How to use polyfit function

   n=csvread('loadextension.csv',3,0);
l=n(:,1); %this is the extension in mm
force=n(:,2); %this is the force in N 
%Part B 
area=6.1*50; %in mm 
stress=force./area; %in N/mm 
strain=l./50; 
figure 
plot(strain, stress) 
xlabel ('Strain')
ylabel ('Stress') 
%Part c 
%part i 
x=strain==linspace(0,.27);
y=stress==linspace(0,.27); 
z=polyfit(x,y,4)  % code
  end

I am trying to estimate the linear portion of a polynomial, and when I am using this polyfit function, I keep getting an error. Can someone help please?

Answer 1

[madhan ravi]

x=linspace(0,.27);
y=linspace(0,.27); 
z=polyfit(x,y,4)

[Sarah Hicks]

This is what the graph looks like now, and it is not correct. Any advice?

[Sarah Hicks]

I also need it to

madhan ravi‘s reply : link doesn’t work

[Sarah Hicks]

I need to get the slope of this linear portion.

[Sarah Hicks]

[madhan ravi]

Try putting 1 instead of 4

[Sarah Hicks]

[madhan ravi]

So provide your datas

[Sarah Hicks]

Here is the data

[madhan ravi]

whats your expected figure

[Sarah Hicks]

It is a stress-strain graph, so it should look exactly like the graph in the pictures above, just without the orange line. I am trying to find the slope of the linear portion of this graph.

[madhan ravi]

perhaps:

n=csvread('LoadExtension.csv',3,0);
l=n(:,1); %this is the extension in mm
force=n(:,2); %this is the force in N 
%Part B 
area=6.1*50; %in mm 
stress=force./area; %in N/mm 
strain=l./50;
plot(strain, stress) 
hold on
xx=linspace(strain(1),strain(2),1000);
yy=polyfit(stress,strain,1);
yy1=polyval(yy,xx);
plot(xx,yy1,'r')
xlabel ('Strain')
ylabel ('Stress')

[Sarah Hicks]

Answer 2

[Star Strider]

Without your file, it is difficult to provide specific code.

However, these lines:

x=strain==linspace(0,.27);
y=stress==linspace(0,.27);

will produce logical vectors, and since there is no guarantee than any of the ‘stress’ or ‘strain’ data will exactly match the values the linspace calls produce, ‘x’ and ‘y’ could be uniformly zero.

A better option is likely:

x = ismembertol(strain, linspace(0,.27), 0.01);
y = ismembertol(stress, linspace(0,.27), 0.01);

and:

z=polyfit(strain(x),stress(y),4)  % code

although the same elements of both vectors would have to be returned, so the elements correspond and the vectors have equal lengths.

[Sarah Hicks]

The vectors are not the same length so how do I make them the same?

[Star Strider]

You have to choose either ‘x’ or ‘y’ for both your ‘stress’ and ‘strain’ vectors.

Either that, or find another way of selecting them, for example:

x = (strain >= 0) & (strain <= 0.27); 
y = (stress >= 0) & (stress <= 0.27);

or whatever works for your data.

You still may have to choose one of the two ‘x’ or ‘y’ logical vectors for both your ‘stress’ and ‘strain’ vectors if they do not exactly match.

EDIT 1 — I would just do a linear approximation of that region. The slope of a linear fit is 3.466.

n = xlsread('LoadExtension.csv');
l=n(:,1); %this is the extension in mm
force=n(:,2); %this is the force in N 
%Part B 
area=6.1*50; %in mm 
stress=force./area; %in N/mm 
strain=l./50; 
figure 
plot(strain, stress) 
xlabel ('Strain')
ylabel ('Stress') 
%Part c 
%part i 
x = (strain >= 0) & (strain <= 0.27); 
y = (stress >= 0) & (stress <= 0.27);
xy = x & y;
z=polyfit(strain(xy),stress(xy),1)  % code
f = polyval(z, strain(xy));

figure
plot(strain, stress)
hold on
plot(strain(xy), f)
hold off
xlim([0 0.1])
text(0.04, 0.15, sprintf('Slope = %7.3f', z(1)))

EDIT 2 — The part of your data that you are selecting is at the very beginning. The plot of that region and the regression line through it are:

Do you intend to regress on a different region?