07962 6.30745 11.3302 4.55826];
L=82.4;
l1=35.7;
L2=39.7;
%syms X1 X2
x1 = zeros(7,1);
x2 = zeros(7,1);
error = zeros(7,1);
for i=1:7
    PX(i)=-Px(i);
    PY(i)=L-Py(i);
    %value = abs((PX(i).^2+PY(i).^2-l1^2-L2^2)/(2*l1*L2))
    %if value > 1
    %   continue  
    %end
    fun = @(x)[l1*cos(x(1))+L2*cos(x(2))-PX(i),l1*sin(x(1))+L2*sin(x(2))-PY(i)];
    [x,fval] = lsqnonlin(fun,[1,2],[0 0],[2*pi,2*pi]);
    %E1 = l1*cos(X1)+L2*cos(X2)-PX(i) == 0;
    %E2 = l1*sin(X1)+L2*sin(X2)-PY(i) == 0;
    %Y = vpasolve([E1,E2],[X1,X2]);
    %Y.X1
    %Y.X2
    %x1(i) = Y.X1;
    %x2(i) = Y.X2;
    x1(i) = x(1);
    x2(i) = x(2);
    error(i) = fval;
end

Local minimum possible.
lsqnonlin stopped because the size of the current step is less than
the value of the step size tolerance.


Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.


Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.


Local minimum possible.
lsqnonlin stopped because the size of the current step is less than
the value of the step size tolerance.


Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.


Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.


Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

error

error = 7×1

39.2914 50.4340 67.0631 48.6821 0.5128 0.0000 19.5560

x1

x1 = 7×1

1.6512 1.5208 1.3341 1.3469 1.5459 1.6666 1.7940

x2

x2 = 7×1

1.6512 1.5208 1.3341 1.3469 1.5459 2.0330 1.7940

[José David Castillo Blanco]

Thanks, just for the record I had an error on the value "l1", also, due to uncertainty in measurement the tolerance can't be too tight.