How to use polyfit function
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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?
回答（2 个）
madhan ravi
2018-11-12
x=linspace(0,.27);
y=linspace(0,.27);
z=polyfit(x,y,4)
12 个评论
Sarah Hicks
2018-11-12
Sarah Hicks
2018-11-12
编辑：Walter Roberson
2018-11-12
I also need it to
madhan ravi‘s reply : link doesn’t work
Sarah Hicks
2018-11-12
Sarah Hicks
2018-11-12
madhan ravi
2018-11-12
Try putting 1 instead of 4
Sarah Hicks
2018-11-12
madhan ravi
2018-11-12
So provide your datas
Sarah Hicks
2018-11-12
madhan ravi
2018-11-12
whats your expected figure
Sarah Hicks
2018-11-12
madhan ravi
2018-11-12
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
2018-11-12
Star Strider
2018-11-12
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.
2 个评论
Sarah Hicks
2018-11-12
Star Strider
2018-11-12
编辑：Star Strider
2018-11-12
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?
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2018-11-12
2018-11-12
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