Combine lsqcurvefit with fsolve

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Gregory Cottone
Gregory Cottone 2021-4-1
评论： Gregory Cottone 2021-4-4
Hi,
my goal is to optimize the length L0, L1, L2 and L3 of the bars of the following mechanism so that the theta2 angle follows the trend of my input data:
This is the mechanism:
And this is the data that I want to fit:
theta1 = linspace(0,pi,n);
theta2 = [0.53,0.40,0.32,0.28,0.27,0.27,0.28,0.29,0.31,0.34,0.36,0.40,0.43,0.47,0.50,0.55,0.59,0.64,0.69,0.75]; %(data to fit)
which they have this trend:
My strategy was to mainly use the lsqcurvefit function and have the kinematic closure equation solved recursively using the fsolve function:
clc;
clear all;
n = 20;
theta1 = linspace(0,pi,n);
theta2 = [0.53,0.40,0.32,0.28,0.27,0.27,0.28,0.29,0.31,0.34,0.36,0.40,0.43,0.47,0.50,0.55,0.59,0.64,0.69,0.75]; %(data to fit)
L0 = [0,0,0,0]; % Initial lenght of the bars, in the order: L0,L1,L2 and L3
for i = 1:n
y0 = [0.5,1.1]; % Initial guess of theta2 and theta3
options = optimset('display', 'off');
y(i,:) = fsolve(@(y)fourbar(y,theta1(i)),y0,options);
end
%Objective: find the L0, L1, L2 and L3 optimized to make the function...
...interpolating the data (theta1,theta2)
L = lsqcurvefit(@fourbar,L0,theta1,theta2);
function Phi = fourbar(y,L,theta1)
u1 = [cos(theta1), sin(theta1)];
u2 = [cos(y(1)), sin(y(1))];
u3 = [cos(y(2)), sin(y(2))];
Phi = L(1)*u1 + L(2)*u2 - L(3)*u3 - [L(4), 0];
end
But unfortunately it doesn't work and I don't know how to solve the problem.
  4 个评论
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Matt J
Matt J 2021-4-3
编辑：Matt J 2021-4-3
I'm still not sure there are enough constraints. For example L1=L2=0, L3=L0 is always a solution.
Also, do you not have theta3 data as well? That would reduce the problem to a set of linear equations.
Gregory Cottone
Gregory Cottone 2021-4-4
Unfortunately I don't have any theta3 data to fit in my problem.

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Matt J
Matt J 2021-4-3
编辑：Matt J 2021-4-3
Ran in:
I would just use lsqnonlin.
theta2 = [0.53,0.40,0.32,0.28,0.27,0.27,0.28,0.29,0.31,0.34,0.36,0.40,0.43,0.47,0.50,0.55,0.59,0.64,0.69,0.75]; %(data to fit)
theta1 = linspace(0,pi,numel(theta2));
L0=1;
opts=optimoptions(@lsqnonlin,'StepTolerance',1e-12,'OptimalityTolerance',1e-12,'FunctionTolerance',1e-12);
[L,resn,res]=lsqnonlin(@(L) resid(L,L0,theta1,theta2), [0.5,0.5,1]*L0, [0,0,0],[1,inf inf]*L0);
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
L
L = 1×3
0.5047 1.4891 1.0920
function fval=resid(L,L0,theta1,theta2)
dX=L(1)*cos(theta1(:))+L(2)*cos(theta2(:));
dY=L(1)*sin(theta1(:))+L(2)*sin(theta2(:));
fval=(L0-dX).^2 + dY.^2-L(3).^2;
end
  3 个评论
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Matt J
Matt J 2021-4-4
编辑：Matt J 2021-4-4
I did not start with fourbar (I don't understand it). I started with your diagram. Basically, the components of u3 are,
u3=[dX-L0,dY]
and the equations we want to solve are
norm(L3)^2=norm(u3)^2
Gregory Cottone
Gregory Cottone 2021-4-4
Now I understand! Thank you Matt.

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