how can i divide a circle into 15 equal sectors in matlab?

Question

akanksha dubey 2016-4-18

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https://ww2.mathworks.cn/matlabcentral/answers/279650-how-can-i-divide-a-circle-into-15-equal-sectors-in-matlab

评论： Star Strider 2018-4-13

采纳的回答： Star Strider

which code is for dividing a circle into 15 equal sectors?

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Answer 1

Star Strider 2016-4-18

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编辑：Star Strider 2018-4-13

在 MATLAB Online 中打开

This works:

a = linspace(0, 2*pi, 150);
r = 1;
x = r*cos(a);
y = r*sin(a);
figure(1)
plot(x, y)
hold on
plot([zeros(1,15); x(1:10:end)], [zeros(1,15); y(1:10:end)])
hold off
axis equal

EDIT — (13 Apr 2018 at 19:45 UCT)

To scale it automatically for any desired number of segments, use this code:

N = 15;                                                         % Number Of Segments
a = linspace(0, 2*pi, N*10);
r = 1;
x = r*cos(a);
y = r*sin(a);
figure(1)
plot(x, y)
hold on
plot([zeros(1,N); x(1:10:end)], [zeros(1,N); y(1:10:end)])
hold off
axis equal

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Eman Bany Salameh 2018-4-13

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I want to decrease the number of sectors to be 8. I tried to change the number in this statment

    % plot([zeros(1,15); x(1:10:end)], [zeros(1,15); y(1:10:end)]) 
but I got this error 
    % Error using vertcat
Dimensions of matrices being concatenated are not consistent.
    Error in sectors (line 8)
    plot([zeros(1,8); x(1:10:end)], [zeros(1,8); y(1:10:end)])

Could you please tell me what this do?

% [zeros(1,8); x(1:10:end)], [zeros(1,8); y(1:10:end)]

Star Strider 2018-4-13

在 MATLAB Online 中打开

You also have to change ‘a’ to be 10 times the number of segments, so:

a = linspace(0, 2*pi, 80);

for 8 segments.

I tweaked my earlier code to be robust to any number of segments, and added it as an edit. You may want to use it instead.

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Answer 2

Azzi Abdelmalek 2016-4-18

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https://ww2.mathworks.cn/matlabcentral/answers/279650-how-can-i-divide-a-circle-into-15-equal-sectors-in-matlab#answer_218433

编辑：Azzi Abdelmalek 2016-4-18

在 MATLAB Online 中打开

alpha=-pi:0.01:pi;
n_sect=7
sect=2*pi/n_sect
clr='bgrycmk';
for k=1:n_sect
  a0=-pi+(k-1)*sect
  a1=-pi+k*sect
  t=a0:0.01:a1
x=cos(t)
y=sin(t)
fill([ 0 x],[0 y],clr(k))
hold on
end
axis equal