Average of two closed curves

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Jullienne Franz
Jullienne Franz on 16 Mar 2019
Commented: Jullienne Franz on 16 Mar 2019
Dear everyone,
Is there a way in Matlab to average two closed curves? say I want to take the average of a unit square and a unit circle? How do I do this in Matlab?
KSSV on 16 Mar 2019
Read about mean

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Answers (2)

Stephan on 16 Mar 2019
Edited: Stephan on 16 Mar 2019
% Calculate Parts to 90°
angle1 = linspace(0,pi/4,180)';
% Radius of circle = length of square = 1
r = 1;
% Calculate the mean
r_mean = ((sqrt(r.*tan(angle1).^2 +r.^2))+r)./2;
% Calculate the coordinates
x = [r_mean.*cos(angle1); r_mean.*cos(angle1)];
y = [r_mean.*sin(angle1); -r_mean.*sin(angle1)];
% Build the vectors to plot
xvals = [x; -x; y; y];
yvals = [y; y; -x; x];
% plot
a further possibility is working with polyshape objects:
% construct the polyshape object
poly = polyshape(xvals, yvals);
pgon = simplify(convhull(poly));
% construct the square
rect = polyshape([-1 1 1 -1], [-1 -1 1 1]);
% construct the circle
t = 0:0.05:2*pi;
x1 = cos(t);
y1 = sin(t);
circ = polyshape(x1,y1);
% plot them all
hold on
hold off
xlim([-1.1 1.1])
ylim([-1.1 1.1])
which gives:
Best regards
Jullienne Franz
Jullienne Franz on 16 Mar 2019
Thank you Stephan for your suggestions. I will try to incorporate your ideas in my code and hope to resolve my problem.

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John D'Errico
John D'Errico on 16 Mar 2019
Edited: John D'Errico on 16 Mar 2019
Given two closed curves, say a circle and a square, I would first formulate the problem as a pair of functions in polar coordinates. Thus, represent these two closed curves in polar form, where we will have r as a function of theta. Once you have those relationships, the problem is absolutely trivial.
So, for a unit curcle, we have
r1 = @(theta) ones(size(theta);
For a comparable square centered at the origin, thus of side length 2, I might get trickier...
r2 = @(theta) min(1./abs(cos(theta)), 1./abs(sin(theta)));
r3 = @(theta) (r1(theta) + r2(theta))/2;
Now, all is, as I said, trivial.
theta = linspace(0,2*pi,1000);
hold on
axis equal
grid on
As I said, trivial.
The problem becomes considerably less trivial if you have some fully general closed curve. For example...
pxy = randn(10,2);
pxy(end+1,:) = pxy(1,:);
You cannot dispute this is a closed curve. Yet interpolation along the curve is now a bit more problematic. Thus conversion to polar coordinates is now a bit less useful. Even a simple figure 8 curve is now less automatic. So the complexity of that closed curve is important. Can it be simply represented as a single valued parametric function in polar form? If so, then all is trivial again. If not, then you need to decide how you wish to interpolate, and what is the meaning of that mean in your complex case.
Otherwise, you are doing something possibly no more meaningful than taking the average of apples and oranges, the result may be just grape juice.
Jullienne Franz
Jullienne Franz on 16 Mar 2019
Thank you John for your suggestions. I will try to work on these ideas and hope to resolve my problem.

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