How do i detect/describe a curvature from a data-set of coordinates

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Conrade Muyambo 2019-5-23

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编辑： Conrade Muyambo 2019-5-24

I have a data-set of coordinates that makes up a 2d body. It's a rectangular shape with one side curved or kinda circular. I would like to describe this curved side with a polyfit line so that I can be able to get the radius. How can I achieve this? Thank you for your very much!

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KALYAN ACHARJYA 2019-5-23

编辑：KALYAN ACHARJYA 2019-5-23

Mathematical language is far better communication way to clear your doubt.

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Answer 1

Star Strider 2019-5-23

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https://ww2.mathworks.cn/matlabcentral/answers/463699-how-do-i-detect-describe-a-curvature-from-a-data-set-of-coordinates#answer_376384

在 MATLAB Online 中打开

The polyfit function will not give you the centre and radius of your arc.

This will:

t = linspace(-pi/4, pi/4, 10);                                  % Create Data
r = 2.5;                                                        % Create Data
x = r*cos(t)+1;                                                 % Create Data
y = r*sin(t)+2;                                                 % Create Data
fcn = @(b) norm(((x(:)-b(1)).^2 + (y(:)-b(2)).^2) - b(3).^2);   % Objective Function
B = fminsearch(fcn, rand(3,1));                                 % Estimate Parameters
Cx = B(1)                                                       % Center X-Coordinate
Cy = B(2)                                                       % Center Y-Coordinate
R  = B(3)                                                       % Radius
% B = fsolve(fcn, rand(3,1))
figure
plot(x, y, 'pg')
hold on
plot(Cx, Cy, 'pb')
plot([Cx x(1)], [Cy y(1)], '-r')
hold off
grid
legend('Data', 'Centre', 'Radius')
axis equal

How do i detect-describe a curvature from a data-set of coordinates - 2019 05 23.png

Experiment with it to get the result you want.

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Star Strider 2019-5-24

在 MATLAB Online 中打开

Try this:

D = load('test2-coordinates.txt');
x = D(:,1);
y = D(:,2);
[xl,xh] = bounds(x);                                                    % X-Coordinate Limits
xr = xh-xl;                                                             % X-Coordinate Range
[yl,yh] = bounds(y);                                                    % Y-Coordinate Limits
yr = yh-yl;                                                             % Y-Coordinate Range
NE = (x>(xh-0.15*xr)) & (y>(yh-0.15*yr));                               % ‘Northeast’ Corner
SE = (x>(xh-0.15*xr)) & (y<(yl+0.15*yr));                               % ‘Southeast’ Corner
SW = (x<(xl+0.15*xr)) & (y<(yl+0.15*yr));                               % ‘Southwest’ Corner
NW = (x<(xl+0.15*xr)) & (y>(yh-0.15*yr));                               % ‘Northwest’ Corner
Lvc = {find(NE),find(SE),find(SW),find(NW)};                            % Vector Cell Array
fcn = @(b,x,y) norm(((x(:)-b(1)).^2 + (y(:)-b(2)).^2) - b(3).^2);       % Objective Function
for k = 1:numel(Lvc)
    xv = x(Lvc{k});
    yv = y(Lvc{k});
    B(:,k) = fminsearch(@(b)fcn(b,xv,yv), rand(3,1));                   % Estimate Parameters
end
figure
plot(x, y)
hold on
for k = 1:size(B,2)
    plot(B(1,k),B(2,k),'xr')
    plot(x(Lvc{k}([1,end])), y(Lvc{k}([1,end])), 'r+')
    text(B(1,k),B(2,k), sprintf('R = %.3f', abs(B(3,k))), 'VerticalAlignment','bottom', 'HorizontalAlignment','center')
end
hold off
grid

The find calls are necessary in order to identify the ends of the arcs. Otherwise, logical indexing alone would suffice in my code. The ends of the arcs are defined as:

xend = x(Lvc{k}([1,end]));
yend = y(Lvc{k}([1,end]));

for each arc, with ‘k’ defined as a for loop index.

My code then produces this figure:

The red ‘x’ markers are the centres of each arc, the red ‘+’ markers are the ends of the arcs (as my code identifies them).

Experiment to get the result you want.

Conrade Muyambo 2019-5-24

编辑：Conrade Muyambo 2019-5-24

Thank you. That helped a lot. But is it now possible, with the start point of one arc and the end point of the second arc to fit a curve to describe the side?

Star Strider 2019-5-24

As always, my pleasure.

No.

The arc parameters describe the arcs only.

I have no idea how to describe the sides, since that is not what you requested initially.

See my oiriginal Answer (the ‘Create Data’ assignments) to understand how to describe the arcs.

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Answer 2

Conrade Muyambo 2019-5-24

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It served the purpose sir! That was only a question. Thank you

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Star Strider 2019-5-24

As always, my pleasure!

You can likely adapt my code to estimate the sides that are not part of the arcs, although since they are more linear that nonlinear, a linear or ‘slightly nonlinear’ regression could work.

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