MATLAB: Converting Python to Matlab

python

Spent a few hours on trying to figure out why the outputs are different, no luck. Python and Matlab are in a txt file along with their outputs.

Suggestions on what I should be looking at to resolve the issue?

Best Answer

Small typo in MATLAB code in nu2

    nu2 = atan2(-(xc(k)-xp(n))*sin(theta(k))+(yc(k)-yp(n))*cos(theta(n))+(xp(n+1)-xp(n))*sin(theta(n))-(yp(n+1)-yp(n))*cos(theta(n)),(xc(k)-xp(n))*cos(theta(n))+(yc(k)-yp(n))*sin(theta(n))-(xp(n+1)-xp(n))*...
  cos(theta(n))-(yp(n+1)-yp(n))*sin(theta(n))); % fixed!

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MATLAB: Newton’s method(calculating theta)

Hi Ahmed,

The condition you placed for the while loop is a series of vector. Update the condition as below and code requires minor modifications to work

while n <= 50 % It runs the loop till n becomes 50
    fe = f(theta_newt(n));
    fpe = fp(theta_newt(n));
    theta_newt(n+1) = theta_newt(n) - fe/fpe;
    if theta_newt(n+1)< 49.73-tol || theta_newt(n+1)>49.73+tol
        return
    else
        theta=theta_newt(n+1);
    end
    n=n+1;
end

Hope this helps.

Regards,

Sriram

MATLAB: How to add to index in for loop

xc = zeros(numPanels, 1);
yc = zeros(numPanels, 1);
for k = 1:numPanels-1
      xc(k) = (xp(k) + xp(k+1))/2
      yc(k) = (yp(k) + yp(k+1))/2
  end

You are in MATLAB..you can avoid loops here.

xc = (xp(1:end-1)+xp(2:end))/2 ;
yc = (yp(1:end-1)+yp(2:end))/2 ;

But the problem is size of xc and yc will be one less then xp and yp.

Best Answer

Related Solutions

MATLAB: Newton’s method(calculating theta)

MATLAB: How to add to index in for loop

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