Find the max in a graph with multiple curves

Hi,

I have a lot of graphs in a figure and i need to know for every abciss the curve that matches the max.

I already know how to do it with only 1 curve.

Here is my graph :

For example while x is between 6 and 11 the program needs to return the value of beta=0° (the blue curve) .

My code :

for b=0:5:30
li=(((b.^3)+1).*(l+0.08*b))./(((b^3)+1)-(0.035.*(l+0.08*b)));
cp=c1.*((c2./li)-c3*b-c4).*exp(-c5./li)+c6.*l;
txt=['Bêta=',num2str(b)];
     plot(l,cp,'DisplayName',txt)
     xlim([2 25])
     ylim([0 0.5])
     hold on;
     %Z=[Z max(cp)];
     [Z,I]=max(cp);
    X=[X Z];
     W=[W I];
     A=[l(W)];
 end
 X; % y max of the cp for every beta 
 A' % x of the max point for every beta 
W; %order of the max of cp for every beta
 hold off; 
 legend show

[Image Analyst]

Your code doesn't run

Undefined function or variable 'l'.

Error in test2 (line 2)

li=(((b.^3)+1).*(l+0.08*b))./(((b^3)+1)-(0.035.*(l+0.08*b)));

Please tell us what l (lower case L) is. Also, l is a very very bad variable to use since it looks so much like 1 (one) and I (upper case i).

[Adam]

You didn't say what the problem is with your current code. What result do you get? what is wrong with it?

Creating a pre-sized array and indexing into it to store results in a loop is better than concatenating in a loop, but that is just semantics and better coding style, the end result will still be the same as concatenating.

Answer 1

[amsel]

The context is that I'm doing mechanical optimization for wind turbine, in x I have wind speed and in y I have the efficiency of the wind turbine. For every speed, I need to choose the right angle in order to have the maximum efficiency .

That's why I'm asking about how can the code return the beta .

Maybe if you have another idea it will be very helpful

Thank you ;)

Here is the entire code :

clear all; 
close all; 
clc;
A=[];
X=[];
W=[];
Z=[];
v=[3:0.2:28];
l=14*4.5./v;
c1=0.5109;
c2=116;
c3=0.4;
c4=5;
c5=21;
c6=0.0068;
for b=0:5:30
li=(((b.^3)+1).*(l+0.08*b))./(((b^3)+1)-(0.035.*(l+0.08*b)));
cp=c1.*((c2./li)-c3*b-c4).*exp(-c5./li)+c6.*l;
txt=['Bêta=',num2str(b)];
     plot(l,cp,'DisplayName',txt)
     xlim([2 25])
     ylim([0 0.5])
     hold on;
     %Z=[Z max(cp)];
     [Z,I]=max(cp);
    X=[X Z];
     W=[W I];
     A=[l(W)];
 end
 X; % y max of the cp for every beta 
 A' % x of the max point for every beta 
W; %order of the max of cp for every beta
 hold off; 
 legend show

Answer 2

[Adam]

b=0:5:30;
li=(((b'.^3)+1).*(l+0.08*b'))./(((b'.^3)+1)-(0.035.*(l+0.08*b')));
cp=c1.*((c2./li)-c3.*b'-c4).*exp(-c5./li)+c6.*l;

will give you all curves in a single matrix. If what you are saying you want is to knnow which curve has the highest y-value for each x-value then you can just use:

[m, idx] = max( cp );

where m will be the amplitude of the max curve and idx will be its index, from 1 to 7.

Is this the code that produces your attached plot in the original question though as the plots I get look nothing like that?