4 equation 2 solution

How can I find two variables from 4 equation, All equation are equal zero. I want fo find Theta_1 and Theta_2. These values are positive and Theta_2 must bigger than Theta_1. Initial value of Theta_1 will find with below solution.

u=deg2rad(15);
hd=24.5395;
R1_p =32.0080;
R2_p =48.0120;

eqn= @(t1, u)  u+atan(sin(t1)./(1-cos(t1)))-pi/2 % theta1 t1 olarak tanımlanmıştır.
Theta_1_initial= (fminsearch(@(t1)norm(eqn(t1, u)),u))

Can Anybody help me ? What is the solution to find Theta_2 and Theta_1

eqn1= @ (Theta_2, Theta_1, u) cos(Theta_2)+sin(Theta_2)*tan(Alpha)-cos(Theta_1)-sin(Theta_1)*tan(u)
eqn2 =@ (R1_p, Phi, L3, Alpha, hd) R1_p-R1_p*cos(Phi)-L3*sin(Alpha+Phi)+hd
Phi  = @( R1_p, R2_p, Theta_2, Theta_1, Alpha) atan(R2_p*(Theta_2-Theta_1)/(R1_p-R2_p*(Theta_2-Theta_1)*tan(Alpha)))
Alpha =@( Theta_1, Theta_2, u) pi/2-atan(sin(Theta_2)/(cos(Theta_1)+sin(Theta_1)*tan(u)-cos(Theta_2)))

Answer 1

[Star Strider]

Since ‘Phi’ and ‘L3’ are not defined, I created them (and ‘Alpha’) as parameters to be estimated hiere.

I am not certain what you are doing, however this produces estimates for the parameters —

u=deg2rad(15);
hd=24.5395;
R1_p =32.0080;
R2_p =48.0120;

% eqn1= @ (Theta_2, Theta_1, u) cos(Theta_2)+sin(Theta_2)*tan(Alpha)-cos(Theta_1)-sin(Theta_1)*tan(u)
% eqn2 =@ (R1_p, Phi, L3, Alpha, hd) R1_p-R1_p*cos(Phi)-L3*sin(Alpha+Phi)+hd
% Phi  = @( R1_p, R2_p, Theta_2, Theta_1, Alpha) atan(R2_p*(Theta_2-Theta_1)/(R1_p-R2_p*(Theta_2-Theta_1)*tan(Alpha)))
% Alpha =@( Theta_1, Theta_2, u) pi/2-atan(sin(Theta_2)/(cos(Theta_1)+sin(Theta_1)*tan(u)-cos(Theta_2)))

eqn= @(t1, u)  u+atan(sin(t1)./(1-cos(t1)))-pi/2 % theta1 t1 olarak tanımlanmıştır.

eqn = function_handle with value:
    @(t1,u)u+atan(sin(t1)./(1-cos(t1)))-pi/2

Theta_1_initial= (fminsearch(@(t1)norm(eqn(t1, u)),u))

Theta_1_initial = 0.5236

fcn = @(Theta_1, Theta_2, u, Alpha, Phi, L3) [cos(Theta_2)+sin(Theta_2)*tan(Alpha)-cos(Theta_1)-sin(Theta_1)*tan(u)
                               R1_p-R1_p*cos(Phi)-L3*sin(Alpha+Phi)+hd
                               atan(R2_p*(Theta_2-Theta_1)/(R1_p-R2_p*(Theta_2-Theta_1)*tan(Alpha)))    
                               pi/2-atan(sin(Theta_2)/(cos(Theta_1)+sin(Theta_1)*tan(u)-cos(Theta_2)))];

B = fminsearch(@(b) norm(fcn(b(1),b(2),u,b(3),b(4),b(5))), [Theta_1_initial; rand(4,1)]);

 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: 0.502807 

fprintf(1,'Theta_1 = %8.4f\nTheta_2 = %8.4f\nAlpha   = %8.4f\nPhi     = %8.4f\nL3      = %8.4f\n',B)

Theta_1 =  -3.1298
Theta_2 =  -9.3933
Alpha   =   1.0667
Phi     =   1.1600
L3      =  55.3248

.

[onur karakurt]

all function are depend on Theta_2 and Theta_1, I am triying to find positive optimum values of these.

[Star Strider]

If you want to constrain ghtm to be positive, use the lsqcurvefit function or fmincon function (that will work with your current code and allows parameter constraints). Both require the Optimization Toolbox.

Using fmincon —

u=deg2rad(15);
hd=24.5395;
R1_p =32.0080;
R2_p =48.0120;

eqn= @(t1, u)  u+atan(sin(t1)./(1-cos(t1)))-pi/2; % theta1 t1 olarak tanımlanmıştır.
Theta_1_initial= (fminsearch(@(t1)norm(eqn(t1, u)),u))

Theta_1_initial = 0.5236

fcn = @(Theta_1, Theta_2, u, Alpha, Phi, L3) [cos(Theta_2)+sin(Theta_2)*tan(Alpha)-cos(Theta_1)-sin(Theta_1)*tan(u)
                               R1_p-R1_p*cos(Phi)-L3*sin(Alpha+Phi)+hd
                               atan(R2_p*(Theta_2-Theta_1)/(R1_p-R2_p*(Theta_2-Theta_1)*tan(Alpha)))    
                               pi/2-atan(sin(Theta_2)/(cos(Theta_1)+sin(Theta_1)*tan(u)-cos(Theta_2)))];

B = fmincon(@(b) norm(fcn(b(1),b(2),u,b(3),b(4),b(5))), [Theta_1_initial; rand(4,1)], [],[],[],[],[0 0 -Inf -Inf -Inf]);

Local minimum possible. Constraints satisfied.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance and constraints are 
satisfied to within the value of the constraint tolerance.

fprintf(1,'Theta_1 = %8.4f\nTheta_2 = %8.4f\nAlpha   = %8.4f\nPhi     = %8.4f\nL3      = %8.4f\n',B)

Theta_1 =   3.1416
Theta_2 =   3.1401
Alpha   =   0.6511
Phi     =   1.0115
L3      =  39.7306

Here, I simply bounded the first two parameters (‘Theta_1’, ‘Theta_2’) to be and let the others range freely. See the fmincon documentation that I linked to, to constrain the other parameters if necessary.

If you have values for ‘Phi’ and ‘L3’, use those values instead of estimating them. Then make appropriate changes to ‘fcn’ and the call to it in fmincon.

The parameter estimates are not reproducable between code runs. Different runs produce different results, although that might improve (stabilise) if you have values for ‘Phi’ and ‘L3’.

.