How can I compute the Area and the Centroid of the following shape?

42 次查看（过去 30 天）
M
M 2022-9-11
评论： Star Strider 2024-11-4
Ran in:
How can I compute the Area and the Centroid of the following shape?
I have used the following code to construct it:
xA = 0;
xB = 1;
xf = 1
xf = 1
x = linspace(0, xf, xf*1e4 + 1);
a = 9.2; %
c = 0.5; % center of function
Y = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
plot(x, Y, 'linewidth', 1.5), grid on, xlim ([0 1.1])
  6 个评论
显示 4更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2022-9-11
Ran in:
The polyshape approach is the easiest way to go on this.
However it is possible to calculate this using the trapz function and the centroid definition —
xA = 0;
xB = 1;
xf = 1
xf = 1
x = linspace(0, xf, xf*1e4 + 1);
a = 9.2; %
c = 0.5; % center of function
Y = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
cx = trapz(x,x.*Y) ./ trapz(x,Y)
cx = 0.7138
cy = trapz(Y,x.*Y) ./ trapz(Y,x)
cy = 0.3934
p = polyshape(x,Y);
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
[xc,yc] = centroid(p)
xc = 0.7176
yc = 0.3974
figure
plot(x, Y, 'linewidth', 1.5), grid on, xlim ([0 1.1])
The values from the two approaches are not exactly the same, however they are reasonably close.
.
  3 个评论
显示 1更早的评论
Star Strider
Star Strider 2022-9-12
Ran in:
As always, my pleasure!
That is esentially the approach I used, with trapz.
The polyshape approach may be more accurate, however the difference is slight —
xA = 0;
xB = 1;
xf = 1;
x = linspace(0, xf, xf*1e4 + 1);
a = 9.2; %
c = 0.5; % center of function
Y = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
cx = trapz(x,x.*Y) ./ trapz(x,Y)
cx = 0.7138
cy = trapz(Y,x.*Y) ./ trapz(Y,x)
cy = 0.3934
p = polyshape(x,Y);
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
[xc,yc] = centroid(p)
xc = 0.7176
yc = 0.3974
x_accuracy = abs(xc-cx)/mean([xc cx])
x_accuracy = 0.0053
y_accuracy = abs(yc-cy)/mean([yc cy])
y_accuracy = 0.0100
figure
plot(x, Y, 'linewidth', 1.5, 'DisplayName','Data'), grid on, xlim ([0 1.1])
hold on
plot(cx,cy,'r+', 'DisplayName','trapz')
plot(xc,yc,'rx', 'DisplayName','polyshape+centroid')
hold off
legend('Location','best')
The relative deviation of the individual values from the mean of each set is about 0.5% for the x-coordinate and about 1% for the y-coordinate. I am not certain how accurate they can be.
.
Star Strider
Star Strider 2024-11-4
I used the expressiion described in By integral formula in the Centroid article iin Wikipedia.
Later, I also wrote an anonymous function for it:
ctrd = @(x,y) [trapz(x, x.*y) / trapz(x, y); trapz(y, x.*y) / trapz(y, x)]; % Returns: [Centroid x-Coordinate; Centroid y-Coordinate]
The (x,y) arguments are the ‘x’ and ‘y’ vectors describing the outline of the region.
.

请先登录，再进行评论。

更多回答（4 个）

Bruno Luong
Bruno Luong 2022-9-12
编辑：Bruno Luong 2022-9-12
Ran in:
xA = 0;
xB = 1;
a = 9.2;
c = 0.5;
sfun = @(x)sigmf(x, [a c]);
Area = integral(@(x)sfun(x),xA,xB);
xc = integral(@(x)sfun(x).*x,xA,xB)/Area
xc = 0.7138
yc = integral(@(x)sfun(x).^2,xA,xB)/(2*Area)
yc = 0.3935
Sam Chak
Sam Chak 2022-9-12
Ran in:
Hi @M
If your intention is to find the defuzzified output value for membership function mf at the interval in x using the centroid method, then you can try this method:
xA = 0;
xB = 1;
x = xA:0.0001:xB;
mf = sigmf(x, [9.2 0.5]);
plot(x, mf), grid on, xlim([-0.2 1.2]), ylim([0 1.2]), xlabel('\it{x}'), ylabel('\mu(\it{x})')
xc = defuzz(x, mf, 'centroid')
xc = 0.7138
xline(xc, '--', sprintf('%.4f', xc), 'LabelVerticalAlignment', 'middle');
Note that there should be no significant difference between
mf = sigmf(x, [9.2 0.5]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
and
mf = sigmf(x, [9.2 0.5]);
Image Analyst
Image Analyst 2022-9-11
centroid
Centroid of polyshape
collapse all in page
Syntax
[x,y] = centroid(polyin)
[x,y] = centroid(polyin,I)
Description
example
[x,y] = centroid(polyin) returns the x-coordinates and the y-coordinates of the centroid of a polyshape.
example
[x,y] = centroid(polyin,I) returns the coordinates of the centroid of the Ith boundary of polyin.
This syntax is only supported when polyin is a scalar polyshape object.
  4 个评论
显示 2更早的评论
Image Analyst
Image Analyst 2022-9-12
You didn't look up the help for polyshape, did you? It's just as trivial as @Star Strider showed:
p = polyshape(x,Y);
[x,y] = centroid(p)

请先登录，再进行评论。

Kwanele
Kwanele 2024-11-2
% Define the functions
f1 = @(x) 0.5 * (80.94 - 0.25 * exp(2 * x.^2));
f2 = @(x) 8.997 - 0.5 * exp(x.^2);
% Define the limits of integration
a = 0;
b = 1.7;
% Use MATLAB's integral function
numerator = integral(f1, a, b);
denominator = integral(f2, a, b);
% Calculate the centroid
centroid = numerator / denominator;
% Display the results
fprintf('Numerator: %.4f\n', numerator);
fprintf('Denominator: %.4f\n', denominator);
fprintf('Centroid: %.4f\n', centroid);

类别

在 Help Center 和 File Exchange 中查找有关 Matrices and Arrays 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by