Lowest value that fulfills a condition

Hello,

I've a script that takes a long time to run due to the small steps I've set up, but i need precision in the calculations so thats why I need the decimal places.

short version script:

T=0;
W=30000;  
while T<=W
      theta_tip=theta_tip+0.001;
      for i=1:100
      T1=int((theta_tip/(y/Rmax))...., y, lim_inf, lim_sup);
      T2=int(T1,0,2*pi());
      T=1/(2*pi)*T2;    
      end
  end

some of the long story:

theta_tip=0; %degrees - angle of atack of blade tip
T=0; %N - Total produced lift
Up=0; %m/s
Rmax=4;
Rmin=0.5;
nbe=5;
ro=0.8;
chord=0.2;
Omega=42;
Cl_alpha=[6.5044    6.6122    6.7820    7.0986    7.5694];
el_size=(Rmax-Rmin)/nbe; %m - element size in meters
lim_inf=Rmin; %integral inferior limit
lim_sup=lim_inf+el_size; %integral superior limit
syms y
tic
while T<=W
  theta_tip=theta_tip+0.00001;
  for i=1:nbe
      if i==1
          lim_inf=lim_inf; %#ok<ASGSL>
          lim_sup=lim_inf+el_size;
      else
          lim_inf=lim_inf+el_size;
          lim_sup=lim_inf+el_size;
      end
  T1=int(1/2*ro*chord*Cl_alpha(i)*((theta_tip/(y/Rmax))*(Omega*y)^2-Up*(Omega*y)), y, lim_inf, lim_sup);
  T2=int(T1,0,2*pi());
  T=1/(2*pi)*T2;    
  end
end
toc

[Edit- more variables given] i'll be grateful for any help! thanks!

[KL]

many variables are missing and we cannot guess them!

read this: https://www.mathworks.com/help/matlab/matlab_prog/techniques-for-improving-performance.html

and this: https://www.mathworks.com/help/matlab/matlab_prog/vectorization.html

vectorization makes a lot of things simpler.

[Diogo Nascimento]

I understand, I'll read those and submit the remaining code, I thought it was best not to put everything in here, but i'll simplify the variables and put it here! thanks for the help!

[Diogo Nascimento]

theta_tip=0; %degrees - angle of atack of blade tip
T=0; %N - Total produced lift
Up=0; %m/s
Rmax=4;
Rmin=0.5;
nbe=5;
ro=0.8;
chord=0.2;
Omega=42;
Cl_alpha=[6.5044    6.6122    6.7820    7.0986    7.5694];
el_size=(Rmax-Rmin)/nbe; %m - element size in meters
lim_inf=Rmin; %integral inferior limit
lim_sup=lim_inf+el_size; %integral superior limit
syms y
tic
while T<=W
    theta_tip=theta_tip+0.00001;
    for i=1:nbe
        if i==1
            lim_inf=lim_inf; %#ok<ASGSL>
            lim_sup=lim_inf+el_size;
        else
            lim_inf=lim_inf+el_size;
            lim_sup=lim_inf+el_size;
        end
    T1=int(1/2*ro*chord*Cl_alpha(i)*((theta_tip/(y/Rmax))*(Omega*y)^2-Up*(Omega*y)), y, lim_inf, lim_sup);
    T2=int(T1,0,2*pi());
    T=1/(2*pi)*T2;    
    end
end
toc

I think this is all that's necessary!

Answer 1

[Star Strider]

Use numeric integration rather that the Symbolic Math Toolbox.

With your Symbolic Math Toolbox code, when I ran it on my laptop, I got:

Elapsed time is 110.680337 seconds.

Running this loop on my laptop:

tic
while T<=W
  theta_tip=theta_tip+0.00001;
  for i=1:nbe
      if i==1
          lim_inf=lim_inf; %#ok<ASGSL>
          lim_sup=lim_inf+el_size;
      else
          lim_inf=lim_inf+el_size;
          lim_sup=lim_inf+el_size;
      end
  T1(i)=integral(@(y) 1/2*ro*chord*Cl_alpha(i)*((theta_tip/(y/Rmax))*(Omega*y)^2-Up*(Omega*y)), lim_inf, lim_sup, 'ArrayValued',1);
  T = T1(i);  
  end
end
toc

I got:

Elapsed time is 4.579191 seconds.

The only difference was in using the anonymous function with integral.

Also, the ‘T2’ and ‘T’ calculations are redundant, since ‘T2’ essentially multiplies ‘T1’ by ‘2*pi’, then ‘T’ divides ‘T2’ by ‘2*pi’. So ‘T’ is simply ‘T1’. I added the subscripts so that now ‘T1’ is saved as a vector for each value of ‘Cl_alpha’.

[Diogo Nascimento]

that did the trick! thank you!!

[Star Strider]

As always, my pleasure!