Polar plot using polyfit
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I'm trying to fit the VALUES to make a polar plot but it seems I couldn't fit one value 1400e2 which the polyfit doesn't take it? what other function should I use to make the polar plot correct so that it contains the values exactly at that degree ?
VGV = [ 2000e2 1800e2 1600e2 1400e2 ]; % [0 45 90 135 ]
angle = (pi/180).* [ 0 45 90 135 ];
alpha = (0:0.1:360);
alpha1 = (pi/180).*(0:0.1:360);
for k=1:1:length(alpha)
p = polyfit(angle,VGV,2);
f = polyval(p,alpha1(1:1:901));
f1 = fliplr(f(1:1:900));
f2 = fliplr(f1);
f3 = fliplr(f2);
Vp = [ f f1 f2 f3];
end
figure;polar(alpha1(1:1:3601),Vp,'o')
Respuesta aceptada
Más respuestas (2)
Alan Stevens
el 14 de Oct. de 2020
I'm somewhat confused by your question and don't understand the need for all the flip commands! However, does the following meet your needs at all (incidentally, angle vs VGV is a straight line, so you don't need to get polyfit to fit a quadratic):
VGV = [ 2000e2 1800e2 1600e2 1400e2 ]; % [0 45 90 135 ]
angle = (pi/180).* [ 0 45 90 135 ];
alpha = (0:0.1:360);
alpha1 = (pi/180).*(0:0.1:360);
for k=1:1:length(alpha)
p = polyfit(angle,VGV,1);
f = polyval(p,alpha1);
% f1 = fliplr(f(1:1:900));
% f2 = fliplr(f1);
% f3 = fliplr(f2);
%
% Vp = [ f f1 f2 f3];
end
figure;
polarplot(alpha1,f,angle,VGV,'o')
This results in

1 comentario
Ramesh Bala
el 14 de Oct. de 2020
Image Analyst
el 14 de Oct. de 2020
First of all, polyfit does not need to be inside a loop since angle and VGV don't change inside the loop.
Next, you can fit a line, order 1, instead of a quadratic of order 2 since your VGV is linear.
Next, you have only 2 coefficients for p instead of 900.
% Create training data.
VGV = [ 2000e2 1800e2 1600e2 1400e2 ]; % [0 45 90 135 ]
angle = (pi/180).* [ 0 45 90 135 ];
% Create more angles to fit between our training angles.
alpha = (pi/180).*(0:0.1:360);
% Plot training data
polar(angle, VGV, 'bo')
% Fit a quadratic
coefficients = polyfit(angle, VGV, 1);
% Get a fit
vFit = polyval(coefficients, alpha);
% Plot fitted data:
hold on;
polar(alpha(1:1:3601), vFit, '-')

3 comentarios
Ramesh Bala
el 14 de Oct. de 2020
Image Analyst
el 14 de Oct. de 2020
Editada: Image Analyst
el 14 de Oct. de 2020
If you want 360 interpolated angles instead of 3601 interpolated angles, you'd have to change the way you define alpha to this:
alpha = (pi/180).*(0 : 359); % Exactly 360 angles
If you really want, you can plot blue circles with 'bo' instead of red lines with 'r-'.
If you really want that weird shape you have instead of a spiral, then you'd have to do that flipping craziness, but be aware that alpha no longer means the angle when you go to plot it.
Ramesh Bala
el 14 de Oct. de 2020
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