How can i divide a circle into 12 parts? 6 sectors having 15 degrees and the rest 6 having 30 degrees?

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MoonPie1 2016-6-10

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https://ww2.mathworks.cn/matlabcentral/answers/289208-how-can-i-divide-a-circle-into-12-parts-6-sectors-having-15-degrees-and-the-rest-6-having-30-degree

回答： Image Analyst 2016-6-10

How can i divide a circle into 12 parts? 6 sectors having 15 degrees and the rest 6 having 45 degrees? How do i mask each 30 degree sector?

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Azzi Abdelmalek 2016-6-10

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/289208-how-can-i-divide-a-circle-into-12-parts-6-sectors-having-15-degrees-and-the-rest-6-having-30-degree#answer_225088

在 MATLAB Online 中打开

t=-pi:0.01:pi;
r=1;
x=cos(t);
y=sin(t);
plot(x,y)
axis square
hold on
t1=linspace(-pi/2,pi/2,7)
x1=cos(t1);
y1=sin(t1);
x2=zeros(size(x1))
y2=zeros(size(x1))
plot([x1;x2],[y1;y2])
t1=linspace(pi/2,3*pi/2,13)
x1=cos(t1);
y1=sin(t1);
x2=zeros(size(x1))
y2=zeros(size(x1))
plot([x1;x2],[y1;y2])

She wanted six 45 degree sectors. The confusing part comes in the next sentence where she then refers to 12 30 degree sectors after she just said that there were supposed to be sectors of only 15 and 45 degrees. That's why I asked for the image.

MoonPie1 2016-6-10

编辑：Image Analyst 2016-6-10

img.PNG

My needs are as follows:

I want to divide the image into 6 sectors named 1 to 6.
I want to obtain a mask of each sectors such that only one sector is visible.
After masking I want to multiply the mask with the original image.

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Image Analyst 2016-6-10

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/289208-how-can-i-divide-a-circle-into-12-parts-6-sectors-having-15-degrees-and-the-rest-6-having-30-degree#answer_225113

So I guess the problem is not solved yet. So, you should look at the FAQ http://matlab.wikia.com/wiki/FAQ#How_do_I_create_an_arc.3F and then once you have the coordinates of the sector, use the poly2mask() function to make your mask. Not hard. Let me know if you have problems.