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How to create a FILLED circle within a matrix

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Edward Villanueva
Edward Villanueva 2019 年 12 月 7 日
Commented: Image Analyst 2019 年 12 月 7 日
this is my code thus far, it is used to create two circles within a matrix and then calculate the values of these circle accordingly. My issue is that I need the circles to be filled, I am not sure how to modify my code to do this. I had previous help implementing the circles but I cannot get the circle to be filled as I need it to be.
cx1 = -.05; %x position of circle
cy1 = 0; %y position of circle
cr1 = .04; %radius of circle
th = 0:pi/500:2*pi;
xunit = cr1 * cos(th) + cx1;
yunit = cr1 * sin(th) + cy1;
cx2 = .03; %x position of circle
cy2 = 0; %y position of circle
cr2 = .02; %radius of circle
xunit2 = cr2 * cos(th) + cx2;
yunit2 = cr2 * sin(th) + cy2;
x= -.1:.001: .1;
[X,Y]=meshgrid(x); % xy space
v1=zeros(size(X)); % previous v
xidx = interp1(x, 1:length(x), xunit, 'nearest');
yidx = interp1(x, 1:length(x), yunit, 'nearest');
xidx2 = interp1(x, 1:length(x), xunit2, 'nearest');
yidx2 = interp1(x, 1:length(x), yunit2, 'nearest');
idx = sub2ind(size(X), yidx, xidx);
idx2 = sub2ind(size(X), yidx2, xidx2);
v1(idx) = 10;
v1(idx2) = 3;
v2=v1; % next v
for n=1:100 %number of iterations
for i=2:length(X(1,:))-1 %not disturbing the boundaries
for j=2:length(Y(1,:))-1
v1=v2; %update v1 for next iteration
pcolor(x,x,v1); colormap jet; colorbar; shading interp;
axis square;
axis tight;

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Image Analyst
Image Analyst 2019 年 12 月 7 日

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Edward Villanueva
Edward Villanueva 2019 年 12 月 7 日
Not exactly what im looking for, I already have a circle made within the matrix, I just need to figure a way out to get that circle filled instead of only having the parimeter
Image Analyst
Image Analyst 2019 年 12 月 7 日
You must have used the wrong FAQ code. use the first code sample:
% Create a logical image of a circle with specified
% diameter, center, and image size.
% First create the image.
imageSizeX = 201;
imageSizeY = 201;
[columnsInImage, rowsInImage] = meshgrid(1:imageSizeX, 1:imageSizeY);
% Next create the circle in the image.
centerX = imageSizeX / 2; % Wherever you want.
centerY = imageSizeY / 2;
radius = 50;
circlePixels = (rowsInImage - centerY).^2 ...
+ (columnsInImage - centerX).^2 <= radius.^2;
% circlePixels is a 2D "logical" array.
% Now, display it.
image(circlePixels) ;
colormap([0 0 0; 1 1 1]);
title('Binary image of a circle');
axis square;
0000 Screenshot.png
As you can see, the circle is solid, not just a perimeter. Adapt as needed.

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