Question

Hello. I need help modifying this MatLab code. This is a basic slider-crank animation. the current program stops the animation and then has the crank move backward. I need the crank to instead to move in full 360 degrees circular motion. Exactly like how a slide-crank mechanism works in real life. thank you and here is the code.

%%Code Begins Here%%

L1=0.2; L2=0.45;
W=0.5;
tt=linspace(0,15);
for i=1:length(tt)
    x2=@(t) L2+L1*sin(W*t);
    y2=0;

    X2=x2(tt(i));
    Y2=y2;

    sol= fsolve(@(x,x2,y2) root2d(x,X2,Y2), [0,0]);
    clf
    hold on
    plot(0,0,'b*','linewidth',20)
    plot(sol(1),sol(2),'ro','linewidth',15)
    plot([0 sol(1)], [0 sol(2)],'linewidth',5,'color','k')
    plot(X2,Y2,'md','linewidth',15)
    plot([X2 sol(1)], [Y2 sol(2)],'linewidth',5,'color','k')

    ylim([-0.5 0.5])
    xlim([-0.3 0.7])
    hold off
    drawnow
    pause(0.025)
  
end

function F = root2d(x,x2,y2)

    F(1) = (x(1)-x2)^2+(x(2)-y2)^2-0.45^2;
    F(2) = (x(1))^2+(x(2))^2-0.2^2;
end

Answer 1

Hi, the code is correct, however you are a victim of a classic problem of the fsolve function which arises when it tries to solve conic equations with it. The problem of the fsolve function is that it always tries to approximate a solution closest to the point of initialization. So a graphical description of the problem is as follows

Point of initialization (0,0) (x,y) Both these points are solutions for x^2+y^2 = r^2, so (x,-Y) fsolve can choose between either r

This was exactly what was happening. Fsolve simply chose to output (x,y) instead of (x,-y) and so your crank kept dancing in the upper half circle.

So the only way out of this problem is to keep a dynamic point of initialization. A graphical depication of the solution would be as follows

inital point set as (0,r) for 0 to 180 (x,y) this point is closer to (0,r), so will always converge on this 0/360 degree 180 degree (x,-y) this point is closer to (0,-r), so will always converge on this initial point set as (0,-r) for 180 to 360 (or 0) degree

Thus we will have to include an algorithm which will automatically detect when the crank pin is crossing 180 or 360 degrees and change the initial point accordingly. I have provided both the text and the screenshot for your better understanding. The screenshot also has some comments for your help. Please note that the screenshot shows the part where the changes are implemented. The rest of the code is the same.

3 $$Code Begins Here Ll=0.2; L2=0.45; W=0.5; tt=linspace (0,15); 4 - 5 6 This is the intial approximation of fsolve. This flips between +ve and -ve value for each half cycle a = 0.2; 7 - 8 9 10 - 11 - 12 - 13 - 14 15 - 16 - *This is a flag to let the program know when the value of a is to be $flipped and which half-cycle is currently under process flag = 0; for i=1:length(tt) x2=@ (t) L2 + Ll*sin(W*t); y2=0; X2=x2 (tt(i)); Y2=y2; sol = fsolve (@(x,x2, y2) root2d(x,X2,42), [0,a]); Initial conditions are to be flipped between (0,2) and (0,-2) for correct solution if sol (1) >= 0.199 && flag == 0 falogrithm to flip the value. Uses flag to make sure the value is flipped only once in a half cycle. a = a* -1; flag = 1; elseif sol(1) <= -0.199 && flag == 1 a = a* -1; flag = 0; end clf hold on 17 - 18 19 20 21 22 23 24 25 26 27 28 Rest of the code same....

%%Code Begins Here%%
L1=0.2; L2=0.45; W=0.5; tt=linspace(0,15);

%This is the intial approximation of fsolve. This flips between +ve and -ve value for each half cycle
a = 0.2;
%This is a flag to let the program know when the value of a is to be
%flipped and which half-cycle is currently under process
flag = 0;
for i=1:length(tt)
x2=@(t) L2 + L1*sin(W*t);
y2=0;
X2=x2(tt(i));
Y2=y2;
sol = fsolve(@(x,x2,y2) root2d(x,X2,Y2), [0,a]); %Initial conditions are to be flipped between (0,2) and (0,-2) for correct solution
if sol(1) >= 0.199 && flag == 0 %alogrithm to flip the value. Uses flag to make sure the value is flipped only once in a half cycle.
a = a * -1;
flag = 1;
elseif sol(1) <= -0.199 && flag == 1
a = a * -1;
flag = 0;
end
clf
hold on
  
%Rest of the code same....

plot(0,0,'b*','linewidth',20);
plot(sol(1),sol(2),'ro','linewidth',15);
plot([0 sol(1)], [0 sol(2)],'linewidth',5,'color','k');
plot(X2,Y2,'md','linewidth',15);
plot([X2 sol(1)], [Y2 sol(2)],'linewidth',5,'color','k');
ylim([-0.5 0.5])
xlim([-0.3 0.7])
hold off
drawnow
pause(0.025)
end

function F = root2d(x,x2,y2)

F(1) = (x(1)-x2)^2+(x(2)-y2)^2-0.45^2;
F(2) = (x(1))^2+(x(2))^2-0.2^2;
end

Naturally it is not possible for me to attach an animation of the output here, but as it worked for me I am sure the code will work for you too.

*A humble request* - If you have any doubt, please use the comment section to communicate. Please be a little patient, it is an honest promise I will reply as soon as possible. This will clarify your doubt, and also help me to get better at answering your next questions. At the same time, If my answer helped you, please consider leaving an upvote. I hope you understand my viewpoint. Thank you :)