Asked by Federico MegaMan
on 10 Nov 2019 at 15:43

Hello,

I have a problem with 13 unknows and 12 equations. I already solve the problem imposing T21=0.

I would to variate 1 of the unknows (I prefer T21) to generate various solution.

How can i do?

I attach you my function.

Thank you in advance.

Answer by Stephan
on 10 Nov 2019 at 16:02

Accepted Answer

Federico MegaMan
on 10 Nov 2019 at 23:33

Oh i'm very sorry i'm new and i didn't know that. I'll keep it in mind for the next times, thank you.

Anyway this is my function:

(the file name is funzmia)

function F=funz(x,T21)

T11=x(1);

T12=x(2);

T13=x(3);

T14=x(4);

T15=x(5);

T22=x(6);

T23=x(7);

r23=x(8);

T31=x(9);

T32=x(10);

T33=x(11);

r32=x(12);

T21=0

F(1)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;

F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);

F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);

F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);

F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);

F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);

F(7)=T12-T21;

F(8)=T31-T14;

F(9)=T13+T22-pi/2;

F(10)=T33-T22-pi;

F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);

F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);

end

and this is the for:

for T21=0:pi/180:5*pi/180

fsolve(@funzmia, [1 2 3 4 5 6 7 8 9 10 11 12])

end

Stephan
on 11 Nov 2019 at 7:56

T21=0:pi/180:5*pi/180;

sol = zeros(12,numel(T21));

for k = 1:numel(T21)

% Results for x are saved in rows 1...12

% every value of T21 corresponds to 1 column

sol(:,k) = fsolve(@(x)funzmia(x,T21(k)), 1:12);

end

function F=funzmia(x,T21)

T11=x(1);

T12=x(2);

T13=x(3);

T14=x(4);

T15=x(5);

T22=x(6);

T23=x(7);

r23=x(8);

T31=x(9);

T32=x(10);

T33=x(11);

r32=x(12);

%T21=0

F(1)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;

F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);

F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);

F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);

F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);

F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);

F(7)=T12-T21;

F(8)=T31-T14;

F(9)=T13+T22-pi/2;

F(10)=T33-T22-pi;

F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);

F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);

end

Federico MegaMan
on 11 Nov 2019 at 16:55

Okay, it works perfectly now. Thank you very much!

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