How to do vector loop equation?
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'input';
for R1=8
R2=5;
R3=7;
R4=6;
end
for angle_2=25
omega_2=24;
alpha_2=-4;
end
'syms, angle_3, angle_4';
[angle_3]='solve(R2*cos(angle2)+R3*cos(angle3)-R4*cos(angle4)-R1=0';
[angle_4] = 'solve(R2*sin(angle2)+R3*sin(angle3)-R4*sin(angle4)=0';
[omega_3] = 'solve(-R2*(omega2)*sin(angle2)-R3*(omega3)*sin(angle3)+R4*(omega4)*sin(angle4)=0';
[omega_4]= 'solve(R2*(omega2)*cos(angle2)+R3*(omega3)*cos(angle3)-R4*(omega4)*cos(angle4)=0';
[alpha_3] = 'solve(-R2*(alpha2*sin(angle2)+(omega2)^2*cos(angle2))-R3*(alpha3*sin(angle3)+(omega3)^2*cos(angle3))+R4*(alpha4*sin(angle4)+(omega4)^2*cos(angle4))=0';
[alpha_4] = 'solve(R2*(alpha2*cos(angle2)-(omega2)^2*sin(angle2))+R3*(alpha3*cos(angle3)-(omega3)^2*sin(angle3))-R4*(alpha4*cos(angle4)-(omega4)^2*sin(angle4))=0';
angle_R3=angle3;
angle_R4=angle4;
R3_angular_velocity=omega_3;
R4_angular_velocity=omega_4;
R3_angular_acceleation=alpha_3;
R4_angular_acceleation=alpha_4;
4 Kommentare
Zubair Ghafoor
am 26 Okt. 2022
Hi you haven't initialized the Angle3 can you please first give the value of this, then your code is going to work.
Jan
am 26 Okt. 2022
@Mhadi Alessa: The code looks very strange:
'input'; % Why? Do you just want to display a message? Then:
disp('input')
for R1=8 % What is the purpose to run a loop over one element?
% easier without FOR:
R1 = 8;
R2=5;
R3=7;
R4=6;
%for angle_2=25 see above
angle_2=25;
omega_2=24;
alpha_2=-4;
% end
% See above: 'syms, angle_3, angle_4';
% Why do you use square brackets here?
[angle_3]='solve(R2*cos(angle2)+R3*cos(angle3)-R4*cos(angle4)-R1=0';
% ^ ^ ?
% Setting angle_3 to a char vector looks strange also.
% Do you mean:
angle_3 = solve('R2*cos(angle2)+R3*cos(angle3)-R4*cos(angle4)-R1 = 0');
Currently the complete code does neither look like meaningful Matlab code nor does it compute anything.
Antworten (1)
Karim
am 26 Okt. 2022
Bearbeitet: Karim
am 26 Okt. 2022
The syntax for the solve function works a bit different, see below for a demonstration.
Notice that if you have a set of two equations with two unkowns, you can pass both of them directly into the solve function.
R1 = 8;
R2 = 5;
R3 = 7;
R4 = 6;
angle_2 = 25;
omega_2 = 24;
alpha_2 = -4;
syms angle_3 angle_4 omega_3 omega_4 alpha_3 alpha_4
% set up equations
Angle_eqs = [R2*cos(angle_2)+R3*cos(angle_3)-R4*cos(angle_4)-R1 == 0;
R2*sin(angle_2)+R3*sin(angle_3)-R4*sin(angle_4) == 0];
Omega_eqs = [-R2*(omega_2)*sin(angle_2)-R3*(omega_3)*sin(angle_3)+R4*(omega_4)*sin(angle_4) == 0;
R2*(omega_2)*cos(angle_2)+R3*(omega_3)*cos(angle_3)-R4*(omega_4)*cos(angle_4) == 0];
Alpha_Eqs = [-R2*(angle_2*sin(angle_2)+(omega_2)^2*cos(angle_2))-R3*(alpha_3*sin(angle_3)+(omega_3)^2*cos(angle_3))+R4*(alpha_4*sin(angle_4)+(omega_4)^2*cos(angle_4)) == 0;
R2*(angle_2*cos(angle_2)-(omega_2)^2*sin(angle_2))+R3*(alpha_3*cos(angle_3)-(omega_3)^2*sin(angle_3))-R4*(alpha_4*cos(angle_4)-(omega_4)^2*sin(angle_4)) == 0];
% solve the equations
Angle = solve(Angle_eqs,[angle_3 angle_4]);
angle_3 = eval(Angle.angle_3);
angle_4 = eval(Angle.angle_4);
% this provides two possible solutions. Below i simply pick the first slution,
% However you need to check if this is the one you want
angle_3 = angle_3(1);
angle_4 = angle_4(1);
Omega = solve(Omega_eqs,[omega_3 omega_4]);
omega_3 = eval(Omega.omega_3);
omega_4 = eval(Omega.omega_4);
Alpha = solve(Alpha_Eqs,[alpha_3 alpha_4]);
alpha_3 = eval(Alpha.alpha_3);
alpha_4 = eval(Alpha.alpha_4);
% print results to the display
fprintf("Angle 3 = %.3f\nAngle 4 = %.3f\n",angle_3,angle_4)
fprintf("Omega 3 = %.3f\nOmega 4 = %.3f\n",omega_3,omega_4)
fprintf("Alpha 3 = %.3f\nAlpha 4 = %.3f\n",alpha_3,alpha_4)
EDIT: i believe this resembles a four bar mechanism, if so we can plot the configuration in the following way:
x = [0;
R2*cos(angle_2);
R2*cos(angle_2)+R3*cos(angle_3)
R2*cos(angle_2)+R3*cos(angle_3)-R4*cos(angle_4)];
y = [0;
R2*sin(angle_2)
R2*sin(angle_2)+R3*sin(angle_3)
R2*sin(angle_2)+R3*sin(angle_3)-R4*sin(angle_4)];
figure
plot(x,y,'LineWidth',1.5,'Marker','o','MarkerFaceColor','red')
grid on
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